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757,656 |
<p>Find the inverse of the function $f(x)= \dfrac{2x-1}{x^2-1}.$</p>
<p>We switch the $x$ and $y$ letters and then solve the the equation, but it became kind of complicated while solving.</p>
|
Haha
| 94,689 |
<p>$\frac {2x-1}{x^2-1}=y<=>yx^2-2x+(1-y)=0$ (1). We have that $D=(-2)^2-4y(1-y)=4y^2-4y+4=4(y^2-y+1)>0$ because for $y^2-y+1$ we have $D*=(-1)^2-4=-3<0$ and the coefficient of $y^2$ is positive. Thus $D>0$ and we have two real roots of (1). You can find them and use one of them by choosing it correctly!</p>
|
1,350,062 |
<p>In almost all of the physics textbooks I have ever read, the author will write the oscillating function as</p>
<blockquote>
<p>$$x(t)=\cos\left(\omega t+\phi\right)$$</p>
</blockquote>
<p>My question is that, is there any practical or historical reason why we should prefer $\cos$ to $\sin$ here? One possible explanation I can think of is that, to trigger a harmonic oscillation movement, we usually push the mass (to the maximum displacement) from the balance point at the initial moment, for which the cosine function will be neater to use than sine ($\phi=0$). But is it really the case?</p>
|
tomi
| 215,986 |
<p>I agree with you that one good reason for using $\cos$ is that it corresponds well with the initial conditions for harmonic motion.</p>
<p>Another good reason is experimental observation. If I come across a physical system that is oscillating, I will want to measure its amplitude and period. Examples include a swinging pendulum, a planet observed rotating around it star, a pulsing wave etc.</p>
<p>Observing the phenomena from a distance, it is possible to note the extreme positions of the motion and determine the amplitude as half the distance between those positions.</p>
<p>To find the period, I want to start my stopwatch at a particular point in the motion and stop it again when that point is repeated, I can choose to do so when the motion is at one extreme or to do so at the moment when the object passes through a particular point. If I try to press the button as the object passes through its central point, I am doing so when it is at its maximum velocity and this can be difficult to observe accurately. It is therefore easier to start timing at one of the extreme values, which corresponds to a $\cos$ function.</p>
|
1,372,779 |
<p>Given that $x$ is a positive integer, find $x$ in $(E)$.</p>
<p>$$\tag{E} j-n=x-n\cdot\left\lceil\frac{x}{n}\right\rceil$$
All $n, j, x$ are positive integers.</p>
|
Ross Millikan
| 1,827 |
<p>Hint: $n\cdot\left\lceil\frac{x}{n}\right\rceil$ is $x$ rounded up to the next multiple of $n$, so the right side of your equation is just $x \bmod n-n$ unless $n|x$ in which case it is zero.</p>
|
832,206 |
<p>I want to approximate the derivative of f(x)</p>
<h2>Finite difference</h2>
<p>$f'(x) \approx \frac{f(x+h)-f(x)}{h}$</p>
<p>I was taught that the error from the subtraction is blown up for small h. This I can verify with MATLAB.</p>
<h2>Smartass method</h2>
<p>So I though maybe the following would fix the problem:</p>
<p>$f'(x) \approx \frac{\log{\frac{\exp(f(x+h))}{\exp(f(x))}}}{h}$</p>
<p>However, I get the same relative errors. (Which also start increasing for h smaller than approx. $\epsilon_{mach}/2$</p>
<p>Why is this?</p>
|
Peter Franek
| 62,009 |
<p>Yes, you can see it by observing that the set of all solutions is a vector space of dimension $d$; this holds because if you choose $q_1,\ldots q_d$, the rest is clearly determined. The solutions $\{r_i^n\}$ are linearly independent (which can be shown by Vandermond determinant, for example), so they generate the whole space.</p>
|
827,072 |
<p>How to find the equation of a circle which passes through these points $(5,10), (-5,0),(9,-6)$
using the formula
$(x-q)^2 + (y-p)^2 = r^2$.</p>
<p>I know i need to use that formula but have no idea how to start, I have tried to start but don't think my answer is right.</p>
|
Américo Tavares
| 752 |
<blockquote>
<p>I know i need to use that formula but have no idea how to start</p>
</blockquote>
<p>\begin{equation*}
\left( x-q\right) ^{2}+\left( y-p\right) ^{2}=r^{2}\tag{0}
\end{equation*}</p>
<p>A possible very elementary way is to use this formula thrice, one for each point. Since the circle passes through the point $(5,10)$, it satisfies $(0)$, i.e.</p>
<p>$$\left( 5-q\right) ^{2}+\left( 10-p\right) ^{2}=r^{2}\tag{1}$$</p>
<p>Similarly for the second point $(-5,0)$:</p>
<p>$$\left( -5-q\right) ^{2}+\left( 0-p\right) ^{2}=r^{2},\tag{2}$$</p>
<p>and for $(9,-6)$:</p>
<p>$$\left( 9-q\right) ^{2}+\left( -6-p\right) ^{2}=r^{2}.\tag{3}$$</p>
<p>We thus have the following system of three simultaneous equations and in the three unknowns $p,q,r$:</p>
<p>$$\begin{cases}
\left( 5-q\right) ^{2}+\left( 10-p\right) ^{2}=r^{2} \\
\left( -5-q\right) ^{2}+p^{2}=r^{2} \\
\left( 9-q\right) ^{2}+\left( 6+p\right) ^{2}=r^{2}
\end{cases}\tag{4}
$$</p>
<p>To solve it, we can start by subtracting the second equation from the first </p>
<p>$$\begin{cases}
\left( 5-q\right) ^{2}+\left( 10-p\right) ^{2}-\left( 5+q\right) ^{2}-p^{2}=0 \\
\left( 5+q\right) ^{2}+p^{2}=r^{2} \\
\left( 9-q\right) ^{2}+\left( 6+p\right) ^{2}=r^{2}
\end{cases}
$$</p>
<p>Expanding now the left hand side of the first equation we get a <em>linear</em> equation</p>
<p>$$\begin{cases}
100-20q-20p=0 \\
\left( 5+q\right) ^{2}+p^{2}=r^{2} \\
\left( 9-q\right) ^{2}+\left( 6+p\right) ^{2}=r^{2}
\end{cases}
$$</p>
<p>Solving the first equation for $q$ and substituting in the other equations, we get</p>
<p>$$\begin{cases}
q=5-p \\
\left( 10-p\right) ^{2}+p^{2}-\left( 4+p\right) ^{2}-\left( 6+p\right) ^{2}=0 \\
\left( 4+p\right) ^{2}+\left( 6+p\right) ^{2}=r^{2}
\end{cases}
$$</p>
<p>If we simplify the second equation, it becomes a <em>linear</em> equation in $p$ only</p>
<p>$$\begin{cases}
q=5-p \\
48-40p=0 \\
\left( 4+p\right) ^{2}+\left( 6+p\right) ^{2}=r^{2}
\end{cases}
$$</p>
<p>We have reduced our quadratic system $(4)$ to two linear equations plus the equation for $r^2$. From the second equation we find $p=6/5$, which we substitute in the first and in the third equations to find $q=19/5$ and $r^2=1972/25$, i.e</p>
<p>$$\begin{cases}
q=5-\frac{6}{5}=\frac{19}{5} \\
p=\frac{6}{5} \\
r^{2}=\left( 4+\frac{6}{5}\right) ^{2}+\left( 6+\frac{6}{5}\right) ^{2}=
\frac{1972}{25}.
\end{cases}\tag{5}
$$</p>
<p>So the equation of the circle is</p>
<p>\begin{equation*}
\left( x-\frac{19}{5}\right) ^{2}+\left( y-\frac{6}{5}\right) ^{2}=\frac{1972}{25}.
\end{equation*}</p>
|
2,322,481 |
<p>Look at this limit. I think, this equality is true.But I'm not sure.</p>
<p>$$\lim_{k\to\infty}\frac{\sum_{n=1}^{k} 2^{2\times3^{n}}}{2^{2\times3^{k}}}=1$$
For example, $k=3$, the ratio is $1.000000000014$</p>
<blockquote>
<p>Is this limit <strong>mathematically correct</strong>?</p>
</blockquote>
|
Mark Fischler
| 150,362 |
<p>$$1\leq \frac{\sum_{n=1}^{k} 2^{2\times3^{n}}}{2^{2\times3^{k}}}=
1+\frac{\sum_{n=1}^{k-1} 2^{2\times3^{n}}}{2^{2\times3^{k}}} =
1+ \sum_{n=1}^{k-1} 2^{-2(3^k-3^{n})}
\leq 1+ \sum_{n=1}^{k-1} 2^{-2(3^k-3^{k-1})}\\=1+(k-1)2^{-4\cdot3^{k-1}}
$$
So for all $k$
$$
1\leq \frac{\sum_{n=1}^{k} 2^{2\times3^{n}}}{2^{2\times3^{k}}}\leq 1+(k-1)2^{-4\cdot3^{k-1}}
$$
and since
$$
\lim_{k\to\infty}(k-1)2^{-4\cdot3^{k-1}}=0
$$
it follows that
$$
\lim_{k\to\infty} \frac{\sum_{n=1}^{k} 2^{2\times3^{n}}}{2^{2\times3^{k}}} = 1
$$</p>
|
2,322,481 |
<p>Look at this limit. I think, this equality is true.But I'm not sure.</p>
<p>$$\lim_{k\to\infty}\frac{\sum_{n=1}^{k} 2^{2\times3^{n}}}{2^{2\times3^{k}}}=1$$
For example, $k=3$, the ratio is $1.000000000014$</p>
<blockquote>
<p>Is this limit <strong>mathematically correct</strong>?</p>
</blockquote>
|
Hitesh Seth
| 460,801 |
<p>I have taken a simpler approach to solve this.</p>
<p>$$\frac{\sum_{n=1}^k 2^{2X3^n}}{2^{2X3^k}}$$ can be written as:</p>
<p>$$\frac{\sum_{n=1}^k 4^{3^n}}{4^{3^k}}$$ which further can be written as :</p>
<p>$$\frac{4^{3^1}}{4^{3^k}} + \frac{4^{3^2}}{4^{3^k}} +\frac{4^{3^3}}{4^{3^k}}......... \frac{4^{3^{k-1}}}{4^{3^k}} +\frac{4^{3^k}}{4^{3^k}}$$</p>
<p>writing this in reverse order:</p>
<p>$$\frac{4^{3^k}}{4^{3^k}} + \frac{4^{3^{k-1}}}{4^{3^k}}......... +\frac{4^{3^3}}{4^{3^k}} + \frac{4^{3^2}}{4^{3^k}} + \frac{4^{3^1}}{4^{3^k}}$$</p>
<p>which becomes </p>
<p>$$\frac{1}{1} + \frac{1}{4^{3^k-3^{k-1}}}......... +\frac{1}{4^{3^k-3^3}} + \frac{1}{4^{3^k-3^2}} + \frac{1}{4^{3^k-3^1}}$$</p>
<p>Now as $k→∞, 3^k →∞ $
which implies, $4^{3^k-3^{k-1}} →∞ $</p>
<p>which imply, $\frac{1}{4^{3^k-3^{k-1}}} + ......... → 0 $</p>
<p>which leaves, $\frac{1}{1} + \frac{1}{4^{3^k-3^{k-1}}}......... +\frac{1}{4^{3^k-3^3}} + \frac{1}{4^{3^k-3^2}} + \frac{1}{4^{3^k-3^1}} → 1 $ as $k→∞ $</p>
|
3,407,474 |
<p>I can understand that The set <span class="math-container">$M_2(2\mathbb{Z})$</span> of <span class="math-container">$2 \times 2$</span>
matrices with even integer entries is an infinite non commutative ring. But why It doesn't have any unity? I think <span class="math-container">$I(2\times2)$</span> is a unity for this ring.
Where is my misunderstanding in subject of <span class="math-container">$I(2\times2)$</span> being a unity?</p>
|
Simon Fraser
| 717,270 |
<p><span class="math-container">$I(2\times 2)=\left[\begin{matrix}1&0\\0&1\end{matrix}\right]\not\in M_2(2\mathbb{Z})$</span> since <span class="math-container">$M_2(2\mathbb{Z})$</span> has only even integers as its entries. It can be shown using an arbitrary matrix <span class="math-container">$B$</span> that <span class="math-container">$I(2\times 2)$</span> is the only possible unity in <span class="math-container">$M_2(2\mathbb{Z}),$</span> but since it is not an element of <span class="math-container">$M_2(2\mathbb{Z}),$</span> there is no unity. There are other rings without a unity, such as <span class="math-container">$2\mathbb{Z},$</span> the set of even integers. </p>
<p>Btw, the notation <span class="math-container">$2\mathbb{Z}$</span> is equivalent to the ideal <span class="math-container">$\langle 2\rangle := \{2x : x\in\mathbb{Z}\},$</span> if you didn't know.</p>
|
1,424,297 |
<p>Why is $a\overline bc + ab = ab + ac$? I think it has something to do with the rule $a + \overline a = 1$, right?</p>
|
robjohn
| 13,854 |
<p>Start with the identity $1-\cos(2x)=2\sin^2(x)$ to get
$$
\begin{align}
\color{#C00000}{n^2}\left(\color{#00A000}{1-\cos\left(\frac1n\right)}\right)
&=\frac{\color{#00A000}{2\sin^2\left(\frac1{2n}\right)}}{\color{#C00000}{\frac1{n^2}}}\\
&=\frac12\left(\frac{\sin\left(\frac1{2n}\right)}{\frac1{2n}}\right)^2\tag{1}
\end{align}
$$
Then use the limit
$$
\lim_{x\to0}\frac{\sin(x)}x=1\tag{2}
$$
to get
$$
\lim_{n\to\infty}n^2\left(1-\cos\left(\frac1n\right)\right)=\frac12\tag{3}
$$</p>
|
2,597,126 |
<p>Given the following problem:</p>
<blockquote>
<p>Applying the division algorithm, prove that all whole numbers that are at the same time a square and a cube have the form $7k$ or $7k+1$.</p>
</blockquote>
<p>I am unable to interpret what it is asking of me and therefore I am unable to provide any solutions. Could someone explain to me what exactly the problem is asking and how do I solve it? </p>
|
fleablood
| 280,126 |
<p>I'm wondering are we to assume the student knows anything. Even that $M =a^2 = b^3$ means that $M$ is 6-power. </p>
<p>We can use the remainder theorem to show that $a^2$ will have remainder $0,1,4,2,2,4,1$ if $a$ has remainder $0,1,2,3,4,5,6$</p>
<p>That is $a = 7m + k$ so $a^2 = 49m^2 + 14mk +k^2$ so $a^2$ will have the same remainder as $k^2$. And we can calculate those to be $0,1,4,2,2,4,1$.</p>
<p>And we can see that $b^3$ will have remainder $0,1,1,6,1,6,6$ if $b$ has remainder of $0,1,2,3,4,5,6$.</p>
<p>If $b = 7n+j$ then $b^3 = 7^3n^3 + 3*7^2n^2j + 3*7nj^2 + j^3$ so $b^3$ will have the same remainder as $j^3$ and we calculate those to be $0,1,1,6,1,6,6$.</p>
<p>As $a^2 = b^3$ then it must have remainder $0,1$.</p>
<p>====== full answer =====</p>
<p>If $M = a^2=b^3$ where $a$ and $b$ are whole numbers then $M = c^6$ for some integer $c$. (Because the prime factors of $M$ must be the prime factors of $b$ and $a$ and must as they divide $a^2$ must be to an even power and as they divide $b^3$ must be to a third power, so they must be to a multiple of $6$ power.)</p>
<p>So $M = c^6$.</p>
<p>No by division algorithm there are unique integers $n,j$ so that $c = 7n + j$ and $0 \le j < 7$.</p>
<p>$(7n + j)^6 = 7^6n^6 + a*7^5n^5*j + b*7^4n^4*j^2 + ... + e7n*j^5 + j^6$ where $a,b,c ....$ are the binomial coefficients.</p>
<p>$7^6n^6 + a*7^5n^5*j + b*7^4n^4*j^2 + ... + e7n*j^5 = 7K$ for an integer $K$ so</p>
<p>$M = 7K + j^6$ where $j = 0,1,....,6$.</p>
<p>$0^6 = 0 = 7*0$.</p>
<p>$1^6 = 1 = 7*0 + 1$</p>
<p>$2^6 = 64 = 7*9 + 1$</p>
<p>$3^6 = 27^2 = (4*7 - 1)^2 = 16*7^2 - 8*7 + 1 = 7(16*7 -8) + 1$</p>
<p>And $4,5,6 = (7-3),(7-2),(7-1)$ and $(7 - k)^6 = 7^6 - a7^5k + b7^4k^2 -.... - e7*k^6 + k^6$. </p>
<p>And as $k^6 = 7m +1 $ for some $m$ $4^6,5^6,6^6$ also equal $7j + 1$ for some integer $j$.</p>
<p>So $M = c^6 = 7K + j^6$. If $j=0$ that is $M = 7K$. If $j=1...6$ then $j^6 = 7m + 1$ for some $m$ and $M = 7(K+m) + 1$.</p>
<p>.... And all of that should incentive enough to learn modulo notation to make the whole thing four lines.</p>
<p>$c^6 \equiv 0 \mod 7$ if $c \equiv 0 \mod 7$</p>
<p>$(\pm 1)^6 \equiv 1 \mod 7$</p>
<p>$(\pm 2)^6 \equiv 64 \equiv 1 \mod 7$</p>
<p>$(\pm 3)^6 \equiv 9^3 \equiv 2^3 \equiv 8 \equiv 1 \mod 7$</p>
<p>So $c^6 \equiv 1 \mod 7$ if $c \not \equiv 0 \mod 7$.</p>
<p>======</p>
<p>Then there is Fermat's Little Theorem:</p>
<p>$a^{p-1} \equiv 1 \mod p$ if $a \not \equiv 0 \mod p$ and $p$ is prime.</p>
<p>So $a^6 \equiv 1 \mod 7$ if $a\not \equiv 0\mod p$ and $a^6\equiv 0\mod 7$ if $a \equiv 0 \mod 7$.</p>
<p>($1$ line.)</p>
|
2,649,756 |
<p>I was given the equation</p>
<p>$x=y^4$</p>
<p>and asked to find the points where the curvature was the biggest and the smallest. I know the curvature equation:</p>
<p>$κ(x)= \frac{|y″|}{\left(1+\left(y′\right)^2\right)^{\frac{3}{2}}}$</p>
<p>and I know $f(x)=y=\sqrt[4]{x}$ </p>
<p>but I am not sure how to minimize or maximize the curvature. Any help would be great!</p>
|
Hw Chu
| 507,264 |
<p>Let $r_i = \frac{p_i}{q_i}$ and $q = \prod_{i=1}^n q_i$.</p>
<p>The group isomorphism $G \to G'$ given by $x \mapsto qx$ sends $G$ to a subgroup of $\mathbb Z$. Then use the fact that every subgroup of $\mathbb Z$ is cyclic.</p>
|
740,323 |
<p>This might be a little open-ended, but I was wondering: are there any natural connections between geometry and the prime numbers? Put differently, are there any specific topics in either field which might entertain relatively close connections?</p>
<p><strong>PS:</strong> feel free to interpret the term <em>natural</em> in a broad sense; I only included it to avoid answers along the lines of "take [fact about the primes] $\to$ [string of connections between various areas of mathematics] $\to$ [geometry!]"</p>
|
Lubin
| 17,760 |
<p>Here’s an example, far from the best, of prime numbers entering into a (relatively) geometric problem. Consider all the points on the unit circle, $X^2+Y^2=1$. Notice that by considering this as the set of complex numbers $a+bi$ of absolute value one, i.e. $a^2+b^2=1$, this has a natural group structure. Explicitly, $(a,b)*(c,d)=(ac-bd,ad+bc)$.</p>
<p>Now, here’s the question: What are the rational points on the circle? That is, what are the points $(a,b)$ on the circle for which both $a$ and $b$ are rational numbers? Your first interesting case is $(3/5,4/5)$. Of course there’s an answer to this question coming from the classical solution to the problem of finding all Pythagorean Triples. But I want to ask an arithmetic question: What are the possible denominators of all the rational points on the circle?</p>
<p>The answer comes out of looking at the “primes” in the ring of Gaussian Integers, but I’ll cut to the chase: a number will appear as the (common) denominator $D$ of a <em>rational</em> pair $(a,b)$ on the unit circle if and only if the only primes dividing $D$ are those of the form $4k+1$. Naturally, I want the rational numbers $a$ and $b$ to be in lowest terms.</p>
|
740,323 |
<p>This might be a little open-ended, but I was wondering: are there any natural connections between geometry and the prime numbers? Put differently, are there any specific topics in either field which might entertain relatively close connections?</p>
<p><strong>PS:</strong> feel free to interpret the term <em>natural</em> in a broad sense; I only included it to avoid answers along the lines of "take [fact about the primes] $\to$ [string of connections between various areas of mathematics] $\to$ [geometry!]"</p>
|
narsep
| 425,570 |
<p><a href="https://i.stack.imgur.com/j0g7j.png" rel="nofollow noreferrer">Events on the horizon of a 2D Universe</a></p>
<p>You may have a look to the graph; it represents all the numbers (red, up to 100) that "see" straight to the origin.
e.g. prime 7 is represented by the seventh vertical point-column, prime 13 by the 13th column. The uniqueness of prime numbers is obvious.</p>
|
2,006,676 |
<p>I'm trying to show that for a non-empty set $X$, the following statements are true for logical statements $P(x)$ and $Q(x)$:</p>
<ul>
<li>$∃x∈X$, ($P(x)$ or $Q(x)$) $\iff$ $(∃x∈X, P(x))$ or $(∃x∈X, Q(x))$</li>
<li>$∃x∈X$, ($P(x)$ and $Q(x)$) $\implies$ $(∃x∈X, P(x))$ and $(∃x∈X, Q(x))$</li>
</ul>
<p>Is it possible to use truth tables to show this? I can't think of any other way to go about it. Any help would be appreciated.</p>
|
Daniel McLaury
| 3,296 |
<p>Not entirely rigorous as stated, but to get an idea of what's going on:</p>
<p>If $\frac{a}{b} \frac{c}{d} = 1$ then $\frac{c}{d} = \frac{b}{a}$. So this inverse $\frac{c}{d}$ exists provided that $a$ is something that's allowed in the denominator of an element of $A$ -- namely, if $a(0) \neq 0$. (Of course we're assuming that $\frac{a}{b}$ is in lowest terms.)</p>
|
4,126,124 |
<p>I managed to come to the end of a proof regarding a determinant of a linear operator. However, I stuck at the end. I know it is kinda simple but I couldn't see it right away.</p>
<p>Here is my previous work:</p>
<p><span class="math-container">$F$</span> is a field. <span class="math-container">$V = M_{nxn}(F)$</span> is a vector space of <span class="math-container">$nxn$</span> matrices over <span class="math-container">$F$</span>, and <span class="math-container">$B$</span> is some arbitrary element in <span class="math-container">$V$</span>. Now, we are given a linear operator <span class="math-container">$T:V \rightarrow V$</span> defined by <span class="math-container">$T(A) = AB - BA$</span>. We wish to show that <span class="math-container">$det(T)=0$</span>.</p>
<p>Here is my attempt:</p>
<p>If <span class="math-container">$T$</span> has an <span class="math-container">$0$</span> as eigenvalue, then by <span class="math-container">$p(x)=det(T-\lambda I)$</span>, <span class="math-container">$p(0)=det(T)=0$</span>. Thus, showing that <span class="math-container">$\lambda = 0$</span> would suffice.</p>
<p>We split into cases.</p>
<p>Case 1: <span class="math-container">$B$</span> is invertible.</p>
<p>Then choose <span class="math-container">$A=B^{-1}$</span>, we have <span class="math-container">$T(A)=0=0*A$</span>, so <span class="math-container">$\lambda = 0$</span> is an eigenvalue.</p>
<p>Case 2: <span class="math-container">$B$</span> is not invertible.</p>
<p>Let <span class="math-container">$A$</span> be a nonzero matrix, and <span class="math-container">$\lambda \in F$</span>
<span class="math-container">$$T(A)=AB-BA= \lambda A$$</span>
<span class="math-container">$$\lambda A - AB = -BA$$</span>
<span class="math-container">$$A \lambda - AB = -BA$$</span>
<span class="math-container">$$A(\lambda I - B) = -BA$$</span>
Take determinants of each side,
<span class="math-container">$$det(A(\lambda I - B)) = det(-BA)$$</span>
<span class="math-container">$$det(A)det(\lambda I - B) = (-1)^n det(B)det(A)$$</span>
Since <span class="math-container">$B$</span> is not invertible, <span class="math-container">$det(B)=0$</span>, so,
<span class="math-container">$$det(A)det(\lambda I - B)=0$$</span></p>
<p>Here we have:
<span class="math-container">$$det(B-\lambda I) = 0$$</span>
It is clear that <span class="math-container">$\lambda$</span> can be zero. However, how can I show that <span class="math-container">$\lambda$</span> must be zero?</p>
|
José Carlos Santos
| 446,262 |
<p><em>By definition</em>, the open sets in a topological space <span class="math-container">$(X,\tau)$</span> <em>are</em> the elements of <span class="math-container">$\tau$</span>. This happens always, and not only when the topology is induced by a metric.</p>
<p>Concerning the last paragraph, I suppose that what your lectures did was something like this: if <span class="math-container">$x\in X$</span>, and <span class="math-container">$X\ne\{x\}$</span>, then <span class="math-container">$X\setminus\{x\}$</span> is not an open subset of <span class="math-container">$X$</span>. However, if <span class="math-container">$d$</span> is any metric on <span class="math-container">$X$</span>, then <span class="math-container">$X\setminus\{x\}$</span> is an open subset of <span class="math-container">$(X,d)$</span>, since<span class="math-container">$$X\setminus\{x\}=\bigcup_{y\in X\setminus\{x\}}D_{d(y,x)}(y).$$</span>What this proves is that there is a set which is not an open subset of <span class="math-container">$X$</span> with respect to the indiscrete topology, but which is open in <span class="math-container">$(X,d)$</span> for any metric <span class="math-container">$d$</span>. So, the indiscrete topology is not induced by a metric.</p>
|
838,631 |
<p>This question may be far too easy for this site but I always seem to get stuck when it comes to the triangle inequality.</p>
<p>For example, I am trying to prove that differentiability implies Lipschitz continuity. Ok, I don't really have a hard time with this except for the step when using the triangle inequality.</p>
<p>Let $| x -x_0| < \epsilon$. Suppose $f$ is differentiable at $x_0$. Then</p>
<p>$ |f(x)-f(x_0)-f'(x_0)(x-x_0) | < | x- x_0|$</p>
<p>The next step is where I am not sure how the triangle inequality allows this,</p>
<p>$|f(x)-f(x_0)| \le |f'(x_0)(x-x_0)|+|x-x_0|$</p>
<p>Why am I allowed to add the $|f'(x_0)(x-x_0)|$ to the right hand side??</p>
<p>Sorry that this question is most likely trivial but I always get stuck with this operation. </p>
|
Ross Millikan
| 1,827 |
<p>The triangle inequality gives you $|f(x)-f(x_0)|-|f'(x_0)(x-x_0) | \le |f(x)-f(x_0)-f'(x_0)(x-x_0) |$ so you have $|f(x)-f(x_0)|-|f'(x_0)(x-x_0) |\le |f(x)-f(x_0)-f'(x_0)(x-x_0) | < | x- x_0|\\|f(x)-f(x_0)|\le|f'(x_0)(x-x_0) |+ | x- x_0|$</p>
|
1,995,471 |
<p>I'm trying to find all of the the roots to the following polynomial with a variable second coefficient:
$$P(x)=4x^3-px^2+5x+6$$
All of the roots are rational, and $p$ is too. It is also given that the difference of 2 roots equals the third, e.g. $r-s=t$. I would like to solve for the roots using relationships between roots & the rational roots theorem.</p>
<p>I know from relationships between roots (Vieta's formula) that $p/4=r+s+t$, which can be reduced to $p/4=2r$ per the previous equation, and therefore $p/8$ is a root. However, I'm not sure where to go from here-- performing the substitution with the other coefficients does not seem to yield anything that lets me solve for a root or $p$. For example, we know from the coefficient of $x^0$ that
$$5/4=rs+rt+st=rs+(r+s)(r-s)$$
but there is no obvious substitution that can be made here that would put things in terms of one variable.</p>
<p>How do I solve for the roots and $p$ using relationships between roots and the rational roots theorem here? Thanks!</p>
|
Parcly Taxel
| 357,390 |
<p>Since $\frac p8$ is a root, we can perform long division on the polynomial and obtain
$$4x^3-px^2+5x+6=\left(x-\frac p8\right)\left(4x^2-\frac p2x+5-\frac{p^2}{16}\right)$$
where
$$\left(5-\frac{p^2}{16}\right)\left(-\frac p8\right)=-\frac{5p}8+\frac{p^3}{128}=6$$
$$p^3-80p-768=0$$
Since $p$ is rational, by the rational root theorem we only need to try the factors of 768. It turns out that $p=12$ is the only rational real root of this equation (the others are $6\pm2\sqrt7i$), so the only possibility for the original cubic is
$$4x^3-12x^2+5x+6=4\left(x-\frac32\right)\left(x+\frac12\right)(x-2)$$
and its roots are $-\frac12,\frac32,2$. Indeed, the difference between $\frac32$ and $-\frac12$ is 2.</p>
|
1,981,553 |
<p>When a person asks: "What is the smallest number (natural numbers) with two digits?"</p>
<p>You answer: "10".</p>
<p>But by which convention 04 is no valid 2 digit number?</p>
<p>Thanks alot in advance</p>
|
Steven Alexis Gregory
| 75,410 |
<p>It depends on the context of the problem. For example. A number $n$ is called "magical" if $n^2$ ends with the digits of $n$. Below is a list of the first six magical numbers whose units digit is $5$. </p>
<pre><code> 5² = 25
25² = 625
625² = 390625
0625² = 390625
90625² = 8212890625
890625² = 793212890625
</code></pre>
<p>In this case, I'd want to say that $0625$ is a magical four-digit number because $0625^2$ end with the digits $0625$.</p>
<p>If I told someone I was making a six-digit yearly salary, it would probably be wrong to say that $\$005489$ was a six-digit income.</p>
|
3,921,335 |
<p>how can I formally proof that for a > 1: <span class="math-container">$$ a> \sqrt a > \sqrt[3]a > \sqrt[4]a ... $$</span>?
Can someone help? ;)</p>
|
Olivier Moschetta
| 369,174 |
<p>Hint: Use logarithms to show that
<span class="math-container">$$a^{\frac{1}{n}}>a^{\frac{1}{n+1}}$$</span></p>
|
58,912 |
<p>In the board game <a href="http://en.wikipedia.org/wiki/Hex_%28board_game%29" rel="nofollow">Hex</a>, players take turns coloring hexagons either red or blue. One player tries to connect the top and bottom edges of the board, colored red; the other tries to connect the left and right edges, colored blue. It is known that a game of Hex will never end in a tie: no matter how it is played, there will always be either a blue path connecting the blue edges, or a red path connecting the red edges.</p>
<p>My question is, if this fact always holds for a finite grid of hexagons, does it also hold on the plane? If the top and bottom edges of a square are colored red, the left and right edges are colored blue, and the interior of the square is colored arbitrarily, must there be either a red path connecting the red edges, or a blue path connecting the blue edges?</p>
<p>More formally, let $S$ be any subset of $[0, 1]^2$. $S$ will represent the points that are red. Must there be either a path within $S$ whose endpoints are of the form $(x, 0)$ and $(x, 1)$, or a path within $[0, 1]^2 - S$ whose endpoints are of the form $(0, y)$ and $(1, y)$?</p>
|
Brian M. Scott
| 12,042 |
<p>If the path is required to be continuous, the answer is <em>no</em>. </p>
<p>Since $[0,1]$ is separable, there are only $2^\omega$ continuous functions from $[0,1]$ into $[0,1]^2$; enumerate those that yield paths connecting opposite sides of the square as $\{\varphi_\xi:\xi < 2^\omega\}$. For $\xi < 2^\omega$ recursively choose points $p_\xi,q_\xi \in [0,1]^2$ as follows. Suppose that $\eta < 2^\omega$, and points $p_\xi$ and $q_\xi$, all distinct, have been chosen for all $\xi<\eta$. Clearly $$|\{p_\xi:\xi<\eta\}\cup\{q_\xi:\xi<\eta\}| < 2^\omega,$$ but $|\operatorname{ran}\varphi_\eta| = 2^\omega$, so we may choose distinct points $$p_\eta,q_\eta \in \operatorname{ran}\varphi_\eta \setminus \left(\{p_\xi:\xi<\eta\}\cup\{q_\xi:\xi<\eta\}\right),$$ and the construction goes through to $2^\omega$.</p>
<p>Now color the points of the set $\{p_\xi:\xi < 2^\omega\}$ blue and those of the set $\{q_\xi:\xi < 2^\omega\}$ red; any remaining points of $[0,1]$ may be colored either blue or red. Then every continuous path in the unit square passes through at least one point of each color.</p>
|
369,104 |
<p>How do I take an algebraic expression and construct a tree out of it?</p>
<p>Sample equation:</p>
<p>((2 + x) - (x * 3)) - ((x - 2) * (3 + y))</p>
<p>If somebody can teach me in steps, that would be really helpful!</p>
|
Brian M. Scott
| 12,042 |
<p>Each internal node (or vertex) of the tree will be labelled by one of the operators appearing in the expression, and each leaf by one of the operands. The algorithm is quite straightforward. Start with the operation that would be performed <strong>last</strong> if you were evaluating the expression. In the case of the expression</p>
<p>$$\big((2 + x)-(x*3)\big)\color{red}{-}\big((x-2)*(3+y)\big)$$</p>
<p>that’s the subtraction that I’ve colored red. The node corresponding to this operation will be the root of your tree. Its first operand is the expression $(2+x)-(x*3)$, and its second is the expression $(x-2)*(3+y)$; these correspond to the left and right children of the root. Thus, at this stage we have this tree:</p>
<pre><code> -
/ \
/ \
/ \
(2+x)-(x*3) (x-2)*(3+y)
</code></pre>
<p>Repeat the process on each of the two leaves of this little tree. On the left the last operation that we’d perform is the subtraction; on the right it’s the multiplication. Thus, the left node gets the label $-$ and children $2+x$ and $x*3$, and the right node gets the label $*$ and the children $x-2$ and $3+y$:</p>
<pre><code> -
/ \
/ \
/ \
- *
/ \ / \
/ \ / \
2+x x*3 x-2 3+y
</code></pre>
<p>Continue in the same way until each node is labelled either with an operation, in which case it has children corresponding to the operands to which that operation is applied, or with an operand, in which case it has no children. Here each leaf requires just one more step, and we end up with the following tree:</p>
<pre><code> -
/ \
/ \
/ \
/ \
/ \
- *
/ \ / \
/ \ / \
+ * - +
/ \ / \ / \ / \
2 x x 3 x 2 3 x
</code></pre>
<p>In this example the leaves are all on the same level of the tree, but that need not happen. Had the original expression been $\big((2+x)-(x*3)\big)-(x-2)$, for instance, the final tree would have been this one:</p>
<pre><code> -
/ \
/ \
/ \
/ \
/ \
- -
/ \ / \
/ \ / \
+ * x 2
/ \ / \
2 x x 3
</code></pre>
<p>Observe that just as we can systematically construct the tree from the expression, so we can also reverse the procedure. Start at the bottom and combine children according to their parent operation, replacing the operation label with the resulting expression. If you apply this process to the last tree above, for instance, you get the following tree after the first two recombinations:</p>
<pre><code> -
/ \
/ \
/ \
/ \
/ \
- -
/ \ / \
/ \ / \
2+x x*3 x 2
</code></pre>
<p>The next two recombinations produce this one:</p>
<pre><code> -
/ \
/ \
/ \
/ \
/ \
(2+x)-(x*3) x-2
</code></pre>
<p>(Note that when you recombine expressions that aren’t single operands, as on the left side of this tree, you enclose them in parentheses.) And one more step yields a tree consisting only of a root labelled by a single expression, that expression being the one that produced the tree in the first place, namely, $\big((2+x)-(x*3)\big)-(x-2)$.</p>
|
2,776,388 |
<p>I tried this problem and I first found the lim of $x^x$ as $x$ approaches zero from right to be 1 (I did this by re-writing $x^x$ as an exponential) and when I repeated the same process to find lim of $x^{(x^x)}$, I found 1 again but the final answer should be ZERO. Could I have an explanation on why it's a zero?</p>
|
Martín-Blas Pérez Pinilla
| 98,199 |
<p>Using $\lim_{x\to 0^+}x^x = 1$, we have:
$$\lim_{x\to 0^+}x^{(x^x)} = 0^1 = 0.$$</p>
|
884,342 |
<p>$N$ is a normal subgroup of $G$ if $aNa^{-1}$ is a subset of $N$ for all elements $a $ contained in $G$. Assume, $aNa^{-1} = \{ana^{-1}|n \in N\}$.</p>
<p>Prove that in that case $aNa^{-1}= N.$</p>
<p>If $x$ is in $N$ and $N$ is a normal subgroup of $G$, for any element $g$ in $G$, $gxg^{-1}$ is in $G$. Suppose $x$ is in $N$, and $y=axa^{-1}$ as is defined. Since $N$ is normal, $aNa^{-1}$ is a subset of N.
$x= a^{-1}ya$. Given that $x$ is in $N$, and $x=a^{-1}ya$, $y$ is also in $N$. If $y$ is in N, then $axa^{-1}$ is also in $N$. $X$ is in $aNa^{-1}$.</p>
<p>Does the proof make sense?</p>
|
Bey
| 2,313 |
<p>By definition, we have $aNa^{-1}\subseteq N$ for all $a\in G$. This is equivalent to saying $aN\subseteq Na$ for all $a\in G$. (Why?) </p>
<p>Thus, we simply need to show $Na\subseteq aN$ for all $a\in G$.</p>
<p>To this end, take $x\in Na$. Then $x=n_1a$ for some $n_1\in N$, and so $a^{-1}x=a^{-1}n_1a=(an_2a^{-1})^{-1}$. (What should $n_2$ be for this to make sense? Why can I do this?)</p>
<p>Now, because $N$ is normal, $an_2a^{-1}\in N$. But this also implies $(an_2a^{-1})^{-1}\in N$ (Why?) and therefore $a^{-1}x\in N$. Hence, we can write $a^{-1}x=n_3$ for some $n_3\in N$. So $x=an_3$ meaning $x\in aN$. </p>
<p>Hence we have $Na\subseteq aN$, and thus $Na=aN$; or, $aNa^{-1}=N$. However, our choice of $a$ was arbitrary so we have proven our claim for all $a\in G$.</p>
|
2,753,504 |
<p>Where $a,b$ and $c$ are positive real numbers.</p>
<p>So far I have shown that $$a^2+b^2+c^2 \ge ab+bc+ac$$ and that $$a^2+b^2+c^2 \ge a\sqrt{bc} + b\sqrt{ac} + c\sqrt{ab}$$ but I am at a loss what to do next... I have tried adding various forms of the two inequalities but always end up with something extra on the side of $ab+bc+ac$. Any help appreciated!</p>
|
Dr. Sonnhard Graubner
| 175,066 |
<p>Use that $$\frac{ab+ac}{2}\geq a\sqrt{bc}$$
$$\frac{ab+bc}{2}\geq b\sqrt{ac}$$
$$\frac{ac+bc}{2}\geq c\sqrt{bc}$$</p>
|
2,647,555 |
<p>In our probability class, we recently learned that the probability measure, $P$, is a set function that takes in a subset of some sample space, $\Omega$, and returns a numerical value that satisfies the probability axioms. We are now learning about conditional probabilities, namely, $P(A|B)$, for two events $A, B \subset \Omega$. To my understanding, $A|B$ only makes sense in a probabilistic context as "the event that $A$ occurs given that $B$ occurs". However, since the probability measure is a set function, does this mean that $A|B$ is also a set somehow? I am unable to see how this makes sense. Thanks.</p>
<p>EDIT: Thanks for some of the answers so far. If $A|B$ is not a set, then what exactly is it's type? If it isn't a set, why can we take the informality of passing it to a set function?</p>
|
mordecai iwazuki
| 167,818 |
<p>Yes, we can see $A|B$ as a set. You can see it as a subset of the set $B$, it is precisely those elements in the set $B$ which are also in the set $A$. Since $B$ has already happened, we know we are "in" the set $B$. Then, we are looking for the elements in which $A$ also occurs, i.e. the elements which are in $B$ and are also in $A$. However, it is worth noting that $A|B$ is somewhat meaningless if it's not inside a probability measure $\mathbb{P}$. We would just write $A\cap B$.</p>
<p>To answer your edit, it's technically just notation. When we pass $A|B$ into a probability measure we are not simply taking the measure of $A\cap B$, we are asking something more complex, because we are requiring $B$ to have occurred, which is meaningful given the context. Writing $\mathbb{P}(A|B)$ is just notation for "the probability that $A$ occurred given that $B$ has already occurred." Nothing more.</p>
|
2,647,555 |
<p>In our probability class, we recently learned that the probability measure, $P$, is a set function that takes in a subset of some sample space, $\Omega$, and returns a numerical value that satisfies the probability axioms. We are now learning about conditional probabilities, namely, $P(A|B)$, for two events $A, B \subset \Omega$. To my understanding, $A|B$ only makes sense in a probabilistic context as "the event that $A$ occurs given that $B$ occurs". However, since the probability measure is a set function, does this mean that $A|B$ is also a set somehow? I am unable to see how this makes sense. Thanks.</p>
<p>EDIT: Thanks for some of the answers so far. If $A|B$ is not a set, then what exactly is it's type? If it isn't a set, why can we take the informality of passing it to a set function?</p>
|
Ethan Bolker
| 72,858 |
<p>Good question. The literal answer is "No, $A|B$ is not a set." But there is good intuition behind your question. $A|B$ is not a set, but $A \cap B$ is. Conditional probability defines a probability function on the subsets of the subset (event) $B$ of the universe $\Omega$. </p>
|
1,548,841 |
<p>As you can see, In the image a rectangle gets translated to another position in the coordinates System.</p>
<p>The origin Coordinates are <code>A1(8,2) B1(9,3)</code> from the length <code>7</code> and the height <code>3</code> you can also guess the vertices of the rectangle.</p>
<p><b>Now the Rectangle gets moved</b>. </p>
<p>Now <code>A1</code> is at <code>A2(16,9)</code> and <code>B1</code> is located at <code>B2(16,11)</code>. </p>
<p>It means that the rectangle got translated, rotated and stretched.</p>
<p><b>How can I calculate the Coordinates, of the left-upper corner?</b></p>
<p><a href="https://i.stack.imgur.com/ajwDY.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ajwDY.jpg" alt="enter image description here"></a></p>
<hr>
<p><i>I first tried to calculate the stretching-factor but then I got stuck when I trying to calculate the angle and translation</i> </p>
<p><b>Thanks for your help</b></p>
|
mvw
| 86,776 |
<p>Your transformation contains translation (2 parameter), rotation (1 parameter) and stretching, which I hope means scaling (1 parameter).</p>
<p>This in general is no linear but an affine transform, except for the case that the origin gets mapped to the origin, which I doubt here.</p>
<p>The transform would be like this, using homogeneous coordinates:
$$
T = \left(
\begin{matrix}
t_{11} & t_{12} & t_{13} \\
t_{21} & t_{22} & t_{23} \\
0 & 0 & 1
\end{matrix}
\right)
\quad (*)
$$</p>
<p>Assuming we first do the translation, then the roation, then the scaling we get a matrix:
\begin{align}
T
&=
T_\text{scale} \, T_\text{rot} \, T_\text{trans} \\
&=
\left(
\begin{matrix}
s & 0 & 0 \\
0 & s & 0 \\
0 & 0 & 1
\end{matrix}
\right)
\left(
\begin{matrix}
\cos \theta & -\sin \theta & 0 \\
\sin \theta & \cos \theta & 0 \\
0 & 0 & 1
\end{matrix}
\right)
\left(
\begin{matrix}
1 & 0 & t_x \\
0 & 1 & t_y \\
0 & 0 & 1
\end{matrix}
\right)
\\
&=
\left(
\begin{matrix}
s \cos \theta & -s \sin \theta & 0 \\
s \sin \theta & s \cos \theta & 0 \\
0 & 0 & 1
\end{matrix}
\right)
\left(
\begin{matrix}
1 & 0 & t_x \\
0 & 1 & t_y \\
0 & 0 & 1
\end{matrix}
\right)
\\
&=
\left(
\begin{matrix}
s \cos \theta & -s \sin \theta & s(t_x \cos \theta - t_y \sin \theta) \\
s \sin \theta & s \cos \theta & s(t_x \sin \theta + t_y \cos \theta) \\
0 & 0 & 1
\end{matrix}
\right)
\end{align}</p>
<p>Your two points give four equations, it might be enough to determine the four parameters $t_x, t_y, \theta, s$.</p>
<p>Inserting the points and their images
$$
\left(
\begin{matrix}
s \cos \theta & -s \sin \theta & s(t_x \cos \theta - t_y \sin \theta) \\
s \sin \theta & s \cos \theta & s(t_x \sin \theta + t_y \cos \theta) \\
0 & 0 & 1
\end{matrix}
\right)
\left(
\begin{matrix}
x_1 \\
y_1 \\
1
\end{matrix}
\right)
=
\left(
\begin{matrix}
x_2 \\
y_2 \\
1
\end{matrix}
\right)
$$
leads to the equations
$$
\cos \theta - \sin \theta = 0 \\
\sin \theta + \cos \theta = 2
$$
where the second one has no real solution. So this is not working.</p>
<p>Looking at the rectangle, the ratio of the sides seems to have changed.</p>
<p>This means we have
\begin{align}
T
&=
T_\text{scale} \, T_\text{rot} \, T_\text{trans} \\
&=
\left(
\begin{matrix}
s & 0 & 0 \\
0 & t & 0 \\
0 & 0 & 1
\end{matrix}
\right)
\left(
\begin{matrix}
\cos \theta & -\sin \theta & 0 \\
\sin \theta & \cos \theta & 0 \\
0 & 0 & 1
\end{matrix}
\right)
\left(
\begin{matrix}
1 & 0 & t_x \\
0 & 1 & t_y \\
0 & 0 & 1
\end{matrix}
\right)
\\
&=
\left(
\begin{matrix}
s \cos \theta & -s \sin \theta & 0 \\
t \sin \theta & t \cos \theta & 0 \\
0 & 0 & 1
\end{matrix}
\right)
\left(
\begin{matrix}
1 & 0 & t_x \\
0 & 1 & t_y \\
0 & 0 & 1
\end{matrix}
\right)
\\
&=
\left(
\begin{matrix}
s \cos \theta & -s \sin \theta & s(t_x \cos \theta - t_y \sin \theta) \\
t \sin \theta & t \cos \theta & t(t_x \sin \theta + t_y \cos \theta) \\
0 & 0 & 1
\end{matrix}
\right)
\end{align}
And we end up with five parameters $t_x, t_y, \theta, s, t$ and only four equations.</p>
<p>This agrees with the six unknowns of equation $(*)$ and making use of the property of the rotation that $1 = \text{det}(R) = \cos^2 \theta + \sin^2 \theta$. I fail to spot a fifth equation so far.</p>
<p>Update: It seems the given data $A1, B1, A2, B2$ does not match the other part of the drawing, especially the original rectangle boundary and its transformed image.</p>
|
871,730 |
<p>Given x and y we define a function as follow : </p>
<pre><code>f(1)=x
f(2)=y
f(i)=f(i-1) + f(i+1) for i>2
</code></pre>
<p>Now given x and y, how to calculate f(n)</p>
<p>Example : If x=2 and y=3 and n=3 then answer is 1</p>
<p>as f(2) = f(1) + f(3), 3 = 2 + f(3), f(3) = 1.</p>
<p>Constraints are : x,y,n all can go upto 10^9.</p>
|
evinda
| 75,843 |
<p>You can use this code:</p>
<pre><code>#include <stdio.h>
int fib(int n,int x,int y){
if (n==1) return x;
else if (n==2) return y;
else return fib(n-1,x,y) + fib(n-2,x,y);
}
int main()
{
int x,y,n,z;
printf("Give a value for x:");
scanf("%d",&x);
printf("Give a value for y:");
scanf("%d",&y);
printf("Give a value for n:");
scanf("%d",&n);
z=fib(n,x,y);
printf("The result is: %d",z);
}
</code></pre>
|
871,730 |
<p>Given x and y we define a function as follow : </p>
<pre><code>f(1)=x
f(2)=y
f(i)=f(i-1) + f(i+1) for i>2
</code></pre>
<p>Now given x and y, how to calculate f(n)</p>
<p>Example : If x=2 and y=3 and n=3 then answer is 1</p>
<p>as f(2) = f(1) + f(3), 3 = 2 + f(3), f(3) = 1.</p>
<p>Constraints are : x,y,n all can go upto 10^9.</p>
|
Vishwa Iyer
| 71,281 |
<p>The general way to solve $$F(i+1)= F(i) - F(i - 1)$$
Is to first to see that if $F(i + 1)$ can be written as the sum of 2 geometric series, so that $$F(n) = ar^n + bs^n$$
So $$ar^{n+1} + bs^{n+1} = ar^{n} + bs^n - ar^{n-1} - bs^{n-1}$$
$$ar^{n-1}(1 - r + r^2) = -bs^{n-1}(1 - s + s^2)$$</p>
<p>Since $r\ne s$, it follows that both $r$ and $s$ must satisfy the quadratic equation $$1-x+x^2 = 0$$
So $$r = \frac{1 + \sqrt{3}i}{2}, s = \frac{1 - \sqrt{3}i}{2}$$
Now we find $a$ and $b$, using $F(1)$ and $F(2)$
$$F(1) = x = a\left(\frac{1 + \sqrt{3}i}{2}\right)^1 + b\left(\frac{1 - \sqrt{3}i}{2}\right)^1$$
$$F(2) = y = a\left(\frac{1 + \sqrt{3}i}{2}\right)^2 + b\left(\frac{1 - \sqrt{3}i}{2}\right)^2$$
And then you find the closed form of your recurrence series.</p>
|
244,489 |
<p>Given:</p>
<p>${AA}\times{BC}=BDDB$</p>
<p>Find $BDDB$:</p>
<ol>
<li>$1221$</li>
<li>$3663$</li>
<li>$4884$</li>
<li>$2112$</li>
</ol>
<p>The way I solved it:</p>
<p>First step - expansion & dividing by constant ($11$):
$AA\times{BC}$=$11A\times{BC}$</p>
<ol>
<li>$1221$ => $1221\div11$ => $111$</li>
<li>$3663$ => $3663\div11$ => $333$</li>
<li>$4884$ => $4884\div11$ => $444$</li>
<li>$2112$ => $2112\div11$ => $192$</li>
</ol>
<p>Second step - each result is now equal to $A\times{BC}$. We're choosing multipliers $A$ and $BC$ manually and in accordance with initial condition. It takes <strong>a lot</strong> of time to pick up a number and check whether it can be a multiplier.</p>
<p>That way I get two pairs:</p>
<p>$22*96$=$2112$</p>
<p>$99*37$=$3663$</p>
<p>Of course $99*37$=$3663$ is the right one.</p>
<p>Is there more efficient way to do this? Am I missing something?</p>
|
Colm
| 252,814 |
<p>Another way of solving this problem is with bounding boxes. A bounding box is the minimum rectangle containing a shape orthogonal to two chosen axes. Consider the bounding boxes of the two squares A and B orthogonal to the main square Q. In each case, the remaining space makes four right angled triangles. Call them the waste triangles.</p>
<p>Now you need two lemmas.</p>
<ol>
<li><p>Given two non-overlapping squares A' and B', A' either doesn't intersect the bounding box of B', or it does so only in one of the waste triangles. Proof: assume that there is an intersection in more than one triangle. Draw a line between the two points in the two distinct triangles. This line intersects B' at some point. But by convexity of the square, this point must also be in A'. Which is impossible. And we're done.</p></li>
<li><p>A square can always be placed in the corner of its bounding box so that it does not intersect the triangle at the opposite corner. For this, use algebra and express everything in terms of the two non-hypotenuse sides of the waste triangle. The bottom left corner of the bounding box is the origin, so all squares in this corner have their top-right corner along the line x = y. The intersection of this line and the line formed by the hypotenuse of the top-right waste triangle gives a value of x that is bigger than the side of the square. And so we're done.</p></li>
</ol>
<p>Now we're ready for the main proof. Take the square A. We can always put it into some corner of its bounding box by our two lemmas. Do this. Repeat the process for B. Now we have two squares orthogonal to the main square. Showing that the sides add up to at most 1 is now trivial.</p>
<hr>
<p>Proof of Lemma 2: We look for the intersection of the diagonal $x = y$ with the line extended from the hypotenuse of the top-right triangle. That is, the green and red lines from the image below:</p>
<p><img src="https://i.stack.imgur.com/I9YOe.png" alt="enter image description here"></p>
<p>Now the equation of the red line is $y = -(\frac{a}{b})x + c$ for some $c$.</p>
<p>To find this c, we just use the fact that that value of $y$ at $x = a$ for the red line is $a + b$. Putting this into the equation we get: $ a + b = -(\frac{a^2}{b}) + c$. From here we get $ c = (\frac{a^2 + b^2 + ab}{b})$. So the equation of our line becomes: $y = -(\frac{a}{b})x + \frac{a^2 + b^2 + ab}{b}$. Intersecting with the line $x = y$, the green line, we get: $x = -(\frac{a}{b})x + \frac{a^2 + b^2 + ab}{b}$, and after some algebraic manipulation, this works out as: $x = \frac{a^2 + b^2 + ab}{a + b}$.</p>
<p>At this point, it's good to reiterate the meaning of $x$. This is the size of the largest square that fits orthogonally into the bottom left corner of the black bounding box without intersecting the top-right waste triangle. Equivalently, it is the largest such square that doesn't intersect the red line. All we have now to prove, to show that the blue square can be placed in this corner, is that the side of the blue square is less than or equal $x$. So our goal is to solve the inequality:</p>
<p>$\sqrt{a^2 + b^2} \leq x$</p>
<p>Substitute our term for $x$:</p>
<p>$\sqrt{a^2 + b^2} \leq \frac{a^2 + b^2 + ab}{a + b}$</p>
<p>Positive terms allows squaring both sides:</p>
<p>$a^2 + b^2 \leq \frac{(a^2 + b^2 + ab)^2}{(a + b)^2} \equiv$
$(a^2 + b^2)(a + b)^2 \leq (a^2 + b^2 + ab)^2$</p>
<p>Now let $u = a^2 + b^2$ yielding:</p>
<p>$u(a + b)^2 \leq (u + ab)^2$</p>
<p>Now multiply it all out:</p>
<p>$ua^2 + ub^2 + 2uab \leq u^2 + a^2b^2 + 2uab \equiv$
$ua^2 + ub^2 \leq u^2 + a^2b^2$</p>
<p>And plug back in our term for $u$ getting:</p>
<p>$a^4 + 2a^2b^2 + b^4 \leq a^4 + b^4 + 2a^2b^2 + a^2b^2 \equiv$
$0 \leq a^2b^2$</p>
<p>Which is clearly true. And we're done proving the lemma.</p>
|
3,304,542 |
<p>Let <span class="math-container">$X,Y$</span> be Banach Spaces, show that <span class="math-container">$x_{n} \xrightarrow{w} x$</span> and <span class="math-container">$T \in BL(X,Y)\Rightarrow Tx_{n} \xrightarrow{w} Tx$</span></p>
<p>Question: Does sequential continuity (which <span class="math-container">$T$</span> clearly has) necessarily imply that <span class="math-container">$T$</span> is weak-sequentially continuous? If so, then the above is trivial. </p>
<p>Otherwise:</p>
<p><span class="math-container">$\vert\ell(Tx_{n})-\ell(Tx)\vert=\vert\ell(Tx_{n}-Tx)\vert=\vert T^{*}\ell(x_{n}-x)\vert\xrightarrow{n\to \infty} 0$</span> since <span class="math-container">$T^{*}\ell \in X^{*}$</span>. I am somewhat unsure about this, since I have not used boundedness of <span class="math-container">$T$</span> anywhere.</p>
|
Kavi Rama Murthy
| 142,385 |
<p>If <span class="math-container">$y^{*} \in Y^{*}$</span> then <span class="math-container">$x^{*}(x)=y^{*}(Tx)$</span> defines a continuous linear functional on <span class="math-container">$X$</span>. Hence <span class="math-container">$y^{*}(Tx_n)\to y^{*}(Tx)$</span>. This implies that <span class="math-container">$Tx_n \to Tx$</span> weakly. </p>
|
3,364,317 |
<p>Points A and B fixed, and point C moves on circle such that ABC acute triangle. <span class="math-container">$AT = BT$</span> and <span class="math-container">$TM \perp AC, \, TN \perp BC$</span>. How can I proove that all the middle perpendiculars (perpendicular bissector) to <span class="math-container">$MN$</span> passes through a fixed point?</p>
|
Batominovski
| 72,152 |
<p>Let the tangent at <span class="math-container">$A$</span> and the tangent at <span class="math-container">$B$</span> of the circumscribing circle <span class="math-container">$\mathcal{C}$</span> of <span class="math-container">$\triangle ABC$</span> meet at <span class="math-container">$U$</span> and let <span class="math-container">$V$</span> be the midpoint of <span class="math-container">$TU$</span>. We assert that <span class="math-container">$V$</span> is the required fixed point through which the perpendicular bisector of <span class="math-container">$MN$</span> passes as <span class="math-container">$C$</span> varies over <span class="math-container">$\mathcal{C}$</span>.</p>
<p>First we claim that <span class="math-container">$CU \perp MN$</span>. This is equivalent to proving that <span class="math-container">$\angle NCT=\angle UCM$</span>. To this end, we must verify that <span class="math-container">$CP$</span> bisects <span class="math-container">$\angle TCU$</span> where <span class="math-container">$P$</span> is the midpoint of the arc <span class="math-container">$AB$</span> of <span class="math-container">$\mathcal{C}$</span> not containing <span class="math-container">$C$</span> (note also that <span class="math-container">$CP$</span> bisects <span class="math-container">$\angle ACB$</span>). This means we have to show that <span class="math-container">$\mathcal{C}$</span> is the circle of Apollonius of the segment <span class="math-container">$TU$</span>. In other words, if <span class="math-container">$Q$</span> is the point such that line segment <span class="math-container">$PQ$</span> is a diameter of <span class="math-container">$\mathcal{C}$</span> we must show that
<span class="math-container">$$\frac{|PU|}{|PT|}=\frac{|QU|}{|QT|}.$$</span></p>
<p>Now if <span class="math-container">$r$</span> is the radius of <span class="math-container">$\mathcal{C}$</span> and <span class="math-container">$h$</span> is the length <span class="math-container">$|TB|$</span>, then it follows easily that <span class="math-container">$|OT|=\sqrt{r^2-h^2}$</span>
and <span class="math-container">$|OU|=\frac{r^2}{\sqrt{r^2-h^2}}$</span> where <span class="math-container">$O$</span> is the center of <span class="math-container">$\mathcal{C}$</span>. So
<span class="math-container">$$|PU|=\frac{r^2}{\sqrt{r^2-h^2}}-r,$$</span>
<span class="math-container">$$|PT|=r-\sqrt{r^2-h^2},$$</span>
<span class="math-container">$$|QU|=\frac{r^2}{\sqrt{r^2-h^2}}+r,$$</span>
<span class="math-container">$$|QT|=r+\sqrt{r^2-h^2}.$$</span>
That is
<span class="math-container">$$\frac{|PU|}{|PT|}=\frac{r}{\sqrt{r^2-h^2}}=\frac{|QU|}{|QT|},$$</span>
finishing the claim. (An alternative proof is to notice that <span class="math-container">$$\angle PBT=90^\circ-\angle OTB = 90^\circ-\left(90^\circ-\frac12\angle TOB\right)=\frac12\angle TOB$$</span>
and <span class="math-container">$$\angle UBP=\angle UBO-\angle OBP=90^\circ-\left(90^\circ-\frac12\angle TOB\right)=\frac12\angle TOB,$$</span>
so <span class="math-container">$PB$</span> bisects <span class="math-container">$\angle UBT$</span> internally. Because <span class="math-container">$Q$</span> is on the line <span class="math-container">$UT$</span> and <span class="math-container">$\angle PBQ=90^\circ$</span>, <span class="math-container">$QB$</span> bisects <span class="math-container">$\angle UBT$</span> externally.)</p>
<p>Now the dilation <span class="math-container">$d$</span> about <span class="math-container">$T$</span> with factor <span class="math-container">$+1/2$</span> sends the straight line <span class="math-container">$CU$</span> to a line <span class="math-container">$\ell$</span> perpendicular to the segment <span class="math-container">$MN$</span>. Since <span class="math-container">$\ell$</span> passes through the midpoint of <span class="math-container">$CT$</span>, which is the center of the circumscribing circle of the cyclic quadrilateral <span class="math-container">$CMTN$</span>, <span class="math-container">$\ell$</span> must be a perpendicular bisector of <span class="math-container">$MN$</span>. Because <span class="math-container">$d(U)=V$</span>, <span class="math-container">$\ell$</span> passes through <span class="math-container">$V$</span> as asserted.</p>
|
2,812,472 |
<p>I am trying to show that</p>
<p>$$
\frac{d^n}{dx^n} (x^2-1)^n = 2^n \cdot n!,
$$ for $x = 1$. I tried to prove it by induction but I failed because I lack axioms and rules for this type of derivatives. </p>
<p>Can someone give me a hint?</p>
|
cansomeonehelpmeout
| 413,677 |
<p><strong>Hint</strong>:</p>
<p>You could do this by induction, with the induction step:</p>
<p>$$\frac{d^{n+1}}{dx^{n+1}}(x^2-1)^{n+1}=\frac{d^n}{dx^{n}}\left (\frac{d}{dx}\left [(x^2-1)^n (x^2-1)\right ]\right )=2x(n+1)\frac{d^n}{dx^n}(x^2-1)^n+(2n+2)\frac{d^{n-1}}{dx^{n-1}}(x^2-1)^n$$</p>
<p>Now you only have to show that $$\frac{d^{n-1}}{dx^{n-1}}(x^2-1)^n |_{x=1}=0$$</p>
|
16,247 |
<p>Given a map $f:B^n \to S^n$, where $B^n$ is the unit ball and $S^n$ is the unit sphere, is it true that the degree of $f|_{S^n}$ is always 0, where $f_{S^n}$ is the restriction of $f$ to $S^n$? If so, why? </p>
<p>Thanks!</p>
|
Sean Tilson
| 627 |
<p>This is just a bit more of an explanation of Mariano's answer:
The homotopy class of a map between spheres of the same dimension is completely determined by its degree, that is, two maps $f,g: S^n \to S^n$ are homotopic if and only if they have the same degree. Now you can compute, using which ever definition you like, that the degree of the constant map is 0. A map out of a sphere is null homotopic, ie. homotopic to a constant map, if and only if it can be extended to a map of the whole unit ball. Since the map in question, $f|_{S^n}$, can indeed be extended to a map on the whole unit ball, namely $f$, it must be null homotopic. So by the above it must have the same degree as the constant map.</p>
<p>Again, this is just an elaboration of Mariano's answer.</p>
|
19,672 |
<p>The <a href="http://en.wikipedia.org/wiki/Stirling_numbers_of_the_second_kind" rel="nofollow noreferrer">Stirling number of the second kind</a> is the number of ways to partition a set of $n$ objects into $k$ non-empty subsets. In Mathematica, this is implemented as <code>StirlingS2</code>. How can I enumerate all the sets? Ideally I would like to get a list of lists, where each list contains $k$ lists.</p>
<p>The question <a href="https://mathematica.stackexchange.com/questions/3044/partition-a-set-into-subsets-of-size-k">Partition a set into subsets of size k</a> seems relevant.</p>
|
Mr.Wizard
| 121 |
<p>This is faster than the Combinatorica function:</p>
<pre><code>KSetP[{}, 0] = {{}};
KSetP[s_List, 0] = {};
KSetP[s_List, k_Integer] /; k > Length@s = {};
KSetP[s_List, k_Integer] /; k > 0 :=
Block[{ikf, s1 = s[[1]]},
ikf[set_] := Array[MapAt[#~Prepend~s1 &, set, #] &, Length@set];
Join[
Prepend[#, {s1}] & /@ KSetP[Rest@s, k - 1],
Join @@ ikf /@ KSetP[Rest@s, k]
]
]
(r1 = KSetPartitions[Range@12, 4]) // Timing // First
(r2 = KSetP[Range@12, 4]) // Timing // First
</code></pre>
<blockquote>
<pre><code>1.529
1.139
</code></pre>
</blockquote>
<p>The output is in a different order but it is equivalent:</p>
<pre><code>Sort[Sort /@ r1] === Sort[Sort /@ r2]
</code></pre>
<blockquote>
<p>True</p>
</blockquote>
|
243,510 |
<p>So this question is seriously flooring me.</p>
<p>Let $G$ be drawn in the plane so that it satisfies: </p>
<ol>
<li>The boundary of the infinite region is a cycle $C$</li>
<li>Every other region has boundary cycle of length $3$</li>
<li>Every vertex of $G$ not in $C$ has even degree</li>
</ol>
<p>Show that $\chi(G) \le 3$, where $\chi(G)$ is the chromatic number.</p>
<p>I know I have to use induction and consider two cases for the first step: Whether some two non-consecutive vertices of $C$ are adjacent and in the second case I would delete an edge of $C$ and apply the induction hypothesis.</p>
<p>Can anyone help?</p>
|
Bumblebee
| 51,408 |
<p>We induct on $|E(G)|$. The base case is trivial.</p>
<p>Suppose first that there are two non-consecutive vertices $u, v$ of $C$ which are adjacent. Let $e$ be the edge joining them. Then, $e$ partitions the graph into two graphs $G_1$ and $G_2$ on either side of $e$ in the plane, both of which satisfy the induction hypothesis, so $\chi(G_1), \chi(G_2) \leq 3$. Permuting the colors of (say) $G_2$ appropriately so that $u$ and $v$ are colored identically in $G_1$ and $G_2$, we obtain a 3-coloring of $G$, so $\chi(G) \leq 3$.</p>
<p>Otherwise, no two nonconsecutive vertices of $C$ are adjacent. Let $w_0, w_1$ be any consecutive vertices on $C$. We know that there is a vertex $u \in V(G) - V(C)$ with $w_0 u, w_1 u \in E(G)$. By the induction hypothesis, the graph $G' = G \setminus w_0 w_1$ has $\chi(G') \leq 3$. I claim that any 3-coloring of $G'$ has $w_0$ and $w_1$ as different colors, so that this coloring extends to a 3-coloring of $G$.</p>
<p>Suppose not, i.e. $w_0$ and $w_1$ are colored identically, and $u$ is colored differently. Consider the triangles incident to $u$, with vertices $u w_i w_{i + 1}$. Since $u$ has even degree, then the path $P = w_1 w_2 \ldots w_{\deg(u) - 1} w_0$ has odd length. Our 3-coloring of $G'$ determines that $P$ is 2-colored, and a 2-colored odd path has opposite colored ends (easily verified), which is a contradiction.</p>
|
3,297,110 |
<p>Evaluate <span class="math-container">$$\int\frac{1}{x^{\frac{25}{25} }\cdot x^{\frac{16}{25}}+x^{\frac{9}{25}}}dx$$</span></p>
<p>I start by factoring <span class="math-container">$$\int\frac{1}{x^{\frac{25}{25} }\cdot x^{\frac{16}{25}}+x^{\frac{9}{25}}}dx=\int\frac{1}{x^{\frac{9}{25}}\left(x^{\frac{32}{25}}+1\right)}dx$$</span></p>
<p>Can I do partial fraction from here? </p>
|
Mnifldz
| 210,719 |
<p>This is a cute <span class="math-container">$u$</span>-substitution problem. The crux of the problem is dividing out by the right factor of <span class="math-container">$x$</span>. Notice that if we factor out one copy of <span class="math-container">$x$</span> we obtain</p>
<p><span class="math-container">$$
\int \frac{1}{x\cdot x^{\frac{16}{25}} + x^{\frac{9}{25}} }dx \;\; =\;\; \int \frac{1}{x \left (x^{\frac{16}{25}} + x^{\frac{-16}{25}} \right )}dx.
$$</span></p>
<p>Now let <span class="math-container">$u = x^{\frac{16}{25}}$</span>, and thus <span class="math-container">$du = \frac{16}{25}x^{-\frac{9}{25}}dx$</span>. What this yields is</p>
<p><span class="math-container">\begin{eqnarray*}
\int \frac{1}{x \left (x^{\frac{16}{25}} + x^{\frac{-16}{25}} \right )}dx & = & \frac{25}{16}\int \frac{x^{\frac{9}{25}}} { x\left (u + \frac{1}{u} \right )} du \\
& = & \frac{25}{16} \int \frac{1}{x^{\frac{16}{25}} \left (u + \frac{1}{u} \right ) }du \\
& = & \frac{25}{16} \int \frac{1}{u\left (u + \frac{1}{u} \right )} du \\
& = & \frac{25}{16} \int \frac{1}{u^2 + 1} du.
\end{eqnarray*}</span></p>
<p>Take <span class="math-container">$u = \tan \theta$</span>. Final answer should be <span class="math-container">$\frac{25}{16}\tan^{-1}\left (x^{\frac{16}{25}}\right ) + c$</span>.</p>
|
2,149,006 |
<p>While learning the power rule, one thing popped up in my mind which is confusing me. We know what the power rule states :</p>
<p>$$\frac{\mathrm{d}}{\mathrm{d}x}(x^n) = nx^{n-1}$$ where $n$ is a real number.</p>
<blockquote>
<p>But instead of $n$, if we have a trig function like $\sin(x)$, <strong>will the power rule still apply?</strong></p>
</blockquote>
<p>Eg. We have a function $y = x^{\sin(x)}$, and thus by the power rule;</p>
<p>$$\frac{dy}{dx} = sin(x)x^{sin(x)-1}$$. </p>
<p>Is this possible? Please tell me if even the function I wrote above really does exist or not.</p>
<p>I know this may seem a stupid question to many, but please help because I cannot find any explanation to this. </p>
|
Khosrotash
| 104,171 |
<p>I think you must start with $y=x^x$ then $y=f(x)^{g(x)}$
$$y=x^x \to \ln y=\ln x^x\\\ln y=x\ln x \to (\ln y=x\ln x')\\
\dfrac{y'}{y}=(1.\ln x+x .\dfrac1x) \\y'=y\times (\ln x+1) \\y'=x^x(\ln x+1) $$then for $$y=f(x)^{g(x)} \to \ln y=\ln f(x)^{g(x)} \\\ln y = g(x)\ln f(x) \to
(\ln y = g(x)\ln f(x))'\\\dfrac{y'}{y}=(g'(x)\ln f(x)+g(x)\dfrac{f'(x)}{f(x)}) \\so\\ y'=y \times (g'(x)\ln f(x)+g(x)\dfrac{f'(x)}{f(x)}) \\y'=f(x)^{g(x)}\times (g'(x)\ln f(x)+g(x)\dfrac{f'(x)}{f(x)})$$
$$y=x^{\sin x} \to \ln y=\sin x \times \ln x \\$$apply derivation
$$\dfrac{y'}{y}=\cos x \ln x+ \sin x \times\dfrac{1}{x} \\ \to
y'=y(\cos x \ln x+ \sin x \times\dfrac{1}{x})\\y'=x^{\sin x}(\cos x \ln x+ \sin x \times \dfrac{1}{x})$$</p>
|
654,263 |
<p>My intention is neither to learn basic probability concepts, nor to learn applications of the theory. My background is at the graduate level of having completed all engineering courses in probability/statistics -- mostly oriented toward the applications without much emphasis on mathematical rigor.</p>
<p>Now I am very interested in learning the core logic and mathematical framework of probability theory, as a math branch. More specifically, I would like to learn answers to the following questions:</p>
<blockquote>
<p>(1) What are the necessary axioms from which we can build probability theory?</p>
<p>(2) What are the core theorems and results in the mathematical theory of probability?</p>
<p>(3) What are the derived rules for reasoning/inference, based on the theorems/results in probability theory?</p>
</blockquote>
<p>So I am seeking a book that covers the "heart" of mathematical probability theory -- not needing much on applications, or discussion on extended topics.</p>
<p>I would like to appreciate your patience for reading my post and any informative responses.</p>
<p>Regards,
user36125</p>
|
Geoff Pointer
| 96,501 |
<p>Get a copy of Elementary Probability by David Stirzaker from the library and take it from there. If you like it then buy a copy. I bought this book on a whim some years after majoring in pure maths and it rekindled my interest in probability.</p>
<p>Brief Review: It starts right from the beginning of Probability so very little previous knowledge of the subject is required. It has a chapter dedicated to Counting which is a particularly favourite topic of mine. It introduces pretty much every branch of Probability and gives ample opportunity for you to decide whether or not you'd want to take any particular area or areas any further.</p>
|
1,209,870 |
<p>I am stuck in a question which says that:
A particle moves on the X- axis according to equation $x=A+Bsin(\omega t)$. The motion is simple harmonic. Find the amplitude of SHM.</p>
<p>The answer of the above problem is "<em>B</em>".<br>
My question is: how the above equation is simple harmonic? I do not know how to reach the answer.
If there were two trig functions in the question then it would be easy to solve for the answer using vector addition method! The "<em>A</em>" in the equation is troubling me.</p>
|
Thomas Olson
| 68,485 |
<p>I made a video on this. Are you the one person that seen it and hit the like button on it?</p>
<p><a href="https://www.youtube.com/watch?v=Y7utC53CNs4" rel="nofollow">https://www.youtube.com/watch?v=Y7utC53CNs4</a></p>
<p>I think these are nd rose curves. Assume x_value is from the spherical coordinates on the wikipedia page</p>
<p><a href="http://en.wikipedia.org/wiki/N-sphere#Spherical_coordinates" rel="nofollow">http://en.wikipedia.org/wiki/N-sphere#Spherical_coordinates</a>
r=x_1+x_2+..x_n</p>
<hr>
<p>For 3d </p>
<p>$x_1=cos(\phi_1)*(x_1+x_2+x_3)$</p>
<p>$x_2=sin(\phi_1)*cos(\phi_2)*(x_1+x_2+x_3)$</p>
<p>$x_3=sin(\phi_1)*sin(\phi_2)*(x_1+x_2+x_3)$</p>
|
1,209,870 |
<p>I am stuck in a question which says that:
A particle moves on the X- axis according to equation $x=A+Bsin(\omega t)$. The motion is simple harmonic. Find the amplitude of SHM.</p>
<p>The answer of the above problem is "<em>B</em>".<br>
My question is: how the above equation is simple harmonic? I do not know how to reach the answer.
If there were two trig functions in the question then it would be easy to solve for the answer using vector addition method! The "<em>A</em>" in the equation is troubling me.</p>
|
Andrew D. Hwang
| 86,418 |
<p>This doesn't exactly answer the question, but with $k$, $m$, and $n$ positive integers, the parametric equations
\begin{alignat*}{3}
x(s, t) &= a\cos(mt) \cos^{k}(ns) &&\cos(t) &&\cos(s), \\
y(s, t) &= a\cos(mt) \cos^{k}(ns) &&\sin(t) &&\cos(s), \\
z(s, t) &= a\cos(mt) \cos^{k}(ns) &&\sin(s) &&
\end{alignat*}
may provide some enjoyable plotting along similar lines.</p>
<p>For example, here's the surface with $m = 4$, $n = 1$, and $k = 8$:</p>
<p><img src="https://i.stack.imgur.com/q5nQs.png" alt="A three-dimensional rose with eight lobes"></p>
<p>The underlying idea is to take $\rho = \cos(m\theta)\cos^{k}(n\phi)$ in spherical coordinates
$$
(x, y, z) = (\rho\cos\theta \cos\phi, \rho\sin\theta \cos\phi, \rho\sin\phi).
$$</p>
<p>You may also enjoy learning about <a href="https://en.wikipedia.org/wiki/Spherical_harmonics" rel="noreferrer">spherical harmonics</a>.</p>
|
1,353,184 |
<p>Let $a$, $b$, and $c$ be positive real numbers. Prove that</p>
<p>$$\sqrt{a^2 - ab + b^2} + \sqrt{a^2 - ac + c^2} \ge \sqrt{b^2 + bc + c^2}$$</p>
<p>Under what conditions does equality occur? That is, for what values of $a$, $b$, and $c$ are the two sides equal?</p>
|
Tae Hyung Kim
| 94,401 |
<p>The expressions $a^2 - ab + b^2$, $a^2 - ac + c^2$, and $b^2 + bc + c^2$ should remind you of the law of cosines. Let $\vec{a}$, $\vec{b}$, $\vec{c}$ be vectors such that $\vec{a}$ and $\vec{b}$ are separated by 60 degrees, and $\vec{a}$ and $\vec{c}$ are separated by 60 degrees. Consider the triangle formed by the three vectors. This triangle has side lengths
$$\sqrt{a^2 - ab + b^2}, \sqrt{a^2 - ac + c^2}, \sqrt{b^2 + bc + c^2}$$
according to the law of cosines. Then the inequality follows by the triangle inequality. </p>
<p>Equality occurs when the three vectors are collinear. The conditions for this shouldn't be too hard to find using basic geometric methods.</p>
|
1,849,809 |
<p>Let $n$ be a positive integer. Define $$\textbf{A}_n(x):= \left[\frac{1}{x+i+j-1}\right]_{i,j\in\{1,2,\ldots,n\}}$$ as a matrix over the field $\mathbb{Q}(x)$ of rational functions over $\mathbb{Q}$ in variable $x$. </p>
<p>(a) Prove that the <em>Hilbert matrix</em> $\textbf{A}_n(0)$ is an invertible matrix over $\mathbb{Q}$ and all entries of the inverse of $\textbf{A}_n(0)$ are integers.</p>
<p>(b) Determine the greatest common divisor (over $\mathbb{Z}$) of all the entries of $\big(\textbf{A}_n(0)\big)^{-1}$.</p>
<p>(c) Show that $\textbf{A}_n(x)$ is an invertible matrix over $\mathbb{Q}(x)$ and every entry of the inverse of $\textbf{A}_n(x)$ is a polynomial in $x$.</p>
<p>(d) Prove that $x+n$ is the greatest common divisor (over $\mathbb{Q}[x]$) of all the entries of $\big(\textbf{A}_n(x)\big)^{-1}$.</p>
<p><strike>Parts (a) and (c) are known.</strike> <strike>Parts (b) and (d) are open.</strike> <strike>Now, Part (d) is known (see i707107's solution below), but Part (b) remains open, although it seems like the answer is $n$.</strike></p>
<hr>
<p>Recall that
$$\binom{t}{r}=\frac{t(t-1)(t-2)\cdots(t-r+1)}{r!}$$
for all $t\in\mathbb{Q}(x)$ and $r=0,1,2,\ldots$. According to i707107, the $(i,j)$-entry of $\big(\textbf{A}_n(x)\big)^{-1}$ is given by
$$\alpha_{i,j}(x)=(-1)^{i+j}\,(x+n)\,\binom{x+n+i-1}{i-1}\,\binom{x+n-1}{n-j}\,\binom{x+n+j-1}{n-i}\,\binom{x+i+j-2}{j-1}\,.\tag{*}$$
This means that, for all integers $k$ such that $k\notin\{-1,-2,\ldots,-2n+1\}$, the entries of $\big(\textbf{A}_n(k)\big)^{-1}$ are integers. <strong><em>I now have a new conjecture, which is the primary target for the bounty award.</em></strong></p>
<blockquote>
<p><strong>Conjecture:</strong> The greatest common divisor $\gamma_n(k)$ over $\mathbb{Z}$ of the entries of $\big(\textbf{A}_n(k)\big)^{-1}$, where $k$ is an integer not belonging in the set $\{-1,-2,\ldots,-2n+1\}$, is given by $$\gamma_n(k)=\mathrm{lcm}(n,n+k)\,.$$</p>
</blockquote>
<p>It is clear from (*) that $n+k$ must divide $\gamma_n(k)$. However, it is not yet clear to me why $n$ should divide $\gamma_n(k)$. I would like to have a proof of this conjecture, or at least a proof that $n \mid \gamma_n(k)$.</p>
<hr>
<p>Let $M_n$ denote the (unitary) cyclic $\mathbb{Z}[x]$-module generated by $\dfrac{1}{\big((n-1)!\big)^2}\,(x+n)$. Then, the (unitary) $\mathbb{Z}[x]$-module $N_n$ generated by the entries of $\big(\textbf{A}_n(x)\big)^{-1}$ is a $\mathbb{Z}[x]$-submodule of $M_n$. </p>
<p>We also denote by $\tilde{M}_n$ for the (unitary) $\mathbb{Z}$-module generated by $\dfrac{1}{\big((n-1)!\big)^2}\,(x+n)\,x^l$ for $l=0,1,2,\ldots,2n-2$. Then, the (unitary) $\mathbb{Z}$-module $\tilde{N}_n$ generated by the entries of $\big(\textbf{A}_n(x)\big)^{-1}$ is a $\mathbb{Z}$-submodule of $\tilde{M}_n$.</p>
<p>For example, $M_2/N_2$ is isomorphic to the (unitary) $\mathbb{Z}[x]$-module $\mathbb{Z}/2\mathbb{Z}$ (in which $x$ acts trivially), and $\tilde{M}_2/\tilde{N}_2$ is isomorphic to the (unitary) $\mathbb{Z}$-module $\mathbb{Z}/2\mathbb{Z}$. Hence, $\left|M_2/N_2\right|=2=\left|\tilde{M}_2/\tilde{N}_2\right|$. For $n=3$, Mathematica yields
$$\tilde{M}_3/\tilde{N}_3\cong (\mathbb{Z}/2\mathbb{Z})\oplus(\mathbb{Z}/3\mathbb{Z})^{\oplus 2}\oplus(\mathbb{Z}/4\mathbb{Z})^{\oplus 3}\,,$$
as abelian groups. That is, $\left|\tilde{M}_3/\tilde{N}_3\right|=1152$. On the other hand,
$$M_3/N_3\cong \mathbb{Z}[x] \big/\left(12,2x^2+6x+4,x^4-x^2\right)$$
as $\mathbb{Z}[x]$-modules, which gives $\left|M_3/N_3\right|=576$. </p>
<blockquote>
<p><strong>Question:</strong> Describe the factor $\mathbb{Z}[x]$-module $M_n/N_n$ and the factor $\mathbb{Z}$-module $\tilde{M}_n/\tilde{N}_n$. It is easily seen that $\left|M_n/N_n\right|\leq\left|\tilde{M}_n/\tilde{N}_n\right|$. What are $\left|M_n/N_n\right|$ and $\left|\tilde{M}_n/\tilde{N}_n\right|$? It can be shown also that the ratio $\dfrac{\left|\tilde{M}_n/\tilde{N}_n\right|}{\left|M_n/N_n\right|}$ is an integer, provided that $\left|\tilde{M}_n/\tilde{N}_n\right|$ is finite. Compute $\dfrac{\left|\tilde{M}_n/\tilde{N}_n\right|}{\left|M_n/N_n\right|}$ for all integers $n>0$ such that $\left|\tilde{M}_n/\tilde{N}_n\right|<\infty$. Is it always the case that $\left|\tilde{M}_n/\tilde{N}_n\right|$ is finite?</p>
</blockquote>
<p><strong><em>Apart from the conjecture above, this question is also eligible for the bounty award.</em></strong> I have not yet fully tried to deal with any case involving $n>3$. However, for $n=4$, the module $\tilde{M}_4/\tilde{N}_4$ is huge:
$$ \tilde{M}_4/\tilde{N}_4\cong (\mathbb{Z}/2\mathbb{Z})^{\oplus 2}\oplus(\mathbb{Z}/3\mathbb{Z})^{\oplus 3}\oplus(\mathbb{Z}/8\mathbb{Z})^{\oplus 2}\oplus(\mathbb{Z}/9\mathbb{Z})^{\oplus 2}\oplus(\mathbb{Z}/16\mathbb{Z})\oplus(\mathbb{Z}/27\mathbb{Z})$$
as abelian groups.</p>
|
Sungjin Kim
| 67,070 |
<p>This is a solution to (d), and partially for (b). </p>
<p>We use <a href="https://en.wikipedia.org/wiki/Cauchy_matrix" rel="nofollow">Cauchy Matrix</a>, and its evaluation of inverse given by Schechter: </p>
<p>If $T$ is a $n\times n$ Cauchy matrix on the sequences $\{x_i\}$, $\{y_j\}$, then
$S=T^{-1}=[s_{ij}]$ is given by:</p>
<p>$$s_{ij} = (x_j - y_i) A_j(y_i) B_i(x_j) $$
where
$$A_i(t) = \frac{A(t)}{A^\prime(x_i)(t-x_i)} \quad\text{and}\quad B_i(t) = \frac{B(t)}{B^\prime(y_i)(t-y_i)}$$
with
$$A(t) = \prod_{i=1}^n (t-x_i) \quad\text{and}\quad B(t) = \prod_{i=1}^n (t-y_i). $$</p>
<p>The matrix $\mathbf{A}_n(x)$ is considered as a Cauchy matrix on the sequences
$$x_i = x+i \quad\text{and} \quad y_j = -j+1.$$</p>
<p>Using the above inverse formula, the $ij$ entry of the inverse up to sign is
$$
\frac{ (x+n)\cdot (x+i) \cdots (x+n-1)}{(n-i)!} \frac{ (x+n+1)\cdots (x+n+i-1)}{(i-1)!} \frac{ (x+j) \cdots (x+j+n-1)}{(x+i+j-1)(n-j)!(j-1)!}
$$</p>
<p>We see that every entry is divisible by $x+n$. </p>
<p>To prove that the GCD is $x+n$, we need to prove that any possible factor other than $x+n$ which is present in one entry is not present in some another entry. This is easy to see.</p>
<p>Moreover, the factor $(x+n)^2$ is present for most of times, but it is possible to have only one $x+n$ in the factorization (in case $x+i+j-1= x+n$).</p>
<p>For (b), we use the well known formula for inverse of Hilbert matrix:
$$
(H^{-1})_{ij}=(-1)^{i+j}(i+j-1){n+i-1 \choose n-j}{n+j-1 \choose n-i}{i+j-2 \choose i-1}^2
$$<br>
Upon rearranging, we obtain that the $ij$ entry up to sign is
$$
n\frac{(n+i-1)\cdots (n+1)}{(i-1)!} \frac{(n-1)\cdots (n-j+1)}{(j-1)!} \binom{n+j-1}{i+j-1} \binom{i+j-2}{i-1}
$$
which is simplified to
$$
n \binom{n+i-1}{i-1} \binom{n-1}{j-1} \binom{n+j-1}{i+j-1} \binom{i+j-2}{i-1}.
$$
This is clearly divisible by $n$, but I was not able to show that $n$ is the GCD which I conjecture to be. </p>
|
3,382,561 |
<p>I just encountered the random variable <span class="math-container">$Y = |X|$</span>, where <span class="math-container">$X \sim \text{N}(\mu, \sigma^2)$</span>. Now, based on what we know about the absolute value function, this random variable is still continuous; however, the absolute value function means that there exists a cusp at <span class="math-container">$X = 0$</span>, and so the derivative is undefined at this point. </p>
<p>This makes me wonder: How does this affect the PDF and CDF? How would we go about calculating such things in this case? </p>
<p>I would greatly appreciate it if people could please take the time to clarify this situation.</p>
|
GReyes
| 633,848 |
<p>New CDF:if <span class="math-container">$z\le 0$</span>, clearly <span class="math-container">$F_Y(z)=0$</span>.Otherwise,
<span class="math-container">$$
F_Y(z)=P\{|X|\le z\}=P\{-z\le X\le z\}=2\Phi(z)-1
$$</span>
where <span class="math-container">$\Phi$</span> is the CDF of the normal variable. Your PDF is zero for <span class="math-container">$z<0$</span>, and <span class="math-container">$2f_X(z)$</span> for <span class="math-container">$z> 0$</span> (just take the derivative). At <span class="math-container">$z=0$</span> it has a jump.</p>
<p>Well, the above is for <span class="math-container">$X\sim \mathcal{N}(0,1)$</span> but you can easily adapt it to the general case.</p>
|
2,063,038 |
<p>Let <span class="math-container">$S$</span> be the region in the plane that is inside the circle <span class="math-container">$(x-1)^2 + y^2 = 1$</span> and outside the circle <span class="math-container">$x^2 + y^2 = 1 $</span>. I want to calculate the area of <span class="math-container">$S$</span>.</p>
<h3>Try:</h3>
<p>first, the circles intersect when <span class="math-container">$x^2 = (x-1)^2 $</span> that is when <span class="math-container">$x = 1/2$</span> and so <span class="math-container">$y =\pm \frac{ \sqrt{3} }{2} $</span>. So, using washer method, we have</p>
<p><span class="math-container">$$Area(S) = \pi \int\limits_{- \sqrt{3}/2}^{ \sqrt{3}/2} [ (1+ \sqrt{1-y^2})^2 - (1-y^2) ] dy $$</span></p>
<p>is this the correct setting for the area im looking for?</p>
|
Christian Blatter
| 1,303 |
<p>The shape $S$ is extremely unwieldy for integration, and has to be cut up in several parts for that purpose. Use elementary geometry instead:</p>
<p>The area in question is a full unit disk minus one third of such a disk, minus two small circular segments. The latter are a sixth of a disc with an equilateral triangle removed. It follows that
$${\rm area}(S)=\pi-{\pi\over3}-2\left({\pi\over6}-{\sqrt{3}\over4}\right)={\pi\over3}+{\sqrt{3}\over2}\ .$$</p>
|
1,170,708 |
<p>What functions satisfy $f(x)+f(x+1)=x$?</p>
<p>I tried but I do not know if my answer is correct.
$f(x)=y$</p>
<p>$y+f(x+1)=x$</p>
<p>$f(x+1)=x-y$</p>
<p>$f(x)=x-1-y$</p>
<p>$2y=x-1$</p>
<p>$f(x)=(x-1)/2$</p>
|
BCLC
| 140,308 |
<p>$f(x) = x - 1 - y$ is wrong. $y$ is a function of $x$. In my opinion, it was such a good try. Wouldn't have thought of that approach. Anyway, is $f(x)$ supposed to be linear? Write out: $f(x) = ax + b$ if so.</p>
<p>Then plug in:</p>
<p>$a(x + 1) + b + ax + b = 1$</p>
<p>Hint: $1 = 0x + 1$.</p>
|
3,860,199 |
<p>What is the integration of <span class="math-container">$\int x^{|x|} dx$</span>?<br />
Actually through several google search finally I have found a solution for the problem <span class="math-container">$\int x^{x} dx$</span> using Gamma function function and I am hardly sure that the same method can be applied to solve the problem if <span class="math-container">$x^{x}$</span> is going to be replaced by <span class="math-container">$x^{|x|}$</span>.</p>
<p>So, what is the appropriate approach to solve the problem <span class="math-container">$\int x^{|x|} dx$</span></p>
|
Henry Lee
| 541,220 |
<p>assuming that this is for <span class="math-container">$x\in\mathbb{R}$</span> you will still run into problems with the output of the function being complex for <span class="math-container">$x<0$</span> since you will have a negative number to a non-integer power, so in that respect this function is very similar to the functions <span class="math-container">$x^x$</span> and <span class="math-container">$x^{-x}$</span> so as others have suggested there are many papers on that integral so I suggest that you use those in this manner:
<span class="math-container">$$\int x^{|x|}dx=\begin{cases}\int x^xdx&x>0\\\int x^{-x}dx&x<0\end{cases}$$</span>
as for at <span class="math-container">$x=0$</span>, both cases will return the same result as it can be proved that <span class="math-container">$\lim_{x\to 0}x^x=1$</span></p>
|
174,075 |
<p>What is the difference when a line is said to be normal to another and a line is said to be perpendicular to other?</p>
|
dtldarek
| 26,306 |
<p>There are different kind of contexts you use term <em>normal</em> in mathematics. You often use <em>perpendicular</em> in case of two or three dimensional geometry, and in hi-dimensional spaces (e.g. infinite-dimensional) a term <em>orthogonal</em> is more common. On the other hand in the context of vectors, <em>normal</em> usually means also that the vector is of unit length, however this is not a must (but it is for example if you speak about orthonormal base). </p>
<p>There is yet another related meaning in computer graphics, where <em>normal</em> is the direction which you would use to reflect the light. This can be a vector perpendicular to rendered face, but for nice effects you use one perpendicular to the original surface, not its approximation. Also, there is <a href="http://en.wikipedia.org/wiki/Bump_mapping" rel="noreferrer">bump mapping</a> technique that lets you change the normal vectors and achieve look of some kind of wrinkles and bumps.</p>
<p>Finally, e.g. in topology or group theory term <em>normal</em> means something completely different, however, I will leave those and others out as I suspect this was not the scope of your question. Hope this helps ;-)</p>
|
174,075 |
<p>What is the difference when a line is said to be normal to another and a line is said to be perpendicular to other?</p>
|
Dovydas Vaiksnys
| 188,561 |
<p>If normal is 90 degree to the surface, that means normal is used in 3D. Perpendicular in that case is more in 2D referring two same entities (line-line) with 90 degree angle between them. Normal in this case could refer 2 different entities (line-surface) making this term valid for 3D case (I definitely remember hearing term "line normal to surface", rather then "line perpendicular to the surface") - line and surface makes 3D case; line and line is one plane - 2D. This I just figured out from one of the answers here which made most sense for me.</p>
|
4,566,926 |
<p><span class="math-container">$a_n$</span> converges to <span class="math-container">$a \in \mathbb{R}$</span>, find <span class="math-container">$\lim\limits_{n \rightarrow \infty}\frac{1}{\sqrt{n}}(\frac{a_1}{\sqrt{1}} + \frac{a_2}{\sqrt{2}} +\frac{a_3}{\sqrt{3}} + \cdots + \frac{a_{n-1}}{\sqrt{n-1}} + \frac{a_n}{\sqrt{n}})$</span></p>
<p>I've tried to use Squeeze theorem. Nothing good. Maybe I've tried it in a wrong way. According to my assumptions, this sequence diverges (I mean "goes" to infinity).</p>
<p>Can we solve this using Stolz–Cesàro theorem? Is there easier/better/more correct approach? If there is, please, share your thoughts</p>
|
Thomas Andrews
| 7,933 |
<p>Define <span class="math-container">$$A_n=\sum_{k=1}^{n}\frac{a_k}{\sqrt{k}}$$</span> and <span class="math-container">$B_n=\sqrt n.$</span></p>
<p>The we want to find the limit of <span class="math-container">$\frac{A_n}{B_n}.$</span></p>
<p>Then Stolz-Cesàro requires us to find the limit:</p>
<p><span class="math-container">$$\frac{A_{n+1}-A_n}{B_{n+1}-B_n}=\frac{a_{n+1}}{\sqrt{n+1}\left(\sqrt{n+1}-\sqrt{n}\right)}$$</span></p>
<p>But <span class="math-container">$$\frac{1}{\sqrt{n+1}-\sqrt n}=\sqrt{n+1}+\sqrt n.$$</span></p>
<p>So you get <span class="math-container">$$\frac{\sqrt{n+1}+\sqrt n}{\sqrt {n+1}}\cdot a_{n+1}\to 2a.$$</span></p>
<p>Now, Stolz requires <span class="math-container">$A_n\to \infty$</span> and <span class="math-container">$B_n\to\infty,$</span> or both converge to <span class="math-container">$0.$</span> So to apply Stolz, you'll probably need <span class="math-container">$a$</span> to be non-zero.</p>
<p>So you might have to prove the case <span class="math-container">$a=0$</span> separately.</p>
|
27,951 |
<p>Something I notice is when there's an advanced/specialized question, it often receives very few upvotes. Even if it is seemingly well written. I try to upvote advanced questions <strong>that I might not even understand</strong>, if they appear well written. </p>
<p>Is this good behaviour? Should we encourage upvoting seemingly well-written questions <strong>even if you don't understand it</strong>? </p>
|
Rob
| 510,296 |
<blockquote>
<p>Should we encourage upvotes on more advanced questions, that are seemingly well written, even if you don't understand it? ... Is this good behaviour? Should we encourage upvoting seemingly well-written questions even if you don't understand it? </p>
</blockquote>
<p>The lack of voters for some questions is discussed here: <a href="https://meta.stackexchange.com/questions/9508/why-arent-people-voting-for-questions">Why aren't people voting for questions?</a> - at length. If it's a bad question it should get voted down, if it's a good question (see: <a href="https://math.meta.stackexchange.com/questions/9959/how-to-ask-a-good-question">How to ask a good question?</a>) it should be voted up. If you don't know don't be embarrassed (no one knows) just move on, you don't want to upvote nonsense.</p>
<p>It would be great if there was an "Interesting Questions" Tab that Mods and a group of designated experts could tag questions for, so the question would get more exposure (without necessarily having many views or upvotes).</p>
<p>If Mods could be given a "Bounty Wallet", so they could add a bounty to a question without pulling the cash from their own pocket, and a question received an "Interesting Question" flag from a single person (from a small group of designated experts), the wallet could supply a couple of hundred rep to at least get the question some views.</p>
<p>The actual upvoting should be reserved for the so-called "good question" (as understood by the voters).</p>
<p>A prime example is some people can take a good photo or are skilled at Mathjax, that doesn't make the question they ask useful or good. Some people need help taking photos or using Mathjax, that doesn't make their question bad or not of general interest; unless it's a FAQ or dupe.</p>
<p>A means to toss a 'good question' with few views or voters into an additional Tab, even to push it into the "Hot Questions" (when it's cold), would be helpful.</p>
<p>Voting for things you don't understand (up or down) puts an odd spin on the ball. You wouldn't wash your neighbour's car if it looked dirty, walk their dog if it looked lonely, take their wife out to dinner if they appeared hungry - if you shouldn't mess with stuff you don't know about then don't. </p>
<p>Those points were made in the links listed, if you're going to answer why not upvote the question, if you choose an answer why not upvote it also. If you don't know that's OK, but adding random votes or answers doesn't help the majority. </p>
<p>Unfortunately the asker will have to ask 'popular' questions for more votes and easier questions for more answers. Spending 10 minutes to get 10 rep is something each must decide for themselves if it's worth it.</p>
|
524,073 |
<p>Hey can some help me with this textbook question</p>
<p>Let $R^{2×2}$ denote the vector space of 2×2 matrices, and let</p>
<p>$S =\left\{
\left[\begin{matrix}
a \space b \\
b \space c \\
\end{matrix}\right]\mid a,b,c \in \mathbb{R}\right\}$</p>
<p>Find (with justication) a basis for $S$ and determine the dimension of $S$.</p>
|
Community
| -1 |
<p><strong>Hint</strong>: $\mathbb{R}^{2 \times 2}$ is a four-dimensional space, so the dimension of $S$ is at most $4$. In fact, it's at most $3$ since the set $S$ does not contain the matrix</p>
<p>$$\left[\begin{array}{c} 0 & 1 \\ -1 & 0 \end{array}\right]$$</p>
<p>So $S$ in fact has dimension at most $3$.</p>
<hr>
<p>Can you find three linearly independent vectors in $S$? Note that you have three variables: $a$, $b$, and $c$; this should guide your construction.</p>
<hr>
<p>As discussed in the comments below, three such matrices are</p>
<p>$$\left[ \begin{array}{c} 1 & 1 \\ 1 & 0 \end{array}\right], \left[ \begin{array}{c} 1 & 0 \\ 0 & 0 \end{array}\right], \left[ \begin{array}{c} 1 & 1 \\ 1 & 1 \end{array}\right]$$</p>
|
3,176,155 |
<blockquote>
<p>Let <span class="math-container">$ABCD$</span> be a parallelogram. I proved that the angle bisectors of <span class="math-container">$A$</span>, <span class="math-container">$B$</span>, <span class="math-container">$C$</span>, <span class="math-container">$D$</span> form a rectangle. How can I prove that the diagonals of this rectangle are parallel to the sides of <span class="math-container">$ABCD$</span>? And is there a relation between the length of these diagonals and <span class="math-container">$AB$</span> or <span class="math-container">$BC$</span>?</p>
</blockquote>
<p>I'm looking for an elementary solution only using parallelograms, congruent triangles.</p>
<p><a href="https://i.stack.imgur.com/uehHJ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/uehHJ.png" alt="enter image description here"></a></p>
|
Blue
| 409 |
<p>Let the angle bisectors at <span class="math-container">$A$</span> and <span class="math-container">$B$</span> meet at <span class="math-container">$P$</span>. Drop perpendiculars from <span class="math-container">$P$</span> to points <span class="math-container">$A^\prime$</span>, <span class="math-container">$B^\prime$</span>, <span class="math-container">$Q$</span> on the sides of the parallelogram as shown:</p>
<p><a href="https://i.stack.imgur.com/AtoVb.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/AtoVb.png" alt="enter image description here"></a></p>
<p>Clearly, we have constructed a few similar right triangles, and, in particular, two pairs of congruent right triangles. We see that <span class="math-container">$P$</span> must be halfway between opposite sides of the parallelogram; likewise for <span class="math-container">$P^\prime$</span>. This guarantees the desired parallelism property. <span class="math-container">$\square$</span></p>
<p>For a relation about the lengths, lop-off the trapezoid on one side and paste it to the other, getting a rectangle whose width is equal to the original base of the parallelogram, <span class="math-container">$\overline{AD}$</span>. </p>
<p><a href="https://i.stack.imgur.com/TAdZ9.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/TAdZ9.png" alt="enter image description here"></a></p>
<p>Then, for the configuration shown (where <span class="math-container">$|AD|>|AB|$</span>):</p>
<blockquote>
<p><span class="math-container">$$|AD| = a + b + d = |AB| + d \qquad\to\qquad d = |AD| - |AB|$$</span></p>
</blockquote>
|
1,748,542 |
<p>So a friend of mine has a little project going, and needs some help.</p>
<p>Basically, we want to create a function that takes two variables; One $X$, and one that we call $DC$ ("Difficulty Class, as this is for a pen-and-paper game).</p>
<p>The output should be $0$ if $X\leq (1/2)DC$, and it should be $100$ if $X\geq 2 DC$.</p>
<p>The rest of the curve should look roughly like an $X^3$ curve, centered around $DC$ (So it goes downwards below $DC$, and upwards over $DC$, with the derivative $F'(DC) = 0$ )</p>
<p>Thank you in advance!</p>
|
bartgol
| 33,868 |
<p>As stated, there is no unique solution. You are looking for a curve of the form $f(x)=ax^3+bx^2+cx+d$. You have three conditions to enforce:</p>
<p>$$
f(DC/2)=0\\
f(2DC) = 100\\
f'(DC)=0
$$</p>
<p>If you plug those conditions in the expression for $f$ (and its derivative) you will get a system of three equations in four unknowns ($a,b,c,d$). You need an additional condition to uniquely fix the solution. Perhaps you can fix $f(DC)$?</p>
|
70,500 |
<p>I am trying to find $\cosh^{-1}1$ I end up with something that looks like $e^y+e^{-y}=2x$. I followed the formula correctly so I believe that is correct up to this point. I then plug in $1$ for $x$ and I get $e^y+e^{-y}=2$ which, according to my mathematical knowledge, is still correct. From here I have absolutely no idea what to do as anything I do gives me an incredibly complicated problem or the wrong answer.</p>
|
Gottfried Helms
| 1,714 |
<p>It may be more helpful to consider the significant hyperbolic identities first. We have in general: </p>
<p>$\small \begin{array} {rcllll}
1)& \exp(z) &=& \cosh(z) + \sinh(z) \\
2)& 1 &=& \cosh(z)^2 - \sinh(z)^2 \\
&&& \implies \\
3)&\sinh(z) &=& \pm \sqrt{\cosh(z)^2-1} & \text{ using 2)}\\
4)& \exp(z)&=& \cosh(z) \pm \sqrt{\cosh(z)^2-1} & \text{ using 1) and 3)}\\
\end{array} $ </p>
<p>Now the given problem is to find another expression for $\small y=\cosh^{-1}(x)$ which means $\small x = \cosh(y) $<br>
We use 4) and insert our current <em>y</em> for the general <em>z</em> to get </p>
<p>$\small \begin{array} {rcllll}
5)& \exp(y)&=& \cosh(y) \pm \sqrt{\cosh(y)^2-1} & \text{ using 4)}\\
6)& \exp(y)&=& x \pm \sqrt{x^2-1} & \text{ inserting x for } \cosh(y)\\
7)& y&=& \log(x \pm \sqrt{x^2-1} ) & \\
8)& \cosh^{-1}(x)&=& \log(x \pm \sqrt{x^2-1} ) &\text{ inserting } \cosh^{-1}(x) \text{ for } y \\
9)& \cosh^{-1}(1)&=& ??? \\
\end{array} $ </p>
<p>Now <em>8)</em> can be used as a new, general hyperbolic identity like that in the list from <em>1)</em> to <em>4)</em> and <em>9)</em> is your remaining little to-do ...</p>
|
2,459,169 |
<p>Let $A$ be an $n \times n$ matrix. Show that if $A^2 = 0$, then $I − A$ is nonsingular and $(I − A)^{−1} = I + A$.</p>
<p>(Matrix Algebra)</p>
|
ajotatxe
| 132,456 |
<p><strong>Hint</strong>:</p>
<p>Verify that $(I-A)(I+A)=I$.</p>
|
3,113,553 |
<p>Let <span class="math-container">$X$</span> be a set, and <span class="math-container">$\mathscr{G}\subset \mathcal{P}(X)$</span> with <span class="math-container">$X\in\mathscr{G}$</span>. Then <span class="math-container">$\sigma(\mathscr{G})$</span> is defined as the “smallest” <span class="math-container">$\sigma$</span>-algebra containing <span class="math-container">$\mathscr{G}$</span>. It would <em>seem</em> that we could construct <span class="math-container">$\sigma(\mathscr{G})$</span> with</p>
<p><span class="math-container">$$\mathscr{G}’:=\left\{\left( \bigcup_{i\in\mathbb{N}}G_i \right),\left( \bigcup_{i\in\mathbb{N}}{G_i} \right)^c \,\, \Big| \,\, G_i,\in\mathscr{G} \text{ or } {G_i}^c\in\mathscr{G} \right\},$$</span></p>
<p>but my measure theory book suggests this is not the case. <strong>But why not?</strong></p>
<p>My first thought was, “maybe those unions generate new sets, and unions of <em>those</em> new sets were not already accounted for.” But this is not a problem: If <span class="math-container">$A$</span> and <span class="math-container">$B$</span> are two sets generated by these countable unions from the sequences <span class="math-container">$(A_i)$</span> and <span class="math-container">$(B_i)$</span>, then <span class="math-container">$A\cup B$</span> is generated by <span class="math-container">$(A_1,B_1,A_2,B_2,\dots)$</span>, and so is already accounted for. The same reasoning can be extended to countable unions of the sets generated by these countable unions of <span class="math-container">$G_i$</span> sets. As for <em>uncountable</em> unions, who cares, the <span class="math-container">$\sigma$</span>-algebra axioms don’t say they need to be included anyway.</p>
|
Set
| 26,920 |
<p>It's actually not terribly difficult, the key observation is that the integral actually splits into a bunch of single variable integrals,</p>
<p><span class="math-container">\begin{align}
\int q_\phi({\bf z})\|{\bf z}\|_2^2d{\bf z}&=\int q_\phi({\bf z})(z_1^2+\ldots+z_n^2)d{\bf z}\notag\\
&=\int q_\phi({\bf z})z_1^2d{\bf z} + \ldots + \int q_\phi({\bf z})z_n^2d{\bf z}\notag\\
&=\int q_\phi(z_1)z_1^2dz_1 + \ldots + \int q_\phi(z_n)z_n^2dz_n\notag.
\end{align}</span></p>
<p>Now since <span class="math-container">$q_\phi(z_i) = N(\mu, \sigma^2)$</span>, the above expression can be rewritten in expected value form as,</p>
<p><span class="math-container">$$\sum_{i=1}^n\text{E}Z_i^2,$$</span></p>
<p>and using the identity that <span class="math-container">$\text{Var}X = \text{E}X^2 - (\text{E}X)^2$</span>, we obtain,</p>
<p><span class="math-container">$$\sum_{i=1}^n\text{E}Z_i^2=\sum_{i=1}^n(\text{E}Z_i)^2 + \text{Var}Z_i=\sum_{i=1}^n\mu_i^2 + \sigma_i^2.$$</span></p>
|
3,113,553 |
<p>Let <span class="math-container">$X$</span> be a set, and <span class="math-container">$\mathscr{G}\subset \mathcal{P}(X)$</span> with <span class="math-container">$X\in\mathscr{G}$</span>. Then <span class="math-container">$\sigma(\mathscr{G})$</span> is defined as the “smallest” <span class="math-container">$\sigma$</span>-algebra containing <span class="math-container">$\mathscr{G}$</span>. It would <em>seem</em> that we could construct <span class="math-container">$\sigma(\mathscr{G})$</span> with</p>
<p><span class="math-container">$$\mathscr{G}’:=\left\{\left( \bigcup_{i\in\mathbb{N}}G_i \right),\left( \bigcup_{i\in\mathbb{N}}{G_i} \right)^c \,\, \Big| \,\, G_i,\in\mathscr{G} \text{ or } {G_i}^c\in\mathscr{G} \right\},$$</span></p>
<p>but my measure theory book suggests this is not the case. <strong>But why not?</strong></p>
<p>My first thought was, “maybe those unions generate new sets, and unions of <em>those</em> new sets were not already accounted for.” But this is not a problem: If <span class="math-container">$A$</span> and <span class="math-container">$B$</span> are two sets generated by these countable unions from the sequences <span class="math-container">$(A_i)$</span> and <span class="math-container">$(B_i)$</span>, then <span class="math-container">$A\cup B$</span> is generated by <span class="math-container">$(A_1,B_1,A_2,B_2,\dots)$</span>, and so is already accounted for. The same reasoning can be extended to countable unions of the sets generated by these countable unions of <span class="math-container">$G_i$</span> sets. As for <em>uncountable</em> unions, who cares, the <span class="math-container">$\sigma$</span>-algebra axioms don’t say they need to be included anyway.</p>
|
MegaNightdude
| 357,588 |
<p>Perhaps a bit late, but may be useful for someone needing the full derivation of KL divergence:</p>
<h2>Worked derivation of KL-divergence</h2>
<p>By definition of KL divergence:
<span class="math-container">$$
\begin {equation}
D_{KL}(Q||P) = \int Q(X) \log \frac{Q(X)}{P(X)}dX
\end {equation}
$$</span></p>
<p>Or, put another way:
<span class="math-container">$$
\begin {equation}
\int Q(X) \bigg[\log P(X) - \log Q(X) \bigg]dX = \int Q(X)\log P(X)dX - \int Q(X) \log Q(X)dX
\end {equation}
$$</span></p>
<p>First, let's expand <span class="math-container">$\log Q(X)$</span>:
<span class="math-container">$$ \log Q(X) = \log \bigg[\frac{1}{\sqrt{2\pi\sigma^2}} e^{-\frac{(x-\mu)^2}{2 \sigma^2}}\bigg] $$</span>
Then using the rule <span class="math-container">$\log(XY) = \log(X) + \log(Y)$</span></p>
<p><span class="math-container">$$
\log Q(X) = \log \frac{1}{\sqrt{2\pi\sigma^2}} - \frac{(x-\mu)^2}{2 \sigma^2}
$$</span></p>
<h3>So we can compute the 1st term in eq. (2) <span class="math-container">$\int Q(X)\log Q(X)$</span>:</h3>
<p><span class="math-container">$$
\int Q(X)\log Q(X) = \int Q(X) \bigg[\log\frac{1}{\sqrt{2\pi\sigma^2}} - \frac{(x-\mu)^2}{2 \sigma^2} \bigg] dX
$$</span></p>
<p>Expand the brackets:
<span class="math-container">$$
\int Q(X)\log Q(X) = \int Q(X) \log\frac{1}{\sqrt{2\pi\sigma^2}} dX - \int Q(X) \frac{(x-\mu)^2}{2 \sigma^2} dX
$$</span></p>
<p>In the first term we take out of the integral the <span class="math-container">$\log\frac{1}{\sqrt{2\pi\sigma^2}}$</span> because it doesn't depend on X, what's left is equal to 1 (by the definition of normal distribution). Because <span class="math-container">$\log\frac{1}{\sqrt{2\pi\sigma^2}} = \log(1)-\log \sqrt{2\pi} + \log\sqrt{\sigma^2} = -\frac{1}{2}\log 2\pi - \frac{1}{2}\log \sigma^2$</span>, and by definition <span class="math-container">$Var(X) = \int Q(X) (x-\mu)^2 dX = \sigma^2 $</span>, we can write:</p>
<p><span class="math-container">$$
-\frac{1}{2}\log 2\pi -\frac{1}{2}\log\sigma^2 - \frac{1}{2 \sigma^2}\sigma^2 =
-\frac{1}{2}\log 2\pi -\frac{1}{2} + \log\sigma^2
$$</span>
Because we have are going to have a batches of <span class="math-container">$\mu$</span> and <span class="math-container">$\sigma$</span>, we need to sum over the batch length <span class="math-container">$J$</span>:</p>
<p><span class="math-container">$$\boxed{-\frac{1}{2}\log 2\pi -\frac{1}{2}\sum_{j=1}^{J} \bigg[1 + \log\sigma_j^2 \bigg]}$$</span></p>
<h3>Now let's do the second term in eq. (2) <span class="math-container">$\int Q(X)\log P(X)dX$</span></h3>
<p>First, let's take the log apart:
<span class="math-container">$$ \log P(X) = \log \bigg[\frac{1}{\sqrt{2\pi\sigma^2}} e^{-\frac{(x-\mu)^2}{2 \sigma^2}}\bigg] $$</span></p>
<p>Since <span class="math-container">$P(X) \stackrel{}{\sim} \mathcal{N} (0, I) $</span> is a normal distribution with <span class="math-container">$\mu = 0$</span> and <span class="math-container">$\sigma = 1$</span>, we can write:</p>
<p><span class="math-container">$$ \log P(X) = \log \bigg[\frac{1}{\sqrt{2\pi}} e^{-\frac{x^2}{2}}\bigg] =
-\frac{1}{2}\log 2\pi -\frac{x^2}{2}||x||^2
$$</span></p>
<p><span class="math-container">$$\int Q(X) \bigg[ -\frac{1}{2}\log 2\pi -\frac{x^2}{2}||x||^2 \bigg]dX$$</span></p>
<p><span class="math-container">$$-\frac{1}{2} \log 2\pi \int Q(X) dX - \frac{1}{2}\int Q(X) ||x||^2 dX
$$</span></p>
<p>Since by definition (see Wikipedia) <span class="math-container">$Var(X) = E[X^2] - E[X]^2 = E[X^2] - \mu^2$</span> and <span class="math-container">$\int Q(X) ||x||^2 dX$</span> is the expectation of <span class="math-container">$||X||^2$</span>, i.e. <span class="math-container">$\int Q(X) ||x||^2 = E[X^2] = Var(X) + E[X]^2 = \sigma^2 + \mu^2 $</span>, we can write:</p>
<p><span class="math-container">$$
-\frac{1}{2}\log 2\pi - \frac{1}{2} E[X^2] = -\frac{1}{2}\log 2\pi - \frac{1}{2} E[\mu^2+\sigma^2]
$$</span></p>
<p>Since <span class="math-container">$\mu$</span> and <span class="math-container">$\sigma$</span> are vectors, we need to replace the expectation with a sum:</p>
<p><span class="math-container">$$
-\frac{1}{2}\log 2\pi - \frac{1}{2} E[X^2] =$$</span></p>
<p><span class="math-container">$$
\boxed {-\frac{1}{2}\log 2\pi - \frac{1}{2} \sum_{j=1}^{J}[\mu_j^2+\sigma_j^2]}
$$</span></p>
<h3>Finally we can put the 1st and 2nd terms together to get <span class="math-container">$D_{KL}$</span></h3>
<p><span class="math-container">$$
D_{KL} = -\frac{1}{2}\log 2\pi -\frac{1}{2}\sum_{j=1}^{J} \big(1 + \log\sigma_j^2 \big) - \Bigg(-\frac{1}{2}\log 2\pi - \frac{1}{2} \sum_{j=1}^{J}\big(\mu_j^2+\sigma_j^2\big) \Bigg)
$$</span></p>
<p><span class="math-container">$$
\boxed {D_{KL} = -\frac{1}{2}
\sum_{j=1}^{J} \big(1 + \log\sigma_j^2 - \mu_j^2 - \sigma_j^2 \big)}
$$</span></p>
|
1,090,658 |
<p>I'm doing some previous exams sets whilst preparing for an exam in Algebra.</p>
<p>I'm stuck with doing the below question in a trial-and-error manner:</p>
<p>Find all $ x \in \mathbb{Z}$ where $ 0 \le x \lt 11$ that satisfy $2x^2 \equiv 7 \pmod{11}$</p>
<p>Since 11 is prime (and therefore not composite), the Chinese Remainder Theorem is of no use? I also thought about quadratic residues, but they don't seem to solve the question in this case.</p>
<p>Thanks in advance</p>
|
Henry
| 6,460 |
<p>$2x^2 \equiv 7 \pmod {11} \iff x^2 \equiv 9 \pmod {11}$ </p>
<p>so you should not be surprised by the solutions $x \equiv \pm3 \pmod {11}$</p>
<p>i.e. $x \equiv 3 \pmod {11}$ or $x \equiv 8 \pmod {11}$</p>
|
203,138 |
<p>I'm starting to take calculus in college (the first offered math class) and I was shown this problem. Find $f(x)$ when$$-\frac{df(x)}{dx}\cdot\frac{x}{f(x)}=1$$Rearranging this to make$$\frac{d\,f(x)}{dx}=-\frac{f(x)}{x}$$gives me an interesting relationship, and very strongly suggests to me a polynomial function, but I think this approach lacks rigor (since I can guess all I want what $f(x)$ is, but that's still heuristic reasoning since my brain can't exhaust all the possibilities).</p>
<p>Can someone walk me through a solution?</p>
<p><strong>edit</strong></p>
<p>This is not a homework problem, as my grade will not depend on its completion. But it's an interesting one. I've seen sorts of problems similar to this before, and they're all very unusual compared to what I've learned before.</p>
|
bartgol
| 33,868 |
<p>Rearrange it like this</p>
<p>$$\frac{df}{f}=-\frac{dx}{x}$$</p>
<p>then integrate both sides each with respect to its variable. You're right, it is a polynomial...an easy one, I would say... =)</p>
<p>Edit: it is NOT a polynomial, as pointed out by Peter Tamaroff. My bad.</p>
|
46,446 |
<p>I'm an highschool graduate who is currently waiting for college.</p>
<p>Meanwhile, I'm trying to do a little project by myself. (Computer stuff) And yesterday, I found that I needed to deal with something called "De Casteljau algorithm"</p>
<p>I know calculus (single-variable, but I'm thinking about learning multi-)
and I don't want an empty 'memorize-without-understanding-or-proving' approach.</p>
<p>Which path will take me there?
(I'm hoping for answers like:
"Calculus -> Differential Eq -> ...")</p>
<p>p.s: I would also appreciate book/video lecture recommendations :)</p>
<p>Thank you!</p>
|
lhf
| 589 |
<p>Try <a href="http://www.farinhansford.com/books/essentials-cagd/" rel="nofollow">The Essentials of CAGD</a> by Farin and Hansford. All books by Farin are very nice.</p>
|
88,880 |
<p>In a short talk, I had to explain, to an audience with little knowledge in geometry or algebra, the three different ways one can define the tangent space $T_x M$ of a smooth manifold $M$ at a point $x \in M$ and more generally the tangent bundle $T M$:</p>
<ul>
<li>Using equivalent classes of smooth curves through $x$</li>
<li>Using derivations near $x$</li>
<li>Using cotangent vectors at $x$</li>
</ul>
<p>Just by looking at the definition, it is not at all clear why they should all define the same object. I went through the proof, but judging from their reaction, it was not very meaningful. I wonder if there is any way I can let them "see", with just intuition, that the three definitions are, in certain sense, the same.</p>
|
Steven Landsburg
| 10,503 |
<p>In each of the three cases, your definition is capturing the intuition of "directions near x"
--- an equivalence class of curves defines a "direction" in which the curves head out from x.
A derivation or a linear functional on the cotangent vectors is a directional derivative,
hence determined by a choice of direction. (Of course it takes proof that these account
for all the derivations, etc. but the intuition is pretty clear.) </p>
|
963,125 |
<p>I'm not exactly sure where to start on this one. Any help would be greatly appreciated.</p>
<p>Show that $4$ does not divide $12x+3$ for any $x$ in the integers.</p>
<p>Here's what I have so far:</p>
<p>There exist c in the integers such that 4 | 12x+3. Then 12x+3 = 4y for some y in the integers. This is a contradiction since y = 3x + 3/4 and y is in the integers.</p>
<p>Is that correct?</p>
|
Community
| -1 |
<p>A couple of hints:</p>
<ol>
<li><p>Any time you want to prove something for all integers a good thing to try would be induction (proving that the truth for $k<n$ implies the truth for $n$).</p></li>
<li><p>Remember that if a number divides $a$ and $b$ then it should divide $a+b$ as well.</p></li>
</ol>
|
330,065 |
<p>What is an example of a manifold <span class="math-container">$M$</span> with <span class="math-container">$\dim(M)>1$</span> whose Lie algebra <span class="math-container">$\chi^{\infty}(M)$</span> of smooth vector fields admit an elliptic operator <span class="math-container">$D:\chi^{\infty}(M)\to \chi^{\infty}(M)$</span> such that <span class="math-container">$D$</span> is a Lie algebra derivation on <span class="math-container">$\chi^{\infty}(M)$</span>?</p>
<p>Does every manifold admit such an operator?</p>
<p>Is there a Riemannian manifold for which the Laplace operator <span class="math-container">$D=\Delta$</span>, naturally defined on <span class="math-container">$\chi^{\infty}(M)\simeq \Omega^1(M)$</span>, would be a derivation of <span class="math-container">$\chi^{\infty}(M)$</span>?</p>
|
Ali Taghavi
| 36,688 |
<p>I would like to apply the hints presented in the answer by Bazin to give the following answer.</p>
<p><a href="http://www.numdam.org/article/CM_1981__43_2_239_0.pdf" rel="nofollow noreferrer">Every derivation of <span class="math-container">$\chi^{\infty}(M)$</span> is inner</a> and it is obvious that every inner derivation is a non elliptic operator. The reason is that every non vanishing vector field is locally look like <span class="math-container">$\partial/\partial_x$</span>, whose corresponding adjoint operator is obviousely non elliptic.</p>
<p>So the non ellipticity of a derivation operator lies in its adjoint-ness not merely on its order.It is first order but this order situation is not enough to conclude it is non elliptic. In fact there is an example of a first order elliptic pde in dim 3 as follows:
<span class="math-container">$$D(u,v, w)= (u_x-v_y, u_y+v_x, w)$$</span></p>
<p>This operator <span class="math-container">$D$</span> satisfies the definition of an elliptic PDE but is of first order.</p>
<p>One can construct a first order elliptic PDE in dimension 4 without a term of order 0. Construct a linear PDE on <span class="math-container">$\chi^{\infty}(\mathbb{R}^4)$</span> whose principal symbol is the <span class="math-container">$4\times 4$</span> matrix representation of quaterniouns <span class="math-container">$$\xi_1+\xi_2 i + \xi_3 j+\xi_4 k$$</span> where <span class="math-container">$(\xi_1,\xi_2,\xi_3,\xi_4)$</span> is a cotangent vector.</p>
<p>So it seems that the first line of the answer by Bazin does not work.</p>
<p><strong>Remark:</strong> The above linked paper contains a reference to a result by F.Takens that every derivation of <span class="math-container">$\chi^{\infty}(M)$</span> is an inner derivation.</p>
|
293,234 |
<p>I recently asked a question about <a href="https://math.stackexchange.com/questions/287116/proof-that-mutual-statistical-independence-implies-pairwise-independence">pairwise versus mutual independence</a> (also related to <a href="https://math.stackexchange.com/questions/281800/example-relations-pairwise-versus-mutual">this</a> and <a href="https://math.stackexchange.com/questions/283579/how-to-model-mutual-independence-in-bayesian-networks">this</a> q). </p>
<p>However, </p>
<p>(1) I inadvertently used incorrect terminology:</p>
<blockquote>
<p>three events, A, B, C are mutually independent when:</p>
<p>P[A,B]=P[A]P[B], P[B,C]=P[B]P[C], P[A,C]=P[A]P[C],
P[A,B,C]=P[A]P[B]P[C]</p>
</blockquote>
<p>Did and others pointed out that</p>
<blockquote>
<p>"Mutual independence means the four identities you copied, pairwise
independence means the first three of these identities." -- Did</p>
</blockquote>
<p>Note that the term <em>mutual</em> has varying definitions across math. For example, <a href="http://en.wikipedia.org/wiki/Mutual_information" rel="nofollow noreferrer">mutual information</a> is a pairwise relation. </p>
<p>(2) Going back to probability, GC Rota said the theory can be approached two ways: focusing on random variables (event algebra) or focusing on distributions. Here I am interested in distributions, where independence can be interpreted as factorization of the probability distribution function. The conditions are the same as above, where P is interpreted as the PDF function. </p>
<p>The following graphic based on a standard example from <a href="http://rads.stackoverflow.com/amzn/click/0412989018" rel="nofollow noreferrer"><em>Counterexamples in Probability and Statistics</em></a> of a 3-dimensional binomial PDF that factorizes pairwise (ie, each of the 3 pairs of random variables are independent and the 2-dim joint distributions can all be written as the product of the respective marginals) but not 3-way independent (the joint distribution cannot be written as the product of the individual marginal distributions)</p>
<p><img src="https://i.stack.imgur.com/IBCra.jpg" alt="enter image description here"></p>
<p>My question is whether the opposite can happen, ie if the 3-dim (or perhaps higher) joint distribution factorizes into the 1-dim marginals, does that imply the pairwise factorization of <em>all</em> 2-dim joint distributions into the marginals? </p>
|
Ewan Delanoy
| 15,381 |
<p>For convenience I will sometimes denote the number $\sqrt[4]{2}$ by $\theta$.</p>
<p>What we have here is a tower of square roots : To $\mathbb Q$ we
can add successively $\sqrt{2},\sqrt[4]{2},i,\alpha,\beta,\gamma$. The main problem
is that in this sequence, some square root might already be contained in the
preceding field.</p>
<p><b> Remark </b> Let $\mathbb A$ be a field and let $a$ be a nonsquare in ${\mathbb A}$ ;
let $b\in {\mathbb A}(\sqrt{a})$. The extension ${\mathbb A}(\sqrt{a}) \subseteq {\mathbb A}(\sqrt{a},\sqrt{b})$ has degree $1$ or $2$.</p>
<p>To know which it is, there are two cases to consider :</p>
<p><b> Extension rule 1 </b> Let ${\mathbb A},a,b$ be as above. If $b\in {\mathbb A}$, then the degree is $1$ iff $ab$ is a square in $\mathbb A$.</p>
<p><b> Extension rule 2 </b> Let ${\mathbb A},a,b$ be as above. If $b\not\in {\mathbb A}$, so that $b=u+v\sqrt{a}$ with $v\neq 0$, then the degree is $1$ iff the equation $x^4-ux^2+\frac{av^2}{4}=0$ has a solution in $\mathbb A$.</p>
<p>(to see why rule 2 holds, not that for $x,y\in {\mathbb A}$ we have
$(x+y\sqrt{a})^2=u+v\sqrt{a}$ iff $y=\frac{v}{2x}$ and $x^4-ux^2+\frac{av^2}{4}=0$).</p>
<p>Note that rule 2 unfortunately raises the degree to $4$ and complicates the recursive analysis, but in the present case we are lucky and will be able to succeed using only a small part of rule 2, namely</p>
<p><b> Extension rule 2' </b> Let ${\mathbb A},a,b,u,v$ as in rule 2 above. If $\sqrt{u^2-av^2}\not\in {\mathbb A}$, then the degree of the extension is $2$. </p>
<p>(to see how this follows from rule $2$, not that $x^4-ux^2+\frac{av^2}{4}=0$ implies
$(2x-u)^2=u^2-av^2$).</p>
<p><b> Fact 1 </b> The four numbers $g_1=2-\sqrt{2},\sqrt{2}g_1=-2+2\sqrt{2},g_2=2+\sqrt{2}$ and $\sqrt{2}g_2=-2-2\sqrt{2}$ are all nonsquares in ${\mathbb Q}(\sqrt{2})$.</p>
<p><b> Proof of fact 1 </b> Use rule 2' with ${\mathbb A}={\mathbb Q}, a=2$, and $(u,v)=(2,1)$ or $(-2,2)$ (yielding $u^2-av^2=2$ or $-4$, both nonsquares in $\mathbb Q$).</p>
<p><b> Fact 2 </b> The seven numbers $g_1=2-\sqrt{2}$, $g_2=6+4(\theta+\theta^2+\theta^3)$, $g_3=6-4(\theta-\theta^2+\theta^3)$, $g_4=4-2\sqrt{2}$,
$g_5=\sqrt{2}+\sqrt[4]{2}$,
$g_6=\sqrt{2}-\sqrt[4]{2}$, and
and $g_7=2+\sqrt{2}$ are all nonsquares in
${\mathbb Q}(\sqrt[4]{2})$.</p>
<p><b> Proof of fact 2 </b> For the numbers $g_1$ and $g_7$, use fact 1 above and rule 1 with ${\mathbb A}={\mathbb Q}(\sqrt{2}), a=\sqrt{2}$. For the numbers $g_5$ and $g_6$, use fact 1 above and rule 2' with ${\mathbb A}={\mathbb Q}(\sqrt{2}), a=\sqrt{2}$. For the other numbers, note that $g_2=g_1(2+\theta+\theta^2+\theta^3)^2, g_3=g_1(2-\theta+\theta^2-\theta^3)^2$ and
$g_4=g_1((\sqrt{2})^2)$.</p>
<p><b> Fact 3 </b> The seven numbers $h_1=-\sqrt{2}+i\sqrt[4]{2}$, $h_2=-2-\sqrt[4]{8}+i(\sqrt{2}+\sqrt[4]{8})$, $h_3=-2+\sqrt[4]{8}+i(-\sqrt{2}+\sqrt[4]{8})$,
$h_4=2-2\sqrt{2}+i(2\sqrt{2}-\sqrt[4]{8})$,
$h_5=\sqrt{2}+\sqrt[4]{2}$,
$h_6=\sqrt{2}-\sqrt[4]{2}$, and
$h_7=2-\sqrt{2}$
are all nonsquares in
${\mathbb Q}(\sqrt[4]{2},i)$. </p>
<p><b> Proof of fact 3 </b> Use fact 2 above and rule 2' with ${\mathbb A}={\mathbb Q}(\sqrt[4]{2}), a=-1$ : for any $k\in \lbrace 1,2,3,4 \rbrace$, we can write $h_k=u_k+v_ki$ where $u_k$ and $v_k$ are both in ${\mathbb Q}(\sqrt[4]{2})$, and
$u_k^2-av_k^2=g_k$. Thus $h_k$ is a nonsquare in ${\mathbb Q}(\sqrt[4]{2},i)$ because
$g_k$ is a nonsquare in ${\mathbb Q}(\sqrt[4]{2})$, when $k\in \lbrace 1,2,3,4 \rbrace$. For numbers $h_5$ to $h_7$ use rule 1 with ${\mathbb A}={\mathbb Q}(\sqrt[4]{2})$ and $a=-1$.</p>
<p><b> Fact 4 </b> The seven numbers $\gamma^2$, $(\alpha\gamma)^2$, $(\beta\gamma)^2$, $(\alpha\beta\gamma)^2$, $\alpha^2$, $\beta^2$ and $(\alpha\beta)^2$ are all nonsquares in ${\mathbb Q}(\sqrt[4]{2},i)$. </p>
<p><b> Proof of fact 4 </b> This is because of fact 3 above and
$\gamma^2=h_1$ ,$(\alpha\gamma)^2=h_2$, $(\beta\gamma)^2=h_3$, $(\alpha\beta\gamma)^2=h_4$, $\alpha^2=h_5$, $\beta^2=h_6$, $(\alpha\beta)^2=h_7$.</p>
<p>Starting from fact 4 and iterating rule 1, we see successively that $\alpha$ has degree $2$ on ${\mathbb Q}(\sqrt[4]{2},i)$, then that $\beta$ has degree $2$ on ${\mathbb Q}(\sqrt[4]{2},i,\alpha)$, then that $\gamma$ has degree $2$ on ${\mathbb Q}(\sqrt[4]{2},i,\alpha,\beta)$. The total degree is therefore $64$.</p>
|
981,748 |
<p>We know a closed-form of the first two powers of the integral of <a href="http://mathworld.wolfram.com/Trilogarithm.html" rel="nofollow noreferrer">trilogarithm function</a> between <span class="math-container">$0$</span> and <span class="math-container">$1$</span>. From the result <a href="https://math.stackexchange.com/q/981650/153012">here</a> we know that</p>
<p><span class="math-container">$$I_1=\int_0^1 \operatorname{Li}_3(x)\,dx = \zeta(3)-\frac{\pi^2}{6}+1.$$</span></p>
<p>From <a href="https://math.stackexchange.com/q/980764/153012">here</a> we also know that</p>
<p><span class="math-container">$$I_2=\int_0^1 \operatorname{Li}_3^2(x)\,dx = 20-8\zeta(2)-10\zeta(3)+\frac{15}{2}\zeta(4)-2\zeta(2)\zeta(3)+\zeta^2(3).$$</span></p>
<p>Is there a closed-form of</p>
<p><span class="math-container">$$I_3=\int_0^1 \operatorname{Li}_3^3(x)\,dx$$</span>
and
<span class="math-container">$$I_4=\int_0^1 \operatorname{Li}_3^4(x)\,dx\,?$$</span></p>
<hr />
<p>Update (by editor after 6 years): By generalizing @Kirill's algorithm one may prove that:</p>
<blockquote>
<p><span class="math-container">$$\int_0^1 \text{Li}_3(x){}^4 \, dx=-51 \pi ^2 \zeta(6,2)+6480 \zeta(6,2)+\frac{4743}{8} \zeta(8,2)+\zeta (3)^4-\frac{2 \pi ^2 \zeta (3)^3}{3}-68 \zeta (3)^3-248 \pi ^2 \zeta (3)^2+16680 \zeta (3)^2+\frac{106 \pi ^6 \zeta (3)}{105}+126 \pi ^4 \zeta (3)+3920 \pi ^2 \zeta (3)+102 \pi ^2 \zeta (5) \zeta (3)-11160 \zeta (5) \zeta (3)-\frac{4743 \zeta (7) \zeta (3)}{4}-114240 \zeta (3)+42 \pi ^4 \zeta (5)+1260 \pi ^2 \zeta (5)-73080 \zeta (5)+\frac{819 \pi ^2 \zeta (7)}{2}-33030 \zeta (7)-9660 \zeta (9)-\frac{4023 \zeta (5)^2}{4}+\frac{563 \pi ^{10}}{39600}+\frac{83 \pi ^8}{42}-1064 \pi ^4-11200 \pi ^2-\frac{184 \pi ^6}{21}+369600$$</span></p>
</blockquote>
<p>Here <span class="math-container">$\zeta(6,2), \zeta(8,2)$</span> are irreducible. One may verify its correctness numerically.</p>
|
Kirill
| 11,268 |
<p>Numerically,
$$ I_3 =
27 \zeta (5) \zeta (2)-3 \zeta (3)^2 \zeta (2)+156 \zeta (3) \zeta (2)-\tfrac{153}{8} \zeta (7)-90 \zeta (5)+\zeta (3)^3+21 \zeta (3)^2-660 \zeta (3)-420 \zeta (2)-315 \zeta (4)-\tfrac{477}{4} \zeta (6)+1680
$$
and $I_4$ doesn't seem to have a similar closed form.</p>
<p>The same approach that I used in
<a href="https://math.stackexchange.com/a/941394/11268">this answer</a> to decompose the integral as a sum of multiple zeta values works here as well because of the integral form
$$ \mathrm{Li}_3(x) = \int_0^x \frac{dt}{t} \int_0^t \frac{du}{u} \int_0^u \frac{du}{1-u}, $$
so it's possible to write $\int_0^1 \mathrm{Li}_3(x)^4\,dx$ as an iterated 13-fold integral, but that looks a little tedious to apply it by hand.</p>
|
146,075 |
<p>Within my limited experience, I have only known free groups to occur through two mechanisms: as fundamental groups of trees (graphs) and ping-pong. And sometimes only through one way: the fact that sufficiently high-powers of hyperbolic elements in a Gromov-hyperbolic group generate a free group arises via ping-pong. I know of no tree to represent the situation. </p>
<p>To the experts, the following question is surely either an obvious yes or no: </p>
<p>Can we explicitly describe all mechanisms by which finitely generated free subgroups of the Artin braid groups $B_n$ (for $n=2,3,\ldots$) arise ? </p>
<p>More specifically, seeing $B_n$ as the isotopy space of all $n$-pointed braids in the closed 2-disk, can we characterize all configurations generating the rank $k$ free group?</p>
<p>Much less specifically, do we know, say, that all finitely-generated free subgroups arise from ping-pong and can we describe all ping-pong games? </p>
|
Ian Agol
| 1,345 |
<p>A couple of remarks: </p>
<p>If a subgroup is torsion-free, and intersects the pure braid group $P_n$ in a free group, then the group is also free. So the question can be interpreted in $P_n$. </p>
<p>There is the Birman exact sequence $F_{n-1} \to P_n \to P_{n-1}$, which is obtained as deleting a puncture, and has kernel the fundamental group of an $n-1$-punctured plane. There is such a free subgroup for each puncture, and these subgroups are not ``quasiconvex" in any sense since they are normal. So I don't think there is some sort of ping-pong description of these from some action of $P_n$ on a nice space. </p>
<p>On the other hand, if the image of the subgroup generated by $g_1,\ldots,g_k$ in $P_{n-1}$ is free of rank $k$, then so is the subgroup of $P_n$. So I could imagine some inductive description of the free subgroups of $P_n$. </p>
<p>Suppose the image in $P_{n-1}$ is free. Then there is a Nielsen transformation of $g_1,\ldots,g_k$ such that $g_1,\ldots,g_j$ have image in $P_{n-1}$ generating a rank $j$ free subgroup, and $g_{j+1},\ldots,g_k$ have trivial image in $P_{n-1}$. Then the question is what is the extension of the free group $\langle g_1,\ldots,g_j\rangle$ by the normal subgroup of $\langle g_1,\ldots,g_n\rangle$ generated by $g_{j+1},\ldots, g_n$? I wouldn't expect a simple answer to this question though, so this is probably a difficult question. Also, the image in $P_{n-1}$ might not be free. </p>
|
2,153,663 |
<p>My discrete book says that the set $A = \{4,5,6\}$ and the relation $R = \{(4,4),(5,5),(4,5),(5,4)\}$ is symmetric and transitive but not reflexive.</p>
<p>I was wondering how this is possible, because if a set $A$ is symmetric, doesn't it also need to include $(5,6),(6,5),(4,6),(6,4)$? </p>
<p>Also if it's transitive, doesn't it have to include $(4,5),(5,6),(4,6)$?
I thought the definition of a transitive relation was that $(x$ $R$ $y)$ $\land$ $(y$ $R$ $z)$ then $(x$ $R$ $z)$.</p>
|
Patrick Stevens
| 259,262 |
<p>A <em>set</em> can't be symmetric; a <em>relation</em> can be. (By the way, it's possible for a set to be "transitive", but that doesn't mean the same thing as a transitive <em>relation</em>: a transitive set is a set $x$ such that if $z \in y$ and $y \in x$, then $z \in x$.)</p>
<p>A relation on a set need not involve every member of the set. For example, the relation on $\mathbb{N}$ given by "is a prime divisor of" doesn't touch $1$ at all: $1$ is not related to anything and nothing is related to it. In your example, $6$ is not related to anything by $R$, and nothing is related to $6$ by $R$.</p>
<p>"Symmetric" just means that <em>if</em> $a \sim b$, <em>then</em> $b \sim a$. Note that it doesn't tell us about any elements of $A$ we haven't seen before: from the mere knowledge that $4 \sim 5$, we can't use symmetry to deduce that anything is related to $6$. Similarly transitivity.</p>
|
2,153,663 |
<p>My discrete book says that the set $A = \{4,5,6\}$ and the relation $R = \{(4,4),(5,5),(4,5),(5,4)\}$ is symmetric and transitive but not reflexive.</p>
<p>I was wondering how this is possible, because if a set $A$ is symmetric, doesn't it also need to include $(5,6),(6,5),(4,6),(6,4)$? </p>
<p>Also if it's transitive, doesn't it have to include $(4,5),(5,6),(4,6)$?
I thought the definition of a transitive relation was that $(x$ $R$ $y)$ $\land$ $(y$ $R$ $z)$ then $(x$ $R$ $z)$.</p>
|
mle
| 66,744 |
<blockquote>
<p><strong>Def.</strong>: let be $A,R$ sets, we define $$ \operatorname{dom}(R):=\{x| \exists y : (x,y) \in R\} \\ \operatorname{cod}(R):=\{x| \exists y :(y,x)\in R\} \\ \operatorname{field}(R):=\operatorname{dom}(R)\cup \operatorname{cod}(R) \\ R \text{ is reflexive in }A \text{ if } \,\forall x \in A:(x,x)\in R \\ R \text{ is symmetric in }A \text{ if } \,\forall x,y \in A:(x,y)\in R \to (y,x)\in R \\ R \text{ is transitive in }A \text{ if } \,\forall x,y,z\in A:(x,y)\in R \wedge (y,z)\in R \to (x,z)\in R \\ R \text{ is reflexive} \text{ if } \,R \text{ is reflexive in}\operatorname{field}(R) \\ R \text{ is symmetric} \text{ if } \,R \text{ is symmetric in}\operatorname{field}(R)\\ R \text{ is transitive } \text{if } \,R \text{ is transitive in}\operatorname{field}(R) $$</p>
</blockquote>
<p>Now, we have $A:=\{4,5,6\}$ and $R:=\{(4,4),(5,5),(4,5),(5,4)\}$, therefore:</p>
<ul>
<li>$R$ is not reflexive in $A$ because $(6,6) \notin R$</li>
<li>$R$ is symmetric in $A$</li>
<li>$R$ is transitive in $A$</li>
</ul>
<p>([$R$ is symmetric in $A$ $\wedge$ $R$ is transitiv in $A$ $\to$ $R$ is reflexiv in $A$] is generally false, and you have an example)</p>
<p>But:</p>
<ul>
<li>$ \operatorname{dom}(R)= \operatorname{cod}(R)= \operatorname{field}(R)=\{4,5\}$</li>
<li>$R$ is reflexiv (in $\operatorname{field}(R)$)</li>
<li>$R$ is symmetric (in $\operatorname{field}(R)$)</li>
<li>$R$ is transitiv (in $\operatorname{field}(R)$)</li>
</ul>
<p>([$R$ is symmetric $\wedge$ $R$ is transitiv $\to$ $R$ is reflexiv] is true, but the converse generally is false, an example $R^´:=\{(1,1),(0,1),(0,0)\}$)</p>
<p>Why $R$ is symmetric and transitive in $A$? For example, let be $4,6 \in A$ and I prove that $(4,6) \in R \to (6,4) \in R$ and $(4,6) \in R \wedge (6,4) \in R \to (4,4)$ are true, but it is vacuously symmetric and transitive by def of "$\to$" (see <a href="https://math.stackexchange.com/questions/1962500/why-is-1-2-3-4-a-transitive-relation?noredirect=1&lq=1">ex.1</a>, <a href="https://math.stackexchange.com/questions/381408/is-any-relation-which-contains-only-one-ordered-pair-transitive">ex.2</a>), similary with $(5,6)$, $(6,5)$...</p>
|
3,717,465 |
<p>I couldn't find a question answering this concept but they seem to be related.</p>
<blockquote>
<p>Extreme Value Theorem (two variables)</p>
<p>If f is a continuous function defined on a closed and bounded set
<span class="math-container">$A⊂\mathbb{R}^2$</span>, then f attains an absolute maximum and absolute
minimum value on A.</p>
</blockquote>
<hr />
<blockquote>
<p>Lagrange Multipliers (two variables)</p>
<p>Extreme values of function f(x, y) subject to constraints g(x, y) = k has solutions in <span class="math-container">$\nabla f=\lambda \nabla g$</span>.</p>
</blockquote>
<p>The constraint in Lagrange Multipliers creates a closed and bounded region that would satisfy EVT, does it not? So does that make Lagrange multipliers a specific case of EVT?</p>
|
Henry Lee
| 541,220 |
<p>If you let:
<span class="math-container">$x=2^t$</span>
then remember we have:
<span class="math-container">$dx=\ln(2)2^tdt$</span> and our original equation is:
<span class="math-container">$$\frac{df(x)}{dx}-f(x/2)=0$$</span>
so plugging in we get:
<span class="math-container">$$\frac{1}{2^t\ln(t)}\frac{df(2^t)}{dt}-f(2^{t-1})=0$$</span>
or we could write it as:
<span class="math-container">$$\frac{df(2^t)}{f(2^{t-1})}=2^t\ln(t)dt$$</span>
<span class="math-container">$$\int\frac{df(2^t)}{f(2^{t-1})}=2^t+C$$</span>
I would assume the best way to approach this would be to assume that:
<span class="math-container">$$f(2^{t-1})=f(2^t)g(2^t)$$</span>
or something like this</p>
|
3,717,465 |
<p>I couldn't find a question answering this concept but they seem to be related.</p>
<blockquote>
<p>Extreme Value Theorem (two variables)</p>
<p>If f is a continuous function defined on a closed and bounded set
<span class="math-container">$A⊂\mathbb{R}^2$</span>, then f attains an absolute maximum and absolute
minimum value on A.</p>
</blockquote>
<hr />
<blockquote>
<p>Lagrange Multipliers (two variables)</p>
<p>Extreme values of function f(x, y) subject to constraints g(x, y) = k has solutions in <span class="math-container">$\nabla f=\lambda \nabla g$</span>.</p>
</blockquote>
<p>The constraint in Lagrange Multipliers creates a closed and bounded region that would satisfy EVT, does it not? So does that make Lagrange multipliers a specific case of EVT?</p>
|
Riemann'sPointyNose
| 794,524 |
<p>Notice</p>
<p><span class="math-container">$${f'(x)=f\left(\frac{x}{2}\right)\Leftrightarrow f''(x) = \frac{1}{2}f'\left(\frac{x}{2}\right), f'''(x)=\frac{1}{2^2}f''\left(\frac{x}{2}\right)}$$</span></p>
<p>Plugging in some numbers we get <span class="math-container">${f(0) = 10, f'(0) = 10, f''(0) = \frac{10}{2}, f'''(0) = \frac{1}{2^2}\left(\frac{10}{2}\right)=\frac{10}{2^3},f''''(0) = \frac{10}{2^6}}$</span>... In general we see</p>
<p><span class="math-container">$${f^{(n)}(0)=\frac{10}{2^{\left(\frac{n(n-1)}{2}\right)}}}$$</span></p>
<p>And now we can create a power series solution using a taylor series centered around <span class="math-container">$0$</span>:</p>
<p><span class="math-container">$${f(x)=10 + \sum_{n\geq 1}\frac{10x^n}{2^{\left(\frac{n(n-1)}{2}\right)}n!}}$$</span></p>
|
2,025,069 |
<p>$$\sum_{n=0}^{\infty} (-1)^{n}(\frac{x^{2n}}{(2n)!}) * \sum_{n=0}^{\infty} C_{n}x^{n} = \sum_{n=0}^{\infty} (-1)^{n}(\frac{x^{2n+1}}{(2n+1)!})$$</p>
<p>How to find coefficients until the $x^{7} $ ? It is basically sin and written in form of sums but i'm not sure how should i use this form to calculate coefficients from this factor sum. How should i precede? </p>
|
Mike Earnest
| 177,399 |
<p>Take the equation
$$
(1-x^2/2+x^4/24-x^6/720+\dots)(C_0+C_1x+\dots+C_7x^7+\dots)=x-x^3/6+x^5/120-x^7/5040+\dots
$$
and expand out the left side. When you collect the powers of $x$, you get
$$
C_0+C_1x+(C_2-C_0/2)x^2+(C_3-C_1/2)x^3+\dots=x-x^3/6+x^5/120-x^7/5040+\dots
$$
On the left side, I only went up to $x^3$, you will need to go up to $x^7$.</p>
<p>Now, since these two power series are equal, their coefficients must be equal, so you get $C_0=0$, $C_1=1$, $C_2-C_0/24=0$, $C_3-C_1/2=-1/6$, etc. Solve.</p>
|
1,374,660 |
<p>The IQ of actuarial science majors is assumed to be normally distributed with mean 112 and standard deviation of 14. In a class of 19 students, find the probability that the mean IQ of all 19 students is greater than 120. </p>
<p>I know this question is based of the CLT, but I don't know what to change when everyone in the group has to be above a certain mean. Shouldn't the mean just be a group average? </p>
|
Jack D'Aurizio
| 44,121 |
<p>If $X_1,X_2,\ldots,X_{19}$ are i.i.d random variables, what it the probability that</p>
<p>$$ \min(X_i)\geq 120 $$
? Obviously, it is given by:
$$ \mathbb{P}[X_1 \geq 120]^{19}.$$
Since $e^{-x^2/2}$ is a fixed point of the Fourier transform, the arithmetic mean of $n$ i.i.d normally distributed $N(\mu,\sigma^2)$ random variables is still a normal variable with the same mean and variance $\frac{\sigma^2}{n}$.</p>
|
3,657,106 |
<blockquote>
<p>Sam was adding the integers from <span class="math-container">$1$</span> to <span class="math-container">$20$</span>. In his rush, he skipped one of the numbers and forgot to add it. His final sum was a multiple of <span class="math-container">$20$</span>. What number did he forget to add?</p>
</blockquote>
<p>My idea was to use Gauss's trick to find this relatively simply so I proceeded as follows.</p>
<p>We have <span class="math-container">$S=1+2+3+ \dots+ 18+19+20$</span>. Using Gauss's trick we get <span class="math-container">$\frac{n(n+1)}{2} = \frac{20(21)}{2} = 210$</span>. Since we want this to equal some multiple of <span class="math-container">$20$</span> we have that <span class="math-container">$210 = 20n$</span>, but solving for <span class="math-container">$n$</span> results in <span class="math-container">$\frac{21}{2} = 10.5$</span>.</p>
<p>The correct answer for this was <span class="math-container">$10$</span>, but it seems that I'm missing something?</p>
|
J. W. Tanner
| 615,567 |
<p><span class="math-container">$210$</span> minus the missing number (call it <span class="math-container">$m$</span>) <span class="math-container">$=20n$</span>.</p>
<p><span class="math-container">$210-m=20n$</span>, where <span class="math-container">$1\le m \le 20$</span>.</p>
<p>Can you take it from here?</p>
|
89,669 |
<p>If we fix a locally profinite group $G$ , we note $R(G)$ the category of smooth representations of $G$, $\mathcal{E}$ the set of equivalence classes of $R(G)$, and finaly $Irr(G)$ the set of irreducible equivalence calasses. I recall that the theorem of Cantor-Bernstein says : If $E$ and $F$ two sets. If there is an injection from $E$ to $F$ and an injection from $F$ to $E$, then there is a bijection between $E$ and $F$. This enables us to define an ordering relation in the set of equipotence classes of sets : If $E$ and $F$ two sets, we note $E\leq F$ if $E$ injects in $F$. </p>
<p>Similarly, we define a relation $\leq$ in $\mathcal{E}$, but in general is not an ordering relation, I think that is an ordering relation if $R(G)$ is semisimple (for example, for compact locally profinite group). </p>
<p>If $L$ a non empty subset of $Irr(G)$, we define a $L$-minimal representation as a smooth representation $\pi$ of $G$ such that :</p>
<p>1) For every $\sigma\in L$, $\sigma \leq \pi$. </p>
<p>2) For every $\tau \in R(G)$, if $\sigma\leq\tau$ for every $\sigma\in L$, then $\pi\leq\tau$.</p>
<p>I ask the following questions: </p>
<p>Q1) An $L$-minimal representation exits ?</p>
<p>Q2) unicity ? </p>
<p>Q3) If $\pi$ an $L$-minimal (if there exist) representation, $dim\mathbf{Hom}_{G}(\sigma,\pi)$, where $\sigma\in L$, is minimal ?</p>
<p>I'm interested of this question for the set $L_{k}$ of equivalence classes of irreducible supercuspidal representation of $PGL(n,F)$ with conductor=$k$.</p>
|
Marc Palm
| 10,400 |
<p>Here are some observations, too long for a comment:</p>
<p>1) Note that cuspidal irreducible representation are compactly induced</p>
<p>$\sigma = c-ind_K^G \tau = Ind_K^G \tau$</p>
<p>2) You have the second adjointness theorem (proposition on pg. 20 Bushnell-Henniart "Local Langlands for GL(2)".)</p>
<p>$Hom_G( c-ind_K^G \tau, \pi) = Hom_K( \tau, Res_K \pi)$</p>
<p>3) Silberger PGL(2) over the $p$ adics assect that $Res_K \pi$ is essentially $Ind_{B(o)}^{GL(2, o)} 1$ except for a finite dimensional part. I expect this to be true for $GL(n)$.</p>
<p>Hence classify the $\tau$ needed for $L_k$ ( I am not sure what your definition is here). $Res_K \pi$ has been described for cuspidal $\pi$ (Bushnell-Kutzko).</p>
<p>In fact, I think that the supercuspidal representation form a semisimple category, so there the question might really reduce to something trivial, very much like for profinite groups. (profinite groups are actually exactly the compact locally profinite groups;)</p>
|
1,602,392 |
<p>I think I have a proof using Pythagoras for $\sqrt{a_1^2} + \sqrt{a_2^2} > \sqrt{a_1^2 + a_2^2}$.</p>
<p><em>I'm interested in whether there's a way to use that proof with Pythagoras to prove the general $a_n$ case (for this, hints are appreciated rather than complete proofs), and <strong>also</strong> in other ways (algebraic, geometric, number theoric, calculusic...anything) that you might know or come up with to prove the general case (for those, either hints or complete proofs are great, up to you).</em></p>
<p><strong><em>Lemma</em></strong>:</p>
<p>Let positive (edited) real numbers $a_1, a_2$ be the legs of a right triangle. </p>
<p>Then $\sqrt{a_1^2 + a_2^2}$ is the length of the hypothenuse of that triangle.</p>
<p>And $\sqrt{a_1^2} + \sqrt{a_2^2}$ is the sum of the length of the two legs.</p>
<p>By the triangular inequality, we know that the length of the hypothenuse has to be less than the length of the sum of the two legs.</p>
<p>Therefore, for any real numbers $a_1, a_2$, $\sqrt{a_1^2} + \sqrt{a_2^2} > \sqrt{a_1^2 + a_2^2}$.</p>
<p><em>I'm stuck here...I was thinking of comparing pairs of elements from each side of the expression using my lemma, but it doesn't seem possible to "extract" pairs of elements from under $\sqrt{a_1^2 + a_2^2 +...+a_n^2}$.
I also thought about summing all elements but $a_1$ into a single number and using my lemma on those simplified expressions, but I run into the same problem.</em></p>
|
Community
| -1 |
<p>It's very easy using induction : </p>
<p>You proved the case when $n=2$ .</p>
<p>Now assume you know it for $n$ and want prove it for $n+1$ :</p>
<p>$$\sqrt{a_1^2}+\ldots+\sqrt{a_n^2}+\sqrt{a_{n+1}^2} >\sqrt{a_1^2+a_2^2+\ldots+a_n^2}+\sqrt{a_{n+1}^2}$$</p>
<p>Now use also the $n=2$ case to finnish it :</p>
<p>$$\sqrt{a_1^2+a_2^2+\ldots+a_n^2}+\sqrt{a_{n+1}^2}>\sqrt{\left ( \sqrt{a_1^2+a_2^2+\ldots+a_n^2} \right )^2}+\sqrt{a_{n+1}^2}>\sqrt{\left ( \sqrt{a_1^2+a_2^2+\ldots+a_n^2} \right )^2+a_{n+1}^2}=\sqrt{a_1^2+a_2^2+\ldots+a_n^2+a_{n+1}^2}$$ as wanted .</p>
<p>You can translate this into a geometric proof : consider an $n$-dimensional box with the sides $a_1,a_2,\ldots,a_n$ . The diagonal of the box is $\sqrt{a_1^2+a_2^2+\ldots+a_n^2}$ and now you can repeatedly apply the triangle's inequality to get your inequality (this is equivalent with the induction proof above )</p>
|
3,107,858 |
<p>I have to find the parametric equation for the line <span class="math-container">$M_1$</span> with the following info: </p>
<ul>
<li><span class="math-container">$M_1$</span> goes through the point <span class="math-container">$P(1,2,2)$</span></li>
<li>Is parallel with the plane <span class="math-container">$x + 3y + z = 1$</span></li>
<li>Has an intersection somewhere with the line <span class="math-container">$M_2 = 1+t, 2-2t, -1+2t$</span></li>
</ul>
<p>So if I understand it correctly, since <span class="math-container">$M_1$</span> is parallel with the plane <span class="math-container">$x + 3y + z = 1$</span>. I can do <span class="math-container">$x = 1-3y-z$</span> and replace <span class="math-container">$y$</span> and <span class="math-container">$z$</span> with <span class="math-container">$t$</span> and <span class="math-container">$s$</span> to get the direction of <span class="math-container">$M_1$</span>. The planes equation will then be:</p>
<p><span class="math-container">$$(x,y,z)=(1-3s-t , s , t)$$</span></p>
<p>So then the <span class="math-container">$M_1$</span> should be with the coordinates from <span class="math-container">$P$</span>:</p>
<p><span class="math-container">$$x = 1 + 1-3s-t$$</span></p>
<p><span class="math-container">$$y = 2 + s$$</span></p>
<p><span class="math-container">$$z = 2 + t$$</span></p>
<p>But it is incorrect because <span class="math-container">$M_1$</span> and <span class="math-container">$M_2$</span> have no intersections. Where did I go wrong here?
Thanks in advance for the answers!</p>
|
Steven Alexis Gregory
| 75,410 |
<p><span class="math-container">$M_1$</span> does not have to be <strong>on</strong> then plane <span class="math-container">$x + 3y + z = 1$</span>. If the line <span class="math-container">$M_1$</span> is parallel to the plane <span class="math-container">$x + 3y + z = 1$</span>, then it must be on the plane
<span class="math-container">$x + 3y + z = k$</span> for some real number <span class="math-container">$k$</span>. Since the line <span class="math-container">$M_1$</span> contains the point <span class="math-container">$P=(1,2,2)$</span>, and since <span class="math-container">$(1) + 3(2) + (2) = 9$</span>, then the line is on the plane <span class="math-container">$x + 3y + z = 9$</span>.</p>
<p>The plane <span class="math-container">$x + 3y + z = 9$</span> intersects the line <span class="math-container">$M_2(t) = (1+t, 2-2t, -1+2t)$</span> when</p>
<p><span class="math-container">\begin{align}
(1+t) + 3(2-2t) + (-1+2t) &= 9 \\
6 - 3t &= 9 \\
t &= -1
\end{align}</span></p>
<p>Hence the point <span class="math-container">$M_2(-1) = (0, 4, -3)$</span> is also on the line <span class="math-container">$M_1$</span>. So the line <span class="math-container">$M_1$</span> is parallel to the direction
<span class="math-container">$(1,2,2) - (0, 4, -3) = (1,-2,5)$</span>, then the equation of <span class="math-container">$M_1$</span> is</p>
<p><span class="math-container">$$M_1(t) = (1,2,2) + t(1,-2,5)$$</span></p>
<p>We can verify your three requirements</p>
<ol>
<li><span class="math-container">$M_1$</span> goes through the point <span class="math-container">$P(1,2,2).$</span>
<ul>
<li>Because <span class="math-container">$M_1(0) = (1,2,2)$</span>.</li>
</ul>
</li>
<li>Is parallel with the plane <span class="math-container">$x + 3y + z = 1.$</span>
<ul>
<li>Because <span class="math-container">$x+3y+z=(1+t)+3(2-2t)+(2+5t) = 9.$</span></li>
</ul>
</li>
<li>Has an intersection somewhere with the line <span class="math-container">$M_2 = 1+t, 2-2t, -1+2t.$</span></li>
<ul>
<li>Because <span class="math-container">$M_2(-1) = (0,4,-3)=M_1(-1)$</span></li>
</ul>
</li>
</ol>
|
181,839 |
<p>Since the specific space $\mathbb{R}^k$ is given, this might be provable in ZF.</p>
<p>Let $\{F_n\}_{n\in \omega}$ be a family of closed subset of $\mathbb{R}^k$, of which the union is $\mathbb{R}^k$.</p>
<p>Suppose $\forall n\in \omega, {F_n}^o=\emptyset$ ($o$ denotes interior)</p>
<p>Fix $x_0 \in \mathbb{R}^k$
Then by assumption, $\forall 0<r\in \mathbb{R}$, there exists $y\in \mathbb{R}^k$ such that $y\in B(r,x_0)\setminus (F_n \cup \{x_0\})$ for every $n\in \omega$.</p>
<p>I tried to show that for every $r$, there exists $y\in \mathbb{Q}^k$ that satisfies the condition.
(Once the existence is guaranteed, since $\mathbb{Q}^k$ can be well-ordered lexicographically, $y$ can be chosen uniquely, hence recursion theorem would be appliable)</p>
<p>But i couldn't. Is there any way to show this?</p>
<p>If it is true, let $y_0 \in B(r,x)\setminus (F_0 \cup \{x_0\})$ and $y_{n+1} \in B(d(y_n,x_0),x_0) \setminus (F_{n+1} \cup \{x_0\})$.</p>
<p>Then, $\forall n\in \omega$, $B(d(y_n,x_0),x_0) \setminus (F_{n+1} \cup \{x_0\})$ is nonempty.</p>
<p>Here, again, i don't know how to show that 'there exists $y\in \mathbb{R}^k$ such that $y\notin \bigcup_{n\in \omega} F_n$.</p>
<p>Please help.</p>
<p>If my argument is wrong, please give me a proper proof.</p>
|
William
| 13,579 |
<p>The Baire Category Theorem implies that complete metric spaces are not first category (meager). That is, they are not a countable union of nowhere dense sets. A set $A$ is nowhere dense if $\text{Int}(\bar{A}) = \emptyset$. </p>
<p>Clearly $\mathbb{R}^n$ is a complete separable metric space. If $\mathbb{R}^n = \bigcup_k A_k$ where each $A_k$ is closed and has empty interior, then this contradicts the Baire Category Theorem. </p>
<p>So you just need that the Baire Category Theorem can be proved without the axiom of choice. This is true for Polish Space, i.e. complete separable metric spaces. Theorem 4 from this pdf from Berkeley has short proof of this result : <a href="http://math.berkeley.edu/~jdahl/104/104Baire.pdf" rel="nofollow">http://math.berkeley.edu/~jdahl/104/104Baire.pdf</a></p>
|
181,839 |
<p>Since the specific space $\mathbb{R}^k$ is given, this might be provable in ZF.</p>
<p>Let $\{F_n\}_{n\in \omega}$ be a family of closed subset of $\mathbb{R}^k$, of which the union is $\mathbb{R}^k$.</p>
<p>Suppose $\forall n\in \omega, {F_n}^o=\emptyset$ ($o$ denotes interior)</p>
<p>Fix $x_0 \in \mathbb{R}^k$
Then by assumption, $\forall 0<r\in \mathbb{R}$, there exists $y\in \mathbb{R}^k$ such that $y\in B(r,x_0)\setminus (F_n \cup \{x_0\})$ for every $n\in \omega$.</p>
<p>I tried to show that for every $r$, there exists $y\in \mathbb{Q}^k$ that satisfies the condition.
(Once the existence is guaranteed, since $\mathbb{Q}^k$ can be well-ordered lexicographically, $y$ can be chosen uniquely, hence recursion theorem would be appliable)</p>
<p>But i couldn't. Is there any way to show this?</p>
<p>If it is true, let $y_0 \in B(r,x)\setminus (F_0 \cup \{x_0\})$ and $y_{n+1} \in B(d(y_n,x_0),x_0) \setminus (F_{n+1} \cup \{x_0\})$.</p>
<p>Then, $\forall n\in \omega$, $B(d(y_n,x_0),x_0) \setminus (F_{n+1} \cup \{x_0\})$ is nonempty.</p>
<p>Here, again, i don't know how to show that 'there exists $y\in \mathbb{R}^k$ such that $y\notin \bigcup_{n\in \omega} F_n$.</p>
<p>Please help.</p>
<p>If my argument is wrong, please give me a proper proof.</p>
|
Carl Mummert
| 630 |
<p>The answer follows from an application of some descriptive set theory and Shoenfield's absoluteness theorem. The rest of this answer works the same way in $\mathbb{R}^k$ as in $\mathbb{R}$, but I will write it up (shortly) in $\mathbb{R}$ for notational simplicity. </p>
<p>Actually, the principle in question is a variant of the Baire category theorem and would be provable in a very weak subsystem of second-order arithmetic, even if generalized to an arbitrary complete separable metric space. Second-order arithmetic is much much weaker than ZF. So showing that the principle is provable in ZF is much easier, although I am going to just use the big hammer of Shoenfield's absoluteness theorem here. </p>
<p>Here is a sketch in ZF, showing how to apply the absoluteness theorem. The key point is that we have to show that the principle of interest can be expressed by a formula at a low level of the analytical hierarchy. </p>
<p>Note that given any open set $U$, ZF can form the set $A(U) = \{ (i,j) : B(q_i, 2^{-j}) \subseteq U\}$ where $(q_i)$ is some fixed enumeration of the rational points and $B(\cdot,\cdot)$ denotes a ball as usual. And this construction is uniform; given a sequence $(U_i)$, ZF is able to form $(A(U_i))$. </p>
<p>Furthermore, a set is by definition closed if and only if its complement is open. </p>
<p>Thus, in ZF, the statement that the union of a sequence $(C_k)$ of closed sets covers all of $\mathbb{R}$ can be written in the following form:</p>
<p>$$
(\forall x \in \mathbb{R})(\exists i, j,k \in\omega)[ (i,j) \in A(U_k) \land d(x, q_i) < 2^{-j}].
$$
where $(U_k)$ is the sequence of the open sets complementary to the sets in thbe sequence $(C_k)$. </p>
<p>The displayed formula is at level $\Pi^1_1$ of the analytical hierarchy if $(A(U_k))$ and the distance function are taken as parameters (the parameters can be taken to be elements of $\omega^\omega$ with routine coding, and the quantifier over $\mathbb{R}$ can be replaced with a quantifier over $\omega^\omega$ with routine methods, and the correctness of these methods can be verified in ZF). </p>
<p>Similarly, the statement that some $C_k$ has nonempty interior can be written as</p>
<p>$$
(\exists k)(\forall x\in \mathbb{R})(\exists i,j)[ d(x, q_i) < 2^{-j} \to x \in C_k]
$$</p>
<p>This is again $\Pi^1_1$ in the analytical hierarchy because we can replace $x \in C_k$ with $x \not \in A(U_k)$ as above. Thus the overall statement "if $\mathbb{R} = \bigcup C_k$ then some $C_k$ has nonempty interior" can be taken, in ZF, to be of the form $(\Pi^1_1) \to (\Pi^1_1)$, and hence this statement is $\Sigma^1_2$ relative to some parameters in $\omega^\omega$. </p>
<p>Shoenfield's absoluteness theorem, which is provable in ZF, states that $\Sigma^1_2$ sentences of the analytical hierarchy with parameters $X \in \omega^\omega$ are absolute - have the same truth value - between $V$ and $L(X)$. Because $L(X)$ provably satisfies ZFC, and ZFC proves the principle in question, ZF can prove that the principle will be true in $L(X)$ and thus also true in $V$ by Shenfield's theorem. </p>
|
181,839 |
<p>Since the specific space $\mathbb{R}^k$ is given, this might be provable in ZF.</p>
<p>Let $\{F_n\}_{n\in \omega}$ be a family of closed subset of $\mathbb{R}^k$, of which the union is $\mathbb{R}^k$.</p>
<p>Suppose $\forall n\in \omega, {F_n}^o=\emptyset$ ($o$ denotes interior)</p>
<p>Fix $x_0 \in \mathbb{R}^k$
Then by assumption, $\forall 0<r\in \mathbb{R}$, there exists $y\in \mathbb{R}^k$ such that $y\in B(r,x_0)\setminus (F_n \cup \{x_0\})$ for every $n\in \omega$.</p>
<p>I tried to show that for every $r$, there exists $y\in \mathbb{Q}^k$ that satisfies the condition.
(Once the existence is guaranteed, since $\mathbb{Q}^k$ can be well-ordered lexicographically, $y$ can be chosen uniquely, hence recursion theorem would be appliable)</p>
<p>But i couldn't. Is there any way to show this?</p>
<p>If it is true, let $y_0 \in B(r,x)\setminus (F_0 \cup \{x_0\})$ and $y_{n+1} \in B(d(y_n,x_0),x_0) \setminus (F_{n+1} \cup \{x_0\})$.</p>
<p>Then, $\forall n\in \omega$, $B(d(y_n,x_0),x_0) \setminus (F_{n+1} \cup \{x_0\})$ is nonempty.</p>
<p>Here, again, i don't know how to show that 'there exists $y\in \mathbb{R}^k$ such that $y\notin \bigcup_{n\in \omega} F_n$.</p>
<p>Please help.</p>
<p>If my argument is wrong, please give me a proper proof.</p>
|
Brian M. Scott
| 12,042 |
<p>Although it’s essentially equivalent to Asaf’s, I prefer the following proof (in ZF) that every separable, completely metrizable space is Baire: I find it easier to work with dense open sets.</p>
<p>Let $\langle X,d\rangle$ be a complete, separable metric space, let $D=\{x_n:n\in\Bbb N\}$ be dense in $X$, let $\{U_n:n\in\Bbb N\}$ be a family of dense open subsets of $X$, and let $V$ be a non-empty open subset of $X$. For $n\in\Bbb N$ and $x\in X$ let $B(x,n)$ be the open $d$-ball of radius $2^{-n}$ centred at $x$, and let $D(x,n)$ be the closed $d$-ball of radius $2^{-n}$ centred at $x$.</p>
<p>Let $$m(0)=\min\left\{n\in\Bbb N:x_n\in G\cap G_0\right\}$$ and $$k(0)=\min\left\{n\in\Bbb N:D\left(x_{m(0)},n\right)\subseteq G\cap G_0\right\}\;.$$</p>
<p>Given $m(n),k(n)\in\Bbb N$, let $$m(n+1)=\min\left\{i\in\Bbb N:x_i\in B\left(x_{m(n)},k(n)\right)\cap G_{n+1}\right\}$$ and $$k(n+1)=\max\left\{2^{n+1},\min\left\{i\in\Bbb N:D\left(x_{m(n+1)},i\right)\subseteq B\left(x_{m(n)},k(n)\right)\cap G_{n+1}\right\}\right\}\;.$$</p>
<p>For $n\in\Bbb N$ let $y_n=x_{m(n)}$. Then</p>
<p>$$B\left(y_n,k(n)\right)\supseteq D\left(y_{n+1},k(n+1)\right)\supseteq B\left(y_{n+1},k(n+1)\right)\supseteq D\left(y_{n+2},k(n+2)\right)\supseteq\ldots$$</p>
<p>for each $n\in\Bbb N$. In particular, $\{y_i:i\ge n\}\subseteq B(y_n,k(n))\subseteq B(y_n,n)$, so $\langle y_n:n\in\Bbb N\rangle$ is a $d$-Cauchy sequence and therefore converges to some $y\in X$. Clearly $y\in D(y_n,k(n))\subseteq G\cap G_n$ for each $n\in\Bbb N$, so $$y\in G\cap\bigcap_{n\in\Bbb N}G_n\ne\varnothing\;.$$ Thus, $\bigcap_{n\in\Bbb N}G_n$ is dense in $X$, and $X$ is Baire.</p>
|
3,510,232 |
<p>I am learning about limits and there is something that I cant quite understand:</p>
<p>If we have the function:</p>
<blockquote>
<p><span class="math-container">$$
f(x) =\frac{x^2-1}{x-1}
$$</span></p>
</blockquote>
<p>Let's say that we want to see which value for y (image) the function approaches as x (domain) gets closer to 1. On a nutshell, we have to take this following limit:</p>
<blockquote>
<p><span class="math-container">$$
\lim_{x\to1}\frac{x^2-1}{x-1}
$$</span></p>
</blockquote>
<p>As soon as we look to this function, we realize that the function is not continuous at x = 1 <strong>(By the way, can I say that?)</strong>. </p>
<p>I know the algorithm to figure out the solution of the limit:</p>
<p>First, there is the need of eliminating the function discontinuity. Usually, it is just a matter of factorizing the function into a new function which the exactly same image as the one before with one crucial difference: <em>The function is continuous for all real numbers</em> </p>
<p>My doubts:Is my way to think about it correct? Can I think like that?</p>
<p>Take the example above: </p>
<blockquote>
<p><span class="math-container">$$
f(x) = \frac{x^2-1}{x-1}
$$</span></p>
</blockquote>
<p>After factorizing, we get:</p>
<p><span class="math-container">$$
f(x) = {x+1}
$$</span></p>
<p>If we plot both functions, they are the same, although the second one has its continuity all along the real numbers domain</p>
<p>Thanks in advance</p>
|
Math1000
| 38,584 |
<p>The rate at which the wasp population is growing is given by
<span class="math-container">$$
P'(t) = \frac{25000000 e^{-\frac{t}{2}}}{\left(1000 e^{-\frac{t}{2}}+1\right)^2}.
$$</span>
We want to find the value of <span class="math-container">$t$</span> that maximizes this function. Since <span class="math-container">$P'$</span> is continuous on <span class="math-container">$[0,25]$</span> and differentiable on <span class="math-container">$(0,25)$</span>, this must occur at a boundary point (<span class="math-container">$t=0$</span> or <span class="math-container">$t=25$</span>) or at a point such that <span class="math-container">$P''(t)=0$</span>. We evaluate <span class="math-container">$P$</span> at the boundary points:
<span class="math-container">\begin{align}
P'(0) &= \frac{25000000}{1002001}\approx 24.95007\\
\quad P'(25) &= \frac{25000000}{\left(1+\frac{1000}{e^{25/2}}\right)^2 e^{25/2}}\approx 92.47579,
\end{align}</span>
then compute the derivative:
<span class="math-container">$$
P''(t) = \frac{12500000 e^{t/2} \left(1000-e^{t/2}\right)}{\left(e^{t/2}+1000\right)^3}.
$$</span>
We see that <span class="math-container">$P''(t)= 0$</span> if and only if <span class="math-container">$12500000 e^{t/2} \left(1000-e^{t/2}\right)=0$</span>, and since the first factor is always positive, this is the case if and only if <span class="math-container">$e^{t/2}=1000$</span>. Taking logs and solving for <span class="math-container">$t$</span> yields <span class="math-container">$t= 2\log1000\approx 13.81551$</span>. Since <span class="math-container">$P''$</span> is positive on <span class="math-container">$[0,2\log1000)$</span> and negative on <span class="math-container">$(2\log1000,25]$</span>, it follows that <span class="math-container">$t=2\log1000$</span> is a local (and hence global) maximum for <span class="math-container">$P'$</span>.</p>
|
3,510,232 |
<p>I am learning about limits and there is something that I cant quite understand:</p>
<p>If we have the function:</p>
<blockquote>
<p><span class="math-container">$$
f(x) =\frac{x^2-1}{x-1}
$$</span></p>
</blockquote>
<p>Let's say that we want to see which value for y (image) the function approaches as x (domain) gets closer to 1. On a nutshell, we have to take this following limit:</p>
<blockquote>
<p><span class="math-container">$$
\lim_{x\to1}\frac{x^2-1}{x-1}
$$</span></p>
</blockquote>
<p>As soon as we look to this function, we realize that the function is not continuous at x = 1 <strong>(By the way, can I say that?)</strong>. </p>
<p>I know the algorithm to figure out the solution of the limit:</p>
<p>First, there is the need of eliminating the function discontinuity. Usually, it is just a matter of factorizing the function into a new function which the exactly same image as the one before with one crucial difference: <em>The function is continuous for all real numbers</em> </p>
<p>My doubts:Is my way to think about it correct? Can I think like that?</p>
<p>Take the example above: </p>
<blockquote>
<p><span class="math-container">$$
f(x) = \frac{x^2-1}{x-1}
$$</span></p>
</blockquote>
<p>After factorizing, we get:</p>
<p><span class="math-container">$$
f(x) = {x+1}
$$</span></p>
<p>If we plot both functions, they are the same, although the second one has its continuity all along the real numbers domain</p>
<p>Thanks in advance</p>
|
Quanto
| 686,284 |
<p>The growth is the fastest when the slope of <span class="math-container">$P(t)$</span> is the largest. Evaluate <span class="math-container">$
P'(t) = \frac{ke^{-t/2}}{\left(1000 e^{-t/2}+1\right)^2}$</span>, with <span class="math-container">$k=25mm$</span>. Then, cast it in the form</p>
<p><span class="math-container">$$P'(t) = \frac k{(1000e^{-t/4}-e^{t/4})^2+2000}\le \frac k{2000}$$</span></p>
<p>where the equality, or, the maximal growth, occurs at <span class="math-container">$1000e^{-t/4}=e^{t/4}$</span>, i.e. <span class="math-container">$t=2\ln 1000$</span>.</p>
|
125,862 |
<p>I have been given that I am working with the space of all 2x2 matrices. The basis $B$ for this space is given as a set of four 2x2 matrices, each with an entry of 1 in a unique position and zeroes everywhere else (sorry about the description in words - I don't know how to format matrices for this site).</p>
<p>I have also been given the basis $B' = ({1, x, x^2})$ for the space of all polynomials of degree 2 or less and the basis $B'' = ({1})$ for $R$.</p>
<p>Then I am given a series of linear transformations and asked to find the matrices associated with them with respect to the bases above. I am completely lost as to how to do this! I would like help with how to achieve one of them so that I can then go and apply what I learn here to the other transformations.</p>
<p>The example I've chosen is the transformation T that maps 2x2 matrices to their transposes. I can't seem to construct a matrix that will bring the element in position '21' up to position '12'.</p>
<p>Can anyone give me some direction with this? Many thanks!!</p>
|
Rasmus
| 367 |
<p>So let's look at the transformation map $T\colon\mathbb M_2\to\mathbb M_2$. Let's write the given basis as $\{e_{11},e_{12},e_{21},e_{22}\}$ and let's fix the order in which we have written it down.</p>
<p>We have $T(e_{ij})=e_{ji}$ for all $i,j\in\{1,2\}$. Hence, the matrix for $T$ in our chosen ordered basis looks as follows:
$$
\pmatrix{1&0&0&0\\0&0&1&0\\0&1&0&0\\0&0&0&1}.
$$</p>
|
125,862 |
<p>I have been given that I am working with the space of all 2x2 matrices. The basis $B$ for this space is given as a set of four 2x2 matrices, each with an entry of 1 in a unique position and zeroes everywhere else (sorry about the description in words - I don't know how to format matrices for this site).</p>
<p>I have also been given the basis $B' = ({1, x, x^2})$ for the space of all polynomials of degree 2 or less and the basis $B'' = ({1})$ for $R$.</p>
<p>Then I am given a series of linear transformations and asked to find the matrices associated with them with respect to the bases above. I am completely lost as to how to do this! I would like help with how to achieve one of them so that I can then go and apply what I learn here to the other transformations.</p>
<p>The example I've chosen is the transformation T that maps 2x2 matrices to their transposes. I can't seem to construct a matrix that will bring the element in position '21' up to position '12'.</p>
<p>Can anyone give me some direction with this? Many thanks!!</p>
|
nrenga
| 363,379 |
<p>I thought I would clarify that the transpose is a linear operation by explicitly giving the set of linear operations that need to be performed on the original matrix to get its transpose. I will give the expression for the case of a square matrix $M_{n \times n}$ but this can be extended to arbitrary matrices too.</p>
<p>Let $I_n$ denote the $n \times n$ identity matrix whose $i^{\rm th}$ column is the standard basis $e_i$. Let $S_{n,n}$ denote the $n^2 \times n^2$ perfect shuffle matrix corresponding to writing an $n^2 \times 1$ vector into an $n \times n$ matrix column-wise and then reading it row-wise. Then, it can be shown that</p>
<p>$$M^T = \sum_{i=1}^{n} \left( e_{i}^{T} \otimes I_n \right) S_{n,n} \left( I_n \otimes M \right) \left( \sum_{j=1}^{n} e_j \otimes e_j \right) e_{i}^T, $$</p>
<p>where $\otimes$ denotes the Kronecker product of two matrices. It should be noted that $\left( \sum_{j=1}^{n} e_j \otimes e_j \right) = vec(I_n)$ is just the vectorized form of $I_n$, $\left( I_n \otimes M \right) vec(I_n) = vec(M)$ is the vectorized form of $M$ and $S_{n,n} vec(M) = vec(M^T)$ is the vectorized form of $M^T$. Hence, this also shows that vectorization is a linear transformation.</p>
|
689,591 |
<p>Today at class, my teacher stated the following proposition saying it is obvious:</p>
<p>Let $x_0 \in U \subset \mathbb{R}^d$, $U$ open, and $f: U \to \mathbb{R}^m$ differentiable at $x_0$, then for any $v \in \mathbb{R}^d$ we have</p>
<p>$$ D_v f(x_0) = D_f (x_0)(v) $$</p>
<p>It is not obvious to me. Can someone explain to me why this is true? Thanks a lot.</p>
|
copper.hat
| 27,978 |
<p>Let $\lambda(t) = x_0+t v$. Then $D_v f(x_0) = (f \circ \lambda)'(0)$.</p>
<p>Since $f$ is differentiable, we can apply the chain rule to get
$(f \circ \lambda)'(0) = D(f \circ \lambda)(0) = Df(\lambda(0)) D \lambda(0) = Df(x_0) v$.</p>
<p>(You can find this many texts, for example in look at the commentary after Definition 3 in Section 6.4 of Marsden's "Elementary Classical Analysis".)</p>
|
29,323 |
<p>You're hanging out with a bunch of other mathematicians - you go out to dinner, you're on the train, you're at a department tea, et cetera. Someone says something like "A group of 100 people at a party are each receive hats with different prime numbers and ..." For the next few minutes everyone has fun solving the problem together.</p>
<p>I love puzzles like that. But there's a problem -- I running into the same puzzles over and over. But there must be lots of great problems I've never run into. So I'd like to hear problems that other people have enjoyed, and hopefully everyone will learn some new ones.</p>
<p>So: What are your favorite dinner conversation math puzzles?</p>
<p>I don't want to provide hard guidelines. But I'm generally interested in problems that are mathematical and not just logic puzzles. They shouldn't require written calculations or a convoluted answer. And they should be fun - with some sort of cute step, aha moment, or other satisfying twist. I'd prefer to keep things pretty elementary, but a cool problem requiring a little background is a-okay.</p>
<p>One problem per answer.</p>
<p>If you post the answer, please obfuscate it with something like <a href="http://www.rot13.com/">rot13</a>. Don't spoil the fun for everyone else.</p>
|
Gerry Myerson
| 3,684 |
<p>Instead of recommending some puzzles, I'll recommend some books containing many puzzles. Peter Winkler, Mathematical Puzzles; Peter Winkler, Mathematical Mind-Benders; Miodrag Petkovic, Famous Puzzles of Great Mathematicians. </p>
|
29,323 |
<p>You're hanging out with a bunch of other mathematicians - you go out to dinner, you're on the train, you're at a department tea, et cetera. Someone says something like "A group of 100 people at a party are each receive hats with different prime numbers and ..." For the next few minutes everyone has fun solving the problem together.</p>
<p>I love puzzles like that. But there's a problem -- I running into the same puzzles over and over. But there must be lots of great problems I've never run into. So I'd like to hear problems that other people have enjoyed, and hopefully everyone will learn some new ones.</p>
<p>So: What are your favorite dinner conversation math puzzles?</p>
<p>I don't want to provide hard guidelines. But I'm generally interested in problems that are mathematical and not just logic puzzles. They shouldn't require written calculations or a convoluted answer. And they should be fun - with some sort of cute step, aha moment, or other satisfying twist. I'd prefer to keep things pretty elementary, but a cool problem requiring a little background is a-okay.</p>
<p>One problem per answer.</p>
<p>If you post the answer, please obfuscate it with something like <a href="http://www.rot13.com/">rot13</a>. Don't spoil the fun for everyone else.</p>
|
Bugs Bunny
| 5,301 |
<p>A room contains 3 bulbs and 3 switches outside controlling the bulbs. Is it possible to determine which switch controls which bulb by entering the room only once and observing bulbs?</p>
|
29,323 |
<p>You're hanging out with a bunch of other mathematicians - you go out to dinner, you're on the train, you're at a department tea, et cetera. Someone says something like "A group of 100 people at a party are each receive hats with different prime numbers and ..." For the next few minutes everyone has fun solving the problem together.</p>
<p>I love puzzles like that. But there's a problem -- I running into the same puzzles over and over. But there must be lots of great problems I've never run into. So I'd like to hear problems that other people have enjoyed, and hopefully everyone will learn some new ones.</p>
<p>So: What are your favorite dinner conversation math puzzles?</p>
<p>I don't want to provide hard guidelines. But I'm generally interested in problems that are mathematical and not just logic puzzles. They shouldn't require written calculations or a convoluted answer. And they should be fun - with some sort of cute step, aha moment, or other satisfying twist. I'd prefer to keep things pretty elementary, but a cool problem requiring a little background is a-okay.</p>
<p>One problem per answer.</p>
<p>If you post the answer, please obfuscate it with something like <a href="http://www.rot13.com/">rot13</a>. Don't spoil the fun for everyone else.</p>
|
Kiochi
| 4,241 |
<p>You are blindfolded, then given a deck of cards in which 23 of the cards have been flipped up, then inserted into the deck randomly (you know this). Without taking the blindfold off, rearrange the deck into two stacks such that both stacks have the same number of up-flipped cards. (You are allowed to flip as many cards as you please.)</p>
|
29,323 |
<p>You're hanging out with a bunch of other mathematicians - you go out to dinner, you're on the train, you're at a department tea, et cetera. Someone says something like "A group of 100 people at a party are each receive hats with different prime numbers and ..." For the next few minutes everyone has fun solving the problem together.</p>
<p>I love puzzles like that. But there's a problem -- I running into the same puzzles over and over. But there must be lots of great problems I've never run into. So I'd like to hear problems that other people have enjoyed, and hopefully everyone will learn some new ones.</p>
<p>So: What are your favorite dinner conversation math puzzles?</p>
<p>I don't want to provide hard guidelines. But I'm generally interested in problems that are mathematical and not just logic puzzles. They shouldn't require written calculations or a convoluted answer. And they should be fun - with some sort of cute step, aha moment, or other satisfying twist. I'd prefer to keep things pretty elementary, but a cool problem requiring a little background is a-okay.</p>
<p>One problem per answer.</p>
<p>If you post the answer, please obfuscate it with something like <a href="http://www.rot13.com/">rot13</a>. Don't spoil the fun for everyone else.</p>
|
BlueRaja
| 2,883 |
<p>Assuming you have unlimited time and cash, is there a strategy that's guaranteed to win at roulette?</p>
|
29,323 |
<p>You're hanging out with a bunch of other mathematicians - you go out to dinner, you're on the train, you're at a department tea, et cetera. Someone says something like "A group of 100 people at a party are each receive hats with different prime numbers and ..." For the next few minutes everyone has fun solving the problem together.</p>
<p>I love puzzles like that. But there's a problem -- I running into the same puzzles over and over. But there must be lots of great problems I've never run into. So I'd like to hear problems that other people have enjoyed, and hopefully everyone will learn some new ones.</p>
<p>So: What are your favorite dinner conversation math puzzles?</p>
<p>I don't want to provide hard guidelines. But I'm generally interested in problems that are mathematical and not just logic puzzles. They shouldn't require written calculations or a convoluted answer. And they should be fun - with some sort of cute step, aha moment, or other satisfying twist. I'd prefer to keep things pretty elementary, but a cool problem requiring a little background is a-okay.</p>
<p>One problem per answer.</p>
<p>If you post the answer, please obfuscate it with something like <a href="http://www.rot13.com/">rot13</a>. Don't spoil the fun for everyone else.</p>
|
Christian Blatter
| 6,958 |
<p>A princess inhabits a flight of 17 rooms in a row. Each room has a door to the outside, and there is a door between adjacent rooms. The princess spends each day in a room that is adjacent to the room she was in the day before. One day a prince arrives from far away to woo for the princess. The guardian explains the habits of the princess and also the rules to him: Each day he may knock at an outside door of his choice. If the princess is behind it she will open and in the end marry him. If not, nothing happens, and he gets another chance the next day. Unfortunately his return ticket expires after 30 days. Does he have enough time to conquer the princess?
(Adapted from "simpler-solutions.net")</p>
|
29,323 |
<p>You're hanging out with a bunch of other mathematicians - you go out to dinner, you're on the train, you're at a department tea, et cetera. Someone says something like "A group of 100 people at a party are each receive hats with different prime numbers and ..." For the next few minutes everyone has fun solving the problem together.</p>
<p>I love puzzles like that. But there's a problem -- I running into the same puzzles over and over. But there must be lots of great problems I've never run into. So I'd like to hear problems that other people have enjoyed, and hopefully everyone will learn some new ones.</p>
<p>So: What are your favorite dinner conversation math puzzles?</p>
<p>I don't want to provide hard guidelines. But I'm generally interested in problems that are mathematical and not just logic puzzles. They shouldn't require written calculations or a convoluted answer. And they should be fun - with some sort of cute step, aha moment, or other satisfying twist. I'd prefer to keep things pretty elementary, but a cool problem requiring a little background is a-okay.</p>
<p>One problem per answer.</p>
<p>If you post the answer, please obfuscate it with something like <a href="http://www.rot13.com/">rot13</a>. Don't spoil the fun for everyone else.</p>
|
Tony Huynh
| 2,233 |
<p>An evil sorceress is holding 100 princes captive. Right now they are all in the same prison cell and can discuss strategy. However, in a moment, they will each be taken to individual prison cells, where no communication is possible. After that, the sorceress will start randomly calling princes to her bedroom (one at a time). This continues indefinitely, so a prince can visit the bedroom many times. The bedroom has two light switches, whose state can be observed only from inside the bedroom. When a prince is called to the bedroom, he can observe the state of the switches, and then must change the state of <em>exactly one</em> of the light switches. The initial state of the light switches is not known. </p>
<p>The princes will be set free if any one of them can determine if all of them have been called to the room. </p>
<p><strong>Puzzle:</strong> Determine a strategy for the princes so that they are guaranteed to be set free eventually. The strategy should never output a false positive. For example, if a prince has been called one million times he can reason that on average, everyone else has been called one million times. Thus it is very likely that all the princes have been called to the bedroom, but it is <em>possible</em> that one prince still hasn't been called. </p>
|
29,323 |
<p>You're hanging out with a bunch of other mathematicians - you go out to dinner, you're on the train, you're at a department tea, et cetera. Someone says something like "A group of 100 people at a party are each receive hats with different prime numbers and ..." For the next few minutes everyone has fun solving the problem together.</p>
<p>I love puzzles like that. But there's a problem -- I running into the same puzzles over and over. But there must be lots of great problems I've never run into. So I'd like to hear problems that other people have enjoyed, and hopefully everyone will learn some new ones.</p>
<p>So: What are your favorite dinner conversation math puzzles?</p>
<p>I don't want to provide hard guidelines. But I'm generally interested in problems that are mathematical and not just logic puzzles. They shouldn't require written calculations or a convoluted answer. And they should be fun - with some sort of cute step, aha moment, or other satisfying twist. I'd prefer to keep things pretty elementary, but a cool problem requiring a little background is a-okay.</p>
<p>One problem per answer.</p>
<p>If you post the answer, please obfuscate it with something like <a href="http://www.rot13.com/">rot13</a>. Don't spoil the fun for everyone else.</p>
|
user7361
| 7,361 |
<p>First, I apologize that this puzzle is not "clean". It's more suitable for the bar rather than dinner (in fact I heard it at a party with mathematicians). I understand if it should be taken down or rephrased. I scanned the previous puzzles and I don't think this puzzle is a duplicate.</p>
<p>Suppose you are male(female) and stranded on an island with three females(males). You wish to have protected sex with all three females(males), but you only have two condoms and no other forms of protection. Clearly you can have protected sex with two of them by using one condom, throwing it away and using the other. Can you have protected sex with all three of them?</p>
<p>By protected I mean there is no exchange of fluids from one person to another, i.e. you can't use a condom with one person and then use it "as is" with another person. Also, despite being surrounded by the ocean you cannot just rinse them off - that's dirty.</p>
|
29,323 |
<p>You're hanging out with a bunch of other mathematicians - you go out to dinner, you're on the train, you're at a department tea, et cetera. Someone says something like "A group of 100 people at a party are each receive hats with different prime numbers and ..." For the next few minutes everyone has fun solving the problem together.</p>
<p>I love puzzles like that. But there's a problem -- I running into the same puzzles over and over. But there must be lots of great problems I've never run into. So I'd like to hear problems that other people have enjoyed, and hopefully everyone will learn some new ones.</p>
<p>So: What are your favorite dinner conversation math puzzles?</p>
<p>I don't want to provide hard guidelines. But I'm generally interested in problems that are mathematical and not just logic puzzles. They shouldn't require written calculations or a convoluted answer. And they should be fun - with some sort of cute step, aha moment, or other satisfying twist. I'd prefer to keep things pretty elementary, but a cool problem requiring a little background is a-okay.</p>
<p>One problem per answer.</p>
<p>If you post the answer, please obfuscate it with something like <a href="http://www.rot13.com/">rot13</a>. Don't spoil the fun for everyone else.</p>
|
jonderry
| 8,013 |
<p>Here's one I saw a while ago:</p>
<p>A prisoner is presented with the following challenge by one of the guards of the jail. The prisoner is to be blindfolded and then the guard will place $n$ coins on a circular turntable with any combination of heads and tails facing up (with at least one tails showing initially). The prisoners goal is to flip over coins until all heads are showing.</p>
<p>This would be easy enough if the guard did not interfere. The prisoner could just try all $2^n$ combinations, and one of them would be guaranteed to result in all heads. However, to complicate matters, the guard may turn the table during this process. More specifically, the following process is repeated. First, the prisoner chooses a set of positions of coins to flip over. Then, before the coins are flipped, the guard turns the turntable so as to try to prevent the prisoner from flipping all of the coins to heads. Finally, the prisoner flips over the coins that are at the positions chosen in the first step. If all heads are showing, the game stops and the prisoner is set free.</p>
<p>The question is, for what values of $n$ does the prisoner have a winning strategy and how many moves does it take?</p>
<p>What if the guard uses 6-sided dice instead of coins with the goal of showing all ones (assuming the orientations of the dice are preserved relative to their positions on the turntable between rounds)?</p>
<p>In general, what values of $n$ allow a solution with $k$-sided dice?</p>
|
29,323 |
<p>You're hanging out with a bunch of other mathematicians - you go out to dinner, you're on the train, you're at a department tea, et cetera. Someone says something like "A group of 100 people at a party are each receive hats with different prime numbers and ..." For the next few minutes everyone has fun solving the problem together.</p>
<p>I love puzzles like that. But there's a problem -- I running into the same puzzles over and over. But there must be lots of great problems I've never run into. So I'd like to hear problems that other people have enjoyed, and hopefully everyone will learn some new ones.</p>
<p>So: What are your favorite dinner conversation math puzzles?</p>
<p>I don't want to provide hard guidelines. But I'm generally interested in problems that are mathematical and not just logic puzzles. They shouldn't require written calculations or a convoluted answer. And they should be fun - with some sort of cute step, aha moment, or other satisfying twist. I'd prefer to keep things pretty elementary, but a cool problem requiring a little background is a-okay.</p>
<p>One problem per answer.</p>
<p>If you post the answer, please obfuscate it with something like <a href="http://www.rot13.com/">rot13</a>. Don't spoil the fun for everyone else.</p>
|
Dave Futer
| 8,183 |
<p>Here's one of my favorites. There are 99 bags, each of which contains some number of apples and some number of oranges. Prove that there's a way to select 50 out of the 99 bags, such that these 50 <i>simultaneously</i> contain at least half the total number of apples and at least half the total number of oranges.</p>
<p>One fun aspect of this problem is that there are a number of distinct solutions, inspired by different areas of math. I know of at least three...</p>
|
29,323 |
<p>You're hanging out with a bunch of other mathematicians - you go out to dinner, you're on the train, you're at a department tea, et cetera. Someone says something like "A group of 100 people at a party are each receive hats with different prime numbers and ..." For the next few minutes everyone has fun solving the problem together.</p>
<p>I love puzzles like that. But there's a problem -- I running into the same puzzles over and over. But there must be lots of great problems I've never run into. So I'd like to hear problems that other people have enjoyed, and hopefully everyone will learn some new ones.</p>
<p>So: What are your favorite dinner conversation math puzzles?</p>
<p>I don't want to provide hard guidelines. But I'm generally interested in problems that are mathematical and not just logic puzzles. They shouldn't require written calculations or a convoluted answer. And they should be fun - with some sort of cute step, aha moment, or other satisfying twist. I'd prefer to keep things pretty elementary, but a cool problem requiring a little background is a-okay.</p>
<p>One problem per answer.</p>
<p>If you post the answer, please obfuscate it with something like <a href="http://www.rot13.com/">rot13</a>. Don't spoil the fun for everyone else.</p>
|
David Richter
| 11,264 |
<p>Can one partition the plane $\mathbb{R}^2$ by closed intervals of equal length? How? The answer to the first question is "yes". In other words, can one cover the plane with translates and rotations of a given closed line segment such that every point lies on exactly one segment?</p>
|
191,373 |
<p>I have a usual mathematical background in vector and tensor calculus. I was trying to use the differential operators of Mathematica, namely <code>Grad</code>, <code>Div</code> and <code>Curl</code>. According to my knowledge, the definitions of Mathematica for <code>Grad</code> and <code>Div</code> coincides with those usually employed in tensor calculus, that is to say</p>
<p><span class="math-container">\begin{align*}
\text{grad}\mathbf{T}&:=\sum_{k=1}^{3}\frac{\partial\mathbf{T}}{\partial x_k}\otimes \mathbf{e}_k\\
\text{div}\mathbf{T}&:=\sum_{k=1}^{3}\frac{\partial\mathbf{T}}{\partial x_k}\cdot\mathbf{e}_k \\
\tag{1}
\end{align*}</span></p>
<p>for any tensor <span class="math-container">$\mathbf{T}$</span> of rank <span class="math-container">$n\ge1$</span>. <span class="math-container">$x_k$</span>'s are Cartesian coordinates and <span class="math-container">$\mathbf{e}_i$</span>'s are the standard basis for <span class="math-container">$\mathbb{R}^3$</span>. <span class="math-container">$\otimes$</span> and <span class="math-container">$\cdot$</span> are the usual generalized outer and inner products which are also defined in Mathematica by <code>Outer</code> and <code>Inner</code>. The usual definition that I know from tensor calculus for the <code>Curl</code> is as follows
<span class="math-container">\begin{align*}
\text{curl}\mathbf{T}&:=\sum_{k=1}^{3}\mathbf{e}_k\times\frac{\partial\mathbf{T}}{\partial x_k}.
\tag{2}
\end{align*}</span>
However, it turns out that Mathematica's definition for curl is totally different. For example, it returns the <code>Curl</code> of a second order tensor as a scalar, while according to <span class="math-container">$(2)$</span> it should be a second order tensor.</p>
<blockquote>
<p>I couldn't find a precise definition of Mathematica for <code>Curl</code> in the documents. I am wondering what this definition is. What is the motivation for this? and How it can be related to the definition given in <span class="math-container">$(2)$</span>?</p>
</blockquote>
<p>Below is a simple piece of code for you to observe the outputs of Mathematica when we apply the <code>Grad</code>, <code>Div</code> and <code>Curl</code> operators to scalar, vector and second order tensor fields. I would like to draw your attention to some observations. <code>Curl</code> of a scalar is returned as a second order tensor, which I don't understand why! <code>Curl</code> of a vector coincides with the usual definition of <code>Curl</code> used in vector calculus. <code>Curl</code> of second order tensor is returned as a scalar, which I don't understand again.</p>
<pre><code>Var={Subscript[x, 1],Subscript[x, 2],Subscript[x, 3]};
Sca=\[Phi][Subscript[x, 1],Subscript[x, 2],Subscript[x, 3]];
Vec={Subscript[v, 1][Subscript[x, 1],Subscript[x, 2],Subscript[x, 3]],Subscript[v, 2][Subscript[x, 1],Subscript[x, 2],Subscript[x, 3]],Subscript[v, 3][Subscript[x, 1],Subscript[x, 2],Subscript[x, 3]]};
Ten=Table[Subscript[T, i,j][Subscript[x, 1],Subscript[x, 2],Subscript[x, 3]],{i,1,3},{j,1,3}];
MatrixForm[Grad[Sca, Var]]
MatrixForm[Grad[Vec, Var]]
MatrixForm[Grad[Ten, Var]]
MatrixForm[Div[Sca, Var]]
MatrixForm[Div[Vec, Var]]
MatrixForm[Div[Ten, Var]]
MatrixForm[Curl[Sca, Var]]
MatrixForm[Curl[Vec, Var]]
MatrixForm[Curl[Ten, Var]]
</code></pre>
<p>I will be happy if someone can reproduce the following result for the curl of a second order tensor with Mathematica's <code>Curl</code> function.</p>
<p><span class="math-container">\begin{align*}
\text{curl}\mathbf{T}&:=\sum_{k=1}^{3}\mathbf{e}_k\times\frac{\partial\mathbf{T}}{\partial x_k}=\sum_{k=1}^{3}\mathbf{e}_k\times\frac{\partial}{\partial x_k}\left(\sum_{i=1}^{3}\sum_{j=1}^{3}T_{ij}\mathbf{e}_i\otimes\mathbf{e}_j\right)\\
&=\sum_{k=1}^{3}\sum_{i=1}^{3}\sum_{j=1}^{3}\frac{\partial T_{ij}}{\partial x_k}(\mathbf{e}_k\times\mathbf{e}_i)\otimes\mathbf{e}_j\\
&=\sum_{k=1}^{3}\sum_{i=1}^{3}\sum_{j=1}^{3}\sum_{m=1}^{3}\epsilon_{kim}\frac{\partial T_{ij}}{\partial x_k}\mathbf{e}_m\otimes\mathbf{e}_j
\tag{3}
\end{align*}</span></p>
<p>where <span class="math-container">$\epsilon_{kim}$</span> is the <code>LeviCivitaTensor</code> for <span class="math-container">$3$</span> dimensions. Consequently, we get</p>
<p><span class="math-container">\begin{align*}
\left(\text{curl}\mathbf{T}\right)_{mj}=\sum_{k=1}^{3}\sum_{i=1}^{3}\epsilon_{kim}\frac{\partial T_{ij}}{\partial x_k}.
\tag{4}
\end{align*}</span></p>
<p>Implementing <span class="math-container">$(4)$</span> in Mathematica, we obtain</p>
<pre><code>CurlTen = Table[
Sum[
LeviCivitaTensor[3][[k, i, m]]
D[Subscript[T, i, j][Subscript[x, 1], Subscript[x, 2], Subscript[x, 3]], {Subscript[x, k]}], {k, 1, 3}, {i, 1, 3}],
{m, 1, 3}, {j, 1, 3}];
MatrixForm[CurlTen]
</code></pre>
|
xzczd
| 1,871 |
<p>I fail to find a reference for the definition used by <code>Curl</code>, but manage to figure out how the <code>Curl</code> is defined.</p>
<p>I'll use <a href="http://www.dr-qubit.org/teaching/summation_delta.pdf" rel="noreferrer">Einstein summation convention</a> for simplicity. The defintion used by <code>Curl</code> is summarized as follows:</p>
<ol>
<li><code>Curl</code> of a scalar / zero-order tensor is defined as</li>
</ol>
<p><span class="math-container">$$\text{curl}\ \phi= \epsilon_{kim}\frac{\partial\phi}{\partial x_m}$$</span></p>
<ol start="2">
<li><code>Curl</code> of a vector / first-order tensor is defined as</li>
</ol>
<p><span class="math-container">$$\text{curl}\ \mathbf{v}= \epsilon_{kim}\frac{\partial v_m}{\partial x_i}$$</span></p>
<ol start="3">
<li><code>Curl</code> of a matrix / second-order tensor is defined as</li>
</ol>
<p><span class="math-container">$$\text{curl}\ \mathbf{T}=\color{red}{\frac{1}{2}} \epsilon_{kim}\frac{\partial T_{im}}{\partial x_k}$$</span></p>
<p>Check</p>
<pre><code>ϵ = LeviCivitaTensor[3];
Table[Sum[ϵ[[k, i, m]] D[Sca, Var[[m]]], {m, 3}], {k, 3}, {i, 3}] ==
ϵ.D[Sca, {Var}] ==
TensorContract[ϵ\[TensorProduct]D[Sca, {Var}], {{3, 4}}] ==
Curl[Sca, Var]
(*True*)
Table[Sum[ϵ[[k, i, m]] D[Vec[[m]], Var[[i]]], {m, 3}, {i, 3}], {k, 3}] ==
TensorContract[ϵ\[TensorProduct]D[Vec, {Var}], {{3, 4}, {2, 5}}] ==
Curl[Vec, Var]
(*True*)
1/2 Sum[ϵ[[k, i, m]] D[Ten[[i, m]], Var[[k]]], {k, 3}, {i, 3}, {m, 3}] ==
1/2 TensorContract[ϵ\[TensorProduct]D[Ten, {Var}], {{2, 4}, {3, 5}, {1, 6}}] ==
Curl[Ten, Var]
(*True*)
</code></pre>
<p>The second question is simple:</p>
<pre><code>Table[Sum[ϵ[[k, i, m]] D[Ten[[i, j]], Var[[k]]], {k, 3}, {i, 3}], {m, 3}, {j, 3}] ==
TensorContract[ϵ\[TensorProduct]D[Ten, {Var}], {{2, 4}, {1, 6}}] ==
(Curl[#, Var] & /@ Transpose@Ten // Transpose)
(* True *)
</code></pre>
|
191,373 |
<p>I have a usual mathematical background in vector and tensor calculus. I was trying to use the differential operators of Mathematica, namely <code>Grad</code>, <code>Div</code> and <code>Curl</code>. According to my knowledge, the definitions of Mathematica for <code>Grad</code> and <code>Div</code> coincides with those usually employed in tensor calculus, that is to say</p>
<p><span class="math-container">\begin{align*}
\text{grad}\mathbf{T}&:=\sum_{k=1}^{3}\frac{\partial\mathbf{T}}{\partial x_k}\otimes \mathbf{e}_k\\
\text{div}\mathbf{T}&:=\sum_{k=1}^{3}\frac{\partial\mathbf{T}}{\partial x_k}\cdot\mathbf{e}_k \\
\tag{1}
\end{align*}</span></p>
<p>for any tensor <span class="math-container">$\mathbf{T}$</span> of rank <span class="math-container">$n\ge1$</span>. <span class="math-container">$x_k$</span>'s are Cartesian coordinates and <span class="math-container">$\mathbf{e}_i$</span>'s are the standard basis for <span class="math-container">$\mathbb{R}^3$</span>. <span class="math-container">$\otimes$</span> and <span class="math-container">$\cdot$</span> are the usual generalized outer and inner products which are also defined in Mathematica by <code>Outer</code> and <code>Inner</code>. The usual definition that I know from tensor calculus for the <code>Curl</code> is as follows
<span class="math-container">\begin{align*}
\text{curl}\mathbf{T}&:=\sum_{k=1}^{3}\mathbf{e}_k\times\frac{\partial\mathbf{T}}{\partial x_k}.
\tag{2}
\end{align*}</span>
However, it turns out that Mathematica's definition for curl is totally different. For example, it returns the <code>Curl</code> of a second order tensor as a scalar, while according to <span class="math-container">$(2)$</span> it should be a second order tensor.</p>
<blockquote>
<p>I couldn't find a precise definition of Mathematica for <code>Curl</code> in the documents. I am wondering what this definition is. What is the motivation for this? and How it can be related to the definition given in <span class="math-container">$(2)$</span>?</p>
</blockquote>
<p>Below is a simple piece of code for you to observe the outputs of Mathematica when we apply the <code>Grad</code>, <code>Div</code> and <code>Curl</code> operators to scalar, vector and second order tensor fields. I would like to draw your attention to some observations. <code>Curl</code> of a scalar is returned as a second order tensor, which I don't understand why! <code>Curl</code> of a vector coincides with the usual definition of <code>Curl</code> used in vector calculus. <code>Curl</code> of second order tensor is returned as a scalar, which I don't understand again.</p>
<pre><code>Var={Subscript[x, 1],Subscript[x, 2],Subscript[x, 3]};
Sca=\[Phi][Subscript[x, 1],Subscript[x, 2],Subscript[x, 3]];
Vec={Subscript[v, 1][Subscript[x, 1],Subscript[x, 2],Subscript[x, 3]],Subscript[v, 2][Subscript[x, 1],Subscript[x, 2],Subscript[x, 3]],Subscript[v, 3][Subscript[x, 1],Subscript[x, 2],Subscript[x, 3]]};
Ten=Table[Subscript[T, i,j][Subscript[x, 1],Subscript[x, 2],Subscript[x, 3]],{i,1,3},{j,1,3}];
MatrixForm[Grad[Sca, Var]]
MatrixForm[Grad[Vec, Var]]
MatrixForm[Grad[Ten, Var]]
MatrixForm[Div[Sca, Var]]
MatrixForm[Div[Vec, Var]]
MatrixForm[Div[Ten, Var]]
MatrixForm[Curl[Sca, Var]]
MatrixForm[Curl[Vec, Var]]
MatrixForm[Curl[Ten, Var]]
</code></pre>
<p>I will be happy if someone can reproduce the following result for the curl of a second order tensor with Mathematica's <code>Curl</code> function.</p>
<p><span class="math-container">\begin{align*}
\text{curl}\mathbf{T}&:=\sum_{k=1}^{3}\mathbf{e}_k\times\frac{\partial\mathbf{T}}{\partial x_k}=\sum_{k=1}^{3}\mathbf{e}_k\times\frac{\partial}{\partial x_k}\left(\sum_{i=1}^{3}\sum_{j=1}^{3}T_{ij}\mathbf{e}_i\otimes\mathbf{e}_j\right)\\
&=\sum_{k=1}^{3}\sum_{i=1}^{3}\sum_{j=1}^{3}\frac{\partial T_{ij}}{\partial x_k}(\mathbf{e}_k\times\mathbf{e}_i)\otimes\mathbf{e}_j\\
&=\sum_{k=1}^{3}\sum_{i=1}^{3}\sum_{j=1}^{3}\sum_{m=1}^{3}\epsilon_{kim}\frac{\partial T_{ij}}{\partial x_k}\mathbf{e}_m\otimes\mathbf{e}_j
\tag{3}
\end{align*}</span></p>
<p>where <span class="math-container">$\epsilon_{kim}$</span> is the <code>LeviCivitaTensor</code> for <span class="math-container">$3$</span> dimensions. Consequently, we get</p>
<p><span class="math-container">\begin{align*}
\left(\text{curl}\mathbf{T}\right)_{mj}=\sum_{k=1}^{3}\sum_{i=1}^{3}\epsilon_{kim}\frac{\partial T_{ij}}{\partial x_k}.
\tag{4}
\end{align*}</span></p>
<p>Implementing <span class="math-container">$(4)$</span> in Mathematica, we obtain</p>
<pre><code>CurlTen = Table[
Sum[
LeviCivitaTensor[3][[k, i, m]]
D[Subscript[T, i, j][Subscript[x, 1], Subscript[x, 2], Subscript[x, 3]], {Subscript[x, k]}], {k, 1, 3}, {i, 1, 3}],
{m, 1, 3}, {j, 1, 3}];
MatrixForm[CurlTen]
</code></pre>
|
jose
| 10,552 |
<p>The definition used (motivated by exterior calculus) is as follows:</p>
<p>Given a rectangular array <span class="math-container">$a$</span> of depth <span class="math-container">$n$</span>, with dimensions <span class="math-container">$\{d, ..., d\}$</span> (so there are <span class="math-container">$n$</span> <span class="math-container">$d$</span>'s) and a list <span class="math-container">$x = \{x_1, ..., x_d\}$</span> of variables, then</p>
<pre><code>Curl[a, x] == (-1)^n (n+1) HodgeDual[Grad[a, x], d]
</code></pre>
<p>If <span class="math-container">$a$</span> has depth <span class="math-container">$n$</span>, then <code>Grad[a, x]</code> has depth <span class="math-container">$n+1$</span>, and therefore <code>HodgeDual[Grad[a, x], d]</code> has depth <span class="math-container">$d-(n+1)$</span>. Clearly then we need <span class="math-container">$n < d$</span>. Note that <span class="math-container">$n = 0$</span> is admitted, i.e. we can take the curl of a scalar function.</p>
<p>In the traditional case of <span class="math-container">$d=3$</span> and <span class="math-container">$n=1$</span> we have <span class="math-container">$d-(n+1)=1$</span> and that's why the curl of a vector is also a vector.</p>
<p>The <code>HodgeDual</code> operation starts by antisymmetrizing its first argument, and hence implicitly we really have something like</p>
<pre><code>Curl[a, x] == (-1)^n (n+1) HodgeDual[Symmetrize[Grad[a, x], Antisymmetric[All]], d]
</code></pre>
<p>Finally, one more comment: The definition given assumes that we work with tensors given in components in a Cartesian coordinated and orthonormal basis. If that wasn't the case we would have to insert some additional metric factors. That's handled by the third argument of <code>Curl</code> for a variety of alternative coordinate systems.</p>
|
2,004,895 |
<p>In my textbook there is a question like below:</p>
<p>If $$f:x \mapsto 2x-3,$$ then $$f^{-1}(7) = $$</p>
<p>As a multiple choice question, it allows for the answers: </p>
<p>A. $11$<br>
B. $5$<br>
C. $\frac{1}{11}$<br>
D. $9$</p>
<p>If what I think is correct and I read the equation as:</p>
<p>$$f(x)=2x-3$$
then,<br>
$$y=2x-3$$
$$x=2y-3$$
$$x+3=2y$$
$$\frac {x+3} {2} = y$$</p>
<p>therefore:</p>
<p>$$f^{-1}(7)=\frac {7+3}{2}$$
$$=5$$</p>
|
lhf
| 589 |
<p>Your mistake is here:
$$
x-3=2y
$$
It should be
$$
x+3=2y
$$</p>
|
2,004,895 |
<p>In my textbook there is a question like below:</p>
<p>If $$f:x \mapsto 2x-3,$$ then $$f^{-1}(7) = $$</p>
<p>As a multiple choice question, it allows for the answers: </p>
<p>A. $11$<br>
B. $5$<br>
C. $\frac{1}{11}$<br>
D. $9$</p>
<p>If what I think is correct and I read the equation as:</p>
<p>$$f(x)=2x-3$$
then,<br>
$$y=2x-3$$
$$x=2y-3$$
$$x+3=2y$$
$$\frac {x+3} {2} = y$$</p>
<p>therefore:</p>
<p>$$f^{-1}(7)=\frac {7+3}{2}$$
$$=5$$</p>
|
Christian Blatter
| 1,303 |
<p>Interpret
$$f:\quad x\mapsto y:=2x-3$$
as a <em>flow diagram</em>: The operation $f$ takes a variable value $x$ as input and produces a variable value $y$ as output, whereby the exact formula for computing $y$ from $x$ is given.</p>
<p>I'm writing here because in your argument you replace without hesitation $y=2x-3$ with $x=2y-3$. This seems to be a recommended procedure, but is a terrible mistake, because now both $x$ and $y$ have two different meanings in the same chain of reasoning. <strong>Keep the names of the variables as long as you are looking at $f$ and $f^{-1}$ simultaneously.</strong></p>
<p>Argue as follows instead: The inverse function $f^{-1}$ should produce the required input value $x$ for a <em>given</em> output value $y$ of $f$. We therefore have to solve the equation $y=2x-3$ for $x$ "as a function of $y$". The result of course is $x={1\over2}(y+3)$, so that the flow diagram for $f^{-1}$ looks as follows:
$$f^{-1}:\quad y\mapsto x:={1\over2}(y+3)\ .$$
In particular $f^{-1}(7)=5$.</p>
|
2,932,799 |
<p><strong>Notations.</strong></p>
<p>Let <span class="math-container">$\xi=(\xi_1,\xi_2,\xi_3,\xi_4)\in\mathbb( R\setminus\mathbb Q)^4$</span> such that <span class="math-container">$\xi_1\xi_4-\xi_2\xi_3\ne 0$</span> and the <span class="math-container">$\xi_i$</span> are linearly independent over <span class="math-container">$\mathbb Q$</span>.</p>
<p>I have the following linear form:</p>
<p><span class="math-container">$$\begin{matrix}L\colon & \mathbb R^6 & \to & \mathbb R \\ &(\eta_1,\ldots,\eta_6) & \mapsto & \eta_1(\xi_1\xi_4-\xi_2\xi_3)-\eta_2\xi_4+\eta_3\xi_3-\eta_4\xi_2+\eta_5\xi_1-\eta_6.\end{matrix}$$</span></p>
<p>We consider the norm <span class="math-container">$\Vert\cdot\Vert$</span> to be the euclidean norm on <span class="math-container">$\mathbb R^6$</span>.</p>
<p><strong>The problem.</strong></p>
<p>I am interesting in finding a constant <span class="math-container">$\gamma>0$</span>, such that if we <em>choose</em> <span class="math-container">$\xi$</span> properly, then the resulting linear form <span class="math-container">$L_\xi$</span> will verify:</p>
<p><span class="math-container">$$\forall \eta\in\mathbb Z^6\setminus\{0\},\quad L_\xi(\eta)\geqslant \frac c{\Vert\eta\Vert^\gamma}\gcd(\eta_1,\ldots,\eta_6),$$</span></p>
<p>where <span class="math-container">$c=c_\xi$</span> is a constant which depends only on <span class="math-container">$\xi$</span>.</p>
<p><strong>The conjecture.</strong></p>
<p>There are hopes for this to be true, since if we choose <span class="math-container">$\xi$</span> properly (for instance badly approximated by rationals), then for <span class="math-container">$\eta\in\mathbb Z^6\setminus\{0\}$</span>, <span class="math-container">$L_\xi(\eta)$</span> will have troubles being too small.</p>
<p>I believe that the constant <span class="math-container">$\gamma=2$</span> would work for a fine choice of <span class="math-container">$\xi$</span>.</p>
<p><strong>Additional remarks.</strong></p>
<p>This is a part of a longer proof, and if this results happens to be true, it would help me a great deal in that other proof. Unfortunately, I don't have any clue on how to start to attack this problem, so any leads would be much appreciated. </p>
<p>I do believe that <span class="math-container">$\gamma=2$</span> would work (and it would be the best), but any proof that would work for a <span class="math-container">$\gamma<4$</span> would be great.</p>
|
Lei Zhang
| 361,807 |
<p>Please see Theorem 8 in epubs.siam.org/doi/10.1137/0719030</p>
<p>The key point is r(A) is a simple eigenvalue. </p>
|
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