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263,983 |
<p>Is there a method to measure <strong>theoretic time load</strong> (hope +theoretic memory load) of an evaluation?</p>
<p>They should not change - <strong>always generate same value after pressing Enter key</strong>(=evaluation).</p>
<p>Hope they are even same in different environment(=different computers).</p>
<p>For example</p>
<pre><code>Timing[1000000!;]
</code></pre>
<p>gives sometimes <code>0.125</code>, sometimes <code>0.140625</code>, sometimes <code>0.15625</code>.</p>
<p>I want an imaginary function <code>TheoreticTiming</code></p>
<pre><code>TheoreticTiming[1000000!;]
</code></pre>
<p>which outputs always <code>0.125</code></p>
<p>There are built-in function <code>AbsoluteTiming</code> but it also changes for each evaluation, just like <code>Timing</code>.</p>
|
John Doty
| 27,989 |
<p>For the sort of computer that you need for <em>Mathematica</em>, it is not possible. Modern high-speed processors use all kinds of shortcuts to speed execution, and these shortcuts depend on the history of everything that has gone on in the computer in its recent execution trajectory, not just <em>Mathematica</em>. A simple microcontroller running a single program can have predictable timing, but the computer you use for <em>Mathematica</em> cannot.</p>
|
135,308 |
<p>Let $G$ be a compact Lie group and $M$ be a smooth manifold on which $G$ acts smoothly and effectively. Then the orbit space $M/G$ admits an Whitney stratification as follows
$$M/G=\bigsqcup_{H<G}(M_{(H)}/G).$$
Where $H$ is a closed subgroup of $G$, and $M_{(H)}$ is the set of the points in $M$ such that the isotroy groups are conjugate to $H$. A result is if $M$ admit an orientation preserved by the $G-$action then
$$codimM_{(H)}\geq2,$$ when $H$ is not trivial subgroup of $G$. Why? </p>
|
Oliver Straser
| 32,972 |
<p>Maybe I add something to Peter Michor's answer. The slice theorem says, that for $y\in M_{(H)}$ with $G_y=H$ there exists a $G$-invariant open neighbourhood $U_y$, such that
$$ U_y\cong G\times_H V_y$$
and in particular $$M_{(H)}\cap U_y= G\times_H V_y^H$$
where $V_y= T_y M/T_y (G\cdot y)$. So $M_{(H)}$ can only have codimension 1 in $M$ if $\dim V_y^H=\dim V_y-1$. This means that $H$ has to act only via one reflection on $V_y$, but this cannot be orientation preserving!</p>
|
306,744 |
<p>So if the definition of continuity is: $\forall$ $\epsilon \gt 0$ $\exists$ $\delta \gt 0:|x-t|\lt \delta \implies |f(x)-f(t)|\lt \epsilon$. However, I get confused when I think of it this way because it's first talking about the $\epsilon$ and then it talks of the $\delta$ condition. Would it be equivalent to say: $\forall$ $\delta \gt 0$ $\exists$ $\epsilon \gt0$ $:|x-t|\lt \delta \implies|f(x)-f(t)|\lt \epsilon$. I guess what I'm asking is whether there is a certain order proofs or more formal statements need to follow. I know I only changed the place where I said there is a $\delta$ but is that permissable in a "formal" way of writing?</p>
|
genepeer
| 50,955 |
<p>It's already been shown how your definition fails but I'll try to explain why it is the way it is. What the definition tries to get at is basically "You can get as close as you want to the limit." $\epsilon$ represents the any closeness to the limit you want to achieve. and the $\delta$ tells you how to achieve it. In other words, if you're less than $\delta$ distance away from $t$, you'll be less than $\epsilon$ distance from the limit. </p>
<p>Clearly, you need to know how close you want to get at first, that is why $\epsilon$ is chosen first. Hope that helps.</p>
|
103,317 |
<p>I have a 3D Plot with 3 separate functions for <code>z</code> in terms of <code>x</code> and <code>y</code>. I would like to convert this to a RegionPlot with the <code>max</code> of the three functions plotted for the variables <code>x</code> and <code>y</code>.</p>
<p>Essentially, this would be the top view of the 3D Plot, as whatever function is visible from the top is the highest and thus the maximum at that point.</p>
<p>However, I also need to add a <code>Manipulate</code> to this so it cannot merely be an image of the top view of the 3D Plot.</p>
<pre><code> Manipulate[
Plot3D[{0.1,
Piecewise[{{α - ((2 α - 1) α)/y + (
x*α^2)/(2 y ), (1 - α) + α^2/(2 y) >=
0.35}}],
Piecewise[{{α + (1 - α)^2/(
2 y), (1 - α) + (((3 α -
1) - ((Abs[1 - x]) (1 - α))) (1 - α))/(
2 y) >= 0.35}}],
Piecewise[{{α - ((2 α - 1) α)/((2 -
a) 2) + ((3 - 2 a - Abs[x - a] ) α^2)/(
2 y*(2 - a)^2), (1 - α) + α^2/(2 y (2 - a)) >=
0.35}}]} /. {a -> ((1 + z)/2), x -> 1} //
Evaluate, {α, 0, 1}, {z, 0, 1},
AxesLabel -> Automatic,
PlotRange -> Automatic,
PlotStyle -> {Black, Green, Orange, Blue}], {{y, 2,
Gamma}, 1, 5}]
</code></pre>
<p><a href="https://i.stack.imgur.com/jgFLS.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/jgFLS.png" alt=""></a></p>
<p>I would like a 2D Plot from this top view</p>
|
Jason B.
| 9,490 |
<p>Here are three example functions,</p>
<pre><code>f1 = Exp[-x^2 - y^2];
f2 = .5 Sin[x - y];
f3 = .002 x^2 + .07 y^2;
</code></pre>
<p>And here is a top-down view of the 3D plot,</p>
<pre><code>Plot3D[{f1, f2, f3}, {x, -6, 6}, {y, -4, 4}, PlotPoints -> 100,
ViewCenter -> {0.5, 0.5, 0.5}, ViewPoint -> {0, 0, 4}]
</code></pre>
<p><a href="https://i.stack.imgur.com/hhftR.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/hhftR.png" alt="enter image description here"></a></p>
<p>And here is the same view, using <code>RegionPlot</code></p>
<pre><code>RegionPlot[{
f1 > f2 && f1 > f3,
f2 > f1 && f2 > f3,
f3 > f1 && f3 > f2},
{x, -6, 6}, {y, -4, 4},
BoundaryStyle -> Dashed]
</code></pre>
<p><a href="https://i.stack.imgur.com/3qG4o.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/3qG4o.png" alt="enter image description here"></a></p>
<p>Or, using your example,</p>
<pre><code>Manipulate[
Module[{flist, f1, f2, f3, f4},
flist = {f1, f2, f3, f4} =
Evaluate[{0.1,
Piecewise[{{α - ((2 α - 1) α)/
y + (x*α^2)/(2 y), (1 - α) + α^2/(2 \
y) >= 0.35}}],
Piecewise[{{α + (1 - α)^2/(2 y), (1 - α) +
(((3 α -
1) - ((Abs[
1 - x]) (1 - α))) (1 - α))/(2 y) >= \
0.35}}],
Piecewise[{{α - ((2 α - 1) α)/((2 -
a) 2) + ((3 - 2 a -
Abs[x - a]) α^2)/(2 y*(2 -
a)^2), (1 - α) + α^2/(2 y (2 - a)) >=
0.35}}]} /. {a -> ((1 + z)/2), x -> 1}];
RegionPlot[{
f1 == Max@flist,
f2 == Max@flist,
f3 == Max@flist,
f4 == Max@flist
}, {α, 0, 1}, {z, 0, 1},
BoundaryStyle -> None
, PlotPoints -> 50, PlotStyle -> {Black, Green, Orange, Blue}]
]
, {{y, 2, Gamma}, 1, 5}]
</code></pre>
<p>Using the <code>PlotPoints</code> option severely slows it down, but makes the plot smoother.</p>
<p><a href="https://i.stack.imgur.com/M0Slf.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/M0Slf.png" alt="enter image description here"></a></p>
|
2,491,881 |
<p>I'm trying to factorise $(4 + 3i)z^2 + 26iz + (-4+3i)$.</p>
<p>I tried to use the quadratic formula to factorise $(4 + 3i)z^2 + 26iz + (-4+3i)$, where $b = 26i$, $a = (4 + 3i)$, $c = (-4+3i)$. This gives us the roots $z = \dfrac{-4i - 12}{25}$ and $z =-4i - 3$.</p>
<p>But $\left( z + \dfrac{4i + 12}{25} \right) \left( z + 4i + 3 \right) \not= (4 + 3i)z^2 + 26iz + (-4+3i)$. </p>
<p>So I did what we do when using the quadratic formula with real numbers and have that $a \not= 0, 1$: $(4 + 3i)\left( z + \dfrac{4i + 12}{25} \right) \left( z + 4i + 3 \right)$. But, as far as I can tell, $(4 + 3i)\left( z + \dfrac{4i + 12}{25} \right) \left( z + 4i + 3 \right) \not= (4 + 3i)z^2 + 26iz + (-4+3i)$.</p>
<p>So what is the correct way to go about this?</p>
<p>I would greatly appreciate it if people could please take the time to explain this.</p>
|
John Griffin
| 466,397 |
<p>Your proof is fine with the slight modification I noted in the comments.</p>
<p>You need to show that $X\cup Y$ is compact. This means that you need to show that every sequence in $X\cup Y$ has a convergent subsequence, so you fix $(x_n)$ in $X\cup Y$. The goal is to now find a subsequence of $(x_n)$ which converges.
As you noted, there is a subsequence of $(x_n)$ which lies entirely in one of the two spaces. Let's assume without loss of generality that $(x_{n_k})$ is a subsequence of $(x_n)$ in $X$. Then, by the compactness of $X$, the sequence $(x_{n_k})$ has a convergent subsequence in $X$. Hence there is a subsequence $(x_{n_{k_j}})$ of $(x_{n_k})$ such that $(x_{n_{k_j}})$ converges to an element in $X$. Since $(x_{n_{k_j}})$ is a subsequence of our original sequence $(x_n)$ which converges to a point in $X\subseteq X\cup Y$, then we are done. Therefore $X\cup Y$ is compact.</p>
<hr>
<p>It was important to add this second subsequence because it isn't true that $(x_{n_k})$ will necessarily converge to something. Compactness only tells us that a subsequence will converge.</p>
<p>Also, in regards to your question about beginning with $x_n\to x$, the answer is no. The definition of compactness says that every sequence has a convergent subsequence. We can't assume our sequence already converges.</p>
|
2,971,980 |
<p>Show that if <span class="math-container">$0<b<1$</span> it follows that
<span class="math-container">$$\lim_{n\to\infty}b^n=0$$</span>
I have no idea how to express <span class="math-container">$N$</span> in terms of <span class="math-container">$\varepsilon$</span>. I tried using logarithms but I don't see how to find <span class="math-container">$N$</span> from this.</p>
|
Peter Szilas
| 408,605 |
<p>Set <span class="math-container">$b=\dfrac{1}{1+x}$</span> , <span class="math-container">$x >0.$</span></p>
<p>Note: <span class="math-container">$ (1+x)^n \gt 1+nx.$</span></p>
<p><span class="math-container">$0<b^n =\dfrac{1}{(1+x)^n} \lt \dfrac{1}{1+nx}\lt$</span></p>
<p><span class="math-container">$(1/x)\dfrac{1}{n}.$</span></p>
<p>Take the limit.</p>
|
2,971,980 |
<p>Show that if <span class="math-container">$0<b<1$</span> it follows that
<span class="math-container">$$\lim_{n\to\infty}b^n=0$$</span>
I have no idea how to express <span class="math-container">$N$</span> in terms of <span class="math-container">$\varepsilon$</span>. I tried using logarithms but I don't see how to find <span class="math-container">$N$</span> from this.</p>
|
with-forest
| 519,097 |
<p>From <span class="math-container">$0<b<1$</span>, we get</p>
<p><span class="math-container">$$0< (n+1)b^n < 1+b+\cdots+b^n = \frac{1-b^{n+1}}{1-b} < \frac{1}{1-b}.$$</span> </p>
<p>Hence, </p>
<p><span class="math-container">$$0< b^n < \frac{1}{n+1} \frac{1}{1-b}.$$</span></p>
<p>And as <span class="math-container">$n$</span> goes to infinity, the limit of <span class="math-container">$b^n$</span> approaches to zero. </p>
|
1,104,163 |
<p>I have come up with the equation in the form $${{dy}\over dx} = axe^{by}$$, where a and b are arbitrary real numbers, for a project I am working on. I want to be able to find its integral and differentiation if possible. Does anyone know of a possible solution for $y$ and/or ${d^2 y}\over {dx^2}$?</p>
|
Dave L. Renfro
| 13,130 |
<p>This is a separable differential equation, as is the case for any equation that can be put into the form $\frac{dy}{dx} = f(x)g(y)$ for some functions $f$ and $g.$ In your case, divide both sides by $e^{by}$ and multiply both sides by $dx,$ then integrate.</p>
|
1,104,163 |
<p>I have come up with the equation in the form $${{dy}\over dx} = axe^{by}$$, where a and b are arbitrary real numbers, for a project I am working on. I want to be able to find its integral and differentiation if possible. Does anyone know of a possible solution for $y$ and/or ${d^2 y}\over {dx^2}$?</p>
|
abel
| 9,252 |
<p>separate the variables like $$ e^{-by}dy = axdx$$ on integrating, you get $$ -\dfrac{1}{b}e^{-by} = \dfrac{1}{2}ax^2 - \dfrac{C}{2b}$$ which can be solved for $y = \dfrac{1}{b}\ln\left( \dfrac{2}{C - ab x^2}\right)$</p>
<p>for $\dfrac{d^2y}{dx^2},$ differencing $$e^{-by}\dfrac{dy}{dx} = ax$$ gives you
$$e^{-by}\dfrac{d^2y}{dx^2} -be^{-by}(\dfrac{dy}{dx})^2= a$$ on further simplification you get
$$\dfrac{d^2y}{dx^2} = axe^{by}(1 + a e^{by})$$</p>
|
3,220,135 |
<p>How many integer numbers, <span class="math-container">$x$</span>, verify that the following</p>
<p><span class="math-container">\begin{equation*}
\frac{x^3+2x^2+9}{x^2+4x+5}
\end{equation*}</span></p>
<p>is an integer?</p>
<p>I managed to do:</p>
<p><span class="math-container">\begin{equation*}
\frac{x^3+2x^2+9}{x^2+4x+5} = x-2 + \frac{3x+19}{x^2+4x+5}
\end{equation*}</span></p>
<p>but I cannot go forward.</p>
|
zwim
| 399,263 |
<p><em>Here is another method that works well when Diophantine quadratics <span class="math-container">$ax^2+bx+c=0$</span> are involved, it consists in saying that since a solution should exists then <span class="math-container">$\exists \delta\in\mathbb Z\mid b^2-4ac=\delta^2$</span>.</em></p>
<p><br></p>
<p>In this problem we'd like <span class="math-container">$\quad\dfrac{3x+19}{x^2+4x+5}=n\quad$</span> to be an integer.</p>
<p>Thus applying the method to <span class="math-container">$$nx^2+(4n-3)x+(5n-19)=0$$</span></p>
<p><br></p>
<p>gives <span class="math-container">$\quad(4n-3)^2-4n(5n-19)=p^2\iff 4n^2-52n+(p^2-9)=0$</span></p>
<p><br>
Which by itself should have solutions in <span class="math-container">$n$</span> : <span class="math-container">$\quad(-52)^2-4(4)(p^2-9)=q^2\iff 2848-16p^2=q^2$</span></p>
<p>We can divide by <span class="math-container">$16$</span> (setting <span class="math-container">$q=4r$</span>) to get <span class="math-container">$$p^2+r^2=178$$</span></p>
<p>This has solutions <span class="math-container">$(\pm 3,\pm 13)$</span>.</p>
<p><br></p>
<ul>
<li>case 1 : <span class="math-container">$p^2=3^2=9$</span></li>
</ul>
<p>Leads to <span class="math-container">$4n^2-52n=0\iff 4n(n-13)=0\iff n=0,\ 13$</span></p>
<p><span class="math-container">$n=0\implies 3x+19=0$</span> impossible</p>
<p><span class="math-container">$n=13\implies 13x^2+49x+46=(x+2)(13x+23)=0$</span> and we verify that <span class="math-container">$x=-2$</span> is solution.</p>
<p><br></p>
<ul>
<li>case 2: <span class="math-container">$p^2=13^2=169$</span></li>
</ul>
<p>Leads to <span class="math-container">$4n^2-52n+160=4(n-5)(n-8)=0\iff n=5,\ 8$</span></p>
<p><span class="math-container">$n=5\implies 5x^2+17x+6=(x+3)(5x+2)=0$</span> and we verify that <span class="math-container">$x=-3$</span> is solution.</p>
<p><span class="math-container">$n=8\implies 8x^2+29x+21=(x+1)(8x+21)=0$</span> and we verify that <span class="math-container">$x=-1$</span> is solution.</p>
<p><br></p>
<p>Finally we have found all solutions <span class="math-container">$x\in\{-3,-2,-1\}$</span></p>
|
140,615 |
<p>A rectangular page is to have a printed area of 62 square inches. If the border is to be 1 inch wide on top and bottom and only 1/2 inch wide on each side find the dimensions of the page that will use the least amount of paper</p>
<p>Can someone explain how to do this?</p>
<p>I started with:</p>
<p>$$A = (x + 2)(y + 1) $$</p>
<p>Then I isolate y and come up with my new equation:</p>
<p>$$A = (x+2)\left(\frac{62}{x + 2}{-1}\right)$$ </p>
<p>Then I think my next step is to create my derivative, but wouldn't it come out to -1?</p>
<p>Anyways, I would appreciate if someone could give me a nudge in the right direction.</p>
<p><strong>EDIT</strong> </p>
<p>How does this look for a derivative?</p>
<p>$$A = \left(\frac{x^2-124}{x^2}\right)$$ </p>
<p>Then to solve:
$$ {x} = 11.1 $$ </p>
<p>$$ y = 98 / 11.1 $$</p>
<p>Does that seem about right?</p>
<p>If not, the only thing I would have left is setting it to 0 and solving.</p>
|
David Mitra
| 18,986 |
<p>If the dimensions of the printed area are $x$ and $y$, where $y$ is the dimension with the $1/2$ inch borders (the "width"),
then the printed area is $$\tag{1}62= x y.$$ You want to minimize the area of the entire page, which is
$$\tag{2}A=(x+2)(y+1).$$
We want $A$ expressed in terms of one variable only; so solve $(1)$ for $y$
$$\tag{3}
y={62\over x }
$$
and substitute into $(2)$, giving
$$\tag{4}
A(x) = (x+2)\cdot\textstyle\bigl( {62\over x }+1\bigr) .
$$
Now you want to minimize $A(x)$ over $x\in(0,\infty)$.
Do this using the normal derivative analysis (remember to examine what happens when $x$ is close to $0$ and when $x$ is big).</p>
<p>Once you've found the value of $x$ that minimizes $(4)$, remember to state the answer to the question explicitly; for example "the dimensions of the paper are $x+2$ inches top to bottom and $y+1$ inches wide" (you can use $(3)$ to find the value of $y$ once you have $x$.</p>
|
3,343,550 |
<p>Show that if <span class="math-container">$(a,15)=1$</span>, then <span class="math-container">$a^4\equiv1 \mod 15$</span>, so that we do not have primitive roots of <span class="math-container">$15$</span>
Please help me with this problem.</p>
|
Community
| -1 |
<p>It is a good idea which you will be able to use in similar questions to consider mod 3 and mod 5 separately.
You will then find that Fermat's Little Theorem will save you a lot of work.</p>
|
1,526,882 |
<p>Let $$r(x,y)=\begin{cases} y &\mbox{ if } 0\leq y\leq x \\ x &\mbox{ if } x\leq y\leq 1\end{cases}$$</p>
<p>Show that $v(x)=\int_0^1r(x,y)f(y) \ dy$ satisfies $-v''(x)=f(x)$, where $0\leq x \leq 1$ and $f$ is continuous. </p>
<p>How can I take the second derivative of this? When I try to do it I feel like differentiating under the integral sign will yield $v''(r)=0$ due to $r_{xx}(x,y)=0$.</p>
|
Samrat Mukhopadhyay
| 83,973 |
<p>Note that $$v(x)=\int_{0}^ x yf(y) dy+\int_{x}^1 xf(y) dy\\\implies v'(x)=xf(x)+\int_{x}^1 f(y) dy-xf(x)\\\implies v''(x)=-f(x)$$ The differentiation results follow from <a href="https://en.wikipedia.org/wiki/Fundamental_theorem_of_calculus" rel="nofollow">fundamental theorem of calculus</a>.</p>
|
2,043,690 |
<p>I'm in the process of studying for an exam and this just popped into my head. Sorry if it's a dumb question</p>
|
Joffan
| 206,402 |
<p>Indeed they are the same, although I'd write them either without brackets or with brackets around the whole expression. </p>
<p>I really wanted to add that that using the $(-1 \bmod 13)$ representation can be very useful, especially when try to find the result of a multiplication. For a simple example, $12\times 7 \bmod 13$ is easily converted to $12\times 7 \equiv -1\times 7 \equiv -7 \equiv 4 \bmod 13$</p>
|
1,258,623 |
<p>Let $a, b, c, d, e$ be distinct positive integers such that $a^4 + b^4 = c^4 + d^4 = e^5$. Show that $ac + bd$ is a composite number.</p>
|
Einar Rødland
| 37,974 |
<p>Assume that $p=ac+bd$ is prime. Without loss of generality, we may assume that $a>c,d$ from which follows that $b<c,d$. Also, $a,b,c,d<p$, and so must all be relatively prime with $p$.</p>
<p>Computing modulo $p$, we have $ac \equiv -bd$ which leads to $(ac)^4\equiv(bd)^4$. Exploiting $a^4+b^4=c^4+d^4$ we get
$$
(a^4+b^4)c^4=(ac)^4+(bc)^4\equiv(db)^4+(bc)^4=(c^4+d^4)b^4=(a^4+b^4)b^4
$$
which means that either $p$ divides $a^4+b^4=c^4+d^4$, or $b^4\equiv c^4$.</p>
<p>We can exclude the alternative $p|a^4+b^4$ since that would make $p|e$, which in turn would require $e\ge p=ac+bd>a+b$ which leads to $e^5>a^4+b^4$. Note that this is the only use for the equality with $e^5$.</p>
<p>So now we know that $b^4\equiv c^4$, which means that
$$
p|c^4-b^4=(c^2+b^2)(c+b)(c-b).
$$
Again, $p>c+b$, so we must have $p|c^2+b^2$. However,
$c^2+b^2<ac+bd=p$ as $c<a$ and $b<d$, which makes $p|c^2+b^2$ impossible.</p>
<p>Thus, no such $a,b,c,d$ exist for which $p$ is prime.</p>
<p>This was written late at night, so I hope I didn't make any glaring mistakes.</p>
|
2,893,388 |
<p>My textbook is confusing me a little. Here is a worked example from my textbook:</p>
<blockquote>
<p>Line <span class="math-container">$l$</span> has the equation <span class="math-container">$\begin{pmatrix}3\\ -1\\ 0\end{pmatrix}+\lambda \begin{pmatrix}1\\ -1\\ 1\end{pmatrix}$</span> and point <span class="math-container">$A$</span> has co-ordinates <span class="math-container">$(3, 9, -2)$</span>.</p>
<p>Find the coordinates of point <span class="math-container">$B$</span> on <span class="math-container">$l$</span> so that <span class="math-container">$AB$</span> is perpendicular to <span class="math-container">$l$</span>.</p>
<p><span class="math-container">$\vec{AB\cdot }\begin{pmatrix}1\\ -1\\ 1\end{pmatrix}=0$</span></p>
<p><span class="math-container">$\vec{OB}=r=\begin{pmatrix}3+\lambda \\ -1-\lambda \\ \lambda \end{pmatrix}$</span></p>
<p><span class="math-container">$\vec{AB}=\begin{pmatrix}\lambda \\ -10-\lambda \\ \lambda +2\end{pmatrix}$</span></p>
<p><span class="math-container">$\begin{pmatrix}\lambda \\ -10-\lambda \\ \lambda +2\end{pmatrix}\cdot \begin{pmatrix}1\\ -1\\ 1\end{pmatrix}=0$</span></p>
<p><span class="math-container">$3\lambda= -12, \lambda = -4$</span></p>
<p>Coordinates of <span class="math-container">$B$</span>: <span class="math-container">$(-1, 3, -4)$</span></p>
</blockquote>
<p>The thing I don't understand is why they found the dot product of the line AB and the direction vector of line l. My textbook does mention that to check whether two vectors are perpendicular, <span class="math-container">$a\cdot b = 0 $</span> and for lines, the dot product of their direction vectors = 0. So why did they mix both here? Didn't they use the entire line <span class="math-container">$AB$</span> and then just the direction vector of line l? Or am I missing something as usual?</p>
|
Henno Brandsma
| 4,280 |
<p>If a space $X$ is <em>hereditarily Lindelöf</em> and $\mathcal{B}$ is any base for the space $X$, then any open set $O$ can be written as a countable union of elements from $\mathcal{B}$.
A space can be hereditarily Lindelöf without there being a countable base for $X$, an example of this is the Sorgenfrey line (or the lower limit topology on $\mathbb{R}$, as Munkres calls it).</p>
|
1,507,519 |
<p>How do I solve this question?</p>
<p><a href="https://i.stack.imgur.com/drfiA.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/drfiA.jpg" alt="Question"></a></p>
<p>I tried using the quadratic formula on the question equation and got</p>
<p>$x_1 = 0.25 +1.089724..i = \ln r$</p>
<p>$x_2 = 0.25 -1.089724..i = \ln s$</p>
<p>I know $\ln x = \log_ex$, but how do I complete the questions with the imaginary numbers?</p>
|
Mythomorphic
| 152,277 |
<p>Actually the key of solving this question is to make use of the <strong>RELATION OF ROOTS</strong>. </p>
<p><strong>Recall:</strong> </p>
<p>If $\forall a,b,c\in\Bbb{R}$, given quadratic of $x$, $ax^2+bx+c=0$, </p>
<p>which can be converted to the form of $x^2+\frac bax+\frac ca=0$, </p>
<p>we say that $-\frac ba$ is the <strong>sum of roots</strong> and $\frac ca$ is the <strong>product of roots</strong>.</p>
<hr>
<p>In this case, $\ln r$ and $\ln s$ are roots, so you can make use of these relations.</p>
|
3,152,532 |
<p>What is meaning of <span class="math-container">$f^2(x)$</span> ? There seems to be confusion in its interpretation.</p>
<p>Is <span class="math-container">$f^2(x)$</span> same as <span class="math-container">$(f(x))^2$</span> or <span class="math-container">$f\circ f$</span>?</p>
|
Community
| -1 |
<p>It can be both, but unless you are in a context where function iteration makes sense, it is more likely to be the square.</p>
<hr>
<p>The confusion is a little worse with the exponent <span class="math-container">$-1$</span>: <span class="math-container">$\sin^{-1}(x)$</span> is more likely to be the arc sine than the cosecant. The functional inverse is not so exotic.</p>
|
4,369,480 |
<p>In's commonly said that in VAE, we use reparameterization trick because "we can't backpropagate through stochastic node"
<a href="https://i.stack.imgur.com/ED72y.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ED72y.png" alt="enter image description here" /></a></p>
<p>It makes sense from the picture, but I found it hard to understand exactly what it means and why. Let's say X ~ N(u, 1).</p>
<p>And we want to compute
<span class="math-container">$$\frac{d X}{d u}$$</span> which is not possible because the sampling operation is non-differentiable. That is, we don't know how changing u a little bit will affect how we got the sample X.</p>
<p>However, say in the MLE for Gaussian. We are trying to estimate the following quantity:</p>
<p><span class="math-container">$$\sum_{i=1}^N \log p(X_i;u)$$</span> for which the derivative <span class="math-container">$$\frac{d \log p(X_i ; u)}{d u}$$</span> can be easily calculated. My confusion comes from the fact that <span class="math-container">$$\frac{d \log p(X_i ; u)}{d u} = \frac{d \log p(X_i ; u)}{d X_i} \frac{d X_i}{d u}$$</span> by the chain rule. If we can't compute <span class="math-container">$\frac{d X_i}{d u}$</span>, why can we compute <span class="math-container">$\frac{d \log p(X_i ; u)}{d u}$</span> ?</p>
|
etal
| 389,777 |
<p>I think your maximum likelihood equation is not correct. In particular, if I understand your setting correctly <span class="math-container">$X_i$</span> is your data, which does not depend on <span class="math-container">$u$</span>. Their log-likelihood does depend on <span class="math-container">$u$</span> since for a constant variance it's essentially the MSE <span class="math-container">$\sum_i (X_i-u)^2$</span> multiplied and added with terms that don't depend on <span class="math-container">$u$</span>. Then taking <span class="math-container">$\sum_i \frac{d\log p(X_i;u)}{du} = O(1)\sum_i (X_i-u) + O(1)$</span>, where <span class="math-container">$O(1)$</span> are terms that don't depend on <span class="math-container">$X_i$</span> nor <span class="math-container">$u$</span>.</p>
<p>Notice the final expression depends on <span class="math-container">$X_i$</span> but not because we differentiated through it. This is because <span class="math-container">$dX_i/du$</span> because is zero, since it's the data and does not depend on <span class="math-container">$u$</span>.</p>
|
2,985,256 |
<blockquote>
<p>Let there are three points <span class="math-container">$(2,5,-3),(5,3,-3),(-2,-3,5)$</span> through which a plane passes. What is the equation of the plane in Cartesian form?</p>
</blockquote>
<p>I know how to find it in using vector form by computing the cross product to get the normal vector and passing through any one of the given points. But I want to do it a bit differently. </p>
<p>We know, the equation of any plane passing through the first point is <span class="math-container">$$a(x-2)+b(y-5)+c(z+3)=0$$</span></p>
<p>This equation must satisfy the other two points. However, this given me two equations with three unknowns <span class="math-container">$a,b,c$</span>. So can I not solve by this method?</p>
|
sree_ranjani
| 492,385 |
<p>After using Cramer's rule, <span class="math-container">$a=-2,b=-3,c=-4$</span>.</p>
|
3,291,190 |
<p>Compute</p>
<p><span class="math-container">$$\text{const} \left( \frac{1}{(1-z_1z_2)(1-z_1^2z_2)(1-z_1)(1-z_2)z_1^{3t}z_2^{2t}} \right) $$</span> </p>
<p>The answer should be a polynomial in <span class="math-container">$t$</span>. This is exercise 2.38 from Beck's Computing the Continuous Discretely, and the correct answer is </p>
<p><span class="math-container">$$\frac{7}{4} t^2 + \frac{5}{2}t + \frac{7+(-1)^t}{8} $$</span></p>
<p>I'm honestly at a loss here. Can someone please explain to me how to compute the constant term here? What about a situation where we have a greater number of variables and products of functions? </p>
<p>This is closely related to an earlier post of mine at <a href="https://math.stackexchange.com/questions/3290516/finding-constant-term-in-product-of-series">Finding Constant Term in Product of Series</a>, which still doesn't have an answer. I would greatly appreciate help on either or both of these problems, and would greatly appreciate a general procedure for doing this.
Thank you so much. </p>
|
amd
| 265,466 |
<p>I don’t know of any real short cuts for attacking this. I think you just need to bash it out one variable at a time. </p>
<p>Starting from <a href="https://math.stackexchange.com/a/3291206/265466">gt6989b’s answer</a>, we’re trying to find <span class="math-container">$[x^{3t}y^{2t}] F(x,y)$</span> with <span class="math-container">$g(x,y)$</span> your generating function (replacing <span class="math-container">$z_1$</span> and <span class="math-container">$z_2$</span> with <span class="math-container">$x$</span> and <span class="math-container">$y$</span> to reduce visual clutter) and <span class="math-container">$F(x,y)=x^{3t}y^{2t}g(x,y)$</span>. Starting with <span class="math-container">$y$</span> because it looks like it’ll produce fewer terms, compute the partial fraction expansion with respect to <span class="math-container">$y$</span>, treating <span class="math-container">$x$</span> as a constant: <span class="math-container">$$\frac1{(1-xy)(1-x^2y)(1-x)(1-y)} = {1 \over (1-x)^3(1+x)(1-y)} - {x \over (1-x)^3(1-xy)} + {x^3 \over (1-x)^3(1+x)(1-x^2y)}$$</span> and extract the required coefficient of <span class="math-container">$y$</span>: <span class="math-container">$$\begin{align} [x^{3t}y^{2t}] F(x,y) &= [x^{3t}]{1 \over (1-x)^3(1+x)} - [x^{3t}]{x^{2t+1} \over (1-x)^3} + [x^{3t}]{x^{4t+3} \over (1-x)^3(1+x)} \\ &= [x^{3t}]{1 \over (1-x)^3(1+x)} - [x^{t-1}]{1 \over (1-x)^3} + [x^{-(t+3)}]{1 \over (1-x)^3(1+x)}. \end{align}$$</span> </p>
<p>Having eliminated <span class="math-container">$y$</span>, we can now proceed with <span class="math-container">$x$</span> in the usual manner. Since <span class="math-container">$t\gt0$</span> we can discard the last term. The second term can be evaluated immediately: it’s equal to <span class="math-container">$-\binom{-3}{t-1} = -\binom{2+t-1}2 = -\frac12t(t+1)$</span>. For the first term, compute the partial fraction expansion and extract coefficients. The resulting sum simplifies to <span class="math-container">$\frac74t^2+\frac52t+\frac{7+(-1)^t}8$</span> as required.</p>
|
4,316,771 |
<p>I'm asked to compute <span class="math-container">$$\sum_{k=-3}^{10} 2k^4$$</span>
I looked up for<a href="https://en.m.wikipedia.org/wiki/Bernoulli_number" rel="nofollow noreferrer">Bernoulli Number</a> on Wikipedia and found a general formula for that. But my teacher has asked me to evaluate this by breaking the summation into the summations whose sum we already know i.e. say <span class="math-container">$\sum k^2$</span>, arithmetic series, geometric series etc.</p>
<p>Can it be solved in this way, can anyone please tell me or give me some hint? It's really urgent and I can't do anything about it.</p>
<p><strong>Edit :</strong> since there is a confusion, I want to clear it by saying that I'm not allowed to use the formula for <span class="math-container">$$\sum_{k=1}^n k^4$$</span>. So I need an answer that doesn't include it.</p>
|
miracle173
| 11,206 |
<p>Assume we have no electronic or mechanical device that assists us in doing the calculations and we are short of time and paper so that we want to avoid excessive calculations by pencil and paper and our mental arithmetic skills are rather limited than I would propose the following way to calculate the value:</p>
<p><span class="math-container">$$(k-1)^4+k^4+(k+1)^4=3k^4+2 {4\choose 2}k^2+2=3k^4+12k^2+2=:f(k)$$</span>
Then</p>
<p><span class="math-container">$$\sum_{k=-3}^{10} k^4\\=2\sum_{k=1}^{3}k^4+\sum_{k=4}^{6}k^4+\sum_{k=7}^{9}k^4+10^4\\
=2f(2)+f(5)+f(8)+10000$$</span>
For the following calculation note that <span class="math-container">$5^4=25^2$</span> can be easily calculated, because
of the well known trick that to square a number with the decimal representation <span class="math-container">$(a5)_{10}$</span> we calculate <span class="math-container">$a(a+1)$</span> and append the digits <span class="math-container">$25$</span> on the resulting decimal number. So to calculate <span class="math-container">$25^2$</span> we calculate <span class="math-container">$2\cdot3=6$</span> and append <span class="math-container">$25$</span> to <span class="math-container">$6$</span> and get <span class="math-container">$625$</span>. We also need <span class="math-container">$2^{12}$</span> which is also simple because <span class="math-container">$2^{12}=2^2\cdot 2^{10}=4\cdot1024$</span> and <span class="math-container">$1024$</span> should be be known at least if one has attended a course in computer science.<br />
<span class="math-container">$$
=2(3\cdot16+12\cdot4+2)+(3\cdot625+12\cdot25+2)+(3\cdot4096+12\cdot64+2)\\
=3(32+625+4096)+12(8+25+64)+8+10000\\
=(4653\cdot3+12\cdot 97+8+10000)\\
=13959+12(100-3)+10008\\
=23967+1200-36\\
=25131$$</span></p>
<p>To get the requested number we have to multiply this by <span class="math-container">$2$</span>, but this is left to the reader.</p>
|
740,294 |
<p>Let $y_n$ satisfy the nonlinear difference equation:</p>
<p>$$(n+1)y_n=(2n)y_{n-1}+n.$$</p>
<p>Let $u_n=(n+1) y_n$. Show that</p>
<p>$$u_n= 2u_{n-1}+n.$$</p>
<p>Solve the linear difference equation for $u_n$. Hence find $y_n$ subject to the initial condition $y_0=4$.</p>
<p>I have showed that $u_n=2u_{n-1}+n$, but I don't know how to do the next step, can anyone help me with this please?</p>
|
Yiyuan Lee
| 104,919 |
<p>Firstly, divide throughout by $2^n$ to get:</p>
<p>$$\frac{u_n}{2^n} = \frac{u_{n-1}}{2^{n-1}} + \frac{n}{2^n}$$</p>
<p>Rearrange to get</p>
<p>$$\frac{u_n}{2^n} - \frac{u_{n-1}}{2^{n-1}} = \frac{n}{2^n}$$</p>
<p>Sum from $1$ to $n$:</p>
<p>$$\sum_{i=1}^n \frac{u_i}{2^i} - \frac{u_{i-1}}{2^{i-1}} = \sum_{i=1}^n\frac{i}{2^i}$$
$$\frac{u_n}{2^n} - \frac{u_0}{2^0} = \left(\frac{1}{2}\right)^1 + 2\cdot\left(\frac{1}{2}\right)^2 + 3\cdot\left(\frac{1}{2}\right)^3 + \dots + n\cdot\left(\frac{1}{2}\right)^n$$</p>
<p>Now, recall that</p>
<p>$$(1 - x^n)(1-x)^{-1} = (1 + x + x^2 + \cdots + x^{n-1})$$</p>
<p>Differentiate both sides to get:</p>
<p>$$(1-x^{n+1})(1-x)^{-2} -(n+1)x^{n}(1-x)^{-1} = 1 + 2x + 3x^2 + \cdots + nx^{n-1}$$</p>
<p>I'll leave the rest to you :)</p>
|
747,816 |
<p>1) Can a non-square matrix have eigenvalues? Why?</p>
<p>2) True or false: If the characteristic polynomial of a matrix A is p($\lambda$)=$\lambda$^2+1, then A is invertible.
Thank you!</p>
|
David
| 119,775 |
<p><strong>Hint</strong>. (1) Try it! For example, can you solve
$$\pmatrix{1&2&3\cr4&5&6\cr}\pmatrix{v_1\cr v_2\cr v_3\cr}
=\lambda\pmatrix{v_1\cr v_2\cr v_3\cr}\ ?$$</p>
<p>(2) You should know</p>
<ul>
<li>the connection between whether or not $A$ is invertible and $\det(A)$;</li>
<li>the connection between $\det(A)$ and the eigenvalues of $A$;</li>
<li>the connection between the eigenvalues of $A$ and its characteristic polynomial.</li>
</ul>
|
3,390,407 |
<p>If <span class="math-container">$t>0,t^2, t+\frac{1}{t},t+t^2,\frac{1}{t}+\frac{1}{t^2}$</span> are all irrational number,
<span class="math-container">$$a_n=n+\left \lfloor \frac{n}{t} \right \rfloor+\left \lfloor \frac{n}{t^2} \right \rfloor,\\
b_n=n+\left \lfloor \frac{n}{t} \right \rfloor +\left \lfloor nt \right \rfloor,\\
c_n=n+\left \lfloor nt \right \rfloor+\left \lfloor nt^2 \right \rfloor, $$</span>
then every positive integer appear exactly once. In other words, the sequences <span class="math-container">$a_1,b_1,c_1,a_2,b_2,c_2,\cdots$</span> together contain all the positive integers without repetition.
I have checked every integer from <span class="math-container">$1$</span> to <span class="math-container">$10^6$</span> for <span class="math-container">$t=2^\frac{1}{4}$</span>:
<span class="math-container">$$a_n=1, 4, 7, 9, 12, 15, 16, 19, 22, 25, 27, 30, 32, 34, 37, 40, 43, 45, 47, 50,\dots \\
b_n=2, 5, 8, 11, 14, 18, 20, 23, 26, 29, 33, 36, 38, 41, 44, 48, 51, 54, 56, 59,\dots \\
c_n=3, 6, 10, 13, 17, 21, 24, 28, 31, 35, 39, 42, 46, 49, 53, 57, 61, 64, 67, 71,\dots $$</span></p>
<p>PS: This is a special case of following statement：</p>
<blockquote>
<p>If <span class="math-container">$t_1,t_2,\cdots,t_k>0$</span>,and <span class="math-container">$\forall i \not =j,\frac{t_j}{t_i}$</span> is irrational,
<span class="math-container">$$a_i(n)=\sum_{j=1}^k{\left \lfloor \frac{t_j}{t_i}n \right \rfloor},i=1,2,\cdots,k,$$</span></p>
<p>then every positive integer appear exactly once in <span class="math-container">$a_1(n),\cdots,a_k(n)$</span>.</p>
</blockquote>
|
user90369
| 332,823 |
<p><em>Often it's crucial how a question is formulated to understand it well. My following considerations involve a functional view. It's a proof for the generalisation.</em></p>
<p><span class="math-container">$k\in\mathbb{N}~$</span> fixed.</p>
<p><span class="math-container">$t_i\in\mathbb{R}^+~$</span> and <span class="math-container">$~\displaystyle t_i,\frac{t_j}{t_i}|_{j\neq i}~$</span> are irrational for all <span class="math-container">$~i,j\in\{1,2,...,k\}~\quad\quad (1)$</span></p>
<p>Let <span class="math-container">$~\displaystyle f(x):=\sum\limits_{j=1}^k\lfloor t_j\,x\rfloor~$</span> with <span class="math-container">$~x\in\mathbb{R}_0^+~$</span> . <span class="math-container">$~~~$</span> <em>Note:</em> <span class="math-container">$~\displaystyle a_i(n) \equiv f\left(\frac{n}{t_i}\right)$</span></p>
<p>It follows <span class="math-container">$~f(x_1)\leq f(x_2)~$</span> for all <span class="math-container">$~x_1\leq x_2~ . \hspace{4.7cm} (2)$</span> </p>
<p>Define <span class="math-container">$~0~$</span> as a infinitesimal small positive value as a simplification for "choosing always a sufficient small value" as it is used in the sense of mathematical border processes for the left side <span class="math-container">$~x\to a-0~$</span> and for the right side <span class="math-container">$~x\to a+0~$</span> . <span class="math-container">$\hspace{4cm} (3)$</span></p>
<p>It means <span class="math-container">$~n=\lfloor n-0\rfloor + 1 ~$</span> for all natural <span class="math-container">$~n~$</span> . </p>
<p>Note: <em>If someone has a problem with such a use of <span class="math-container">$~0~$</span> then it's better to define <span class="math-container">$~\delta>0~$</span> as a infinite small value so that <span class="math-container">$~n=\lfloor n-\delta\rfloor + 1 ~$</span> and substitute <span class="math-container">$~0~$</span> by <span class="math-container">$~\delta~$</span>. But then the argumentation always has to be supplemented by <span class="math-container">$~\delta\to 0 ~$</span>.</em></p>
<p><span class="math-container">$\text{(A)}$</span></p>
<p>Because of <span class="math-container">$~(1)~$</span> we have <span class="math-container">$~\displaystyle\frac{n_1}{t_{i_1}}\neq\frac{n_2}{t_{i_2}}~$</span> for <span class="math-container">$~i_1\neq i_2~$</span> and <span class="math-container">$n_1, n_2\in\mathbb{N}~$</span>, <span class="math-container">$~$</span> and </p>
<p>together with <span class="math-container">$~(2)~$</span> and <span class="math-container">$~(3)~$</span> we get <span class="math-container">$~\displaystyle f\left(\frac{n}{t_i}\right) = f\left(\frac{n-0}{t_i}\right) + 1~$</span>; <span class="math-container">$~$</span> it follows:</p>
<blockquote>
<p><span class="math-container">$$\{f(x)\,|\,x\in\mathbb{R}_0^+\}=\mathbb{N}_0$$</span> </p>
</blockquote>
<p>In words: <span class="math-container">$~f(x)~$</span> grows always by <span class="math-container">$~1~$</span>, <span class="math-container">$\,$</span>there is never a jump of <span class="math-container">$~2~$</span> or more.</p>
<p>And <span class="math-container">$~\displaystyle f\left(\frac{n}{t_i}\right)-n = f\left(\frac{n-0}{t_i}\right)-\lfloor n-0\rfloor~$</span> tells us that <span class="math-container">$~f(x)~$</span> can only grow, </p>
<p>if at least one of it's components <span class="math-container">$~\left\lfloor t_j\,x\right\rfloor~$</span> grows:</p>
<blockquote>
<p><span class="math-container">$$\left\{f\left(\frac{n}{t_i}\right)|\,i\in\{1,2,...,k\}\right\}=\{f(x)\,|\,x\in\mathbb{R}_0^+\}$$</span></p>
</blockquote>
<p><span class="math-container">$\text{(B)}$</span></p>
<p>Because of <span class="math-container">$~(1)~$</span> we have <span class="math-container">$~\displaystyle\left|\frac{n_1}{t_{i_1}}-\frac{n_2}{t_{i_2}}\right|>0~$</span> for <span class="math-container">$~i_1\neq i_2~$</span> and <span class="math-container">$n_1, n_2\in\mathbb{N}~$</span> .</p>
<p>E.g. we choose <span class="math-container">$~\displaystyle\frac{n_1}{t_{i_1}}<\frac{n_2}{t_{i_2}}~$</span> and with <span class="math-container">$~(3)~$</span> we get <span class="math-container">$~\displaystyle\frac{n_1}{t_{i_1}}<\frac{n_2-0}{t_{i_2}}~$</span> .</p>
<p>Assume that <span class="math-container">$~\displaystyle f\left(\frac{n_1}{t_{i_1}}\right) = f\left(\frac{n_2}{t_{i_2}}\right)~$</span> . </p>
<p>Then we get <span class="math-container">$~\displaystyle f\left(\frac{n_1}{t_{i_1}}\right) = f\left(\frac{n_2-0}{t_{i_2}}\right) + 1~$</span>
which means <span class="math-container">$~\displaystyle f\left(\frac{n_1}{t_{i_1}}\right) > f\left(\frac{n_2-0}{t_{i_2}}\right)~$</span> .</p>
<p>But this is a <span class="math-container">$\,$</span><em>contradiction</em><span class="math-container">$\,$</span> to <span class="math-container">$~(2)~$</span> , <span class="math-container">$~$</span> so that we have</p>
<blockquote>
<p><span class="math-container">$$\left\{f\left(\frac{n_1}{t_{i_1}}\right)\right\}\cap\left\{f\left(\frac{n_2}{t_{i_2}}\right)\right\} = \emptyset$$</span> </p>
</blockquote>
<p>for all <span class="math-container">$~i_1\neq i_2~$</span> and <span class="math-container">$n_1, n_2\in\mathbb{N}~$</span> . </p>
<p>With <span class="math-container">$\text{(A)}$</span> and <span class="math-container">$\text{(B)}$</span> follows, that the claim is correct.</p>
|
983,849 |
<p>Given</p>
<ul>
<li>$G$ be a finite group</li>
<li>$X$ is a subset of group $G$</li>
<li>$|X| > \frac{|G|}{2}$</li>
</ul>
<p>I noticed that any element in $G$ can be expressed as the product of 2 elements in $X$. Is there a valid way to prove this？</p>
<p>If the third condition was $|X| = \frac{|G|}{2}$ instead, does the above statement still hold?</p>
<p>Thank you.</p>
|
David
| 119,775 |
<p>Answer for the additional question: the statement need not be true if $|X|=|G|/2$.</p>
<p>For example, let $G$ be the cyclic group of order $2$ and $X$ consist of the non-identity element.</p>
<p>A (slightly) more general example: let $G$ be the cyclic group of order $2n$, let $g$ be a generator, and let
$$X=\{g^{2k}\mid k=0,1,\ldots,n-1\}\ .$$</p>
<p>And a bit more general again: if $X$ is a proper subgroup (not just a subset) of $G$ then the product of elements of $X$ is still in $X$, and hence not all elements of $G$ will be obtained.</p>
|
1,148,043 |
<p>Is $2\sqrt{12}$ or $4\sqrt{3}$ a better representation? Also, for $\sqrt{675}$, is $3\sqrt{75}$ or $15\sqrt{3}$ considered more simplified? Why is one more simplified than the others?</p>
|
Asinomás
| 33,907 |
<p>I would say it should be $4\sqrt 3$ or $\sqrt {48}$ . </p>
<p>Why should it be $\sqrt{48}$? You can get a very easy approximation of it just by knowing $\sqrt{49}=7$.</p>
<p>Why should it be $4\sqrt{3}$? You can calculate just by knowing $\sqrt{3}$. Which most mathematicians should know up to at least two places. (I don't though)</p>
|
398,371 |
<p>How to calculate $$\lim_{t\rightarrow1^+}\frac{\sin(\pi t)}{\sqrt{1+\cos(\pi t)}}$$? I've tried to use L'Hospital, but then I'll get</p>
<p>$$\lim_{t\rightarrow1^+}\frac{\pi\cos(\pi t)}{\frac{-\pi\sin(\pi t)}{2\sqrt{1+\cos(\pi t)}}}=\lim_{t\rightarrow1^+}\frac{2\pi\cos(\pi t)\sqrt{1+\cos(\pi t)}}{-\pi\sin(\pi t)}$$
and this doesn't get me further. Any ideas?</p>
|
Bill Kleinhans
| 73,675 |
<p>In the original expression, use the half angle equation in the denominator, and the double angle equation in the numerator. Then $ cos(\pi t/2) $ cancels, and the original expression equals minus $\sqrt 2 sin(\pi t/2) $</p>
|
3,958,142 |
<p>The following problem is in Mathematical Methods in physical science CH8 miscellaneous problems
<span class="math-container">$$
(2x-y\sin(2x))dx = (\sin^2x-2y)dy
$$</span>
it isnt an exact equation the only difference is <span class="math-container">$-$</span> sign
because it not an exact equation I tried to rearrange it to produce linear first order but it did not work
I dont know what to do
I assume there an integration factor for such problem to make it exact or methods</p>
<p>Also I tried to use Mathematica DSolve function to find a solution</p>
<pre><code>DSolve[{y'[x] == (2 x - y[x] sin (2 x))/(sin^2 x - 2 y[x])}, y[x], x]
</code></pre>
<p>This what it produced</p>
<pre><code>DSolve::deqn: Equation or list of equations expected instead of True in the first argument {True}.
DSolve[{True}, y[x], x]
</code></pre>
<p>thank you</p>
|
J.G.
| 56,861 |
<p>Since <span class="math-container">$d[x^2+y^2-y\sin^2x]=(2x-y\sin2x)dx+(2y-\sin^2x)dy=0$</span>, the solution is <span class="math-container">$x^2+y^2-y\sin^2x=c$</span>.</p>
|
3,958,142 |
<p>The following problem is in Mathematical Methods in physical science CH8 miscellaneous problems
<span class="math-container">$$
(2x-y\sin(2x))dx = (\sin^2x-2y)dy
$$</span>
it isnt an exact equation the only difference is <span class="math-container">$-$</span> sign
because it not an exact equation I tried to rearrange it to produce linear first order but it did not work
I dont know what to do
I assume there an integration factor for such problem to make it exact or methods</p>
<p>Also I tried to use Mathematica DSolve function to find a solution</p>
<pre><code>DSolve[{y'[x] == (2 x - y[x] sin (2 x))/(sin^2 x - 2 y[x])}, y[x], x]
</code></pre>
<p>This what it produced</p>
<pre><code>DSolve::deqn: Equation or list of equations expected instead of True in the first argument {True}.
DSolve[{True}, y[x], x]
</code></pre>
<p>thank you</p>
|
user577215664
| 475,762 |
<p><span class="math-container">$$(2x-y\sin(2x))dx = (\sin^2x-2y)dy$$</span>
<span class="math-container">$$(2x-y\sin(2x))= (\sin^2x-2y)y'$$</span>
<span class="math-container">$$2x+2yy'=y'\sin^2 x +y \sin (2x)$$</span>
<span class="math-container">$$2x +(y^2)'=(y \sin^2 x)'$$</span>
Integrate.</p>
|
855,329 |
<p>$$
\mbox{Question: Evaluate}\quad
\tan^{2}\left(\pi \over 16\right) + \tan^{2}\left(2\pi \over 16\right) + \tan^{2}\left(3\pi \over 16\right) + \cdots + \tan^{2}\left(7\pi \over 16\right)
$$</p>
<p>What I did:
Well I know that $\tan^{2}\left(7\pi/16\right)$ is the same as
$\cot^{2}\left(\pi/16\right)$. Thus this will repeat for all values up to
$\tan^{2}\left(4\pi/16\right)$.</p>
<p>However, I don't understand where to proceed from there. </p>
|
user8268
| 8,268 |
<p>The 15 numbers $\tan k\pi/16$, $-7\leq k\leq7$, are the roots of the polynomial (of degree 15)
$(1+ix)^{16}-(1-ix)^{16}$ (using the geometric interpretation of complex numbers and their multiplication). The polynomial is of the form $-32ixp(x^2)$, where $p(t)=t^7-35t^6+\dots$ is a polynomial of degree 7, the roots of $p$ are thus $\tan^2 k\pi/2$, $1\leq k\leq7$. Your sum is therefore $35$.</p>
|
855,329 |
<p>$$
\mbox{Question: Evaluate}\quad
\tan^{2}\left(\pi \over 16\right) + \tan^{2}\left(2\pi \over 16\right) + \tan^{2}\left(3\pi \over 16\right) + \cdots + \tan^{2}\left(7\pi \over 16\right)
$$</p>
<p>What I did:
Well I know that $\tan^{2}\left(7\pi/16\right)$ is the same as
$\cot^{2}\left(\pi/16\right)$. Thus this will repeat for all values up to
$\tan^{2}\left(4\pi/16\right)$.</p>
<p>However, I don't understand where to proceed from there. </p>
|
lab bhattacharjee
| 33,337 |
<p>Like <a href="https://math.stackexchange.com/questions/718346/find-sum-limits-k-112-tan-frack-pi13-cdot-tan-frac3k-pi13/720170#720170">Find $\sum\limits_{k=1}^{12}\tan \frac{k\pi}{13}\cdot \tan \frac{3k\pi}{13}$</a>,</p>
<p>$$\tan16\theta=\frac{\binom{16}1t-\binom{16}3t^3+\cdots+\binom{16}{13}t^{13}-\binom{16}{15}t^{15}}{\cdots}\text{ where }t=\tan\theta$$</p>
<p>If $\displaystyle\tan16\theta=0,16\theta=r\pi$ where $r$ is any integer</p>
<p>$\displaystyle\implies\theta=\frac{r\pi}{16}$ where $0\le r\le15$</p>
<p>So, the roots of $\displaystyle\binom{16}1t-\binom{16}3t^3+\cdots+\binom{16}{13}t^{13}-\binom{16}{15}t^{15}=0$ are $\displaystyle\tan r\theta$ where $\displaystyle0\le r\le15,r\ne8$ as $\displaystyle\tan\frac{8\cdot\pi}{16}$ is not finitely defined</p>
<p>So, the roots of $\displaystyle\binom{16}1-\binom{16}3t^2+\cdots+\binom{16}{13}t^{12}-\binom{16}{15}t^{14}=0$</p>
<p>or of $\displaystyle\binom{16}{15}t^{14}-\binom{16}{13}t^{12}+\cdots+\binom{16}3t^2-\binom{16}1=0$</p>
<p>are $\displaystyle\tan r\theta$ where $\displaystyle1\le r\le15,r\ne8$</p>
<p>But as $\displaystyle\tan\frac{(16-n)\pi}{16}=\tan\left(\pi-\frac{n\pi}{16}\right)=-\tan\frac{n\pi}{16},\tan^2\frac{(16-n)\pi}{16}=\tan^2\frac{n\pi}{16};$</p>
<p>the roots of $\displaystyle\binom{16}{15}u^7-\binom{16}{13}u^6+\cdots+\binom{16}3u-\binom{16}1=0$ are $\displaystyle \tan^2\frac{(16-n)\pi}{16}=\tan^2\frac{n\pi}{16}$ where $1\le n\le 7$ or $9\le r\le15$ </p>
<p>Now Vieta's formula is inviting </p>
|
500,579 |
<p>Is there a formula for the coefficients of $x^n$ for </p>
<p>$$
\prod_{i=1}^N(x+x_i)
$$</p>
<p>in terms of $x_i$?</p>
|
bubba
| 31,744 |
<p>Another approach that might be easier to program:</p>
<p>Let's call your function $f(x)$. We know it's a polynomial of degree $n$, so it can be written in the form $p(x) = \sum_{i=0}^n a_ix^i$. Choose $n+1$ values $z_0, \ldots, z_n$. By equating values of $f$ and $p$ at $x = z_0, \ldots, z_n$, we get a system of $n+1$ linear equations
$$
\sum_{i=0}^n a_iz_j^i = f(z_j) \quad (j=0,1,\ldots,n)
$$
Solve this linear system to get $a_0, \ldots, a_n$.</p>
<p>If you already have (or can find) a function to solve a system of linear equations, then there's not much code for you to write.</p>
<p>The numerical conditioning of the system of equations is bad, so be careful. It can be improved by judicious choice of $z_0, \ldots, z_n$. Certainly you should not try to compute the inverse of the coefficient matrix. If you're interested in this approach, and you need more details, please feel free to ask again.</p>
|
2,775,087 |
<blockquote>
<p>Without using a calculator, what is the sum of digits of the numbers from $1$ to $10^n$?</p>
</blockquote>
<p>Now, I'm familiar with the idea of pairing the numbers as follows:</p>
<p>$$\langle 0, 10^n -1 \rangle,\, \langle 1, 10^n - 2\rangle , \dotsc$$</p>
<p>The sum of digits of each pair is $9n$. How can I prove this property retains for all pairs?</p>
|
Christian Blatter
| 1,303 |
<p>From $0000$ till $9999$ we have $10\,000$ four digit words. The $40\,000$ digits have an average value of $4.5$ each. Add $1$ for the extra number $10\,000$ and obtain $180\,001$.</p>
|
1,344,883 |
<p>I had an idea that passes by declaring a new type of <strong>computer variable</strong> (like Integer, Double, etc.) <strong>that represents a statistical probability distribution (PDF)</strong>, for that I would need to define the basic operations; sum, multiplication, inverse and negation.</p>
<p>The problem is that I have no idea of how to define such operations or if that is even possible. (I'm electrical engineer...)</p>
<p>So the question is; How to <code>sum</code>, <code>multiply</code>, <code>inverse</code> and <code>negate</code> probability distributions, or where can I learn how to do it. A numerical approach would be sufficient.</p>
<p>Thanks.</p>
|
Tolaso
| 203,035 |
<p><strong>There is no closed solution</strong>. However note that:</p>
<p>$$e^x = \frac{1}{x}$$</p>
<p>has one real root. To see this you can consider it as a function , differentiate , determine the range etc. Therefore we have a root. </p>
<p>This root is a famous constant denoted as $\Omega$. Its approximate value is $0.5671$. </p>
<p>Otherwise, this root can be expressed via Lambert W.That is if $r$ denotes the root then we have $r=W(1)$.</p>
<p><strong>P.S</strong> There exists an integral represantation of $\Omega$. </p>
|
482,596 |
<p>How do I evaluate $ sin(20)$ exactly? [in degrees]</p>
<p>I derived the relationship between $sin(x) $ and $sin(3x)$ where $x = sin(x)$ and $ y = sin(3x)$</p>
<p><a href="http://www.wolframalpha.com/input/?i=-4x%5E3+%2B+3x+%3D+y%2C+solve+for+x" rel="nofollow">http://www.wolframalpha.com/input/?i=-4x%5E3+%2B+3x+%3D+y%2C+solve+for+x</a></p>
<p>Now I am interested in subsituting $sin(60) $ and moving along but I am not sure which formula will result in me getting a real solution.</p>
<p><a href="http://www.wolframalpha.com/input/?i=-4x%5E3+%2B+3x+%3D+%283%29%5E%281%2F2%29%2F2%2C+solve+for+x" rel="nofollow">http://www.wolframalpha.com/input/?i=-4x%5E3+%2B+3x+%3D+%283%29%5E%281%2F2%29%2F2%2C+solve+for+x</a></p>
<p>I am curious how to de-nest this mess into something cleaner. I would really like to get rid of all the imaginary numbers but if that is not possible I still feel that this can definitely be de-nested into a simpler looking form even if the radical count does not go down. </p>
<p>I have tried a couple attempts at substituting stuff back in but my answer seems to change whenever I move thing in and out of the cube root</p>
|
André Nicolas
| 6,312 |
<p>If $7s+11t=1$, then $7(s+11k)+11(t-7k)=1$ for all $k$. So there are infinitely many choices.</p>
<p>As for <em>finding</em> one pair $(s,t)$ that work, though there are general procedures, for such small numbers it is easier to go by experimentation. Note that we can take $s=8$ and $t=-5$.</p>
<p><strong>Remark:</strong> Suppose that we know that $7s_0+11t_0=1$. We want to find all $(s,t)$ such that $7s+11t=1$. If $s$ and $t$ have this property, then $7s+11t=7s_0+11t_0$.
this can be rewritten as
$$7(s-s_0)=11(t_0-t).$$
The prime $11$ divides the right-hand side. Since $11$ does not divide $7$, it divides $s-s_0$. Let $s-s_0=11k$. Then from $(7)(11k)=11(t_0-t)$ we get $t=t_0-7k$. So in fact <strong>all</strong> solutions are of the shape $s=s_0+11k$, $t=t_0-7k$.</p>
|
3,215,556 |
<p>I have a question in my paper, Express 4225 as the product of its prime factors in index notation. That was easy to answer, but the next question is express the square root of 42250000 using prime factorisation. Apparently there is a way to use my answer in the first question to do the second, but how do I?</p>
|
Arthur
| 15,500 |
<p>By way of
<span class="math-container">$$
42\,250\,000 = 4225\cdot 10\,000
$$</span>
the two numbers have a lot of factorisation in common.</p>
|
324,219 |
<p>Given an urn with $M$ unique balls, how many times do I need to draw with replacement before the probability that I have seen each ball at least once is greater than $\epsilon$?</p>
|
Coiacy
| 62,460 |
<p>Order the $M$ balls form $1$ to $M$ and suppose you have drawn $n$ balls with replacement. Among the $n$ balls, the number of ball$1$ is $x_1$, the number of ball$2$ is $x_2,\ \cdots$, the number of ball$M$ is $x_M$, then
$$
x_1+x_2+\cdots +x_M=n
$$
you want to see each ball at least once, that is the restiction:
$$ x_i\ge 1,\ for\ 1\le n\le M
$$
we have ${n-M \choose M-1}$ different choices, so I claim:
$$
P(the\ desired\ event)={n-M \choose M-1}\cdot \frac{1}{M^n}
$$</p>
|
41,183 |
<p>Is it true that any manifold homotopy equivalent to a k-dimensional CW-complex admits a proper Morse function with critical points all of index <= k? I believe this is not true, so I would like to see a counterexample.</p>
|
Ryan Budney
| 1,465 |
<p>Take a contractible $3$-manifold which is not homeomorphic to $\mathbb R^3$ -- like the Whitehead manifold. </p>
<p>If such a Morse function existed on the Whitehead manifold, it would be a Morse function with only one critical point, the minimum, and therefore the Whitehead manifold would be an open 3-ball. The proof of this has two steps: (1) $f$ can have at most one critical point, WLOG a minimum by homotopy-type considerations, and it can't have less than one critical point by the "edit" below. Step (2) if it has one critical point let it be $f(p)=0$, the minimum. By the Morse Lemma, $f^{-1}[0,\epsilon]$ is diffeomorphic to a closed 3-ball. The flow the the gradient of $f$ gives a diffeomorphism between $f^{-1}[\epsilon,\infty)$ and $S^2 \times [\epsilon, \infty)$. Pasting these two diffeomorphisms together gives you a diffeomorphism between the manifold and $\mathbb R^3$. </p>
<p>edit: Well, I guess there's the special case to consider that the Morse function could have no critical points but then you could deal with this by the argument that the Whitehead manifold isn't a product of a surface and $\mathbb R$, which by the classification of surfaces amounts to saying that the Whitehead manifold isn't homeo/diffeomorphic to $\mathbb R^3$. </p>
<p>Welcome to MO, Victor!</p>
|
1,752,506 |
<p>Question: $ \sqrt{x^2 + 1} + \frac{8}{\sqrt{x^2 + 1}} = \sqrt{x^2 + 9}$</p>
<p>My solution: $(x^2 + 1) + 8 = \sqrt{x^2 + 9} \sqrt{x^2 + 1}$</p>
<p>$=> (x^2 + 9) = \sqrt{x^2 + 9} \sqrt{x^2 + 1}$</p>
<p>$=> (x^2 + 9) - \sqrt{x^2 + 9} \sqrt{x^2 + 1} = 0$</p>
<p>$=> \sqrt{x^2 + 9} (\sqrt{x^2 + 9} - \sqrt{x^2 + 1}) = 0$</p>
<p>So, either $\sqrt{x^2 + 9} = 0$ or $(\sqrt{x^2 + 9} - \sqrt{x^2 + 1}) = 0$</p>
<p>From the first expression, I get $x = \pm 3 i$ and from the second expression, I get nothing.</p>
<p>Now, notice how in the 2nd step, I could've divided both the sides by $\sqrt{x^2 + 9}$, but I didn't because I learned here that we must never do that and that we should always factor: <a href="https://math.stackexchange.com/questions/1751410/how-to-square-both-the-sides-of-an-equation#comment3574205_1751435">Why one should never divide by an expression that contains a variable.</a></p>
<p>So, my question is: is the solution above correct? Would it have been any harm had I divided both the sides by $\sqrt{x^2 + 9}$?</p>
|
brainst
| 309,526 |
<p>It would indeed , because then you wont have $x=\pm 3i$ as the root because that would make the denominator zero , which is the greatest offense one can do in algebra!! or more broadly in Mathematics (:P)</p>
|
1,187,142 |
<p>Let a(n) be an arithmetic sequence and, as usual let $A(n)=a(1) + a(2) + \dots + a(n).$ If $A(5) = 50$ and $A(20) = 650$, find $A(15).$</p>
<p>I'm not so sure how to solve that.
So </p>
|
Jr Antalan
| 207,778 |
<p>Let $A(n)$ be an arithmetic sequence, then: </p>
<p>$A(n)=\frac{n}{2}(2a+(n-1)d)$</p>
<p>where $a$ is $A(1)$ and $d$ is the common difference.</p>
<p>Using the formula above we yield the system of 2 equations below:</p>
<p>$\frac{5}{2}(2a+4d)=50$</p>
<p>$\frac{20}{2}(2a+19d)=650$</p>
<p>Now with this two equations I can find $a$ and $d$, So I can now find $A(15)$ using the formula above.</p>
<p>Hope that this helps.</p>
|
2,066,501 |
<p>If Q = \begin{bmatrix}1&1/√6&1/3√3\\1&-1/√6&-1/3√3\\0 & √2/√3 & -1/3√3\end{bmatrix}</p>
<p>and R = \begin{bmatrix}2√2&1/√2&1/√2\\0&4/√6&1/√6\\0&0&1/3√3\end{bmatrix}.</p>
<p>Is A invertible? (no computation required) Is the system Ax=b solvable for each b in $R^3$ (give the formula for its solutions).</p>
|
TheGeekGreek
| 359,887 |
<p><strong>Hint.</strong> Consider the determinant of $A$. I assume you mean the $QR$-decomposition of $A$, i.e. $A = QR$.</p>
|
1,470,039 |
<p>This problem will deal with the group <span class="math-container">$G = D_4 \times S_3$</span>.</p>
<p>How many elements of each order do the groups have <span class="math-container">$D_4$</span> and <span class="math-container">$S_3$</span> have? Using this info, determine how
many elements of each order <span class="math-container">$G$</span> has. (Of course, it would be extremely tedious to compute all 48
orders one by one. You should NOT be doing that. Perhaps organize your work in a table.)</p>
<p>I can't figure out a good way to do this. Any help would be appreciated, thank you.</p>
<p>EDIT: To find the subgroups of G which are isomorphic to <span class="math-container">$Z_2 \times Z_2$</span>, do I have to find all the elements with an order of 4, which are non-abelian?</p>
|
Groups
| 28,017 |
<p>Elements of order $2$ will come from produc of an element of order $\leq 2$ in $D_4$ and an element of order $\leq 2$ in $S_3$ (not both taken identity). Count them.</p>
<p>Elements of order $4$ will be product of elements of order $4$ inside $D_4$, with elements of order $\leq 2$ in $S_3$ (count them).</p>
<p>Elements of order $3$ lie in Sylow-3 subgroup, which is unique here, inside $S_3$, count them.</p>
<p>Elements of order $6$ will come from product of an element of order $3$ and commuting element of order $2$. The elements of order $2$ commuting with element of order $3$ lie in $D_4$, count how many you get.</p>
<p>Sylow-2 subgroup has exponent 4, so no element of order 8.</p>
<p>Elements of order 12 will come from element of order $3$ and commuting element of order $4$. Elements of order $3$ are in $S_3$ and with them commuting elements of order $4$ are in $D_4$, count how many you get.</p>
<p>The maximum order of any element in $D_4\times S_3$ is product of maximum order of element of $D_4$ (which is 4) and of $S_3$ (which is 3). So we covered all elements (except identity :P)</p>
|
1,470,039 |
<p>This problem will deal with the group <span class="math-container">$G = D_4 \times S_3$</span>.</p>
<p>How many elements of each order do the groups have <span class="math-container">$D_4$</span> and <span class="math-container">$S_3$</span> have? Using this info, determine how
many elements of each order <span class="math-container">$G$</span> has. (Of course, it would be extremely tedious to compute all 48
orders one by one. You should NOT be doing that. Perhaps organize your work in a table.)</p>
<p>I can't figure out a good way to do this. Any help would be appreciated, thank you.</p>
<p>EDIT: To find the subgroups of G which are isomorphic to <span class="math-container">$Z_2 \times Z_2$</span>, do I have to find all the elements with an order of 4, which are non-abelian?</p>
|
Nicky Hekster
| 9,605 |
<p>Hint: if <span class="math-container">$x \in G, y \in H$</span>, then in <span class="math-container">$G \times H$</span>, the <span class="math-container">$ord((x,y))=lcm(ord(x),ord(y))$</span></p>
|
411,875 |
<p>This is an exam question I encountered while studying for my exam for our topology course:</p>
<blockquote>
<p>Give two continuous maps from $S^1$ to $S^1$ which are not homotopic. (Of course, provide a proof as well.)</p>
</blockquote>
<p>The only continuous maps from $S^1$ to $S^1$ I can think of are rotations, and I thought rotations on a circle can be continuously morphed into one another. </p>
|
Karl Kroningfeld
| 67,848 |
<p>The identity is not homotopic to a constant map; otherwise, $S^1$ would be contractible, which would imply $\pi_1(S^1)=0$.</p>
|
3,733,428 |
<p>I would like to understand why <span class="math-container">$\mathbb{R}^2\setminus \mathbb{Q}^2$</span> endowed with the subspace topology is not a topological manifold. It seems to me it is Hausdorff and second countable. So I am wondering. Why is it not locally Euclidian?</p>
|
Alessandro Codenotti
| 136,041 |
<p>Any neighbourhood of any point in <span class="math-container">$\Bbb R^2\setminus\Bbb Q^2$</span> is not simply connected, so no it cannot be homeomorphic to <span class="math-container">$\Bbb R^n$</span>.</p>
|
1,232,143 |
<p>If
$$f(x)=\sum_{n=0}^\infty\frac{x^n}{n!}, x\in\mathbb{R}$$
and
$$
g(x) = 1 + \int_0^x f(t) \,dt
$$</p>
<p>prove that $g(x)=f(x)$ for all $x\in\mathbb{R}$
and prove that $f$ is differentiable on $\mathbb{R}$ as well as show that $f'(x)=f(x)$ for all $x\in\mathbb{R}$.</p>
<p>I know that $f$ is continuous on $\mathbb{R}$ and the series $f$ converges uniformly on $[-r,r]$ for all $r>0$, but I'm unsure of how to approach these questions. Any help is greatly appreciated! </p>
|
Paramanand Singh
| 72,031 |
<p>The problem is trivial if we use uniform convergence because then we can do term by term differentiation and get $f'(x) = f(x)$ and by definition of $g(x)$ we get $g'(x) = f(x)$ so that $h'(x) = 0$ where $h(x) = f(x) - g(x)$. Then $h(x)$ is a constant and $$h(x) = h(0) = f(0) - g(0) = 1 - 1 = 0$$ and therefore $f(x) = g(x)$.</p>
<p>However I believe the crux of the problem is to establish it without using term by term differentiation. A simple approach is to show that $f(x + y) = f(x)f(y) $ for all $x, y \in \mathbb{R}$ using multiplication of infinite series. Next we can show that $$\lim_{x \to 0}\frac{f(x) - 1}{x} = 1$$ (without using term by term limits). These two facts will establish that $f'(x) = f(x)$ and hence the job is done (as in the first paragraph).</p>
<p>I am wondering if there is any other approach which shows that $f(x) = g(x)$ (without somehow using $f'(x) = f(x)$).</p>
|
1,655,884 |
<blockquote>
<p>How many integer-sided right triangles exist whose sides are combinations of the form $\displaystyle \binom{x}{2},\displaystyle \binom{y}{2},\displaystyle \binom{z}{2}$?</p>
</blockquote>
<p><strong>Attempt:</strong></p>
<p>This seems like a hard question, since I can't even think of one example to this. Mathematically we have,</p>
<p>$$\left(\dfrac{x(x-1)}{2} \right)^2+\left (\dfrac{y(y-1)}{2} \right)^2 = \left(\dfrac{z(z-1)}{2} \right)^2\tag1$$ </p>
<p>where we have to find all positive <em>integer</em> solutions $(x,y,z)$. </p>
<p>I find this hard to do. But here was my idea. Since we have $x^2(x-1)^2+y^2(y-1)^2 = z^2(z-1)^2$, we can try doing $x = y+1$. If we can prove there are infinitely many solutions to,</p>
<p>$$(y+1)^2y^2+y^2(y-1)^2 = z^2(z-1)^2\tag2$$ </p>
<p>then we are done.</p>
|
vonbrand
| 43,946 |
<p><strong>Hint:</strong> Start the other way around, by the formula to generate all <a href="https://en.wikipedia.org/wiki/Pythagorean_triple" rel="nofollow">Pythagorean triples</a>.</p>
|
901,357 |
<p>Let there be $T:R^3 \rightarrow R^3$
<br>
$T(0,-1,1)=(3,3,3)$
<br>
$T(1,0,-1)=(0,1,1)$
<br>
$T(1,1,0)=(1,2,-1)$</p>
<p>Is (1,2,3) is the only image of the vector $(1, \frac{-7}{9}, \frac{-8}{9})$?</p>
<p>I have thought to create a matrix $[T]^T_E$*$[T]^E_C$=$[T]^T_C$
so I will have a matrix that does the transformation and the I can check the calim, but that did not work </p>
|
Adam Hughes
| 58,831 |
<p>You want to write</p>
<p>$${1\over 1-z}={1\over 1-5i-(z-5i)}$$</p>
<p>Then this is just</p>
<p>$${1\over 1-5i}\left({1\over 1-{z-5i\over 1-5i}}\right)$$</p>
<p>Which, by geometric series, instantly gives</p>
<p>$${1\over 1-5i}\sum_n {(z-5i)^n\over (1-5i)^n}=\sum_n{(z-5i)^n\over (1-5i)^{n+1}}$$</p>
<p>and moreover we know geometric series have convergence if and only if the common ratio, $r$, has $|r|<1$ i.e. for</p>
<p>$$\left|{z-5i\over 1-5i}\right|<1\iff |z-5i|<|1-5i|=\sqrt{26}$$</p>
|
869,358 |
<p>This is a seemingly simple induction question that has me confused about perhaps my understanding of how to apply induction</p>
<p>the question;</p>
<p>$$\frac{1}{1^2}+ \cdots+\frac{1}{n^2}\ \le\ 2-\frac{1}{n},\ \forall\ n \ge1.$$</p>
<p>this true for $n=1$, so assume the expression is true for $n\le k$. which produces the expression, </p>
<p>$$\frac{1}{1^2} + \cdots + \frac{1}{k^2} \le\ 2-\frac{1}{k}.$$ now to show the expression is true for $k+1$,</p>
<p>$$\frac{1}{1^2}+\cdots+ \frac{1}{k^2} + \frac{1}{(k+1)^2} \le\ 2-\frac{1}{k}+\frac{1}{(k+1)^2}.$$</p>
<p>this the part I am troubled by, because after some mathemagical algebraic massaging, I should be able to equate,</p>
<p>$$2-\frac{1}{k}+\frac{1}{(k+1)^2}=2-\frac{1}{(k+1)},$$</p>
<p>which would prove the expression is true for $k+1$ and I'd be done. right? but these two are not equivalent for even $k=1$, because setting $k=1$ you wind up with $\frac{5}{4}=\frac{3}{2}$, so somewhere i am slipping up and I'm not sure how else to show this if someone has some insight into this induction that I'm not getting. thanks.</p>
|
Martin Sleziak
| 8,297 |
<p>Despite the fact that such solution is given in other posts (for example in this question about similar infinite series: <a href="https://math.stackexchange.com/questions/1000642/proof-that-sum-1-infty-frac1n2-2">Proof that $\sum_{1}^{\infty} \frac{1}{n^2} <2$</a> it is used as an auxiliary result in some answers) it might be useful to mention the solution using <a href="http://en.wikipedia.org/wiki/Telescoping_series" rel="nofollow noreferrer">telescoping sum</a> in a question asking about finite sum.</p>
<p>Let us look at the sum starting with $k=2$:
$$\sum\limits_{k=2}^n \frac1{k^2} \le \sum\limits_{k=2}^n \frac1{k(k-1)}.$$
After rewriting this as $\frac1{k(k-1)}=\frac1{k-1}-\frac1k$we see that on the RHS we get a telescoping sum, where many terms will cancel out:
$$\sum_{k=2}^n \left(\frac1{k-1}-\frac1k\right)=(1-\frac12)+\left(\frac12-\frac13\right)+\dots+\left(\frac1{n-1}-\frac1n\right)=1-\frac1n$$
(We started with $k=2$ so that we do not get zero in the denominator in the expressions $\frac1{k-1}$ and $\frac1{k(k-1)}$.)</p>
|
3,007,785 |
<p><span class="math-container">$\lim\limits_{x\to 0}\frac{e^x-\sqrt{1+2x+2x^2}}{x+\tan (x)-\sin (2x)}$</span> I know how to count this limit with the help of l'Hopital rule. But it is very awful, because I need 3 times derivate it. So, there is very difficult calculations. I have the answer <span class="math-container">$\frac{2}{5}$</span>. </p>
<p>I want to know if there is other ways to calculate it, without 3 times using l'Hopital rule? (I could write my steps, but they are very big. I just took third derivative of numerator and denominator)</p>
|
user
| 505,767 |
<p><strong>HINT</strong></p>
<p>By Taylor's series we have that</p>
<p><span class="math-container">$$\frac{e^x-\sqrt{1+2x+2x^2}}{x+\tan (x)-\sin (2x)}=\frac{\overbrace{1+x+\frac12 x^2+\frac16x^3+o(x^3)}^{\color{red}{e^x}}-(\overbrace{1+x+\frac12x^2-\frac12 x^3+o(x^3)}^{\color{red}{\sqrt{1+2x+2x^2}}})}{x+\underbrace{x+\frac13 x^3o(x^3)}_{\color{red}{\tan x}}-(\underbrace{2x-\frac43 x^3 + o(x^3)}_{\color{red}{\sin 2x}})}$$</span></p>
|
2,874,840 |
<blockquote>
<p>If $P\left(A\right)=0.8\:$ and $P\left(B\right)=0.4$, find the maximum and minimum values of $\:P(A|B)$.</p>
</blockquote>
<p>My textbook says the answer is $0.5$ to $1$. But I think the answer should be $0$ to $1$.</p>
<p>The textbook claims $P(A∩B)$ is $0.2$ when $P(A'∩B')=0$</p>
<p>I think that the minimum value arises when $A$ and $B$ are mutually exclusive. So there isn't a chance of both happening and so you have $\frac{0}{0.4} = P(A|B) = 0$ (right?)</p>
<p>I agree with the textbook in saying the maximum value is one.</p>
|
Clarinetist
| 81,560 |
<p>I think a visualization helps most with understanding these probability bounds.</p>
<p>To begin,
$$P(A \mid B) = \dfrac{P(A \cap B)}{P(B)} = \dfrac{P(A \cap B)}{0.4}\text{.}$$
To finish this problem, we need to find bounds for $P(A \cap B)$.
Let's draw a Venn diagram for these two events.
<a href="https://i.stack.imgur.com/i66Fb.png" rel="noreferrer"><img src="https://i.stack.imgur.com/i66Fb.png" alt="enter image description here"></a>
Notice that $P(A) = x+y$, $P(B) = y+z$, and $P(A \cap B) = y$.</p>
<p>Since $P(A) = 0.8$ and $P(B) = 0.4$, we know that
$$\begin{align}
x+y &= 0.8 \\
y+z &= 0.4\text{.}
\end{align}$$
Furthermore, since $x$, $y$, and $z$ all are greater than or equal to $0$, we know immediately that $y$ must be less than or equal to $0.4$ by the equations above. ($y$ can't exceed $0.4$, otherwise $y+z$ would exceed $0.4$, which is a contradiction with the above system of equations.)</p>
<p>Lastly, we have to find the smallest value of $y$. $y$ cannot be $0$, because otherwise $x + z = 1.2$, and one cannot have probabilities summing to exceed $1$ in a sample space.</p>
<p>So, the way I would approach this is as follows: we know $P(A)$ and $P(B)$. What's the largest value that I can subtract from their sum $P(A) + P(B)$ so that I get probabilities summing up to $1$?</p>
<p>Well, we know that
$$0.8 + 0.4 - 0.2 = 1$$
so we would be subtracting $0.2$. This amounts to $0.2$ being the smallest possible value of $y$, hence
$$0.2 \leq y = P(A \cap B) \leq 0.4$$
or
$$\dfrac{0.2}{0.4} = \dfrac{1}{2} \leq \dfrac{P(A \cap B)}{P(B)} \leq \dfrac{0.4}{0.4} = 1$$
or
$$\dfrac{1}{2} \leq P(A \mid B) \leq 1\text{.}$$</p>
|
786,329 |
<p>$$f(x,y)=\begin{cases}
\frac{xy}{x^2+y^2}, \text{ if } x^2+y^2\neq 0
\\
0, \text{ if } x^2+y^2=0
\end{cases}$$</p>
<p>Is it continuous at $(0,0)$?</p>
|
Jlamprong
| 105,847 |
<p>No, since along line $y=x$
$$
\lim_{x,y\to0}f(x,y)=\frac12\neq0=f(0,0).
$$</p>
|
2,317,496 |
<p><strong>Definition:</strong> An ordinal number $\alpha$ is called a <strong><em>limit ordinal number</em></strong> if there is no ordinal number immediately preceding $\alpha$. </p>
<p>Now my lecture notes say that $\omega, 2\omega, \omega^2, \omega^\omega$ are limit ordinal numbers whereas $\omega+3,2^\omega+5$ are not which is intuitively clear. But is there a characterization of limit ordinals that may work when proving the case with some rigor? Or is it "just look and identify" sort of a thing using the ordering of the family of ordinal numbers?</p>
|
Asaf Karagila
| 622 |
<p>There can be several answers, depending on what you mean by "identify".</p>
<p>The simplest answer would be to look at the <a href="https://en.wikipedia.org/wiki/Ordinal_arithmetic#Cantor_normal_form" rel="noreferrer">Cantor normal form</a> of $\alpha$, and see if it has any finite ordinal there. If the answer is no, then $\alpha$ is a limit ordinal (or $0$, which may or may not be a limit ordinal depending on you convention) and otherwise it is a successor ordinal.</p>
<p>Another answer would be that $\alpha$ is a limit ordinal if and only if for every $\beta<\alpha$, $\beta+1<\alpha$ (with the same caveat about $0$ as before). Although it seems not to be exactly what you are looking for.</p>
|
2,484,855 |
<p>How can I prove that these three sets have no common values:</p>
<ul>
<li>A: {prime numbers} </li>
<li>B: {Fibonacci numbers}</li>
<li>C: {8|n+1} </li>
</ul>
<blockquote>
<ul>
<li>C: for example 15: 15+1 = 16 => 8|16 <= 16/8 = 2</li>
<li>C: for example 23: 23+1 = 24 => 8|24 <= 24/8 = 3</li>
</ul>
</blockquote>
|
Barry Cipra
| 86,747 |
<p>Consider the Fibonacci numbers mod $8$:</p>
<p>$$1,1,2,3,5,0,5,5,2,7,1,0,\ldots$$</p>
<p>We see that $F_m\equiv7$ mod $8$ if and only if $m\equiv10$ mod $12$, which implies $m$ is even (and greater than $4$). But $F_n\mid F_{2n}$ for all $n$, so $F_m$ cannot be a prime if $8\mid F_m+1$.</p>
<p>Remark: The theorem $F_n\mid F_{2n}$ for all $n$ (and, more generally, $F_n\mid F_{kn}$ for all $k$ and $n$) boils down to the banal observation that $F_n\equiv0$ mod $F_n$. This means that the sequence of Fibonacci numbers mod $F_n$ begins</p>
<p>$$1,1,2,3,\ldots,F_{n-1},0,F_{n-1},F_{n-1},2F_{n-1},3F_{n-1},\ldots,F_{n-1}^2,0,\ldots$$</p>
<p>That is, once you hit the first $0$, the sequence repeats itself, multiplied by (powers of) $F_{n-1}$.</p>
|
2,733,142 |
<p>This question has been asked a <a href="https://math.stackexchange.com/questions/207029/a-b2-for-which-matrix-a">few</a> <a href="https://math.stackexchange.com/questions/583442/square-root-of-nilpotent-matrix">times</a>. In the former case I noticed that there was some argument trending towards using the Jordan form. For the sake of completeness, I was hoping to see if I'm on the right track.</p>
<p>Let $N$ be a nilpotent matrix such that $N^n=0$ and $N^{n-1}\neq0$. Suppose there exists $A^2=N$. Then I can write $A^2=\lambda I+N'$ where $N'$ is a nilpotent matrix. Using the Taylor expansion, I can see that $A^2=(\lambda I+N')^{1/2}=\sqrt{\lambda}\left(I+\frac{N'}{2\lambda}-\frac{N'^2}{8\lambda^2}+\frac{N'^3}{16\lambda^3}+....\right)$. This series is convergent since it only has finitely many terms. But, $A^2=N$ and $N$ is nilpotent. So $\lambda^i=0$ for all $i\in \{0,...,n\}.$ Hence such an $A^2$ doesn't exist.</p>
|
Will Jagy
| 10,400 |
<p>trying 3 by 3</p>
<p>$$
\left(
\begin{array}{ccc}
a & b & c \\
d & e & f \\
g & h & i
\end{array}
\right)^2 =
\left(
\begin{array}{ccc}
0 & 1 & 0 \\
0 & 0 & 1 \\
0 & 0 & 0
\end{array}
\right) \; ,
$$
then
$$
\left(
\begin{array}{ccc}
a^2 + bd+cg & ab+be+ch & ac+bf+ci \\
ad+de+fg & e^2+bd+fh & cd +ef +fi \\
ag+dh+gi & bg+eh+hi & i^2+cg +fh
\end{array}
\right) =
\left(
\begin{array}{ccc}
0 & 1 & 0 \\
0 & 0 & 1 \\
0 & 0 & 0
\end{array}
\right) \; ,
$$</p>
|
2,727,237 |
<p>$$
\begin{matrix}
1 & 0 & -2 \\
0 & 1 & 1 \\
0 & 0 & 0 \\
\end{matrix}
$$</p>
<p>I am told that the span of vectors equal $R^m$ where $m$ is the rows which has a pivot in it. So when describing the span of the above vectors, is it correct it saying that they don't span $R^3$ but only span $R^2$?.</p>
<p>Thanks</p>
|
P Vanchinathan
| 28,915 |
<p>The point is the distinction between vectors and their representation as linear combination with respect to a basis. Orthogonality as a concept is a condition <em>on the given two vectors</em>. Of course this can be translated into a condition on their representation. But that translated condition will vary depending on the basis under discussion.</p>
<p>As an analogy think of positive integers as opposed to their representation in base 2, or base 10 or base 5 or base 7. In usual base 10 number system a number $\ne2$ being prime means its rightmost digit has to be 1, 3, 7 or 9 necessarily (but not sufficient). In base 7, a prime number can have 1, 2, 3,4 or 5 or 6 as the examples below show:</p>
<p>The following two-digit numbers: <strong>41, 32, 23, 14, 25,16</strong> interpreted as base 7 numbers are the prime numbers 29, 23, 17, 11, 19, 13 (base 10).</p>
<p>ASIDE: If you choose a prime number $p$ as base (as we did with 7 above), every number from 1 to $p-1$ can occur as the right most digit of a suitable prime number. (This is simply Dirichlet's theorem on Prime Numbers on Arithmetic Progressions, from the year 1837).</p>
|
2,145,098 |
<p>This was part of a three part question where I was supposed to prove two sets have equal cardinality by finding bijections. I've created a bijection $f: \Bbb Z \Rightarrow 2\Bbb Z$ by $f(x)=2x$. I've created a bijection $g: (0,1) \Rightarrow (4,50)$ by $g(x)=46x+4$. I think those are both correct. My last question is finding a a bijection between (0,1) and [0,1]. I've seen this question several times on this board, but I've yet to understand them and I can't really go back and ask any questions to the original posters. I know a bijective function exists between (0,1) and $\Bbb R$, but I don't think that helps me here. </p>
|
fleablood
| 280,126 |
<p>It's counter intuitive but push 0 to $a_1$. And push $1$ to $a_2$. Then create an infinite sequence $a_i $ and push $a_i $ to $a_{i+3}$. Meanwhile for any $x $ not in the sequence, map $x $ to $x $. </p>
<p>Example: Let $a_n = 1/n$. Let $f (0)=1/2$. Let $f (1)= 1/3$. Let $f (1/n)=\frac 1 {n+2} $. If $x \ne \frac 1n $ for any natural $n $, let $f (x)=x $.</p>
<p>That's easy to be shown to be bijective.</p>
<p>Another, perhaps more common, example is to let $\{a_n\}=\mathbb Q $ be an sequence of all the rationals. Map $0\rightarrow a_1$ and map $1\rightarrow a_2$ and $a_i \rightarrow a_{i+2} $ while every irrational gets mapped mapped to itself.</p>
|
1,642,427 |
<p>If the stem of a mushroom is modeled as a right circular cylinder with diameter $1$, height $2$, its cap modeled as a hemisphere of radius $a$ the mushroom has axial symmetry, is of uniform density,and its center of mass lies at center of plane where the cap and stem join, then find $a$.</p>
<p>I really need help.</p>
|
Wojowu
| 127,263 |
<p>The problem with this argument, and many other alike, is that the arc length of a curve is not a continuous function of a curve, which means that if sequence of curves $\gamma_i$ converges to a curve $\gamma$ (notion of convergence of functions like that can be made formal, but let me not do that here), then there is no guarantee the sequence of arclengths $l(\gamma_i)$ to converge to the arclength $l(\gamma)$. Indeed, if you look at the integrals $\int_0^1\sqrt{1-\gamma_i'(t)^2}dt$, then you might expect that the expression under the integral converges to $\sqrt{1-\gamma'(t)^2}$, which however isn't the case, because convergent sequence of differentiable functions (I am ignoring the "cusps" where the semicircles meet here; they are important as well, but that's beyond my point right now), then the sequence of derivatives need not be convergent.</p>
|
2,071,828 |
<p>Find the splitting field of $x^6-2x^4-8x^2+16$ over $\mathbb {F}_3$ and list the intermediate fields between the base camp and the splitting field.</p>
|
lhf
| 589 |
<p>$x^6-2x^4-8x^2+16 = x^6+x^4+x^2+1$ in $\mathbb {F}_3[x]$.</p>
<p>$x^6+x^4+x^2+1 = \dfrac{x^8-1}{x^2-1} = (x^2 + 1) (x^4 + 1)$</p>
<p>Therefore, the splitting field of $x^6-2x^4-8x^2+16$ is the same as the splitting field of $x^4 + 1=(x^2 + x + 2) (x^2 + 2 x + 2)$, which is $\mathbb {F}_9$.</p>
|
46,726 |
<p>In many proofs I see that some variable is "fixed" and/or "arbitrary". Sometimes I see only one of them and I miss a clear guideline for it. Could somebody point me to a reliable source (best a well-known standard book) which explains, when and how to use both in proofs?</p>
<p>EDIT: A little add-on to the question: Take a usual induction on natural numbers and assume that you are teaching it to students in their first semester. How do you explain "fixed" and "arbitrary" in this scenario?</p>
|
ItsNotObvious
| 9,450 |
<p>When you encounter the term "arbitrary", it usually just means that a given statement is specified for any element from a given set of elments. For example, if I say, let $x$ be an arbitrary element of the interval $[0, 1]$ I just mean that $x$ can take on any value within that interval. </p>
<p>The term <em>fixed</em> connotes a similar but more specific meaning. If I say let $x$ be a fixed element of the interval $[0, 1]$ I mean that, firstly, $x$ is an arbitrary element of $[0,1]$ but it's value is unchanging througout its usage.</p>
<p>The difference in meaning is subtle but sometimes important. If you start out with a "fixed" element you can't, for example, choose its value at some later point in the argument. On the other hand, if I prove that something holds for an <em>arbitrary</em> element of a given class, then it will hold for any particular element of that class which can often be very useful.</p>
|
2,046,492 |
<p>Let's think of two events $1$ and $2$.</p>
<p>Both events happen randomly $n_1$/$n_2$-times during a given time $T$ and last for a time of $t_1$/$t_2$.</p>
<p><strong>What is probability $P$, that both events happen simultaneously at some moment?</strong></p>
<hr>
<hr>
<p><em>EXAMPLE 1:</em></p>
<p>$T = 60$ min</p>
<p>Event $1$ - looking out of the office window: $n_1 = 8$ and $t_1 = 1$ min </p>
<p>Event $2$ - a green car is on the street visible: $n_2 = 20$ and $t_2 = 0.5$ min</p>
<p>$P$: How likely do I see a green car during these $60$ min?</p>
|
leonbloy
| 312 |
<p>[Not an answer, but some pointers]</p>
<p>In statistical physics parlance, you have a pair of independent <a href="http://www.sklogwiki.org/SklogWiki/index.php/1-dimensional_hard_rods" rel="nofollow noreferrer">1D hard-rods</a> (or Tonk gas). In the limit $L\to \infty$ the "canonical ensemble" (fixed volumen and number of particles) can be replaced by other ensembles, for example, a grand-canonical in which the number of particles is variable (only the average is prescribed).
I turn, this ensemble (en 1D only) is equivalent to a causal stochastic process, or a <a href="https://en.wikipedia.org/wiki/Point_process" rel="nofollow noreferrer">point process</a>. This can be useful, no only for theoretical analysis but also for simulation [*].</p>
<p>Specifically, the distance among particles in our model, in which we prescribe a "density" $\rho = \overline n/L$ and a particle "diameter" $d$, would follow a density</p>
<p>$$f_X(x)=\lambda e^{-\lambda(x-d)} u(x-d)$$</p>
<p>where $u(\cdot)$ is the <a href="https://en.wikipedia.org/wiki/Heaviside_step_function" rel="nofollow noreferrer">unit step function</a> and $\lambda$ is given by $E(X)=d+\lambda^{-1}=1/\rho$ or</p>
<p>$$ \lambda=\frac{\rho}{1-\rho d}$$</p>
<p>To estimate $P_o=$ probability of "overlap" (there exist particles of both different point processses separated by less than $d^*=(d_1+d_2)/2$) I can think of two approaches:</p>
<p>First, compute $P_1=$ probability that a single point of the process 1 overlaps with some point of process 2, and then set $P_o=1-(1-P_1)^{n_1}$ . This would amount to assume that the events "first particle of process 1 overlaps with process 2" and "second particle of process 1 overlaps with process 2" are independent, which only could work as an approximation. $P_1$ should be computable from the theory of point processes, I think.</p>
<p>Alternatively, perhaps it's possible to build a "marked" point process that is equivalent to the pair of processes...</p>
<p>[*] You can play here <a href="https://jsfiddle.net/vdo1boue/" rel="nofollow noreferrer">https://jsfiddle.net/vdo1boue/</a></p>
|
1,630,074 |
<p>My brother needs help coming up with a formula for a problem that I already did but failed to write out the formula for. </p>
<p>The problem is:</p>
<p>Consider a circle with the point (5,4) and a radius of 3.
Determine if a vertical line segment with the points (3,5) being one end of the segment and (3,1) being the other end of the segment, intersects this circle. List the intersecting points if so.</p>
<p>:
Again I need help coming up with a simple formula to determine where the line intersects the circle.</p>
<p>Thanks</p>
|
Plutoro
| 108,709 |
<p>The problem arises when applying $H_{t}$ to the point $(-1,0,0,...)$. We get $$H_t(-1,0,0,...)=\frac{(2t-1,0,0,...)}{\Vert (2t-1,0,0,...)\Vert}=\begin{cases} (-1,0,0,...) & t<1/2\\
(1,0,0,...) & t>1/2\\
\text{undefined} & t=1/2.\end{cases}$$
This is not continuous. We do not need a two step description for a homotopy, but the one given in this example is a simple one which includes no singularities associated with dividing by zero.</p>
|
376,530 |
<p>Let <span class="math-container">$G$</span> be a graph which does not contain a simple cycle <span class="math-container">$v_1\ldots v_k$</span> and two "crossing" chords <span class="math-container">$v_iv_j$</span> and <span class="math-container">$v_pv_q$</span>, <span class="math-container">$i<p<j<q$</span>. An example of such graph is a triangulation of the convex polygon. Is it true that the number of edges in <span class="math-container">$G$</span> does not exceed <span class="math-container">$2n-3$</span>, where <span class="math-container">$n$</span> denotes the number of vertices?</p>
|
David Wood
| 25,980 |
<p>Thomassen and Toft [<a href="https://doi.org/10.1016/S0095-8956(81)80025-1" rel="noreferrer">JCTB 31(2):199-224, 1981</a>] showed that any graph with minimum degree at least 3 contains a cycle with two crossing chords from neighbouring vertices on the cycle. The <span class="math-container">$2n-3$</span> upper bound follows by induction on <span class="math-container">$n$</span>, since we may delete a vertex of degree at most <span class="math-container">$2$</span>.</p>
|
33,743 |
<p>I have a lot of sum questions right now ... could someone give me the convergence of, and/or formula for, $\sum_{n=2}^{\infty} \frac{1}{n^k}$ when $k$ is a fixed integer greater than or equal to 2? Thanks!!</p>
<p>P.S. If there's a good way to google or look up answers to these kinds of simple questions ... I'd love to know it...</p>
<p>Edit: Can I solve it by integrating $\frac{1}{x^k}$ ? I can show it converges, but to find the formula? Is my question just the Riemann Zeta function?</p>
<p>(edit)</p>
<p>Thanks guys! This got me the following result:</p>
<p>$\sum_{p} \frac{1}{p} \log \frac{p}{p-1} ~ ~ \leq ~ \zeta(2)$</p>
<p>summing over all primes $p$. (And RHS is Riemann zeta function.)</p>
<p>First, sum over all integers $p$ instead of primes. Then transform the log into $\sum_{m=1}^{\infty} \frac{1}{m p^{m}}$ (<a href="http://en.wikipedia.org/wiki/Natural_logarithm#Derivative.2C_Taylor_series" rel="nofollow">reference: wikipedia</a>. I know.). Now we have (with rearranging):</p>
<p>$\leq ~ \sum_{m=1}^{\infty} \frac{1}{m} \sum_{p=2}^{\infty} \frac{1}{p^{m+1}}$</p>
<p>By the result of this question (Arturo's answer), this inner sum, which is $\zeta(m+1)$ is at most $\frac{1}{m+1-1} = \frac{1}{m}$. So we have</p>
<p>$\leq ~ ~ \sum_{m=1}^{\infty} \frac{1}{m} \frac{1}{m} = \zeta(2)$</p>
<p>I think this is a very pretty little proof. Thanks again, hope you math people enjoyed reading this....</p>
|
Qiaochu Yuan
| 232 |
<p>Yes, this sum converges, and yes, you can prove this using the integral test. For $k$ even the exact value (plus $1$) turns out to be a rational multiple of $\pi^k$ (see, for example, <a href="http://en.wikipedia.org/wiki/Bernoulli_number#Asymptotic_approximation" rel="nofollow">Wikipedia</a>), but for odd $k$ the exact values are much more mysterious. </p>
<p>The best way to find answers to these simple questions is to learn the basic techniques for solving them. In the context of discussing the convergence of series this means learning the basic convergence tests, as well as the knowledge of several specific examples such as might be covered in a typical calculus course. Beyond that, questions like this are hard to google for, so your best option is to ask them here! </p>
|
2,990,642 |
<p><span class="math-container">$\lim_{n\to \infty}(0.9999+\frac{1}{n})^n$</span></p>
<p>Using Binomial theorem:</p>
<p><span class="math-container">$(0.9999+\frac{1}{n})^n={n \choose 0}*0.9999^n+{n \choose 1}*0.9999^{n-1}*\frac{1}{n}+{n \choose 2}*0.9999^{n-2}*(\frac{1}{n})^2+...+{n \choose n-1}*0.9999*(\frac{1}{n})^{n-1}+{n \choose n}*(\frac{1}{n})^n=0.9999^n+0.9999^{n-1}+\frac{n-1}{2n}*0.9999^{n-2}+...+n*0.9999*(\frac{1}{n})^{n-1}+(\frac{1}{n})^n$</span></p>
<p>A limit of each element presented above is 0. How should I prove that limit of "invisible" elements (I mean elements in "+..+") is also 0?</p>
|
user
| 505,767 |
<p><strong>HINT</strong></p>
<p>We have that</p>
<p><span class="math-container">$$\left(0.9999+\frac{1}{n}\right)^n=(0.9999)^n\left(1+\frac{\frac{1}{0.9999}}{n}\right)^n$$</span></p>
|
1,651,922 |
<p>I'm solving this equation:</p>
<p>$$\sin(3x) = 0$$</p>
<p>The angle is equal to 0, therefore:</p>
<p>$$3x=0+2k\pi \space\vee\space3x= (\pi-0)+2k\pi$$ </p>
<p>$$x = \frac {2}{3}k\pi \space \vee \space x = \frac {\pi + 2k\pi}{3}$$</p>
<p>Though, the answer is </p>
<p>$$x = k\frac {\pi}{3}$$</p>
<p>It looks like the two trigonometric equations have been combined into one. I must have made a mistake. Any hints?</p>
|
fosho
| 166,258 |
<p>All we need is for $3x$ to be an integer multiple of $\pi$. In other words</p>
<p>$$3x = k\pi \Rightarrow x = \frac{k\pi}{3}$$</p>
|
1,651,922 |
<p>I'm solving this equation:</p>
<p>$$\sin(3x) = 0$$</p>
<p>The angle is equal to 0, therefore:</p>
<p>$$3x=0+2k\pi \space\vee\space3x= (\pi-0)+2k\pi$$ </p>
<p>$$x = \frac {2}{3}k\pi \space \vee \space x = \frac {\pi + 2k\pi}{3}$$</p>
<p>Though, the answer is </p>
<p>$$x = k\frac {\pi}{3}$$</p>
<p>It looks like the two trigonometric equations have been combined into one. I must have made a mistake. Any hints?</p>
|
StackTD
| 159,845 |
<p>You're not missing anything: sometimes it's possible to efficiently combine sets of solutions.</p>
<p>Notice that for $\sin x=0$, the solutions $x=0+2k\pi$ ($0$ and then 'adding full circles') $\vee \; x=\pi+2k\pi$ ($\pi$ and then 'adding full circles') can be combined as $x=k\pi$ ($0$ and 'adding <strong>half</strong> circles'), where always $k \in \mathbb{Z}$.</p>
<p>Draw the solutions and realise that you're not 'missing' anything: both ways of writing down the solutions contain the exact same angles; you 'run through' the same angles.</p>
<p><strong>Addendum</strong></p>
<p>This is not always possible for equations of the form $\sin x = c$ (only if $c=k\pi$) or $\cos x = c$ (only if $c=\pi/2+k\pi$), but it <em>is</em> always possible for $\tan x = c$ since the solutions
$$x = \arctan c + 2k\pi \, \vee x = \pi + \arctan c + 2k\pi$$can alwayes be combined as
$$x = \arctan c + k\pi$$
You can easily see this by drawing a trigonometric circle and visualising the solutions.</p>
|
708,633 |
<p>My question is as in the title: is there an example of a (unital but not necessarily commutative) ring $R$ and a left $R$-module $M$ with nonzero submodule $N$, such that $M \simeq M/N$?</p>
<p>What if $M$ and $N$ are finitely-generated? What if $M$ is free? My intuition is that if $N$ is a submodule of $R^n$, then $R^n/N \simeq R^n$ implies $N=0$. It seems like $N\neq 0$ implies $R^n/N$ has nontrivial relations, so $R^n/N$ can't be free.</p>
<p>If $R^n/N \simeq R^n$, we'd have an exact sequence</p>
<p>$0 \rightarrow N \hookrightarrow R^n \twoheadrightarrow R^n/N \simeq R^n \rightarrow 0$</p>
<p>which splits since $R^n$ is free, so $R^n \simeq R^n \oplus N$. Does this imply $N=0$? What if we assume $R$ is commutative, or even local? Maybe Nakayama can come in handy.</p>
<p>I'm interested in noncommutative examples too. Thanks!</p>
|
Mariano Suárez-Álvarez
| 274 |
<p>One keyword which should bring up many useful results: Leavitt álgebras.</p>
|
1,672,882 |
<p>Why we always see extended operations, like arbitrary unions, products, etc. in different parts of mathematics in the form of extensions of finite ones upon arbitrary <em>sets</em> (called index set)? Why never use an index-<em>class</em> for a proper class?</p>
<p>EDIT:
Of course I think this a reason, for example, that in the context of category theory, some theorems only hold for small/locally small categories: we are not allowed to operate upon arbitrary proper classes.</p>
|
Martín-Blas Pérez Pinilla
| 98,199 |
<p>The greatest operation that you will see is the power of the universe $V^A$ with $A$ set. This is a proper class, but the <em>elements</em> are sets: functions $f:A\longrightarrow V$. In the case $\text{set}^{\text{proper class}}$ the possible "elements" are proper classes.</p>
|
159,510 |
<p>I have data points of the form <code>{x,y,z}</code>:</p>
<pre><code>s = {{0, 1, 2}, {3, 4, 5}, {6, 7, 8}}
</code></pre>
<p>And I need to remove <em>entire data points</em> based on a condition. For example, let's say that I want to remove any data points where <code>z > 6</code>. The result should be this:</p>
<pre><code>s2 = {{0, 1, 2}, {3, 4, 5}}
</code></pre>
<p>How do I do this? I think <code>DeleteCases</code> might be the way to go, but I'm still fairly inexperienced with Mathematica and am not sure how to use this function to make this work.</p>
|
Jack LaVigne
| 10,917 |
<p>I copied the 10 lines in your question and pasted them into a text file called <strong><em>out.txt</em></strong>.</p>
<pre><code>data = Import["D:\\at_work\\mathematica\\stack_exchange\out.txt",
"text"]
</code></pre>
<p>The basic problem with the way you were working is that what appears to be numbers are strings and you have to convert them.</p>
<p>I did that as follows. First I split them into separate lines.</p>
<pre><code>dataN = StringSplit[data, EndOfLine]
</code></pre>
<p>This produces a list of strings of the form you indicated for each line, one per line.</p>
<p>Now I used <code>StringCases</code> on each of these lines converting the strings to numbers.</p>
<pre><code>data2 = Map[
StringCases[#,
"[" ~~ Whitespace ~~ x__ ~~ "," ~~ Whitespace ~~ y__ ~~ "," ~~
Whitespace ~~ z__ ~~ Whitespace ~~ "]" :>
Sequence[ToExpression[x], ToExpression[y], ToExpression[z]]] &,
dataN
]
(* {{-495.012, -158.36, 2705.}, {-489.156, -127.23,
2673.}, {-487.692, -97.6799, 2665.}, {-487.326, -68.4594,
2663.}, {-485.862, -39.1942, 2655.}, {-485.313, -10.1231,
2652.}, {-484.032, 18.8537, 2645.}, {-482.751, 47.6774,
2638.}, {-481.653, 76.3768, 2632.}, {-481.653, 105.185, 2632.}} *)
</code></pre>
<p>The output is a list of lists with three real numbers in each sub-list.</p>
<pre><code>ListPlot3D[data2]
</code></pre>
<p><img src="https://i.stack.imgur.com/bq7lm.png" alt="Mathematica graphics"></p>
|
159,510 |
<p>I have data points of the form <code>{x,y,z}</code>:</p>
<pre><code>s = {{0, 1, 2}, {3, 4, 5}, {6, 7, 8}}
</code></pre>
<p>And I need to remove <em>entire data points</em> based on a condition. For example, let's say that I want to remove any data points where <code>z > 6</code>. The result should be this:</p>
<pre><code>s2 = {{0, 1, 2}, {3, 4, 5}}
</code></pre>
<p>How do I do this? I think <code>DeleteCases</code> might be the way to go, but I'm still fairly inexperienced with Mathematica and am not sure how to use this function to make this work.</p>
|
george2079
| 2,079 |
<p>another approach.</p>
<pre><code>ImportString[
StringReplace[
Import["test.txt", "Text"], {"[", "]"} -> ""], "CSV"]
</code></pre>
|
1,129,052 |
<p>How many 3-digits numbers possess the following property: </p>
<blockquote>
<p>After subtracting $297$ from such a number, we get a $3$-digit number consisting of the same digits in the reverse order.</p>
</blockquote>
|
Pp..
| 203,995 |
<p>A $3$-digit number is of the form $$a_0+10a_1+100a_2\ \ \ \ \ \ \ \ \text{with }\ \ \ \ \ \ 0\leq a_0,a_1\leq9,\ \ \ \ \ \ 1\leq a_2\leq9$$
Then the condition to impose is $$a_0+10a_1+100a_2-297=a_2+10a_1+100a_0\ \ \ \ \ \text{ and }\ \ \ \ a_0\neq0$$
Therefore
$$99a_2-99a_0=297$$</p>
<p>or equivalently $$a_2-a_0=3$$</p>
<p>We get the solutions $$\begin{align}a_0&=1& a_2=4\\a_0&=2&a_2=5\\a_0&=3&a_2=6\\a_0&=4&a_2=7\\a_0&=5&a_2=8\\a_0&=6&a_2=9\end{align}$$</p>
<p>For each of these solutions for $a_0,a_2$, the $a_1$ can take any value $0,1,2,...,9$.</p>
|
1,129,052 |
<p>How many 3-digits numbers possess the following property: </p>
<blockquote>
<p>After subtracting $297$ from such a number, we get a $3$-digit number consisting of the same digits in the reverse order.</p>
</blockquote>
|
Community
| -1 |
<p>Following the perfect explanation given by Arian, there are a total of 70 solutions -- 60 of them are <em>true</em> solutions (> 390):</p>
<p>$$ \left[
\begin{array}{cccccccccc}
300&310&320&330&340&350&360&370&380&390\\
\hline
401&411&421&431&441&451&461&471&481&491\\
502&512&522&532&542&552&562&572&582&592\\
603&613&623&633&643&653&663&673&683&693\\
704&714&724&734&744&754&764&774&784&794\\
805&815&825&835&845&855&865&875&885&895\\
906&916&926&936&946&956&966&976&986&996\\
\end{array}
\right] $$</p>
|
4,285,448 |
<p>For training I have decided to solve this limit of a succession</p>
<p><span class="math-container">$$\lim\limits_{n\to \infty }\left(3^{n+1}-3^{\sqrt{n^2-1}}\right).$$</span></p>
<p><strong>My first attempt</strong>:</p>
<p><span class="math-container">\begin{split}
\lim_{n\to \infty }\left(3^{n+1}-3^{\sqrt{n^2-1}}\right) =\\
&=\lim_{n\to \infty } 3^{n+1}\left(1-\frac{3^{n\cdot \sqrt{1-\frac{1}{n^2}}}}{3\cdot 3^{n}}\right)\\
&=\ldots
\end{split}</span></p>
<p>I have abandoned this approach because I wanted to use a notable limit (if <span class="math-container">$\{b_n\}\to 0$</span>) that could probably be useful; but seeing the rounded brackets I have thought that I have occurred many times:</p>
<p><span class="math-container">$$\lim_{n\to \infty} \frac{a^{b_n}-1}{b_n}$$</span></p>
<p><strong>My second attempt</strong>:</p>
<p><span class="math-container">\begin{split}
\lim_{n\to \infty }\left(3^{n+1}-3^{\sqrt{n^2-1}}\right) =\\
&=\lim_{n\to \infty } 3^{\sqrt{n^2-1}}\left(\frac{3^{n+1}}{3^{\sqrt{n^2-1}}}-1\right)\\
&=\ldots
\end{split}</span></p>
<p>Considering only</p>
<p><span class="math-container">$$\frac{3^{n+1}}{3^{\sqrt{n^2-1}}}=3^{(n+1-\sqrt{n^2-1})}$$</span></p>
<p>Taking only the exponent I have:</p>
<p><span class="math-container">$$n+1-\sqrt{n^2-1}=\frac{((n+1)-\sqrt{n^2-1})\cdot ((n+1)+\sqrt{n^2-1})}{(n+1)+\sqrt{n^2-1}}$$</span>
<span class="math-container">$$=\frac{2n+2}{n+1+\sqrt{n^2+1}} \tag 1$$</span></p>
<p>If I take the limit of the <span class="math-container">$(1)$</span>, easily:</p>
<p><span class="math-container">$$\lim_{n\to \infty}\frac{2n+2}{n+1+\sqrt{n^2+1}}=1$$</span></p>
<p>i.e.</p>
<p><span class="math-container">$$\lim_{n\to \infty}3^{(n+1-\sqrt{n^2-1})}=3$$</span></p>
<p>Definitely I think that this limit is <span class="math-container">$\infty\cdot 2=+\infty$</span>.
I think that is correct. Do you users think there are better alternatives? Thanks in advance.</p>
|
Matsmir
| 685,805 |
<p>The obvious way to obtain <span class="math-container">$\infty$</span> as the limit:</p>
<p><span class="math-container">$$
3^{n+1} - 3^{\sqrt{n^2 - 1}} \ge 3^{n+1} - 3^{n} = 2(3^n) \rightarrow \infty.
$$</span>
Your second attempt seems correct but is more complicated.</p>
|
3,902,501 |
<p>Over <span class="math-container">$\mathbb R$</span>, the only linear maps are those of the form <span class="math-container">$ax$</span>.</p>
<p>If we discuss rational functions over <span class="math-container">$\mathbb R$</span>, this extra structure would allow us to describe a wider variety of linear maps.</p>
<p>But the obvious maps such as limits, differentiation, summation, doing <span class="math-container">$f(x)\mapsto f(x+k)$</span> seem too easy. Is there a linear map which specifically takes advantage of the fact that we have a rational function?</p>
<p>For instance, something like "double the coefficient of <span class="math-container">$x$</span> of the numerator and triple that of <span class="math-container">$x$</span> in the denominator". I know this is not a good example because it isn't linear, but it illustrates what I mean by "taking advantage of the fact that we have a rational function", i.e., something which doesn't generalise easily to a wider class of functions (say <span class="math-container">$C^1$</span>).</p>
|
GEdgar
| 442 |
<p>A example of a linear functional on <span class="math-container">$\mathbb R(X)$</span> ... write it in partial fraction form, and take the coefficient of <span class="math-container">$1/(X-3)^2$</span>. (Of course <span class="math-container">$0$</span> if that terms does not appear.)</p>
|
2,697,251 |
<p>In my c++ application I have a 3d plane, with a point above it. The plane is rotated, and the point is at 0,0,0. </p>
<p>In the image below, the plane is represented by the triangle side B - C. (S3)</p>
<p>A is the point that is at zero.
B is the mid-point of the plane.</p>
<p>What I need to calculate is the distance from Point A, to where it intersects the plane (Point C).</p>
<p>I know the angle of the plane, and the coordinates of point A and Point B.</p>
<p>Point A is at (xyz) 0,0,0;
Point B is at: -1,2.7,-19.7;</p>
<p>From this, i know the length of Side S2 (19.7).
The angle of Angle B is 17.675 degrees.</p>
<p>So I now have one side and one angle. I know that the corner at point C is 90 degrees. Is this enough to get me the length of the the side S1?</p>
<p><a href="https://i.stack.imgur.com/jFdnP.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/jFdnP.jpg" alt="enter image description here"></a></p>
|
user
| 505,767 |
<p>Yes we can, indeed recall that</p>
<ul>
<li>$S_3=S_2 \cos B$</li>
<li>$S_1=S_2 \sin B$</li>
</ul>
<p><a href="https://i.stack.imgur.com/5WWvx.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/5WWvx.gif" alt="enter image description here"></a></p>
|
2,697,251 |
<p>In my c++ application I have a 3d plane, with a point above it. The plane is rotated, and the point is at 0,0,0. </p>
<p>In the image below, the plane is represented by the triangle side B - C. (S3)</p>
<p>A is the point that is at zero.
B is the mid-point of the plane.</p>
<p>What I need to calculate is the distance from Point A, to where it intersects the plane (Point C).</p>
<p>I know the angle of the plane, and the coordinates of point A and Point B.</p>
<p>Point A is at (xyz) 0,0,0;
Point B is at: -1,2.7,-19.7;</p>
<p>From this, i know the length of Side S2 (19.7).
The angle of Angle B is 17.675 degrees.</p>
<p>So I now have one side and one angle. I know that the corner at point C is 90 degrees. Is this enough to get me the length of the the side S1?</p>
<p><a href="https://i.stack.imgur.com/jFdnP.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/jFdnP.jpg" alt="enter image description here"></a></p>
|
Mohammad Riazi-Kermani
| 514,496 |
<p>Yes, the right triangle is uniquely determined by its hypotenuse and an angle $\alpha.$</p>
<p>The other two sides are $$ BC=AB cos( \alpha) \\AC= AB sin( \alpha)$$ </p>
|
2,624,272 |
<p>It is well known that, in order to track constant reference values with zero asymptotic error, in the presence of uncertainty, an integrator is required in the open loop (either by the plant or the controller itself).</p>
<p>Lets for simplicity assume the plant $P$ has no integral components.</p>
<p>In order to track constants, $r(t) = 1,$ one integrator in the controller is needed. Tracking ramps, $r(t) = t$, needs two integrators, while tracking parabolas, $r(t) = t^2$ needs 3 integrators, and so on.</p>
<p>So, generally speaking, in order to track $r(t) = t^n$ with zero asymptotic error for $P$, you need $n + 1$ integrators in the controller.</p>
<p>However, we also know that the exponential function grows faster than any polynomial. So, what is for references like $r(t) = \exp(t)$? Wouldn't it then require an <em>infinite</em> amount of integrators in order to track such a reference with zero asymptotic error? Is that even possible at all?</p>
|
Kwin van der Veen
| 76,466 |
<p>The reason why this can be done in the first place can be derived from the <a href="https://en.wikipedia.org/wiki/Final_value_theorem" rel="nofollow noreferrer">final value theorem</a>. For this one needs the Laplace transform of the tracking error, which can be obtained by multiplying the sensitivity transfer function by the Laplace transform of the reference</p>
<p>$$
\lim_{t\to\infty} e(t) = \lim_{s\to 0} s\,\frac{1}{1+C(s)\,G(s)}R(s).
$$</p>
<p>When considering the limit then $C(s)\,G(s)$ can just be simplified to their total number of integrators, denoted by $m$. Of course this result only hold if the closed loop is stable, so just picking a few integrators for $C(s)$ wouldn't be a good controller. Using the Laplace transform of $t^n$ <a href="http://tutorial.math.lamar.edu/pdf/Laplace_Table.pdf" rel="nofollow noreferrer">from here</a> and substituting this all into the first equation yields</p>
<p>$$
\lim_{t\to\infty} e(t) = \lim_{s\to 0} s\,\frac{s^m}{s^m + 1}\frac{n!}{s^{n+1}} = n!\,\lim_{s\to 0} \frac{s^{m-n}}{s^m + 1}.
$$</p>
<p>When $m<n$ this limit is undefined (this corresponds to the error blowing up to infinity), when $m=n$ the error goes to $n!$ and when $m>n$ the error goes to zero.</p>
<p>Changing the reference to $e^{a\,t}$, with $a$ some positive real number, would change its Laplace transform to $(s-a)^{-1}$. Now you might say that plugging this into the limit will always yields zero. However the theorem can only be used when the Laplace transform only has poles in the <a href="http://mathworld.wolfram.com/LeftHalf-Plane.html" rel="nofollow noreferrer">open left-half plane</a> (OLHP) and the only reason it can be used for polynomials (whose Laplace transforms poles lie outside the OLHP) is because of pole-zero cancellation.</p>
<p>So from this it can be shown that for a reference $e^{a\,t}$ it is impossible to have a zero steady state tracking error when using a finite number of integrators in the feedback loop. Instead feedforward could be used to achieve tracking and use feedback as disturbance rejection. However you stated that you have (model) uncertainty, in which case simple feedforward would not be a good option. In that case you might have to resort to some adaptive control for the feedforward.</p>
|
1,558,665 |
<p>I am looking for examples of finitely generated solvable groups that are not polycyclic. In <a href="http://groupprops.subwiki.org/wiki/Finitely_generated_and_solvable_not_implies_polycyclic#Some_examples_based_on_the_general_construction_and_otherwise" rel="nofollow">Wikipedia</a> Baumslag-Solitar group $BS(1,2)$ is an example. But how to prove this fact?</p>
|
anomaly
| 156,999 |
<p>The Baumslag-Solitar group $BS(1, 2) = \mathbb{Z} \ltimes \mathbb{Z}[\frac{1}{2}]$ under the isomorphism $a \to (0, 1), b \to (1, 0)$, where the generator $g$ of the left term acts by $g.1 = 2$. But $\mathbb{Z}[\frac{1}{2}]\subset BS(1, 2)$ is clearly not finitely generated, so $BS(1, 2)$ is not polycyclic.</p>
|
3,702,309 |
<p>The question reads,
Find all differentiable functions <span class="math-container">$f$</span> such that <span class="math-container">$$f(x+y)=f(x)+f(y)+x^2y$$</span> for all <span class="math-container">$x,y \in \mathbb{R} $</span>. The function <span class="math-container">$f$</span> also satisfies <span class="math-container">$$\lim_{x \rightarrow 0}\frac {f(x)}x = 0$$</span></p>
<p>To solve the problem I wrote the expression for <span class="math-container">$\frac{df}{dx}$</span> using first principle and found that <span class="math-container">$\frac{df}{dx} = x^2.$</span>(Due to the given conditions.)</p>
<p>Using this and calculating <span class="math-container">$f(0)$</span> as zero I got the implication that <span class="math-container">$f(x)$</span> must be <span class="math-container">$\frac{x^3}{3}.$</span></p>
<p>But,
Clearly the calculated <span class="math-container">$f(x)$</span> does not satisfy the required condition for the problem. Thinking about where I had committed the mistake, I realized that <span class="math-container">$f(x+y)=f(x)+f(y)+x^2y$</span> cannot be true for all real <span class="math-container">$x,y$</span>. </p>
<p>So how is it that these operations that implied <span class="math-container">$f(x)=\frac{x^3}{3}$</span> went wrong?</p>
<p>And why is it that while doing these operations I could not 'see' that the conditions cannot be satisfied for all real <span class="math-container">$x$</span>?</p>
<p>EDIT:Reading the comments, I want to make the clarification that I realise the fact that there can not be any <span class="math-container">$ f(x) $</span> that satisfies these conditions using a different method. However I fail to understand why the method elaborated in the question does not reflect this fact to me??</p>
|
Kavi Rama Murthy
| 142,385 |
<p>There cannot be any function satisfying this equation for the simple reason that <span class="math-container">$f(x+y)-f(x)-f(y)$</span> does not change if you interchange <span class="math-container">$x$</span> and <span class="math-container">$y$</span>. So if such a function exists we must have <span class="math-container">$x^{2}y=y^{2}x$</span> for all <span class="math-container">$x,y \in \mathbb R$</span> which is absurd.</p>
<p>Now as to what went wrong in your approach: From <span class="math-container">$x^{2} =-1$</span> you can deduce that <span class="math-container">$x^{4}= 1$</span> but <span class="math-container">$x^{4} =1$</span> is not the final answer. In fact there is no real number <span class="math-container">$x$</span> such that <span class="math-container">$x^{2} =-1$</span>. The simple answer to your question is converse of result is not always true. If you got <span class="math-container">$f(x)=\frac {x^{3}} 3$</span> from the given equation it doesn't mean that there is a solution. You should always go back to the original equation and check if the function you obtained is indeed a solution. If it is not then you have not arrived at any contradiction because converse of result is not always true.</p>
|
1,099,214 |
<p>I'm having trouble calculating some probabilities for a simple card game I'm into </p>
<p>1) So say we have a deck of 13 cards total. You draw 3 cards so there are 10 cards left in the deck. Say there is 1 special card in the deck that we want to figure out the probability of getting it in the hand. How do we figure it out? Would it be:</p>
<p>$$ \dbinom{1}{1} * \dbinom{12}{2} / \dbinom{13}{3} $$</p>
<p>1 choose 1 for the special card. 12 choose 2 for the other 2 cards which we don't care what they are. I was thinking I might have to multiply by 3! because there are 3! ways to order the 3 cards in the hand but that would make the probability greater than 1.</p>
<p>2) Now let's say in the deck there is that 1 special card, and also 3 copies of a different type of card that we want. How would we calculate the probability of forming a hand that contains the 1 special card AND 1 or more of any of the 3 copies of the 2nd type of card? </p>
<p>3) Now let's say we have a deck with 2 copies of card type A, and 2 copies of card type B. How would we calculate the probability of choosing 1 or more from type A AND 1 or more from type B assuming a 13 card deck where we draw 3 cards? (for example: 1 type A + 1 type B + 1 any other card, 1 type A + 2 type B, etc).</p>
<p>I remember doing math like this in high school and it being really basic but don't quite remember exactly how to solve them. Thanks!</p>
|
BCLC
| 140,308 |
<p>Looks right.</p>
<p>"I was thinking I might have to multiply by 3! because there are 3! ways to order the 3 cards in the hand " --> No need. Order doesn't matter in this case.</p>
<p>"Now let's say in the deck there is that 1 special card, and also 3 copies of a different type of card that we want. How would we calculate the probability of forming a hand that contains the 1 special card AND 1 or more of any of the 3 copies of the 2nd type of card?"</p>
<p>1 special card, 1 of 2nd type</p>
<p>$\binom{1}{1} \binom{3}{1} \binom{9}{1} / same$</p>
<p>1 special card, 2 of 2nd type</p>
<p>$\binom{1}{1} \binom{3}{2} \binom{9}{0} / same$</p>
<p>1 special card, 3 of 2nd type</p>
<p>impossible since we draw 3</p>
<p>"Now let's say we have a deck with 2 copies of card type A, and 2 copies of card type B. How would we calculate the probability of choosing 1 or more from type A AND 1 or more from type B assuming a 13 card deck where we draw 3 cards? (for example: 1 type A + 1 type B + 1 any other card, 1 type A + 2 type B, etc)."</p>
<p>X type A, Y type B</p>
<p>$\binom{2}{X} \binom{2}{Y} \binom{10}{3-X-Y} / same$</p>
<p>where X = 1 or 2, Y = 1 or 2 and X + Y $\leq 3$</p>
|
687,897 |
<p>I need help with a differential equation, the trouble is I don't think it's separable and I have tried and failed to apply the method of characteristics to figure it out. z is also bound between zero and one.</p>
<p>$$
\frac{\partial u}{\partial x}+z\frac{\partial u}{\partial y}-Cz\frac{\partial u}{\partial z}=0
$$</p>
<p>The problem with characteristics is I couldn't figure out how to apply the boundary conditions.</p>
<p>$$
u(0,y,z)=\delta(y)\delta(z-1)
$$</p>
<p>Any and all help would be greatly appreciated.</p>
|
user127096
| 127,096 |
<p>Parametrized by $x$, characteristic curves have equation
$$x=x,\quad y = y_0+\frac{z_0}{C}(1-e^{-Cx}),\quad z = z_0 e^{-Cx} \tag{1}$$
where $(0,y_0,z_0)$ is the point of crossing the plane $x=0$. To check this, differentiate (1) with respect to $x$:
$$x'=1,\quad y' =z_0 e^{-Cx} = z ,\quad z' = -C z_0 e^{-Cx} = -Cz \tag{2}$$
The PDE says that $u$ is constant along each curve with equation (1). This implies that $u=0$ outside of the characteristic curve that begins at $y_0=0$, $z_0=1$, which is
$$x=x,\quad y = \frac{1}{C}(1-e^{-Cx}),\quad z = e^{-Cx} \tag{3}$$
Along the curve (3), $u$ is something quite singular. </p>
<p>I am inclined to suspect that this PDE+BC problem does not capture the phenomenon you were trying to describe.</p>
|
238,673 |
<p><a href="http://vihart.com/doodling/" rel="nofollow">Vi Hart's doodling videos</a> and a 4 year old son interested in mazes has made me wonder:</p>
<p>What are some interesting mathematical "doodling" diversions/games that satisfy the following criteria:</p>
<p>1) They are "solitaire" games, i.e. require only one player;
2) They require only a pencil and blank sheet of paper;
3) They don't rely on abstract mathematical language/substitutions. More precisely, I'm interested in "picture" type games with very simple rules, and not the cheeky answer: "mathematics".</p>
<p>What I'd really like to see is a game with simple rules that, out of the planarity of a sheet of paper, somehow "generates" a maze for the player from the "automaton" type rules...A good starting point may be a solitaire version of dots-and-boxes or something like this...</p>
<p>If such a thing is impossible, under certain simple desirable assumptions, I'd like to see proofs of such a fact, too.</p>
|
Egor Skriptunoff
| 50,643 |
<p>@Karolis Juodelė<br>
... or emulate <a href="http://en.wikipedia.org/wiki/Busy_beaver" rel="nofollow">4-state, 2-symbol busy beaver</a> :-)</p>
|
1,038,060 |
<p>Can anyone help me with this question: I know it before, but I have tried to solve it myself and didnt succeed.
what is the regular expression for this language: L=all words that have 00 or 11 but not both.</p>
<p>Thank you!</p>
|
student
| 379,844 |
<p>Assume that $A_n$ is a simple group for $n\neq 4$. If $H$ is any subgroup of index $2$ in $S_n$, then $H$ must be normal in $S_n$ (because the only non-trivial coset in $S_n/H$ must be the set-theoretic complement of $H$, and must therefore be both the right and left coset obtained by multiplying on the right, or the left by any particular $g\notin H$), and so $(H\cap A_n)$ must be a normal subgroup of $A_n$. Since $A_n$ is simple for $n\neq 4$, it follows that $(H\cap A_n)$ is either equal to $\{1\}$ (the subgroup of $S_n$ consisting only of the identity element) or $A_n$ itself, these two possibilities being mutually exclusive. If $(H\cap A_n)=\{1\}$, then ${\it every}$ element of $H$ (except for the identity permutation) must be an odd permutation. Furthermore, since $H$ has index $2$ in $S_n$, $|H|={\frac{n!}{2}}> 2$ for $n\geq 3$. Let $\sigma$ and $\tau$ be any two distinct odd permutations in $H$. Then, since ${\sigma}^2$, and ${\sigma}{\tau}$ are both even permutations in $H$, we would be forced to conlcude that both are equal to the identity permutation which is the only even permutation in $H$ by hypothesis. This in turn forces us to conclude that $\sigma = \tau$, which is a contradiction. Thus, $(H\cap A_n) = A_n$. In particular, $A_n\subset (H\cap A_n)$ and as both $A_n$ and $H$ have the same cardinality, they must be equal.</p>
|
136,808 |
<p>Let $A$ be a nonempty subset of $\omega$, the set of natural numbers.
I want to prove this statement:</p>
<blockquote>
<p>If $\bigcup A=A$ then $n\in A \implies n^+\in A$.</p>
</blockquote>
<p>Help...</p>
|
lesnikow
| 30,013 |
<p>Oh yes, this can be pretty confusing! Here is how I think of it:</p>
<p>$\bigcup A = A$ really means $\bigcup A \subseteq A$ and $\bigcup A \supseteq A$.</p>
<p>Now assume $a \in A$. From $A \subseteq \bigcup A$, we get $a \in \bigcup A$,
which, looking at the definition of $\bigcup A$, means that $a$ is an element of some element of $A$, say $b$. That is, $a \in b \in A$. But $b$ is a natural number, and since $a \in b$, $b$ is a $strictly$ $bigger$ natural than $a$. If $b = a^+$, we have $a^+ \in A$, and we are done. Otherwise, $a^+$ is an element of $b$. But then from $\bigcup A \subseteq A$, we get $a^+ \in A$ again. </p>
<p>And looking at the proof, we see that $A$ must have actually been $\omega$ all along! </p>
|
2,220,105 |
<p>Hatcher makes the following definition in exercise 7 on page 258. </p>
<p>For a map $f: M \rightarrow N $ between connected, closed, orientable $n$-manifolds with fundamental classes $[M]$ and $[N]$, the <em>degree</em> of $f$ is defined to be the integer $d$ such that $f_{*}[M]=d[N]$ (so the sign of the degree depends on the choice of fundamental classes).</p>
<blockquote>
<p>I'm confused as to why $f_{*}[M]$ has be of the form $d[N]$ for arbitrary coefficient ring $R$. I know that $[M]$ will be a generator in the group that it lies in, namely $H_{n}(M;R) \approx R$ and similarly $[N]$ is a generator for $H_{n}(N,R) \approx R$. I showed that for any isomorphism $f:R\rightarrow S$ a generator of $R$ must map to a generator of $S$, where my definition of generator for $R$ is an element $u$ such that $Ru=R$. But with this I still can't see why $f_{*}[M]=d[N]$. </p>
</blockquote>
|
Doug M
| 317,162 |
<p>This sort of sorting problem is called pancake sorting. Bill Gates made important contributions to the pancake sorting problem. Then he dropped out of school. He did all right for himself, though.</p>
<p>Base case:</p>
<p>$n = 2$</p>
<p>Either the sequence is already sorted, or it needs to be inverted and can be sorted with one inversion.</p>
<p>Inductive hypothesis:</p>
<p>Suppose a list of $n$ elements can be pancake-sorted in $2n-3$ (or fewer) flips.</p>
<p>We will need to show that a list of $n+1$ elements can be sorted in $2n-1$ (or fewer) flips.</p>
<p>With the first inversion, move the $(n+1)^{th}$ element to the front of the list.
With the second inversion, invert the entire list, moving the $(n+1)^{th}$ element to the $(n+1)^{th}$ place.</p>
<p>Now you have $n$ elements still to sort.
By the inductive hypothesis we can sort a list of n elements in $2n-3$ inversions or fewer.</p>
<p>We have shown that a list of $n+1$ elements can be sorted in $2n-1$ (or fewer) flips.</p>
<p>QED</p>
|
1,745,832 |
<p>It is stated in Kolmogorov & Fomin's <em>Introductory Real Analysis</em> that every normal space is Hausdorff. I cannot seem to find an explanation for this anywhere, and don't see why this is obviously true... since it is not necessarily the case in an arbitrary topological space that a singleton is closed. </p>
<p>Any help in understanding this would be appreciated! </p>
|
DanielWainfleet
| 254,665 |
<p>It is taken as part of the def'n of a normal space that it has the $T_1$ property, and it follows that a normal space is $T_2$. Engelking, in General Topology, calls a space in which disjoint closed subsets are completely separated a pseudo-normal space, as the $T_1$ property may fail. The simplest examples of a non-normal pseudo-normal space are (1) the topology $\{\phi, S\}$ on any $S$ with at least $2$ members, and (2) Sierpinski space : the topology $\{\phi,S,\{0\}\}$ on $S=\{0,1\}.$</p>
|
991,530 |
<p>I am trying to prove the following:
For any positive real number $x>0$, there exist a positive integer N such that $x>1/N>0$. </p>
<p>What I know is:
since x is a real number, it can be expressed as the formal limit of a Cauchy sequence
$$x:=LIM (b_n)_{n=1}^\inf$$
since $x>0$, the sequence $(b_n)$ is bounded away from zero, i.e. $|b_n|>c$, $c>0$
Also, since $(b_n)$ is Cauchy then its bounded by M ($|b_n|\leq M$)</p>
<p>However, I dont know how to further proceed. Would appreciate your help. </p>
|
Community
| -1 |
<p><strong>Hint:</strong>
$$x>0$$ There will be infinitely many positive integers $A$ such that $$A>\frac{1}{x}.$$</p>
|
1,485,463 |
<p>At a school function, the ratio of teachers to students is $5:18$. The ratio of female students to male students is $7:2$. If the ratio of the female teachers to female students is $1:7$, find the ratio of the male teachers to male students. </p>
|
eyqs
| 280,815 |
<p>Let's say that there are $T_f$ female teachers, $T_m$ male teachers, $S_f$ female students, and $S_m$ male students. You are given that,</p>
<p>$$
(T_f + T_m) / (S_f + S_m) = 5/18\\
S_f / S_m = 7/2\\
T_f / S_f = 1/7
$$</p>
<p>And what you want to find is,</p>
<p>$$
T_m / S_m
$$</p>
<p>You can solve this now by substituting in the things you already know,</p>
<p>$$
T_f = 1/7 \times S_f \\
S_m = 2/7 \times S_f \\
T_m = 5/18 \times (S_f + S_m) - T_f
$$</p>
<p>So,</p>
<p>$$
\begin{aligned}
T_m/S_m &= T_m / S_f \times S_f / S_m \\
&= (5/18 \times (S_f + S_m) - T_f) / S_f \times 7/2 \\
&= (5/18 \times (S_f + 2/7 \times S_f) - 1/7 \times S_f) / S_f \times 7/2 \\
&= (5/18 \times (1 + 2/7) - 1/7) \times 7/2 \\
&= 3/4
\end{aligned}
$$</p>
|
113,960 |
<p>When trying to compute the (Serre-generalized) intersection number of two varieties at a closed point, I came to a need to compute the following $\operatorname{Tor}$:</p>
<blockquote>
<p>Let $k$ be an algebrically closed field, $A=k[x_1,x_2,x_3,x_4]$ and $\mathfrak m=(x_1,x_2,x_3,x_4)$. Let $M = k[x_1,x_2,x_3,x_4]/(x_1x_3,x_1x_4,x_2x_3,x_2x_4)$, $N=k[x_1,x_2,x_3,x_4]/(x_1-x_3,x_2-x_4)$. I want to compute $\operatorname{Tor}^i_{A_{\mathfrak m}}(M_{\mathfrak m},N_{\mathfrak m})$.</p>
</blockquote>
<p>Any ideas how to do this?</p>
<p>I first noted that $N$ is gotten from $A$ by quotient by a a regular sequence, so that the Koszul complex of $(x_1-x_3,x_2-x_4)$ is a free resolution of $N$ over $A$. However, tensoring with $M$ computations became too hard, and I was not able to find the cohomology of the resulting complex.</p>
<p>Any ideas?</p>
<p>Thanks!</p>
|
Mariano Suárez-Álvarez
| 274 |
<p>You can do this computation without thinking much, really.</p>
<p>The two generators of the ideal which defines $N$ form a regular sequence, so you can use a Koszul complex to projectively resolve $N$. Tensor it with $M$ and just compute the homology of the resulting complex.</p>
|
1,412,091 |
<p>The typewriter sequence is an example of a sequence which converges to zero in measure but does not converge to zero a.e.</p>
<p>Could someone explain why it does not converge to zero a.e.?</p>
<blockquote>
<p><span class="math-container">$f_n(x) = \mathbb 1_{\left[\frac{n-2^k}{2^k}, \frac{n-2^k+1}{2^k}\right]} \text{, where } 2^k \leqslant n < 2^{k+1}.$</span></p>
</blockquote>
<p>Note: the <a href="http://terrytao.wordpress.com/2010/10/02/245a-notes-4-modes-of-convergence/" rel="nofollow noreferrer">typewriter sequence</a> (Example 7).</p>
|
Ben Grossmann
| 81,360 |
<p>Note that at any choice of $x$ and for any integer $N$, there is an $n>N$ with $f_n(x)=1$. So, the numerical sequence $f_n(x)$ cannot converge to $0$.</p>
<p>Note, however, that we can certainly select a <em>subsequence</em> of this sequence of functions that converges pointwise a.e.</p>
|
3,839,244 |
<p>Is <span class="math-container">$\{3\}$</span> a subset of <span class="math-container">$\{\{1\},\{1,2\},\{1,2,3\}\}$</span>?</p>
<p>If the set contained <span class="math-container">$\{3\}$</span> plain and simply I would know but does the element <span class="math-container">$\{1,2,3\}$</span> include <span class="math-container">$\{3\}$</span> such that it would be a subset?</p>
|
user
| 505,767 |
<p>No it is not, we have that <span class="math-container">$\{1\}\in \{\{1\},\{1,2\},\{1,2,3\}\}$</span> but <span class="math-container">$\{3\}$</span> is not an element of the set, what is true is that <span class="math-container">$3\in\{1,2,3\}$</span>.</p>
|
1,935,320 |
<p>If $a^2-b^2=2$ then what is the least possible value of: \begin{vmatrix} 1+a^2-b^2 & 2ab &-2b\\ 2ab & 1-a^2+b^2&2a\\2b&-2a&1-a^2-b^2 \end{vmatrix}</p>
<p>I tried to express the determinant as a product of two determinants but could not do so. Seeing no way out, I tried expanding it but that took too long and was difficult to evaluate. Please help me with this one, thanks.</p>
|
MJD
| 25,554 |
<p>Do you know that 2×2 matrices correspond to linear transformations of the plane, and that composing the transformations multiplies the corresponding matrices?</p>
<p>Can you think of a linear transformation of the plane which, repeated three times, is the identity transformation?</p>
<p>Mouse over for hint:</p>
<blockquote class="spoiler">
<p> How about a rotation?</p>
</blockquote>
|
599,656 |
<p>If the objects of a category are algebraic structures in their own right, this often places additional structure on the homsets. Is there somewhere I can learn more about this general idea?</p>
<hr>
<p><strong>Example 1.</strong> Let $\cal M$ denote the category of magmas and <strong>functions</strong>, not necessarily structure-preserving. Then given objects $X$ and $Y$ of $\cal M$, the hom"set" $\mathrm{Hom}(X,Y)$ is probably best viewed as a <em>magma</em> as opposed to a set, with the pointwise operation inherited from $Y$. Furthermore we have a right (but not left) distributivity law. In particular, given objects $X, Y$ and $Z$ and arrows $f:X \rightarrow Y$ and $g,g' : Y \rightarrow Z,$</p>
<p>$$(g+g')\circ f = (g \circ f) + (g' \circ f).$$</p>
<p><strong>Example 2.</strong> Now let $\cal N$ denote the category of <a href="http://en.wikipedia.org/wiki/Medial_magma" rel="nofollow">medial</a> magmas and magma <strong>homomorphisms</strong>. Then given objects $X$ and $Y$ of $\cal N$, the hom"set" $\mathrm{Hom}(X,Y)$ can be viewed as a medial magma in its own right. Furthermore, we have both left and right distributivity laws. In the sense that given objects $X, Y$ and $Z$ and arrows $f,f':X \rightarrow Y$ and $g,g' : Y \rightarrow Z,$</p>
<p>$$(g+g')\circ f = (g \circ f) + (g' \circ f)$$</p>
<p>$$g \circ (f + f') = (g \circ f) + (g \circ f').$$</p>
|
mainak
| 114,270 |
<p>obviously 8 is one of the solutions with no doubts at all.</p>
|
2,101,894 |
<p>As part of a larger proof, I need to use the fact that for a homomorphism $\phi : A \rightarrow B$, where $A$ is cyclic and $a \in A$ its generator, $\phi(a^{-n}) = \phi(a^n)^{-1}$. All I came up with is that $\phi(a^{-n}) = \phi((a^n)^{-1})$ and then I hit a wall. Is it true in general that $\phi(a^{nm}) = \phi(a^n)^m$, even for $m < 0$?</p>
<p>I'm also not sure if it's important that $A$ is cyclic and $a$ is its generator.</p>
<p>Thanks in advance for any help.</p>
|
Mustafa
| 400,050 |
<p>because $\phi$ is homomorphism group then
$\phi ( (a^n)^{-1}))=\phi((a.a...a)^{-1})=\phi(a^{-1}...a^{-1})=\phi(a^{-1})...\phi(a^{-1})= \phi(a)^{-1}...\phi(a)^{-1}=(\phi(a)...\phi(a))^{-1}=(\phi(a^n))^{-1} $</p>
|
2,992,477 |
<p>Can u help me prove that every simple module is cyclic? Ive already proved that a cyclic module is isomorphic to A quotiented by I ( I being a left ideal of A) using homomorphisms, but i cant seem to make any advance in proving this.Thx in advance.</p>
|
sandoh
| 1,083,786 |
<p>Let M be a simple module. Then the only submodules of M are the trivial submodule and M itself.
Suppose M is not cyclic, that is, it is not generated by any of its elements. Because M is a module, we know it is an additive subgroup and so the sets generated by each of its non-zero elements will form a subgroup of M but this contradicts our assumption that the only submodules of M are the trivial submodule and M itself</p>
|
548,563 |
<p>Calculate $$\sum \limits_{k=0}^{\infty}\frac{1}{{2k \choose k}}$$</p>
<p>I use software to complete the series is $\frac{2}{27} \left(18+\sqrt{3} \pi \right)$</p>
<p>I have no idea about it. :|</p>
|
Ron Gordon
| 53,268 |
<p>Consider the function</p>
<p>$$f(x) = \frac{\arcsin{x}}{\sqrt{1-x^2}}$$</p>
<p>$f(x)$ has a Maclurin expansion as follows:</p>
<p>$$f(x) = \sum_{n=0}^{\infty} \frac{2^{2 n}}{\displaystyle (2 n+1) \binom{2 n}{n}} x^{2 n+1}$$</p>
<p>Differentiating, we get</p>
<p>$$f'(x) = \frac{x \, \arcsin{x}}{(1-x^2)^{3/2}} + \frac{1}{1-x^2} = \sum_{n=0}^{\infty} \frac{2^{2 n}}{\displaystyle \binom{2 n}{n}} x^{2 n}$$</p>
<p>Evaluate at $x=1/2$</p>
<p>$$\sum_{n=0}^{\infty} \frac{1}{\displaystyle \binom{2 n}{n}} = \frac{\frac12 \arcsin{\frac12}}{3 \sqrt{3}/8} + \frac{4}{3} = \frac{2\sqrt{3} \pi+36}{27}$$</p>
<p><strong>ADDENDUM</strong></p>
<p>There are many derivations here of the above result for the Maclurin series for $f(x)$; I refer you to <a href="https://math.stackexchange.com/questions/534736/evaluate-this-power-series/534756#534756">this one</a>.</p>
|
3,450,692 |
<p>How to solve this equation?</p>
<p><span class="math-container">$$
\frac{dy}{dx}=\frac{x^{2}+y^{2}}{2xy}
$$</span></p>
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emonHR
| 565,609 |
<p><span class="math-container">$
\frac{dy}{dx}=\frac{x^{2}\left(1+\frac{y^2}{x^{2}}\right)}{2xy}=\frac{1+\left(\frac{y}{x}\right)^2}{2\frac{y}{x}}
$</span> Can you go from here<span class="math-container">$?$</span></p>
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2,238,830 |
<p>I have just come across "Foundations of Mathematics". Wikipedia describes it as </p>
<blockquote>
<p>Foundations of mathematics can be conceived as the study of the basic
mathematical concepts (number, geometrical figure, set, function,
etc.) and how they form hierarchies of more complex structures and
concepts, especially the fundamentally important structures that form
the language of mathematics (formulas, theories and their models
giving a meaning to formulas, definitions, proofs, algorithms, etc.)
also called metamathematical concepts, with an eye to the
philosophical aspects and the unity of mathematics.</p>
</blockquote>
<p>Now this seems a bit strange to me. If foundations of mathematics really do deal with the study of basic mathematical concepts shouldn't they be taught to us as the first thing ? Also in theory wouldn't this mean that someone who has a good grasp of foundations of mathematics will have a very good understanding of mathematics as a whole ?</p>
<p>If that is true then why isn't foundation of mathematics taught at a earlier stage in life and more often ?</p>
<p>I apologize for the over simplification, but I am not an ace mathematician. Hence I want some insight into what foundations of mathematics are. </p>
|
J.G.
| 56,861 |
<p>In ancient Greece, Euclid proposed an idea mathematicians have liked (in principle) ever since: have assumptions called "axioms" we take for granted, then prove "theorems" from them. In theory, you could write a list of all the axioms of modern mathematics, then take it from there. Euclid had a few axioms of geometry, as if to show us how it's done.</p>
<p>However, until c. 1900 mathematicians rarely took that approach. They were very careful about the way results were proven, but pretty much intuited the starting points. Nowadays, they can say, "Number theory has the Peano axioms" or "All mainstream mathematics has a model in ZFC" (a version of set theory), or something like that; but such axioms were designed to capture what we were already doing, and today when you teach people these axioms it's the familiarity of the implications that makes them palatable.</p>
<p>There was actually a teach-foundations-to-children experiment called the <a href="https://en.wikipedia.org/wiki/New_Math" rel="nofollow noreferrer">New Math</a>, but it went badly. The basic problem was that, even though theoretically you'd think the easiest way to understand the subject is to start from axioms then see where they lead, in practice that's not a particularly intuitive way to do it, if only because you need some quite heavy-handed machinery even to state the starting points when they're done properly. For example, how do you found the idea of real numbers as a continuum? One option is as Dedekind cuts on rational numbers; another is as equivalence classes of Cauchy sequences of rational numbers. Yeah: good luck explaining that in primary school, even if rational numbers have been explained or taken for granted.</p>
<p>We've had much more success teaching mathematics a very different way that more closely (though not exactly) parallels the history of ideas' development. First you teach natural numbers, then you introduce fractions, then you get to surds and complex numbers and so on. (I'm skipping topics other than "types of numbers" to simplify the timetable.) The reason this works, presumably, is because each level of the resulting education provides the right context to understand the problems that motivated the next development in mathematical theory. But to motivate the foundation of mathematics the same way, it turns out you have to natter about some very abstract and complicated ideas like consistency and decidability.</p>
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