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1,397,646
<blockquote> <p>Given $f_n:= e^{-n(nx-1)^2} $, any suggested approaches for showing that $\displaystyle \lim_{n \to \infty}\int^1_0 f_n(x)dx = 0$? </p> </blockquote> <p>I have already shown that $f_n \to 0$ pointwise (and not uniformly) on $[0,1]$ and have been trying to show that $$ \lim_{n \to \infty}\int^1_0 f_n(x)dx = 0$$ I know that $ \forall n \in \mathbb {N}, $ $\exists x_n = \frac {1}{n} \in [0,1]$ such that $f_n(x_n) = 1$, which has me thinking of doing something where I split the integral in such a way that isolates the interval to which the point $x_n$ belongs. Something along the lines of $$ \lim_{n \to \infty}\int^1_0 f_n(x)dx = \lim_{n \to \infty}\Big(\int^1_{x_n}f_n(x)dx + \int_0^{x_n} f_n(x)dx\Big)$$ However, this particular split doesn't help much because I can't really do anything with the first integral like I could with the second, bounding it with the sup norm and then observing that $x_n \to 0 $ as $ n \to \infty$ since $\sup_{[0,1]}|f_n| = 1$ (or so I think I can't). Any hints to other approaches I could use or to any modifications to my initial approach?</p> <p>Thanks in advance.</p>
Claude Leibovici
82,404
<p>In the same spirit as Cameron Williams's answer, assuming you know the error function, $$ \int e^{-n(nx-1)^2}\,dx=\frac{\sqrt{\pi } }{2 n^{3/2}}\text{erf}\left(\sqrt{n} (n x-1)\right)$$ which makes $$ \int_0^1 e^{-n(nx-1)^2}\,dx=\frac{\sqrt{\pi } }{2 n^{3/2}}\Big(\text{erf}\left(\sqrt{n}\right)+\text{erf}\left((n-1) \sqrt{n}\right)\Big)$$ For large values of the argument, a very fast convergence expansion of the error function is $$\text{erf}(x)=1-\frac{e^{-x^2}}{\sqrt{\pi}}\sum_{n=0}^\infty \frac{(2n-1)!!}{2^n}\frac 1{x^{n+1}}$$ So, limited to the first term $$ \int_0^1 e^{-n(nx-1)^2}\,dx \approx\frac{\sqrt{\pi } }{ n^{3/2}}$$ which is an extremely close approximation; for $n=5$, the exact value is $\approx \color{red}{0.158}4090079$ and the approximation is $\approx \color{red}{0.158}5330919$; for $n=10$, these values are respectively $\approx \color{red}{0.056049}69513$ and $\approx\color{red}{0.056049}91216$;for $n=15$, these values are respectively $\approx \color{red}{0.030509707}10$ and $\approx\color{red}{0.030509707}76$.</p> <p>For sure, much better approximation could be obtained introducing the second term of the expansion.</p>
1,634
<p>I am not too certain what these two properties mean geometrically. It sounds very vaguely to me that finite type corresponds to some sort of "finite dimensionality", while finite corresponds to "ramified cover". Is there any way to make this precise? Or can anyone elaborate on the geometric meaning of it?</p>
David E Speyer
297
<p><a href="https://mathoverflow.net/questions/1634/finite-type-finite-morphism/1643#1643">Peter-arndt</a> gives exactly the right intuition. I will add a cautionary example: It is not literally true that finite fibers implies finite. For example, consider Spec k[x, y]/(xy-1) ---> Spec k[x]. The way I think about this intuitively is that xy=1 has a vertical asymptote at 0 so, even though the fiber over 0 is empty, the map acts like there is an infinite fiber over 0 and is not a finite map. So a more precise intuition is that finite maps have finite fibers, and none of the preimages go off to infinity.</p> <p>I believe that a precise statement is that finite fibers + proper implies finite.</p>
758,466
<blockquote> <p>A fair 6-sided die is tossed 8 times. The sequence of 8 results is recorded to form an 8-digit number. For example if the tosses give {3, 5, 4, 2, 1, 1, 6, 5}, the resultant number is $35421165$. What is the probability that the number formed is a multiple of 8.</p> </blockquote> <p>I solved this by listing all possibilities for the last 3 digits that give multiples of 8, and found this to be $\frac{1}{8}$. </p> <p>The solution key agrees with my answer, but also says that "There are quicker ways to solve the problem using a more advanced knowledge of number theory"</p> <hr> <p>What would be a faster way to solve this using number theory?</p>
Zhuli
141,184
<p>We only care about the last three digits obviously, so the possible sums of just those three die are $100a + 10b + c$, where $a$, $b$, and $c$ are all between $1$ and $6$, inclusive.</p> <p>We check the divisibility by taking this modulo $8$. If the result is equivalent to $0$, then it is divisible.</p> <p>Setting up the equivalence condition: $$ 100a+10b+c \equiv 0 \pmod{8} \\ 4a + 2b + c \equiv 0 \pmod{8} \\ 2(2a+b) + c \equiv 0 \pmod{8} \\ 2(2a+b) \equiv -c \pmod{8} \\ 2a + b \equiv -\frac{c}{2} \pmod{4} $$ The range of $2a+b$ is between $3$ and $18$, inclusive. Each of the values can be obtained in three different ways, except $\{3,4,17,18\}$ each can only be obtained one way, and $\{5,6,15,16\}$ each can be done two ways (there are $8$ integers that can be obtained $3$ ways).</p> <p>Three cases: $-\frac{c}{2}$ must be $3$, $2$, or $1$ modulo $8$. Because the consecutive $8$ integers in the middle can be done $3$ ways, each case has exactly $2$ numbers that satisfy the equivalence, giving us $3\cdot{3}\cdot{2} = 18$ cases. Taking the other integers modulo $4$, only $1$ number from each of the other sets is not congruent to any of $\{3,2,1\}$, giving us $3\times{3} = 9$ more cases, totalling $18+9 = 27$.</p> <p>Final probability: $\frac{27}{6^3} = \frac{1}{8}$.</p>
758,466
<blockquote> <p>A fair 6-sided die is tossed 8 times. The sequence of 8 results is recorded to form an 8-digit number. For example if the tosses give {3, 5, 4, 2, 1, 1, 6, 5}, the resultant number is $35421165$. What is the probability that the number formed is a multiple of 8.</p> </blockquote> <p>I solved this by listing all possibilities for the last 3 digits that give multiples of 8, and found this to be $\frac{1}{8}$. </p> <p>The solution key agrees with my answer, but also says that "There are quicker ways to solve the problem using a more advanced knowledge of number theory"</p> <hr> <p>What would be a faster way to solve this using number theory?</p>
Neat Math
843,178
<p>Let's use Zhuli's notation.</p> <p><span class="math-container">$c$</span> has a probability of <span class="math-container">$\frac{1}{2}$</span> to be even.</p> <p>Now if <span class="math-container">$\frac{c}{2}$</span> is even (i.e., <span class="math-container">$4|c$</span>), then <span class="math-container">$4|2b+c \iff b$</span> is even. If <span class="math-container">$\frac{c}{2}$</span> is odd, then <span class="math-container">$4|2b+c \iff b$</span> is odd. Both conditional probabilities are <span class="math-container">$\frac{1}{2}$</span> therefore the marginal probability <span class="math-container">$4|2b+c$</span> is <span class="math-container">$\frac{1}{4}$</span>.</p> <p>Next we apply the same reasoning on <span class="math-container">$2a + (b+\frac{c}{2})$</span> we get the marginal probability that <span class="math-container">$4 | 2a + (b+\frac{c}{2})$</span>, or equivalently, <span class="math-container">$8 | 4a + 2b +c$</span>, is <span class="math-container">$\frac{1}{8}$</span>. Q.E.D.</p> <p>============================</p> <p>Alternatively we could avoid conditional probabilities and just use independence and marginal probabilities.</p> <p>First note that <span class="math-container">$\Pr(2|a)=\Pr(2|b)=\Pr(2|c)=\frac 12$</span> and <span class="math-container">$a,b,c$</span> are independent.</p> <p>Next, <span class="math-container">$$ \Pr(4|2b+c)=\Pr(2|c,4|c,2|b) + \Pr(2|c, 4\nmid c, 2\nmid b) $$</span> <span class="math-container">$$ =\Pr(2|c, 4|c) \Pr(2|b)+\Pr(2|c, 4\nmid c)\Pr(2\nmid b) \tag{independence}\\ = \frac 12 \left( \Pr(2|c, 4|c) + \Pr (2|c, 4\nmid c)\right) = \frac 12 \Pr(2|c)=\frac 14. $$</span></p> <p>Finally, <span class="math-container">$$ \Pr(8|4a+2b+c)=\Pr(4|2b+c,8|2b+c,2|a) + \Pr(4|2b+c, 8\nmid 2b+c, 2\nmid a)\\ =\Pr(4|2b+c,8|2b+c)\Pr(2|a) + \Pr(4|2b+c, 8\nmid 2b+c)\Pr(2\nmid a)\\ =\frac 12 \left(\Pr(4|2b+c,8|2b+c) + \Pr(4|2b+c, 8\nmid 2b+c)\right)\\ =\frac 12 \Pr(4|2b+c)=\frac 12 \cdot \frac 14 = \frac 18.\blacksquare $$</span></p> <p><strong>Remark:</strong> We can prove, via induction, the generalization proposed by Anant using the technique above.</p>
1,470,244
<p>Prove that the sum of the moduli of the roots of $x^4-5x^3+6x^2-5x+1=0$ is $4$.<br></p> <hr> <p>I tried various methods to find its roots,but no luck.Is there any method to solve it without actually finding the roots of the equation.Please help me.<br> Thanks.</p>
implicati0n
204,739
<p>I'll give you an answer which is the application of a simple method for finding the solutions of symmetric polynomials:</p> <p>Because $0$ is not a root, divide by $x^2$. Then use the substitution $$ t=x+\frac{1}{x}. $$</p> <p>This yields</p> <p>$$ t^2-5t+4=0, $$</p> <p>which you can solve in a simple matter and then put the solutions back to the substitution.</p> <p>EDIT: This is the method for when such equations are of even degree. When it's of odd degree, then $-1$ is always one solution. Then use polynomial division (Horner's algorithm or something you prefer) and you always get a symmetric polynomial of even degree as a result of division, so then you can use the method from my answer again.</p> <p>In general, when your equation is of even degree, you divide it by the "middle" power of $x$, i.e. with $x$ to the power of the highest degree divided by $2$.</p>
4,070,810
<p>I'm trying to find the number of subgroups of <span class="math-container">$\mathbb{Z}^n$</span> such that the quotient is a <span class="math-container">$\bf{cyclic}$</span> group of order <span class="math-container">$p^k$</span> (<span class="math-container">$p$</span> a prime).</p> <p>I guess I don't know enough about the finitely generated abelian groups, though I consulted some results about that. So I hope for a hint, thank you very much.</p>
Limestone
866,771
<p>Observe that, <span class="math-container">$\triangle AEP\sim \triangle BEQ$</span> and hence <span class="math-container">$\angle APE=\angle BQE$</span>. From here, it is seen that quadrilateral <span class="math-container">$PCQE$</span> is cyclic.</p> <p>Thus, <span class="math-container">$\angle CPQ=\angle CEQ=\angle CED=\angle CAD$</span> and therefore <span class="math-container">$PQ\parallel AD$</span>.</p>
1,002,191
<p>I have a problem:</p> <blockquote> <p>A family has six children, A,B,C,D,E,F. The ages of these six children are added together to make an integer less than $32$ where none of the children are less than one year old.</p> </blockquote> <p>What is the chances of guessing the ages of all of the children.</p> <p>I thought stars and bars, and this is equivalent to choosing $7$ children for $$x_1+\dots+x_7=31$$</p> <p>So we get (1 in)${31-1 \choose 7-1}$</p> <p>But they said that $x_1+\dots+\dots+x_7=32$ is correct. I feel like that would be the case if the original question said $x_1+\dots+x_6\leq32$, but it said $x_1+\dots+x_6\lt32$</p> <p>So apparently that makes it (1 in)${32-1 \choose 7-1}$</p> <p>Which is right?(Probably them I know, but I want to know why mine is wrong please)</p>
André Nicolas
6,312
<p>We want to solve $x_1+\cdots+x_7=32$, where all the $x_i$ are $\ge 1$. For $x_7\ge 1$, which means $x_1+\cdots +x_6\lt 32$.</p> <p>If you use $31$, then since $x_7\ge 1$, the ages of the real children would be forced to add up to less than $31$.</p> <p><strong>Remark:</strong> This answers the Stars and Bars question. However, that question has little connection with the motivating question about the chances of guessing the ages of the children. For one thing the ages are a monotone sequence of integers, with equalities relatively unlikely. </p>
306,337
<p>I have a question. Is this integral improper? $$\int_0^\infty \frac{5x}{e^x-e^{-x}} \, dx = \int_0^a \frac{5x}{e^{x}-e^{-x}} \, dx+ \int_a^\infty \frac{5x}{e^x-e^{-x}} \, dx$$</p> <p>Why is $\displaystyle\int_0^a \frac{5x}{e^{x}-e^{-x}}dx$ an improper integral at point $x=0$? And $\displaystyle\int_a^\infty \frac{5x}{e^{x}-e^{-x}} \, dx$ is improper integral at $x \to \infty $?</p> <p>Thanks!</p>
Jack Zega
62,659
<p>Maybe you can try a Taylor series expansion. Make the lower limit of your first integral on the RHS $L$ such that $0 &lt; L \ll 1$, then let $L &lt; a \ll 1$. You can do a Taylor series expansion of the denominator which will be $2x + O(x^3)$. You can cancel a factor of $x$ giving the integrand $(5/2)\cdot(1+O(x^2))^{-1}$. Using the Taylor series of $\frac{1}{1+x}$, your integral will be $\frac{5x}{2} + O(x^3)$. Take the limit as $L$ goes to zero.</p>
2,820,842
<p>I am trying to prove that if $Q$ is an algebraic extension of $F$ ($F$ is a field), if $R$ is a subring of $Q$, and we have $F\subseteq R$, then $R$ is a subfield of $Q$.</p> <p>My first approach was to show every element has an inverse. But there was no result.</p> <p>Any help would be great.</p>
nonuser
463,553
<p>Since $p\mid a-b$ we have $a\equiv _p b$, so</p> <p>$$a^p-b^p = \underbrace{(a-b)}_{m\cdot p^n}(\underbrace{a^{p-1}+a^{p-2}b+...+b^{p-1}}_{\equiv p\cdot a^{p-1}}) = m\cdot p^n\cdot p \cdot k$$</p> <p>where $a^{p-1}+a^{p-2}b+...+b^{p-1} =p\cdot k$ and obviously $p\not{|}\; mk$ </p>
167,536
<p>It has been years since I took calculus and I am trying to figure out if it is possible to calculate the derivative of a mutli variable equation and what the result would be. This equation is from the video game Diablo 3 that represents your damage based on your critical chance and critical damage. I am trying to determine where are the optimal points (or best bang for buck stat) by using derivatives. </p> <p>x = critical chance, y = critical damage, z = damage output</p> <p>Equation: $z = 1 + xy$</p>
Argon
27,624
<p>There are two variables in $z$, $x$ and $y$. Taking the <a href="http://en.wikipedia.org/wiki/Partial_derivative" rel="nofollow">partial derivative</a> in respect to $x$ (treating $y$ like a constant), we get:</p> <p>$$\frac{\partial z}{\partial x}=y$$</p> <p>similarly, the partial derivative in terms of $y$ is</p> <p>$$\frac{\partial z}{\partial y}=x$$</p>
167,536
<p>It has been years since I took calculus and I am trying to figure out if it is possible to calculate the derivative of a mutli variable equation and what the result would be. This equation is from the video game Diablo 3 that represents your damage based on your critical chance and critical damage. I am trying to determine where are the optimal points (or best bang for buck stat) by using derivatives. </p> <p>x = critical chance, y = critical damage, z = damage output</p> <p>Equation: $z = 1 + xy$</p>
Micah
30,836
<p>As the other answer says, your partial derivatives are $$\frac{\partial z}{\partial x} = y, \quad \frac{\partial z}{\partial y} = x \, . $$</p> <p>What does this mean in terms of actual gameplay? It means that, if you currently have stats $x=x_0$, $y=y_0$, you should (approximately) value small changes in $x$ by multiplying them by $y_0$, and value small changes in $y$ by multiplying them by $x_0$.</p> <p>For example, say that at some point in the game your stats are $x=0.5, y=0.25$ (you have a $50$% chance of scoring a critical, and it does one-quarter again as much damage as an ordinary hit). Then, if you're given the option of increasing $x$ by $0.04$ at the cost of decreasing $y$ by $0.01$, you should value this at $0.04(0.25)-0.01(0.5)=0.01-0.005=0.005$. Since this is positive, that option is beneficial.</p> <p>Since your initial formula for $z$ was fairly simple, this is a little bit superfluous. You could just plug everything into the original $z$-formula. Your initial damage output would be $$z_0=1+(0.5)(0.25)=1.125 \, .$$ Your new damage output would be $$z_1=1+(0.5+0.04)(0.25-0.01)=1.1296 \, .$$ Again, since $z_1&gt;z_0$, we can see that taking this new option is beneficial.</p> <p>Note that $z_1-z_0=0.0046$: very close to, but not actually equal to, the $0.005$ we computed by using partial derivatives. </p> <p>Why would you ever use partial derivatives at all, if you could just use the original function to get an exact answer? In this case, I would say you probably shouldn't unless you're in a big hurry. If $z$ were more complicated, though, it might save you a bunch of time, especially if you've got a bunch of different possible changes to try from the same starting point. Since all you care about is which change leads to the biggest (positive) shift in $z$, you won't need to worry that the your partial-derivative computations are approximations unless you have two different potential options that are <em>really</em> close in value.</p> <p>One thing to be careful about: since $x$ is a probability, presumably it's capped at $1$ (or maybe at some lower value if D3 doesn't like the idea that all your hits could be criticals). So your formula for $z$ probably isn't universal. But it's pretty obvious what you should do when it breaks down (maximize $y$ at all costs, stop worrying about $x$).</p>
47,188
<p>I am an amateur mathematician, and I had an idea which I worked out a bit and sent to an expert. He urged me to write it up for publication. So I did, and put it on arXiv. There were a couple of rounds of constructive criticism, after which he tells me he thinks it ought to go a "top" journal (his phrase, and he went on to name two). </p> <p>His opinion is that my outsider status will have no effect on the reviewing process and my paper will be taken seriously. I am pleased and quite flattered to hear this, and my inclination is to do as he suggests. But I have to say it sounds too good to be true. Does this match other people's experience? I understand it's rare enough for undergraduates to publish anywhere, and I am not even an undergraduate! Surely a "non-mathematician submitting to top mathematical journal" must instantly rank high on the crackpot scale. How often does this actually occur successfully these days? Any advice?</p>
Community
-1
<p>First of all, congratulations for making it as far as you have. I have a similar story, in 1986 I met with Stephen Wolfram and discussed with him my research. Wolfram was very supportive and even offered to edit and publish my research if I would just write it up. Then Wolfram left academia, forcing me to reach out to others to publish my work. Having only taken a few college level math classes and not knowing TeX, I was unprepared for how much work it would take to write a good paper. My first problem was that I was over-reaching in both the breath and depth of what I was trying to communicate. So my main advice is that small is beautiful. There are mathematicians of significance who's dissertations had a strong impact on mathematics, but if you engage in "empire building" as a novice, people are likely not to take you seriously. </p> <p>In general no one is going to know who you are. As my knowledge of mathematics and my writing ability have grown I can now submit articles for publication and the publishers now refer to me as Doctor unless I correct them. So there is no barrier to a non-professional publishing except being unable to write at a professional level. But it took me fifteen years to get to this place by myself.</p> <p>Know what you want and what you are willing to pay for it. I ended up going back to school in mathematics just so I could understand math better and I'm glad I did.</p> <p>Enjoy the journey. It is now twenty five years later and I am still unpublished, but that is ok because I'm enjoying the journey. </p> <hr> <p>Note: I recommend when publishing in research journals that not only does your research need to be solid, but it should discuss the relevance of other important research in the field of your publication. You are not just adding another block of knowledge, you cementing it into place in the edifice of mathematics. Educational journals might be more accepting of stand alone material. </p>
737,212
<p>Please how can I easily remember the following trig identities: $$ \sin(\;\pi-x)=\phantom{-}\sin x\quad \color{red}{\text{ and }}\quad \cos(\;\pi-x)=-\cos x\\ \sin(\;\pi+x)=-\sin x\quad \color{red}{\text{ and }}\quad \cos(\;\pi+x)=-\cos x\\ \sin(\frac\pi2-x)=\phantom{-}\cos x\quad \color{red}{\text{ and }}\quad \cos(\frac\pi2-x)=\phantom{-}\sin x\\[12pt] \text{and similar things where we add a radian angle inside cos or sin as you can see} $$</p> <p>So how can I remember all of them? They're pretty confusing! I can easily interchange some and do mistakes thus...</p>
Blue
409
<p>I <a href="http://daylateanddollarshort.com/bloog/proof-without-words-angle-sum-and-difference-formulas/" rel="noreferrer">created the images</a> that "inspired" <a href="http://en.wikipedia.org/wiki/List_of_trigonometric_identities#Angle_sum_and_difference_identities" rel="noreferrer">the illustration of the Angle-Sum and Difference identities</a> on Wikipedia, so I'm all in favor of learning those relations. However, this is how <em>I</em> remember your particular class of identities.</p> <hr> <p>First, in the Unit Circle, I imagine the "sine-cosine-$1$" triangle for a generic first-quadrant angle $\theta$, taking $\theta$ small enough that the triangle is considerably wider than it is tall, creating an obvious "long leg" and "short leg". Because trig values are positive in the first quadrant, I can say that the "long leg" of the triangle has <em>length</em> $\cos\theta$, and that the "short leg" has <em>length</em> $\sin\theta$.</p> <p><img src="https://i.stack.imgur.com/AdWzGm.png" alt="enter image description here"></p> <p>Then, I imagine what happens when I rotate that triangle through multiples of $90^\circ$, getting a nice windmill:</p> <p><img src="https://i.stack.imgur.com/SSVWQm.png" alt="enter image description here"></p> <p>The "(co-)reference triangle" for each of the compound angles is a rotation of the original triangle, and we can read off sine and cosine values by paying attention to the positions of the short legs and long legs (and assigning signs, as appropriate). For instance, the point labeled "$\theta+90^\circ$" is at distance "$\cos\theta$" above the $x$-axis, and at distance "$\sin\theta$" to the left of the $y$-axis; therefore, $$\sin\left(\theta+\phantom{1}90^\circ\right) = \phantom{-}\color{red}{\cos\theta} \qquad\qquad \cos\left(\theta+\phantom{1}90^\circ\right) = -\color{blue}{\sin\theta}$$</p> <p>Likewise, $$\sin\left(\theta+180^\circ\right) = -\color{blue}{\sin\theta} \qquad\qquad \cos\left(\theta+180^\circ\right) = -\color{red}{\cos\theta}$$ $$\sin\left(\theta-\phantom{1}90^\circ\right) = -\color{red}{\cos\theta} \qquad\qquad \cos\left(\theta-\phantom{1}90^\circ\right) = \phantom{-}\color{blue}{\sin\theta}$$</p> <p>Similarly, there's a windmill for compound angles involving $-\theta$:</p> <p><img src="https://i.stack.imgur.com/Ac33Tm.png" alt="enter image description here"></p> <p>And we have $$\sin\left(\phantom{180^\circ}-\theta\right) = -\color{blue}{\sin\theta} \qquad\qquad \cos\left(\phantom{180^\circ}-\theta\right) = \phantom{-}\color{red}{\cos\theta}$$ $$\sin\left(\phantom{1}90^\circ-\theta\right) = \phantom{-}\color{red}{\cos\theta} \qquad\qquad \cos\left(\phantom{1}90^\circ-\theta\right) = \phantom{-}\color{blue}{\sin\theta}$$ $$\sin\left(180^\circ-\theta\right) = \phantom{-}\color{blue}{\sin\theta} \qquad\qquad \cos\left(180^\circ-\theta\right) = -\color{red}{\cos\theta}$$ $$\sin\left(-90^\circ-\theta\right) = -\color{red}{\cos\theta} \qquad\qquad \cos\left(-90^\circ-\theta\right) = -\color{blue}{\sin\theta}$$</p> <hr> <p>It's worthwhile to point out that the identity $$\cos\theta = \sin\left(90^\circ - \theta\right)$$ is, for some (such as myself), <em>definitional</em>: </p> <blockquote> <p><em>The co-sine of the angle is the sine of the co-angle.</em></p> </blockquote> <p>where "co-angle" means "complementary angle", just as "co-sine" literally <em>means</em> "complementary sine".</p> <p>Also, the identities</p> <p>$$\cos(-\theta) = \cos\theta \qquad \sin(-\theta) = -\sin\theta$$</p> <p>have significance in establishing that cosine is an "even function" (it acts on negative arguments ---killing the sign--- the way an even exponent would) and sine is an "odd function" (it acts on negative arguments ---preserving the sign--- the way an odd exponent would). These are handy properties, which say interesting things about how the graphs are drawn. So, you might want to reserve a special area of your brain-space for them.</p>
737,212
<p>Please how can I easily remember the following trig identities: $$ \sin(\;\pi-x)=\phantom{-}\sin x\quad \color{red}{\text{ and }}\quad \cos(\;\pi-x)=-\cos x\\ \sin(\;\pi+x)=-\sin x\quad \color{red}{\text{ and }}\quad \cos(\;\pi+x)=-\cos x\\ \sin(\frac\pi2-x)=\phantom{-}\cos x\quad \color{red}{\text{ and }}\quad \cos(\frac\pi2-x)=\phantom{-}\sin x\\[12pt] \text{and similar things where we add a radian angle inside cos or sin as you can see} $$</p> <p>So how can I remember all of them? They're pretty confusing! I can easily interchange some and do mistakes thus...</p>
colormegone
71,645
<p>Attempting to memorize all the relations you'll see in trigonometry is a daunting task, at least until you've worked with them for some time. For the sort of identities you're talking about, it can be helpful to go back to the unit circle and picture the angles there.</p> <p><img src="https://i.stack.imgur.com/Fay9j.jpg" alt="enter image description here"></p> <p>Starting with the angle $ \ x \ , $ the unit radius pointing in that direction has its tip at a "vertical distance" from the $ \ x-$ axis which we call $ \ \sin x \ $ , and a "horizontal distance" from the $ \ y-$ axis which we call $ \ \cos x \ . $ The supplementary angle $ \ \pi - x \ $ completes a semi-circle from the positive $ \ x-$ direction to the negative $ \ x-$ direction.</p> <p><img src="https://i.stack.imgur.com/FAPCi.jpg" alt="enter image description here"></p> <p>We can now flip this around to look at the corresponding distances for the angle $ \ \pi - x \ $ . We see that we have the same triangle as before, but now it is reversed, on the other side of the $ \ y-$ axis. The "vertical distance" for the tip of the unit radius is the same as it is for angle $ \ x \ , $ so we say $ \ \sin(\pi - x) \ = \ \sin x \ . $ The "horizontal distance", however, is now in the opposite direction, so it is the negative of what it is for angle $ \ x \ , $ or $ \ \cos(\pi - x) \ = \ -\cos x \ . $ Since $ \ \tan x \ = \ \frac{\sin x}{\cos x} \ , $ we can also say that </p> <p>$$ \ \tan (\pi - x) \ = \ \frac{\sin (\pi - x)}{\cos (\pi - x)} \ = \ \frac{\sin x}{-\cos x} \ = \ -\tan x \ \ . $$</p> <p>$$ \\ $$</p> <p><img src="https://i.stack.imgur.com/XCg2b.jpg" alt="enter image description here"></p> <p>The next identities you list apply to the angle $ \ \pi + x \ , $ for which the unit radius points in <em>exactly the opposite direction</em> to the angle $ \ x \ . $ The triangle for $ \ \pi + x \ $ is "upside down and backwards" from the triangle for angle $ \ x \ $ so <strong>both</strong> the "vertical" and "horizontal distances" have their signs changed. Thus, we have $ \ \sin(\pi + x) \ = \ -\sin x \ $ and $ \ \cos(\pi + x) \ = \ -\cos x \ ; $ the relation for tangent will then be</p> <p>$$ \ \tan (\pi + x) \ = \ \frac{\sin (\pi + x)}{\cos (\pi + x)} \ = \ \frac{-\sin x}{-\cos x} \ = \ \tan x \ \ . $$</p> <p>$$ \\ $$</p> <p><img src="https://i.stack.imgur.com/LW4NM.jpg" alt="enter image description here"></p> <p>The third pair you have applies to the complementary angle to $ \ x \ , $ which is $ \ \frac{\pi}{2} - x \ . $ (If two of the angles in a right triangle are $ \ \frac{\pi}{2} \ $ (or 90º) and $ \ x \ $ , the third angle is $ \ \frac{\pi}{2} - x \ . $ ) In the unit circle, that angle fills in the corner of an imaginary box we can draw as shown in the graph above. Now we'll flip this box around in the unit circle like this:</p> <p><img src="https://i.stack.imgur.com/gjOwN.jpg" alt="enter image description here"></p> <p>This has the effect of making the "vertical distance" for angle $ \ x \ $ the "<em>horizontal</em> distance" for angle $ \ \frac{\pi}{2} - x \ , $ and <em>vice versa</em>. So this tells us that $ \ \sin(\frac{\pi}{2} - x) \ = \ \cos x \ $ and $ \ \cos(\frac{\pi}{2} - x) \ = \ \sin x \ . $ (Because these formulas concern complementary angles, they are sometimes referred to as the " <strong>co-</strong> relations". This is also the origin of the "co-" prefix in the names of trig functions: the <strong>co</strong>-sine of an angle is the sine of the <em>complementary</em> angle, etc.) For tangent, we then get a complementary function,</p> <p>$$ \ \tan (\frac{\pi}{2} - x) \ = \ \frac{\sin (\frac{\pi}{2} - x )}{\cos (\frac{\pi}{2} - x)} \ = \ \frac{\cos x}{\sin x} \ = \ \frac{1}{\tan x} \ = \ \cot x \ \ . $$</p>
40,463
<p>If a rectangle is formed from rigid bars for edges and joints at vertices, then it is flexible in the plane: it can flex to a parallelogram. On any smooth surface with a metric, one can define a linkage (e.g., a rectangle) whose edges are geodesics of fixed length, and whose vertices are joints, and again ask if it is rigid or flexible on the surface. This leads to my first, specific question:</p> <blockquote> <p><b>Q1</b>. Is a rhombus, or a rectangle, always flexible on a sphere?</p> </blockquote> <p><br /><img src="https://i.stack.imgur.com/FRW0R.jpg" alt="alt text" /><br /></p> <p>It seems the answer should be <em>Yes</em> but I am a bit uncertain if there must be a restriction on the edge lengths. (In the above figure, the four arcs are each <span class="math-container">$49^\circ$</span> in length, comfortably short.)</p> <blockquote> <p><b>Q2</b>. The same question for other surfaces: Arbitrary convex surfaces? A torus?</p> </blockquote> <p>I am especially interested to learn if there are situations where a linkage that is flexible in the plane is rendered rigid when embedded on some surface. It seems this should be possible...?</p> <blockquote> <p><b>Q3</b>. More generally, <a href="https://mathworld.wolfram.com/LamansTheorem.html" rel="nofollow noreferrer">Laman's theorem</a> provides a combinatorial characterization of the rigid linkages in the plane. The <span class="math-container">$n{=}4$</span> rectangle is not rigid because it has fewer than <span class="math-container">$2n-3 = 5$</span> bars: it needs a 5th diagonal bar to rigidify. Has Laman's theorem been extended to arbitary (closed, smooth) surfaces embedded in <span class="math-container">$\mathbb{R}^3$</span>? Perhaps at least to spheres, or to all convex surfaces?</p> </blockquote> <p>Thanks for any ideas or pointers to relevant literature!</p> <p><b>Addendum</b>. I found one paper related to my question: &quot;Rigidity of Frameworks Supported on Surfaces&quot; by A. Nixon, J.C. Owen, S.C. Power. <a href="https://arxiv.org/abs/1009.3772v1" rel="nofollow noreferrer" title="Journal version at doi:10.1137/110848852. zbMATH review at https://zbmath.org/?q=an:1266.52018">arXiv:1009.3772v1 math.CO</a> In it they prove an analog of Laman's theorem for the circular cylinder in <span class="math-container">$\mathbb{R}^3$</span>. If one phrases Laman's theorem as requiring for rigidity that the number of edges <span class="math-container">$E \ge 2 V - 3$</span> in both the graph and in all its subgraphs, then their result (Thm. 5.3) is that, on the cylinder, rigidity requires <span class="math-container">$E \ge 2 V -2$</span> in the graph and in all its subgraphs. This is not the precise statement of their theorem. They must also insist that the graph be <em>regular</em> in a sense that depends on the rigidity matrix achieving maximal rank (Def. 3.3). They give as examples of <em>irregular</em> linkages on a sphere one that contains an edge with antipodal endpoints, or one that includes a triangle all three of whose vertices lie on a great circle. But modulo excluding irregular graphs and other minor technical details, they essentially replace the constant 3 in Laman's theorem for the plane with 2 for the cylinder.</p> <p>Theirs is a very recent paper but contains few citations to related work on surfaces, suggesting that perhaps the area of linkages embedded on surfaces is not yet well explored. In light of this apparent paucity of information, it seems appropriate that I 'accept' one of the excellent answers received. Thanks!</p> <p><b>Addendum [31Jan11].</b> I just learned of a 2010 paper by Justin Malestein and Louis Theran, &quot;Generic combinatorial rigidity of periodic frameworks&quot; <a href="https://arxiv.org/abs/1008.1837v2" rel="nofollow noreferrer" title="Journal version at doi:10.1016/j.aim.2012.10.007. zbMATH review at https://zbmath.org/?q=an:1268.52021">arXiv:1008.1837v2 (math.CO)</a>, which pretty much completely solves the problem of linkages on a flat 2-torus, generalizing to flat orbifolds. They obtain a combinatorial characterization for generic minimal rigidity for &quot;planar periodic frameworks,&quot; which encompass these surfaces.</p>
Pierre Dehornoy
9,248
<p>This might be a cheat, but if you consider a linkage turning once around the meridian of a torus, it might be not flexible. Actually, a linkage homeomorphic to a circle that minimizes length in its homology class is not flexible.</p>
3,127,795
<p>So I’m struggling to understand how to find the angle in this circle, we’ve recently learnt about trigonometry and like finding the area of a circle and all that but I can’t seem to remember which formula I have to use to find this angle. Can anyone lend a helping hand? <img src="https://i.stack.imgur.com/V2Sm7.jpg" alt=""></p>
Jonathan Levy
379,205
<p>You have an isosceles triangle with two sides as the radius of 7. This is because they tell you O is the center. Then you have <span class="math-container">$\sin(\theta/2)=5/7$</span>. So the angle is <span class="math-container">$\sin^{-1}(5/7)*2$</span></p>
3,752,167
<p>Freshman question, really, but the more I think about it, the more I doubt.</p> <p>Suppose that two sets belong to the same equivalence class. Are they in effect interchangeable? (I understand that there is no axiom of `interchangeability' in the definition of an equivalence relation.)</p> <p>For example, consider the equivalence class of all people who are 30 years old. This equivalence class contains both men and women who are 30; and men and women are different 'objects' if I may say. Yet if I consider the class of people who are 30 for some analysis, it does not matter if I pick a man or a woman. They are interchangeable as long as what matter is their age.</p> <p>I just wonder if this is characteristic of <em>all</em> equivalence classes one can encounter in mathematics.</p>
Daniel
391,594
<p>I may be just rephrasing some of the answers, but I like to think that an equivalence relation is simply a problem specific notion of equality; the relation tries to summarize all the important properties that are relevant to the problem in question. In this sense, if any two objects are equivalent they can be used interchangeably without consequence.</p> <p>To exemplify, you can consider that a sphere is equivalent to a cylinder if all that matters to your problem is purely topological. As long as distances start to matter, this topological notion of 'equality' is not suitable anymore you should look for a more stringent notion of 'equality'.</p>
2,009,134
<p>Suppose $$\frac{{{{\sin }^4}(\alpha )}}{a} + \frac{{{{\cos }^4}(\alpha )}}{b} = \frac{1}{{a + b}}$$ for some $a,b\ne 0$. </p> <p>Why does $$\frac{{{{\sin }^8}(\alpha )}}{{{a^3}}} + \frac{{{{\cos }^8}(\alpha )}}{{{b^3}}} = \frac{1}{{{{(a + b)}^3}}}$$</p>
Yuriy S
269,624
<p>There is a very direct way to solve this. First, use the given equation to find, for example, $\sin^2 \alpha$:</p> <p>$$x=\sin^2 \alpha$$</p> <p>$$\frac{1}{a} x^2+\frac{1}{b}(1-x)^2=\frac{1}{a+b}$$</p> <p>Solving this easy quadratic, we get a single root:</p> <p>$$x=\frac{a}{a+b}=\sin^2 \alpha$$</p> <p>It follows:</p> <p>$$1-x=\frac{b}{a+b}=\cos^2 \alpha$$</p> <p>Now substitute these values into the second equation and see that it holds.</p>
749,097
<p>In any calculus course, one of the first thing we learn is that $(uv)'=u'v+v'u$ rather than the what I've written in the title. This got me wondering: when is this <em>dream product rule</em> true? There are of course trivial examples, and also many instances where the equality is true at a handful of points. Less obvious though, is the following:</p> <blockquote> <p>Are there non-constant $u,v$ such that there exists an interval $I$ where $(uv)'=u'v'$ over $I?$</p> </blockquote> <p>I have a feeling there should be, but I am having trouble constructing such a pair.</p>
David
119,775
<p>There are many cases where this will happen. By simple algebra, the relation $(uv)'=u'v'$ can be written as $$\Bigl(\frac{u'}{u}-1\Bigr)\Bigl(\frac{v'}{v}-1\Bigr)=1$$ and this leads to $$\ln u(t)=\int\frac{v'}{v'-v}\,dt\ .$$ You will have to make sure that the integral exists and gives a suitable result, but this should give you a solution for many functions $v$.</p>
209,728
<p>I'm attempting to calculate the gradient of a function defined by two variables. After following a tutorial I found, I have the following code: </p> <pre><code>f[x_, y_] = (2 x^4 - x^2 + 3 x - 7 y + 1)/(e^(2 x^2/3 + 3 y^2/4)); (*t is gradient of f*) t[{x_, y_}] := { D[f[x, y], x], D[f[x, y], y] }; (* Trying to use NestList to basically calculate gradient descent 'n' times/ use Euler's method starting at the point (.510, -.445) *) NestList[t, {.51, -.445}, n] </code></pre> <p>However, the output I'm getting is ".51" and "-.445" are not valid variables. I understand this is because I when I apply NestList, I am essentially taking the derivative with respect to ".51" and "-.445" which isn't possible obviously. However, I am unsure of how to apply NestList with respect to the variables <em>x</em> and <em>y</em> at the point <strong>(.51,-.445)</strong>.</p>
H. Khan
67,854
<p>Via @kglr: </p> <pre><code>ClearAll[t]; t[{x_, y_}] = { D[f[x, y], x], D[f[x, y], y] }; NestList[t, {.51, -.445}, n] </code></pre>
268,482
<p>One has written a paper, the main contribution of which is a few conjectures. Several known theorems turned out to be special cases of the conjectures, however no new case of the conjectures was proven in the paper. In fact, no new theorem was proven in the paper. </p> <p>The work was reported on a few seminars, and several experts found the conjectures interesting. </p> <p>One would like to publish this paper in a refereed journal. The paper was rejected from a certain journal just two days after its submission because "this genre of article does not fit the journal".</p> <blockquote> <p><strong>QUESTION.</strong> Are there examples of publications of this genre in refereed journals?</p> </blockquote> <p><strong>ADD:</strong> The mentioned paper explains the background, states the conjectures, discusses various special cases and consequences, and lists known cases. It is 20 pages long. </p>
Gil Kalai
1,532
<p>Yes, there are quite a few papers of this kind. Of course, if proving the implication to known theorems or between the various conjectures is interesting and challenging then this is "ordinary". If the paper contains only a statement of the conjectures then this is indeed more special. There are journals which have short note sections or problem sections appropriate for such papers. (I published a couple of papers with just conjectures in such sections.) </p> <p>As a general rule, it always makes sense to try other journals, to consult with friends, colleagues and teachers, to put the paper on the arxiv, and to try to improve the paper. </p>
268,482
<p>One has written a paper, the main contribution of which is a few conjectures. Several known theorems turned out to be special cases of the conjectures, however no new case of the conjectures was proven in the paper. In fact, no new theorem was proven in the paper. </p> <p>The work was reported on a few seminars, and several experts found the conjectures interesting. </p> <p>One would like to publish this paper in a refereed journal. The paper was rejected from a certain journal just two days after its submission because "this genre of article does not fit the journal".</p> <blockquote> <p><strong>QUESTION.</strong> Are there examples of publications of this genre in refereed journals?</p> </blockquote> <p><strong>ADD:</strong> The mentioned paper explains the background, states the conjectures, discusses various special cases and consequences, and lists known cases. It is 20 pages long. </p>
Piyush Grover
30,684
<p>Depending upon the content of the paper, you may look at the new Arnold Mathematical Journal (<a href="http://www.springer.com/mathematics/journal/40598" rel="noreferrer">http://www.springer.com/mathematics/journal/40598</a>).</p> <p>From their site, "Problems, objectives, work in progress. Most scholarly publications present results of a research project in their “final" form, in which all posed questions are answered. Some open questions and conjectures may be even mentioned, but the very process of mathematical discovery remains hidden. Following Arnold, publications in AMJ will try to unhide this process and made it public by encouraging the authors to include informal discussion of their motivation, possibly unsuccessful lines of attack, experimental data and close by research directions. AMJ publishes well-motivated research problems on a regular basis. Problems do not need to be original; an old problem with a new and exciting motivation is worth re-stating. Following Arnold's principle, a general formulation is less desirable than the simplest partial case that is still unknown."</p>
268,482
<p>One has written a paper, the main contribution of which is a few conjectures. Several known theorems turned out to be special cases of the conjectures, however no new case of the conjectures was proven in the paper. In fact, no new theorem was proven in the paper. </p> <p>The work was reported on a few seminars, and several experts found the conjectures interesting. </p> <p>One would like to publish this paper in a refereed journal. The paper was rejected from a certain journal just two days after its submission because "this genre of article does not fit the journal".</p> <blockquote> <p><strong>QUESTION.</strong> Are there examples of publications of this genre in refereed journals?</p> </blockquote> <p><strong>ADD:</strong> The mentioned paper explains the background, states the conjectures, discusses various special cases and consequences, and lists known cases. It is 20 pages long. </p>
Ofir Gorodetsky
31,469
<p>Supplementing Carlo's answer, I also want to recommend the journal 'Mathematics of Computation'. Several important number-theoretic conjectures were first stated there, supported with numerical data. So if some computation was involved in providing evidence for your conjectures, it seems like a suitable choice.</p> <p>Here are some notable examples:</p> <ol> <li><a href="http://www.jstor.org/stable/2004584?seq=1#page_scan_tab_contents" rel="noreferrer">"The Riemann Hypothesis and Pseudorandom Features of the Möbius Sequence"</a> by Good and Churchhouse</li> <li><a href="http://www.ams.org/journals/mcom/1962-16-079/S0025-5718-1962-0148632-7/home.html" rel="noreferrer">"A heuristic asymptotic formula concerning the distribution of prime numbers "</a> by Bateman and Horn - This is a conjecture which generalizes Dirichlet's theorem on primes in arithmetic progression.</li> </ol>
2,407,183
<blockquote> <p>Calculate $\int_{\lambda} dz/(z^2-1)^2$, where $\lambda$ is the path in $\mathbb{R^2}-\{1,-1\}$ plotted below:</p> <p><a href="https://i.stack.imgur.com/T1uFH.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/T1uFH.png" alt="enter image description here"></a></p> </blockquote> <p>This may seem like an ordinary calculus integral but I'm studying $1$-forms and homotopic paths, but I can't connect all the theory I've been viewing in order to solve this integral in this context. </p> <p>For example, I know that this is an integral over a closed path. There are theorems relating closed paths with the form being exact, and therefore the integral should be $0$, I guess? However, I'm not integrating an $1$-form... So I wonder what this exercise is askign me to use. Could somebody help me?</p>
Simply Beautiful Art
272,831
<p>I noticed that the integrand is faily quadratic between integer values, hence</p> <p>$$-I=\underbrace{\int_0^2\frac{\sin(\pi x)}{\ln(x)}~\mathrm dx}_{\approx-3.64061894952}+\sum_{n=2}^\infty\int_n^{n+1}\frac{\sin(\pi x)}{\ln(x)}~\mathrm dx$$</p> <p>$$\small\frac{\sin(\pi x)}{\ln(x)}=\frac{(-1)^{n+1}}{t^2\ln^3(t)}\left[t^2\ln^2(t)+t(x-t)\ln(t)+\bigg[\ln(t)+2-\pi^2t^2\ln^2(t)\bigg]\frac{(x-t)^2}2\right]+\mathcal O((x-t)^3),\\t=n+0.5$$</p> <p>This is a pretty close approximation and gives us the series</p> <p>$$I\approx3.64061894952+\sum_{n=2}^\infty\underbrace{\frac{(-1)^n}{t^2\ln^3(t)}\left[\left(1-\frac{\pi^2}{24}\right)t^2\ln^2(t)+\frac{\ln(t)}{24}+\frac1{12} \right]}_{a_n}$$</p> <p>Which converges really slowly since $|a_n|\in\mathcal O(1/\ln(n))$, but by applying an Euler sum, convergence is rapidly sped up:</p> <p>$$I\approx3.64061894952+\sum_{k=0}^\infty\frac1{2^{k+1}}\sum_{n=0}^k\binom kna_{n+2}$$</p> <hr> <p>If one wishes to allow integrals into the sum,</p> <p>$$I=3.64061894952-\sum_{k=0}^\infty\frac1{2^{k+1}}\sum_{n=0}^k\binom kn\int_{n+2}^{n+3}\frac{\sin(\pi x)}{\ln(x)}~\mathrm dx$$</p> <p>Evaluating this up to $k=20$, I get</p> <pre><code>k I_k ---- ------------ 0 3.2884372794 1 3.2399068661 2 3.2264174379 3 3.2219217488 4 3.2202756011 5 3.2196366110 6 3.2193784711 7 3.2192711120 8 3.2192254632 9 3.2192057126 10 3.2191970462 11 3.2191931992 12 3.2191914749 13 3.2191906955 14 3.2191903408 15 3.2191901783 16 3.2191901035 17 3.2191900689 18 3.2191900528 19 3.2191900453 20 3.2191900418 </code></pre> <p>By approximating further out, I get</p> <p>\begin{align}I&amp;\approx\int_0^{50}\frac{\sin(\pi x)}{\ln(x)}~{\rm d}x+\sum_{n=0}^{19}\frac1{2^{n+1}}\sum_{k=0}^n\binom nk(-1)^k\int_0^1\frac{\sin(\pi t)}{\ln(t+k+50)}~{\rm d}t\\&amp;\approx\underline{-3.219190038646}743\end{align}</p>
423,479
<p>I know that $$\lim_{n\rightarrow \infty}\frac{\Gamma(n+\frac{1}{2})}{\sqrt{n} \Gamma(n)}=1,$$ but I'm interested in the exact behaviour of </p> <p>$$a_n =1- \left( \frac{\Gamma(n+\frac{1}{2})}{\sqrt{n} \Gamma(n)} \right) ^2$$</p> <p>particularily compared to $$b_n = \frac{1}{4n}$$</p> <p>I haven't studied asymptotics yet, so I have no idea how to approach this, but I need this particular result in a statistics problem I'm working on.</p>
Start wearing purple
73,025
<p>Using <a href="http://en.wikipedia.org/wiki/Stirling%27s_approximation#Stirling.27s_formula_for_the_gamma_function">Stirling formula</a> you can show that $$\frac{\Gamma(x+\frac12)}{\sqrt{x}\,\Gamma(x)}=1-\frac{1}{8x}+\frac{1}{128x^2}+O(x^{-3})$$ as $x\rightarrow \infty$. In principle one can obtain as many terms in the asymptotics as one wants using the formula with Bernoulli numbers in the above link.</p>
2,161,463
<p>I am in the final stages of a proof and need help. I have simplified my starting expression expression down to $\dfrac{v\Gamma(-1+\frac{v}{2})}{2\Gamma(\frac{v}{2})}$</p> <p>I know the above expression is to equal $\frac{v}{v-2}$</p> <p>I am having ahard time getting there.</p> <p>I know $\Gamma(n) = (n-1)!$</p> <p>So I thought something like $\dfrac{v*(\dfrac{v}{2}-2)(\dfrac{v}{2}-1)(\dfrac{v}{2}-0)}{2(\dfrac{v}{2}-1)(\dfrac{v}{2}-0)}$ Would work but the terms only cancel to produce $\dfrac{v*(\dfrac{v}{2}-2)}{2}$</p> <p>Can anyone give me a push in the right direction?</p>
Kenny Wong
301,805
<p>Use $\Gamma(z+1) = z \Gamma(z)$.</p> <p>See <a href="https://en.wikipedia.org/wiki/Gamma_function#Main_definition" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Gamma_function#Main_definition</a>.</p>
357,138
<blockquote> <p>If $a_1,a_2,\dotsc,a_n $ are positive real numbers, then prove that</p> </blockquote> <p>$$\lim_{x \to \infty} \left[\frac {a_1^{1/x}+a_2^{1/x}+.....+a_n^{1/x}}{n}\right]^{nx}=a_1 a_2 \dotsb a_n.$$ </p> <p>My Attempt: </p> <p>Let $P=\lim_{x \to \infty} \left[\dfrac {a_1^{\frac{1}{x}}+a_2^{\frac {1}{x}}+.....+a_n^{\frac {1}{x}}}{n}\right]^{nx} \implies \ln P=\lim_{x \to \infty} \ln \left[\frac {a_1^{\frac{1}{x}}+a_2^{\frac {1}{x}}+.....+a_n^{\frac {1}{x}}}{n}\right]^{nx} =\lim_{x \to \infty} nx \ln \left[\frac {a_1^{\frac{1}{x}}+a_2^{\frac {1}{x}}+.....+a_n^{\frac {1}{x}}}{n}\right]= \lim_{x \to \infty} n \left[\frac {\ln (a_1^{1/x}+a_2^{1/x}+...+a_n^{1/x})-\ln n}{1/x}\right]$ </p> <p>and this is $0/0$ form and so I have to apply L'Hospital's rule. Now things get a bit complicated during derivative. </p> <p>Can someone point me in the right direction? Thanks in advance for your time.</p>
learner
33,640
<p>Let $$\begin{align}P &amp;=\lim_{x \to \infty} \Big[\dfrac {a_1^{\dfrac{1}{x}}+a_2^{\dfrac {1}{x}}+.....+a_n^{\dfrac {1}{x}}}{n}\Big]^{nx}\\ \implies \ln P &amp;=\lim_{x \to \infty} \ln \Big[\dfrac {a_1^{\dfrac{1}{x}}+a_2^{\dfrac {1}{x}}+.....+a_n^{\dfrac {1}{x}}}{n}\Big]^{nx} \\&amp;=\lim_{x \to \infty} nx \ln \Big[\dfrac {a_1^{\dfrac{1}{x}}+a_2^{\dfrac {1}{x}}+.....+a_n^{\dfrac {1}{x}}}{n}\Big]\\&amp;= \lim_{x \to \infty} n \Big[\dfrac {\ln (a_1^{1/x}+a_2^{1/x}+...+a_n^{1/x})-\ln n}{1/x}\Big]\\&amp;=\lim_{z \to 0} n \Big[\dfrac {\ln (a_1^{z}+a_2^{z}+...+a_n^{z})-\ln n}{z}\Big]\end{align}$$ </p> <p>and this is $0/0$ form and so I have to apply L'Hospital's rule. So,$$\begin{align}\ln P &amp;=n \lim_{z \to 0}\dfrac {1}{ (a_1^{z}+a_2^{z}+...+a_n^{z})} \times \{a_1^z \ln a_1+a_2^z \ln a_2+.....+a_n^z \ln a_n\} \\&amp;=n \times \dfrac {1}{n}\{\ln a_1+\ln a_2+....+\ln a_n\}\\&amp;=\ln (a_1.a_2...a_n)\\\implies P&amp;=a_1.a_2...a_n \end{align}$$. </p> <p>This completes the proof.</p>
478,523
<p>I'm trying to reason through whether $\int_{x=2}^\infty \frac{1}{xe^x} dx$ converges.</p> <p>Intuitively, It would seem that since $\int_{x=2}^\infty \frac{1}{x} dx$ diverges, then multiplying the denominator by something to make it smaller would make it converge.</p> <p>If I apply a Taylor expansion to the $\frac{1}{e^x}$, then I get</p> <p>$\displaystyle \int_{x=2}^\infty \frac{1}{x} e^{-x} dx = \int_{x=2}^\infty \frac{1}{x}(1 + (-x) + \frac{(-x)^2}{2} + \cdots) dx$</p> <p>My initial thought was to multiply everything out and just look at the leading order behavior, $ x^{n-1}$, but this would seem to mean the integral diverges.</p> <p>What is wrong with this approach?</p>
Mhenni Benghorbal
35,472
<p>Just notice that,</p> <p>$$ x \geq 2 \implies \frac{1}{x}\leq \frac{1}{2}\implies \frac{1}{xe^{x}}\leq \frac{1}{2e^{x}}= \frac 1 2 e^{-x}. $$</p> <p>So you can compare with the function $ \frac 1 2 e^{-x} $.</p>
815,418
<p>Ok, so I've been playing around with radical graphs and such lately, and I discovered that if the </p> <pre><code>nth x = √(1st x √ 2nd x ... √nth x); </code></pre> <p>Then</p> <p>$$\text{the "infinith" } x = x$$</p> <p>Example: </p> <p>$$\sqrt{4\sqrt{4\sqrt{4\sqrt{4\ldots}}}}=4$$<br> Try it yourself, type calc in Google search, hit then a number, such as $4$, and repeat, ending with $4$, (or press the buttons instead).</p> <p>I'm a math-head, not big enough though, I think this sequence is divergent or convergent or whatever, too lazy to search up the difference.</p> <p>However, can this be explained to me? Like how the Pythagorean Theorem can be explained visually.</p>
Emily
31,475
<p>Suppose $y = \sqrt{x\sqrt{x\sqrt{x\cdots}}}$.</p> <p>Multiply both sides by $x$ and take the square root:</p> <p>$$\sqrt{xy} = \sqrt{x\sqrt{x\sqrt{x\cdots}}} = y$$</p> <p>Therefore, $\sqrt{xy} = y$, and solving we have $xy = y^2 \implies x = y$.</p>
803,608
<p>I believe that the statement is True, and this is my argument:</p> <p>Since there exists an element $((1-\sqrt{2})/(1-(\sqrt{2})^2)\in\mathbb{Q}[\sqrt{2}]$ such that their products gives $1$ (multiplicative identity), therefore ( $1+\sqrt{2}$ ) is a unit. </p> <p>A $unit$ is an element $a\in R|ab=1$ where as $b\in R$. Where $b$ is known as the multiplicative inverse of $a$.</p> <p>Is my argument reasonable enough? And if there is/are any other possible argument, It would be really helpful to know.</p>
Bill Dubuque
242
<p>Probably this exercise is a warmup for the proof that $\,\Bbb Q[\sqrt{2}]\,$ is a field. The key idea is simple: to invert $\,w = 1+\sqrt{2}\,$ simply <em>rationalize the denominator</em> of $\,\dfrac{1}{1+\sqrt 2},\,$ i.e. multiply the numerator and denominator by the conjugate $\,w' = 1-\sqrt{2}\,$ to force the denominator to be <em>rational</em> $\,ww' = r.\,$ Being a nonzero rational, $\,r\,$ is invertible, which yields the sought inverse $\, \dfrac{1}w = \dfrac{w'}{ww'}= r^{-1} w'$ </p> <p>Thus, by taking norms $\,w\to ww',$ this method transforms the problem of inverting a quadratic <em>irrational</em> to the simpler problem of inverting a <em>rational</em>. Since the same method works for any irrational $\,w = a + b\sqrt{2} \in \Bbb Q[\sqrt{2}]\,$ we infer that $\Bbb Q[\sqrt{2}]$ is a field. Note that in this general proof it is crucial to show that the denominator $\,ww'= a^2-2b^2 \ne 0.\,$ But if it $=0\,$ then $\, 2 = (a/b)^2\Rightarrow\!\Leftarrow$</p>
4,077,223
<p><span class="math-container">$\exists x \phi(x)$</span> in mathematics means there is something <span class="math-container">$c$</span>, s.t. <span class="math-container">$\phi(c)$</span> holds. What is the formulation of that same idea as a inference rule, an axiom, or something else in a proof system in FOL?</p> <p>Particularly what is the elimination rule for <span class="math-container">$\exists$</span>? (It is not correct that if <span class="math-container">$\Phi \vdash \exists x \phi$</span>, there exists a term <span class="math-container">$t$</span> so <span class="math-container">$\Phi \vdash \phi[t/x]$</span>.)</p> <p>What is the similar one for <span class="math-container">$\forall$</span>?</p>
Mauro ALLEGRANZA
108,274
<p>In the <a href="https://iep.utm.edu/nat-ded/#H7" rel="nofollow noreferrer">Natural Deduction</a> proof system we have a pair of inference rules for the existential (as well as the universal) quantifier:</p> <blockquote> <p><span class="math-container">$(\exists \text I) \ \varphi [x/a] \vdash \exists x \varphi$</span></p> </blockquote> <blockquote> <p><span class="math-container">$(\exists \text E) \ \text { if } \Gamma \vdash \exists x \varphi \text { and } \Delta, \varphi [x/a] \vdash \psi, \text { then } \Gamma, \Delta \vdash \psi$</span>.</p> </blockquote> <p>In an Hilbert-style proof system we have at least one axiom for it, unless it is defined in terms of the universal one.</p> <p>See e.g. J. Shoenfield, <a href="https://books.google.it/books?id=Vl1ZDwAAQBAJ&amp;pg=PA21" rel="nofollow noreferrer">Mathematical Logic</a> for the axiom</p> <blockquote> <p><span class="math-container">$A[x/a] \to \exists x A$</span></p> </blockquote> <p>and the rule:</p> <blockquote> <p><span class="math-container">$\dfrac {A \to B}{\exists x A \to B}$</span>, provided that <span class="math-container">$x$</span> is not free in <span class="math-container">$B$</span>.</p> </blockquote> <p>In <a href="https://books.google.it/books?id=dVncCl_EtUkC&amp;pg=PA78" rel="nofollow noreferrer">Enderton's system (see page 78)</a> the existential quantifier is an abbreviation: <span class="math-container">$∃x α$</span> abbreviates <span class="math-container">$(¬∀x(¬α))$</span>.</p>
4,077,223
<p><span class="math-container">$\exists x \phi(x)$</span> in mathematics means there is something <span class="math-container">$c$</span>, s.t. <span class="math-container">$\phi(c)$</span> holds. What is the formulation of that same idea as a inference rule, an axiom, or something else in a proof system in FOL?</p> <p>Particularly what is the elimination rule for <span class="math-container">$\exists$</span>? (It is not correct that if <span class="math-container">$\Phi \vdash \exists x \phi$</span>, there exists a term <span class="math-container">$t$</span> so <span class="math-container">$\Phi \vdash \phi[t/x]$</span>.)</p> <p>What is the similar one for <span class="math-container">$\forall$</span>?</p>
supinf
168,859
<p>There are different ways to introduce <span class="math-container">$\exists$</span> in a FOL proof system.</p> <p>I think the most common approach is that <span class="math-container">$\exists$</span> is defined via <span class="math-container">$\forall$</span>:</p> <p><span class="math-container">$$\exists x \phi(x) \stackrel{\text{def}}\iff \lnot\forall x \lnot \phi(x).$$</span></p> <p>And <span class="math-container">$\forall$</span> is introduced axiomatically before that.</p>
438,818
<p>I am looking for all two consecutive integers $A$ and $A+1$, which can be represented as sums of two squares $A=a^2+b^2$ and $A+1=c^2+d^2$, $a,b,c,d&gt;0$.</p>
Gerry Myerson
8,269
<p>Not a complete answer --- I doubt there is one --- but $$(n^2-n)^2+(n^2-n)^2,(n^2-2n)^2+(n^2-1)^2,(n^2-n-1)^2+(n^2-n+1)^2$$ gives three consecutive numbers, each a sum of two non-zero squares. This example is taken from <a href="http://www.math.ksu.edu/~cochrane/research/sumsquare.pdf" rel="nofollow">Cochrane and Dressler, Consecutive triples of sums of two squares</a>, which also cites earlier results about consecutive pairs of sums of two squares. </p>
2,214,231
<blockquote> <p>Find the derivative of $\tan^3[\sin(2x^2-17)]$. </p> </blockquote> <p>Sorry if my question is a little too specific but I am confused on this trig equation. After completing the derivative I was wondering why does the $3tan^2$ not distribute to $sec^2$? Is there a rule for this? How come the exponents and the power of $2$ don't get placed onto $sec^2$? Am I missing out on some of the properties of the chain rule? </p>
Alex Robinson
366,219
<p>Consider x = 0, at this point ln(o) is undefined, but you are really solving for ln(x) = 1, and the only thing that fits the bill is taking x to be e, which gives your solution as the coordinate (e,0)</p>
20,802
<p>Look at the following example:</p> <p>Which picture has four apples?</p> <p>A<a href="https://i.stack.imgur.com/Tpm46.png" rel="noreferrer"><img src="https://i.stack.imgur.com/Tpm46.png" alt="enter image description here" /></a></p> <hr /> <p>B <a href="https://i.stack.imgur.com/AOv29.png" rel="noreferrer"><img src="https://i.stack.imgur.com/AOv29.png" alt="enter image description here" /></a></p> <hr /> <p>C <a href="https://i.stack.imgur.com/lZNmQ.png" rel="noreferrer"><img src="https://i.stack.imgur.com/lZNmQ.png" alt="enter image description here" /></a></p> <hr /> <p>D <a href="https://i.stack.imgur.com/BWqpH.png" rel="noreferrer"><img src="https://i.stack.imgur.com/BWqpH.png" alt="enter image description here" /></a></p> <p>B is the expected answer but should not the correct answer be BCD? Technically if a set has <strong>exactly</strong> <span class="math-container">$m$</span> elements, then it has <span class="math-container">$k$</span> elements if <span class="math-container">$k\leq m$</span>. This is also how we talk in everyday language:</p> <blockquote> <p>&quot;Do you have three dollars?&quot; &quot;Yes.&quot;</p> </blockquote> <p>The second speaker is not indicating he has exactly three dollars. He simply indicates that he has <strong>at least</strong> three dollars.</p> <p>So I am wondering if we are teaching children correct logic here. Shouldn't the original question be rephrased as &quot;which picture has <strong>exactly</strong> four apples&quot;?</p>
Aeryk
401
<p>Perhaps &quot;shows&quot; instead of &quot;has&quot;. If you asked me to show you 4 apples, I can't think of a logical argument in favor of me grabbing 5 apples and smiling smugly.</p>
20,802
<p>Look at the following example:</p> <p>Which picture has four apples?</p> <p>A<a href="https://i.stack.imgur.com/Tpm46.png" rel="noreferrer"><img src="https://i.stack.imgur.com/Tpm46.png" alt="enter image description here" /></a></p> <hr /> <p>B <a href="https://i.stack.imgur.com/AOv29.png" rel="noreferrer"><img src="https://i.stack.imgur.com/AOv29.png" alt="enter image description here" /></a></p> <hr /> <p>C <a href="https://i.stack.imgur.com/lZNmQ.png" rel="noreferrer"><img src="https://i.stack.imgur.com/lZNmQ.png" alt="enter image description here" /></a></p> <hr /> <p>D <a href="https://i.stack.imgur.com/BWqpH.png" rel="noreferrer"><img src="https://i.stack.imgur.com/BWqpH.png" alt="enter image description here" /></a></p> <p>B is the expected answer but should not the correct answer be BCD? Technically if a set has <strong>exactly</strong> <span class="math-container">$m$</span> elements, then it has <span class="math-container">$k$</span> elements if <span class="math-container">$k\leq m$</span>. This is also how we talk in everyday language:</p> <blockquote> <p>&quot;Do you have three dollars?&quot; &quot;Yes.&quot;</p> </blockquote> <p>The second speaker is not indicating he has exactly three dollars. He simply indicates that he has <strong>at least</strong> three dollars.</p> <p>So I am wondering if we are teaching children correct logic here. Shouldn't the original question be rephrased as &quot;which picture has <strong>exactly</strong> four apples&quot;?</p>
Mark Walker
15,960
<p>This was not the question:</p> <p>&quot;Which pictures have four apples?&quot;</p> <p>This test is a multiple choice question that does not appear to have other than the 4 unique choices: a, b, c, d.</p> <p>There is no all of the above or b-d choice which can be selected.</p> <p>If the test was a fill in the blank write in, there is a bit of ambiquity.</p> <p>More importantly though, examining the response with the student would be far better than pushing a black and white mindset.</p>
3,580,258
<p>Hi: The definition I'll use is this: Let <span class="math-container">$F$</span> be an abelian group and <span class="math-container">$X$</span> a subset of <span class="math-container">$F$</span>. Then <span class="math-container">$F$</span> is a free abelian group on <span class="math-container">$X$</span> if for every abelian group <span class="math-container">$G$</span> and every function <span class="math-container">$f$</span> from <span class="math-container">$X$</span> to <span class="math-container">$G$</span> there is a homomorphism <span class="math-container">$\phi$</span> from <span class="math-container">$F$</span> to <span class="math-container">$G$</span> that extends <span class="math-container">$f$</span>.</p> <p>Let <span class="math-container">$G$</span> be a finite group and <span class="math-container">$X$</span> a subset of <span class="math-container">$G$</span>. Let <span class="math-container">$F$</span> be the free abelian group on <span class="math-container">$X$</span>. Then <span class="math-container">$F=\langle X\rangle$</span> and so <span class="math-container">$F\subseteq G$</span>. That is, every finite group has an infinite subgroup. What am I doing wrong?</p> <p>EDIT: It will be easier to make myself clear working with free groups. I'll quote from Derek Robinson, A Course in the Theory of Groups, 2nd ed.</p> <p><a href="https://i.stack.imgur.com/BJioC.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/BJioC.png" alt=" "></a></p> <p>From this a free group is not only always free on a subset but additionally that subset generates it. If <span class="math-container">$G$</span> is a group and <span class="math-container">$X$</span> is a subset, however, indeed there will exist a free group on <span class="math-container">$X$</span> but I am unable to show it will be generated by <span class="math-container">$X$</span> based in the above quote. Which is very natural, of course. Thanks for the posts. Honestly none of the feedback, up to now, throws light in the paradox (paradox for me, of course).</p>
Shaun
104,041
<p>The <span class="math-container">$X$</span> in the free abelian group definition is a collection of formal symbols. They are assumed to have the minimum properties necessary for <span class="math-container">$\langle X\rangle$</span> to be an abelian group. That is what it means to be "free" in this context.</p> <p>However, for any <span class="math-container">$\Xi\subseteq G$</span> for a finite group <span class="math-container">$G$</span>, there are extra constraints on the <em>subgroup</em> <span class="math-container">$\langle \Xi\rangle_G$</span> of <span class="math-container">$G$</span>; for instance, each element of <span class="math-container">$\langle \Xi\rangle_G $</span> is of finite order. Thus it is not free.</p>
546,239
<p>Myself and my Math teacher are at a disagreement in to what the proper method of solving the question <em>In how many ways can four students be chosen from a group of 12 students?</em> is.</p> <p>The question comes straight out of a Math revision sheet from a Math book distributed under the national curriculum. The options it gives for answers are:</p> <ol> <li>12</li> <li>48</li> <li>495</li> <li>11880</li> <li>40320</li> </ol> <p>As we are currently learning Permutations and Combinations, my above interpretation is that it is asking for a <strong>Combination without repetition</strong> or $\frac{(n+r-1)!}{r!(n-1)!}$ which gives you the amount of combinations without repetition (as you cannot pick the same student twice.) Now my teacher argues that the answer the book provides is correct. The books answer simply says to use $^{n}C_{r}$ or $\frac{n!}{r!(n-r)!}$.</p> <p>What is the correct method of answering this? The book states <em>3. 495</em> is the answer.</p>
Benjamin Dickman
37,122
<p>You have 12 choices for the first student chosen, 11 choices for the next, then 10, then 9. However, this over-counts everything by a factor of 4! (the number of ways in which four objects can be arranged with regard to order).</p> <p>Thus, the answer is</p> <p>$$\frac{12\cdot 11 \cdot 10 \cdot 9}{4!} = 495$$</p>
2,203,770
<p>If I am asked to find all values of $z$ such that $z^3=-8i$, what is the best method to go about that?</p> <p>I have the following formula: $$z^{\frac{1}{n}}=r^\frac{1}{n}\left[\cos\left(\frac{\theta}{n}+\frac{2\pi k}{n}\right)+i\sin\left(\frac{\theta}{n}+\frac{2\pi k}{n}\right)\right]$$</p> <p>for $k=0,\pm1, \pm2,...$</p> <p>Applying this formula, I find the cubed root of $8$, which is $2$. And then when I apply it to the formula, I get the following:</p> <p>$$z = 2\left[\cos\left(\frac{\pi}{3}+\frac{2\pi k}{3}\right)+i\sin\left(\frac{\pi}{3}+\frac{2\pi k}{3}\right)\right]$$ for $k=0,\pm1, \pm2,...$</p> <p>I am confused, because the given solution is as follows: $$z = 2\left[\cos\left(\frac{\pi}{2}+\frac{2\pi k}{3}\right)+i\sin\left(\frac{\pi}{2}+\frac{2\pi k}{3}\right)\right]$$ for $k=0,\pm1, \pm2,...$</p> <p>Where did I go wrong? How would my approach changed if I was asked to find all values for $-8$, or $8i$?</p>
haqnatural
247,767
<p>Short way write it as $$z^{ 3 }=-8i\\ z^{ 3 }=8{ e }^{ i\frac { 3\pi }{ 2 } }\\ z=\sqrt [ 3 ]{ 8 } { e }^{ i\left( \frac { \frac { 3\pi }{ 2 } +2\pi k }{ 3 } \right) }=2{ e }^{ i\left( \frac { 3\pi +4\pi k }{ 6 } \right) },k=0,1,2\quad $$</p>
2,203,770
<p>If I am asked to find all values of $z$ such that $z^3=-8i$, what is the best method to go about that?</p> <p>I have the following formula: $$z^{\frac{1}{n}}=r^\frac{1}{n}\left[\cos\left(\frac{\theta}{n}+\frac{2\pi k}{n}\right)+i\sin\left(\frac{\theta}{n}+\frac{2\pi k}{n}\right)\right]$$</p> <p>for $k=0,\pm1, \pm2,...$</p> <p>Applying this formula, I find the cubed root of $8$, which is $2$. And then when I apply it to the formula, I get the following:</p> <p>$$z = 2\left[\cos\left(\frac{\pi}{3}+\frac{2\pi k}{3}\right)+i\sin\left(\frac{\pi}{3}+\frac{2\pi k}{3}\right)\right]$$ for $k=0,\pm1, \pm2,...$</p> <p>I am confused, because the given solution is as follows: $$z = 2\left[\cos\left(\frac{\pi}{2}+\frac{2\pi k}{3}\right)+i\sin\left(\frac{\pi}{2}+\frac{2\pi k}{3}\right)\right]$$ for $k=0,\pm1, \pm2,...$</p> <p>Where did I go wrong? How would my approach changed if I was asked to find all values for $-8$, or $8i$?</p>
Joe
107,639
<p>The problem with you solution is the following: you didn't wrote properly the modulus and the exponential part of $w:=-8i$.</p> <p>Clearly $$ r=|w|=8\\ -i=\exp{\left(\frac{3}{2}\pi i\right)}=\exp{\left(\frac{3}{2}\pi i+2ki\pi\right)}\;,\;\;k\in\Bbb Z $$ so that $$ z^3=r\exp[{i\pi\left(3/2+2k\right)}]\\ \Longrightarrow z_k =\sqrt[3]r\exp\left[{\frac{i\pi}3\left(\frac32+2k\right)}\right] =2\exp\left[{\frac{i\pi}2+\frac{2ki\pi}3}\right] $$ and as $k\in\Bbb Z$ you will notice that $z_k$ assumes only three different values, which are the solution of your initial equation.</p>
199,199
<p>Suppose a box contains 5 white balls and 5 black balls.</p> <p>If you want to extract a ball and then another:</p> <p>What is the probability of getting a black ball and then a black one?</p> <p>I think that this is the answer:</p> <p>Let $A:$ get a black ball in the first extraction, $B:$ get a black ball in the second extraction. Then: $P(A)=\frac{5}{10}$ and $$P(B|A)=\frac{P(A\cap B)}{P(A)}=\frac{\frac{4}{9}}{\frac{1}{2}}=\frac{8}{9}$$</p> <p>Is this correct? </p> <p>Now, If you want to extract two balls at the same time: what is the probability that both are black? Is exactly the same probability of the first question? Why?</p> <p>Thanks for your help.</p>
wnvl
23,122
<p>If the first ball is not replaced after extraction, both probabilities are the same.</p> <p>$P(\text{first ball black and second ball black})$</p> <p>$=P(\text{first ball black}) \cdot P(\text{second ball black } | \text{ first ball black})$</p> <p>$=\frac{5}{10}\cdot\frac{4}{9}$</p> <p>$=\frac{2}{9}$</p>
3,474,847
<p><strong>Prop:</strong> The union of two open sets is open. </p> <p><strong>Pf:</strong> Let <span class="math-container">$(X,d)$</span> be a metric space, and <span class="math-container">$U$</span> and <span class="math-container">$V$</span> are non empty and open sets such that <span class="math-container">$U\subset X$</span> and <span class="math-container">$V \subset X$</span>. Let <span class="math-container">$p$</span> be a point such that <span class="math-container">$d(p,q)&lt; \epsilon_1 $</span> for some <span class="math-container">$\epsilon_1 &gt; 0$</span> and <span class="math-container">$q \in U$</span>. And <span class="math-container">$d(p,q)&lt;e_2$</span> for some <span class="math-container">$\epsilon_2 &gt;0$</span> and <span class="math-container">$q \in V$</span>. Since we know that <span class="math-container">$q\in V$</span> and <span class="math-container">$q \in U$</span>, we notice <span class="math-container">$q \in( U \bigcup V)$</span>. Therefore, <span class="math-container">$q$</span> is an interior point of the union of those sets <span class="math-container">$\square$</span>.</p> <p>I want to know if my proof is well-phrased and proves the proposition.</p>
Fred
380,717
<p>With all due respect, your proof is chaotic. For <span class="math-container">$x \in X$</span> and <span class="math-container">$r&gt;0$</span> let us denote by <span class="math-container">$B_r(x)$</span> the set</p> <p><span class="math-container">$$B_r(x)= \{y \in X: d(y,x) &lt;r\}.$$</span></p> <p>If <span class="math-container">$U$</span> and <span class="math-container">$V$</span> are open subsets of <span class="math-container">$X$</span>, let <span class="math-container">$W:=U \cup V.$</span> We have to show that <span class="math-container">$W$</span> is open.</p> <p>To this end take <span class="math-container">$x \in W$</span>. Then <span class="math-container">$x \in U$</span> or <span class="math-container">$x \in V.$</span> WLOG we can assume that <span class="math-container">$x \in U.$</span> Then there is <span class="math-container">$r&gt;0$</span> such that</p> <p><span class="math-container">$$ B_r(x) \subseteq U.$$</span></p> <p>Then we derive that</p> <p><span class="math-container">$$ B_r(x) \subseteq W.$$</span></p> <p>The proof is complete.</p>
3,606,255
<p>I want to ask you something about the following quadratic inequality: <span class="math-container">$$9x^2+12x+4\le 0.$$</span> Let us find the "=0" points. Here we have <span class="math-container">$9x^2+12x+4=9(x+\frac{2}{3})^2\ge0$</span> (we can factor this by calculating the discriminant <span class="math-container">$D=36-36=0$</span> and using <span class="math-container">$ax^{2}+bx+c=a(x-x_1)(x-x_2)$</span>). I am practising the wavy curve method (the method of intervals) to solve quadratic inequalities, so I would want to solve it using this way. What is the general method to approach the quadratic inequality when the discriminant is zero, and we want to use the wavy curve method? <a href="https://i.stack.imgur.com/YjIB4.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/YjIB4.png" alt="enter image description here"></a></p>
Matteo
686,644
<p>In this case there is no need to find the intersection with <span class="math-container">$y=0$</span>, because you re considering the inequality: <span class="math-container">$$9(x+\frac{2}{3})^2\ge0$$</span> which is always true for every <span class="math-container">$x$</span> in <span class="math-container">$R$</span>. On the other side, when you have an inequality in the form: <span class="math-container">$$ax^2+bx+c\geq0$$</span> you can always rewite it as: <span class="math-container">$$a(x-x_1)(x-x_2)\geq0$$</span> and then study the sign of <span class="math-container">$x-x_1$</span> and <span class="math-container">$x-x_2$</span>. The interval where both factors are positive or both are negative are correct and are solutions to the inequality.</p> <p>Also, I recommend you to know this theorem:</p> <blockquote> <p>Give a polynomial <span class="math-container">$P(x)=a^+bx+c$</span>, the sign of <span class="math-container">$P(x)$</span> betweehn the roots <span class="math-container">$(P(x)=0)$</span> is opposite respect to the sign outsider the interval of the roots.</p> </blockquote>
1,351,881
<p>This argument comes up once every while on Lambda the Ultimate. I want to know where the flaw is.</p> <p><em>Take a countable number of TMs all generating different bitstreams. Construct a Cantor TM which runs every nth TM upto the nth bit and outputs the reversed bit.</em></p> <p><em>Now I have established there are uncountably many TMs.</em></p> <p>(The argument is that every bitstream is a unique real in the set [0,1]. Since every TM can be seen to be represented by that real, we can use Cantor's construction to prove that there are uncountably many TMs.)</p> <p>Where does it go wrong?</p>
Marco
247,273
<p>There are only a countable number of constructible reals. The Cantor TM only proves that any enumeration can be extended with an extra element. Since it didn't start of with a bijection between the natural numbers and the reals, the construction doesn't prove that that bijection doesn't exist.</p> <p>(I answered my own question by summarizing the answers.)</p>
1,351,881
<p>This argument comes up once every while on Lambda the Ultimate. I want to know where the flaw is.</p> <p><em>Take a countable number of TMs all generating different bitstreams. Construct a Cantor TM which runs every nth TM upto the nth bit and outputs the reversed bit.</em></p> <p><em>Now I have established there are uncountably many TMs.</em></p> <p>(The argument is that every bitstream is a unique real in the set [0,1]. Since every TM can be seen to be represented by that real, we can use Cantor's construction to prove that there are uncountably many TMs.)</p> <p>Where does it go wrong?</p>
Russell Easterly
51,106
<p>There are several problems with this argument. If you want to use contradiction to prove the set of all TM's is uncountable you must first assume there is an enumeration of all TM's. This "proof" only assumes we have a countably infinite set of TM's with unique output. There are a lot of such sets. One such is the set of TM's that output the binary representation of an unique rational number. Using a diagonal argument to prove there is a TM that outputs the representation of an irrational number proves nothing about the size of this set of rational output TM's.</p> <p>The argument fails even if we assume we have an enumeration of all TM's. To make things simpler, assume we have an enumeration of all <a href="http://www.drivenbycode.com/turing-machines-and-the-busy-beaver-game/" rel="nofollow">busy beavers</a> (BB) and want to prove there is an uncountable number of BB's. There are $((4*(n+1))^{2n}$ possible 2 symbol BB's with $n$ states. We can encode all $n$ states of a specific machine into a natural number. Once we specify an encoding we can enumerate all the valid numbers that encode a BB. This gives us an enumeration of all BB's.</p> <p>The argument says to construct a BB that runs the BB encoded as $n$ for $n$ steps and reverses the output. A BB has to output something on every step. If the BB halts before step $n$ we can assume it outputs $0$ on step $n$. </p> <p>This is the second major flaw in this argument. There is no way to construct such a BB. We are assuming on step $n$ this BB outputs the opposite of the $nth$ BB. Obviously, no BB can emulate the first $n$ steps of another BB in less than $n$ steps. This magical BB would have to know in advance what to output on step $n$.</p> <p>The argument still fails if we try to construct a BB that outputs the anti-diagonal. We can construct such a BB for the first $n$ machines. A BB that outputs the anti-diagonal for all BB's would have to have an infinite number of states. By definition, a BB can only have a finite number of states.</p>
2,606,999
<p>$f: X \to Y$ prove that if $A \subseteq X$, then A is a subset of the pre-image of the image of $A$, which is shown by these symbols respectively: $f^{-1}[f[A]]$</p> <p>This is my proof:</p> <p>If $A \nsubseteq f^{-1}[f[A]]$, then there exists "$x$" as an element of $A$, such that $f^{-1}[f[A]] \neq x$ which is a contradiction.</p> <p>Is this the correct way of going about doing this? I'm also not sure if we can assume that $f^{-1}[f[A]] = x$ is always true</p>
egreg
62,967
<p>You might prove it by contradiction like in your attempt. However $$ A\not\subseteq f^{-1}[f[A]] $$ means that</p> <blockquote> <p>there exists $x\in A$ such that $x\notin f^{-1}[f[A]]$</p> </blockquote> <p><em>There's a big difference between “$\notin$” and “$\ne$”.</em></p> <blockquote> <p>Since $x\notin f^{-1}[f[A]]$, we have $f(x)\notin f[A]$. However, as $x\in A$, we have $x\in f[A]$: a contradiction.</p> </blockquote> <hr> <p>Now we can transform this in a proof without contradiction.</p> <blockquote> <p>Suppose $x\notin f^{-1}[f[A]]$. Then $f(x)\notin f[A]$ and therefore $x\notin A$.</p> <p>By contrapositive, if $x\in A$, we must have also $x\in f^{-1}[f[A]]$.</p> </blockquote> <p>The main fact used is that, for $B$ a subset of $Y$ and $x\in X$, $$ x\in f^{-1}[B]\quad\text{if and only if}\quad f(x)\in B $$ applying this to $B=f[A]$.</p>
1,506,763
<p>This is not true for infinite measures (<a href="https://math.stackexchange.com/questions/342039/pointwise-convergence-but-not-in-measure">Pointwise convergence, but not in measure</a>). Is it true for a finite measure? Namely, let a finite (probability) measure $\mu(\cdot)$. Does a point-wise convergence of $\mu-$measurable functions, $\{f_n(x)\}$, imply convergence in measure $\mu(\cdot)$?</p> <p>Equivalently, does $$ \forall x~\lim_{n\to \infty} f_n(x) \to f(x)\,, $$ imply $$ \forall \varepsilon ~\lim_{n\to \infty} \mu\big{(}\left\{x~\big{|}~|(f_n(x) - f(x)|&gt;\varepsilon\right\}\big{)} \to 0\,~? $$</p>
Reveillark
122,262
<p>If $f_n \overset{a.e.}{\longrightarrow} f$ and the space has finite measure, by Egoroff's theorem $f_n$ converges to $f$ almost uniformly, and therefore converges in measure to $f$.</p>
1,506,763
<p>This is not true for infinite measures (<a href="https://math.stackexchange.com/questions/342039/pointwise-convergence-but-not-in-measure">Pointwise convergence, but not in measure</a>). Is it true for a finite measure? Namely, let a finite (probability) measure $\mu(\cdot)$. Does a point-wise convergence of $\mu-$measurable functions, $\{f_n(x)\}$, imply convergence in measure $\mu(\cdot)$?</p> <p>Equivalently, does $$ \forall x~\lim_{n\to \infty} f_n(x) \to f(x)\,, $$ imply $$ \forall \varepsilon ~\lim_{n\to \infty} \mu\big{(}\left\{x~\big{|}~|(f_n(x) - f(x)|&gt;\varepsilon\right\}\big{)} \to 0\,~? $$</p>
shalop
224,467
<p>Yeah this is true. You can use Egoroff, but a direct proof using only DCT is not difficult either.</p> <p>Let $\epsilon&gt;0$. Define $A_n:=\{x: |f(x)-f_n(x)|&gt;\epsilon\}$. Since $f_n \to f$ pointwise, it follows that $1_{A_n} \to 0$ pointwise. Then notice that all of the functions $1_{A_n}$ are bounded above by the constant function $1$, which is in $ L^1$ (since $\mu$ is finite). By DCT, it follows that $\mu(A_n) = \int 1_{A_n} d\mu \to 0$. But $\mu(A_n) \to 0$ precisely means that $f_n \to f$ in measure.</p>
1,515,417
<p>I understand the idea that some infinities are "bigger" than other infinities. The example I understand is that all real numbers between 0 and 1 would not be able to "fit" on an infinite list.</p> <p>I have to show whether these sets are countable or uncountable. If countable, how would you enumerate the set? If uncountable, how would you prove using diagonalization? </p> <p>Set 1. All real numbers represented only by 1's. EX) 1, .11, 111.11, 1.111...</p> <p>Set 2. All real numbers represented only by 2's and 3's. EX) .2, 23.2, 22.2232...</p>
ncmathsadist
4,154
<p>Set 1 is very countable; it can be placed into 1-1 correspondence with the integers. Set 2 is not countable. It is extremally disconnected and its intersection with $[0,1]$ is homeomorphic to a Cantor set.</p>
335,258
<p>Find the domain of the function: $$f(x)= \sqrt{x^2 - 4x - 45}$$</p> <p>I'm just guessing here; how about if I square everything and then put it in the graphing calculator? Thanks, Lauri</p>
amWhy
9,003
<p>The function $f(x)$ is not defined when $$x^2 - 4x - 45 \lt 0,\tag{1}$$ as the square root function is defined in the reals for non-negative reals only. The only valid "input" for the square root function is non-negative real numbers.</p> <p>If a function $f(x)$ is not defined on an interval(s) = $I \subseteq \mathbb R$, or is not defined for any particular $c\in \mathbb R$, then the interval(s) and/or points at which $f(x)$ is <strong><em>not defined</em></strong> are <strong><em>excluded</em></strong> from the function's <strong><em>domain</em></strong>.</p> <p>So what are you left with if you exclude all $x\in \mathbb R$ such that $x^2 - 4x - 45 &lt; 0$? This amounts to excluding your solutions to the inequality given in $(1)$. What remains in $\mathbb R$ is then your domain. </p> <p>Alternatively, your domain consists of all real $x$ such that $$x^2 - 4x - 45 \geq 0\tag{2}$$</p> <p>Then your task would then be, essentially, to determine the solutions to the equality given by $(2)$. its solution will be your domain.</p>
3,901,613
<p>Determine <span class="math-container">$\int_Axy\space\space d(x,y)$</span>, when <span class="math-container">$A$</span> is a closed set, bounded by the curve <span class="math-container">$y=x-1$</span>, and the parabel <span class="math-container">$y^2 = 2x+6.$</span></p> <p>I believe we only need to know the range, where <span class="math-container">$A$</span> is actually bounded by these two curves. I think this will be <span class="math-container">$-3\leq x \leq 5$</span>. But to be specific, isn't <span class="math-container">$A\subset Int ( G_h \cap G_g)$</span>, s.t <span class="math-container">$g(x)=x-1$</span>, and <span class="math-container">$h(x)=y^2 = 2x+6$</span>? Where <span class="math-container">$-3\leq x \leq 5$</span>. Can we determine the integral as <span class="math-container">$$\int_{-3}^5\Bigg(\int_{x-1}^{2x+6}xy\space\space dy\Biggr)dx\space\space?$$</span></p> <p>Also, I'm assuming that <span class="math-container">$(x,y)\in \Bbb{R^2}$</span>.</p>
Z Ahmed
671,540
<p>Take log on base 2, then <span class="math-container">$$16^x \log_2 2=2^x \log_2 16. \implies 16^x= 4. 2^x$ \implies 2^{4x}=2^{x+2} \implies 4x=x+2 \implies x=2/3.$$</span></p>
3,901,613
<p>Determine <span class="math-container">$\int_Axy\space\space d(x,y)$</span>, when <span class="math-container">$A$</span> is a closed set, bounded by the curve <span class="math-container">$y=x-1$</span>, and the parabel <span class="math-container">$y^2 = 2x+6.$</span></p> <p>I believe we only need to know the range, where <span class="math-container">$A$</span> is actually bounded by these two curves. I think this will be <span class="math-container">$-3\leq x \leq 5$</span>. But to be specific, isn't <span class="math-container">$A\subset Int ( G_h \cap G_g)$</span>, s.t <span class="math-container">$g(x)=x-1$</span>, and <span class="math-container">$h(x)=y^2 = 2x+6$</span>? Where <span class="math-container">$-3\leq x \leq 5$</span>. Can we determine the integral as <span class="math-container">$$\int_{-3}^5\Bigg(\int_{x-1}^{2x+6}xy\space\space dy\Biggr)dx\space\space?$$</span></p> <p>Also, I'm assuming that <span class="math-container">$(x,y)\in \Bbb{R^2}$</span>.</p>
user2661923
464,411
<p>Alternative approach</p> <p><span class="math-container">$$16^{2^x} = [2^4]^{2^x} = 2^{4 \times 2^x}.$$</span></p> <p>Since this is equal to <span class="math-container">$2^{16^x}$</span> you have</p> <p><span class="math-container">$$16^x = 4 \times 2^x \implies 8^x = 4 \implies x = (2/3).$$</span></p>
167,440
<blockquote> <p>Let $\Bbb F$ be a field of characteristic $p\gt 0$ and $f(x)=x^{p^n}-c \in\Bbb F[x]$ where $n$ is a positive integer. If $c \notin \{a^p:a\in \Bbb F \}$, show that $f$ is irreducible in $\Bbb F[x]$.</p> </blockquote> <p>I recently started studying Field theory, so I don't know How to approach this problem? Any help would be appreciated.</p>
Georges Elencwajg
3,217
<p>Let $\gamma\in \mathbb F^{alg}$ be a root of $f(x)=x^{p^n}-c \in\Bbb F[x]$ in an algebraic closure of $\mathbb F$.<br> Thus we have $f(x)=x^{p^n}-c=(x-\gamma)^{p^n}$.<br> On the other hand let $m(x)\in \Bbb F[x]$ be the minimal polynomial of $\gamma$ over $\Bbb F$. </p> <p>If $n(x)$ is an irreducible monic factor of $f(x)$ over $\Bbb F$, its only root in $\mathbb F^{alg}$ is $\gamma$, so that $n(x)$ must also be the minimal polynomial of $\gamma$ in $\Bbb F[x]$. Hence $n(x)=m(x)$. </p> <p>This means that the decomposition of $f(x)$ into irreducibles in $\mathbb F[x]$ is $f(x)=m(x)^{p^e}$ for some integer $e\geq 0$ [the exponent is a power of $p$ because it divides $p^n=deg (f(x))$].<br> More explicitly we have $f(x)=x^{p^n}-c =m(x)^{p^e}=(x^s+...+m_0)^{p^e}$.<br> But then $c=-m_0^{p^e}=(-m_0)^{p^e}$ and since $c$ is supposed not to be a $p$th-power in $\Bbb F$, this forces $e=0$ and so $f(x)=m(x)^{p^e}=m(x)$ is irreducible in $\Bbb F[x]$</p>
3,317,088
<blockquote> <p>Let <span class="math-container">\begin{align*} \mathcal{P} : &amp;\min_{x}~&amp; (x + a)^2 \\ &amp; s.t. &amp; x + b \leq 0 \end{align*}</span></p> </blockquote> <p>where <span class="math-container">$ a, b \in \mathbb{R} $</span>. I would like to find <span class="math-container">$ x $</span>.</p> <p><strong>My attempt:</strong> The Lagrangian is <span class="math-container">$ L_\pi = |x + a|^2 + \pi (x + b) $</span>.</p> <p>Stationarity: <span class="math-container">$ x + a + \pi = 0 $</span></p> <p>Complementary slackness: <span class="math-container">$ \pi (x + b) = 0 $</span></p> <ul> <li>Replacing <span class="math-container">$ x = -\pi - a $</span> in <span class="math-container">$ \pi (x + b) = 0 $</span>, we obtain that <span class="math-container">$ \pi (-\pi - a + b) = 0 $</span>.</li> </ul> <p>This is the point where I am not able to continue. I get two roots por <span class="math-container">$ \pi $</span> but I am unable to realize which one I should use to get <span class="math-container">$ x $</span>.</p>
Alireza Shamsian
573,454
<p>That's the point! As <span class="math-container">$R$</span> is a domain, we conclude <span class="math-container">$\langle0\rangle$</span> is a prime ideal of <span class="math-container">$R$</span>. So <span class="math-container">$\dim R=r$</span> means that there is a maximal chain like <span class="math-container">$\langle0\rangle=P_0\subsetneq P_1\subsetneq\cdots\subsetneq P_r$</span> such that <span class="math-container">$\forall 0\leq i\leq r$</span> we have <span class="math-container">$P_i \in \operatorname{Spec}(R)$</span>. By definition we have: <span class="math-container">$$\dim\dfrac{R}{P}=\max\{s\in\mathbb{N}: \left&lt; 0 \right&gt;=Q_0\subsetneq Q_1\subsetneq\cdots\subsetneq Q_s\quad\text{such that for each } i,Q_i\in\operatorname{Spec}(\dfrac{R}{P})\}.$$</span> But we know that <span class="math-container">$$\operatorname{Spec}(\dfrac{R}{P})=\{Q\in\operatorname{Spec}(R) : P\subseteq Q\}.$$</span> As <span class="math-container">$P\neq0$</span>, so <span class="math-container">$s\leq r-1$</span>.</p>
30,402
<p>The envelope of parabolic trajectories from a common launch point is itself a parabola. In the U.S. soon many will have a chance to observe this fact directly, as the 4th of July is traditionally celebrated with fireworks.</p> <p>If the launch point is the origin, and the trajectory starts off at angle $\theta$ and velocity $v$, then under unit gravity it follows that the parabola $$ y = x \tan \theta - [x^2 /(2 v^2)] (1 + \tan^2 \theta) $$ and the envelope of all such trajectories, is another parabola: $$ y = v^2 /2 - x^2 / (2v^2) $$ <hr> &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; <img src="https://i.stack.imgur.com/tn5pg.jpg" alt="envelope"></p> <hr> <p>These equations are not difficult to derive. I have two questions. First, is there a way to see that the envelope of parabolic trajectories is itself a parabola, without computing these equations? Is there a purely geometric argument? Perhaps there is a way to nest cones and obtain the above picture through conic sections, but I couldn't see it.</p> <p>Second, of course the trajectories are actually pieces of ellipses, not parabolas, if we follow the true inverse-square law of gravity. Is the envelope of these elliptical trajectories also an ellipse? (I didn't try to work out the equations.) Perhaps the same geometric viewpoint (if it exists) could apply, e.g. by slightly tilting the sections.</p>
Andrey Rekalo
5,371
<ul> <li><p>E. Torricelli, who was the last Galileo's secretary, suggested a purely geometrical method to find the envelope in his <em>De motu Proiectorum</em>. He also coined the term `parabola of safety'. Apparently it was the first example of computation of an envelope. The method is briefly described in <a href="http://www.jstor.org/stable/3620177?seq=1">this note</a>.</p></li> <li><p>Another approach is to launch identical missiles with the same velocity at all possible angles simultaneously. At time $t$, their positions describe a circle $$x^2+\left(y-\frac{t^2}{2}\right)^2=(vt)^2.$$ The latter equation has a unique solution in $t$ provided $(x,y)$ belongs to the parabola $$y=\frac{v^2}{2}-\frac{x^2}{2v^2}.$$</p></li> <li><p>In the case of missiles moving in a Kepler field (with the attractive potential $\sim -1/r$), the envelope of elliptic trajectories is indeed an ellipse. A web search gave the nice short <a href="http://iopscience.iop.org/0143-0807/25/6/016/pdf/0143-0807_25_6_016.pdf">article</a> which contains several elementary geometric proofs of this and related results.</p></li> </ul> <p><strong>Edit.</strong> A free version of J.-M. Richard article can be found <a href="http://arxiv.org/abs/physics/0410034">here</a>.</p>
1,119,524
<p>Find $\lim_{x \rightarrow 3} \frac{x^2 - 9}{x - 3}$</p> <p>My professor showed us a few ways to compute the limit</p> <p>1) Factor the numerator</p> <p>$\lim_{x \rightarrow 3} \frac{x^2 - 9}{x - 3} = \lim_{x \rightarrow 3} (x + 3) = 6$</p> <p>2)</p> <p>$\lim_{x \rightarrow 3} \frac{f(x) - f(3)}{x - 3} = f'(3) = 2 \cdot 3 = 6$</p> <p>Here I'm a little confused, where did $\frac{f(x) - f(3)}{x - 3}$ come from? Also, how does $\lim_{x \rightarrow 3} f(x) = f'(3)$?</p>
davidlowryduda
9,754
<p>Let's get you started.</p> <p>Suppose without loss of generality that $\mu_1 \leq \mu_2$. Then we have that $$ \mu = x_1\mu_1 + x_2\mu_2 \leq (x_1 + x_2)\mu_2 = \mu_2.$$ Each other subcomponent of the proof looks extremely similar to this line.</p>
1,119,524
<p>Find $\lim_{x \rightarrow 3} \frac{x^2 - 9}{x - 3}$</p> <p>My professor showed us a few ways to compute the limit</p> <p>1) Factor the numerator</p> <p>$\lim_{x \rightarrow 3} \frac{x^2 - 9}{x - 3} = \lim_{x \rightarrow 3} (x + 3) = 6$</p> <p>2)</p> <p>$\lim_{x \rightarrow 3} \frac{f(x) - f(3)}{x - 3} = f'(3) = 2 \cdot 3 = 6$</p> <p>Here I'm a little confused, where did $\frac{f(x) - f(3)}{x - 3}$ come from? Also, how does $\lim_{x \rightarrow 3} f(x) = f'(3)$?</p>
Brian M. Scott
12,042
<p>HINT: Without loss of generality assume that $\mu_1\le\mu_2$. Then</p> <p>$$\mu=x_1\mu_1+x_2\mu_2=x_1\mu_1+x_2\big(\mu_1+(\mu_2-\mu_1)\big)=(x_1+x_2)\mu_1+\ldots$$</p> <p>Finish this off, and it gives you half of what you want, and a similar idea applied to $\mu_2$ gets you the other hald.</p> <p>Then try to apply the basic idea to the general case.</p>
141,112
<p>I'm trying to put the following equation in determinant form: $12h^3 - 6ah^2 + ha^2 - V = 0$, where $h, a, V$ are variables (this is a volume for a pyramid frustum with $1:3$ slope, $h$ is the height and $a$ is the side of the base, $V$ is the volume). </p> <p>The purpose of identifying the determinant is to construct a nomogram. I'm not sure if it actually can be placed in determinant form, and I'm curious if there is a Mathematica function that can do this? I've been trying a pen and pencil approach as listed <a href="http://www.projectrho.com/nomogram/determinant06.html" rel="nofollow">here</a>. But this approach has hopefully been automated. </p> <p>Any tips are appreciated!</p>
user326210
326,210
<p>You can prove, using Warmus's criteria, that the function can't be nomographed as it's currently written. Roughly speaking, it has four independent <span class="math-container">$h$</span> terms <span class="math-container">$(h^0, h^1, h^2, h^3)$</span> that can't be reduced further, but you can only directly nomograph functions where each variable has up to three independent terms ("rank 3").</p> <p>Specifically, the fact that <span class="math-container">$h$</span> has rank 4 proves that there are no functions <span class="math-container">$f_i,g_i,h_i$</span> such that: </p> <p><span class="math-container">$$F(a,h,V) = \det\left|\begin{array}{ccc}f_1(a) &amp;f_2(a) &amp;f_3(a)\\g_1(h)&amp;g_2(h)&amp;g_3(h)\\h_1(V) &amp; h_2(V) &amp; h_3(V)\end{array}\right|$$</span></p> <p>That said, Warmus's criteria <em>only</em> cover the existing form of an expression. They do cover all ways of grouping terms together differently, but don't account for even easy manipulations such as taking logarithms or changing variables, and they don't account for straightforward compound nomograms. </p> <p>So we haven't ruled out everything. A nomogram is possible wherever you can find functions such that: <span class="math-container">$$F(a,h,V) = 0 \iff\det \left|\begin{array}{ccc}f_1(a) &amp;f_2(a) &amp;f_3(a)\\g_1(h)&amp;g_2(h)&amp;g_3(h)\\h_1(V) &amp; h_2(V) &amp; h_3(V)\end{array}\right|=0$$</span></p> <p>and that's a much more expansive category than what we've ruled out. It occurs when the determinant on the right has the same zeroes as the function on the left, even if the two sides don't agree at nonzero values.</p>
4,215,824
<p>Is there a formal definition of &quot;almost always less than&quot; or &quot;almost always greater than&quot;? I think one could define it using probabilities but not sure how to go about it. If one could show the following, then I think you could say <span class="math-container">$X$</span> is almost always less than a value <span class="math-container">$x$</span>. Is there other ways of going about this?</p> <p><span class="math-container">$$ P(X&lt;x)=1 $$</span></p>
User5678
632,875
<p>Yes, a more typical terminology would be to say “almost surely (a.s.) less than <em>value</em>”</p> <p><span class="math-container">$$X &lt; c\;a.s \iff P(X&lt;c)=1$$</span></p> <p>Now, if the above holds <em>and</em> <span class="math-container">$\{\omega \in \Omega:X(\omega)\geq c\} = \emptyset$</span> then we can say “surely” since there isn’t even a set of probability <span class="math-container">$0$</span> that results in <span class="math-container">$X\geq c$</span></p>
1,932,599
<p>Find the solution of the differential equation that satisfies the given initial condition <span class="math-container">$$~y'=\frac{x~y~\sin x}{y+1}, ~~~~~~y(0)=1~$$</span></p> <p>When I integrate this function I get <span class="math-container">$$y+\ln(y)= -x\cos x + \sin x + C.$$</span></p> <p>Have I integrated the function correctly? </p> <p>How do I complete the second part of the question <span class="math-container">$~y(0)=1~$</span></p>
ComplexF
369,196
<p>What you can do is to use the so-called <em>Separation of variables</em> (see for example <a href="https://en.wikipedia.org/wiki/Separation_of_variables" rel="nofollow">https://en.wikipedia.org/wiki/Separation_of_variables</a>):</p> <ol> <li>Write $y' = \frac{dy}{dx}$</li> <li>Do the following transformation: $\frac{dy}{dx} = \frac{xy \sin(x)}{y+1} \Longleftrightarrow \frac{y+1}{y} dy = x\sin(x) dx$. Now integrate both sides (with respect to the corresponding variable), arriving at</li> <li>$\int (1 + \frac{1}{y}) dy = \int x \sin(x) dx \Longleftrightarrow y + \ln(y) = \sin(x) - x \cos(x) + C$.</li> <li>To get the integration constant $C$, set $x = 0$ and $y = 1$, then: $1 = 0 + C \Longleftrightarrow C = 1$.</li> <li>You'll not be able to solve the equation in 3 for the function $y = y(x)$ in terms of <em>elementary</em> functions (such as trigonometric functions, exponential/logarithmic functions etc.); But what you can do is to use the so-called <em>Lambert W function</em>, see <a href="https://en.wikipedia.org/wiki/Lambert_W_function" rel="nofollow">https://en.wikipedia.org/wiki/Lambert_W_function</a>. Then you will get $y(x) = W(e^{\sin(x) - x \cos(x) + 1})$.</li> </ol>
3,102,218
<p>Given a fraction:</p> <p><span class="math-container">$$\frac{a}{b}$$</span></p> <p>I now add a number <span class="math-container">$n$</span> to both numerator and denominator in the following fashion:</p> <p><span class="math-container">$$\frac{a+n}{b+n}$$</span></p> <p>The basic property is that the second fraction is suppose to closer to <span class="math-container">$1$</span> than the first one. My question is how can we prove that?</p> <p>What I have tried:</p> <p>I know <span class="math-container">$\frac{n}{n} = 1$</span> so now adding numbers <span class="math-container">$a$</span> and <span class="math-container">$b$</span> to it would actually "move it away" from <span class="math-container">$1$</span>. But I cannot understand why <span class="math-container">$\frac{a}{b}$</span> is actually farther away from <span class="math-container">$1$</span> than <span class="math-container">$\frac{a+n}{b+n}$</span>.</p> <p>Why is that? What does it mean to add a number to both the numerator and denominator?</p>
Wuestenfux
417,848
<p>Well, <span class="math-container">$\frac{a+n}{b+n} = \frac{\frac{a}{n}+1}{\frac{b}{n}+1}$</span>. So if <span class="math-container">$n\rightarrow \infty$</span>, then <span class="math-container">$\frac{a}{n}\rightarrow 0$</span> and <span class="math-container">$\frac{b}{n}\rightarrow 0$</span>. Thus <span class="math-container">$\frac{a+n}{b+n}\rightarrow 1$</span>.</p> <p>As said in the comments, the answer is incorrect in that it does not address precisely what the OP asks, but gives some intuition as to why it is true.</p>
3,102,218
<p>Given a fraction:</p> <p><span class="math-container">$$\frac{a}{b}$$</span></p> <p>I now add a number <span class="math-container">$n$</span> to both numerator and denominator in the following fashion:</p> <p><span class="math-container">$$\frac{a+n}{b+n}$$</span></p> <p>The basic property is that the second fraction is suppose to closer to <span class="math-container">$1$</span> than the first one. My question is how can we prove that?</p> <p>What I have tried:</p> <p>I know <span class="math-container">$\frac{n}{n} = 1$</span> so now adding numbers <span class="math-container">$a$</span> and <span class="math-container">$b$</span> to it would actually "move it away" from <span class="math-container">$1$</span>. But I cannot understand why <span class="math-container">$\frac{a}{b}$</span> is actually farther away from <span class="math-container">$1$</span> than <span class="math-container">$\frac{a+n}{b+n}$</span>.</p> <p>Why is that? What does it mean to add a number to both the numerator and denominator?</p>
José Carlos Santos
446,262
<p>You should start by thinking about particular cases. For instance, <span class="math-container">$\dfrac{3+2}{7+2}=\dfrac59$</span>, which is indeed closer to <span class="math-container">$1$</span> than <span class="math-container">$\dfrac37$</span>.</p> <p>Anyway, note that, if <span class="math-container">$a&lt;b$</span> (and consequently, <span class="math-container">$a+n&lt;b+n$</span>, for which <span class="math-container">$\frac ab&lt;1$</span> and <span class="math-container">$\frac{a+n}{b+n} &lt; 1$</span>), then<span class="math-container">$$\frac{a+n}{b+n}-\frac ab=\frac{(a+n)b-a(b+n)}{(b+n)b}=\frac{n(b-a)}{(b+n)b}&gt;0$$</span> This shows <span class="math-container">$\frac{a+n}{b+n}-\frac ab&gt;0$</span>, and we already know both are <span class="math-container">$&lt;1$</span>, so: <span class="math-container">$$\frac ab&lt;\frac{a+n}{b+n}&lt;1.$$</span>So, yes, <span class="math-container">$\dfrac{a+n}{b+n}$</span> is closer to <span class="math-container">$1$</span> than <span class="math-container">$\dfrac ab$</span>.</p> <p>Can you deal with the case <span class="math-container">$a&gt;b$</span> now?</p>
3,102,218
<p>Given a fraction:</p> <p><span class="math-container">$$\frac{a}{b}$$</span></p> <p>I now add a number <span class="math-container">$n$</span> to both numerator and denominator in the following fashion:</p> <p><span class="math-container">$$\frac{a+n}{b+n}$$</span></p> <p>The basic property is that the second fraction is suppose to closer to <span class="math-container">$1$</span> than the first one. My question is how can we prove that?</p> <p>What I have tried:</p> <p>I know <span class="math-container">$\frac{n}{n} = 1$</span> so now adding numbers <span class="math-container">$a$</span> and <span class="math-container">$b$</span> to it would actually "move it away" from <span class="math-container">$1$</span>. But I cannot understand why <span class="math-container">$\frac{a}{b}$</span> is actually farther away from <span class="math-container">$1$</span> than <span class="math-container">$\frac{a+n}{b+n}$</span>.</p> <p>Why is that? What does it mean to add a number to both the numerator and denominator?</p>
Steven Alexis Gregory
75,410
<p>Suppose <span class="math-container">$a,b,n \in \mathbb Q$</span>, <span class="math-container">$0 &lt; a &lt; b$</span> and <span class="math-container">$n &gt; 0$</span>.</p> <p><span class="math-container">$$\color{red}{\dfrac ab} = \dfrac{a(b+n)}{b(b+n)} = \dfrac{ab+an}{b(b+n)} \color{red}{&lt;} \dfrac{ab+bn}{b(b+n)} = \dfrac{b(a+n)}{b(b+n)} = \color{red}{\dfrac{a+n}{b+n}} \color{red}{&lt;} \dfrac{b+n}{b+n} = \color{red}1$$</span></p>
3,102,218
<p>Given a fraction:</p> <p><span class="math-container">$$\frac{a}{b}$$</span></p> <p>I now add a number <span class="math-container">$n$</span> to both numerator and denominator in the following fashion:</p> <p><span class="math-container">$$\frac{a+n}{b+n}$$</span></p> <p>The basic property is that the second fraction is suppose to closer to <span class="math-container">$1$</span> than the first one. My question is how can we prove that?</p> <p>What I have tried:</p> <p>I know <span class="math-container">$\frac{n}{n} = 1$</span> so now adding numbers <span class="math-container">$a$</span> and <span class="math-container">$b$</span> to it would actually "move it away" from <span class="math-container">$1$</span>. But I cannot understand why <span class="math-container">$\frac{a}{b}$</span> is actually farther away from <span class="math-container">$1$</span> than <span class="math-container">$\frac{a+n}{b+n}$</span>.</p> <p>Why is that? What does it mean to add a number to both the numerator and denominator?</p>
Milan
503,397
<p><span class="math-container">$$ f(x)=\frac{a+x}{b+x} $$</span> <span class="math-container">$ b &gt; 0$</span></p> <p><span class="math-container">$$\lim_{x→ ∞} \frac{a+x}{b+x}=1$$</span> <span class="math-container">$$ f'(x)= \frac{b+x-a-x}{(b+x)^2}=\frac{b-a}{(b+x)^2}$$</span></p> <p>We can conclude if <span class="math-container">$b &gt; a$</span> the function is monotonically increasing to 1.</p> <p>If <span class="math-container">$b &lt; a $</span> function is monotonically decreasing to 1</p> <p>If <span class="math-container">$b&lt;0$</span> the conclusion doesnt follow because there exists a vertical asymptote at <span class="math-container">$x=-b$</span></p>
3,194,249
<p>If we are given <span class="math-container">$d$</span> an integer, then I've seen it written that the floor of <span class="math-container">$\sqrt{d^2-1}=d-1$</span>, but how is this found?</p> <p>It's not immediately obvious, at least not to me.It makes me feel there must be an algorithm that can be used . </p> <p>Would anyone be able to impart their knowledge on how to find the floor of such an expression?</p> <p>I read it here : <a href="https://math.stackexchange.com/questions/2230529/show-that-the-simple-continued-fraction-of-sqrtd2-1-is-d-1-overline1">Show that the simple continued fraction of $\sqrt{d^2-1}$ is $[d-1; \overline{1, 2d-1}]$ for $d \geq 2$</a></p>
timon92
210,525
<p>It is because <span class="math-container">$$(d-1)^2 \le d^2-1 &lt; d^2,$$</span> so <span class="math-container">$$d-1 \le \sqrt{d^2-1} &lt; d,$$</span> so <span class="math-container">$$\lfloor \sqrt{d^2-1}\rfloor = d-1.$$</span></p>
2,259,099
<blockquote> <p><em><strong>Theorem 29.2.</strong></em> Let <span class="math-container">$X$</span> be a Hausdorff space. Then <span class="math-container">$X$</span> is locally compact if and only if given <span class="math-container">$x \in X$</span>, and given a neighborhood <span class="math-container">$U$</span> of <span class="math-container">$x$</span>,there is a neighborhood <span class="math-container">$V$</span> of <span class="math-container">$x$</span> such that <span class="math-container">$\overline{V}$</span> is compact and <span class="math-container">$\overline{V} \subseteq U$</span>.</p> <p><em>Proof</em>: Clearly this new formulation implies local compactness; the set <span class="math-container">$C = \overline{V}$</span> is the desired compact set containing a neighborhood of <span class="math-container">$x$</span>. To prove the converse, suppose <span class="math-container">$X$</span> is locally compact, let <span class="math-container">$x$</span> be a point of <span class="math-container">$X$</span> and let <span class="math-container">$U$</span> be a neighborhood of <span class="math-container">$x$</span>. Take the one-point compactification <span class="math-container">$Y$</span> of <span class="math-container">$X$</span>, and let <span class="math-container">$C$</span> be the set <span class="math-container">$Y - U$</span>. Then <span class="math-container">$C$</span> is closed in <span class="math-container">$Y$</span>, so that <span class="math-container">$C$</span> is a compact subspace of <span class="math-container">$Y$</span>. Apply Lemma 26.4 to choose disjoint open sets <span class="math-container">$V$</span> and <span class="math-container">$W$</span> containing <span class="math-container">$x$</span> and <span class="math-container">$C$</span>, respectively. Then the closure <span class="math-container">$\overline{V}$</span> of <span class="math-container">$V$</span> in <span class="math-container">$Y$</span> is compact, furthermore, <span class="math-container">$\overline{V}$</span> is disjoint from <span class="math-container">$C$</span>, so that <span class="math-container">$\overline{V} \subset U$</span>, as desired.</p> </blockquote> <p>I didn't understand how exactly the fact of <span class="math-container">$V$</span> and <span class="math-container">$W$</span> are disjoint opent sets containing <span class="math-container">$x$</span> and <span class="math-container">$C$</span>, respectively imply that <span class="math-container">$\overline{V}$</span> of <span class="math-container">$V$</span> in <span class="math-container">$Y$</span> is compact, someone can help me? Thanks in advance!</p>
Chickenmancer
385,781
<p>The subspace $\overline{V}$ is compact since $Y$ is a compact space, and since closed subspaces of compact spaces are compact, you have the result.</p>
2,081,612
<p>I'd like to evaluate the following integral:</p> <p>$$\int \frac{\cos^2 x}{1+\tan x}dx$$</p> <p>I tried integration by substitution, but I was not able to proceed. </p>
lab bhattacharjee
33,337
<p>$$\dfrac{2\cos^2x}{1+\tan x}=\dfrac2{(1+\tan x)(1+\tan^2x)} =\dfrac{1+\tan^2x+1-\tan^2x}{(1+\tan x)(1+\tan^2x)}$$</p> <p>$$=\dfrac1{1+\tan x}+\dfrac{1-\tan x}{1+\tan^2x}$$</p> <p>$$=\dfrac{\cos x}{\cos x+\sin x}+\cos^2x(1-\tan x)$$</p> <p>$\cos^2x(1-\tan x)$ can be managed easily.</p> <p>Now $\dfrac{2\cos x}{\cos x+\sin x}=1+\dfrac{\cos x-\sin x}{\cos x+\sin x}$</p> <p>$\dfrac{d(\cos x+\sin x)}{dx}=?$</p>
6,929
<p>For example, I have posted quite a few questions regarding mathematics in general, because I want to learn more about the field and I just want to make sure I get an answer or an opinion from someone who has good knowledge about my question. I tag it with the advice tag and I have seen, more than once, that people want to close it or it gets downvoted. It's not like it's a bad question but people don't want it on the mathSE website. So if that's the case and that's how people are going to be, what's the point of the advice tag if people just want to close down the question? And I have also noticed that it's not just my questions, but others as well.</p>
Tom Oldfield
45,760
<p>I think the problem is one of the kind of advice being sought. From the FAQ:</p> <p>"You should only ask practical, answerable questions based on actual problems that you face."</p> <p>All questions asked on the main site should be definitively answerable, in the sense that they should not be open to interpretation, where different opinions may be equally valid. I think that most "advice" questions that get closed are closed because they fail to meet this criteria.</p> <p>An example of an advice question that might be closed is "What is your favourite method of studying linear algebra?" Whereas the similar question "What are some good books to learn linear algebra from?" would probably not be closed.</p>
978,399
<p>A friend of mine posed this question to me a couple days ago and it's been bugging me ever since. He told me to take the square root of 5 twenty times, subtract 1 from it, and then multiply it by 2^20, giving me a rough approximation of ln (5). I can't for the life of my figure out why this is.</p> <p>Also: How can you find the natural log of any positive number X using only arithmetic operations (+, x, -, /) and square root?</p>
Travis Willse
155,629
<p>There is no expression for $\ln$, which is a transcendental function, in terms of radical functions and the usual arithmetic operations, whose combinations are, by contrast, algebraic functions.</p> <p>You can, however, produce arbitrarily good approximations of $\log$ (or any other continuous function for that matter) on any closed interval $[a, b] \subset (0, \infty)$.</p> <p>For a relatively simple example, for $x \in (-1, 1)$ we have the <em>Maclaurin series</em> $$\ln(1 + x) = \sum_{k = 1}^{\infty} (-1)^{k + 1} \frac{x^k}{k},$$ and we can produce variously good approximations to $\ln(1 + x)$ on that interval by simply discarding all of the terms of the series after some point. (Notice that this always produces a polynomial approximation.)</p>
1,721,345
<p>Found a game related to knight's tour <a href="http://www.emathhelp.net/math-games-and-logic-puzzles/knight-score/" rel="nofollow" title="Knight&#39;s Score">Knight's Score</a></p> <p>The basic idea is to gather as much points as possible, avoiding some cells.</p> <p>What is the math of this? Can algorithm be constructed to visit all cells except certain ones (to avoid negative values in the game)?</p> <p>Is the problem always solvable?</p>
Robert Israel
8,508
<p>As Hagen noted, because there are only finitely many cells and you can't visit a cell more than once, the game is finite. Thus there is a "brute-force" algorithm to find an optimal path by examining all possible paths; of course, depending on the particular instance, the optimal value may turn out to be positive, negative or zero. However, on a reasonably-sized board (the one given is $10 \times 10$) the number of possible paths is so huge that brute force is not a realistic option. </p> <p>It is highly unlikely that there is any efficient (polynomial-time) algorithm. However, heuristic optimization techniques such as simulated annealing and tabu search will probably be able to obtain very good, if not optimal, solutions.</p>
27,666
<p>I have reading a book myself; more than 90 % of the book is ok to me; but I have some problems on it. Is it ok to ask my questions in a related chat-room? Because I believe that they are so easy, and it seems shameful to me to ask such an absurd question?</p>
hardmath
3,111
<p>Yes, one of the functions of the chatrooms is to allow discussion of problems in a "chatty" fashion that is not appropriate to the Q&amp;A style of main Math.SE. Back and forth discussions often start in the Comment section of a Question and get migrated to Chat.</p>
2,736,286
<p>Just wondering if I am anywhere near the correct answer.</p> <p>We have the set {z $\in \mathbb{C} : |z - 1| &lt; |z + i|$}</p> <p>So we let $0 &lt; \delta &lt; |z + i| - |z - 1|$ then use the triangle inequality to prove $\delta + |z - 1| &lt; |z + i|$.</p> <p>Thanks!</p>
Mohammad Riazi-Kermani
514,496
<p>So far so good.</p> <p>Now you consider the open ball around your $z$ with radius $\delta /2$ and show that it is included in the region $|z-1|&lt; |z+i|$ using the triangle inequality. </p>
2,913,156
<p>I'm working through the definitions given in <strong>2.18</strong> in Baby Rudin, and I was wondering how to prove that <span class="math-container">$E = [0,1] \cap \Bbb{Q}$</span> is closed in <span class="math-container">$\Bbb{Q}$</span>.</p> <p>Clearly, the set <span class="math-container">$[0,1] \cap \Bbb{R}$</span> is closed in <span class="math-container">$\Bbb{R}$</span>, since for all neighborhoods of a point <span class="math-container">$p \in [0,1]$</span>, we can always find a <span class="math-container">$q \neq p$</span> such that <span class="math-container">$q \in [0,1]$</span>. Intuitively, this is because for all <span class="math-container">$p \in [0,1] \cap \Bbb{R}$</span>, we can always find a <span class="math-container">$q \in [0,1] \cap \Bbb{R}$</span> such that <span class="math-container">$p - \epsilon &lt; q &lt; p$</span> or <span class="math-container">$p &lt; q &lt; p+\epsilon$</span> for all <span class="math-container">$\epsilon &gt;0$</span>.</p> <p>However, that the set <span class="math-container">$E = [0,1] \cap \Bbb{Q}$</span> is closed in <span class="math-container">$\Bbb{Q}$</span> is sort of clear, but not intuitive to me, since <span class="math-container">$\Bbb{Q}$</span> is not uncountably infinite as <span class="math-container">$\Bbb{R}$</span> is. Does one have to use that fact that it is closed in <span class="math-container">$\Bbb{R}$</span> and that <span class="math-container">$\Bbb{Q}$</span> is dense in <span class="math-container">$\Bbb{R}$</span>? How would one solve it from first principles? i.e. Something along the lines of: Consider an arbitrary <span class="math-container">$p \in [0,1] \cap \Bbb{Q}$</span>, we show that it is a limit point by...</p> <p>Any help would be greatly appreciated!</p>
Peter Szilas
408,605
<p>A bit formal :</p> <p>Baby Rudin: Theorem 2.30.</p> <p>Suppose $Y \subset X$. $A$ subset $E$ of $Y$ is open relative to $Y$</p> <p>$\iff$</p> <p>$E=Y \cap G$ for some open subset $G$ of $X$.</p> <p>$B= [0,1]$ is closed in $\mathbb{R}$.</p> <p>$B^c:= \mathbb{R}$\ $[0,1]$ is open in $\mathbb{R}$. </p> <p>Hence </p> <p>$(B^c \cap \mathbb{Q})$ is open in $\mathbb{Q}$.</p> <p>The complement of $(B^c \cap \mathbb{Q})$ in $\mathbb{Q}$ is closed in $\mathbb{Q}$ :</p> <p>$\mathbb{Q} - (B^c \cap \mathbb{Q})=$</p> <p>$\mathbb{Q} \cap (B^c \cap \mathbb{Q})^c =$</p> <p>$\mathbb{Q} \cap (B \cup \mathbb{Q}^c)=$</p> <p>$(\mathbb{Q} \cap B)\cup (\mathbb{Q} \cap \mathbb{Q}^c)=$</p> <p>$\mathbb{Q} \cap B$.</p>
897,955
<p>For a function $f:X\rightarrow Y$ we have $f(U)\subset V\iff U\subset f^{-1}(V)$ proof</p> <p>I'm following a book and it just uses this, it doesn't say anything about the function, so I've not assumed it's one-to-one. I've spent to much time on this (as in it should be 4 lines for both ways at most) so I hope you can help me.</p> <p><strong>What have I tried?</strong></p> <p>For $\implies$</p> <p>$x\in U\implies f(x)\in f(U)$ then by hypothesis $f(x)\in V$<br> Here's where I get stuck, because $f^{-1}(\{f(x)\})\subset f^{-1}(V)$ isn't true.</p> <p>Wait, It is because $f(x)\in V$ everything that maps to $f(x)$ will be in $f^{-1}(V)$, writing this really does help!</p> <p>I then say $x\in f^{-1}(\{f(x)\})$ thus $x\in f^{-1}(V)$ - is that right?</p> <p>For $\impliedby$</p> <p>$y\in f(U)\implies \exists x\in U:f(x)=y$ thus $x\in f^{-1}(V)$ by hypothesis, so $f(x)\in f(f^{-1}(V))$ which is the same as $y\in V$</p> <p>I don't like these sort of proves largely because they're glossed over in the first chapter of all books, and the term "function" is used very ambiguously (is it one-to-one implicitly for example?)</p> <p>Are these proofs okay?</p>
Ishfaaq
109,161
<p>Well the proofs seem okay. But it can be considerably shortened. I'm sure somewhere in the book you use, you will find the following definitions, </p> <p>$$ f^{-1}(A) = \{ x \ | \ f(x) \in A \} $$ $$ f(B) = \{f(x) \ | \ x \in B\} $$</p> <p>$\implies$</p> <p>$x \in U \implies f(x) \in f(U) \implies f(x) \in V$ from which we can directly concluse according to the definition above that $x \in f^{-1}(V)$ </p> <p>$ \Leftarrow$</p> <p>Lets use a trick. Suppose $f(x) \in f(U)$ which is entirely fair again by the definition. Then we have $x \in U \implies x \in f^{-1}(V) \implies f(x) \in V$</p> <p>And we are done. The term "function" is rarely used ambiguously. The function $f$ need not be one-to-one for this result to work. I believe you were thrown off by the notation $f^{-1}(V)$ but its definition does not stipulate the existence of an inverse function. </p>
54,856
<p>I'm trying to simplify an expression with sums of <code>ArcTan</code>, for which I've written a transformation rule that works for simple sums, so that <code>ArcTan[a] + ArcTan[b] -&gt; ArcTan[(a+b)/(1-ab)]</code>. However, it fails for terms of the type <code>5 ArcTan[a] + ArcTan[b]</code>. One way to simplify this expression would be to expand the integer into a sum of units, then transform repeatedly until the expression consists of only one <code>ArcTan</code>, but I haven't been able so far to find the way.</p> <p>I've tried <code>Hold</code>,<code>Distribute</code>, <code>Sum</code>,<code>Fold</code> with <code>Plus</code>to no avail.</p>
Michael
18,486
<p>How about this?</p> <pre><code>ArcTan[Simplify[TrigExpand[Tan[5 ArcTan[a] + ArcTan[b]]]]] </code></pre> <p><img src="https://i.stack.imgur.com/CC08o.png" alt="ArcTan Simplify"></p> <p>Could save yourself some work!</p> <h1>Explanation:</h1> <p>The trick here is to understand the identity you mentioned is about Tan, not ArcTan:</p> <p>Tan[x+y] = (Tan[x]+Tan[y])/(1 -Tan[x]Tan[y]).</p> <p>TrigExpand does the above type of expansion for trig functions. Replace x and y with ArcTan[a] and ArcTan[b], then take ArcTan of both sides and you have your identity. Simplify collects the terms of the expansion into one.</p> <p>The same method works for any linear combination of x=Arctan[a] and y=Arctan[b].</p>
1,578,378
<p>Why does $$\int_{-L}^{L} \sum_{n=1}^{\infty}a_n\cos \frac{n\pi x}{L}=\sum_{n=1}^{\infty}a_n\int_{-L}^{L}\cos \frac{n\pi x}{L}$$</p> <p>This is used in a derivation of the Fourier coefficients. I see why they flip them but I don't understand why this is true and when we are allowed to do such a thing.</p> <p>Thanks</p>
Disintegrating By Parts
112,478
<p>If $f$ is square integrable on $[-L,L]$, then $$ \lim_{N\rightarrow\infty}\int_{-L}^{L}\left|f(x)-a_0-\sum_{n=1}^{N}\{a_n\sin(n\pi x/L)+b_n\cos(n\pi x/L)\}\right|^2dx =0, $$ where $a_n$, $b_n$ are the Fourier coefficients. That's enough to say that, for all square integrable $g$ on $[-L,L]$, the following holds \begin{align} \int_{-L}^{L}f(x)g(x)dx &amp;= a_0\int_{-L}^{L}g(x)dx \\ &amp;+ \sum_{n=1}^{\infty}a_n\int_{-L}^{L}\cos(n\pi x/L)g(x)dx \\ &amp;+ \sum_{n=1}^{\infty}b_n\int_{-L}^{L}\sin(n\pi x/L)g(x)dx. \end{align} That includes $g(x)=1$, $g(x)=\cos(n\pi x/L)$ and $g(x)=\sin(n\pi x/L)$, all of which are square integrable on $[-L,L]$. So the basic answer to your questions is that you can always interchange summation and integration in the cases where the functions involved are square integrable. You couldn't know that <em>a priori</em>, but it does come out that way in the end. It's really quite a remarkable fact of Fourier analysis because discussions of pointwise convergence are not needed at all.</p>
1,270,107
<p>How many real roots does the below equation have?</p> <p>\begin{equation*} \frac{x^{2000}}{2001}+2\sqrt{3}x^2-2\sqrt{5}x+\sqrt{3}=0 \end{equation*}</p> <p>A) 0 B) 11 C) 12 D) 1 E) None of these</p> <p>I could not come up with anything.</p> <p>(Turkish Math Olympiads 2001)</p>
Clement C.
75,808
<p><strong>Hint:</strong> $2\sqrt{3} x^2 - 2\sqrt{5} x + \sqrt{3}$ is positive and has no roots.</p>
489,528
<p>I need to check for convergence here. Why is every example completely different and nothing which I've learned from the examples before helps me in the next? Hate it!</p> <p>$$ \sum_{n\geq1} \frac{n^{n-1}}{n!} $$</p> <p>so I tried the ratio test and I got</p> <p>$$ \frac{1}{1+\frac{1}{n}} $$ which converges to $1$ which isn't $&gt;$ or $&lt;$ then $1,$ so that doesn't work.</p> <p>which test will work and how can I transform the series?</p>
Nick Peterson
81,839
<p><strong>Hint:</strong> Note that you can write $$ \frac{n^{n-1}}{n!}=\frac{n\cdot n\cdots n}{2\cdot3\cdots n} $$ where the numerator contains $n-1$ copies. This can be written as $$ \frac{n^{n-1}}{n!}=\frac{n}{2}\cdot\frac{n}{3}\cdots\frac{n}{n}. $$ Can this possibly tend to $0$ as $n\rightarrow\infty$?</p>
489,528
<p>I need to check for convergence here. Why is every example completely different and nothing which I've learned from the examples before helps me in the next? Hate it!</p> <p>$$ \sum_{n\geq1} \frac{n^{n-1}}{n!} $$</p> <p>so I tried the ratio test and I got</p> <p>$$ \frac{1}{1+\frac{1}{n}} $$ which converges to $1$ which isn't $&gt;$ or $&lt;$ then $1,$ so that doesn't work.</p> <p>which test will work and how can I transform the series?</p>
DonAntonio
31,254
<p>Ratio test:</p> <p>$$\frac{(n+1)^n}{(n+1)!}\frac{n!}{n^{n-1}}=\left(1+\frac1n\right)^n\frac n{n+1}\xrightarrow[n\to\infty]{}e&gt;1$$</p> <p>so the series...</p> <p><strong>Added on request:</strong></p> <p>Since</p> <p>$$\frac{k!}{(k-1)!}=k\;,\;\;\frac1{n^{n-1}}=\frac n{n^n}\;,\;\;\text{we get}$$</p> <p>$$\frac{(n+1)^n}{(n+1)!}\frac{n!}{n^{n-1}}=\frac{(n+1)^n}{n^n}\frac{n\cdot n!}{(n+1)!}=\left(1+\frac1n\right)^n\frac n{n+1}\ldots$$</p>
207,382
<p>I was interested in exploiting the Terminal` package, as demonstrated in <a href="https://mathematica.stackexchange.com/questions/13961/plotting-graphics-as-ascii-plots">this post</a>, to show some ASCII plots straight through the terminal of my Raspberry Pi Model 3 B+. However, it doesn't seem that I can get a plot to show when using the package either in the terminal or <em>even in the front-end</em>. Here's what happens in either case when I try to replicate a simple example:</p> <pre><code>In[1]:= &lt;&lt; Terminal` -- Terminal graphics initialized -- In[2]:= Plot[Sinc[x],{x,0,20},PlotRange-&gt;All] # StringJoin[MapThread[System`ConvertersDump`RasterDataPacket[#"#", #"#", #"#", #"+", ColorFunction -&gt; #"+"] &amp; ]] Out[2]= -Graphics- </code></pre> <p>The package seems to load fine, the stuff between the In[2] line and the Out[2] line is stuff Mathematica outputted (I didn't type it), and the graphics the system outputs in both cases are fully formatted correct plots; just not in ASCII. <strong>Is the package broken, or am I missing something</strong>?</p>
bobthechemist
7,167
<p><strong>Note</strong> After looking through the referenced python code, it stands to reason that all of the tools are available from within Mathematica. I have yet to fully debug this sub-answer. You can take advantage of <code>$DisplayFunction</code> to estimate what <code>Terminal</code> did:</p> <pre><code>$DisplayFunction = ExportString[ StringJoin /@ Flatten /@ (ImageData@ ColorConvert[Rasterize[#, RasterSize -&gt; 100], "Grayscale"] /. {1. -&gt; " ", _Real :&gt; "x"}), "Text"] &amp; </code></pre> <p><strong>Inspiration</strong> <em>Original answer is below</em></p> <p>It looks like <code>Terminal.m</code> uses <code>ExportString[...,"TTY"]</code> in a <code>$DisplayFunction</code> call, which is not (no longer?) a valid export type. A rudimentary hack involves some python code.</p> <pre><code>wget raw.githubusercontent.com/RameshAditya/asciify/master/asciify.py </code></pre> <p>Towards the end of the python script, you'll note that there is an option to grab images from the web. This library does not appear to be installed by default on the RPi so I removed that 3-4 lines of code as it is irrelevant here. Now creating an asciiplot requires 2 steps:</p> <pre><code>o=Plot[Sinc[x],{x,0,10},AspectRatio-&gt;1/2,ImageSize-&gt;100,Axes-&gt;False,Frame-&gt;{True,True,False,False},FrameTicks-&gt;{{0,10},{-0.2,1}}]; Export["out.png", o]; Run["python asciify.py out.png"]; </code></pre> <p><a href="https://i.stack.imgur.com/bY7T0.png" rel="noreferrer"><img src="https://i.stack.imgur.com/bY7T0.png" alt="enter image description here"></a></p> <p>One needs to play around with image features (aspect ratio, size, tick marks) to make this work reasonably. The nice feature is that it is raster-based so one can in principle view any graphical mathematica output in the terminal.</p> <p>FWIW, the python code seems straightforward enough that it can probably be incorporated into the newish language connectors that are available.</p>
3,034,374
<p>Two cyclists start from the same place to ride in the same direction.A starts at noon with a speed of 8km/hr and B starts at 2pm with a speed of 10km/hr.At what times A and B will be 5km apart ? My thought process: As A starts early at 12 so it will have already covered 16km(8*2). so S relative=V relative*t or say 16=2*t1 and thus t1=8. Now we want Srelative to be 5 so 5=2*t2 and t2=2 and a half hour so they will meet at t1-t2 . Is this correct process ?</p>
AgentS
168,854
<p>I'll give an alternative, which might look a bit less tedious.</p> <p>Assume <span class="math-container">$t=0$</span> is at 12 noon.<br> Since <span class="math-container">$A$</span> starts at <span class="math-container">$t=0$</span> and goes at a speed of 8, we can express the distance traveled by him as <span class="math-container">$$a(t) = 8t$$</span></p> <p>But <span class="math-container">$B$</span> starts with a delay of 2 units; this means his graph shifts to the <code>right</code> by 2 units : <span class="math-container">$$b(t) = 10(t-2)$$</span></p> <p>Now that equations are setup, you simply have to solve <span class="math-container">$|b(t)-a(t)|=5$</span></p>
2,944,397
<blockquote> <p>In one plane <span class="math-container">$Oxy$</span>, given the circle <span class="math-container">$(\!{\rm C}\!): x^{2}+ y^{2}- 2x+ 4y- 4= 0$</span> around point <span class="math-container">$O$</span> for <span class="math-container">$60^{\circ}$</span> and it maps in the circle <span class="math-container">$(\!{\rm C}'\!)$</span>. Find the loci of the circle <span class="math-container">$(\!{\rm C}'\!)$</span>.</p> </blockquote> <p>Firstly, we let <span class="math-container">${\rm M}(x, y)\in (\!{\rm C}\!), {\rm M}'({x}',{y}')\in (\!{\rm C}'\!)$</span>. We always have the following system of equations <span class="math-container">$$\left\{\begin{matrix} {x}'= x\cos\alpha- y\sin\alpha\\ {y}'= x\sin\alpha+ y\cos\alpha \end{matrix}\right.$$</span> or <span class="math-container">$$\left\{\begin{matrix} {x}'= x\cos 60^{\circ}- y\sin 60^{\circ}\\ {y}'= x\sin 60^{\circ}+ y\cos 60^{\circ} \end{matrix}\right.$$</span> so <span class="math-container">$$x= \frac{1}{2}{x}'+ \frac{\sqrt{3}}{2}{y}'$$</span> <span class="math-container">$$y= \frac{1}{2}{y}'- \frac{\sqrt{3}}{2}{x}'$$</span> Secondly <em><a href="https://www.wolframalpha.com/input/?i=(1%2F2a%2Bsqrt(3)%2F2b)%5E2%2B(-sqrt(3)%2F2a%2B1%2F2b)%5E2-2(1%2F2a%2Bsqrt(3)%2F2b)%2B4(-sqrt(3)%2F2a%2B1%2F2b)-4%3D0" rel="nofollow noreferrer">https://www.wolframalpha.com/input/?i=(1%2F2a%2Bsqrt(3)%2F2b)%5E2%2B(-sqrt(3)%2F2a%2B1%2F2b)%5E2-2(1%2F2a%2Bsqrt(3)%2F2b)%2B4(-sqrt(3)%2F2a%2B1%2F2b)-4%3D0</a></em>. So, how can I prove the given system of equations as follow ? <span class="math-container">$$\left\{\begin{matrix} x'= x\cos\alpha- y\sin\alpha\\ y'= x\sin\alpha+ y\cos\alpha \end{matrix}\right.$$</span> Finally, I need to the helps! Thanks for all the nice comments!</p>
Community
-1
<p>We have <span class="math-container">$x^{2}+ y^{2}- 2x+ 4y- 4= 0 \therefore {x}'^{2}- (\!2\sqrt{3}+ 1\!){x}'+ {y}'^{2}- (\!\sqrt{3}- 2\!){y}'- 4= 0$</span>. Using twice <span class="math-container">$$ax^{\,2}+ bx+ c= \frac{(\!2ax+ b\!)^{2}+ 4ac- b^{2}}{4a}\,\therefore\,\frac{1}{4}(\!-\,2{x}'+ 2\sqrt{3}+ 1\!)^{2}+ \frac{1}{4}(\!-2{y}'+ \sqrt{3}- 2\!)^{2}- 9= 0$$</span></p>
2,260,975
<p>Problem: Let $\mu : \mathcal{B}(\mathbb{R}^k) \rightarrow \mathbb{R}$ be a nonnegative and finitely additive set function with $\mu(\mathbb{R}^k) &lt; \infty $. Suppose that \begin{equation} \mu(A) = \sup \{ \mu(K) : K \subset A, K \text{ compact}\} \end{equation} for each $A \in \mathcal{B}(\mathbb{R}^k)$. Then $\mu$ is a finite measure.</p> <p>I'm stuck showing that $\mu$ is countably additive. First, let $A_{i}$ disjoint, and $K_i \subset A_i$ compact such that $\mu(A_i) \leq \mu(K_i)+\epsilon/2^i$ for $\epsilon &gt; 0$. Then \begin{equation} \mu\left(\bigcup_{i=1}^{\infty} A_i \right) \geq \mu\left(\bigcup_{i=1}^{n} A_i \right) \geq \mu\left(\bigcup_{i=1}^{n} K_i \right) = \sum_{i=1}^n \mu(K_i) \geq \sum_{i=1}^n \mu(A_i) + \frac{\epsilon}{2^i} , \end{equation} which follows from monotonicity (obvious) and finite additivity (on the class of compact sets). Hence, letting $n \rightarrow \infty$, and since $\epsilon$ was arbitrary, \begin{equation} \mu\left(\bigcup_{i=1}^{\infty} A_i \right) \geq \sum_{i=1}^\infty \mu(A_i) . \end{equation} Next, I want to establish the reverse inequality. This is where I'm stuck. Any ideas?</p> <p>Thanks in advance!</p> <p>Chris</p>
Veridian Dynamics
408,632
<p>This is enough:</p> <p><strong>Claim</strong>: Let $(K_{n})_{k\in\mathbb{N}}$ a decreasing sequence of compacts subsets such that $\bigcap_{n\in\mathbb{N}}K_{n}=\emptyset$. Then, $\lim_{m\to+\infty}\mu(K_{m})=0$.</p> <p>If not, there is some $\delta&gt;0$ such that $\mu(K_{n})&gt;\delta$, for all $n$. In particular, every $K_{n}$ is non-empty and so, by the finite intersection property, $\bigcap_{n\in\mathbb{N}}K_{n}\not=\emptyset$.</p>
2,900,014
<p>How would solve for $a$ in this equation without using an approximation ? is it possible?</p> <p>where $x&gt;0$ and $0&lt;a&lt;\infty$</p> <p>$x=\Sigma _{i=1}^{n} i^a$</p> <p>for example $120=\Sigma _{i=1}^{6} i^a$ what is $a$ in this equation?</p>
Claude Leibovici
82,404
<p>If you want solve for $a$ the equation $$x=\sum_{i=1}^{n} i^a=H_n^{(-a)}$$ you will need a numerical method (as already said in answers) and, for that, you would need some initial estimate.</p> <p>To get such an estimate, you could consider the asymptotics $$H_n^{(-a)}=n^a \left(\frac{n}{a+1}+\frac{1}{2}+\frac{a}{12 n}+O\left(\frac{1}{n^3}\right)\right)+\zeta (-a)$$ and set the problem as $$x \approx \frac{n^{a+1}}{a+1}$$ for which the solution is given in terms of <a href="https://en.wikipedia.org/wiki/Lambert_W_function" rel="nofollow noreferrer">Lambert function</a>. This will be $$a=-1-\frac{1}{\log (n)}\,W_{-1}\left(-\frac{\log (n)}{k}\right)$$ Since the argument of the function is supposed to be small, you could use for its evaluation $$W(t)=L_1-L_2+\frac{L_2}{L_1}+\frac{L_2(L_2-2)}{2L_1^2}+\frac{L_2(6-9L_2+2L_2^2)}{6L_1^3}+\cdots$$ where $L_1=\log(-t)$ and $L_2=\log(-L_1)$ (as given in the linked page).</p> <p>For your specific example $(n=6,x=120)$, this would give $a\approx 2.34602$ which is not too bad for an estimate.</p> <p>Now to polish the root, consider that you look for the zero of function $$f(a)=\sum_{i=1}^{n} i^a-x$$ $$f'(a)=\sum_{i=1}^{n}i^a \log(i)$$ and use Newton method which, starting from the guess $a_0$, will update it according to $$a_{n+1}=a_n-\frac{f(a_n)}{f'(a_n)}$$ Applied to the example, this will give as iterates $$\left( \begin{array}{cc} n &amp; a_n \\ 0 &amp; 2.34602 \\ 1 &amp; 2.19989 \\ 2 &amp; 2.17961 \\ 3 &amp; 2.17927 \end{array} \right)$$</p> <p>If you do not want to use Lambert function, you could consider that $$\sum_{i=1}^{n} i^a &lt;\sum_{i=1}^{n} n^a=n^{a+1}\implies a_0=\frac{\log (x)}{\log (n)}-1$$ This would be simpler but will require more iterations as shown below $$\left( \begin{array}{cc} n &amp; a_n \\ 0 &amp; 1.67195 \\ 1 &amp; 2.45312 \\ 2 &amp; 2.23205 \\ 3 &amp; 2.18146 \\ 4 &amp; 2.17927 \end{array} \right)$$</p> <p><strong>Edit</strong></p> <p>Since function $f(a)$ varies very fast with $a$, it could be better to make it more linear considering instead $$g(a)=\log\left(\sum_{i=1}^{n} i^a \right)-\log(x)$$ $$g'(a)=\frac{\sum_{i=1}^{n} i^a \log(i) }{\sum_{i=1}^{n} i^a }$$ Using the simple estimate, for the worked example, the iterates would be $$\left( \begin{array}{cc} n &amp; a_n \\ 0 &amp; 1.67195 \\ 1 &amp; 2.18953 \\ 2 &amp; 2.17927 \end{array} \right)$$ Just for the fun, let us use $(n=20,x=123456789)$; for this case, the iterates would be $$\left( \begin{array}{cc} n &amp; a_n \\ 0 &amp; 5.21931 \\ 1 &amp; 5.80609 \\ 2 &amp; 5.80465 \end{array} \right)$$</p>
4,295,212
<p>I have a statics problem in which my approach differs from the suggested solution, so I was wondering if anyone can point out my flawed logic. Below is the question.</p> <p>Two identical uniform rods <span class="math-container">$ab$</span> and <span class="math-container">$bc$</span>, each of length <span class="math-container">$2l$</span> and weight <span class="math-container">$W$</span>, are smoothly hinged at <span class="math-container">$b$</span>. Ends <span class="math-container">$a$</span> and <span class="math-container">$c$</span> rest on a smooth horizontal floor. The system is kept in equilibrium in a vertical plane by a light inextensible string joining <span class="math-container">$a$</span> to the midpoint of <span class="math-container">$bc$</span>. the and between <span class="math-container">$ab$</span> and <span class="math-container">$bc$</span> is <span class="math-container">$2\alpha$</span>. Show the tension of the string is <span class="math-container">$T = \frac{W}{4} \sqrt{9\tan^2\alpha + 1}$</span>.</p> <p>So below is the force diagram I have created.</p> <p><a href="https://i.stack.imgur.com/rzxc9.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/rzxc9.jpg" alt="enter image description here" /></a></p> <p>The suggested solution for this problem resolves the <span class="math-container">$\vec{j}$</span> components of the system as:</p> <p><span class="math-container">$$ R_a + R_c = 2W $$</span></p> <p>without taking into consideration the resolved tension force, <span class="math-container">$T$</span>, of the light inextensible string.</p> <p>Also taking moments about the point <span class="math-container">$a$</span> of the system abc the solution is:</p> <p>$R_c \cdot 4l \sin \alpha = l \sin\alpha \cdot W + 3l\sin \alpha \cdot W $$</p> <p>Once again no mention of the tension.</p> <p>So I am I wrong to include the tension force for the <span class="math-container">$\vec{j}$</span> components of the system and the moments about <span class="math-container">$a$</span>.</p> <p>Any help appreciated.</p>
YNK
587,353
<p><a href="https://i.stack.imgur.com/R6M7c.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/R6M7c.png" alt="RodsinEquilibrium" /></a></p> <p>It is true that the tension force <span class="math-container">$T$</span> in the string is a single force. But, this tension in the string exerts two equal forces of <span class="math-container">$T$</span> on the system at both ends, i.e. at <span class="math-container">$a$</span> and at <span class="math-container">$d$</span>, as shown in the diagram.</p> <p>If you still have doubts, remove the string and apply two unequal forces, say <span class="math-container">$T_1$</span> and <span class="math-container">$T_2$</span>, but acting along the same straight line at <span class="math-container">$a$</span> and <span class="math-container">$d$</span>. When you resolve the forces acting on the system in <span class="math-container">$\underline{i}$</span> and <span class="math-container">$\underline{j}$</span> directions assuming that the system is in stable equilibrium, you get the two equation shown below. <span class="math-container">$$R_a + R_c = 2W +T_2\sin(\alpha) – T_1\sin(\alpha)$$</span> <span class="math-container">$$T_1\cos(\alpha) - T_2\cos(\alpha) = 0 \qquad\rightarrow\qquad T_1 = T_2$$</span></p> <p>Since <span class="math-container">$T_1 = T_2$</span>, the first equation reduces to <span class="math-container">$$R_a + R_c = 2W$$</span></p> <p>When you are taking moments about a given point, you can ignore all the forces passing through that point, because such forces generate zero moments. Therefore, you have to consider neither <span class="math-container">$T_1$</span> nor <span class="math-container">$T_2$</span> as they both pass through <span class="math-container">$a$</span>.</p>
2,349,476
<p>Suppose $n$ and $k$ are positive integers. Are there any known conditions on $n$ and $k$ such that the polynomial $x^{k+1}-x^k-n$ has rational or integer roots?</p>
Bob Krueger
228,620
<p>It is a well-known fact (attribution?) that the rational roots of a polynomial with leading coefficient $a$ and constant term $b$ must be of the form $\pm \frac{p}{q}$, where $p$ divides $b$ and $q$ divides $a$. In your equation, the only possible rational roots are plus/minus factors of $n$. Let $m$ divide $n$ be a solution to your polynomial. We then have</p> <p>$$ m^k(m-1) = n .$$</p> <p>This equation states that $m^k$ is also a factor of $n$, along with $m-1$. We conclude that $n$ must have the form $m^k(m-1)$. As @Bernard notes, on top of this, there isn't too much to say.</p>
244
<p>I know that Hilbert schemes can be very singular. But are there any interesting and nontrivial Hilbert schemes that are smooth? Are there any necessary conditions or sufficient conditions for a Hilbert scheme to be smooth?</p>
Daniel Erman
4
<p>I don't know of many global conditions for a Hilbert scheme to be smooth/singular. Ben's answer probably gives the most interesting example of a smooth Hilbert scheme, namely the Hilbert scheme of n points on a smooth surface.</p> <p>Here are two more examples of smooth Hilbert schemes.</p> <p>1) The Hilbert scheme of hypersurfaces of degree d in PP^n. Such hypersurfaces are parametrized by homogeneous degree d polynomials in n+1 variables, and hence this Hilbert scheme is a projective space of dimension n+d choose d.</p> <p>2) The Hilbert scheme of linear subpsace of dimension d of PP^n. This is just the Grassmanian Gr(d+1,n+1).</p>
3,807,504
<p>Suppose that <span class="math-container">$z=f(x,y)$</span> is given. where <span class="math-container">$f(x,y)=x^3+y^3$</span>. I am a bit confused about the notation of <span class="math-container">$z_x$</span> and <span class="math-container">$f_x$</span>.</p> <p>A lot of people use the below notation for <span class="math-container">$z_x$</span>. <span class="math-container">$$z_x=3x^2$$</span></p> <p>But what if <span class="math-container">$y$</span> is related to <span class="math-container">$x$</span> in some equation. For example <span class="math-container">$$y=x.$$</span> Then <span class="math-container">$${\partial z\over\partial x}$$</span> should also consider variable <span class="math-container">$y$</span> since <span class="math-container">$y$</span> can be represented by <span class="math-container">$x$</span>. But in a similar situation, I saw in a book <span class="math-container">$${\partial f\over\partial x }=3x^2$$</span> which doesn't consider y as a variable.</p> <p>Summarizing, given that <span class="math-container">$z=f(x,y)$</span>, <span class="math-container">$f(x,y)=x^3+y^3$</span>, <span class="math-container">$y=x$</span> what is <span class="math-container">$z_x$</span> and <span class="math-container">$f_x$</span>?</p>
Isaac Browne
429,987
<p>Probably the best way of going about this one is to write it as an infinite sum, as there does not appear to be an easy closed form.</p> <p>So, what we can do is write <span class="math-container">$$\int_0^R \frac{bx}{x^a-b}dx = -\int_0^R \frac{x}{1-x^a/b}dx = -\int_0^R \sum_{n=0}^\infty x\left(\frac{x^a}{b}\right)^n$$</span> And now the trick is that we can switch the order of integration and summation, so we get: <span class="math-container">$$-\sum_{n=0}^\infty \int_0^R x^{an+1}/b^n = -\sum_{n=0}^\infty \frac{R^{an+2}}{(n+2)b^n}$$</span> Note that this only works if <span class="math-container">$x^a &lt; b$</span> all the time, i.e. <span class="math-container">$R^a &lt; b$</span>, as otherwise, we integrate across a singularity and the infinite sum diverges (another way of seeing this is <span class="math-container">$x^a=b$</span> causes division by <span class="math-container">$0$</span>).</p> <p>Finally, there are likely some special cases where you can get more of a closed form, but if you want a solution for all <span class="math-container">$a,b,R$</span> with <span class="math-container">$R^a&lt;b$</span>, this is probably the best we can do.</p>
310,513
<p>I haven't found this yet and I'm somehow not sure if my idea is correct.</p> <p>The Problem: Let $X$ be a separable Banach-Space, let $x_k\to x$ weakly and such that for every $\lambda_k \to \lambda$ weakly-* there holds $\lambda_k(x_k)\to \lambda(x)$. Then $x_k\to x$ strongly (in the norm of $X$).</p> <p>My idea was to give a proof with contradiction. Hence assume there holds for some $\epsilon &gt;0$ and a subsequence of $x_k$ denoted again by $k$: $$ \epsilon &lt;||x_k-x||=|\lambda^*_k(x_k-x)| $$ for a functional provided by Hahn-Banach theorem with norm 1. From that and the separabilty we conclude that there is a further subsequence such that $\lambda_{k_l}^*\to \lambda^*$ weakly-* . Since each subsequence of $x_k$ also converges weakly to $x$ we use the "weakly-*" assumption to receive a contradiction since for the previous subsequence $$|\lambda^*_{k_l}(x_{k_l}-x)|=|\lambda^*_{k_l}(x_{k_l})-\lambda^*_{k_l}(x)|\to 0$$</p> <p>Somehow this seems to easy and I feel like I'm not using especially the weak convergence in the right way. </p>
Harald Hanche-Olsen
23,290
<p>I think your proof looks just fine. There is nothing to add and nothing to subtract. Oh, except you need to take a subsequence to get the first inequality. But you must already have intended that, or else why did you say <em>further</em> subsequence? Hahn–Banach and compactness are powerful tools. You should not be surprised that they can yield strong results with little effort.</p>
2,672,497
<p>$$\lim _{n\to \infty }\sum _{k=1}^n\frac{1}{n+k+\frac{k}{n^2}}$$ I unsuccessfully tried to find two different Riemann Sums converging to the same value close to the given sum so I could use the Squeeze Theorem. Is there any other way to solve this?</p>
Felix Marin
85,343
<p><span class="math-container">$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$</span> <span class="math-container">\begin{align} &amp;\bbox[10px,#ffd]{\lim _{n \to \infty }\sum _{k = 1}^{n}{1 \over n + k + k/n^{2}}} = \lim _{n \to \infty }\bracks{{n^{2} \over n^{2} + 1} \sum _{k = 1}^{n}{1 \over k + n^{3}/\pars{n^{2} + 1}}} \\[5mm] = &amp;\ \lim _{n \to \infty }\braces{{n^{2} \over n^{2} + 1} \sum _{k = 1}^{\infty}\bracks{{1 \over k + n^{3}/\pars{n^{2} + 1}} - {1 \over k + n^{3}/\pars{n^{2} + 1} + n}}} \\[5mm] = &amp;\ \lim _{n \to \infty }\braces{{n^{2} \over n^{2} + 1} \bracks{H_{\large n^{3}/\pars{n^{2} + 1} + n}\ -\ H_{\large n^{3}/\pars{n^{2} + 1}}}} \end{align}</span> where <span class="math-container">$\ds{H_{z}}$</span> is a <a href="https://en.wikipedia.org/wiki/Harmonic_number" rel="nofollow noreferrer">Harmonic Number</a>.</p> <blockquote> <p>In using the <a href="https://en.wikipedia.org/wiki/Harmonic_number#Calculation" rel="nofollow noreferrer"><span class="math-container">$\ds{H_{z}}$</span> asymptotic behavior as <span class="math-container">$\ds{\verts{z} \to \infty}$</span></a>, it's straightforward found:</p> </blockquote> <p><span class="math-container">$$ \bbx{\lim _{n \to \infty }\sum _{k = 1}^{n}{1 \over n + k + k/n^{2}} = \ln\pars{2}} $$</span></p>
19,290
<pre><code>MouseAppearance[Plot[Sin[x], {x, 0, 5}], "LinkHand"] </code></pre> <p><img src="https://i.stack.imgur.com/XjHbo.png" alt="hand"></p> <p>Unfortunately I cannot get a screen grab to show the link hand but it is a cartoon hand with a finger pointing. In most/many graphics packages there is also an open hand (open carton hand) and a closed hand (closed cartoon hand) used for sliding a graphics within a region.</p> <p><em>I realize that if I can find some open source clip art</em> then I can use this as a mouse appearance:</p> <pre><code>MouseAppearance[Plot[Sin[x], {x, 0, 5}], clip art] </code></pre> <p>But was wondering if undocumented open hand and grab hand exists within Mathematica?</p> <p>Intended usage: I have an application for sliding a graphic. I wanted to use an open hand over the graphic and grab hand/closed hand for sliding the graphic.</p> <p>My reason for seeking something built in is solely aesthetic so that it matches the "LinkHand" appearance. If I cannot have a built in solution I'll look for clip art for the full set of icons -- including the link hand.</p>
kglr
125
<p>On Windows (Vista 64bit), you can find several hand icons in native cursor format <a href="http://reference.wolfram.com/mathematica/ref/format/CUR.html" rel="nofollow noreferrer"><code>CUR</code></a> in the directory <code>C:\Windows\Cursors</code> </p> <pre><code> {hand1, hand2, hand3, hand4, hand5} = {Import["C:\\Windows\\Cursors\\hmove.cur"][[1]], Import["C:\\Windows\\Cursors\\hnwse.cur"][[1]], Import["C:\\Windows\\Cursors\\harrow.cur"][[1]], Import["C:\\Windows\\Cursors\\hcross.cur"][[1]], Import["C:\\Windows\\Cursors\\hnodrop.cur"][[1]]} </code></pre> <p><img src="https://i.stack.imgur.com/GBMcg.png" alt="enter image description here"></p> <pre><code> MouseAppearance[Plot[Sin[x], {x, 0, 5}], hand2] </code></pre> <p><img src="https://i.stack.imgur.com/Q1dzN.png" alt="enter image description here"></p> <p><strong>Update:</strong> Mathematica cell expressions for the cursor graphics:</p> <p>hand1:</p> <pre><code> Cell[BoxData[ GraphicsBox[ TagBox[RasterBox[CompressedData[" 1:eJxTTMoPSmNiYGAo5gASQYnljkVFiZXBLECOU2VJahIzkKEAxSDB////j+JR PIrJwA0NDf9BeCDshGTd//8PHDgAZtPDLVDz4XbiYQ+o/bQIC2xhTq+wANmN bjYsLECAQLhQJczBBkHNxgewuIVq9iOzCQFa2E8KGLWfuul/oOweDPaT6oaB tJ+WdQAxbqBXHTQQdg8G+xkYQPUgpC7EhWlr938obsCDaeOGUftHtv3DEQMA AMvl3Q== "], {{0, 32}, {32, 0}}, {0, 255}, ColorFunction -&gt; RGBColor], BoxForm`ImageTag["Byte", ColorSpace -&gt; "RGB", Interleaving -&gt; True], Selectable -&gt; False], BaseStyle -&gt; "ImageGraphics", ImageSizeRaw -&gt; {32, 32}, PlotRange -&gt; {{0, 32}, {0, 32}}]], "Output"] // CellPrint </code></pre> <p>hand2:</p> <pre><code> Cell[BoxData[ GraphicsBox[ TagBox[RasterBox[CompressedData[" 1:eJztlVEKwjAMhof64DH0KHqEeYIN9HUwBdlbj9abxY4aLDFddaSRFQv/2nXQ 7wtr6b7t6suqqqrr1j3q5n7o+2Y4bdzLcbid27Ub7J4ZJwHgn4XEGANc/G/M zw6btRa54VilZoabxYHWnKhd1AHrxLU/ccBewgFrD9ekPpRLe0n+1LeS+ZwD x4/sDbG9P64ZnkGc45rUOYidv1QrhT/XoST+HAfNu0CL/41D7vs41Zj7Qs3B Mw2JjgPP1nGYZvtI89HBs0GdjXnxeYec7Hc+qLJ5PqixYw6abOrwC/ZS8wD+ AMqd "], {{0, 32}, {32, 0}}, {0, 255}, ColorFunction -&gt; RGBColor], BoxForm`ImageTag["Byte", ColorSpace -&gt; "RGB", Interleaving -&gt; True], Selectable -&gt; False], BaseStyle -&gt; "ImageGraphics", ImageSizeRaw -&gt; {32, 32}, PlotRange -&gt; {{0, 32}, {0, 32}}]], "Output"] // CellPrint </code></pre> <p>hand3:</p> <pre><code> Cell[BoxData[ GraphicsBox[ TagBox[RasterBox[CompressedData[" 1:eJxTTMoPSmNiYGAo5gASQYnljkVFiZXBLECOU2VJahIzkKEAxSDB////Dync 0NDwH4QHym4QOHDgwH9I0I3aP5Ltp3daxOJ/uroFn/30iBdi7ac0LGD60f1C gv8pCgtc6RwmDgLY2NSMC2xuQLYTF6BWHkW2C2YWzG/EAGqkRXQ3gPBA2U8q oFZeJNcNA2k/tcshUt0w3OwnxQ20rAOIcQO96qCBsHsw2M/AAKoXG+DlMTZM W7v/Q3EDHkwbN4zaP7LtH44YAM543sY= "], {{0, 32}, {32, 0}}, {0, 255}, ColorFunction -&gt; RGBColor], BoxForm`ImageTag["Byte", ColorSpace -&gt; "RGB", Interleaving -&gt; True], Selectable -&gt; False], BaseStyle -&gt; "ImageGraphics", ImageSizeRaw -&gt; {32, 32}, PlotRange -&gt; {{0, 32}, {0, 32}}]], "Output"] // CellPrint </code></pre> <p>hand4:</p> <pre><code> Cell[BoxData[ GraphicsBox[ TagBox[RasterBox[CompressedData[" 1:eJztlUEOgjAQRSfqwmPoUfQIeAJIdGuCJoZdj9abja0yOph2WrBUFyV5pDQw 708KZducq9MCAC5rc6rq265t6+6wMhf77npslmaw6bGTiFgoFArRKKUwlin1 QdiX+pqDQ2uNz0e84yR+l3uODC4/d1NNix2nzsD9SgFayEW1aJ5D94zNIByv 26hf6ZdFrsD6jOqfe3n/Pvh6pPBL60/fGM+Ww8/rhfpO5fdlCPU9xS8h7QEx fgD1IHcG7s6dweVOlcEivQeS+1s/4doD3uuNs7rlPMjI4x5mQGeGud1+P2Zx +zLkdH9m+IW78P/cAe6De2M= "], {{0, 32}, {32, 0}}, {0, 255}, ColorFunction -&gt; RGBColor], BoxForm`ImageTag["Byte", ColorSpace -&gt; "RGB", Interleaving -&gt; True], Selectable -&gt; False], BaseStyle -&gt; "ImageGraphics", ImageSizeRaw -&gt; {32, 32}, PlotRange -&gt; {{0, 32}, {0, 32}}]], "Output"] // CellPrint </code></pre> <p>hand5:</p> <pre><code> Cell[BoxData[ GraphicsBox[ TagBox[RasterBox[CompressedData[" 1:eJxTTMoPSmNiYGAo5gASQYnljkVFiZXBLECOU2VJahIzkKEAxSDB////Dync 0NDwH4QHym4QOHDgwH9I0I3aP5Ltp3daxOJ/uroFn/30iBdi7ac0LGD60f1C gv8pCgtc6RwmDgLY2NSMC2xuQLYTF6BWHkW2C2YWzG9gAGIjYzRAjbSI7gYQ htkNZ0Pl0N1AbfuR/Y3LPmQxauVFDDegxwEOObrYj8M91C6HUNyAFvcoYvSw H4sbaBX3hNyALf/Rsg4gpvyhVx00EHYPBvsZGED1YgO8PMaGaWs3LMk14MG0 ccOo/SPb/uGIAbULqQ4= "], {{0, 32}, {32, 0}}, {0, 255}, ColorFunction -&gt; RGBColor], BoxForm`ImageTag["Byte", ColorSpace -&gt; "RGB", Interleaving -&gt; True], Selectable -&gt; False], BaseStyle -&gt; "ImageGraphics", ImageSizeRaw -&gt; {32, 32}, PlotRange -&gt; {{0, 32}, {0, 32}}]], "Output"] // CellPrint </code></pre>
19,290
<pre><code>MouseAppearance[Plot[Sin[x], {x, 0, 5}], "LinkHand"] </code></pre> <p><img src="https://i.stack.imgur.com/XjHbo.png" alt="hand"></p> <p>Unfortunately I cannot get a screen grab to show the link hand but it is a cartoon hand with a finger pointing. In most/many graphics packages there is also an open hand (open carton hand) and a closed hand (closed cartoon hand) used for sliding a graphics within a region.</p> <p><em>I realize that if I can find some open source clip art</em> then I can use this as a mouse appearance:</p> <pre><code>MouseAppearance[Plot[Sin[x], {x, 0, 5}], clip art] </code></pre> <p>But was wondering if undocumented open hand and grab hand exists within Mathematica?</p> <p>Intended usage: I have an application for sliding a graphic. I wanted to use an open hand over the graphic and grab hand/closed hand for sliding the graphic.</p> <p>My reason for seeking something built in is solely aesthetic so that it matches the "LinkHand" appearance. If I cannot have a built in solution I'll look for clip art for the full set of icons -- including the link hand.</p>
John Fultz
309
<p>There is an undocumented "Hand" named cursor that should work on all platforms.</p> <pre><code>MouseAppearance[Plot[Sin[x], {x, 0, 6}], "Hand"] </code></pre> <p>Unfortunately, there isn't anything that resembles a grabbed hand. Here are some other undocumented ones. Nothing here is secret, but most of it is comparatively boring and, being undocumented, is potentially subject to change in future versions. <code>"Hand"</code> is a system cursor, but most of the rest of these are hand-drawn cursor resources bundled with Mathematica:</p> <ul> <li><code>"MathEditxx"</code> where xx = "", 45, 90, 135, 180, 225, 270, 315</li> <li><code>"Crosshair"</code></li> <li><code>"Formatted"</code></li> <li><code>"Cellxx"</code> where xx = Select, Shade, Sound, Insert</li> <li><code>"NameSelect"</code></li> <li><code>"SimpleWait"</code></li> <li><code>"FramexxRotate"</code> for xx = LC, RC, CT, CB, LT, RB, RT, LB</li> <li><code>"MeasureGraphic"</code></li> <li><code>"xxGraphics"</code> for xx = Delete, Test, Select</li> <li><code>"DragGraphicsxxy"</code> for xx = "", Highlight, Direct, HighlightDirect and Y = "", H, V</li> <li><code>"DragSegment"</code></li> <li><code>"SampleStyle"</code></li> <li><code>"DoNothing"</code></li> <li><code>"Drawxx"</code> for xx = "", Line, Polyline, Polygon, Rectangle, Arrow, Circle, ArcCursor, Point, Freehand, Pixel</li> <li><code>"Placexx"</code> for xx = Text, Math</li> <li><code>"xxView"</code> for xx = Rotate, Pan, Zoom, as well as <code>"RotateViewVertical"</code></li> <li><code>"Obscure"</code></li> <li><code>"NetworkGraphics"</code></li> <li><code>"NetworkGraphicsEdit"</code></li> </ul> <p>You can try them all with the following:</p> <pre><code>cursors = {"LinkHand", "Hand", "MathEdit45", "MathEdit90", "MathEdit135", "MathEdit180", "MathEdit225", "MathEdit270", "MathEdit315", "Crosshair","Formatted", "CellSelect", "CellShade", "CellSound", "CellInsert", "NameSelect", "SimpleWait", "FrameLCRotate", "FrameRCRotate", "FrameCTRotate", "FrameCBRotate", "FrameLTRotate", "FrameRBRotate", "FrameRTRotate" , "FrameLBRotate", "MeasureGraphic", "DeleteGraphics", "TestGraphics", "SelectGraphics", "DragGraphics", "DragGraphicsHighlight", "DragGraphicsDirect", "DragGraphicsHighlightDirect", "DragGraphicsH", "DragGraphicsHighlightH", "DragGraphicsDirectH", "DragGraphicsHighlightDirectH", "DragGraphicsV", "DragGraphicsHighlightV", "DragGraphicsDirectV", "DragGraphicsHighlightDirectV","DragSegment", "SampleStyle", "DoNothing", "DrawLine", "DrawPolyline", "DrawPolygon", "DrawRectangle", "DrawArrow", "DrawCircle", "DrawArcCursor", "DrawPoint", "DrawFreehand", "DrawPixel", "PlaceText", "PlaceMath", "RotateView", "PanView", "ZoomView", "RotateViewVertical", "Obscure", "NetworkGraphics", "NetworkGraphicsEdit"}; (MouseAppearance[#, #] &amp; /@ cursors) ~Partition~ 3 // TableForm </code></pre>
3,118,238
<p>I know there is a pretty simply way to check if a 2D point is inside a circle, I wanted to know how to do there same but in the 3rd dimension. The variables are the point's X, Y, Z and the sphere's 3D point and radius.</p> <p>Let's say we know that X and Y is inside the circle, can we rotate it 90 degrees and pretend the Z is the X or Y again and use the same algorithm?</p> <p>I will check in 2 for loops for X and Y if each point is inside circle, I will use the X and Y to plot the pixel and the Z to determine the colour.</p> <p>EDIT:</p> <p>Thanks for the answer, I didn't realize I could just include the Z into the algorithm I was talking about. In case you were interested here is the code, it works great! It's in F#.</p> <pre><code>open System.IO open System.Drawing type Vector = { X: int; Y: int; Z: int } type Sphere = { Center: Vector; Radius: int } let intersect v s = let x0, y0, z0 = float(v.X), float(v.Y), float(v.Z) let x1, y1, z1 = float(s.Center.X), float(s.Center.Y), float(s.Center.Z) sqrt(pown(x0 - x1) 2 + pown(y0 - y1) 2 + pown(z0 - z1) 2) &lt; float(s.Radius) let sphere = { Center = { X = 127; Y = 127; Z = 127 }; Radius = 127 } let bitmap = new Bitmap(256, 256) for x in 0 .. 255 do for y in 0 .. 255 do for z in 0 .. 255 do if intersect { X = x; Y = y; Z = z } sphere then bitmap.SetPixel(x, y, Color.FromArgb(z, 0, 0, 255)) bitmap.Save(Path.Combine(__SOURCE_DIRECTORY__, "bitmap.png")) </code></pre>
PackSciences
588,260
<p>If a point <span class="math-container">$M$</span> of cartesian coordinates <span class="math-container">$x_0,y_0,z_0$</span> is inside a circle of center <span class="math-container">$x_c,y_c,z_c$</span> of radius <span class="math-container">$R$</span>, it means that <span class="math-container">$\sqrt{(x_0-x_c)^2+(y_0-y_c)^2 + (z_0-z_c)^2} \leq R$</span>.</p>
1,018,672
<blockquote> <p><span class="math-container">$$\int_0^{\infty} \frac{1}{x^3-1}dx$$</span></p> </blockquote> <p>What I did:</p> <p><span class="math-container">$$\lim_{\epsilon\to0}\int_0^{1-\epsilon} \frac{1}{x^3-1}dx+\lim_{\epsilon\to0}\int_{1+\epsilon}^{\infty} \frac{1}{x^3-1}dx$$</span></p> <hr /> <p><span class="math-container">$$\lim_{\epsilon\to0}\int_0^{1-\epsilon}\frac{1}{3(x-1)}-\frac{2x+1}{6(x^2+x+1)}-\frac{1}{2(x^2+x+1)}dx+\lim_{\epsilon\to0}\int_{1+\epsilon}^{\infty}\frac{1}{3(x-1)}-\frac{2x+1}{6(x^2+x+1)}-\frac{1}{2(x^2+x+1)}dx$$</span></p> <hr /> <p><span class="math-container">$$\lim_{\epsilon\to0}\int_0^{1-\epsilon}\frac{1}{3(x-1)}-\frac{2x+1}{6(x^2+x+1)}-\frac{1}{2[(x+\frac{1}{2})^2+\frac{3}{4}]}dx+\lim_{\epsilon\to0}\int_{1+\epsilon}^{\infty}\frac{1}{3(x-1)}-\frac{2x+1}{6(x^2+x+1)}-\frac{1}{2[(x+\frac{1}{2})^2+\frac{3}{4}]}dx$$</span></p> <hr /> <p><span class="math-container">$$[\frac{1}{3}ln(x-1)-\frac{1}{6}ln(x^2+x+1)-\frac{1}{\sqrt3}\arctan(\frac{2x+1}{\sqrt3})]_{0}^{1-\epsilon}+[\frac{1}{3}ln(x-1)-\frac{1}{6}ln(x^2+x+1)-\frac{1}{\sqrt3}\arctan(\frac{2x+1}{\sqrt3})]_{1+\epsilon}^{\infty}$$</span></p> <hr /> <p><span class="math-container">$$[\frac{1}{6}(2ln(x-1)-ln(x^2+x+1))-\frac{1}{\sqrt3}\arctan(\frac{2x+1}{\sqrt3})]_{0}^{1-\epsilon}+[\frac{1}{6}ln{2(x-1})-\frac{1}{6}ln(x^2+x+1)-\frac{1}{\sqrt3}\arctan(\frac{2x+1}{\sqrt3})]_{1+\epsilon}^{\infty}$$</span></p> <hr /> <p><span class="math-container">$$[\frac{1}{6}ln(\frac{(x-1)^2}{x^2+x+1})-\frac{1}{\sqrt3}\arctan(\frac{2x+1}{\sqrt3})]_{0}^{1-\epsilon}+[\frac{1}{6}ln(\frac{(x-1)^2}{x^2+x+1})-\frac{1}{\sqrt3}\arctan(\frac{2x+1}{\sqrt3})]_{1+\epsilon}^{\infty}$$</span></p> <hr /> <p><span class="math-container">$$[\frac{1}{6}ln(\frac{x^2-2x+1}{x^2+x+1})-\frac{1}{\sqrt3}\arctan(\frac{2x+1}{\sqrt3})]_{0}^{1-\epsilon}+[\frac{1}{6}ln(\frac{x^2-2x+1}{x^2+x+1})-\frac{1}{\sqrt3}\arctan(\frac{2x+1}{\sqrt3})]_{1+\epsilon}^{\infty}$$</span></p> <hr /> <p><span class="math-container">$$\lim_{\epsilon\to0}[\frac{1}{6}ln(\frac{(1-\epsilon)^2-2(1-\epsilon)+1}{(1-\epsilon)^2+1-\epsilon+1})-\frac{1}{\sqrt3}\arctan(\frac{2(1-\epsilon)+1}{\sqrt3})+\frac{1}{\sqrt3}\arctan(\frac{1}{\sqrt3})]+\lim_{\epsilon\to 0} [ \frac{1}{6}ln(\frac{(\infty)^2-2(\infty)+1}{(\infty)^2+(\infty)+1})+\cdots]$$</span></p> <hr /> <blockquote> <p>This is where my problem is, what is :</p> <p><span class="math-container">$$ \frac{1}{6}ln(\frac{(\infty)^2-2(\infty)+1}{(\infty)^2+(\infty)+1})$$</span></p> </blockquote> <p>^^^ If I know past this, I know how to proceed. The only thing stopping me is this ^^^. Please help.</p>
Ron Gordon
53,268
<p>We can in fact evaluate the Cauchy principal value of the integral as follows (which is what I think you were trying to do).</p> <p>Consider the following contour integral:</p> <p>$$\oint_C dz \frac{\log{z}}{z^3-1}$$</p> <p>$C$ is a modified keyhole contour about the positive real axis of outer radius $R$ and inner radius $\epsilon$. The modification lies on small semicircular bumps above and below $z=1$ of radius $\epsilon$, and we will consider the limits as $\epsilon \to 0$ and $R\to\infty$.</p> <p>Let's evaluate this integral over the contours. There are $8$ pieces to evaluate, as follows:</p> <p>$$\int_{\epsilon}^{1-\epsilon} dx \frac{\log{x}}{x^3-1} + i \epsilon \int_{\pi}^0 d\phi \, e^{i \phi} \frac{\log{\left (1+\epsilon e^{i \phi}\right )}}{(1+\epsilon e^{i \phi})^3-1} \\ + \int_{1+\epsilon}^R dx \frac{\log{x}}{x^3-1} + i R \int_0^{2 \pi} d\theta \, e^{i \theta} \frac{\log{\left (R e^{i \theta}\right )}}{R^3 e^{i 3 \theta}-1} \\ + \int_R^{1+\epsilon} dx \frac{\log{x}+i 2 \pi}{x^3-1} + i \epsilon \int_{2 \pi}^{\pi} d\phi \, e^{i \phi} \frac{\log{\left (1+\epsilon e^{i \phi}\right ) }+i 2 \pi}{(1+\epsilon e^{i \phi})^3-1} \\ + \int_{1-\epsilon}^{\epsilon} dx \frac{\log{x}+i 2 \pi}{x^3-1} + i \epsilon \int_{2 \pi}^0 d\phi \, e^{i \phi} \frac{\log{\left (\epsilon e^{i \phi}\right )}}{\epsilon^3 e^{i 3 \phi}-1} $$ </p> <p>(To see this, draw the contour out, including the bumps about $z=1$.)</p> <p>As $R \to \infty$, the fourth integral vanishes as $\log{R}/R^2$. As $\epsilon \to 0$, the second integral vanishes as it is $O(\epsilon^2)$, while the eighth integral vanishes as $\epsilon \log{\epsilon}$. This leaves the first, third, fifth, sixth and seventh integrals, which in the above limits, become</p> <p>$$PV \int_0^{\infty} dx \frac{\log{x} - (\log{x}+i 2 \pi)}{x^3-1} + \frac{2 \pi^2}{3}$$</p> <p>It should be appreciated that, in the fifth, sixth, and seventh integrals, the $i 2 \pi $ factor appears because, on the lower branch of the real axis, we write $z=x \, e^{i 2 \pi}$. In the sixth integral, in fact, $z = e^{i 2 \pi} + \epsilon \, e^{i \phi + 2 \pi}$.</p> <p>The $PV$ denotes the Cauchy principal value of the integral. As it stands, the integral does not actually converge. Nevertheless, we are not actually considering the integral straight through the pole at $z=1$, but a very small detour around the pole. Thus, in the limit, we get the Cauchy PV. A little rearranging cancels the $\log$ term, and we now have:</p> <p>$$-i 2 \pi PV \int_0^{\infty} \frac{dx}{x^3-1} + \frac{2 \pi^2}{3}$$</p> <p>The contour integral is also equal to $i 2 \pi$ times the sum of the residues at the poles. The poles here are at $z=e^{i 2 \pi/3}$ and $z=e^{i 4 \pi/3}$. Note that the pole at $z=1$ is <em>not</em> inside the contour $C$ because of the detour around that "pole". It should be appreciated that the poles <strong>must</strong> have their arguments between $[0,2 \pi]$ because of the way we defined $C$.</p> <p>In any case, we now have that the above 1D integrals over the positive real line are equal to</p> <p>$$i 2 \pi \left [\frac{i 2 \pi/3}{3 e^{i 4 \pi/3}} + \frac{i 4 \pi/3}{3 e^{i 8 \pi/3}} \right ] = \frac{2 \pi ^2}{3}+i \frac{2\pi ^2}{3 \sqrt{3}} $$</p> <p>We may now solve for the principal value and get:</p> <p>$$ PV \int_0^{\infty} \frac{dx}{x^3-1} = -\frac{\pi}{3 \sqrt{3}} $$</p>
1,056,045
<blockquote> <p>Show that </p> <p>$2^{3n}-1$ is divisble by $7$ for all $n$ $\in \mathbb N$</p> </blockquote> <p>I'm not really sure how to get started on this problem, but here is what I have done so far:</p> <p>Base case $n(1)$:</p> <p>$\frac{2^{3(1)}-1}{7} = \frac{8-1}{7} = \frac{7}{7}$</p> <p>But not sure where to go from here. Tips?</p>
barak manos
131,263
<p>For simplicity, please note that $2^{3n}=8^n$.</p> <hr> <p>First, show that this is true for $n=1$:</p> <ul> <li>$\frac{8^1-1}{7}=1\in\mathbb{N}$</li> </ul> <p>Second, assume that this is true for $n$:</p> <ul> <li>$\frac{8^n-1}{7}=k\in\mathbb{N}$</li> </ul> <p>Third, prove that this is true for $n+1$:</p> <ul> <li><p>$\frac{8^{n+1}-1}{7}=\frac{8\cdot8^n-1}{7}$</p></li> <li><p>$\frac{8\cdot8^n-1}{7}=\frac{8\cdot8^n-8+7}{7}$</p></li> <li><p>$\frac{8\cdot8^n-8+7}{7}=\frac{8(8^n-1)+7}{7}$</p></li> <li><p>$\frac{8(8^n-1)+7}{7}=\frac{8\cdot7k+7}{7}$ <code>assumption used here</code></p></li> <li><p>$\frac{8\cdot7k+7}{7}=\frac{7(8k+1)}{7}$</p></li> <li><p>$\frac{7(8k+1)}{7}=8k+1\in\mathbb{N}$</p></li> </ul>
3,744,610
<p>The following puzzle was introduced by the psychologist Peter Wason in 1966, and is one of the most famous subject tests in the psychology of reasoning.</p> <blockquote> <p>Four cards are placed on the table in front of you. You are told (truthfully) that each has a letter printed on one side and a digit on the other, but of course you can only see one face of each. What you see is:</p> <pre><code> B E 4 7 </code></pre> <p>You are now told that the cards you are looking at were chosen to follow the rule \If there is a vowel on one side, then there is an odd number on the other side.&quot; What is the least number of cards you have to turn over to verify this rule, and which cards do you in fact have to turn over?</p> </blockquote> <p>My solution:</p> <blockquote> <p>Assuming A: Vowel on one side B: Odd number on one side</p> <p>by the data given to us, A(conditional)B {A implies B}. We have to verify this rule. We can do this by proving that the contrapositive of the above given statement is true, which says (NO ODD IMPLIES NO VOWEL). By doing this, We'll have to turn the &quot;E&quot; and the &quot;4&quot; card.</p> </blockquote> <p>The correct answer is : &quot;E&quot; and &quot;7&quot; card.</p> <p>Working:</p> <blockquote> <p>You need to turn over 2 cards (the E and the 7). The rule can be (simply) equivalently stated like so: There isn't a vowel on one side OR there is an odd number on one side. Thus, if you see a letter and it isn't a vowel, you don't need to verify the rule; likewise, if you see a number and it's odd, you don't need to verify the rule.</p> </blockquote> <p>I have two concerns, 1. How is my reasoning wrong. and 2. How is the statement -&quot;There isn't a vowel on one side OR there is an odd number on one side&quot; equivalent to - &quot;If there is a vowel on one side, then there is an odd number on the other side.&quot;</p>
Shubham Johri
551,962
<p>The statements <span class="math-container">$(1)~V\implies O$</span> and <span class="math-container">$(2)~V'\cup O$</span> are equivalent. See the truth table:<span class="math-container">$$\begin{matrix}V&amp;O&amp;(1)&amp;(2)\\0&amp;0&amp;1&amp;1\\0&amp;1&amp;1&amp;1\\1&amp;0&amp;0&amp;0\\1&amp;1&amp;1&amp;1\end{matrix}$$</span>But you arrived at the correct answer. The book's explanation is also correct: since <span class="math-container">$B$</span> is not a vowel and <span class="math-container">$7$</span> is odd, these two cards already satisfy the rule. The ones you will need to check are, as you found out, <span class="math-container">$E,4$</span>, to check whether the other side is having an odd number and non-vowel respectively. There seems to be a misprint.</p>
541,926
<p>I've been wrecking my brain with this problem and I really hope you can help me. You see I have a triangle that is either an isosceles or equilateral or right and I have to find a way to: 1)Convert it to a right one by moving one of its vertices, 2)Convert it to an isosceles one by moving one of its vertices, 3)Convert it to an equilateral one by moving two of its vertices. I am trying to write a program in C that will do this but unfortunately I can't figure out the math. Hope you can help. Thank you in advance.</p>
Han de Bruijn
96,057
<p>Perhaps your question can be answered by a standard piece of Finite Element theory. Here comes.<P>Let's consider the simplest non-trivial finite element shape in two dimensions: the linear triangle. Function behaviour is approximated inside such a triangle by a <I>linear</I> interpolation between the function values at the vertices, also called: nodal points. Let $T$ be such a function, and $(x,y)$ coordinates, then: $$ T = A.x + B.y + C $$ Where the constants A, B, C are to be determined. Substitute $ x=x_k $ and $ y=y_k $ with $ k=1,2,3 $: $$ \left[ \begin{array}{c} T_1 \\ T_2 \\ T_3 \end{array} \right] = \left[ \begin{array}{ccc} 1 &amp; x_1 &amp; y_1 \\ 1 &amp; x_2 &amp; y_2 \\ 1 &amp; x_3 &amp; y_3 \end{array} \right] \left[ \begin{array}{c} C \\ A \\ B \end{array} \right] $$ The first of these equations can already be used to eliminate the constant $C$, once and forever: $$ T_1 = A.x_1 + B.y_1 + C $$ Resulting in: $$ T - T_1 = A.(x - x_1) + B.(y - y_1) $$ Hence the constants $A$ and $B$ are determined by: $$ \begin{array}{ll} T_2 - T_1 = A.(x_2 - x_1) + B.(y_2 - y_1) \\ T_3 - T_1 = A.(x_3 - x_1) + B.(y_3 - y_1) \end{array} $$ Two equations with two unknowns. The solution is found by straightforward elimination, or by applying Cramer's rule: $$ \begin{array}{ll} A = [ (y_3 - y_1).(T_2 - T_1) - (y_2 - y_1).(T_3 - T_1) ] / \Delta \\ B = [ (x_2 - x_1).(T_3 - T_1) - (x_3 - x_1).(T_2 - T_1) ] / \Delta \end{array} $$ There are several forms of the determinant $\Delta$, which should be memorized when it is appropriate: \begin{eqnarray*} &amp;&amp; \Delta = (x_2 - x_1).(y_3 - y_1) - (x_3 - x_1).(y_2 - y_1) \\ &amp;&amp; \Delta = 2 \times \mbox{ area of triangle } \\ &amp;&amp; \Delta = x_1.y_2 + x_2.y_3 + x_3.y_1 - y_1.x_2 - y_2.x_3 - y_3.x_1 \\ &amp;&amp; \Delta = x_1.(y_2 - y_3) + x_2.(y_3 - y_1) + x_3.(y_1 - y_2) \\ &amp;&amp; \Delta = y_1.(x_3 - x_2) + y_2.(x_1 - x_3) + y_3.(x_2 - x_1) \\ &amp;&amp; \Delta = \left\| \begin{array}{ccc} 1 &amp; x_1 &amp; y_1 \\ 1 &amp; x_2 &amp; y_2 \\ 1 &amp; x_3 &amp; y_3 \end{array} \right\| \end{eqnarray*} Anyway, it is concluded that: $$ T - T_1 = \xi.(T_2 - T_1) + \eta.(T_3 - T_1) $$ Where: $$ \begin{array}{ll} \xi = [ (y_3 - y_1).(x - x_1) - (x_3 - x_1).(y - y_1) ] / \Delta \\ \eta = [ (x_2 - x_1).(y - y_1) - (y_2 - y_1).(x - x_1) ] / \Delta \end{array} $$ Or: $$ \left[ \begin{array}{c} \xi \\ \eta \end{array} \right] = \left[ \begin{array}{cc} + (y_3 - y_1) &amp; - (x_3 - x_1) \\ - (y_2 - y_1) &amp; + (x_2 - x_1) \end{array} \right] / \Delta \left[ \begin{array}{c} x - x_1 \\ y - y_1 \end{array} \right] $$ The inverse of the following problem is recognized herein: $$ \left[ \begin{array}{c} x - x_1 \\ y - y_1 \end{array} \right] = \left[ \begin{array}{cc} (x_2 - x_1) &amp; (x_3 - x_1) \\ (y_2 - y_1) &amp; (y_3 - y_1) \end{array} \right] \left[ \begin{array}{c} \xi \\ \eta \end{array} \right] $$ Or: $$ \begin{array}{ll} x - x_1 = \xi .(x_2 - x_1) + \eta.(x_3 - x_1) \\ y - y_1 = \xi .(y_2 - y_1) + \eta.(y_3 - y_1) \end{array} $$ But also: $$ T - T_1 = \xi .(T_2 - T_1) + \eta.(T_3 - T_1) $$ Therefore the <I>same</I> expression holds for the function $T$ as well as for the coordinates $x$ and $y$ . This is precisely what people mean by an <I> isoparametric</I> ("same parameters") transformation, a terminology which is quite common in Finite Element contexts. Now recall the formulas which express $\xi$ and $\eta$ into $x$ and $y$ : $$ \begin{array}{ll} \xi = [ (y_3 - y_1).(x - x_1) - (x_3 - x_1).(y - y_1) ]/\Delta \\ \eta = [ (x_2 - x_1).(y - y_1) - (y_2 - y_1).(x - x_1) ]/\Delta \end{array} $$ Thus $\xi$ can be interpreted as: area of the sub-triangle spanned by the vectors $ (x - x_1 , y - y_1) $ and $ (x_3 - x_1 , y_3 - y_1) $ divided by the whole triangle area. And $\eta$ can be interpreted as: area of the sub-triangle spanned by the vectors $ (x - x_1 , y - y_1) $ and $ (x_2 - x_1 , y_2 - y_1) $ divided by the whole triangle area. This is the reason why $ \xi $ and $ \eta $ are sometimes called <I>area-coordinates</I>; see the figure below, where (two times) the area of the triangle as a whole is denoted as $\Delta$. There exist even <I>three</I> of these coordinates in literature. But the third area-coordinate is, of course, dependent on the other two, being equal to $(1-\xi-\eta)$. Instead of area-coordinates, we prefer to talk about <I>local coordinates</I> $\xi$ and $\eta$ of an element, in contrast to the <I>global coordinates</I> $x$ and $y$. It is possible that local coordinates coincide with the global coordinates. A triangle for which such is the case is called a <I>parent element</I>. The portrait of the parent triangle is also depicted in the same figure: the two sides of it are coincident with the unit vectors in a Cartesian coordinate system.<BR> So far so good about standard FEM. Now about your real problem (I hope).<P>Suppose you have two triangles, one with coordinates $\left\{ (x_1,y_1), (x_2,y_2), (x_3,y_3) \right\} $ and one with coordinates $\left\{ (x'_1,y'_1), (x'_2,y'_2), (x'_3,y'_3) \right\} $, then it is always possible to transform the first triangle into the second via the parent element. As follows. $$ \left[ \begin{array}{c} x - x'_1 \\ y - y'_1 \end{array} \right] = \left[ \begin{array}{cc} (x'_2 - x'_1) &amp; (x'_3 - x'_1) \\ (y'_2 - y'_1) &amp; (y'_3 - y'_1) \end{array} \right] \left[ \begin{array}{c} \xi \\ \eta \end{array} \right] $$ $$ \left[ \begin{array}{c} \xi \\ \eta \end{array} \right] = \left[ \begin{array}{cc} (x_2 - x_1) &amp; (x_3 - x_1) \\ (y_2 - y_1) &amp; (y_3 - y_1) \end{array} \right]^{-1} \left[ \begin{array}{c} x - x_1 \\ y - y_1 \end{array} \right] $$ The resulting transformation is independent of rotation and scaling of the global coordinates, i.e. it leaves similarities intact at both sides. So if you have two sample triangles that are transformed into each other, then all triangles that are similar to those two are transformed in the same way.<img src="https://i.stack.imgur.com/l8A3e.jpg" alt="enter image description here"></p>
362
<p>The Math Stack Exchange chat room has had some heated arguments/suspensions in the past and I was wondering if there were any plans for the newly opened Math Overflow chat room to have its own moderators to provide the required special attention to prevent similar problem situations?</p>
Community
-1
<p>The site <a href="http://mathoverflow.net">http://mathoverflow.net</a> is <em>not</em> a new site; it exists since more than three and a half years (end of september it will be four). Only it joined the SE-network recently. Then like now it has moderators (the same as before). Who they are can be seen on the already provided page (in analogy with other sites) <a href="https://mathoverflow.net/users?tab=moderators">https://mathoverflow.net/users?tab=moderators</a> </p> <p>Some of these moderators were elected some others not, as they started the site "from scratch" and there was not yet anybody to elect them. In the old version of MO, Anton Geraschenko, the main founder of the site, was not only moderator but also administrator.</p> <p>Please, do not think of this site as some site in "beta" or something along these lines.</p>
1,326,816
<p>How to find the Maclaurin series of the function $$f(x)=\frac{1}{(9-x^2)^2}$$ I guess we are gonna use derivatives but i have no idea how the final answer should be formed.</p>
JimmyK4542
155,509
<p>Start with the familiar geometric series: $\dfrac{1}{1-y} = \displaystyle\sum_{n = 0}^{\infty}y^n$.</p> <p>Next, differentiate both sides to get this series: $\dfrac{1}{(1-y)^2} = \displaystyle\sum_{n = 0}^{\infty}ny^{n-1}$.</p> <p>Now, what can you plug in for $y$ to make $\dfrac{1}{(1-y)^2}$ more like $\dfrac{1}{(9-x^2)^2}$? </p> <p>Hint: $\dfrac{1}{(9-x^2)^2} = \dfrac{1}{9^2(1-\tfrac{x^2}{9})^2}$.</p>
2,485,447
<p>My attempt:</p> <p>3x≡1 mod 7 (1)</p> <p>4x≡1 mod 9 (2)</p> <p>Multiply (1) by 5</p> <p>Multiply (2) by 7</p> <p>x≡5 mod 7</p> <p>x≡7 mod 9</p> <p>So x≡9k+7</p> <p>9k+7=5(mod7)</p> <p>k=5(mod7)</p> <p>k=7j+5</p> <p>x=9(7j+5)+7</p> <p>=63j+52</p> <p>x≡52(mod63)</p>
Ozzy
473,926
<p>Summing two equations you get $7x=2$, or $0=2$ in mod 7. So no solutions exist.</p>