I had a go reading through the paper and I think I found the error. The main argument in the paper can be summarized as follows: The Riemann $\Xi$-function $\Xi(t) = \xi\left(\frac{1}{2} + it\right)... View answer 31 votes I found a bug in the code given in Hooked's answer (which means that my original reanalysis was also flawed): one also have to check for insufficient material when assessing a draw, i.e. int(board.... View answer Accepted answer 30 votes Fourier series (as with infinite series in general) cannot always be term-by-term differentiated. For general series we have the following theorem Theorem: Term-by-term differentiation: If a series ... View answer Accepted answer 25 votes It's a nice identity, but this one is fairly simple compared to some of the other magical results in the notebook. The function$f(a) = \int_0^\infty \frac{e^{ax}-e^{-ax}}{e^{\pi x}-e^{-\pi x}} \frac{...

You can prove that $f(x) \equiv x$ in four fairly simple steps: Show that $f$ is one-to-one. Assume $f(x) = f(y)$ and show that this implies $x=y$ by applying $f$ two times to each side of the ...

First the wave-equation: let's start with a small narrow pulse located close to $x=0$ ($u$ nonzero close to $x=0$ and zero elsewhere) and then let it go at $t=0$. Now let's see how long time does it ...

The result found by OP turns out to be quite generic; it holds for a wide range of sequences $\rho_n$ and not just the zeros of the $\zeta$-function. A precise formulation of what is observed is the ...

Here is another solution using the residue theorem. If we try to integrate the function $\frac{\sinh(az)}{\sinh(\pi z)}\frac{1}{1+n^2z^2}$ directly then we obtain the result $$\int_0^\infty \frac{\... View answer Accepted answer 14 votes Make a change of variables u = e^x$$\int_0^1 f(x) e^{nx}dx = \int_{1}^e g(u) u^{n}du = 0$$where g(u) = \frac{f(\log(u))}{u} is a continuous function. You can now apply Weierstrass ... View answer Accepted answer 14 votes The substitution u = \log(1+x) gives$$I = \int_0^\infty \frac{\log(1+x)}{x(x+1)} dx = \int_0^\infty \frac{udu}{e^u-1}$$Now by using \frac{1}{1-x} = 1+x+x^2+\ldots with x=e^{-u} we get$$I = ...
Put $$a_n = \frac{n!^{1/n}}{n}$$ then $$\log a_n = \frac{1}{n}\left(\log n! - n\log n\right)= \frac{1}{n}\sum_{k=1}^n\log\left(\frac{k}{n}\right)$$ where we have used $\log n! = \log 1 + \log 2 + \... View answer Accepted answer 14 votes Just in case it is not assumed that$\alpha\in\mathbb{R}$, let$\alpha=a+bi$with$a,b\in\mathbb{R}. Then \begin{align} \cos(a\pi+b\pi i)&=\cos(a\pi)\sin(b\pi i)+\sin(a\pi)\cos(b\pi i)\\ \... View answer Accepted answer 12 votes Just so this question is fully answered (i.e. getting the bound 8/3) here I add a condensed proof of x_n^2 \leq \frac{8}{3} from the answer given here as Community Wiki. DefineE_n = (x_n - x_{... View answer 11 votes A perfect square can be written on the formn=(p_1^{\alpha_1}\cdots p_k^{\alpha_k})^2$where$p_i$are some prime numbers and$\alpha_i> 0$are integers. The sum of the divisors of$n$is given ... View answer Accepted answer 11 votes I will here derive an expression for the more general problem of evaluating$\sum_{n=1}^\infty \frac{H_n z^{2n+1}}{(2n+1)^2}$for$|z| < 1$in terms of polylogarithms. This derivation uses the same ... View answer 11 votes From the definition$f'(x) = \frac{1}{x}$with$f(0)=1$we have $$f(x) \equiv \int_1^x \frac{dt}{t}$$ By a simple change of variables$z = ty$to the integral above we get the result $$f(x) = \int_y^... View answer Accepted answer 11 votes This is a theorem knows as Lerchs theorem. I will give a standard proof of this below. I will restrict myself to continuous functions, but the generalization to integrable functions (continuous almost ... View answer Accepted answer 10 votes Your approach looks fine. The theorem you need to justify your claims is the following (see for example Theorem 2 in this note): Term by term differentiation theorem: If \sum f_k(x_0) converges at ... View answer Accepted answer 10 votes TLDR; We can get$$x_n^2 \leq \frac{8}{3}$$directly and with some computation we can establish the best bound x_n^2 \leq \frac{9}{4} for all n. The argument below can be made much shorter, I just ... View answer Accepted answer 10 votes Rough estimate: We have \log(1+x) \approx x as |x| \ll 1 so for large N the sum is approximately$$ \frac{1}{N}\sum_{k=1}^N \frac{1}{k^{\phi^2}\log\left(1 + \frac{k^{\phi^{-2}}}{N}\right)} \... View answer Accepted answer 10 votes To show$e^{(A+B)t} = e^{At}e^{Bt}$for all$t\implies [A,B] = 0$compare the coefficient of$t^2$in the power-series expansion of both sides of the equation. We have $$e^{(A+B)t} = 1+ (A+B)t + \... View answer Accepted answer 10 votes The sum of the first three terms is$$1000\left[\frac{1}{2^{1}}+\frac{1}{2^{4}}+\frac{1}{2^{9}}\right] \simeq 564.453$$The sum of the rest of the terms is bounded by (using a geometrical series)$$... View answer Accepted answer 10 votes Using the series expansion of$\arcsin^2(x)\$ (see this question) given by $$\arcsin^2(x) = \frac{1}{2}\sum_{n=1}^\infty\frac{(2x)^{2n}}{n^2{2n\choose n}}$$ we see that your sum is $$\sum_{n=0}^\... View answer Accepted answer 10 votes The operator e^{t(x+D)} is an exponential generating function for the sequence (x+D)^n. The commutator of the two operators x and D is given by [x,D] = xD - Dx = -1 and [[x,D],x] = [[x,D],D]... View answer 9 votes This is a bit overkill for solving the question at hand, but for completness I'll add an answer for how one can evaluate the integral for a general value of n using residue calculus giving us the ... View answer Accepted answer 9 votes Let's first start with defining what we mean by the expression a = x^{x^{x^{\cdot^{\cdot^{\cdot}}}}}: the infinite power tower is the limit of the recursion$$a_{n+1} = x^{a_n},~~~~~~~ a_0 = x$$... View answer 9 votes This problem has an interesting history from the early days of real analysis attached to it. I really like this story and the proof it contains of the problem in this question is also different from ... View answer Accepted answer 9 votes The complex logarithm \log(z) is defined as the inverse function to the exponential function, i.e. it satisfies e^{\log(z)} \equiv z. Since re^{i\theta} = re^{i\theta + 2\pi k i} for all k\in\... View answer 9 votes Hint: A convex function always satisfy$$f(tx+(1-t)y)\leq tf(x) + (1-t)f(y),~~~~~~~t\in[0,1]$$Take f(x) = |x|^p, show that this is convex for p\geq 1 (for example by the second derivative test)... View answer 8 votes A neat method to construct a continued fraction for \pi is to use the addition formula for \arctan:$$\arctan(x) + \arctan(y) = \arctan\left(\frac{x+y}{1-xy}\right)$$which can also be written$$...