Welcome to my blog, a place for mathematics, physics, programming, and more!
Problem Evaluate [\sum_{k=1}^\infty\frac{(-1)^{k+1}}{k}\sum_{n=0}^\infty\frac{1}{k2^n+1}.] Preliminary Ideas Before jumping right in and solving the problem, lets look at this double sum and se...
A problem given by Mathologer is the following: Pick any positive integer, say $n=4$, then write out all of its ordered partitions [\begin{align}4 &=4 \ 4 &=3+1 \ 4 &=1+3 \ 4 &=2+2...
I enjoy archiving media as a hobby and a serious pursuit. I mainly focus on saving books, course notes, and old documentaries. Besides archiving for the sake of archiving itself, it serves a practi...
One of my favorite little oddities are integrals of infinite products. Definite integrals of infinite sums are usually straightforward but those of infinite products can be striking, in my opinion....
It has been known since the time of the Babylonians that it is possible to parametrize Pythagorean triples using a rather simple collection of expressions — their understanding of this is re...
In July of 2010 I was introduced to Minecraft by a friend and he showed me a redstone mechanism he had constructed to operate a door. Although the device itself was quite simple, we were able to se...
In a previous post we proved that [\int_0^\alpha \frac{du}{\sqrt{1+u^4}}=\frac{3\Gamma^2\left(\frac{1}{4}\right)}{16\sqrt{\pi}}] where $\alpha=\sqrt{1+\sqrt{2}+\sqrt{2+2\sqrt{2}}}$. Here we will ...
Consider the contour integral [\frac{1}{2\pi i}\oint_C\frac{(1+z)^n}{z^{k+1}}\,\mathrm dz] where $C$ is a positively oriented circular contour centered at $z=0$. If we expand the term in the nume...
A long time ago, I saw an intriguing question posed by Shobhit, asking to show that [\begin{align}\int_0^\alpha \frac{du}{\sqrt{1+u^4}}=\frac{3\Gamma^2\left(\frac{1}{4}\right)}{16\sqrt{\pi}}\tag{1...
A quick way to make work of some challenging integrals is the following: Glasser’s Master Theorem: Let $f(x)$ be a Riemann integrable function over $(-\infty,\,\infty)$ then \[\displaystyle{\i...
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