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Lapsed engineer/scientist (now a patent practitioner) seeking to expand his math skills. Most of those skills lay in areas useful in optics such as sums & integrals (like the one below), complex analysis, differential equations, transforms, and data analysis.

My greatest hits page. (a work in progress)

"Wir müssen wissen, wir werden wissen." - David Hilbert

The picture on the left represents the integration region for the following integral of functions of finite support:

$$T(\mathbf{u}',\mathbf{u}'') = \int_{\mathbb{R}^2} d^2 \mathbf{\sigma} \; S(\mathbf{\sigma}) P(\mathbf{u}'+\mathbf{\sigma}) P^*(\mathbf{u}''+\mathbf{\sigma})$$


Aug
18
comment Need help with $\int_0^\infty\frac{\log(1+x)}{\left(1+x^2\right)\,\left(1+x^3\right)}dx$
@Superabound: it's due to the form of the summand. In other words, write out the terms of the sum and see how you can make the replacement.
Aug
14
awarded  Guru
Aug
11
comment Interesting log sine integrals $\int_0^{\pi/3} \log^2 \left(2\sin \frac{x}{2} \right)dx= \frac{7\pi^3}{108}$
@ChantryCargill: much appreciated! The identity comes from the Taylor series for $(1-x)^{-1/2}$ about $x=0$.
Aug
3
comment The limit of the product $\prod_{i=1}^n\frac{1 - (2i + 1)a/(2n)}{1 - ia/n}$ as $n\to\infty$
@did: thanks for pushing me to improve the post.
Aug
3
comment The limit of the product $\prod_{i=1}^n\frac{1 - (2i + 1)a/(2n)}{1 - ia/n}$ as $n\to\infty$
@did: I added a bit to the answer.
Aug
3
revised The limit of the product $\prod_{i=1}^n\frac{1 - (2i + 1)a/(2n)}{1 - ia/n}$ as $n\to\infty$
added 951 characters in body
Aug
3
comment The limit of the product $\prod_{i=1}^n\frac{1 - (2i + 1)a/(2n)}{1 - ia/n}$ as $n\to\infty$
@did: if you take the next term in the log and sum over $i$, you get an error term $$\frac{a^2}{4 (1-a)} \frac1{n}$$ resulting from another Riemann sum.
Aug
3
comment The limit of the product $\prod_{i=1}^n\frac{1 - (2i + 1)a/(2n)}{1 - ia/n}$ as $n\to\infty$
@did: fair question and I admit to being informal here. I meant that the $log$ term may be replaced by the $a/(2 n)$... piece for sufficiently large $n$.
Aug
2
comment The limit of the product $\prod_{i=1}^n\frac{1 - (2i + 1)a/(2n)}{1 - ia/n}$ as $n\to\infty$
@Did: I don't think I am doing that. Rather, I am saying that $$\log{\left( 1-\frac{a/(2 n)}{1-a i/n}\right)} \sim -\frac{a}{2} \frac{1}{n} \frac{1}{1-a i/n} $$ This holds for sufficiently large $n$. If I were doing the approximation you stated, then I would have that this approximation would behave as $-a/(2 n)$ which is indeed wrong.
Aug
2
comment Examples of functions with this property
$f(x) = 1-e^{-x}$
Aug
2
comment The limit of the product $\prod_{i=1}^n\frac{1 - (2i + 1)a/(2n)}{1 - ia/n}$ as $n\to\infty$
@Did: maybe I am missing something, but since $0<a<1$, $0 < a i/n < 1$ as well. Thus, even when $i=n$, one can always find some $n$ sufficiently large such that the above expansions are true. But maybe you are seeing something else that I am not.
Aug
1
comment Show that $\int^{\infty}_0 \frac{\sin^4 x}{x^4}=\frac{\pi} 3$.
Here's an approach that uses Fourier transforms: math.stackexchange.com/questions/318037/…
Jul
27
awarded  Enlightened
Jul
27
awarded  Nice Answer
Jul
21
awarded  Revival
Jul
18
awarded  Nice Answer
Jul
14
comment Correct Dampening Formula
Oh my, didn't mean to scare you. There may be a much simpler way, but you have to assume a model (e.g., exponential) with fewer parameters. Lagrange is simply a sure-fire way if you can assume that your data fits a polynomial.
Jul
14
comment Correct Dampening Formula
en.wikipedia.org/wiki/Lagrange_polynomial
Jul
14
awarded  Notable Question
Jul
11
comment Integral $I:=\int_0^1 \frac{\log^2 x}{x^2-x+1}dx=\frac{10\pi^3}{81 \sqrt 3}$
This integral is evaluated in math.stackexchange.com/questions/305124/…