# User talk:GFauxPas

## Contents

## Source Review

Just a quick heads up to draw your attention to Definition:Random Variable, which has an open SourceReview call listed on your name. — Lord_Farin (talk) 11:42, 5 April 2014 (UTC)

- I have that book packed away; I'll have to dig it up. --GFauxPas (
**talk**) 11:20, 8 April 2014 (UTC)

## zeta 2

If you've been watching my sandbox, you might have seen me work on this:

- $\displaystyle \int_{\to 0}^{\to 1} \ln x \ln \left({1-x}\right) \, \mathrm dx = 2 - \zeta\left({2}\right)$

Should I put this up as a theorem, or is it too specific an exercise to be worth sharing?

--GFauxPas (**talk**) 11:42, 9 April 2014 (UTC)

- I think it's worth sharing, specially after all the hard work you put into it. The name needs to be considered carefully. Might want to put it into an "/Examples" subpage of either Logarithms or Integral Calculus (or both) ... once I've reached that part of my Spiegel handbook where there are hundreds of examples of integrations I will be in there to flesh it out, but it's something I haven't got to yet. --prime mover (talk) 04:26, 10 April 2014 (UTC)

- Or maybe under Basel Problem... --GFauxPas (
**talk**) 11:03, 10 April 2014 (UTC)

- Or maybe under Basel Problem... --GFauxPas (

- No, rather not under Basel problem -- the latter is a specific problem with a specific solution (i.e. the one Euler came up with, which explains its name), this just happens to have a result which includes the same zeta instance. --prime mover (talk) 16:43, 10 April 2014 (UTC)

## Real and Imaginary Parts

Are these theorems?

- $\displaystyle \int \operatorname{Im}\,\left({f\left({s}\right)}\right)\, \mathrm ds = \operatorname{Im}\,\left({\int f\left({s}\right)\,\mathrm ds}\right)$

- $\displaystyle \int \operatorname{Re}\,\left({f\left({s}\right)}\right)\, \mathrm ds = \operatorname{Re}\,\left({\int f\left({s}\right)\,\mathrm ds}\right)$

for real $s$? For complex $s$?

--GFauxPas (**talk**) 12:14, 5 May 2014 (UTC)

- If $f$ is real then it's true trivially. If $f$ is complex then I believe (although I would have to go back to my source works to prove it) that it's not.

- The question for whether it is true for real or complex $s$ (as opposed to $f$) I'm not sure how much sense it makes, but consider the fact that while $s$ may be real then it may not be the case that so is $f(s)$.

- I don't know without thinking about it for more than 5 seconds but, as I say, I don't believe so. --prime mover (talk) 12:52, 5 May 2014 (UTC)

## Gamma function Proof

Regarding the proof for:

- $\displaystyle \mathcal L \left \{ {t^p} \right \} = \frac {\Gamma \left({p+1}\right)} { s^{p+1} }$

for $\operatorname{Re}\left({p}\right) > -1$, some sites are doing this. Is it justified?

- $\displaystyle \int_0^{\to +\infty} t^p e^{-st} \, \mathrm dt = \int_{0}^{?} t^p e^{-u} \frac {\mathrm dt}{\mathrm du}\, \mathrm du$

where $u = st$, and then doing various stuff with the limits of integration. But it seems to me like a mistake; if $u$ is complex, then how can the limits of integration be complex? If $u$ is real, how can we replace a complex variable with a real variable?

I asked someone about this and he suggested that maybe they're making the integral into a contour integral, but that doesn't help me make it look like $\Gamma$... or does it? --GFauxPas (**talk**) 14:19, 26 May 2014 (UTC)

- It's justified. You can prove by using the contour of two rays going to infinity, connected by an ever larger arc, that the "direction" of the infinity does not influence the value of the integral. This is because the integral over the arc goes to zero, because of the exponential. 't May be beyond the current scope of PW, though. — Lord_Farin (talk) 17:46, 26 May 2014 (UTC)

- Once I get to it, I intend this *not* to be beyond the current scope. It's all in my complex analysis source notes -- but these are all buried way down in the substrata of the intellectual sedimentary rock that my office resembles ... and anyway, I'm mucking about with primitives at the moment. --prime mover (talk) 19:35, 26 May 2014 (UTC)

- Ah, so, it's not $u = st$ so much as $u = rt$ where $s = re^{i\theta}$? And with $\displaystyle \int_C t^p e^{-u}\, \mathrm du$, $C$ is a curve connecting
*any*two rays connected by an arc, as the length of the rays go to $+\infty$? and $\arg s$ doesn't affect anything? --GFauxPas (**talk**) 11:09, 27 May 2014 (UTC)

- Ah, so, it's not $u = st$ so much as $u = rt$ where $s = re^{i\theta}$? And with $\displaystyle \int_C t^p e^{-u}\, \mathrm du$, $C$ is a curve connecting

- Yes, I think that's about it. — Lord_Farin (talk) 20:48, 27 May 2014 (UTC)

## Modeling question

So I'm taking a class on modeling. What's the explanation or justification of this process?:

Given a sequence $\langle a \rangle : \N \to \R$, we create a function $f:\R \to \R$ such that $f \restriction_{\N} = \langle a \rangle$.

The objective is to find $\sup (\langle a \rangle)$ or $\inf(\langle a \rangle)$.

To do so, we find the local extrema of $f$ (say at $f(c)$), and the assumption is that $\langle a \rangle$ attains its supremum/infimum at either $a_i$ or $a_{i+i}$, where $i \le c \le i + 1$.

Why and when is this method kosher? --GFauxPas (**talk**) 22:57, 11 September 2014 (UTC)

- Very quickly, thinking as I type ...

- If you've identified *all* the local extrema of $f$ (assuming that there are a finite number), then it can intuitively seen that one of the points either side of the *most* extreme one is going to be "as close as you can get" to it. This of course requires that the function does not have great big peaks at the $f \left({n + \dfrac 1 2}\right)$ points. Consider $f = \sin \left({\pi x}\right)$ as an awkward case which does not fit this intuitive notion but which *does* fulfil the conditions of the statement. --prime mover (talk) 05:29, 12 September 2014 (UTC)

## Number theory test question

I had a number theory test today. We had to choose 6 out of 8 questions, and I did fine, but I want to know how to solve this one I didn't know how to do:

Show that there are $100$ consecutive integers that are not square free. Hint: Take $100$ different primes $p$ and arrange for $p^2$ to divide an appropriate member of the sequence.

How does one do this? --GFauxPas (**talk**) 02:20, 18 November 2014 (UTC)

- Choose $p_1, \ldots p_{100} > 5$. Because $p_i^2$ is coprime to $100$, it has a multiplicative inverse $q_i$ (with $1 < q_i < 100$) so that $p_i^2 q_i \equiv 1 \bmod 100$. Then $p_i^2 q_i i \equiv i \bmod 100$.

- It follows that we can find adequate multiples $m_i p_i^2$ of each $p_i^2$ such that $100 m_i p_i^2 + p_i^2 q_i i = 100 \prod_i p_i^2 + i$. The result follows. — Lord_Farin (talk) 19:24, 18 November 2014 (UTC)

- Thanks, but I lost you at the RHS of the last equation. --GFauxPas (
**talk**) 22:29, 18 November 2014 (UTC)

- Thanks, but I lost you at the RHS of the last equation. --GFauxPas (

- That's not so strange, because I went too quickly and arrived at a wrong conclusion. Let me try again:

We are looking for an adequate $n$ such that $p_i^2 \mid n + i$. That is, such that for each $i$:

- $n \equiv - i \bmod p_i^2$

This has solutions $n$ by virtue of the Chinese remainder theorem. The result follows. — Lord_Farin (talk) 17:36, 21 November 2014 (UTC)

Nice. --GFauxPas (**talk**) 17:40, 21 November 2014 (UTC)

- Cut some excess fat. — Lord_Farin (talk) 17:56, 21 November 2014 (UTC)

## Enhancement to Eqn template

I have added an enhancement to the eqn template to enhance the formatting of expressions of the form $a \le x \le b$, which you asked for some time back but I didn't spend any time thinking about at the time.

I have added the parameters "m" and "mo" to the template so you can do something like this:

\(\displaystyle a\) | \(\le\) | \(\displaystyle x - y\) | \(\le\) | \(\displaystyle b\) | |||||||||

\(\displaystyle a + y\) | \(\le\) | \(\displaystyle x\) | \(\le\) | \(\displaystyle b + y\) |

I'm not sure I like it very much, and it may need more work, but it's there, feel free to experiment. --prime mover (talk) 07:42, 27 November 2014 (UTC)

- Interesting. --GFauxPas (
**talk**) 12:13, 27 November 2014 (UTC)