something I don't understand to do with Dirichlets theorem

wildrabbitt · 2020-02-23, 17:31

Hi,

the following is something I've been reading.

He started from the power series

$\sum_{n=1}^\infty \big(\frac{n}{q}\big)x^n=\frac{1}{1-x^q}\sum_{m=1}^{q-1}\big(\frac{m}{q}\big)x^m=\frac{xf(x)}{1-x^q}$

say, and by putting this in the formula

$\Gamma(s)n^{-1}=\int_0^1 x^{n-1}(\log x^{-1})^{s-1} \mathrm{d}x$

he obtained

$\Gamma(s)L(s)=-\int_0^1\frac{f(x)}{x^q-1}(\log x^{-1})^{s-1}\mathrm{d}x$

I'm stuck because I can't see how he put what he put in the formula.

Can anyone explain it step by step?

R.D. Silverman · 2020-02-23, 20:17

Quote:

Originally Posted by wildrabbitt

Hi,

the following is something I've been reading.

He started from the power series

$\sum_{n=1}^\infty \big(\frac{n}{q}\big)x^n=\frac{1}{1-x^q}\sum_{m=1}^{q-1}\big(\frac{m}{q}\big)x^m=\frac{xf(x)}{1-x^q}$

say, and by putting this in the formula

$\Gamma(s)n^{-1}=\int_0^1 x^{n-1}(\log x^{-1})^{s-1} \mathrm{d}x$

he obtained

$\Gamma(s)L(s)=-\int_0^1\frac{f(x)}{x^q-1}(\log x^{-1})^{s-1}\mathrm{d}x$

I'm stuck because I can't see how he put what he put in the formula.

Can anyone explain it step by step?

I am getting "math processing error". Your post is not being parsed correctly.

Dr Sardonicus · 2020-02-23, 23:58

Quote:

Originally Posted by wildrabbitt

Hi,

the following is something I've been reading.

He started from the power series

$\sum_{n=1}^\infty \big(\frac{n}{q}\big)x^n=\frac{1}{1-x^q}\sum_{m=1}^{q-1}\big(\frac{m}{q}\big)x^m=\frac{xf(x)}{1-x^q}$

say, and by putting this in the formula

$\Gamma(s)n^{-1}=\int_0^1 x^{n-1}(\log x^{-1})^{s-1} \mathrm{d}x$

he obtained

$\Gamma(s)L(s)=-\int_0^1\frac{f(x)}{x^q-1}(\log x^{-1})^{s-1}\mathrm{d}x$

I'm stuck because I can't see how he put what he put in the formula.

Can anyone explain it step by step?

I took the liberty of TEXing your equations. That makes them a whole lot easier to read, for me at least. You can probably drop the big parens.

Is that supposed to be (log(x)^-1^(s-1)? That would be log(x)^1-s, yes?

wildrabbitt · 2020-02-24, 07:52

I just wrote what was written in the book.

I'm pretty sure it's supposed to be

$\log \frac{1}{x} $

otherwise it would have been written the way you wrote.

wildrabbitt · 2020-02-24, 09:51

I aught to mention that

$ L(s)=\sum_{n=1}^{\infty}\big(\frac{n}{q}\big)n^{-s}$

for which

$\big(\frac{n}{q}\big)$ is the Legendre symbol as is it is the first post.

Don't know what to do if my latex is not parsing correctly. It comes out right when I preview the post.

I can give more info if necessary.

Dr Sardonicus · 2020-02-24, 12:49

Quote:

Originally Posted by wildrabbitt

I just wrote what was written in the book.

I'm pretty sure it's supposed to be

$\log \frac{1}{x} $

otherwise it would have been written the way you wrote.

Ah, log(1/x), not 1/log(x). OK, now I see a Gamma-function integral in prospect in your second equation. (x = exp^-u, scribble scribble, u = t/n, scribble scribble...)

Your second equation is wrong. It should be

$\Gamma(s)n^{-s}=\int_0^1 x^{n-1}(\log x^{-1})^{s-1} \mathrm{d}x$

(It is n^-s on the left side, not n^-1)

wildrabbitt · 2020-02-24, 20:40

Thanks.

Are you substituting $e^{-u}$ for $x$ ? What are you doin with $u$?

Also, in case there's an obvious remedy to my latex not parsing properly, here's an example of what I wrote in the first post without the \$ and \$.

\sum_{n=1}^\infty \big(\frac{n}{q}\big)x^n=\frac{1}{1-x^q}\sum_{m=1}^{q-1}\big(\frac{m}{q}\big)x^m=\frac{xf(x)}{1-x^q}

Nick · 2020-02-25, 08:21

Quote:

Originally Posted by wildrabbitt

Also, in case there's an obvious remedy to my latex not parsing properly, here's an example of what I wrote in the first post without the \$ and \$.

\sum_{n=1}^\infty \big(\frac{n}{q}\big)x^n=\frac{1}{1-x^q}\sum_{m=1}^{q-1}\big(\frac{m}{q}\big)x^m=\frac{xf(x)}{1-x^q}

LaTeX: just put backslash then open-square-bracket before it, and backslash then close-square-bracket afterwards:

\[\sum_{n=1}^\infty \big(\frac{n}{q}\big)x^n=\frac{1}{1-x^q}\sum_{m=1}^{q-1}\big(\frac{m}{q}\big)x^m=\frac{xf(x)}{1-x^q}\]

Dr Sardonicus · 2020-02-25, 13:32

Quote:

Originally Posted by wildrabbitt

Thanks.

Are you substituting $e^{-u}$ for $x$ ? What are you doin with $u$?

Also, in case there's an obvious remedy to my latex not parsing properly, here's an example of what I wrote in the first post without the \$ and \$.
<snip>

My problem with LaTex in posts is mine alone. Viewing on other computers, everything looks OK. The fact that everything looked OK on my system when I simply slapped tex tags around the equations says the expressions parse OK.

If you don't know how to make substitutions in integrals, it's high time you learn. Unfortunately, I'm probably not the one to teach you.

wildrabbitt · 2020-02-25, 17:01

I didn't really get the answer I was hoping for so I thought I'd just ask the first things that it occurred to me to ask.

I do know how to make substitutions in integrals I just wasn't sure if that's what you were doing and it looked like you were making two substitutions.

Thanks though. So you're substiting for x. I don't know what you're doing with u.

2020-02-23, 17:31	#1
wildrabbitt Jul 2014 110111111₂ Posts	something I don't understand to do with Dirichlets theorem Hi, the following is something I've been reading. He started from the power series \(\sum_{n=1}^\infty \big(\frac{n}{q}\big)x^n=\frac{1}{1-x^q}\sum_{m=1}^{q-1}\big(\frac{m}{q}\big)x^m=\frac{xf(x)}{1-x^q}\) say, and by putting this in the formula \(\Gamma(s)n^{-1}=\int_0^1 x^{n-1}(\log x^{-1})^{s-1} \mathrm{d}x\) he obtained \(\Gamma(s)L(s)=-\int_0^1\frac{f(x)}{x^q-1}(\log x^{-1})^{s-1}\mathrm{d}x\) I'm stuck because I can't see how he put what he put in the formula. Can anyone explain it step by step? Last fiddled with by wildrabbitt on 2020-02-23 at 17:32

2020-02-24, 07:52	#4
wildrabbitt Jul 2014 1BF₁₆ Posts	I just wrote what was written in the book. I'm pretty sure it's supposed to be \(\log \frac{1}{x} \) otherwise it would have been written the way you wrote.

2020-02-24, 09:51	#5
wildrabbitt Jul 2014 1BF₁₆ Posts	clarification I aught to mention that \( L(s)=\sum_{n=1}^{\infty}\big(\frac{n}{q}\big)n^{-s}\) for which \(\big(\frac{n}{q}\big)\) is the Legendre symbol as is it is the first post. Don't know what to do if my latex is not parsing correctly. It comes out right when I preview the post. I can give more info if necessary. Last fiddled with by wildrabbitt on 2020-02-24 at 09:52

2020-02-24, 20:40	#7
wildrabbitt Jul 2014 110111111₂ Posts	Thanks. Are you substituting \(e^{-u}\) for \(x\) ? What are you doin with \(u\)? Also, in case there's an obvious remedy to my latex not parsing properly, here's an example of what I wrote in the first post without the \\( and \\). \sum_{n=1}^\infty \big(\frac{n}{q}\big)x^n=\frac{1}{1-x^q}\sum_{m=1}^{q-1}\big(\frac{m}{q}\big)x^m=\frac{xf(x)}{1-x^q} Last fiddled with by wildrabbitt on 2020-02-24 at 20:44 Reason: needed to access page source

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2020-02-25, 17:01	#10
wildrabbitt Jul 2014 3×149 Posts	I didn't really get the answer I was hoping for so I thought I'd just ask the first things that it occurred to me to ask. I do know how to make substitutions in integrals I just wasn't sure if that's what you were doing and it looked like you were making two substitutions. Thanks though. So you're substiting for x. I don't know what you're doing with u.