/sci/ - Science & Math

>>10325148
>you just can't do it -- using microscopic laws you can't derive boltzmann entropy.
>What is Sakur-Tetrode equation
>What is stat mech

>>8123447
ur right and whatnot but I had to stop and lmao there

>>8117042
To continue on how he derives the reflection formula for the analytic continuation on the left of s=1/2... it's worth writing down because it's pretty cute:

Start with the Gamma funciton defined as

[math] \Gamma(s) := \int_0^\infty x^{s-1} {\mathrm e}^{-x}\,{\mathrm d}x [/math]

Substitution x to n·x let's you write

[math] \Gamma(s) = n^s \int_0^\infty x^{s-1} {\mathrm e}^{-nx}\,{\mathrm d}x [/math]

or

[math] {\frac {1} {n^s}} = {\frac {1} {\Gamma(s)}} \int_0^\infty x^{s-1} {\mathrm e}^{-nx}\,{\mathrm d}x [/math]

Recognizing the geometric series

[math] \sum_{n=1}^\infty ({\mathrm e}^{-x})^n = \frac{1} { {\mathrm e}^{-x}-1} [/math]

(which btw. expands as [math] \frac{1} {x} - \frac{1}{2} + \frac{1}{12}x+O(x^2) [/math])

gives you

[math] \sum_{n=1}^\infty \frac{1}{n^s} = \frac{1} { \Gamma(s) } \int_0^\infty \frac{x^{s-1}} { {\mathrm e}^x-1} \, {\mathrm d}x [/math]

Which is the famous integral representation.

The next step is a bit more elaborate, but you can see where the sinus pops in.
He takes the integral into the complex plane, where the [math] \frac{1} { {\mathrm e}^x-1}[/math] diverges periodically in steps of [math]2\pi\,i [/math].

He pics up the sum again on the other side and that's how the ingrediences to

[math] \zeta(s) = 2\,(2\pi)^{s-1}\sin{\left(\pi\,s/2\right)}\,\Gamma(1-s)\,\zeta(1-s) [/math]

come together.

The integer values are all determined by Bernoulli numbers, essentially stemming from the integrand above.
If you look at the more general polylog [math] \sum_{n=1}^\infty n^{-s}\,z^n [/math], there is a more general integral representation just as well.

>>8117042
To continue on how he derives the reflection formula for the analytic continuation on the left of s=1/2... it's worth writing down because it's pretty cute:

Start with the Gamma funciton defined as

[math] \Gamma(s) := \int_0^\infty x^{s-1} {\mathrm e}^{-x}\,{\mathrm d}x [/math]

Substitution x to n·x let's you write

[math] \Gamma(s) = n^s \int_0^\infty x^{s-1} {\mathrm e}^{-nx}\,{\mathrm d}x [/math]

or

[math] {\frac {1} {n^s}} = {\frac {1} {\Gamma(s)}} \int_0^\infty x^{s-1} {\mathrm e}^{-nx}\,{\mathrm d}x [/math]

Recognizing the geometric series

[math] \sum_{n=1}^\infty ({\mathrm e}^{-x})^n = \frac{1} { {\mathrm e}^{-x}-1} [/math]

(which btw. expands as [math] \frac{1} {x} - \frac{1}{2} + \frac{1}{12}x+O(x^2) [/math])

gives you

[math] \sum_{n=1}^\infty \frac{1}{n^s} = \frac{1} { \Gamma(s) } \int_0^\infty \frac{x^{s-1}} { {\mathrm e}^x-1} \, {\mathrm d}x [/math]

Which is the famous integral representation.

The next step is a bit more elaborate, but you can see where the sinus pops in.
He takes the integral into the complex plane, where the [math] \frac{1} { {\mathrm e}^x-1}$ diverges periodically in steps of $2\pi\,i [/math].

He pics up the sum again on the other side and that's how the ingrediences to

[math] \zeta(s) = 2\,(2\pi)^{s-1}\sin{\left(\pi\,s/2\right)}\,\Gamma(1-s)\,\zeta(1-s) [/math]

come together.

The integer values are all determined by Bernoulli numbers, essentially stemming from the integrand above.
If you look at the more general polylog [math] \sum_{n=1}^\infty n^{-s}\,z^n [/math], there is a more general integral representation just as well.

No Pre-Calculus, Calculus I-III, Differential Equations or elementary Linear Algebra questions allowed.

No Calculus I-III, Differential Equations or elementary Linear Algebra questions allowed.

No Pre-Calculus/Calculus I-III questions allowed.

Anything else is fair game.

No Pre-Calculus, Calculus I-III questions allowed. Anything else is fair game.

The title says it all. Post your mathematical problems or thoughts and have people discuss them with you.
No college advice, no textbook recs.