- #1

- 391

- 0

[tex] \sum_{p} p^{-s}=P(s) [/tex]

* the functional equation relating P(s) and P(1-s)

* the relation with Riemann zeta

- Thread starter zetafunction
- Start date

- #1

- 391

- 0

[tex] \sum_{p} p^{-s}=P(s) [/tex]

* the functional equation relating P(s) and P(1-s)

* the relation with Riemann zeta

- #2

- 1,838

- 7

[tex]\zeta(s) = \sum_{r_{1},r_{2}\ldots}\prod_{j}p_{j}^{-sr_{j}}[/tex]

where [itex]p_{j}[/itex] is the jth prime and the [itex]r_{j}[/itex] in the summation range from zero to infinity. Summing over the [itex]r_{j}[/itex] gives:

[tex]\zeta(s)= \prod_{p}\frac{1}{1-p^{-s}}[/tex]

Take the log of both sides:

[tex]\log\left[\zeta(s)\right]= -\sum_{p}\log\left(1-p^{-s}\right)[/tex]

Expand the logarithm and sum over the primes p:

[tex]\log\left[\zeta(s)\right]=\sum_{k=1}^{\infty}\frac{P(ks)}{k}[/tex]

You can then invert this relation to find the [itex]P(s)[/itex] using Möbius inversion.

- #3

- 1,838

- 7

[tex]P(s) = \log\left[\zeta(s)\right] - \sum_{p}\frac{\log\left[\zeta(ps)\right]}{p} + \sum_{p_{1}<p_{2}}\frac{\log\left[\zeta(p_{1}p_{2}s)\right]}{p_{1}p_{2}}- \sum_{p_{1}<p_{2}<p_{3}}\frac{\log\left[\zeta(p_{1}p_{2}p_{3}s)\right]}{p_{1}p_{2}p_{3}}+\cdots[/tex]

- #4

- 54

- 0

[tex]P(s)= \log \zeta(s)+\sum_{m=0}^{\infty}\sum_{n=1}^{\infty}(-1)^{m+1}\frac{\log\zeta\Big(s\prod_{k=0}^{m}p_{n+k}\Big)}{\prod_{k=0}^{m} p_{n+k}}.

[/tex]

I assume that's for [tex]\textnormal{Re}(s)>1[/tex].

- Replies
- 2

- Views
- 5K

- Last Post

- Replies
- 2

- Views
- 3K

- Last Post

- Replies
- 3

- Views
- 3K

- Last Post

- Replies
- 12

- Views
- 5K

- Last Post

- Replies
- 1

- Views
- 2K

- Last Post

- Replies
- 10

- Views
- 8K

- Last Post

- Replies
- 8

- Views
- 10K

- Last Post

- Replies
- 8

- Views
- 4K

- Last Post

- Replies
- 3

- Views
- 2K

- Last Post

- Replies
- 1

- Views
- 3K