- #1

- 107

- 0

[tex]m\pi=\sum^{\infty}_{n=0} \frac{(-1^{n}) \Gamma_{2n+1}}{(2n+1) r^{2n+1}}[/tex]

m = 0,1,2,....

r is real number

is there an explicit expression for [tex]\Gamma_{2n+1} [/tex] in terms of n ?

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter mmzaj
- Start date

- #1

- 107

- 0

[tex]m\pi=\sum^{\infty}_{n=0} \frac{(-1^{n}) \Gamma_{2n+1}}{(2n+1) r^{2n+1}}[/tex]

m = 0,1,2,....

r is real number

is there an explicit expression for [tex]\Gamma_{2n+1} [/tex] in terms of n ?

- #2

mathman

Science Advisor

- 7,956

- 498

- #3

Mute

Homework Helper

- 1,388

- 10

[itex]\Gamma_{2n+1}[/itex] need not have anything to do with the gamma function! It's just a coefficient labelled with a capital gamma.

As for the form of gamma... it will obviously depend on m, and it will have to depend on r, too. I'm guessing that you'll need to add more conditions on the form you want, otherwise you may not get a unique answer. (And I don't know if you can get an answer independent of r).

(Following formulae lifted from wikipedia. check external reference to ensure correctness).

For instance,

[tex]\frac{\pi}{4} = \sum_{n=0}^\infty = \frac{(-1)^n}{2n+1}[/tex],

so we could to choose [itex]\Gamma_{2n+1} = 4 m r^{2n+1}[/itex].

Another series is

[tex]\frac{\pi}{2} = \sum_{n=0}^\infty \frac{n!}{(2n+1)!!}[/tex]

(where the (2n+1)!! = 1*3*5*...(2n+1) ), so we could choose [itex]\Gamma_{2n+1} = 2(-1)^n m r^{2n+1} (2n+1)n!/(2n+1)!![/itex]. (Admittedly, though, the factor of (-1)^n in there sort of breaks with the subscript notation of the coefficient).

As for general conditions on Gamma, I guess I would run through the convergence tests for series, like the ratio test. http://en.wikipedia.org/wiki/Ratio_test

Last edited:

- #4

- 2,112

- 18

http://en.wikipedia.org/wiki/Alternating_series The Leibniz' test could be more useful.

- #5

- 107

- 0

[tex]tan^{-1}(\frac{s}{r})=\sum^{\infty}_{n=0} \frac{(-1^{n}) s^{2n+1} }{(2n+1) r^{2n+1}}[/tex]

where

[tex]\frac{s}{r}<1[/tex]

=> [tex]\Gamma_{2n+1} = s^{2n+1}[/tex]

Share: