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Super Half Vamp Riku

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Well if the question was:

 

infinity

______ _ 2 q

\ (2q + 1 + pi\/n)(pi n)

\ _____________________

/

/_____ (2q + 1)!

 

q=0

 

(where n is a positive integer), but have not been able to find the process for evaluating this power series. Any and all information on this problem would be greatly appreciated.

 

This summation can be broken up and rewritten. Splitting the fraction into two pieces by writing (2q + 1 + pi sqrt(n)) (pi^2 n)^q / (2q+1)! as the sum of (2q + 1) (pi^2 n)^q / (2q+1)! and (pi sqrt(n)) (pi^2 n)^q / (2q+1)!, then rewriting (pi^2 n)^q as (pi sqrt(n))^(2q) and simplifying, gives

infinity infinity

______ _ 2q _ 2q+1 ______ _ k

\ (pi \/n) (pi \/n) \ (pi \/n)

\ __________ + _______ = \ _________

/ /

/_____ (2q)! (2q+1)! /_____ k!

 

q=0 k=0

 

Will the answer be 42?

 

EDIT: chances are it doesn't look right in the post but I still doubt you would figure it out.

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Well if the question was:

 

infinity

______ _ 2 q

\ (2q + 1 + pi\/n)(pi n)

\ _____________________

/

/_____ (2q + 1)!

 

q=0

 

(where n is a positive integer)' date=' but have not been able to find the process for evaluating this power series. Any and all information on this problem would be greatly appreciated.

 

This summation can be broken up and rewritten. Splitting the fraction into two pieces by writing (2q + 1 + pi sqrt(n)) (pi^2 n)^q / (2q+1)! as the sum of (2q + 1) (pi^2 n)^q / (2q+1)! and (pi sqrt(n)) (pi^2 n)^q / (2q+1)!, then rewriting (pi^2 n)^q as (pi sqrt(n))^(2q) and simplifying, gives

infinity infinity

______ _ 2q _ 2q+1 ______ _ k

\ (pi \/n) (pi \/n) \ (pi \/n)

\ __________ + _______ = \ _________

/ /

/_____ (2q)! (2q+1)! /_____ k!

 

q=0 k=0

 

Will the answer be 42?

 

EDIT: chances are it doesn't look right in the post but I still doubt you would figure it out.

[/quote']im only in 8th grade


What do you prefer Kyubi or Hachibi.

kyubi,you are talking about tailed beast's right

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