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andrei.frit02
Nov 15, 2019

A little power series.

in Math Problems



can anyone help me find the sum of this power series?


5 comments
0
Ian Fowler
Nov 16, 2019

This is a power series which produces an infinite polynomial in x. It does not converge to a single value. The polynomial is:

1 + (2/3)x + (4/5)x^2 + (8/7)x^3 + (16/9)x^4 + ....


Using the ratio test we evaluate the limit as n ---> infinity of the ratio a(n+1)/a(n) where

a(n) = 2^n*x^n/(2n+1) . This ratio simplifies to: 2x(2n+1)/(2n+3)

As n ---> infinity this limit becomes 2x. Therefore this series will converge to a single value when the value of x lies in the interval 0 < x < 1/2.


Other than that I am not sure what you are asking for.


0
andrei.frit02
Nov 16, 2019

Thank you for the replay. Sorry if I wasn't clear enough.I must find the function that has this series expansion.

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Ian Fowler
Nov 16, 2019

Fair point. I should have figured that out. I got this far:

1/(1-y) = 1 + y + y^2 + y^3 + y^4 + ... for |y| < 1

Replace y with 2x and we get:

1/(1-2x) = 1 + 2x + 4x^2 + 8x^3 + 16x^4 + ... for |x| < 1/2

This gives us the numerators in the terms of the series (i. e. the numerators for a Geometric Series)


I can't see a way to alter 1/(1-2x) to get the odd numbers in the denominator. generating functions maybe? I'll keep at it.

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Aneesh Srinivas
Feb 05, 2020

;okay 1 slight correction on the radius of convergence i said on the last step,

it should be for -1/2≤x<1/2 and x≠0(arctanh(√2x)/√(2x) is undefined there).


andrei.frit02
Feb 10, 2020

Thank you so much. After all it was quite simple, but I lack creativity, I guess. :))

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5 comments