How do you find the power series for #f(x)=int(ln(1+2t^2)dt# from [0,x] and determine its radius of convergence?

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Answer 1 · 2017-02-28T19:42:19

Series is:

#f(x) = sum _{n=1}^{\infty }(-1)^{n+1}2^n{x^{2n+1}}/{n(2n+1)} #

Converges absolutely for: # x in (- 1/sqrt 2, 1/sqrt 2)#

For #f(x)=int_0^x ln(1+2t^2) \ dt#, we have from the FTC:

#f'(x)=ln(1+2x^2)#

We note the following power series:

#ln(1+z) = sum _{n=1}^{\infty }(-1)^{n+1}{x^{n}}/{n} = x - x^2/2 +x^3 /3 - ...# for #|x|<1#

So:

#f'(x) = sum _{n=1}^{\infty }(-1)^{n+1}{(2x^2)^{n}}/{n} #

#= sum _{n=1}^{\infty }(-1)^{n+1}2^n{x^{2n}}/{n} #

Integrating this:

#f(x) = sum _{n=1}^{\infty }(-1)^{n+1}2^n{x^{2n+1}}/{n(2n+1)} + C #

We know from the integral that #f(0) = 0 implies C = 0#, so:

#f(x) = sum _{n=1}^{\infty }(-1)^{n+1}2^n{x^{2n+1}}/{n(2n+1)} #

For convergence we apply the ratio test:

#abs( ((-1)^{n+2}2^(n+1){x^{2n+3}}/{(n+1)(2n+3)} )/((-1)^{n+1}2^n{x^{2n+1}}/{n(2n+1)} ) )_(n to oo)#

# = abs(-2 x^2* ( n(2n+1) )/((n+1)(2n+3)) )_(n to oo)#

# = abs(-2 x^2* ( 2+1/n) /(2 + 5/n + 3/n^2) )_(n to oo)#

#implies abs(2x^2) < 1#

This will definitely converge for: # x in (- 1/sqrt 2, 1/sqrt 2)#