Question

How do you find the power series for `f'(x)` and `int f(t)dt` from [0,x] given the function `f(x)=Sigma 1/n^2x^(2n)` from `n=[1,oo)`?

Answer 1

For `x in (-1,1)`

`f(x) = sum_(n=1)^oo x^(2n)/n^2 =x^2+x^4/4+x^6/9+...`

`f'(x) = 2 sum_(n=1)^oo x^(2n-1)/n = 2x+x^3+2/3x^5+...`

`int_0^x f(t)dt = sum_(n=1)^oo x^(2n+1)/(n^2(2n+1)) = x^3/3+x^5/20+x^7/63+...`

Explanation:

Consider the series:

`sum_(n=1)^oo x^(2n)/n^2`

and evaluate its radius of convergence using the ratio test:

`lim_(n->oo) abs(a_(n+1)/a_n) = lim_(n->oo) abs ((x^(2(n+1))/(n+1)^2)/(x^(2n)/n^2)) = lim_(n->oo) abs (x^(2n+2)/x^(2n)) n^2/(n+1)^2 = x^2`

So the series is absolutely convergent for `absx<1`, and we can see that it is also absolutely convergent for `abs x = 1` since:

`sum_(n=0)^oo 1/n^2 = pi^2/6`

Thus for `x in(-1,1)` we can differentiate and integrate term by term:

`f(x) = sum_(n=1)^oo x^(2n)/n^2`

`f'(x) = sum_(n=1)^oo d/dx (x^(2n)/n^2) = sum_(n=1)^oo (2nx^(2n-1))/n^2 = 2 sum_(n=1)^oo x^(2n-1)/n`

`int_0^x f(t)dt = sum_(n=1)^oo int_0^x (t^(2n)/n^2)dt = sum_(n=1)^oo x^(2n+1)/(n^2(2n+1))`