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 $n=[1,oo)$ ?

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Answer 1 · 2017-02-15T00:17:34

For $x in (-1,1)$

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

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

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

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:

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

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

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

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