# 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)?

Feb 15, 2017

For $x \in \left(- 1 , 1\right)$

$f \left(x\right) = {\sum}_{n = 1}^{\infty} {x}^{2 n} / {n}^{2} = {x}^{2} + {x}^{4} / 4 + {x}^{6} / 9 + \ldots$

$f ' \left(x\right) = 2 {\sum}_{n = 1}^{\infty} {x}^{2 n - 1} / n = 2 x + {x}^{3} + \frac{2}{3} {x}^{5} + \ldots$

${\int}_{0}^{x} f \left(t\right) \mathrm{dt} = {\sum}_{n = 1}^{\infty} {x}^{2 n + 1} / \left({n}^{2} \left(2 n + 1\right)\right) = {x}^{3} / 3 + {x}^{5} / 20 + {x}^{7} / 63 + \ldots$

#### Explanation:

Consider the series:

${\sum}_{n = 1}^{\infty} {x}^{2 n} / {n}^{2}$

and evaluate its radius of convergence using the ratio test:

${\lim}_{n \to \infty} \left\mid {a}_{n + 1} / {a}_{n} \right\mid = {\lim}_{n \to \infty} \left\mid \frac{{x}^{2 \left(n + 1\right)} / {\left(n + 1\right)}^{2}}{{x}^{2 n} / {n}^{2}} \right\mid = {\lim}_{n \to \infty} \left\mid {x}^{2 n + 2} / {x}^{2 n} \right\mid {n}^{2} / {\left(n + 1\right)}^{2} = {x}^{2}$

So the series is absolutely convergent for $\left\mid x \right\mid < 1$, and we can see that it is also absolutely convergent for $\left\mid x \right\mid = 1$ since:

${\sum}_{n = 0}^{\infty} \frac{1}{n} ^ 2 = {\pi}^{2} / 6$

Thus for $x \in \left(- 1 , 1\right)$ we can differentiate and integrate term by term:

$f \left(x\right) = {\sum}_{n = 1}^{\infty} {x}^{2 n} / {n}^{2}$

$f ' \left(x\right) = {\sum}_{n = 1}^{\infty} \frac{d}{\mathrm{dx}} \left({x}^{2 n} / {n}^{2}\right) = {\sum}_{n = 1}^{\infty} \frac{2 n {x}^{2 n - 1}}{n} ^ 2 = 2 {\sum}_{n = 1}^{\infty} {x}^{2 n - 1} / n$

${\int}_{0}^{x} f \left(t\right) \mathrm{dt} = {\sum}_{n = 1}^{\infty} {\int}_{0}^{x} \left({t}^{2 n} / {n}^{2}\right) \mathrm{dt} = {\sum}_{n = 1}^{\infty} {x}^{2 n + 1} / \left({n}^{2} \left(2 n + 1\right)\right)$