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)#?

Question

Isaiah Allen · Accepted Answer

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)) #

Scarlett Allison · Answer

undefined To find the power series for $f'(x)$, differentiate the given function term by term, then find the power series for $\int f(t) dt$ from $0$ to $x$ by integrating the function term by term.