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

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Answer 1 · 2017-08-30T12:39:56

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

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

and both series converge absolutely for every $x in RR$

Write the series as:

$sum_(n=1)^oo n^-nx^n = sum_(n=1)^oo (x/n)^n$

For any $x in RR$ we can see that choosing $N > 2x$ we have:

$x/n < 1/2$ for $n > N$

and then:

$(x/n)^n < (1/2)^n$ for $n > N$

so that the series is convergent. As the radius of convergence is $R=oo$ , we can differentiate and integrate the series term by term for any $x$ :

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

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

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 n^-nx^nf(x)=Sigma n^-nx^n from n=[1,oo)n=[1,oo)?