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 n^-nx^n` from `n=[1,oo)`?

Answer 1

`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`

Explanation:

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)`