Question

How do you find the power series for `f'(x)` and `int_0^x f(t)dt` given the function `f(x)=sum_(n=0)^oo x^(2n)` ?

Answer 1

`f'(x) = 0 + 2x^1 + 4x^3 + 6x^5 + ...`
`" " = sum_(n=1)^oo (2n)x^(2n-1)`

`int_0^x \ f(t) \ dt = x + x^3/3 + x^5/5 + x^7/7 + ...`
`" " = sum_(n=0)^oo x^(2n+1)/(2n+1)`

Explanation:

We have `f(x)` defined by its power series:

`f(x) = sum_(n=0)^oo x^(2n)`

We can expand the fist few terms to get an idea of how the series behaves:

`f(x) = x^0 + x^2 + x^4 + x^6 + ...`
`" " = 1 + x^2 + x^4 + x^6 + ...`

So first we find the derivative by differentiating term by term:

`f'(x) = 0 + 2x^1 + 4x^3 + 6x^5 + ...`
`" " = sum_(n=1)^oo (2n)x^(2n-1)`

And similarly we find the integral by integrating term by term:

`int_0^x \ f(t) \ dt = int_0^x \ 1 + t^2 + t^4 + t^6 + ... \ dt`
`" " = [t + t^3/3 + t^5/5 + t^7/7 + ... ]_0^x`

`" " = {x + x^3/3 + x^5/5 + x^7/7 + ... } - {0}`
`" " = x + x^3/3 + x^5/5 + x^7/7 + ...`

`" " = sum_(n=0)^oo x^(2n+1)/(2n+1)`