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

Answer 1

`f'(x) = sum_(n=1)^oo \ (n+1)x^(n-1)`

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

Explanation:

We have:

`f(x) = sum_(1=0)^oo \ (n+1)/nx^n`

And so:

`f'(x) = d/dx sum_(n=1)^oo \ (n+1)/n \ x^n`

`" " = sum_(n=1)^oo \ d/dx \ (n+1)/n \ x^n`

`" " = sum_(n=1)^oo \ (n+1)/n \ nx^(n-1)`

`" " = sum_(n=1)^oo \ (n+1)x^(n-1)`

And:

`int_0^x \ f(t) \ dt = int_0^x \ sum_(n=1)^oo \ (n+1)/nt^n \ dt`
`" "= sum_(n=1)^oo \ (n+1)/n \ int_0^x \ t^n \ dt`

`" "= sum_(n=1)^oo \ (n+1)/n \ { \ [ t^(n+1)/(n+1)]_0^x }`

`" "= sum_(n=1)^oo \ (n+1)/n \ { \ x^(n+1)/(n+1)- 0 }`

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

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