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 10^nx^n` from `n=[0,oo)`?

Answer 1

`f'(x) = sum_(n=0)^oo \ n10^n x^(n-1)`

`int_0^x \ f(t) \ dt=sum_(n=0)^oo \ 10^n / (n+1) \ x^(n+1)`

Explanation:

We have:

`f(x) = sum_(n=0)^oo \ 10^nx^n`

And so:

`f'(x) = d/dx sum_(n=0)^oo \ 10^nx^n`
`" " = sum_(n=0)^oo \ d/dx10^nx^n`
`" " = sum_(n=0)^oo \ 10^nd/dxx^n`
`" " = sum_(n=0)^oo \ 10^n nx^(n-1)`
`" " = sum_(n=0)^oo \ n10^n x^(n-1)`

And:

`int_0^x \ f(t) \ dt= int_0^x \ sum_(n=0)^oo \ 10^nt^n \ dt`
`" "= sum_(n=0)^oo \ int_0^x \ 10^nt^n \ dt`
`" "= sum_(n=0)^oo \ 10^n \ int_0^x \ t^n \ dt`
`" "= sum_(n=0)^oo \ 10^n \ [ t^(n+1)/(n+1) ]_0^x`

`" "= sum_(n=0)^oo \ 10^n \ { x^(n+1)/(n+1) - 0 }`
`" "= sum_(n=0)^oo \ 10^n \ x^(n+1)/(n+1)`
`" "= sum_(n=0)^oo \ 10^n / (n+1) \ x^(n+1)`