Question

Question #7f415

Answer 1

`f(x)=sum_(n=0)^(infty)(-1)^n (x-1)^(n+1)/(n+1)`

Explanation:

Let `g(t)=ln(1+t) Rightarrow g(0)=0`

By differentiating w.r.t. `t`,

`g'(t)=1/(1+t)=1/(1-(-t))`

By the sum of geometric series `a/(1-r)=sum_(n=0)^(infty)ar^n`,

`g'(t)=sum_(n=0)^(infty)1cdot(-t)^n=sum_(n=0)^(infty)(-1)^n t^n`

By integrating w.r.t. `t`,

`g(t)=sum_(n=0)^(infty)(-1)^n t^(n+1)/(n+1)+C`,

where `C=0` since `g(0)=0`

So, we have

`ln(1+t)=sum_(n=0)^(infty)(-1)^n t^(n+1)/(n+1)`

By setting `t=x-1`,

`ln(x)=sum_(n=0)^(infty)(-1)^n (x-1)^(n+1)/(n+1)`

I hope that this was clear.