Question

How do you use the Maclaurin series for `f(x) = ln abs((1+x)/(1-x))`?

Answer 1

`ln abs ((1+x)/(1-x)) = 2 sum_(n=0)^oo x^(2n+1)/(2n+1)`

for `abs(x) < 1`

Explanation:

Differentiate the function:

`d/dx ln abs ((1+x)/(1-x)) = d/dx ln abs (1+x) -d/dx ln abs (1-x) = 1/(1+x)+1/(1-x) = (1-x+1+x)/((1-x)(1+x))= 2/(1-x^2)`

Now note that:

`1/(1-x^2)`

is the sum a geometric series with ratio `x^2`, so we have:

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

for `abs(x) < 1`.

We can now integrate term by term:

`ln abs ((1+x)/(1-x)) = 2int_0^x (dt)/(1-t^2) = 2 sum_(n=0)^oo int_0^x t^(2n)dt =2 sum_(n=0)^oo [t^(2n+1)/(2n+1)]_0^x`

So, always for `abs(x) < 1`:

`ln abs ((1+x)/(1-x)) = 2 sum_(n=0)^oo x^(2n+1)/(2n+1)`