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

1 Answer
Apr 15, 2017

#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)#