How do you use the Maclaurin series for f(x)=ln1+x1x?

1 Answer
Apr 15, 2017

ln1+x1x=2n=0x2n+12n+1

for |x|<1

Explanation:

Differentiate the function:

ddxln1+x1x=ddxln|1+x|ddxln|1x|=11+x+11x=1x+1+x(1x)(1+x)=21x2

Now note that:

11x2

is the sum a geometric series with ratio x2, so we have:

21x2=2n=0(x2)n=2n=0x2n

for |x|<1.

We can now integrate term by term:

ln1+x1x=2x0dt1t2=2n=0x0t2ndt=2n=0[t2n+12n+1]x0

So, always for |x|<1:

ln1+x1x=2n=0x2n+12n+1