How do you find the Maclaurin series of #f(x)=ln(1+x^2)# ?

1 Answer
Nov 5, 2015

Answer:

Plug in #x^2# for every x in the Maclaurin series for #f(x)=ln(1+x)#

Explanation:

So, you have the Maclaurin series for #f(x)=ln(1+x)# defined as:

#f(x)=ln(1+x) = x-x^2/2+x^3/3-... =sum_(k=1)^n(((-1)^(k+1)x^k) / k)#

The question is essentially asking you to plug in #x^2# for every x here, so:

#(x^2)^1-(x^2)^2/2+(x^2)^3/3-(x^2)^4/4+...#
which gives... #x^2 - x^4/2+x^6/3-x^8/4+... = sum_(k=1)^n(((-1)^(k+1)(x^2)^k) / k)#