Question

How do you find the maclaurin series expansion of `ln(1+x^2)`?

Answer 1

`ln (1+x^2)= x^2-1/2x^4+1/3x^6-1/4x^8+1/5x^10 - ...`

Explanation:

The Maclaurin series for ln(1+x) is as follows:

`ln (1+x) = x-1/2x^2+1/3x^3-1/4x^4+1/5x^5 - ...`

Or in sum notation:
`ln (1+x) = sum_(n=1)^oo(-1)^(n+1)x^n/n`

Replacing "`x`" in this series with "`x^2`" we get:

`ln (1+x^2) = (x^2)-1/2(x^2)^2+1/3(x^2)^3-1/4(x^2)^4+1/5(x^2)^5 - ...`

`= x^2-1/2x^4+1/3x^6-1/4x^8+1/5x^10 - ...`

Or in sum notation:
`ln (1+x^2) = sum_(n=1)^oo(-1)^(n+1)x^(2n)/n`