# Thread: Taylor series ln(1 + x^2)

1. ## Taylor series ln(1 + x^2)

ln(1 + x^2)
how would I do this?
Remember $\ln (1 + t) = \sum_{n=0}^{\infty} \frac{(-1)^n t^n}{n}$ for $|t|<1$.
Therefore if $x^2 < 1$ then $\ln (1+x^2) = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{n}$