# Taylor series ln(1 + x^2)

• October 8th 2008, 09:37 AM
khuezy
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 $\ln (1+x^2) = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{n}$