Maclaurin series questions

Question

How can I find the Maclaurin series for $\ln(1+x^2+x)$? I'm not sure if I should try writing 1+x+x^2 in some other way or..Also if I want to find $h^{(3k)}(0)$ (the derivative of order 3k) where should I start?

Answer 1

Use

$$\ln(1+x+x^2)=\ln(1-x^3)-\ln(1-x)=-\sum_{k=1}^\infty\frac{x^{3k}}k+\sum_{k=1}^\infty\frac{x^k}k=\sum_{k=1,3\nmid k}^\infty\frac{x^k}k.$$

Answer 2

Take $f(x)=\ln(x^2+x+1)$ therefore $f'(x)=\frac{2x+1}{1+x+x^2}=\frac{1+x-2x^2}{1-x^3}$. Also we have:$$\frac{1}{1-x^3}=\sum_{n=0}^{\infty}x^{3n}$$then we have$$f'(x)=(1+x-2x^2)\sum_{n=0}^{\infty}x^{3n}=\sum_{n=0}^{\infty}x^{3n}+\sum_{n=0}^{\infty}x^{3n+1}+\sum_{n=0}^{\infty}-2x^{3n+2}=\sum_{n=0}^{\infty}a_nx^n$$where $a_n=1$ for $n=3k$ or $3k+1$ and $a_n=-2$ for $n=3k+2$. Therefore we have:$$f(x)=C+\sum_{n=0}^{\infty}\frac{a_n}{n+1}x^{n+1}$$and $C=0$ obviously. Therefore we obtain:$$f(x)=\sum_{n=1}^{\infty}\frac{a_{n-1}}{n}x^{n}$$with the same definition of $a_n$ or equivalently:$$f(x)=\sum_{n=0}^{\infty}\frac{1}{3n+1}x^{3n+1}+\sum_{n=0}^{\infty}\frac{1}{3n+2}x^{3n+2}-2\sum_{n=1}^{\infty}\frac{1}{3n}x^{3n}$$