Compute Taylor Series $\ln(x^2-x+1)$

Question

Please share if you have a faster approach, but I’ll share what I have so far. I think it reduces to some combinatorial problem, I’d like to see a combinatorial argument if possible.

From computing for small $n$ in wolfram alpha, I think the answer is \begin{align*} \ln(x^2-x+1)&=\sum_{n=1}^\infty (-1)^n \frac{a_n}{n} x^n \\ &= -\frac{1}{1}x^1+\frac{1}{2}x^2-\frac{2}{3}x^3+\frac{1}{4}x^4-\frac{1}{5}x^5+\dots \end{align*} where $a_n=2$ if $n=3,9,27,81,\dots$ and $a_n=1$ otherwise.

We know $$ \ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\dots $$

So $$ \ln(1+(x^2-x))=\frac{x}{1}(x-1)-\frac{x^2}{2}(x -1)^2+\frac{x^3}{3}(x-1)^3-\dots$$

Now we compute the coefficient of $x^n$. \begin{align*} x^8 \to -\frac{1}{4}\binom{4}{4}+\frac{1}{5}\binom{5}{3}-\frac{1}{6}\binom{6}{2}+\frac{1}{7}\binom{7}{1}-\frac{1}{8}\binom{8}{8}=\frac{1}{8}\cdot8! \\ x^9 \to \frac{1}{5}\binom{5}{4}-\frac{1}{6}+\frac{1}{7}\binom{7}{2}-\frac{1}{8}\binom{8}{1}+\frac{1}{9}\binom{9}{0}=-\frac{2}{9}\cdot 9! \\ x^{10} \to \frac{1}{5}\binom{5}{5}-\frac{1}{6}\binom{6}{4}+\frac{1}{7}\binom{7}{3}-\frac{1}{8}\binom{8}{2}+\frac{1}{9}\binom{9}{1}-\frac{1}{10}\binom{10}{0}=\frac{1}{10} \cdot 10! \end{align*} In other words, I want this identity: $$ \sum_{r+k=n} (-1)^{r+1} \frac{1}{r} \binom{r}{k} = (-1)^n \frac{a_n}{n} n! $$

with $a_n$ defined as above. Any help to prove the identity would be appreciated. Any other faster approaches would also be appreciated. Thanks!

Also, this LaTeX was a pain to type. Is there a way to reduce it? (Like \frac to \f, \binom to \b)

Answer 1

Tips: $$\ln(1-x+x^2)=\ln(1+x^3)-\ln(1+x)$$

Answer:

$$\ln(1+x)=x-\dfrac{x^2}{2}+\dfrac{x^3}{3}-\dfrac{x^4}{4}+\cdots \\ \ln(1+x^3)=x^3-\dfrac{x^6}{2}+\dfrac{x^9}{3}-\dfrac{x^{12}}{4}+\cdots \\ \ln(1-x+x^2)=\ln(1+x^3)-\ln(1+x)=\sum_{k=1}^\infty a_kx^k \\ a_k=\begin{cases} \dfrac{2(-1)^{k+1}}{k} & \text{when }3|k \\ \dfrac{(-1)^k}{k} & \text{otherwise}\end{cases}$$