Find the Taylor series of function $\frac{-x^2-x-1}{x^3 - 2x^2 + 3x -6}$

Question

I've got a function $f(x)=\dfrac{-x^2-x-1}{x^3-2x^2+3x-6}$ and I need to find Taylor series at $x = 0$ for this function $f$ and find $f^{(88)}(0)$.

Please help me ...

Answer 1

$$\frac{-x^2-x-1}{x^3 - 2x^2 + 3x -6}=\frac{1}{2-x}-\frac{1}{3+x^2}=\frac{1}{2(1-x/2)}-\frac{1}{3(1+x^2/3)}$$

and then use the following geometric series $$\frac{1}{1-x}=\sum_{n=0}^{\infty }x^n$$ $$\frac{1}{1+x}=\sum_{n=0}^{\infty }(-1)^nx^n$$ so $$\frac{1}{2(1-x/2)}-\frac{1}{3(1+x^2/3)}=\frac{1}{2}\sum_{n=0}^{\infty }(x/2)^n-\frac{1}{3}\sum_{n=0}^{\infty }(-1)^n(x^2/3)^n$$ $$f(x)=\frac{1}{2}\sum_{n=0}^{\infty }(\frac{x}{2})^n-\frac{1}{3}\sum_{n=0}^{\infty }(-1)^n(\frac{x}{\sqrt{3}})^{2n}$$ so $$f^1(x)=\frac{1}{2}\sum_{n=0}^{\infty }\frac{n}{2^1}(\frac{x}{2})^{n-1}-\frac{1}{3}\sum_{n=0}^{\infty }\frac{2n}{\sqrt{3}^1}(-1)^n(\frac{x}{\sqrt{3}})^{2n-1}$$ $$f^2(x)=\frac{1}{2}\sum_{n=0}^{\infty }\frac{n(n-1)}{2^2}(\frac{x}{2})^{n-2}-\frac{1}{3}\sum_{n=0}^{\infty }\frac{2n(2n-1)}{\sqrt{3}^2}(-1)^n(\frac{x}{\sqrt{3}})^{2n-2}$$ the $$f^{88}(0)=\frac{1}{2}\frac{n(n-1)(n-2)...(n-87)}{2^{88}}-\frac{1}{3}\frac{2n(2n-1)(2n-2)...(2n-87)}{3^{44}}$$