Find $f^{(22)}(0)$ for $f(x)=\frac{x}{x^{3}-x^{2}+2x-2}$

Question

Find $f^{(22)}(0)$ for $f(x)=\frac{x}{x^{3}-x^{2}+2x-2}$

I know that I should use Taylor's theorem and create power series.

However I don't have idea how I can find $a_{n}$ such that $f(x)=\sum_{n=1}^{+\infty} a_{n}x^{n}$. I know only that $f(x)=\frac{x}{(x-1)(x^{2}+2)}$ but don't know how I can continue it because $f(x)=\frac{A}{(x^2+2)}+\frac{B}{(x-1)}$ and I get that $A=1, B=0, A=2B$ so it is conflict.

Have you some tips how I can write this function and finish this task?

Updated:

Thanks to @gt6989b I know that $f(x)=\frac{\frac{-1}{3}x+\frac{2}{3}}{x^{2}+2}+\frac{1}{3} \cdot \frac{1}{1-x}=\frac{\frac{-1}{3}x+\frac{2}{3}}{x^{2}+2}+\frac{1}{3} \cdot (1+x+x^{2}+...)$ but I still have a problem with $\frac{\frac{-1}{3}x+\frac{2}{3}}{x^{2}+2}$

Answer 1

$$ \begin{split} f(x) &= \frac{x}{x^3-x^2+2x-2} = \frac{x}{x^2(x-1)+2(x-1)} = \frac{x}{(x^2+2)(x-1)} \end{split} $$ and use partial fractions. Then use $$ \frac{1}{1-u} = 1 + u + u^2 + \ldots $$ and $$ \frac{1}{1+u} = \frac{1}{1-(-u)} = 1 -u + u^2-u^3 \pm \ldots $$

UPDATE

From partial fractions, $$ \frac{Ax+B}{x^2+2} + \frac{C}{x-1} = \frac{(Ax+B)(x-1) + C(x^2+2)}{(x^2+2)(x-1)} $$ expanding the numerator and equating to the desired expression you get $$ x = (Ax+B)(x-1) + C(x^2+2) = x^2(A+C) + x(-A+B) -B+2C $$ which results in the system of 3 equations and 3 unknowns $$ \begin{cases} A & & + C & = 0 \\ -A& +B & & = 1 \\ & -B & +2C & = 0 \end{cases} $$ Adding all three together yields $C = 1/3$ which implies $A = -1/3$ and $B = 2/3$.

Can you now finish this problem?

UPDATE 2

Another hint: $$ \frac{Ax+B}{x^2+2} = \left[\frac{Ax}{2} + \frac{B}{2}\right] \frac{1}{1 + (x/\sqrt{2})^2} $$ and now the fraction expands as $(1+u)^{-1}$ and then expand the bracket by multiplying by the resulting Taylor series, getting 2 different series, i.e. $$ (ax+b) \sum_{k=0}^\infty a_k x^k = ax \sum_{k=0}^\infty a_k x^k + b \sum_{k=0}^\infty a_k x^k = \sum_{k=0}^\infty (a a_k) x^{k+1} + \sum_{k=0}^\infty (b a_k) x^k $$ and now change the index on the left summation and combine to get $\sum_{j=0}^\infty c_j x^j$ and you need $c_{22}$...

Answer 2

By way of a contrast, here's an intuitve method that lets you see the bigger picture rather than getting bogged down in technical details, $$f(x)=\frac{x}{x^{3}-x^{2}+2x-2}$$ $$=\frac{1}{1-x}\times \frac{x}{2+x^2}$$ $$=\frac{1}{1-x}\times \frac{x}{2(1+(\frac{x}{\sqrt{2}})^2}$$ $$=\frac{1}{1-x}\times \frac{1}{\sqrt{2}}\times\frac{\frac{x}{\sqrt{2}}}{(1+(\frac{x}{\sqrt{2}})^2}$$ Now, standard result, $$\frac{1}{1+x^2}=1-x^2+x^4-x^6+x^8-x^{10}+\dots$$ $$\frac{x}{1+x^2}=x-x^3+x^5-x^7+x^9-x^{11}+\dots$$ $$\frac{\frac{x}{\sqrt{2}}}{(1+(\frac{x}{\sqrt{2}})^2}=\frac{x}{\sqrt{2}}-\big(\frac{x}{\sqrt{2}}\big)^3+\big(\frac{x}{\sqrt{2}}\big)^5-\big(\frac{x}{\sqrt{2}}\big)^7+\big(\frac{x}{\sqrt{2}}\big)^9-\big(\frac{x}{\sqrt{2}}\big)^{11}+\dots$$ $$\frac{1}{\sqrt{2}}\times\frac{\frac{x}{\sqrt{2}}}{(1+(\frac{x}{\sqrt{2}})^2}=\frac{x}{2}-\frac{x^3}{4}+\frac{x^5}{8}-\frac{x^7}{16}+\dots$$ Another standard result; multiplying by $\frac{1}{1-x}$ forms the partial sums, $$\frac{x}{x^{3}-x^{2}+2x-2}=\frac{x}{2}+\frac{x^3}{4}+\frac{3x^5}{8}+\frac{5x^7}{16}+\frac{11x^9}{32}+\dots$$ and I'll leave it to the reader to fight through to the end of the question.