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Using $n$-th Taylor polynomial for $f_1(x)=\frac{1}{1-x}$ with center in $0$, find $4$-th derivative of $f_2(x)=\frac{1+x+x^2}{1-x+x^2}$ in the point $0$ without calculating it's $1$,$2$ or $3$ derivative.

I'm looking for hints, it's a homework. I've tried using the Taylor expansion with Lagrange rest, but it won't work because of restriction of calculating $1-3$ derivatives. Also I don't see connection between $f_1$ and $f_2$

Thanks in advance for help!

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$$f_2(x)=\frac{1+x+x^2}{1-x+x^2}=\frac{(1+x+x^2)(1+x)}{(1-x+x^2)(1+x)}=\\ =\frac{(1+x+x^2)(1+x)}{1+x^3}= =\frac{1+2x+2x^2+x^3}{1+x^3}=\\ =1+2\frac{1+x}{1+x^3}=1+2 \frac{1}{1+x^3}+2x\frac{1}{1+x^3}.$$ Added:
From Taylor's polynomial for $$\frac{1}{1-x}=\sum\limits_{k=1}^{n}{\;x^k}+o(x^n)$$ we have $$\frac{1}{1+x^3}=\sum\limits_{k=1}^{n}{(-1)^k (x^3)^k}+o((x^3)^n)=\sum\limits_{k=1}^{n}{(-1)^k x^{3k}}+o(x^{3n}),$$ so $$f_2(x)=1+2 \sum\limits_{k=1}^{n}{(-1)^k x^{3k}}+2x \sum\limits_{k=1}^{n}{(-1)^k x^{3k}}=\\ =1+2 \sum\limits_{k=1}^{n}{(-1)^k x^{3k}} + 2\sum\limits_{k=1}^{n}{(-1)^k x^{3k+1}}+o(x^{3n+1}).$$ In this expansion coefficient $a_4$ at $x^4$ equals $-2.$ On the other hand, $a_4=\frac{f_2^{(4)}(0)}{4!},$ therefore, $$f_2^{(4)}(0)=(-2)\cdot 4!=-48.$$

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  • $\begingroup$ Now, how to apply Taylor polynomial? I see how to (moderate difficutly) calculate 4-th derivative, but I need to use Taylor teorem here. $\endgroup$ – Joggi Mar 26 '13 at 19:55
  • $\begingroup$ I included this in my answer. $\endgroup$ – M. Strochyk Mar 26 '13 at 20:29
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Hint: $$\frac{1}{1-x+x^2}=\frac{1+x}{1+x^3}$$ So $$f_2(x)=(1+x+x^2)(1+x)f_1(-x^3)$$

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  • $\begingroup$ the same question as above, how to use Taylor polynomial to calculate 4th derivative of this? $\endgroup$ – Joggi Mar 26 '13 at 19:55

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