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I've found this task: evaluate $n$-th derivative at $x=0$ without finding the general formula:

  1. $f(x)=\dfrac{1+x+x^2}{1-x+x^2}$, $n=4$
  2. $f(x)= \sqrt[3]{\sin(x^3)}$, $n=9$

It's interesting, how the fact that we need exactly $f^{(n)}(0)$ affects that we don't have to find all derivatives?

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    I'm sorry but I still don't understand how to expand first function in to series.. in 2. can I rewrite $f(x)=exp(\frac13\ln\sin x^3)=1+\frac13\ln \sin x^3 + \frac{\ln^2\sin x^3}{18}+..$ and see it will be $f^{n}(x)=0$? or is it wrong?2012-03-22

2 Answers 2

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For the first question, it may be useful to simplify a little, and observe that $\frac{1+x+x^2}{1-x+x^2}=\frac{2x}{1-x+x^2}+1.$ Note that $\frac{2x}{1-x+x^2}=\frac{2x(1+x)}{1+x^3}.$ The series expansion of $\frac{1}{1+x^3}$ is standard: use the fact that $\frac{1}{1-t}=1+t+t^2+\cdots,$ and set $t=-x^3$. Unfortunately, this will give us a general formula for the $n$-th derivative. But we can refuse to notice that, and just calculate for $n=4$.

We have $\frac{1}{1+x^3}=1-x^3+x^6-x^9+\cdots$. So the $x^4$ term in the power series expansion of our original function is just $-2x^4$.

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For (2), $\sin(x^3) = x^3 - x^9/6 + O(x^{15}) = x^3 (1 - x^6/6 + O(x^{12}))$ so $\left(\sin(x^3)\right)^{1/3} = x (1 - x^6/6 + O(x^{12}))^{1/3} = x (1 - x^6/18 + O(x^{12}))$ and the answer is $0$.