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It's $g(x)={{x^{2}-1}\over{x^{2}+2}}$ and i have to calculate $g^{13}(0)$.

I can't calculate all the derivates so i think to use power series.

$g(x)={{x^2\over{x^{2}+2}}-{1\over{x^2+2}}}$

Can i use the geometric series?

Giulia B.
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3 Answers3

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First perform a simple arithmetic trick $$g(x) = \frac{x^2-1}{x^2+2}=1-\frac{3}{x^2+2}$$ then, since $g'(x) = \frac{6x}{(2 + x^2)^2}$ is a odd function, then odd-order derivatives must be odd functions so they are like $$x \cdot p(x)$$ for a certain polynomial function such that $\deg[p(x)]$ is even. Then conclude and find $g^{13}(0) = 0$ since $0$ is a root of the polynomial-rational odd-degree derivative.

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sirfoga
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  • and for $g^{14}(0)$ is different? – Giulia B. Nov 02 '15 at 21:36
  • Yes, because as proved in the link provided, derivative of odd function is even and vice-versa, so derivative of $g^{13}(0)$ is even and thus $0$ is no more a root of the polynomial-rational derivative, so you can't be sure that $g^{14}(0) = 0$ – sirfoga Nov 02 '15 at 21:38
  • I can use the power series? – Giulia B. Nov 02 '15 at 21:40
  • Of course, but since the function is rational and not at all difficult, you'd better not use them! :) – sirfoga Nov 02 '15 at 21:41
  • but if i want use the power series how can i do? – Giulia B. Nov 02 '15 at 21:45
  • series of given function is $-\frac12 + \frac{3x^2}{4} - \frac{3x^4}{8} + O(x^6)$ is even function, then since derivative of even function is odd and vice-versa, $g'(x)$ is odd, $g''(x)$ is even .. $\Rightarrow g^{13}(x)$ is odd – sirfoga Nov 02 '15 at 21:48
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In principle yes, but it would be easier if you were to perform a polynomial division $$ \frac{x^2-1}{x^2+2}=1-\frac{3}{x^2+2}. $$ Perhaps you can even divine from the symmetry of the function what odd derivatives are likely to be?

Lutz Lehmann
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$$g(x)={{x^{2}-1}\over{x^{2}+2}}=\frac{(x^2+2)-2-1}{x^2+2}=1-\frac{3}{x^2+2}$$

$$=1-\frac{3}{2}(\frac{1}{1+(\frac{x}{\sqrt{2}})^2})$$ depending on the geometric series $$=1-\frac{3}{2}(1-(\frac{x}{\sqrt{2}})^2+(\frac{x}{\sqrt{2}})^4-(\frac{x}{\sqrt{2}})^6+.........$$ so every odd derivative at $x=0$ will be $0$ because all remaining terms will contain $x$ with power one or more

E.H.E
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