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?
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3 Answers
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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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$\begingroup$ 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 $g^{14}(0) = 0$ $\endgroup$– sirfogaNov 2, 2015 at 21:38
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$\begingroup$ Of course, but since the function is rational and not at all difficult, you'd better not use them! :) $\endgroup$– sirfogaNov 2, 2015 at 21:41
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$\begingroup$ but if i want use the power series how can i do? $\endgroup$ Nov 2, 2015 at 21:45
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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?
answered Nov 2, 2015 at 21:24
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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
