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I want to find $f^{(10)}(0)$ where $f(x)=\ln(2+x^2)$.

I know that it can be done "by hand", but I believe there is a smarter way.

I think I should use Taylor series and the fact that $f^{(n)}(0)=a_n*n!$ , but I'm not sure how.

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  • $\begingroup$ You're on the right track; so then, what is the Taylor series for $f$ at $x = 0$? $\endgroup$ – Travis Willse Aug 4 '15 at 14:06
  • $\begingroup$ Maple says $$ 928972800\,{\frac {{x}^{8}}{ \left( {x}^{2}+2 \right) ^{9}}}-812851200 \,{\frac {{x}^{6}}{ \left( {x}^{2}+2 \right) ^{8}}}+290304000\,{\frac {{x}^{4}}{ \left( {x}^{2}+2 \right) ^{7}}}-36288000\,{\frac {{x}^{2}}{ \left( {x}^{2}+2 \right) ^{6}}}+725760\, \left( {x}^{2}+2 \right) ^{- 5}-$$ $$371589120\,{\frac {{x}^{10}}{ \left( {x}^{2}+2 \right) ^{10}}} $$ $\endgroup$ – user64494 Aug 4 '15 at 14:06
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    $\begingroup$ i get $$22680$$ $\endgroup$ – Dr. Sonnhard Graubner Aug 4 '15 at 14:10
  • $\begingroup$ I know the Taylor series for $ln(1+t)$ but I'm don't know how to convert it to $t=1+x^2$ in a way that can be helpful. $\endgroup$ – alex kur Aug 4 '15 at 14:12
  • $\begingroup$ It's certianly larger and more computationally rigorous then considering $f(x)=\ln(-2+x^{2})$; in that example $f^{(10)}=2^{-4}$. $\endgroup$ – Autolatry Aug 4 '15 at 14:22
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\begin{align} f(x) &= \ln(2 + x^{2}) = \ln 2 + \ln\left( 1 + \frac{x^{2}}{2}\right) \\ &= \ln 2 + \sum_{k=1}^{\infty} \frac{(-1)^{k-1} \, x^{2k}}{2^{k} \, k} \\ &= \ln 2 + \frac{x^{2}}{2} - \frac{x^{4}}{8} + \frac{x^{6}}{24} - \frac{x^{8}}{64} + \frac{x^{10}}{160} - \cdots \\ f(x) &= \ln 2 + \frac{x^{2}}{2!} - 3 \, \frac{x^{4}}{4!} + 30 \, \frac{x^{6}}{6!} - 630 \, \frac{x^{8}}{8!} + 22680 \, \frac{x^{10}}{(10)!} - 1247400 \, \frac{x^{12}}{(12)!} + \cdots \end{align} From this series expansion it can be determined that \begin{align} f^{(2n+1)}(0) &= 0 \hspace{10mm} \text{for} \quad n \geq 0 \\ f^{(2n)}(0) &= \frac{(-1)^{n-1} \, (2n)!}{2^{n} \, n} = \frac{(-1)^{n-1} \, (2n-1)!}{2^{n-1}} \hspace{10mm} \text{for} \quad n \geq 1 \\ f(0) &= \ln 2. \end{align}

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We have $$ \frac{2x}{x^2+2} = \frac{x}{1+\frac{x^2}{2}} = x\bigg(1-\frac{x^2}{2} + \frac{x^4}{4} - \frac{x^6}{8} + \frac{x^8}{16}+\cdots \bigg). $$ Then integrating we get $$ \log(x^2+2)-\log2 = \int_0^x \frac{2t}{t^2+2}\; dt = \frac{x^2}{2}-\frac{x^4}{2\cdot 4} + \frac{x^6}{4\cdot 6} - \frac{x^8}{8\cdot 8} + \frac{x^{10}}{10\cdot 16}+\ldots $$

Answer: $22680$ :)

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\begin{align} f'(x)=\frac{x}{1+\frac{x^{2}}{2}}=\sum_{k=0}^{\infty }(-1)^{k}\left ( \frac{x^{2k+1}}{2^{k}} \right )=x-\frac{x^{3}}{2}+\cdots+\frac{x^{9}}{2^{4}}-\frac{x^{11}}{2^{5}}+\cdots \end{align}

Therefore if we differentiate this series nine times, we see that $f^{(10)}(0)$ is the constant term, which is $\frac{9!}{16}=22680$

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