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Question Number 212177 by MrGaster last updated on 05/Oct/24

$$f\left(x\right)=\arctan \left(\frac{1-x}{1+x}\right)=?$$$$\mathrm{ask}:f^{2023}\left(0\right)$$

Answered by a.lgnaoui last updated on 05/Oct/24

$$\mathrm{f}\left(0\right)=\frac{\pi }{4}\Rightarrow {\mathrm{f}}^{2023}\left(0\right)=\frac{{\pi }^{2023}}{4^{2023}}$$

Answered by Ghisom last updated on 05/Oct/24

$$\mathrm{if}\mathrm{you}\mathrm{mean}\frac{d^{2023}f}{{df}^{2023}}\mathrm{the}\mathrm{answer}\mathrm{is}2022!$$$$\frac{d^{2n}f\left(0\right)}{{df}^{2n}}=9$$$$\frac{d^{2n+1}f\left(0\right)}{{df}^{2n+1}}=\{\begin{array}{l}-\left(2n\right)!;n=2k\\ \left(2n\right)!;n=2k+1\end{array}$$$$$$$$[2023=2n+1=2(2k+1)+1=4k+3\Rightarrow k=505;n=1011]$$

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