Question Number 212177 by MrGaster last updated on 05/Oct/24
$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{f}\left({x}\right)=\mathrm{arctan}\left(\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}\right)=? \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{ask}:{f}^{\mathrm{2023}} \left(\mathrm{0}\right) \\ $$
Answered by a.lgnaoui last updated on 05/Oct/24
$$\mathrm{f}\left(\mathrm{0}\right)=\frac{\pi}{\mathrm{4}}\Rightarrow\:\:\:\mathrm{f}^{\mathrm{2023}} \left(\mathrm{0}\right)=\frac{\pi^{\mathrm{2023}} }{\mathrm{4}^{\mathrm{2023}} } \\ $$
Answered by Ghisom last updated on 05/Oct/24
![if you mean (d^(2023) f/df^(2023) ) the answer is 2022! ((d^(2n) f(0))/df^(2n) )=9 ((d^(2n+1) f(0))/df^(2n+1) )= { ((−(2n)!; n=2k)),(((2n)!; n=2k+1)) :} [2023=2n+1=2(2k+1)+1=4k+3 ⇒ k=505; n=1011]](https://www.tinkutara.com/question/Q212180.png)
$$\mathrm{if}\:\mathrm{you}\:\mathrm{mean}\:\frac{{d}^{\mathrm{2023}} {f}}{{df}^{\mathrm{2023}} }\:\mathrm{the}\:\mathrm{answer}\:\mathrm{is}\:\mathrm{2022}! \\ $$$$\frac{{d}^{\mathrm{2}{n}} {f}\left(\mathrm{0}\right)}{{df}^{\mathrm{2}{n}} }=\mathrm{9} \\ $$$$\frac{{d}^{\mathrm{2}{n}+\mathrm{1}} {f}\left(\mathrm{0}\right)}{{df}^{\mathrm{2}{n}+\mathrm{1}} }=\begin{cases}{−\left(\mathrm{2}{n}\right)!;\:{n}=\mathrm{2}{k}}\\{\left(\mathrm{2}{n}\right)!;\:{n}=\mathrm{2}{k}+\mathrm{1}}\end{cases} \\ $$$$ \\ $$$$\:\:\:\:\:\left[\mathrm{2023}=\mathrm{2}{n}+\mathrm{1}=\mathrm{2}\left(\mathrm{2}{k}+\mathrm{1}\right)+\mathrm{1}=\mathrm{4}{k}+\mathrm{3}\:\Rightarrow\:{k}=\mathrm{505};\:{n}=\mathrm{1011}\right] \\ $$