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Question Number 143381 by Mathspace last updated on 13/Jun/21

$$letf\left(x\right)=arctan\left(\sqrt{2}{x}^2\right)$$$$1)calculate{f}^{\left(n\right)}\left(x\right)and{f}^{\left(n\right)}\left(0\right)$$$$2)iff\left(x\right)=\mathrm{\Sigma }{a}_n{x}^{n}findthe$$$$sequence{a}_n$$

Answered by TheHoneyCat last updated on 13/Jun/21

$$$$$$\mathrm{not}\mathrm{in}\mathrm{the}\mathrm{right}\mathrm{order},\mathrm{to}\mathrm{make}\mathrm{it}\mathrm{more}\mathrm{simple}$$$$\mathrm{\forall }x\in ]-1,1[$$$$\frac{1}{1-x}=\underset{n=0}{\sum ^{\mathrm{\infty }}}{x}^n$$$$thus:\frac{1}{1+x^2}=\underset{n=0}{\sum ^{\mathrm{\infty }}}{(-1)}^n{x}^{2n}$$$$knowingthat{\arctan }^{\prime }=x\to \frac{1}{1+x^2}andthat\arctan \left(0\right)=0$$$$\arctan \left(x\right)=\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^n}{2n+1}{x}^{2n+1}$$$$andtherefor:$$$$\arctan \left(\sqrt{2}{x}^2\right)=2\sqrt{2}\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^n}{2n+1}{x}^{4n+2}$$$$$$$$so\mathrm{\forall }n\in \mathbb{N}\mathrm{if}n\equiv 2\left[4\right]a_n=\frac{{(-1)}^{\frac{n-2}{4}}}{\frac{n-2}{2}+1}$$$$\mathrm{otherwise}{a}_n=0$$$$$$$$\mathrm{and}\mathrm{\forall }n\in \mathbb{N}{f}^{\left(n\right)}\left(0\right)=a_n$$$$$$$$asfor{f}^{\left(n\right)}\left(x\right)Iamquitenotcourageousenought$$$$f^{\left(0\right)}\left(x\right)=\frac{2\sqrt{2}x}{1+2x^4}$$$$fromthatpointon,onemustuse{Leibnietz}^{\prime }sformulaforthederivationofaproduct$$$${\left(fg\right)}^{\left(n\right)}=\underset{k=0}{\sum ^n}\left(\begin{array}{c}n\\ k\end{array}\right)f^{(n-k)}{g}^{\left(k\right)}$$$$withf:x\to 2\sqrt{2}xandg:x\to \frac{1}{1+2x^2}$$$$$$$$\mathrm{it}\mathrm{obviously}\mathrm{simplifies}\mathrm{due}\mathrm{to}{\mathrm{f}}^{″}=x\to 0$$$$\mathrm{but}\mathrm{like}\mathrm{I}\mathrm{said},\mathrm{I}\mathrm{am}\mathrm{not}\mathrm{courrageous}\mathrm{enougth}$$$$$$$$\mathrm{maybe}\mathrm{there}\mathrm{is}\mathrm{a}\mathrm{faster}\mathrm{method}$$$$(\mathrm{for}\mathrm{instance},\mathrm{if}\mathrm{you}\mathrm{guess}\mathrm{a}\mathrm{solution},\mathrm{you}\mathrm{only}\mathrm{have}\mathrm{to}\mathrm{verify}\mathrm{it}\mathrm{with}\mathrm{a}\mathrm{recurence})$$$$\mathrm{good}\mathrm{luck}\mathrm{to}\mathrm{you}\mathrm{with}\mathrm{that}\mathrm{last}\mathrm{bit}.$$

Answered by mathmax by abdo last updated on 14/Jun/21

$$1)\mathrm{f}\left(\mathrm{x}\right)=\arctan \left(\sqrt{2}{\mathrm{x}}^2\right)\Rightarrow {\mathrm{f}}^{{}^{\prime }}\left(\mathrm{x}\right)=\frac{2\sqrt{2}\mathrm{x}}{1+{2\mathrm{x}}^4}=\frac{2\sqrt{2}\mathrm{x}}{(\sqrt{2}{\mathrm{x}}^2-\mathrm{i})(\sqrt{2}{\mathrm{x}}^2+\mathrm{i})}$$$$=\frac{2\sqrt{2}\mathrm{x}}{2({\mathrm{x}}^2-\frac{\mathrm{i}}{\sqrt{2}})({\mathrm{x}}^2+\frac{\mathrm{i}}{\sqrt{2}})}=\frac{\sqrt{2}\mathrm{x}}{(\mathrm{x}-\frac{1}{\sqrt{\sqrt{2}}}{\mathrm{e}}^{\frac{\mathrm{i}\pi }{4}})(\mathrm{x}+\frac{1}{\sqrt{\sqrt{2}}}{\mathrm{e}}^{\frac{\mathrm{i}\pi }{4}})(\mathrm{x}-\frac{1}{\sqrt{\sqrt{2}}}{\mathrm{e}}^{-\frac{\mathrm{i}\pi }{4}})(\mathrm{x}+\frac{1}{\sqrt{\sqrt{2}}}{\mathrm{e}}^{-\frac{\mathrm{i}\pi }{4}})}$$$$(\lambda =\frac{1}{\sqrt{\sqrt{2}}})=\frac{\mathrm{a}}{\mathrm{x}-\lambda {\mathrm{e}}^{\frac{\mathrm{i}\pi }{4}}}+\frac{\mathrm{b}}{\mathrm{x}+\lambda {\mathrm{e}}^{\frac{\mathrm{i}\pi }{4}}}+\frac{\mathrm{c}}{\mathrm{x}-\lambda {\mathrm{e}}^{-\frac{\mathrm{i}\pi }{4}}}+\frac{\mathrm{d}}{\mathrm{x}+\lambda {\mathrm{e}}^{-\frac{\mathrm{i}\pi }{4}}}$$$$\mathrm{a}=\frac{\lambda \sqrt{2}{\mathrm{e}}^{\frac{\mathrm{i}\pi }{4}}}{2\lambda {\mathrm{e}}^{\frac{\mathrm{i}\pi }{4}}({\lambda }^2\mathrm{i}+\frac{\mathrm{i}}{\sqrt{2}})}=\frac{1}{\sqrt{2}\left(\frac{2\mathrm{i}}{\sqrt{2}}\right)}=\frac{1}{2\mathrm{i}}\mathrm{its}\mathrm{eazy}\mathrm{to}\mathrm{find}\mathrm{other}\mathrm{coefficient}$$$$\Rightarrow {\mathrm{f}}^{\left(\mathrm{n}\right)}\left(\mathrm{x}\right)=\frac{\mathrm{a}{(-1)}^{\mathrm{n}-1}(\mathrm{n}-1)!}{{(\mathrm{x}-\lambda {\mathrm{e}}^{\frac{\mathrm{i}\pi }{4}})}^{\mathrm{n}}}+\frac{\mathrm{b}{(-1)}^{\mathrm{n}-1}(\mathrm{n}-1)!}{(\mathrm{x}+\lambda {\mathrm{e}}^{\frac{\mathrm{i}\pi }{4}})}$$$$+\frac{\mathrm{c}{(-1)}^{\mathrm{n}-1}(\mathrm{n}-1)!}{{(\mathrm{x}-\lambda {\mathrm{e}}^{-\frac{\mathrm{i}\pi }{4}})}^{\mathrm{n}}}+\frac{\mathrm{d}{(-1)}^{\mathrm{n}-1}(\mathrm{n}-1)!}{(\mathrm{x}+\lambda {\mathrm{e}}^{-\frac{\mathrm{i}\pi }{4}})}$$$$={(-1)}^{\mathrm{n}-1}(\mathrm{n}-1)!\{\frac{\mathrm{a}}{{(\mathrm{x}-\lambda {\mathrm{e}}^{\frac{\mathrm{i}\pi }{4}})}^{\mathrm{n}}}+\frac{\mathrm{b}}{{(\mathrm{x}+\lambda {\mathrm{e}}^{\frac{\mathrm{i}\pi }{4}})}^{\mathrm{n}}}+\frac{\mathrm{c}}{{(\mathrm{x}-\lambda {\mathrm{e}}^{-\frac{\mathrm{i}\pi }{4}})}^{\mathrm{n}}}$$$$+\frac{\mathrm{d}}{{(\mathrm{x}+\lambda {\mathrm{e}}^{-\frac{\mathrm{i}\pi }{4}})}^{\mathrm{n}}}\}\Rightarrow$$$${\mathrm{f}}^{\left(\mathrm{n}\right)}\left(0\right)={(-1)}^{\mathrm{n}-1}(\mathrm{n}-1)!\{\frac{{\mathrm{ae}}^{-\frac{\mathrm{in}\pi }{4}}}{{(-\lambda )}^{\mathrm{n}}}+\frac{{\mathrm{be}}^{-\frac{\mathrm{in}\pi }{4}}}{{\lambda }^{\mathrm{n}}}+\frac{{\mathrm{ce}}^{\frac{\mathrm{in}\pi }{4}}}{{(-\lambda )}^{\mathrm{n}}}+\frac{{\mathrm{de}}^{\frac{\mathrm{in}\pi }{4}}}{{\lambda }^{\mathrm{n}}}\}(\lambda =\frac{1}{\left({}^4\sqrt{2}\right)})$$$$2)\mathrm{if}\mathrm{f}\left(\mathrm{x}\right)=\mathrm{\Sigma }{\mathrm{a}}_{\mathrm{n}}{\mathrm{x}}^{\mathrm{n}}\Rightarrow {\mathrm{a}}_{\mathrm{n}}=\frac{{\mathrm{f}}^{\left(\mathrm{n}\right)}\left(0\right)}{\mathrm{n}!}$$

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