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Question Number 147007 by Ar Brandon last updated on 17/Jul/21

Answered by Olaf_Thorendsen last updated on 17/Jul/21

$$\left(1\right)$$$$f\left(x\right)=\underset{n=0}{\sum ^{\mathrm{\infty }}}{a}_n{x}^n$$$$f^{\prime }\left(x\right)=\underset{n=1}{\sum ^{\mathrm{\infty }}}{na}_n{x}^{n-1}=\underset{n=0}{\sum ^{\mathrm{\infty }}}(n+1)a_{n+1}{x}^n$$$$f^{″}\left(x\right)=\underset{n=1}{\sum ^{\mathrm{\infty }}}n(n+1)a_{n+1}{x}^{n-1}$$$$f^{″}\left(x\right)=\underset{n=0}{\sum ^{\mathrm{\infty }}}(n+1)(n+2)a_{n+2}{x}^n$$$${xf}^{″}\left(x\right)+2f^{\prime }\left(x\right)+xf\left(x\right)=0$$$$\underset{n=0}{\sum ^{\mathrm{\infty }}}(n+1)(n+2)a_{n+2}{x}^{n+1}+2\underset{n=0}{\sum ^{\mathrm{\infty }}}(n+1)a_{n+1}{x}^n$$$$+\underset{n=0}{\sum ^{\mathrm{\infty }}}{a}_n{x}^{n+1}=0$$$$\underset{n=0}{\sum ^{\mathrm{\infty }}}(n+1)(n+2)a_{n+2}{x}^{n+1}+2a_1+2\underset{n=0}{\sum ^{\mathrm{\infty }}}(n+2)a_{n+2}{x}^{n+1}$$$$+\underset{n=0}{\sum ^{\mathrm{\infty }}}{a}_n{x}^{n+1}=0$$$$\mathrm{Necessairement}{a}_1=0$$$$\mathrm{Et}\mathrm{par}\mathrm{identification}\mathrm{des}\mathrm{puissances}$$$$\mathrm{comparables}:$$$$(n+1)(n+2)a_{n+2}+2(n+2)a_{n+2}+a_n=0$$$$a_{n+2}=-\frac{a_n}{(n+2)(n+3)}$$$$$$$$\left(2\right)$$$$\mathrm{Comme}{a}_1=0\Rightarrow a_{2p+1}=0$$$$\mathrm{Et}\mathrm{pour}\mathrm{les}\mathrm{indices}\mathrm{pairs}:$$$$a_2=-\frac{a_0}{2.3}=-\frac{1}{2.3}=-\frac{1}{3!}$$$$a_4=-\frac{a_2}{4.5}={(-1)}^2\frac{1}{2.3.4.5}=\frac{1}{5!}$$$$...\mathrm{etc}...$$$$a_{2p}=\frac{{(-1)}^p}{(2p+1)!}\mathrm{et}{a}_{2p+1}=0$$$$$$$$\left(3\right)$$$$\mathrm{Compte}\mathrm{tenu}\mathrm{des}\mathrm{criteres}\mathrm{sur}\mathrm{les}\mathrm{series}$$$$\mathrm{alternees},f\left(x\right)\mathrm{converge}\mathrm{partout}$$$$\mathrm{sur}\mathbb{R}\mathrm{et}:$$$$f\left(x\right)=\underset{n=0}{\sum ^{\mathrm{\infty }}}{(-1)}^p\frac{x^{2p}}{(2p+1)!}$$$$xf\left(x\right)=\underset{n=0}{\sum ^{\mathrm{\infty }}}{(-1)}^p\frac{x^{2p+1}}{(2p+1)!}=\sin x$$$$\mathrm{et}\mathrm{donc}f\left(x\right)=\frac{\sin x}{x}$$$$\mathrm{On}\mathrm{pourrait}\mathrm{verifier},\mathrm{avec}\mathrm{sa}\mathrm{forme}$$$$\mathrm{explicite}\mathrm{que}f\mathrm{verifie}\mathrm{bien}{\mathrm{l}}^{\prime }\mathrm{equation}$$$$\mathrm{differentielle}.$$

Commented by Ar Brandon last updated on 17/Jul/21

$$\mathrm{Merci}\mathrm{beaucoup}\mathrm{monsieur}.$$$$\mathrm{Bien}\mathrm{le}\mathrm{jour}!$$

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