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Question Number 98185 by abdomathmax last updated on 12/Jun/20

$$\mathrm{developp}\mathrm{at}\mathrm{fourier}\mathrm{serie}$$$$\mathrm{g}\left(\mathrm{x}\right)=\frac{2}{3+{\sin }^2\mathrm{x}}$$

Answered by mathmax by abdo last updated on 12/Jun/20

$$\mathrm{g}\left(\mathrm{x}\right)=\frac{2}{3+{\sin }^2\mathrm{x}}=\frac{2}{3+\frac{1-\cos \left(2\mathrm{x}\right)}{2}}=\frac{4}{7-\cos \left(2\mathrm{x}\right)}=\frac{4}{7-\frac{{\mathrm{e}}^{\mathrm{i}2\mathrm{x}}+{\mathrm{e}}^{-\mathrm{i}2\mathrm{x}}}{2}}$$$$=\frac{8}{14-{\mathrm{e}}^{2\mathrm{ix}}-{\mathrm{e}}^{-2\mathrm{ix}}}{=}_{{\mathrm{e}}^{2\mathrm{ix}}=\mathrm{z}}\frac{8}{14-\mathrm{z}-{\mathrm{z}}^{-1}}=\frac{8\mathrm{z}}{14\mathrm{z}-{\mathrm{z}}^2-1}=\frac{-8\mathrm{z}}{{\mathrm{z}}^2-14\mathrm{z}+1}=\mathrm{h}\left(\mathrm{z}\right)$$$${\mathrm{z}}^2-14\mathrm{z}+1=0\to {\mathrm{\Delta }}^{{}^{\prime }}=7^2-1=48\Rightarrow {\mathrm{z}}_1=7+\sqrt{48}=7+4\sqrt{3}$$$${\mathrm{z}}_2=7-4\sqrt{3}\mathrm{so}\frac{-8\mathrm{z}}{{\mathrm{z}}^2-14\mathrm{z}+1}=\frac{-8\mathrm{z}}{(\mathrm{z}-{\mathrm{z}}_1)(\mathrm{z}-{\mathrm{z}}_2)}=\frac{\mathrm{a}}{\mathrm{z}-{\mathrm{z}}_1}+\frac{\mathrm{b}}{\mathrm{z}-{\mathrm{z}}_2}$$$$\mathrm{a}=\frac{-{8\mathrm{z}}_1}{8\sqrt{3}}=-\frac{7+4\sqrt{3}}{\sqrt{3}}=-\frac{7}{\sqrt{3}}+4$$$$\mathrm{b}=\frac{-{8\mathrm{z}}_2}{-8\sqrt{3}}=\frac{7-4\sqrt{3}}{\sqrt{3}}=\frac{7}{\sqrt{3}}-4\Rightarrow \mathrm{h}\left(\mathrm{z}\right)=\frac{-\mathrm{a}}{{\mathrm{z}}_1-\mathrm{z}}-\frac{\mathrm{b}}{{\mathrm{z}}_2-\mathrm{z}}=\frac{-\mathrm{a}}{{\mathrm{z}}_1(1-\frac{\mathrm{z}}{{\mathrm{z}}_1})}-\frac{\mathrm{b}}{{\mathrm{z}}_2(1-\frac{\mathrm{z}}{{\mathrm{z}}_2})}$$$$\mathrm{so}\mathrm{for}\mid \mathrm{z}\mid <\inf (\mid {\mathrm{z}}_1\mid ,\mid {\mathrm{z}}_2\mid )\mathrm{h}\left(\mathrm{z}\right)=-\frac{\mathrm{a}}{{\mathrm{z}}_1}\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{\mathrm{z}}^{\mathrm{n}}}{{\mathrm{z}}_1^{\mathrm{n}}}-\frac{\mathrm{b}}{{\mathrm{z}}_2}\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{\mathrm{z}}^{\mathrm{n}}}{{\mathrm{z}}_2^{\mathrm{n}}}$$$$=-\frac{\mathrm{a}}{{\mathrm{z}}_1}\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{1}{{\mathrm{z}}_1^{\mathrm{n}}}{\mathrm{e}}^{2\mathrm{inx}}-\frac{\mathrm{b}}{{\mathrm{z}}_2}\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{1}{{\mathrm{z}}_2^{\mathrm{n}}}{\mathrm{e}}^{2\mathrm{inx}}$$$$=-\sum _{\mathrm{n}=0}^{\mathrm{\infty }}(\frac{\mathrm{a}}{{\mathrm{z}}_1^{\mathrm{n}+1}}+\frac{\mathrm{b}}{{\mathrm{z}}_2^{\mathrm{n}+1}}){\mathrm{e}}^{2\mathrm{inx}}=\mathrm{g}\left(\mathrm{x}\right)$$

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