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Question Number 98430 by mathmax by abdo last updated on 13/Jun/20

$$\mathrm{let}\mathrm{g}\left(\mathrm{x}\right)=\frac{2}{\cos \left(\pi \mathrm{x}\right)}\mathrm{developp}\mathrm{g}\mathrm{at}\mathrm{fourier}\mathrm{serie}$$

Answered by abdomathmax last updated on 13/Jun/20

$$\mathrm{we}\mathrm{have}\mathrm{g}\left(\mathrm{x}\right)=\frac{2}{\cos \left(\pi \mathrm{x}\right)}=\frac{4}{{\mathrm{e}}^{\mathrm{i}\pi \mathrm{x}}+{\mathrm{e}}^{-\mathrm{i}\pi \mathrm{x}}}$$$$\mathrm{changement}{\mathrm{e}}^{\mathrm{i}\pi \mathrm{x}}=\mathrm{z}\mathrm{give}$$$$\mathrm{g}\left(\mathrm{x}\right)=\frac{4}{z+z^{-1}}=\frac{4z}{z^2+1}=\frac{4z}{(z-i)(z+i)}$$$$=\frac{\mathrm{a}}{\mathrm{z}-\mathrm{i}}+\frac{\mathrm{b}}{\mathrm{z}+\mathrm{i}}\Rightarrow \mathrm{a}=\frac{4\mathrm{i}}{2\mathrm{i}}=2$$$$\mathrm{b}=\frac{-4\mathrm{i}}{-2\mathrm{i}}=2\Rightarrow \mathrm{g}\left(\mathrm{x}\right)=\frac{1}{2}\{\frac{1}{\mathrm{z}-\mathrm{i}}+\frac{1}{\mathrm{z}+\mathrm{i}}\}$$$$=\frac{1}{2}\{\frac{\mathrm{i}}{\mathrm{iz}+1}+\frac{\mathrm{i}}{\mathrm{iz}-1}\}=\frac{\mathrm{i}}{2}\{\frac{1}{1+\mathrm{iz}}-\frac{1}{1-\mathrm{iz}}\}$$$$=\frac{\mathrm{i}}{2}\{\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{(-\mathbf{i})}^{\mathbf{n}}{\mathbf{z}}^{\mathbf{n}}-\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{\mathrm{i}}^{\mathrm{n}}{\mathrm{z}}^{\mathrm{n}}\}$$$$=-\frac{\mathrm{i}}{2}\sum _{\mathrm{n}=0}^{\mathrm{\infty }}({\mathrm{i}}^{\mathrm{n}}-{(-\mathrm{i})}^{\mathrm{n}}){\mathrm{z}}^{\mathrm{n}}$$$$=-\frac{\mathrm{i}}{2}\sum _{\mathrm{n}=0}^{\mathrm{\infty }}2\mathrm{isin}\left(\frac{\mathrm{n}\pi }{2}\right){\mathrm{e}}^{\mathrm{in}\pi \mathrm{x}}$$$$=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\sin \left(\frac{\mathrm{n}\pi }{2}\right)\{\cos \left(\mathrm{n}\pi \mathrm{x}\right)+\mathrm{isin}\left(\mathrm{n}\pi \mathrm{x}\right)\}$$$$\mathrm{but}\mathrm{g}\left(\mathrm{x}\right)\mathrm{is}\mathrm{real}\Rightarrow$$$$\frac{2}{\cos \left(\pi \mathrm{x}\right)}=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\sin \left(\frac{\mathrm{n}\pi }{2}\right)\cos \left(\mathrm{n}\pi \mathrm{x}\right)$$$$=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\sin \left(\frac{(2\mathrm{n}+1)\pi }{2}\right)\cos \left((2\mathrm{n}+1)\pi \mathrm{x}\right)$$$$=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\sin (\mathrm{n}\pi +\frac{\pi }{2})\cos \left((2\mathrm{n}+1)\pi \mathrm{x}\right)$$$$=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{(-1)}^{\mathrm{n}}\cos \left\{(2\mathrm{n}+1)\pi \mathrm{x}\right\}$$$$$$

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