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Question Number 96656 by mathmax by abdo last updated on 03/Jun/20

let g(x) =(2/(cosx)) developp f at fourier serie

$$\mathrm{let}\:\mathrm{g}\left(\mathrm{x}\right)\:=\frac{\mathrm{2}}{\mathrm{cosx}}\:\:\:\mathrm{developp}\:\mathrm{f}\:\mathrm{at}\:\mathrm{fourier}\:\mathrm{serie} \\ $$

Answered by mathmax by abdo last updated on 04/Jun/20

we have g(x) =(2/(cosx)) ⇒g(x) =(4/(e^(ix) +e^(−ix) )) changement e^(ix) =z give g(x) =h(z) =(4/(z +z^(−1) )) =((4z)/(z^2 +1)) =4z Σ_(n=0) ^∞ z^(2n) =4z Σ_(n=0) ^∞ e^(i2nx) =4 Σ_(n=0) ^∞ e^(i(2n+1)x) =4 Σ_(n=0) ^∞ cos(2n+1)x +4iΣ_(n=0) ^∞ sin(2n+1)x but g(x)∈R ⇒g(x) =4Σ_(n=0) ^∞ cos(2n+1)x

$$\mathrm{we}\:\mathrm{have}\:\mathrm{g}\left(\mathrm{x}\right)\:=\frac{\mathrm{2}}{\mathrm{cosx}}\:\Rightarrow\mathrm{g}\left(\mathrm{x}\right)\:=\frac{\mathrm{4}}{\mathrm{e}^{\mathrm{ix}} \:+\mathrm{e}^{−\mathrm{ix}} }\:\mathrm{changement}\:\mathrm{e}^{\mathrm{ix}} \:=\mathrm{z}\:\mathrm{give} \\ $$$$\mathrm{g}\left(\mathrm{x}\right)\:=\mathrm{h}\left(\mathrm{z}\right)\:=\frac{\mathrm{4}}{\mathrm{z}\:+\mathrm{z}^{−\mathrm{1}} }\:=\frac{\mathrm{4z}}{\mathrm{z}^{\mathrm{2}} \:+\mathrm{1}}\:=\mathrm{4z}\:\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\mathrm{z}^{\mathrm{2n}} \\ $$$$=\mathrm{4z}\:\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\mathrm{e}^{\mathrm{i2nx}} \:=\mathrm{4}\:\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\mathrm{e}^{\mathrm{i}\left(\mathrm{2n}+\mathrm{1}\right)\mathrm{x}} \:=\mathrm{4}\:\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\mathrm{cos}\left(\mathrm{2n}+\mathrm{1}\right)\mathrm{x}\:+\mathrm{4i}\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\mathrm{sin}\left(\mathrm{2n}+\mathrm{1}\right)\mathrm{x} \\ $$$$\mathrm{but}\:\mathrm{g}\left(\mathrm{x}\right)\in\mathrm{R}\:\Rightarrow\mathrm{g}\left(\mathrm{x}\right)\:=\mathrm{4}\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\mathrm{cos}\left(\mathrm{2n}+\mathrm{1}\right)\mathrm{x}\: \\ $$

Commented by 1549442205 last updated on 04/Jun/20

There is a place that isn′t correct: (1/(z^2 +1))=Σ_(n=0) ^(∞) (−1)^n z^(2n) (!?)

$$\mathrm{There}\:\mathrm{is}\:\mathrm{a}\:\mathrm{place}\:\mathrm{that}\:\mathrm{isn}'\mathrm{t}\:\mathrm{correct}: \\ $$$$\frac{\mathrm{1}}{\mathrm{z}^{\mathrm{2}} +\mathrm{1}}=\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\Sigma}}\left(−\mathrm{1}\right)^{\mathrm{n}} \mathrm{z}^{\mathrm{2n}} \left(!?\right) \\ $$

Commented by Rio Michael last updated on 04/Jun/20

i have to study in dept this fourier series. sorry i don′t know much about it.

$$\mathrm{i}\:\mathrm{have}\:\mathrm{to}\:\mathrm{study}\:\mathrm{in}\:\mathrm{dept}\:\mathrm{this}\:\mathrm{fourier}\:\mathrm{series}. \\ $$$$\mathrm{sorry}\:\mathrm{i}\:\mathrm{don}'\mathrm{t}\:\mathrm{know}\:\mathrm{much}\:\mathrm{about}\:\mathrm{it}. \\ $$

Commented by abdomathmax last updated on 04/Jun/20

ooh yes i forgot (−1)^n thank you for this remark!

$$\mathrm{ooh}\:\mathrm{yes}\:\mathrm{i}\:\mathrm{forgot}\:\left(−\mathrm{1}\right)^{\mathrm{n}} \:\mathrm{thank}\:\mathrm{you}\:\mathrm{for}\:\mathrm{this}\:\mathrm{remark}! \\ $$

Commented by abdomathmax last updated on 04/Jun/20

sorry g(z) =((4z)/(1+z^2 )) =4z Σ_(n=0) ^∞ (−1)^n z^(2n) =4 Σ_(n=0) ^∞ (−1)^n z^(2n+1) =4 Σ_(n=0) ^∞ (−1)^n e^(i(2n+1)x) =4 Σ_(n=0) ^∞ (−1)^n cos(2n+1)x +4iΣ_(n=0) ^∞ (−1)^n sin(2n+1)x g(x)real ⇒★g(x) =4 Σ_(n=0) ^∞ (−1)^n cos(2n+1)x★

$$\mathrm{sorry}\:\mathrm{g}\left(\mathrm{z}\right)\:=\frac{\mathrm{4z}}{\mathrm{1}+\mathrm{z}^{\mathrm{2}} }\:=\mathrm{4z}\:\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\left(−\mathrm{1}\right)^{\mathrm{n}} \:\mathrm{z}^{\mathrm{2n}} \\ $$$$=\mathrm{4}\:\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\left(−\mathrm{1}\right)^{\mathrm{n}} \:\mathrm{z}^{\mathrm{2n}+\mathrm{1}} \\ $$$$=\mathrm{4}\:\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\left(−\mathrm{1}\right)^{\mathrm{n}} \mathrm{e}^{\mathrm{i}\left(\mathrm{2n}+\mathrm{1}\right)\mathrm{x}} \\ $$$$=\mathrm{4}\:\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\left(−\mathrm{1}\right)^{\mathrm{n}} \:\mathrm{cos}\left(\mathrm{2n}+\mathrm{1}\right)\mathrm{x}\:+\mathrm{4i}\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \left(−\mathrm{1}\right)^{\mathrm{n}} \:\mathrm{sin}\left(\mathrm{2n}+\mathrm{1}\right)\mathrm{x} \\ $$$$\mathrm{g}\left(\mathrm{x}\right)\mathrm{real}\:\Rightarrow\bigstar\mathrm{g}\left(\mathrm{x}\right)\:=\mathrm{4}\:\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\left(−\mathrm{1}\right)^{\mathrm{n}} \:\mathrm{cos}\left(\mathrm{2n}+\mathrm{1}\right)\mathrm{x}\bigstar \\ $$

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