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Question Number 38208 by prof Abdo imad last updated on 22/Jun/18

let f(x)=ch(αx) developp f at fourier serie. (f 2π periodic even)

$${let}\:{f}\left({x}\right)={ch}\left(\alpha{x}\right)\: \\ $$$${developp}\:{f}\:{at}\:{fourier}\:{serie}. \\ $$$$\left({f}\:\mathrm{2}\pi\:{periodic}\:{even}\right) \\ $$

Commented by prof Abdo imad last updated on 24/Jun/18

f(x)=(a_0 /2) +Σ_(n=1) ^∞ a_n cos(nx) with a_n = (2/T) ∫_([T]) f(x)cos(nx)dx =(2/(2π)) ∫_(−π) ^π ch(αx)cos(nx)dx =(2/π) ∫_0 ^π ch(αx)cos(nx)dx ⇒ (π/2) a_n = Re( ∫_0 ^π ((e^(αx) +e^(−αx) )/2) e^(inx) dx)=Re(A_n ) A_n = (1/2) ∫_0 ^π e^((α+in)x) dx +(1/2) ∫_0 ^π e^((−α +in)x) dx 2A_n = [(1/(α+in)) e^((α+in)x) ]_0 ^π +[ (1/(−α +in)) e^((−α+in)x) ]_0 ^π =((e^(απ) (−1)^n −1)/(α+in)) + (1/(−α +in))((e^(−απ) (−1)^n −1)/1) = (({(−1)^n e^(απ) −1}(α−in))/(α^2 +n^2 )) −(({(−1)^n e^(−απ) −1}(α+in})/(α^2 +n^2 ))

$${f}\left({x}\right)=\frac{{a}_{\mathrm{0}} }{\mathrm{2}}\:+\sum_{{n}=\mathrm{1}} ^{\infty} \:{a}_{{n}} {cos}\left({nx}\right)\:{with} \\ $$$${a}_{{n}} =\:\frac{\mathrm{2}}{{T}}\:\int_{\left[{T}\right]} {f}\left({x}\right){cos}\left({nx}\right){dx} \\ $$$$=\frac{\mathrm{2}}{\mathrm{2}\pi}\:\int_{−\pi} ^{\pi} \:{ch}\left(\alpha{x}\right){cos}\left({nx}\right){dx} \\ $$$$=\frac{\mathrm{2}}{\pi}\:\int_{\mathrm{0}} ^{\pi} \:{ch}\left(\alpha{x}\right){cos}\left({nx}\right){dx}\:\Rightarrow \\ $$$$\frac{\pi}{\mathrm{2}}\:{a}_{{n}} =\:{Re}\left(\:\int_{\mathrm{0}} ^{\pi} \:\:\:\frac{{e}^{\alpha{x}} \:+{e}^{−\alpha{x}} }{\mathrm{2}}\:{e}^{{inx}} \:{dx}\right)={Re}\left({A}_{{n}} \right) \\ $$$${A}_{{n}} =\:\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\pi} \:\:{e}^{\left(\alpha+{in}\right){x}} {dx}\:+\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\pi} \:\:{e}^{\left(−\alpha\:+{in}\right){x}} {dx} \\ $$$$\mathrm{2}{A}_{{n}} =\:\left[\frac{\mathrm{1}}{\alpha+{in}}\:{e}^{\left(\alpha+{in}\right){x}} \right]_{\mathrm{0}} ^{\pi} \:+\left[\:\frac{\mathrm{1}}{−\alpha\:+{in}}\:{e}^{\left(−\alpha+{in}\right){x}} \right]_{\mathrm{0}} ^{\pi} \\ $$$$=\frac{{e}^{\alpha\pi} \left(−\mathrm{1}\right)^{{n}} −\mathrm{1}}{\alpha+{in}}\:+\:\frac{\mathrm{1}}{−\alpha\:+{in}}\frac{{e}^{−\alpha\pi} \left(−\mathrm{1}\right)^{{n}} −\mathrm{1}}{\mathrm{1}} \\ $$$$=\:\frac{\left\{\left(−\mathrm{1}\right)^{{n}} {e}^{\alpha\pi} −\mathrm{1}\right\}\left(\alpha−{in}\right)}{\alpha^{\mathrm{2}} \:+{n}^{\mathrm{2}} }\:\:−\frac{\left\{\left(−\mathrm{1}\right)^{{n}} {e}^{−\alpha\pi} −\mathrm{1}\right\}\left(\alpha+{in}\right\}}{\alpha^{\mathrm{2}} \:+{n}^{\mathrm{2}} } \\ $$

Commented by prof Abdo imad last updated on 25/Jun/18

2A_n = ((α{ (−1)^n e^(απ) −1}−in{(−1)^n e^(απ) −1}−α{(−1)^n e^(−απ) −1}−in{(−1)^n e^(−απ) −1))/(α^2 +n^2 )) Re( 2A_n ) = ((α(−1)^n e^(απ) −α −α(−1)^n e^(−απ) +α)/(α^2 +n^2 )) =((α(−1)^n ( e^(απ) −e^(−απ) ))/(α^2 +n^2 )) = ((2α(−1)^n sh(απ))/(α^2 +n^2 )) ⇒ (π/2)a_n = ((α(−1)^n sh(απ))/(α^2 +n^2 )) ⇒ a_n =((2α (−1)^n sh(απ))/(π( α^2 +n^2 ))) ⇒ a_0 = ((2αsh(απ))/(πα^2 )) ⇒ (a_0 /2) = ((sh(απ))/(απ)) ⇒ ch(αx)= ((sh(πα))/(πα)) +((sh(πα))/π)Σ_(n=1) ^∞ ((2α(−1)^n )/(α^2 +n^2 ))

$$\mathrm{2}{A}_{{n}} =\:\frac{\alpha\left\{\:\left(−\mathrm{1}\right)^{{n}} \:{e}^{\alpha\pi} −\mathrm{1}\right\}−{in}\left\{\left(−\mathrm{1}\right)^{{n}} {e}^{\alpha\pi} −\mathrm{1}\right\}−\alpha\left\{\left(−\mathrm{1}\right)^{{n}} {e}^{−\alpha\pi} −\mathrm{1}\right\}−{in}\left\{\left(−\mathrm{1}\right)^{{n}} \:{e}^{−\alpha\pi} −\mathrm{1}\right)}{\alpha^{\mathrm{2}} \:+{n}^{\mathrm{2}} } \\ $$$${Re}\left(\:\mathrm{2}{A}_{{n}} \right)\:=\:\frac{\alpha\left(−\mathrm{1}\right)^{{n}} \:{e}^{\alpha\pi} −\alpha\:−\alpha\left(−\mathrm{1}\right)^{{n}} {e}^{−\alpha\pi} \:+\alpha}{\alpha^{\mathrm{2}} \:+{n}^{\mathrm{2}} } \\ $$$$=\frac{\alpha\left(−\mathrm{1}\right)^{{n}} \left(\:{e}^{\alpha\pi} \:−{e}^{−\alpha\pi} \right)}{\alpha^{\mathrm{2}} \:+{n}^{\mathrm{2}} }\:=\:\frac{\mathrm{2}\alpha\left(−\mathrm{1}\right)^{{n}} {sh}\left(\alpha\pi\right)}{\alpha^{\mathrm{2}} \:+{n}^{\mathrm{2}} }\:\Rightarrow \\ $$$$\frac{\pi}{\mathrm{2}}{a}_{{n}} =\:\frac{\alpha\left(−\mathrm{1}\right)^{{n}} {sh}\left(\alpha\pi\right)}{\alpha^{\mathrm{2}} \:+{n}^{\mathrm{2}} }\:\Rightarrow \\ $$$${a}_{{n}} =\frac{\mathrm{2}\alpha\:\left(−\mathrm{1}\right)^{{n}} {sh}\left(\alpha\pi\right)}{\pi\left(\:\alpha^{\mathrm{2}} \:+{n}^{\mathrm{2}} \right)}\:\Rightarrow \\ $$$${a}_{\mathrm{0}} =\:\frac{\mathrm{2}\alpha{sh}\left(\alpha\pi\right)}{\pi\alpha^{\mathrm{2}} }\:\Rightarrow\:\frac{{a}_{\mathrm{0}} }{\mathrm{2}}\:=\:\:\frac{{sh}\left(\alpha\pi\right)}{\alpha\pi}\:\Rightarrow \\ $$$${ch}\left(\alpha{x}\right)=\:\frac{{sh}\left(\pi\alpha\right)}{\pi\alpha}\:+\frac{{sh}\left(\pi\alpha\right)}{\pi}\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\:\:\frac{\mathrm{2}\alpha\left(−\mathrm{1}\right)^{{n}} }{\alpha^{\mathrm{2}} \:+{n}^{\mathrm{2}} } \\ $$

Commented by prof Abdo imad last updated on 25/Jun/18

ch(αx)=((sh(πα))/(πα)) +2α ((sh(πα))/π)Σ_(n=1) ^∞ (((−1)^n )/(α^2 +n^2 ))cos(nx)

$${ch}\left(\alpha{x}\right)=\frac{{sh}\left(\pi\alpha\right)}{\pi\alpha}\:+\mathrm{2}\alpha\:\frac{{sh}\left(\pi\alpha\right)}{\pi}\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{\alpha^{\mathrm{2}} \:+{n}^{\mathrm{2}} }{cos}\left({nx}\right) \\ $$

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