Question Number 38208 by prof Abdo imad last updated on 22/Jun/18
$${let}\:{f}\left({x}\right)={ch}\left(\alpha{x}\right)\: \\ $$$${developp}\:{f}\:{at}\:{fourier}\:{serie}. \\ $$$$\left({f}\:\mathrm{2}\pi\:{periodic}\:{even}\right) \\ $$
Commented by prof Abdo imad last updated on 24/Jun/18
$${f}\left({x}\right)=\frac{{a}_{\mathrm{0}} }{\mathrm{2}}\:+\sum_{{n}=\mathrm{1}} ^{\infty} \:{a}_{{n}} {cos}\left({nx}\right)\:{with} \\ $$$${a}_{{n}} =\:\frac{\mathrm{2}}{{T}}\:\int_{\left[{T}\right]} {f}\left({x}\right){cos}\left({nx}\right){dx} \\ $$$$=\frac{\mathrm{2}}{\mathrm{2}\pi}\:\int_{−\pi} ^{\pi} \:{ch}\left(\alpha{x}\right){cos}\left({nx}\right){dx} \\ $$$$=\frac{\mathrm{2}}{\pi}\:\int_{\mathrm{0}} ^{\pi} \:{ch}\left(\alpha{x}\right){cos}\left({nx}\right){dx}\:\Rightarrow \\ $$$$\frac{\pi}{\mathrm{2}}\:{a}_{{n}} =\:{Re}\left(\:\int_{\mathrm{0}} ^{\pi} \:\:\:\frac{{e}^{\alpha{x}} \:+{e}^{−\alpha{x}} }{\mathrm{2}}\:{e}^{{inx}} \:{dx}\right)={Re}\left({A}_{{n}} \right) \\ $$$${A}_{{n}} =\:\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\pi} \:\:{e}^{\left(\alpha+{in}\right){x}} {dx}\:+\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\pi} \:\:{e}^{\left(−\alpha\:+{in}\right){x}} {dx} \\ $$$$\mathrm{2}{A}_{{n}} =\:\left[\frac{\mathrm{1}}{\alpha+{in}}\:{e}^{\left(\alpha+{in}\right){x}} \right]_{\mathrm{0}} ^{\pi} \:+\left[\:\frac{\mathrm{1}}{−\alpha\:+{in}}\:{e}^{\left(−\alpha+{in}\right){x}} \right]_{\mathrm{0}} ^{\pi} \\ $$$$=\frac{{e}^{\alpha\pi} \left(−\mathrm{1}\right)^{{n}} −\mathrm{1}}{\alpha+{in}}\:+\:\frac{\mathrm{1}}{−\alpha\:+{in}}\frac{{e}^{−\alpha\pi} \left(−\mathrm{1}\right)^{{n}} −\mathrm{1}}{\mathrm{1}} \\ $$$$=\:\frac{\left\{\left(−\mathrm{1}\right)^{{n}} {e}^{\alpha\pi} −\mathrm{1}\right\}\left(\alpha−{in}\right)}{\alpha^{\mathrm{2}} \:+{n}^{\mathrm{2}} }\:\:−\frac{\left\{\left(−\mathrm{1}\right)^{{n}} {e}^{−\alpha\pi} −\mathrm{1}\right\}\left(\alpha+{in}\right\}}{\alpha^{\mathrm{2}} \:+{n}^{\mathrm{2}} } \\ $$
Commented by prof Abdo imad last updated on 25/Jun/18
$$\mathrm{2}{A}_{{n}} =\:\frac{\alpha\left\{\:\left(−\mathrm{1}\right)^{{n}} \:{e}^{\alpha\pi} −\mathrm{1}\right\}−{in}\left\{\left(−\mathrm{1}\right)^{{n}} {e}^{\alpha\pi} −\mathrm{1}\right\}−\alpha\left\{\left(−\mathrm{1}\right)^{{n}} {e}^{−\alpha\pi} −\mathrm{1}\right\}−{in}\left\{\left(−\mathrm{1}\right)^{{n}} \:{e}^{−\alpha\pi} −\mathrm{1}\right)}{\alpha^{\mathrm{2}} \:+{n}^{\mathrm{2}} } \\ $$$${Re}\left(\:\mathrm{2}{A}_{{n}} \right)\:=\:\frac{\alpha\left(−\mathrm{1}\right)^{{n}} \:{e}^{\alpha\pi} −\alpha\:−\alpha\left(−\mathrm{1}\right)^{{n}} {e}^{−\alpha\pi} \:+\alpha}{\alpha^{\mathrm{2}} \:+{n}^{\mathrm{2}} } \\ $$$$=\frac{\alpha\left(−\mathrm{1}\right)^{{n}} \left(\:{e}^{\alpha\pi} \:−{e}^{−\alpha\pi} \right)}{\alpha^{\mathrm{2}} \:+{n}^{\mathrm{2}} }\:=\:\frac{\mathrm{2}\alpha\left(−\mathrm{1}\right)^{{n}} {sh}\left(\alpha\pi\right)}{\alpha^{\mathrm{2}} \:+{n}^{\mathrm{2}} }\:\Rightarrow \\ $$$$\frac{\pi}{\mathrm{2}}{a}_{{n}} =\:\frac{\alpha\left(−\mathrm{1}\right)^{{n}} {sh}\left(\alpha\pi\right)}{\alpha^{\mathrm{2}} \:+{n}^{\mathrm{2}} }\:\Rightarrow \\ $$$${a}_{{n}} =\frac{\mathrm{2}\alpha\:\left(−\mathrm{1}\right)^{{n}} {sh}\left(\alpha\pi\right)}{\pi\left(\:\alpha^{\mathrm{2}} \:+{n}^{\mathrm{2}} \right)}\:\Rightarrow \\ $$$${a}_{\mathrm{0}} =\:\frac{\mathrm{2}\alpha{sh}\left(\alpha\pi\right)}{\pi\alpha^{\mathrm{2}} }\:\Rightarrow\:\frac{{a}_{\mathrm{0}} }{\mathrm{2}}\:=\:\:\frac{{sh}\left(\alpha\pi\right)}{\alpha\pi}\:\Rightarrow \\ $$$${ch}\left(\alpha{x}\right)=\:\frac{{sh}\left(\pi\alpha\right)}{\pi\alpha}\:+\frac{{sh}\left(\pi\alpha\right)}{\pi}\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\:\:\frac{\mathrm{2}\alpha\left(−\mathrm{1}\right)^{{n}} }{\alpha^{\mathrm{2}} \:+{n}^{\mathrm{2}} } \\ $$
Commented by prof Abdo imad last updated on 25/Jun/18
$${ch}\left(\alpha{x}\right)=\frac{{sh}\left(\pi\alpha\right)}{\pi\alpha}\:+\mathrm{2}\alpha\:\frac{{sh}\left(\pi\alpha\right)}{\pi}\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{\alpha^{\mathrm{2}} \:+{n}^{\mathrm{2}} }{cos}\left({nx}\right) \\ $$