Question Number 67381 by mathmax by abdo last updated on 26/Aug/19
$${let}\:{f}\left({x}\right)\:={x}^{\mathrm{2}} \:\:\:\:\:\:\:\mathrm{2}\pi\:{periodic}\:\:{even}\:\:{develop}\:{f}\:{at}\:{fourier}\:{serie} \\ $$
Commented by mathmax by abdo last updated on 27/Aug/19
![f even ⇒f(x) =(a_0 /2) +Σ_(n=1) ^∞ a_n cos(nx) with a_n =(2/T)∫_([T]) x^2 cos(nx)dx =(2/(2π))∫_(−π) ^π x^2 coz(nx)dx =(2/π)∫_0 ^π x^2 cos(nx)dx ⇒(π/2)a_n =∫_0 ^π x^2 cos(nx)dx by parts ∫_0 ^π x^2 cos(nx)dx =[(x^2 /n)sin(nx)]_0 ^π −∫_0 ^π ((2x)/n)sin(nx)dx −(2/n) ∫_0 ^π xsin(nx)dx =−(2/n)[ [−(x/n)cos(nx)]_0 ^π −∫_0 ^π −(1/n)cos(nx)dx} =−(2/n){−(π/n) (−1)^n +(1/n^2 )[sinnx]_0 ^π } =((2π)/n^2 )(−1)^n ⇒ a_n =(2/π)×((2π)/n^2 )(−1)^(n ) =((4π)/(πn^2 ))(−1)^n a_0 =(2/π) ∫_0 ^π x^2 dx =(2/π)[(x^3 /3)]_0 ^π =(2/π)×(π^3 /3) =((2π^2 )/3) ⇒ ★x^2 =(π^2 /3) +4Σ_(n=1) ^∞ (((−1)^n )/n^2 )cos(nx)★](https://www.tinkutara.com/question/Q67460.png)
$${f}\:{even}\:\Rightarrow{f}\left({x}\right)\:=\frac{{a}_{\mathrm{0}} }{\mathrm{2}}\:+\sum_{{n}=\mathrm{1}} ^{\infty} \:{a}_{{n}} {cos}\left({nx}\right)\:\:{with} \\ $$$${a}_{{n}} =\frac{\mathrm{2}}{{T}}\int_{\left[{T}\right]} \:\:{x}^{\mathrm{2}} \:{cos}\left({nx}\right){dx}\:=\frac{\mathrm{2}}{\mathrm{2}\pi}\int_{−\pi} ^{\pi} \:{x}^{\mathrm{2}} {coz}\left({nx}\right){dx} \\ $$$$=\frac{\mathrm{2}}{\pi}\int_{\mathrm{0}} ^{\pi} \:{x}^{\mathrm{2}} {cos}\left({nx}\right){dx}\:\Rightarrow\frac{\pi}{\mathrm{2}}{a}_{{n}} =\int_{\mathrm{0}} ^{\pi} \:{x}^{\mathrm{2}} \:{cos}\left({nx}\right){dx}\:\:{by}\:{parts} \\ $$$$\int_{\mathrm{0}} ^{\pi} \:{x}^{\mathrm{2}} \:{cos}\left({nx}\right){dx}\:=\left[\frac{{x}^{\mathrm{2}} }{{n}}{sin}\left({nx}\right)\right]_{\mathrm{0}} ^{\pi} \:−\int_{\mathrm{0}} ^{\pi} \:\frac{\mathrm{2}{x}}{{n}}{sin}\left({nx}\right){dx} \\ $$$$−\frac{\mathrm{2}}{{n}}\:\int_{\mathrm{0}} ^{\pi} \:{xsin}\left({nx}\right){dx}\:=−\frac{\mathrm{2}}{{n}}\left[\:\:\:\:\left[−\frac{{x}}{{n}}{cos}\left({nx}\right)\right]_{\mathrm{0}} ^{\pi} \:−\int_{\mathrm{0}} ^{\pi} −\frac{\mathrm{1}}{{n}}{cos}\left({nx}\right){dx}\right\} \\ $$$$=−\frac{\mathrm{2}}{{n}}\left\{−\frac{\pi}{{n}}\:\left(−\mathrm{1}\right)^{{n}} \:+\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\left[{sinnx}\right]_{\mathrm{0}} ^{\pi} \right\}\:=\frac{\mathrm{2}\pi}{{n}^{\mathrm{2}} }\left(−\mathrm{1}\right)^{{n}} \:\Rightarrow \\ $$$${a}_{{n}} =\frac{\mathrm{2}}{\pi}×\frac{\mathrm{2}\pi}{{n}^{\mathrm{2}} }\left(−\mathrm{1}\right)^{{n}\:} \:=\frac{\mathrm{4}\pi}{\pi{n}^{\mathrm{2}} }\left(−\mathrm{1}\right)^{{n}} \\ $$$${a}_{\mathrm{0}} =\frac{\mathrm{2}}{\pi}\:\int_{\mathrm{0}} ^{\pi} \:{x}^{\mathrm{2}} \:{dx}\:=\frac{\mathrm{2}}{\pi}\left[\frac{{x}^{\mathrm{3}} }{\mathrm{3}}\right]_{\mathrm{0}} ^{\pi} \:=\frac{\mathrm{2}}{\pi}×\frac{\pi^{\mathrm{3}} }{\mathrm{3}}\:=\frac{\mathrm{2}\pi^{\mathrm{2}} }{\mathrm{3}}\:\Rightarrow \\ $$$$\bigstar{x}^{\mathrm{2}} \:=\frac{\pi^{\mathrm{2}} }{\mathrm{3}}\:+\mathrm{4}\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{\mathrm{2}} }{cos}\left({nx}\right)\bigstar \\ $$$$ \\ $$