Question Number 28072 by abdo imad last updated on 20/Jan/18
$${let}\:{give}\:{the}\:{function}\:\:{f}\left({x}\right)={x}^{\mathrm{4}} \:\:\:\mathrm{2}\pi\:{periodic}\:{and}\:{even} \\ $$$${developp}\:\:\:{f}\:{atfourier}\:{series}. \\ $$
Commented by abdo imad last updated on 26/Jan/18
![f(−x)=f(x) and f 2π periodic so f(x)=(a_0 /2) + Σ_(n=1) ^(+∞) a_n cos(nx) with a_(n ) = (2/T) ∫_([T]) f(x)cos(nx)dx ( T=2π) a_n = (1/π) ∫_(−π) ^π x^4 cos(nx)dx = (2/π) ∫_0 ^π x^4 cos(nx)dx let put A_p = ∫_0 ^π x^p cos(nx)dx we know that A_(2p) = (1/n^2 )( 2pπ^(2p−1) (−1)^n −2p(2p−1)A_(2p−2) ) so A_4 = (1/n^2 )( 4 π^3 (−1)^n −12 A_2 ) but A_(2 ) = (1/n^2 )( 2π(−1)^n −2A_0 )=((2π)/n^2 )(−1)^n ( A_0 =0) A_4 = (1/n^2 )( 4 π^3 (−1)^n −((24π)/n^2 )(−1)^n ) so a_4 =(2/π) A_4 = (2/(πn^2 ))( 4π^3 (−1)^n −((24π)/n^2 )(−1)^n ) = ((8π^2 )/n^2 )(−1)^n −((48)/n^4 )(−1)^n let find a_(0 ) ? a_0 =(2/π) ∫_0 ^π x^4 dx= (2/π) (1/5) π^5 =((2π^4 )/5) and (a_0 /2) =(π^4 /5) and x^4 = (π^5 /5) +8π^2 Σ_(n=1) ^(+∞) (((−1)^n )/n^2 )cos(nx) −48 Σ_(n=1) ^(+∞) (((−1)^n )/n^4 ) cos(nx).](https://www.tinkutara.com/question/Q28493.png)
$${f}\left(−{x}\right)={f}\left({x}\right)\:{and}\:{f}\:\mathrm{2}\pi\:{periodic}\:{so} \\ $$$${f}\left({x}\right)=\frac{{a}_{\mathrm{0}} }{\mathrm{2}}\:+\:\sum_{{n}=\mathrm{1}} ^{+\infty} \:{a}_{{n}} \:{cos}\left({nx}\right)\:\:{with} \\ $$$${a}_{{n}\:} =\:\frac{\mathrm{2}}{{T}}\:\int_{\left[{T}\right]} \:{f}\left({x}\right){cos}\left({nx}\right){dx}\:\:\:\left(\:\:{T}=\mathrm{2}\pi\right) \\ $$$${a}_{{n}} \:=\:\frac{\mathrm{1}}{\pi}\:\int_{−\pi} ^{\pi} {x}^{\mathrm{4}} \:{cos}\left({nx}\right){dx}\:=\:\frac{\mathrm{2}}{\pi}\:\int_{\mathrm{0}} ^{\pi} \:{x}^{\mathrm{4}} \:{cos}\left({nx}\right){dx}\:{let}\:{put} \\ $$$${A}_{{p}} =\:\int_{\mathrm{0}} ^{\pi} \:{x}^{{p}} {cos}\left({nx}\right){dx}\:{we}\:{know}\:{that} \\ $$$${A}_{\mathrm{2}{p}} =\:\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\left(\:\mathrm{2}{p}\pi^{\mathrm{2}{p}−\mathrm{1}} \left(−\mathrm{1}\right)^{{n}} \:\:−\mathrm{2}{p}\left(\mathrm{2}{p}−\mathrm{1}\right){A}_{\mathrm{2}{p}−\mathrm{2}} \right)\:{so} \\ $$$${A}_{\mathrm{4}} =\:\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\left(\:\mathrm{4}\:\pi^{\mathrm{3}} \left(−\mathrm{1}\right)^{{n}} \:−\mathrm{12}\:{A}_{\mathrm{2}} \:\:\right)\:{but} \\ $$$${A}_{\mathrm{2}\:} =\:\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\left(\:\mathrm{2}\pi\left(−\mathrm{1}\right)^{{n}} \:−\mathrm{2}{A}_{\mathrm{0}} \:\:\right)=\frac{\mathrm{2}\pi}{{n}^{\mathrm{2}} }\left(−\mathrm{1}\right)^{{n}} \:\:\:\:\:\:\:\:\:\left(\:{A}_{\mathrm{0}} =\mathrm{0}\right) \\ $$$${A}_{\mathrm{4}} \:=\:\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\left(\:\mathrm{4}\:\pi^{\mathrm{3}} \left(−\mathrm{1}\right)^{{n}} \:−\frac{\mathrm{24}\pi}{{n}^{\mathrm{2}} }\left(−\mathrm{1}\right)^{{n}} \right)\:{so} \\ $$$${a}_{\mathrm{4}} =\frac{\mathrm{2}}{\pi}\:{A}_{\mathrm{4}} =\:\frac{\mathrm{2}}{\pi{n}^{\mathrm{2}} }\left(\:\mathrm{4}\pi^{\mathrm{3}} \left(−\mathrm{1}\right)^{{n}} \:−\frac{\mathrm{24}\pi}{{n}^{\mathrm{2}} }\left(−\mathrm{1}\right)^{{n}} \right) \\ $$$$=\:\frac{\mathrm{8}\pi^{\mathrm{2}} }{{n}^{\mathrm{2}} }\left(−\mathrm{1}\right)^{{n}} \:\:\:−\frac{\mathrm{48}}{{n}^{\mathrm{4}} }\left(−\mathrm{1}\right)^{{n}} \:\:\:{let}\:{find}\:{a}_{\mathrm{0}\:} ? \\ $$$${a}_{\mathrm{0}} =\frac{\mathrm{2}}{\pi}\:\int_{\mathrm{0}} ^{\pi} \:{x}^{\mathrm{4}} {dx}=\:\frac{\mathrm{2}}{\pi}\:\frac{\mathrm{1}}{\mathrm{5}}\:\pi^{\mathrm{5}} =\frac{\mathrm{2}\pi^{\mathrm{4}} }{\mathrm{5}}\:\:{and}\:\frac{{a}_{\mathrm{0}} }{\mathrm{2}}\:=\frac{\pi^{\mathrm{4}} }{\mathrm{5}}\:\:\:{and} \\ $$$${x}^{\mathrm{4}} \:=\:\frac{\pi^{\mathrm{5}} }{\mathrm{5}}\:+\mathrm{8}\pi^{\mathrm{2}} \:\sum_{{n}=\mathrm{1}} ^{+\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{\mathrm{2}} }{cos}\left({nx}\right)\:−\mathrm{48}\:\sum_{{n}=\mathrm{1}} ^{+\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{\mathrm{4}} }\:{cos}\left({nx}\right). \\ $$$$ \\ $$