Question Number 134915 by BHOOPENDRA last updated on 08/Mar/21
Answered by Olaf last updated on 08/Mar/21
![Fourier : a_0 (f) = (1/T)∫_(−(T/2)) ^(+(T/2)) f(t)dt a_0 (f) = (1/2)∫_(−1) ^(+1) (1−t^2 )dt a_0 (f) = (1/2)[t−(t^3 /3)]_(−1) ^(+1) = (1/2)((4/3)) = (2/3) a_n (f) = (2/T)∫_(−(T/2)) ^(+(T/2)) f(t)cos(nt((2π)/T))dt a_n (f) = ∫_(−1) ^(+1) (1−t^2 )cos(πnt)dt a_n (f) = [(1−t^2 )((sin(πnt))/(πn))]_(−1) ^(+1) −∫_(−1) ^(+1) (−2t)((sin(πnt))/(πn))dt a_n (f) = (2/(πn)) ∫_(−1) ^(+1) tsin(πnt)dt a_n (f) = (2/(πn))[t(−((cos(πnt))/(πn)))]_(−1) ^(+1) −(2/(πn)) ∫_(−1) ^(+1) (−((cos(πnt))/(πn)))dt a_n (f) = (−1)^(n+1) (4/(π^2 n^2 ))+(2/(π^2 n^2 )) ∫_(−1) ^(+1) cos(πnt)dt a_n (f) = (−1)^(n+1) (4/(π^2 n^2 )) b_n (f) = ∫_(−1) ^(+1) (1−t^2 )sin(πnt)dt b_n (f) = 0 (f is a even function) f(t) = a_0 (f)+Σ_(k=1) ^∞ a_n (f)cos(πnt) ...](https://www.tinkutara.com/question/Q134927.png)
$$\mathrm{Fourier}\:: \\ $$$${a}_{\mathrm{0}} \left({f}\right)\:=\:\frac{\mathrm{1}}{\mathrm{T}}\int_{−\frac{\mathrm{T}}{\mathrm{2}}} ^{+\frac{\mathrm{T}}{\mathrm{2}}} {f}\left({t}\right){dt} \\ $$$${a}_{\mathrm{0}} \left({f}\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}}\int_{−\mathrm{1}} ^{+\mathrm{1}} \left(\mathrm{1}−{t}^{\mathrm{2}} \right){dt} \\ $$$${a}_{\mathrm{0}} \left({f}\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}}\left[{t}−\frac{{t}^{\mathrm{3}} }{\mathrm{3}}\right]_{−\mathrm{1}} ^{+\mathrm{1}} \:=\:\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{4}}{\mathrm{3}}\right)\:=\:\frac{\mathrm{2}}{\mathrm{3}} \\ $$$${a}_{{n}} \left({f}\right)\:=\:\frac{\mathrm{2}}{\mathrm{T}}\int_{−\frac{\mathrm{T}}{\mathrm{2}}} ^{+\frac{\mathrm{T}}{\mathrm{2}}} {f}\left({t}\right)\mathrm{cos}\left({nt}\frac{\mathrm{2}\pi}{\mathrm{T}}\right){dt} \\ $$$${a}_{{n}} \left({f}\right)\:=\:\int_{−\mathrm{1}} ^{+\mathrm{1}} \left(\mathrm{1}−{t}^{\mathrm{2}} \right)\mathrm{cos}\left(\pi{nt}\right){dt} \\ $$$${a}_{{n}} \left({f}\right)\:=\:\left[\left(\mathrm{1}−{t}^{\mathrm{2}} \right)\frac{\mathrm{sin}\left(\pi{nt}\right)}{\pi{n}}\right]_{−\mathrm{1}} ^{+\mathrm{1}} −\int_{−\mathrm{1}} ^{+\mathrm{1}} \left(−\mathrm{2}{t}\right)\frac{\mathrm{sin}\left(\pi{nt}\right)}{\pi{n}}{dt} \\ $$$${a}_{{n}} \left({f}\right)\:=\:\frac{\mathrm{2}}{\pi{n}}\:\int_{−\mathrm{1}} ^{+\mathrm{1}} {t}\mathrm{sin}\left(\pi{nt}\right){dt} \\ $$$${a}_{{n}} \left({f}\right)\:=\:\frac{\mathrm{2}}{\pi{n}}\left[{t}\left(−\frac{\mathrm{cos}\left(\pi{nt}\right)}{\pi{n}}\right)\right]_{−\mathrm{1}} ^{+\mathrm{1}} −\frac{\mathrm{2}}{\pi{n}}\:\int_{−\mathrm{1}} ^{+\mathrm{1}} \left(−\frac{\mathrm{cos}\left(\pi{nt}\right)}{\pi{n}}\right){dt} \\ $$$${a}_{{n}} \left({f}\right)\:=\:\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} \frac{\mathrm{4}}{\pi^{\mathrm{2}} {n}^{\mathrm{2}} }+\frac{\mathrm{2}}{\pi^{\mathrm{2}} {n}^{\mathrm{2}} }\:\int_{−\mathrm{1}} ^{+\mathrm{1}} \mathrm{cos}\left(\pi{nt}\right){dt} \\ $$$${a}_{{n}} \left({f}\right)\:=\:\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} \frac{\mathrm{4}}{\pi^{\mathrm{2}} {n}^{\mathrm{2}} } \\ $$$${b}_{{n}} \left({f}\right)\:=\:\int_{−\mathrm{1}} ^{+\mathrm{1}} \left(\mathrm{1}−{t}^{\mathrm{2}} \right)\mathrm{sin}\left(\pi{nt}\right){dt} \\ $$$${b}_{{n}} \left({f}\right)\:=\:\mathrm{0}\:\left({f}\:\mathrm{is}\:\mathrm{a}\:\mathrm{even}\:\mathrm{function}\right) \\ $$$$ \\ $$$${f}\left({t}\right)\:=\:{a}_{\mathrm{0}} \left({f}\right)+\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}{a}_{{n}} \left({f}\right)\mathrm{cos}\left(\pi{nt}\right) \\ $$$$… \\ $$
Commented by BHOOPENDRA last updated on 08/Mar/21
$${thanks}\:{sir} \\ $$