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Question Number 134915 by BHOOPENDRA last updated on 08/Mar/21

Answered by Olaf last updated on 08/Mar/21

$$\mathrm{Fourier}:$$$$a_0\left(f\right)=\frac{1}{\mathrm{T}}{\int }_{-\frac{\mathrm{T}}{2}}^{+\frac{\mathrm{T}}{2}}f\left(t\right)dt$$$$a_0\left(f\right)=\frac{1}{2}{\int }_{-1}^{+1}(1-t^2)dt$$$$a_0\left(f\right)=\frac{1}{2}{[t-\frac{t^3}{3}]}_{-1}^{+1}=\frac{1}{2}\left(\frac{4}{3}\right)=\frac{2}{3}$$$$a_n\left(f\right)=\frac{2}{\mathrm{T}}{\int }_{-\frac{\mathrm{T}}{2}}^{+\frac{\mathrm{T}}{2}}f\left(t\right)\cos \left(nt\frac{2\pi }{\mathrm{T}}\right)dt$$$$a_n\left(f\right)={\int }_{-1}^{+1}(1-t^2)\cos \left(\pi nt\right)dt$$$$a_n\left(f\right)={\left[(1-t^2)\frac{\sin \left(\pi nt\right)}{\pi n}\right]}_{-1}^{+1}-{\int }_{-1}^{+1}(-2t)\frac{\sin \left(\pi nt\right)}{\pi n}dt$$$$a_n\left(f\right)=\frac{2}{\pi n}{\int }_{-1}^{+1}t\sin \left(\pi nt\right)dt$$$$a_n\left(f\right)=\frac{2}{\pi n}{\left[t(-\frac{\cos \left(\pi nt\right)}{\pi n})\right]}_{-1}^{+1}-\frac{2}{\pi n}{\int }_{-1}^{+1}(-\frac{\cos \left(\pi nt\right)}{\pi n})dt$$$$a_n\left(f\right)={(-1)}^{n+1}\frac{4}{{\pi }^2{n}^2}+\frac{2}{{\pi }^2{n}^2}{\int }_{-1}^{+1}\cos \left(\pi nt\right)dt$$$$a_n\left(f\right)={(-1)}^{n+1}\frac{4}{{\pi }^2{n}^2}$$$$b_n\left(f\right)={\int }_{-1}^{+1}(1-t^2)\sin \left(\pi nt\right)dt$$$$b_n\left(f\right)=0\left(f\mathrm{is}\mathrm{a}\mathrm{even}\mathrm{function}\right)$$$$$$$$f\left(t\right)=a_0\left(f\right)+\underset{k=1}{\sum ^{\mathrm{\infty }}}{a}_n\left(f\right)\cos \left(\pi nt\right)$$$$...$$

Commented by BHOOPENDRA last updated on 08/Mar/21

$$thankssir$$

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