Question Number 136899 by BHOOPENDRA last updated on 27/Mar/21
Answered by Olaf last updated on 28/Mar/21
$$\mathrm{3}. \\ $$$${f}\left({x}\right)\:=\:{x}^{\mathrm{2}} ,\:−\mathrm{2}\leqslant{x}\leqslant\mathrm{2} \\ $$$${a}_{\mathrm{0}} \left({f}\right)\:=\:\frac{\mathrm{1}}{\mathrm{T}}\int_{−\frac{\mathrm{T}}{\mathrm{2}}} ^{+\frac{\mathrm{T}}{\mathrm{2}}} {f}\left({x}\right){dx} \\ $$$${a}_{\mathrm{0}} \left({f}\right)\:=\:\frac{\mathrm{1}}{\mathrm{4}}\int_{−\mathrm{2}} ^{+\mathrm{2}} {x}^{\mathrm{2}} {dx}\:=\:\frac{\mathrm{1}}{\mathrm{4}}\left[\frac{{x}^{\mathrm{3}} }{\mathrm{3}}\right]_{−\mathrm{2}} ^{+\mathrm{2}} \:=\:\frac{\mathrm{4}}{\mathrm{3}} \\ $$$${a}_{{n}} \left({f}\right)\:=\:\frac{\mathrm{2}}{\mathrm{T}}\int_{−\frac{\mathrm{T}}{\mathrm{2}}} ^{+\frac{\mathrm{T}}{\mathrm{2}}} {f}\left({x}\right)\mathrm{cos}\left(\frac{\mathrm{2}\pi{nx}}{\mathrm{T}}\right){dx} \\ $$$${a}_{{n}} \left({f}\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}}\int_{−\mathrm{2}} ^{+\mathrm{2}} {x}^{\mathrm{2}} \mathrm{cos}\left(\frac{\pi{nx}}{\mathrm{4}}\right){dx} \\ $$$${a}_{{n}} \left({f}\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}}\left[\frac{\left(\left(\mathrm{2}\pi^{\mathrm{2}} {n}^{\mathrm{2}} {x}^{\mathrm{2}} −\mathrm{16}\right)\mathrm{sin}\left(\frac{\pi{nx}}{\mathrm{2}}\right)+\mathrm{8}\pi{nx}\mathrm{cos}\left(\frac{\pi{nx}}{\mathrm{2}}\right)\right.}{\pi^{\mathrm{3}} {n}^{\mathrm{3}} }\right]_{−\mathrm{2}} ^{+\mathrm{2}} \\ $$$${a}_{{n}} \left({f}\right)\:=\:\frac{\mathrm{4}\left(−\mathrm{1}\right)^{{n}} }{\pi^{\mathrm{2}} {n}^{\mathrm{2}} } \\ $$$${b}_{{n}} \left({f}\right)\:=\:\frac{\mathrm{2}}{\mathrm{T}}\int_{−\frac{\mathrm{T}}{\mathrm{2}}} ^{+\frac{\mathrm{T}}{\mathrm{2}}} {f}\left({x}\right)\mathrm{sin}\left(\frac{\mathrm{2}\pi{nx}}{\mathrm{T}}\right){dx} \\ $$$${b}_{{n}} \left({f}\right)\:=\:\mathrm{0}\:\mathrm{because}\:{f}\:\mathrm{is}\:\mathrm{even} \\ $$$${f}\left({x}\right)\:=\:{a}_{\mathrm{0}} +\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{a}_{{n}} \mathrm{cos}\left(\frac{\mathrm{2}\pi{nx}}{\mathrm{T}}\right) \\ $$$${f}\left({x}\right)\:=\:\frac{\mathrm{4}}{\mathrm{3}}+\frac{\mathrm{4}}{\pi^{\mathrm{2}} }\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{\mathrm{2}} }\mathrm{cos}\left(\frac{\pi{nx}}{\mathrm{2}}\right) \\ $$
Commented by greg_ed last updated on 28/Mar/21
$$\boldsymbol{\mathrm{i}}\:\boldsymbol{\mathrm{like}}\:\boldsymbol{\mathrm{it}}\:! \\ $$
Answered by Olaf last updated on 28/Mar/21
$$ \\ $$$$\mathrm{4}. \\ $$$${f}\left({x}\right)\:=\:\begin{cases}{\mathrm{2}+{x},\:−\mathrm{2}\leqslant{x}\leqslant\mathrm{0}}\\{\mathrm{2}−{x},\:\mathrm{0}\:\leqslant{x}\leqslant\mathrm{2}}\end{cases} \\ $$$$\mathrm{or}\:{f}\left({x}\right)\:=\:\mathrm{2}−\mid{x}\mid,\:−\mathrm{2}\leqslant{x}\leqslant\mathrm{2} \\ $$$${a}_{\mathrm{0}} \left({f}\right)\:=\:\frac{\mathrm{1}}{\mathrm{T}}\int_{−\frac{\mathrm{T}}{\mathrm{2}}} ^{+\frac{\mathrm{T}}{\mathrm{2}}} {f}\left({x}\right){dx} \\ $$$${a}_{\mathrm{0}} \left({f}\right)\:=\:\frac{\mathrm{1}}{\mathrm{4}}\int_{−\mathrm{2}} ^{+\mathrm{2}} \left(\mathrm{2}−\mid{x}\mid\right){dx} \\ $$$${a}_{\mathrm{0}} \left({f}\right)\:=\:\frac{\mathrm{1}}{\mathrm{4}}\int_{−\mathrm{2}} ^{\mathrm{0}} \left(\mathrm{2}+{x}\right){dx}+\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\mathrm{2}} \left(\mathrm{2}−{x}\right){dx} \\ $$$${a}_{\mathrm{0}} \left({f}\right)\:=\:\frac{\mathrm{1}}{\mathrm{4}}\left[\mathrm{2}{x}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\right]_{−\mathrm{2}} ^{\mathrm{0}} +\frac{\mathrm{1}}{\mathrm{4}}\left[\mathrm{2}{x}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\right]_{\mathrm{0}} ^{\mathrm{2}} \\ $$$${a}_{\mathrm{0}} \left({f}\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\:=\:\mathrm{1} \\ $$$${a}_{{n}} \left({f}\right)\:=\:\frac{\mathrm{2}}{\mathrm{T}}\int_{−\frac{\mathrm{T}}{\mathrm{2}}} ^{+\frac{\mathrm{T}}{\mathrm{2}}} {f}\left({x}\right)\mathrm{cos}\left(\frac{\mathrm{2}\pi{nx}}{\mathrm{T}}\right){dx} \\ $$$${a}_{{n}} \left({f}\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}}\int_{−\mathrm{2}} ^{+\mathrm{0}} \left(\mathrm{2}+{x}\right)\mathrm{cos}\left(\frac{\pi{nx}}{\mathrm{2}}\right){dx} \\ $$$$+\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{2}} \left(\mathrm{2}−{x}\right)\mathrm{cos}\left(\frac{\pi{nx}}{\mathrm{2}}\right){dx} \\ $$$${a}_{{n}} \left({f}\right)\:=\:\int_{−\mathrm{2}} ^{+\mathrm{2}} \mathrm{cos}\left(\frac{\pi{nx}}{\mathrm{2}}\right){dx}−\int_{\mathrm{0}} ^{\mathrm{2}} {x}\mathrm{cos}\left(\frac{\pi{nx}}{\mathrm{2}}\right){dx} \\ $$$${a}_{{n}} \left({f}\right)\:=\:\frac{\mathrm{4}}{\pi^{\mathrm{2}} {n}^{\mathrm{2}} }\left[\left(−\mathrm{1}\right)^{{n}} −\mathrm{1}\right] \\ $$$${a}_{\mathrm{2}{n}} \left({f}\right)\:=\:\mathrm{0}\:\mathrm{and}\:{a}_{\mathrm{2}{n}+\mathrm{1}} \left({f}\right)\:=\:−\frac{\mathrm{8}}{\pi^{\mathrm{2}} \left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${b}_{{n}} \left({f}\right)\:=\:\frac{\mathrm{2}}{\mathrm{T}}\int_{−\frac{\mathrm{T}}{\mathrm{2}}} ^{+\frac{\mathrm{T}}{\mathrm{2}}} {f}\left({x}\right)\mathrm{cos}\left(\frac{\mathrm{2}\pi{nx}}{\mathrm{T}}\right){dx} \\ $$$${b}_{{n}} \left({f}\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}}\int_{−\mathrm{2}} ^{+\mathrm{0}} \left(\mathrm{2}+{x}\right)\mathrm{sin}\left(\frac{\pi{nx}}{\mathrm{2}}\right){dx} \\ $$$$+\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{2}} \left(\mathrm{2}−{x}\right)\mathrm{sin}\left(\frac{\pi{nx}}{\mathrm{2}}\right){dx} \\ $$$${b}_{{n}} \left({f}\right)\:=\:\int_{−\mathrm{2}} ^{+\mathrm{2}} \mathrm{sin}\left(\frac{\pi{nx}}{\mathrm{2}}\right){dx}\:=\:\mathrm{0} \\ $$$${f}\left({x}\right)\:=\:{a}_{\mathrm{0}} +\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{a}_{{n}} \mathrm{cos}\left(\frac{\pi{nx}}{\mathrm{2}}\right) \\ $$$${f}\left({x}\right)\:=\:\mathrm{1}−\frac{\mathrm{8}}{\pi^{\mathrm{2}} }\underset{{p}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{cos}\left(\frac{\pi\left(\mathrm{2}{p}+\mathrm{1}\right){x}}{\mathrm{2}}\right)}{\left(\mathrm{2}{p}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$
Commented by BHOOPENDRA last updated on 28/Mar/21
$${thankyou}\:{sir} \\ $$