Question Number 136899 by BHOOPENDRA last updated on 27/Mar/21
Answered by Olaf last updated on 28/Mar/21
![3. f(x) = x^2 , −2≤x≤2 a_0 (f) = (1/T)∫_(−(T/2)) ^(+(T/2)) f(x)dx a_0 (f) = (1/4)∫_(−2) ^(+2) x^2 dx = (1/4)[(x^3 /3)]_(−2) ^(+2) = (4/3) a_n (f) = (2/T)∫_(−(T/2)) ^(+(T/2)) f(x)cos(((2πnx)/T))dx a_n (f) = (1/2)∫_(−2) ^(+2) x^2 cos(((πnx)/4))dx a_n (f) = (1/2)[((((2π^2 n^2 x^2 −16)sin(((πnx)/2))+8πnxcos(((πnx)/2)))/(π^3 n^3 ))]_(−2) ^(+2) a_n (f) = ((4(−1)^n )/(π^2 n^2 )) b_n (f) = (2/T)∫_(−(T/2)) ^(+(T/2)) f(x)sin(((2πnx)/T))dx b_n (f) = 0 because f is even f(x) = a_0 +Σ_(n=1) ^∞ a_n cos(((2πnx)/T)) f(x) = (4/3)+(4/π^2 )Σ_(n=1) ^∞ (((−1)^n )/n^2 )cos(((πnx)/2))](https://www.tinkutara.com/question/Q136941.png)
$$\mathrm{3}. \\ $$$${f}\left({x}\right)\:=\:{x}^{\mathrm{2}} ,\:−\mathrm{2}\leqslant{x}\leqslant\mathrm{2} \\ $$$${a}_{\mathrm{0}} \left({f}\right)\:=\:\frac{\mathrm{1}}{\mathrm{T}}\int_{−\frac{\mathrm{T}}{\mathrm{2}}} ^{+\frac{\mathrm{T}}{\mathrm{2}}} {f}\left({x}\right){dx} \\ $$$${a}_{\mathrm{0}} \left({f}\right)\:=\:\frac{\mathrm{1}}{\mathrm{4}}\int_{−\mathrm{2}} ^{+\mathrm{2}} {x}^{\mathrm{2}} {dx}\:=\:\frac{\mathrm{1}}{\mathrm{4}}\left[\frac{{x}^{\mathrm{3}} }{\mathrm{3}}\right]_{−\mathrm{2}} ^{+\mathrm{2}} \:=\:\frac{\mathrm{4}}{\mathrm{3}} \\ $$$${a}_{{n}} \left({f}\right)\:=\:\frac{\mathrm{2}}{\mathrm{T}}\int_{−\frac{\mathrm{T}}{\mathrm{2}}} ^{+\frac{\mathrm{T}}{\mathrm{2}}} {f}\left({x}\right)\mathrm{cos}\left(\frac{\mathrm{2}\pi{nx}}{\mathrm{T}}\right){dx} \\ $$$${a}_{{n}} \left({f}\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}}\int_{−\mathrm{2}} ^{+\mathrm{2}} {x}^{\mathrm{2}} \mathrm{cos}\left(\frac{\pi{nx}}{\mathrm{4}}\right){dx} \\ $$$${a}_{{n}} \left({f}\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}}\left[\frac{\left(\left(\mathrm{2}\pi^{\mathrm{2}} {n}^{\mathrm{2}} {x}^{\mathrm{2}} −\mathrm{16}\right)\mathrm{sin}\left(\frac{\pi{nx}}{\mathrm{2}}\right)+\mathrm{8}\pi{nx}\mathrm{cos}\left(\frac{\pi{nx}}{\mathrm{2}}\right)\right.}{\pi^{\mathrm{3}} {n}^{\mathrm{3}} }\right]_{−\mathrm{2}} ^{+\mathrm{2}} \\ $$$${a}_{{n}} \left({f}\right)\:=\:\frac{\mathrm{4}\left(−\mathrm{1}\right)^{{n}} }{\pi^{\mathrm{2}} {n}^{\mathrm{2}} } \\ $$$${b}_{{n}} \left({f}\right)\:=\:\frac{\mathrm{2}}{\mathrm{T}}\int_{−\frac{\mathrm{T}}{\mathrm{2}}} ^{+\frac{\mathrm{T}}{\mathrm{2}}} {f}\left({x}\right)\mathrm{sin}\left(\frac{\mathrm{2}\pi{nx}}{\mathrm{T}}\right){dx} \\ $$$${b}_{{n}} \left({f}\right)\:=\:\mathrm{0}\:\mathrm{because}\:{f}\:\mathrm{is}\:\mathrm{even} \\ $$$${f}\left({x}\right)\:=\:{a}_{\mathrm{0}} +\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{a}_{{n}} \mathrm{cos}\left(\frac{\mathrm{2}\pi{nx}}{\mathrm{T}}\right) \\ $$$${f}\left({x}\right)\:=\:\frac{\mathrm{4}}{\mathrm{3}}+\frac{\mathrm{4}}{\pi^{\mathrm{2}} }\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{\mathrm{2}} }\mathrm{cos}\left(\frac{\pi{nx}}{\mathrm{2}}\right) \\ $$
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
![4. f(x) = { ((2+x, −2≤x≤0)),((2−x, 0 ≤x≤2)) :} or f(x) = 2−∣x∣, −2≤x≤2 a_0 (f) = (1/T)∫_(−(T/2)) ^(+(T/2)) f(x)dx a_0 (f) = (1/4)∫_(−2) ^(+2) (2−∣x∣)dx a_0 (f) = (1/4)∫_(−2) ^0 (2+x)dx+(1/4)∫_0 ^2 (2−x)dx a_0 (f) = (1/4)[2x+(x^2 /2)]_(−2) ^0 +(1/4)[2x−(x^2 /2)]_0 ^2 a_0 (f) = (1/2)+(1/2) = 1 a_n (f) = (2/T)∫_(−(T/2)) ^(+(T/2)) f(x)cos(((2πnx)/T))dx a_n (f) = (1/2)∫_(−2) ^(+0) (2+x)cos(((πnx)/2))dx +(1/2)∫_0 ^2 (2−x)cos(((πnx)/2))dx a_n (f) = ∫_(−2) ^(+2) cos(((πnx)/2))dx−∫_0 ^2 xcos(((πnx)/2))dx a_n (f) = (4/(π^2 n^2 ))[(−1)^n −1] a_(2n) (f) = 0 and a_(2n+1) (f) = −(8/(π^2 (2n+1)^2 )) b_n (f) = (2/T)∫_(−(T/2)) ^(+(T/2)) f(x)cos(((2πnx)/T))dx b_n (f) = (1/2)∫_(−2) ^(+0) (2+x)sin(((πnx)/2))dx +(1/2)∫_0 ^2 (2−x)sin(((πnx)/2))dx b_n (f) = ∫_(−2) ^(+2) sin(((πnx)/2))dx = 0 f(x) = a_0 +Σ_(n=1) ^∞ a_n cos(((πnx)/2)) f(x) = 1−(8/π^2 )Σ_(p=0) ^∞ ((cos(((π(2p+1)x)/2)))/((2p+1)^2 ))](https://www.tinkutara.com/question/Q136944.png)
$$ \\ $$$$\mathrm{4}. \\ $$$${f}\left({x}\right)\:=\:\begin{cases}{\mathrm{2}+{x},\:−\mathrm{2}\leqslant{x}\leqslant\mathrm{0}}\\{\mathrm{2}−{x},\:\mathrm{0}\:\leqslant{x}\leqslant\mathrm{2}}\end{cases} \\ $$$$\mathrm{or}\:{f}\left({x}\right)\:=\:\mathrm{2}−\mid{x}\mid,\:−\mathrm{2}\leqslant{x}\leqslant\mathrm{2} \\ $$$${a}_{\mathrm{0}} \left({f}\right)\:=\:\frac{\mathrm{1}}{\mathrm{T}}\int_{−\frac{\mathrm{T}}{\mathrm{2}}} ^{+\frac{\mathrm{T}}{\mathrm{2}}} {f}\left({x}\right){dx} \\ $$$${a}_{\mathrm{0}} \left({f}\right)\:=\:\frac{\mathrm{1}}{\mathrm{4}}\int_{−\mathrm{2}} ^{+\mathrm{2}} \left(\mathrm{2}−\mid{x}\mid\right){dx} \\ $$$${a}_{\mathrm{0}} \left({f}\right)\:=\:\frac{\mathrm{1}}{\mathrm{4}}\int_{−\mathrm{2}} ^{\mathrm{0}} \left(\mathrm{2}+{x}\right){dx}+\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\mathrm{2}} \left(\mathrm{2}−{x}\right){dx} \\ $$$${a}_{\mathrm{0}} \left({f}\right)\:=\:\frac{\mathrm{1}}{\mathrm{4}}\left[\mathrm{2}{x}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\right]_{−\mathrm{2}} ^{\mathrm{0}} +\frac{\mathrm{1}}{\mathrm{4}}\left[\mathrm{2}{x}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\right]_{\mathrm{0}} ^{\mathrm{2}} \\ $$$${a}_{\mathrm{0}} \left({f}\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\:=\:\mathrm{1} \\ $$$${a}_{{n}} \left({f}\right)\:=\:\frac{\mathrm{2}}{\mathrm{T}}\int_{−\frac{\mathrm{T}}{\mathrm{2}}} ^{+\frac{\mathrm{T}}{\mathrm{2}}} {f}\left({x}\right)\mathrm{cos}\left(\frac{\mathrm{2}\pi{nx}}{\mathrm{T}}\right){dx} \\ $$$${a}_{{n}} \left({f}\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}}\int_{−\mathrm{2}} ^{+\mathrm{0}} \left(\mathrm{2}+{x}\right)\mathrm{cos}\left(\frac{\pi{nx}}{\mathrm{2}}\right){dx} \\ $$$$+\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{2}} \left(\mathrm{2}−{x}\right)\mathrm{cos}\left(\frac{\pi{nx}}{\mathrm{2}}\right){dx} \\ $$$${a}_{{n}} \left({f}\right)\:=\:\int_{−\mathrm{2}} ^{+\mathrm{2}} \mathrm{cos}\left(\frac{\pi{nx}}{\mathrm{2}}\right){dx}−\int_{\mathrm{0}} ^{\mathrm{2}} {x}\mathrm{cos}\left(\frac{\pi{nx}}{\mathrm{2}}\right){dx} \\ $$$${a}_{{n}} \left({f}\right)\:=\:\frac{\mathrm{4}}{\pi^{\mathrm{2}} {n}^{\mathrm{2}} }\left[\left(−\mathrm{1}\right)^{{n}} −\mathrm{1}\right] \\ $$$${a}_{\mathrm{2}{n}} \left({f}\right)\:=\:\mathrm{0}\:\mathrm{and}\:{a}_{\mathrm{2}{n}+\mathrm{1}} \left({f}\right)\:=\:−\frac{\mathrm{8}}{\pi^{\mathrm{2}} \left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${b}_{{n}} \left({f}\right)\:=\:\frac{\mathrm{2}}{\mathrm{T}}\int_{−\frac{\mathrm{T}}{\mathrm{2}}} ^{+\frac{\mathrm{T}}{\mathrm{2}}} {f}\left({x}\right)\mathrm{cos}\left(\frac{\mathrm{2}\pi{nx}}{\mathrm{T}}\right){dx} \\ $$$${b}_{{n}} \left({f}\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}}\int_{−\mathrm{2}} ^{+\mathrm{0}} \left(\mathrm{2}+{x}\right)\mathrm{sin}\left(\frac{\pi{nx}}{\mathrm{2}}\right){dx} \\ $$$$+\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{2}} \left(\mathrm{2}−{x}\right)\mathrm{sin}\left(\frac{\pi{nx}}{\mathrm{2}}\right){dx} \\ $$$${b}_{{n}} \left({f}\right)\:=\:\int_{−\mathrm{2}} ^{+\mathrm{2}} \mathrm{sin}\left(\frac{\pi{nx}}{\mathrm{2}}\right){dx}\:=\:\mathrm{0} \\ $$$${f}\left({x}\right)\:=\:{a}_{\mathrm{0}} +\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{a}_{{n}} \mathrm{cos}\left(\frac{\pi{nx}}{\mathrm{2}}\right) \\ $$$${f}\left({x}\right)\:=\:\mathrm{1}−\frac{\mathrm{8}}{\pi^{\mathrm{2}} }\underset{{p}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{cos}\left(\frac{\pi\left(\mathrm{2}{p}+\mathrm{1}\right){x}}{\mathrm{2}}\right)}{\left(\mathrm{2}{p}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$
Commented by BHOOPENDRA last updated on 28/Mar/21
$${thankyou}\:{sir} \\ $$