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Question Number 136899 by BHOOPENDRA last updated on 27/Mar/21

	Answered by Olaf last updated on 28/Mar/21

	$3 .$ $f (x) = x^{2}, - 2 ⩽ x ⩽ 2$ $a_{0} (f) = \frac{1}{T} \int_{- \frac{T}{2}}^{+ \frac{T}{2}} f (x) d x$ $a_{0} (f) = \frac{1}{4} \int_{- 2}^{+ 2} x^{2} d x = \frac{1}{4} {[\frac{x^{3}}{3}]}_{- 2}^{+ 2} = \frac{4}{3}$ $a_{n} (f) = \frac{2}{T} \int_{- \frac{T}{2}}^{+ \frac{T}{2}} f (x) \cos (\frac{2 π n x}{T}) d x$ $a_{n} (f) = \frac{1}{2} \int_{- 2}^{+ 2} x^{2} \cos (\frac{π n x}{4}) d x$ $a_{n} (f) = \frac{1}{2} {[\frac{((2 π^{2} n^{2} x^{2} - 16) \sin (\frac{π n x}{2}) + 8 π n x \cos (\frac{π n x}{2})}{π^{3} n^{3}}]}_{- 2}^{+ 2}$ $a_{n} (f) = \frac{4 {(- 1)}^{n}}{π^{2} n^{2}}$ $b_{n} (f) = \frac{2}{T} \int_{- \frac{T}{2}}^{+ \frac{T}{2}} f (x) \sin (\frac{2 π n x}{T}) d x$ $b_{n} (f) = 0 because f is even$ $f (x) = a_{0} + \underset{n = 1}{\sum^{\infty}} a_{n} \cos (\frac{2 π n x}{T})$ $f (x) = \frac{4}{3} + \frac{4}{π^{2}} \underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n}}{n^{2}} \cos (\frac{π n x}{2})$
	Commented by greg_ed last updated on 28/Mar/21

	$i like it!$
	Answered by Olaf last updated on 28/Mar/21
	$4. f(x) = { ((2+x, −2≤x≤0)),((2−x, 0 ≤x≤2)) :} or f(x) = 2−∣x∣, −2≤x≤2 a_0 (f) = (1/T)∫_(−(T/2)) ^(+(T/2)) f(x)dx a_0 (f) = (1/4)∫_(−2) ^(+2) (2−∣x∣)dx a_0 (f) = (1/4)∫_(−2) ^0 (2+x)dx+(1/4)∫_0 ^2 (2−x)dx a_0 (f) = (1/4)[2x+(x^2 /2)]_(−2) ^0 +(1/4)[2x−(x^2 /2)]_0 ^2 a_0 (f) = (1/2)+(1/2) = 1 a_n (f) = (2/T)∫_(−(T/2)) ^(+(T/2)) f(x)cos(((2πnx)/T))dx a_n (f) = (1/2)∫_(−2) ^(+0) (2+x)cos(((πnx)/2))dx +(1/2)∫_0 ^2 (2−x)cos(((πnx)/2))dx a_n (f) = ∫_(−2) ^(+2) cos(((πnx)/2))dx−∫_0 ^2 xcos(((πnx)/2))dx a_n (f) = (4/(π^2 n^2 ))[(−1)^n −1] a_(2n) (f) = 0 and a_(2n+1) (f) = −(8/(π^2 (2n+1)^2 )) b_n (f) = (2/T)∫_(−(T/2)) ^(+(T/2)) f(x)cos(((2πnx)/T))dx b_n (f) = (1/2)∫_(−2) ^(+0) (2+x)sin(((πnx)/2))dx +(1/2)∫_0 ^2 (2−x)sin(((πnx)/2))dx b_n (f) = ∫_(−2) ^(+2) sin(((πnx)/2))dx = 0 f(x) = a_0 +Σ_(n=1) ^∞ a_n cos(((πnx)/2)) f(x) = 1−(8/π^2 )Σ_(p=0) ^∞ ((cos(((π(2p+1)x)/2)))/((2p+1)^2 ))$
	$4 .$ $f (x) = {\begin{cases} 2 + x, - 2 ⩽ x ⩽ 0 \\ 2 - x, 0 ⩽ x ⩽ 2 \end{cases}$ $or f (x) = 2 - ∣ x ∣, - 2 ⩽ x ⩽ 2$ $a_{0} (f) = \frac{1}{T} \int_{- \frac{T}{2}}^{+ \frac{T}{2}} f (x) d x$ $a_{0} (f) = \frac{1}{4} \int_{- 2}^{+ 2} (2 - ∣ x ∣) d x$ $a_{0} (f) = \frac{1}{4} \int_{- 2}^{0} (2 + x) d x + \frac{1}{4} \int_{0}^{2} (2 - x) d x$ $a_{0} (f) = \frac{1}{4} {[2 x + \frac{x^{2}}{2}]}_{- 2}^{0} + \frac{1}{4} {[2 x - \frac{x^{2}}{2}]}_{0}^{2}$ $a_{0} (f) = \frac{1}{2} + \frac{1}{2} = 1$ $a_{n} (f) = \frac{2}{T} \int_{- \frac{T}{2}}^{+ \frac{T}{2}} f (x) \cos (\frac{2 π n x}{T}) d x$ $a_{n} (f) = \frac{1}{2} \int_{- 2}^{+ 0} (2 + x) \cos (\frac{π n x}{2}) d x$ $+ \frac{1}{2} \int_{0}^{2} (2 - x) \cos (\frac{π n x}{2}) d x$ $a_{n} (f) = \int_{- 2}^{+ 2} \cos (\frac{π n x}{2}) d x - \int_{0}^{2} x \cos (\frac{π n x}{2}) d x$ $a_{n} (f) = \frac{4}{π^{2} n^{2}} [{(- 1)}^{n} - 1]$ $a_{2 n} (f) = 0 and a_{2 n + 1} (f) = - \frac{8}{π^{2} {(2 n + 1)}^{2}}$ $b_{n} (f) = \frac{2}{T} \int_{- \frac{T}{2}}^{+ \frac{T}{2}} f (x) \cos (\frac{2 π n x}{T}) d x$ $b_{n} (f) = \frac{1}{2} \int_{- 2}^{+ 0} (2 + x) \sin (\frac{π n x}{2}) d x$ $+ \frac{1}{2} \int_{0}^{2} (2 - x) \sin (\frac{π n x}{2}) d x$ $b_{n} (f) = \int_{- 2}^{+ 2} \sin (\frac{π n x}{2}) d x = 0$ $f (x) = a_{0} + \underset{n = 1}{\sum^{\infty}} a_{n} \cos (\frac{π n x}{2})$ $f (x) = 1 - \frac{8}{π^{2}} \underset{p = 0}{\sum^{\infty}} \frac{\cos (\frac{π (2 p + 1) x}{2})}{{(2 p + 1)}^{2}}$
	Commented by BHOOPENDRA last updated on 28/Mar/21

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