1) find a and b wich verify ∫_0 ^π (at^2 +bt)cos(nx) =(1/n^2 ) 2) find the value of Σ


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Question Number 82286 by mathmax by abdo last updated on 19/Feb/20

$1) f i n d a a n d b w i c h v e r i f y \int_{0}^{π} ({a t}^{2} + b t) c o s (n x) = \frac{1}{n^{2}}$ $2) f i n d t h e v a l u e o f \sum_{n = 1}^{\infty} \frac{1}{n^{2}}$
Commented by john santu last updated on 22/Feb/20

$(2) \sin x = x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} - . . .$ $(3) \frac{\sin x}{x} = \underset{n = 1}{\prod^{\infty}} (1 - \frac{x^{2}}{n^{2} π^{2}}) = (1 - \frac{x^{2}}{π^{2}}) (1 - \frac{x^{2}}{4 π^{2}}) . . .$ $(2) = (3)$ $1 - \frac{x^{2}}{3!} + \frac{x^{4}}{5!} - \frac{x^{6}}{7!} + . . . = (1 - \frac{x^{2}}{π^{2}}) (1 - \frac{x^{2}}{4 π^{2}}) . . .$ $c o e f f i c i e n t t e r m o f x^{2} m u s t b e s a m e$ $- \frac{1}{3!} = - (\frac{1}{π^{2}} + \frac{1}{4 π^{2}} + \frac{1}{9 π^{2}} + . . .)$ $- \frac{1}{6} = - \frac{1}{π^{2}} \underset{n = 1}{\sum^{\infty}} \frac{1}{n^{2}}$ $t h e r e f o r e \underset{n = 1}{\sum^{\infty}} \frac{1}{n^{2}} = \frac{π^{2}}{6} .$
Commented by mathmax by abdo last updated on 20/Feb/20

$s i r j o h n c a n y o u p r o v e t h e s e c o n d l i n e i t s n o t c l e a r . . .$
Commented by mind is power last updated on 20/Feb/20

$n i c e B u t i t h i n k 1 & 2 a r e r e l a t e d q u a t i o n$
	Answered by mind is power last updated on 20/Feb/20
	$∫_0 ^π t^2 cos(nt)dt =−2∫_0 ^π ((tsin(nt))/n)dt =(2/n)[((tcos(nt))/n)]_0 ^π =((2π(−1)^n )/n^2 ) ∫tcos(nt)dt=−∫_0 ^π ((sin(nt))/n)dt=(([cos(nt)]_0 ^π )/n^2 )=(((−1)^n −1)/n^2 ) ⇒∫_0 ^π (at^2 +bt)cos(nt)dt=(((−1)^n )/n^2 )(2πa+b)−(b/n^2 )=(1/n^2 ) ⇒ { ((−b=1)),((2πa+b=0)) :}⇒b=−1,a=(1/(2π)) 2) let f(t)=(t^2 /(2π))−t 2π periodic odd f(x)=f(−x)paire f∈C_0 ]−π,π[ serie of fourier of f is S_n (f)=a_0 +Σ_(n≥1) a_n (f)cos(nx) a_0 =(1/(2π))∫_(−π) ^π f(x)dx=(1/π)∫_0 ^π ((x^2 /(2π))−x)dx =(π/6)−(π/2)=−(π/3) a_k =(1/π)∫_(−π) ^π f(x)cos(kx)dx=(2/π)∫_0 ^π f(x)cos(kx)dx by one a_k =(2/π)(1/k^2 ) S_n (f)(x)=−(π/3)+Σ_(n≥1) ((2cos(kx))/k^2 ) by direchlet f(0)=S_n (f)(0)⇒ 0=−(π/3)+(2/π)Σ_(k≥1) ((cos(k.0))/k^2 )⇒ (π/3)=(2/π)Σ_(k≥1) (1/k^2 )⇒Σ_(k≥1) (1/k^2 )=(π^2 /6)$
	$\int_{0}^{π} t^{2} c o s (n t) d t$ $= - 2 \int_{0}^{π} \frac{t s i n (n t)}{n} d t$ $= \frac{2}{n} {[\frac{t c o s (n t)}{n}]}_{0}^{π} = \frac{2 π {(- 1)}^{n}}{n^{2}}$ $\int t c o s (n t) d t = - \int_{0}^{π} \frac{s i n (n t)}{n} d t = \frac{{[c o s (n t)]}_{0}^{π}}{n^{2}} = \frac{{(- 1)}^{n} - 1}{n^{2}}$ $\Rightarrow \int_{0}^{π} ({a t}^{2} + b t) c o s (n t) d t = \frac{{(- 1)}^{n}}{n^{2}} (2 π a + b) - \frac{b}{n^{2}} = \frac{1}{n^{2}}$ $\Rightarrow {\begin{cases} - b = 1 \\ 2 π a + b = 0 \end{cases} \Rightarrow b = - 1, a = \frac{1}{2 π}$ $2)$ $l e t f (t) = \frac{t^{2}}{2 π} - t 2 π p e r i o d i c o d d f (x) = f (- x) p a i r e$ $f \in C_{0}] - π, π [$ $s e r i e o f f o u r i e r o f f i s$ $S_{n} (f) = a_{0} + \sum_{n ⩾ 1} a_{n} (f) c o s (n x)$ $a_{0} = \frac{1}{2 π} \int_{- π}^{π} f (x) d x = \frac{1}{π} \int_{0}^{π} (\frac{x^{2}}{2 π} - x) d x$ $= \frac{π}{6} - \frac{π}{2} = - \frac{π}{3}$ $a_{k} = \frac{1}{π} \int_{- π}^{π} f (x) c o s (k x) d x = \frac{2}{π} \int_{0}^{π} f (x) c o s (k x) d x b y o n e$ $a_{k} = \frac{2}{π} \frac{1}{k^{2}}$ $S_{n} (f) (x) = - \frac{π}{3} + \sum_{n ⩾ 1} \frac{2 c o s (k x)}{k^{2}}$ $b y d i r e c h l e t f (0) = S_{n} (f) (0) \Rightarrow$ $0 = - \frac{π}{3} + \frac{2}{π} \sum_{k ⩾ 1} \frac{c o s (k . 0)}{k^{2}} \Rightarrow$ $\frac{π}{3} = \frac{2}{π} \sum_{k ⩾ 1} \frac{1}{k^{2}} \Rightarrow \sum_{k ⩾ 1} \frac{1}{k^{2}} = \frac{π^{2}}{6}$
	Commented by mathmax by abdo last updated on 20/Feb/20

	$t h a n k s s i r .$
	Commented by mind is power last updated on 21/Feb/20

	$w i t h e p l e a s u r$
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