1-find-a-and-b-wich-verify-0-pi-at-2-bt-cos-nx-1-n-2-2-find-the-value-of-n-1-1-n-2-

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!} - \dots

(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}}) \dots

(2) = (3)

1 - \frac{x^{2}}{3!} + \frac{x^{4}}{5!} - \frac{x^{6}}{7!} + \dots = (1 - \frac{x^{2}}{π^{2}}) (1 - \frac{x^{2}}{4 π^{2}}) \dots

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}} + \dots)

- \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 \dots

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

\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