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Question Number 33894 by math khazana by abdo last updated on 26/Apr/18

1) l e t f R \to C 2 π p e r i o d i c e v e n / f (x) = x

\forall x \in [0, π [d e v e l o p p f a t f o u r i e r s e r i e

2) c a l c u l a t e \sum_{p = 0}^{\infty} \frac{1}{{(2 p + 1)}^{2}} .

Commented by abdo imad last updated on 28/Apr/18

f i s e v e n s o f (x) = \frac{a_{0}}{2} + \sum_{n = 1}^{\infty} a_{n} c o s (n x) w i t h

a_{n} = \frac{2}{T} \int_{[T]} f (x) c o s (n x) d x = \frac{2}{2 π} \int_{- π}^{π} x c o s (n x) d x

= \frac{2}{π} \int_{0}^{π} x c o s (n x) d x \Rightarrow \frac{π}{2} a_{n} = \int_{0}^{π} x c o s (n x) d x l e t i n t e g r a t e

b y p a r t s u^{} = x a n d v^{^{'}} = c o s (n x) \Rightarrow

\frac{π}{2} a_{n} = {(\frac{x}{n} s i n (n x)]}_{0}^{π} - \int_{0}^{π} \frac{1}{n} s i n (n x) d x

= - \frac{1}{n} \int_{0}^{π} s i n (n x) = \frac{1}{n^{2}} {[c o s (n x)]}_{0}^{π} = \frac{1}{n^{2}} ({(- 1)}^{n} - 1) \Rightarrow

a_{n} = \frac{2}{π n^{2}} ({(- 1)}^{n} - 1) \Rightarrow a_{2 n} = 0 a n d a_{2 n + 1} = \frac{- 4}{π {(2 n + 1)}^{2}}

\frac{π}{2} a_{0} = \int_{0}^{π} x d x = \frac{π^{2}}{2} \Rightarrow a_{0} = π \Rightarrow

f (x) = \frac{π}{2} - \frac{4}{π} \sum_{n = 0}^{\infty} \frac{c o s (2 n + 1) x}{{(2 n + 1)}^{2}} (d)

2) l e t t a k e x = 0 i n (d) w e g e t

0 = \frac{π}{2} - \frac{4}{π} \sum_{n = 0}^{\infty} \frac{1}{{(2 n + 1)}^{2}} \Rightarrow \frac{4}{π} \sum_{n = 0}^{\infty} \frac{1}{{(2 n + 1)}^{2}} = \frac{π}{2} \Rightarrow

\sum_{n = 0}^{\infty} \frac{1}{{(2 n + 1)}^{2}} = \frac{π^{2}}{8} .