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Question Number 45044 by maxmathsup by imad last updated on 07/Oct/18

l e t f (x) = x^{2}, f u n c t i o n 2 π p e r i d i c e v e n

1) d e v e l o p p f a t f o u r i e r s e r i e

2) f i n d t h e v a l u e o f \sum_{n = 1}^{\infty} \frac{1}{n^{4}}

Commented by maxmathsup by imad last updated on 08/Oct/18

f i s e v e n \Rightarrow 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) = \frac{2}{2 π} \int_{- π}^{π} x^{2} c o s (n x) d x

= \frac{2}{π} \int_{0}^{π} x^{2} c o s (n x) \Rightarrow \frac{π}{2} a_{n} = \int_{0}^{π} x^{2} c o s (n x) b y p s r t s w e g e t

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

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

= \frac{2}{n^{2}} (π {(- 1)}^{n}) - \frac{2}{n^{2}} \int_{0}^{π} c o s (n x) d x = \frac{2 π {(- 1)}^{n}}{n^{2}} \Rightarrow \frac{π}{2} a_{n} = \frac{2 π {(- 1)}^{n}}{n^{2}} \Rightarrow

a_{n} = \frac{4 {(- 1)}^{n}}{n^{2}} a l s o a_{0} = \frac{2}{π} \int_{0}^{π} x^{2} d x = \frac{2}{π} \frac{π^{3}}{3} = \frac{2 π^{2}}{3} \Rightarrow \frac{a_{0}}{2} = \frac{π^{2}}{3} \Rightarrow

★ x^{2} = \frac{π^{2}}{3} + 4 \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n^{2}} c o s (n x) ★

2) t h e p a r s e v a l f o r m u l a e g i v e \frac{1}{T} \int_{[T]} f^{2} (x) d x = {(\frac{a_{o}}{2})}^{2} + \frac{1}{2} \sum_{n ⩾ 1} (a_{n}^{2} + b_{n}^{2}) \Rightarrow

\frac{1}{2 π} \int_{- π}^{π} x^{4} d x = \frac{π^{4}}{9} + 8 \sum_{n = 1}^{\infty} \frac{1}{n^{4}} \Rightarrow \frac{1}{π} \int_{0}^{π} x^{4} d x = \frac{π^{4}}{9} + 8 \sum_{n = 1}^{\infty} \frac{1}{n^{4}} \Rightarrow

\frac{π^{4}}{5} - \frac{π^{4}}{9} = 8 \sum_{n = 1}^{\infty} \frac{1}{n^{4}} \Rightarrow \sum_{n = 1}^{\infty} \frac{1}{n^{4}} = \frac{1}{8} (\frac{4 π^{4}}{45}) = \frac{π^{4}}{90}

★ \sum_{n = 1}^{\infty} \frac{1}{n^{4}} = \frac{π^{4}}{90} ★