let f(x) =x^2 , function 2π peridic even 1) developp f at fourier serie 2)find the value of Σ


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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
$f is even ⇒f(x)=(a_0 /2) +Σ_(n=1) ^∞ a_n cos(nx) with a_n =(2/T)∫_([T]) f(x)cos(nx) =(2/(2π)) ∫_(−π) ^π x^2 cos(nx)dx =(2/π) ∫_0 ^π x^2 cos(nx) ⇒(π/2)a_n = ∫_0 ^π x^2 cos(nx) by psrts we get ∫_0 ^π x^2 cos(nx)dx =[(x^2 /n)sin(nx)]_0 ^π −∫_0 ^π ((2x)/n)sin(nx)dx =−(2/n) ∫_0 ^π x sin(nx)dx =−(2/n){ [−(x/n)cos(nx)]_0 ^π −∫_0 ^π −(1/n)cos(nx)dx} =(2/n^2 ) ( π(−1)^n ) −(2/n^2 )∫_0 ^π cos(nx)dx =((2π(−1)^n )/n^2 ) ⇒(π/2)a_n =((2π(−1)^n )/n^2 ) ⇒ a_n = ((4(−1)^n )/n^2 ) also a_0 =(2/π) ∫_0 ^π x^2 dx =(2/π) (π^3 /3) =((2π^2 )/3) ⇒(a_0 /2) =(π^2 /3) ⇒ ★x^2 =(π^2 /3) +4 Σ_(n=1) ^∞ (((−1)^n )/n^2 ) cos(nx)★ 2) the parseval formulae give (1/T)∫_([T]) f^2 (x)dx =((a_o /2))^2 +(1/2)Σ_(n≥1) (a_n ^2 +b_n ^2 ) ⇒ (1/(2π)) ∫_(−π) ^π x^4 dx =(π^4 /9) +8 Σ_(n=1) ^∞ (1/n^4 ) ⇒(1/π) ∫_0 ^π x^4 dx =(π^4 /9) +8 Σ_(n=1) ^∞ (1/n^4 ) ⇒ (π^4 /5) −(π^4 /9) =8Σ_(n=1) ^∞ (1/n^4 ) ⇒ Σ_(n=1) ^∞ (1/n^4 ) =(1/8)(((4π^4 )/(45))) = (π^4 /(90)) ★ Σ_(n=1) ^∞ (1/n^4 ) =(π^4 /(90)) ★$
$f i s e v e n \Rightarrow f (x) = \frac{a_{0}}{2} + \sum_{n = 1}^{\infty} a_{n} c o s (n x) w i t h$ $a_{n} = \frac{2}{T} \int_{[T]} f (x) c o s (n x) = \frac{2}{2 π} \int_{- π}^{π} x^{2} c o s (n x) d x$ $= \frac{2}{π} \int_{0}^{π} x^{2} c o s (n x) \Rightarrow \frac{π}{2} a_{n} = \int_{0}^{π} x^{2} c o s (n x) b y p s r t s w e g e t$ $\int_{0}^{π} x^{2} c o s (n x) d x = {[\frac{x^{2}}{n} s i n (n x)]}_{0}^{π} - \int_{0}^{π} \frac{2 x}{n} s i n (n x) d x$ $= - \frac{2}{n} \int_{0}^{π} x s i n (n x) d x = - \frac{2}{n} {{[- \frac{x}{n} c o s (n x)]}_{0}^{π} - \int_{0}^{π} - \frac{1}{n} c o s (n x) d x}$ $= \frac{2}{n^{2}} (π {(- 1)}^{n}) - \frac{2}{n^{2}} \int_{0}^{π} c o s (n x) d x = \frac{2 π {(- 1)}^{n}}{n^{2}} \Rightarrow \frac{π}{2} a_{n} = \frac{2 π {(- 1)}^{n}}{n^{2}} \Rightarrow$ $a_{n} = \frac{4 {(- 1)}^{n}}{n^{2}} a l s o a_{0} = \frac{2}{π} \int_{0}^{π} x^{2} d x = \frac{2}{π} \frac{π^{3}}{3} = \frac{2 π^{2}}{3} \Rightarrow \frac{a_{0}}{2} = \frac{π^{2}}{3} \Rightarrow$ $★ x^{2} = \frac{π^{2}}{3} + 4 \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n^{2}} c o s (n x) ★$ $2) t h e p a r s e v a l f o r m u l a e g i v e \frac{1}{T} \int_{[T]} f^{2} (x) d x = {(\frac{a_{o}}{2})}^{2} + \frac{1}{2} \sum_{n ⩾ 1} (a_{n}^{2} + b_{n}^{2}) \Rightarrow$ $\frac{1}{2 π} \int_{- π}^{π} x^{4} d x = \frac{π^{4}}{9} + 8 \sum_{n = 1}^{\infty} \frac{1}{n^{4}} \Rightarrow \frac{1}{π} \int_{0}^{π} x^{4} d x = \frac{π^{4}}{9} + 8 \sum_{n = 1}^{\infty} \frac{1}{n^{4}} \Rightarrow$ $\frac{π^{4}}{5} - \frac{π^{4}}{9} = 8 \sum_{n = 1}^{\infty} \frac{1}{n^{4}} \Rightarrow \sum_{n = 1}^{\infty} \frac{1}{n^{4}} = \frac{1}{8} (\frac{4 π^{4}}{45}) = \frac{π^{4}}{90}$ $★ \sum_{n = 1}^{\infty} \frac{1}{n^{4}} = \frac{π^{4}}{90} ★$
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