let f(x) = x∣x∣ , 2π periodic odd developp f at fourier serie .


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Question Number 53963 by maxmathsup by imad last updated on 27/Jan/19

$l e t f (x) = x ∣ x ∣, 2 π p e r i o d i c o d d$ $d e v e l o p p f a t f o u r i e r s e r i e .$
Commented by maxmathsup by imad last updated on 31/Jan/19
$f(x) = Σ_(n=1) ^∞ a_n sin(nx) with a_n =(2/T) ∫_([T]) f(x)sin(nx)dx =(2/(2π)) ∫_(−π) ^π x∣x∣ sin(nx)dx =(2/π) ∫_0 ^π x^2 sin(nx)dx ⇒(π/2)a_n =∫_0 ^π x^2 sin(nx)dx by parts u =x^2 and v =sin(nx) we get ∫_0 ^π x^2 sin(nx)dx =[−(1/n)x^2 cos(nx)]_0 ^π −∫_0 ^π 2x(−(1/n)cos(nx))dx =−(π^2 /n)(−1)^n +(2/n) ∫_0 ^π x cos(nx)dx also by parts ∫_0 ^π x cos(nx)dx =[(x/n)sin(nx)]_0 ^π −∫_0 ^π (1/n)sin(nx)dx =−(1/n)[−(1/n)cos(nx)]_0 ^π =(1/n^2 )((−1)^n −1) ⇒ (π/2) a_n =−(π^2 /n)(−1)^n +(2/n^3 ){ (−1)^n −1} ⇒ a_n =−(2/π) (π^2 /n)(−1)^n +(4/(πn^3 )){ (−1)^n −1} =((2π)/n)(−1)^(n−1) +(4/(πn^3 )){ (−1)^n −1} ⇒ x ∣x∣ =2π Σ_(n=1) ^∞ (((−1)^(n−1) )/n)sin(nx) +(4/π) Σ_(n=1) ^∞ (((−1)^n −1)/n^3 ) sin(nx) ⇒ x∣x∣ =2π Σ_(n=1) ^∞ (((−1)^(n−1) )/n) sin(nx) −(8/π) Σ_(n=1) ^∞ ((sin(2n+1)x)/((2n+1)^3 )) .$
$f (x) = \sum_{n = 1}^{\infty} a_{n} s i n (n x) w i t h a_{n} = \frac{2}{T} \int_{[T]} f (x) s i n (n x) d x$ $= \frac{2}{2 π} \int_{- π}^{π} x ∣ x ∣ s i n (n x) d x = \frac{2}{π} \int_{0}^{π} x^{2} s i n (n x) d x \Rightarrow \frac{π}{2} a_{n} = \int_{0}^{π} x^{2} s i n (n x) d x$ $b y p a r t s u = x^{2} a n d v = s i n (n x) w e g e t$ $\int_{0}^{π} x^{2} s i n (n x) d x = {[- \frac{1}{n} x^{2} c o s (n x)]}_{0}^{π} - \int_{0}^{π} 2 x (- \frac{1}{n} c o s (n x)) d x$ $= - \frac{π^{2}}{n} {(- 1)}^{n} + \frac{2}{n} \int_{0}^{π} x c o s (n x) d x a l s o b y p a r t s$ $\int_{0}^{π} x c o s (n x) d x = {[\frac{x}{n} s i n (n x)]}_{0}^{π} - \int_{0}^{π} \frac{1}{n} s i n (n x) d x$ $= - \frac{1}{n} {[- \frac{1}{n} c o s (n x)]}_{0}^{π} = \frac{1}{n^{2}} ({(- 1)}^{n} - 1) \Rightarrow$ $\frac{π}{2} a_{n} = - \frac{π^{2}}{n} {(- 1)}^{n} + \frac{2}{n^{3}} {{(- 1)}^{n} - 1} \Rightarrow$ $a_{n} = - \frac{2}{π} \frac{π^{2}}{n} {(- 1)}^{n} + \frac{4}{π n^{3}} {{(- 1)}^{n} - 1} = \frac{2 π}{n} {(- 1)}^{n - 1} + \frac{4}{π n^{3}} {{(- 1)}^{n} - 1} \Rightarrow$ $x ∣ x ∣ = 2 π \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n} s i n (n x) + \frac{4}{π} \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n} - 1}{n^{3}} s i n (n x) \Rightarrow$ $x ∣ x ∣ = 2 π \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n} s i n (n x) - \frac{8}{π} \sum_{n = 1}^{\infty} \frac{s i n (2 n + 1) x}{{(2 n + 1)}^{3}} .$
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