let f(x) =xsinx ,2π periodic even developp f at Fourier serie .


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Question Number 53966 by maxmathsup by imad last updated on 27/Jan/19

$l e t f (x) = x s i n x, 2 π p e r i o d i c e v e n$ $d e v e l o p p f a t F o u r i e r s e r i e .$
Commented by maxmathsup by imad last updated on 30/Jan/19
$f(x)=(a_0 /2) +Σ_(n=1) ^∞ a_n cos(nx) and a_n =(2/T) ∫_([T]) f(x) cos(nx)dx =(2/(2π)) ∫_(−π) ^π xsin(x) cos(nx)dx =(2/π) ∫_0 ^π x sin(x)cos(nx) dx but sin(a+b) =sina cosb +cosa sinb sin(a−b)=sina cosb −cosa sinb ⇒sina cosb=(1/2){sin(a+b)+sin(a−b)} ⇒ (π/2) a_n =(1/2) ∫_0 ^π x{sin(n+1)x−sin(n−1)x}dx ⇒ π a_n =∫_0 ^π x sin(n+1)x dx −∫_0 ^π x sin(n−1)xdx let find I=∫_0 ^π x sin(αx)dx by parts I =[−(x/α)cos(αx)]_0 ^π −∫_0 ^π −(1/α) cos(αx)dx =−(1/α)[ xcos(αx)]_0 ^π +(1/α)[(1/α)sin(αx)]_0 ^π =−(π/α) cos(πα) +(1/α^2 ) sin(απ) ⇒ πa_n =−(π/(n+1))(−1)^(n+1) −(−(π/(n−1))(−1)^(n−1) ) =((π(−1)^n )/(n+1)) −((π(−1)^n )/(n−1)) =π(−1)^n {(1/(n+1)) −(1/(n−1))} =((−2π(−1)^n )/(n^2 −1)) =((2π(−1)^(n−1) )/(n^2 −1)) with n≥2 a_0 =(2/π) ∫_0 ^π xsin(x)dx =(2/π){π}=2 ⇒(a_0 /2) =1 (π/2) a_1 =∫_0 ^π xsin(2x)dx =−(π/2) ⇒a_1 =−1 ⇒ xsin(x) =2Σ_(n=2) ^∞ (((−1)^(n−1) )/(n^2 −1)) cos(nx).$
$f (x) = \frac{a_{0}}{2} + \sum_{n = 1}^{\infty} a_{n} c o s (n x) a n d a_{n} = \frac{2}{T} \int_{[T]} f (x) c o s (n x) d x$ $= \frac{2}{2 π} \int_{- π}^{π} x s i n (x) c o s (n x) d x = \frac{2}{π} \int_{0}^{π} x s i n (x) c o s (n x) d x b u t$ $s i n (a + b) = s i n a c o s b + c o s a s i n b$ $s i n (a - b) = s i n a c o s b - c o s a s i n b \Rightarrow s i n a c o s b = \frac{1}{2} {s i n (a + b) + s i n (a - b)} \Rightarrow$ $\frac{π}{2} a_{n} = \frac{1}{2} \int_{0}^{π} x {s i n (n + 1) x - s i n (n - 1) x} d x \Rightarrow$ $π a_{n} = \int_{0}^{π} x s i n (n + 1) x d x - \int_{0}^{π} x s i n (n - 1) x d x l e t f i n d$ $I = \int_{0}^{π} x s i n (α x) d x b y p a r t s I = {[- \frac{x}{α} c o s (α x)]}_{0}^{π} - \int_{0}^{π} - \frac{1}{α} c o s (α x) d x$ $= - \frac{1}{α} {[x c o s (α x)]}_{0}^{π} + \frac{1}{α} {[\frac{1}{α} s i n (α x)]}_{0}^{π} = - \frac{π}{α} c o s (π α) + \frac{1}{α^{2}} s i n (α π) \Rightarrow$ $π a_{n} = - \frac{π}{n + 1} {(- 1)}^{n + 1} - (- \frac{π}{n - 1} {(- 1)}^{n - 1}) = \frac{π {(- 1)}^{n}}{n + 1} - \frac{π {(- 1)}^{n}}{n - 1}$ $= π {(- 1)}^{n} {\frac{1}{n + 1} - \frac{1}{n - 1}} = \frac{- 2 π {(- 1)}^{n}}{n^{2} - 1} = \frac{2 π {(- 1)}^{n - 1}}{n^{2} - 1} w i t h n ⩾ 2$ $a_{0} = \frac{2}{π} \int_{0}^{π} x s i n (x) d x = \frac{2}{π} {π} = 2 \Rightarrow \frac{a_{0}}{2} = 1$ $\frac{π}{2} a_{1} = \int_{0}^{π} x s i n (2 x) d x = - \frac{π}{2} \Rightarrow a_{1} = - 1 \Rightarrow$ $x s i n (x) = 2 \sum_{n = 2}^{\infty} \frac{{(- 1)}^{n - 1}}{n^{2} - 1} c o s (n x) .$
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