let f(x) =e^(−iαx) ,2π periodic .developp f at fourier serie.


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Question Number 68035 by mathmax by abdo last updated on 03/Sep/19

$l e t f (x) = e^{- i α x}, 2 π p e r i o d i c . d e v e l o p p f a t f o u r i e r s e r i e .$
Commented by mathmax by abdo last updated on 08/Sep/19
$f(x) =Σ_(n=−∞) ^(+∞) a_n e^(inwx) with w =((2π)/T) =1 ⇒ f(x)=Σ_(n=−∞) ^(+∞) a_n e^(inx) and a_n =(1/T)∫_(−(T/2)) ^(T/2) f(x)e^(−inx) dx =(1/(2π)) ∫_(−π) ^π e^(−iαx) e^(−inx) dx =(1/(2π)) ∫_(−π) ^π e^(−i(α+n)x) dx ⇒ 2π a_n =∫_(−π) ^π e^(−i(α+n)x) dx =[(1/(−i(α+n)))e^(−i(α+n)x) ]_(−π) ^π =((−1)/(i(α+n))){ e^(−i(α+n)π) −e^(−i(α+n)(−π)) } =(i/(α+n)){ (−1)^n e^(−iπα) −(−1)^n e^(iπα) } =((−i(−1)^n )/(α+n)){ e^(iπα) −e^(−iπα) } =((−i(−1)^n )/(n+α)) (2isin(πα)) =((2(−1)^n sin(πα))/(n+α)) ⇒ e^(−iαx) =Σ_(n=−∞) ^(+∞) ((2(−1)^n sin(πα))/(n+α)) e^(inx) =2sin(πα) Σ_(n=−∞) ^(+∞) (((−1)^n )/(n+α)) e^(inx) .$
$f (x) = \sum_{n = - \infty}^{+ \infty} a_{n} e^{i n w x} w i t h w = \frac{2 π}{T} = 1 \Rightarrow$ $f (x) = \sum_{n = - \infty}^{+ \infty} a_{n} e^{i n x} a n d a_{n} = \frac{1}{T} \int_{- \frac{T}{2}}^{\frac{T}{2}} f (x) e^{- i n x} d x$ $= \frac{1}{2 π} \int_{- π}^{π} e^{- i α x} e^{- i n x} d x = \frac{1}{2 π} \int_{- π}^{π} e^{- i (α + n) x} d x \Rightarrow$ $2 π a_{n} = \int_{- π}^{π} e^{- i (α + n) x} d x = {[\frac{1}{- i (α + n)} e^{- i (α + n) x}]}_{- π}^{π}$ $= \frac{- 1}{i (α + n)} {e^{- i (α + n) π} - e^{- i (α + n) (- π)}}$ $= \frac{i}{α + n} {{(- 1)}^{n} e^{- i π α} - {(- 1)}^{n} e^{i π α}}$ $= \frac{- i {(- 1)}^{n}}{α + n} {e^{i π α} - e^{- i π α}} = \frac{- i {(- 1)}^{n}}{n + α} (2 i s i n (π α))$ $= \frac{2 {(- 1)}^{n} s i n (π α)}{n + α} \Rightarrow$ $e^{- i α x} = \sum_{n = - \infty}^{+ \infty} \frac{2 {(- 1)}^{n} s i n (π α)}{n + α} e^{i n x}$ $= 2 s i n (π α) \sum_{n = - \infty}^{+ \infty} \frac{{(- 1)}^{n}}{n + α} e^{i n x} .$
Commented by mathmax by abdo last updated on 08/Sep/19

$2 π a_{n} = \frac{2 {(- 1)}^{n} s i n (π α)}{n + α} \Rightarrow a_{n} = \frac{s i n (π α)}{π} \frac{{(- 1)}^{n}}{n + α} \Rightarrow$ $e^{- i α x} = \frac{s i n (π α)}{π} \sum_{n = - \infty}^{+ \infty} \frac{{(- 1)}^{n}}{n + α} e^{i n x}$
Commented by mathmax by abdo last updated on 08/Sep/19
$remark we have Σ_(n=−∞) ^(+∞) (((−1)^n )/(n+α))e^(inx) =Σ_(n=−∞) ^o (((−1)^n )/(n+α))e^(inx) +Σ_(n=1) ^∞ (((−1)^n )/(n+α)) e^(inx) =(1/α) +Σ_(n=1) ^∞ (((−1)^n )/(α−n))e^(−inx) +Σ_(n=1) ^∞ (((−1)^n )/(n+α)) e^(inx) =(1/α) +Σ_(n=1) ^∞ (((−1)^n )/(α−n))(cos(nx)−isin(nx))+Σ_(n=1) ^∞ (((−1)^n )/(n+α))(cos(nx)+isin(nx)) =(1/α)+Σ_(n=1) ^∞ (−1)^n cos(nx){(1/(α−n))+(1/(α+n))} +iΣ_(n=1) ^∞ (−1)^n sin(nx){(1/(α+n))−(1/(α−n))} =(1/α) +Σ_(n=1) ^∞ (((−1)^n 2α)/(α^2 −n^2 )) cos(nx) −i Σ_(n=1) ^∞ (((−1)^n (2n))/(α^2 −n^2 )) =cos(αx)−isin(αx) ⇒ cos(αx) =(1/α) +2α Σ_(n=1) ^∞ (((−1)^n cos(nx))/(α^2 −n^2 )) and sin(αx) =2 Σ_(n=1) ^∞ ((n(−1)^n )/(α^2 −n^2 ))$
$r e m a r k w e h a v e \sum_{n = - \infty}^{+ \infty} \frac{{(- 1)}^{n}}{n + α} e^{i n x}$ $= \sum_{n = - \infty}^{o} \frac{{(- 1)}^{n}}{n + α} e^{i n x} + \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n + α} e^{i n x}$ $= \frac{1}{α} + \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{α - n} e^{- i n x} + \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n + α} e^{i n x}$ $= \frac{1}{α} + \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{α - n} (c o s (n x) - i s i n (n x)) + \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n + α} (c o s (n x) + i s i n (n x))$ $= \frac{1}{α} + \sum_{n = 1}^{\infty} {(- 1)}^{n} c o s (n x) {\frac{1}{α - n} + \frac{1}{α + n}}$ $+ i \sum_{n = 1}^{\infty} {(- 1)}^{n} s i n (n x) {\frac{1}{α + n} - \frac{1}{α - n}}$ $= \frac{1}{α} + \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n} 2 α}{α^{2} - n^{2}} c o s (n x) - i \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n} (2 n)}{α^{2} - n^{2}}$ $= c o s (α x) - i s i n (α x) \Rightarrow$ $c o s (α x) = \frac{1}{α} + 2 α \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n} c o s (n x)}{α^{2} - n^{2}} a n d$ $s i n (α x) = 2 \sum_{n = 1}^{\infty} \frac{n {(- 1)}^{n}}{α^{2} - n^{2}}$
Commented by mathmax by abdo last updated on 08/Sep/19
$error at final line e^(−iαx) =((sin(απ))/π){(1/α)+2α Σ_(n=1) ^∞ (((−1)^n cos(nx))/(𝛂^2 −n^2 ))} −i((sin(απ))/π) Σ_(n=1) ^∞ (((−1)^n (2n))/(α^2 −n^2 )) =cos(αx)−isin(αx) ⇒ cos(αx) =((sin(απ))/(απ)) +((2α)/π) sin(απ)Σ_(n=1) ^∞ (((−1)^n cos(nx))/(α^2 −n^2 )) sin(αx)=((2sin(απ))/π)Σ_(n=1) ^∞ (((−1)^n )/(α^2 −n^2 ))$
$e r r o r a t f i n a l l i n e$ $e^{- i α x} = \frac{s i n (α π)}{π} {\frac{1}{α} + 2 α \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n} c o s (n x)}{α^{2} - n^{2}}}$ $- i \frac{s i n (α π)}{π} \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n} (2 n)}{α^{2} - n^{2}} = c o s (α x) - i s i n (α x) \Rightarrow$ $c o s (α x) = \frac{s i n (α π)}{α π} + \frac{2 α}{π} s i n (α π) \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n} c o s (n x)}{α^{2} - n^{2}}$ $s i n (α x) = \frac{2 s i n (α π)}{π} \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{α^{2} - n^{2}}$
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