let f(x)=ch(αx) developp f at fourier serie. (f 2π periodic even)


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Question Number 38208 by prof Abdo imad last updated on 22/Jun/18

$l e t f (x) = c h (α x)$ $d e v e l o p p f a t f o u r i e r s e r i e .$ $(f 2 π p e r i o d i c e v e n)$
Commented by prof Abdo imad last updated on 24/Jun/18
$f(x)=(a_0 /2) +Σ_(n=1) ^∞ a_n cos(nx) with a_n = (2/T) ∫_([T]) f(x)cos(nx)dx =(2/(2π)) ∫_(−π) ^π ch(αx)cos(nx)dx =(2/π) ∫_0 ^π ch(αx)cos(nx)dx ⇒ (π/2) a_n = Re( ∫_0 ^π ((e^(αx) +e^(−αx) )/2) e^(inx) dx)=Re(A_n ) A_n = (1/2) ∫_0 ^π e^((α+in)x) dx +(1/2) ∫_0 ^π e^((−α +in)x) dx 2A_n = [(1/(α+in)) e^((α+in)x) ]_0 ^π +[ (1/(−α +in)) e^((−α+in)x) ]_0 ^π =((e^(απ) (−1)^n −1)/(α+in)) + (1/(−α +in))((e^(−απ) (−1)^n −1)/1) = (({(−1)^n e^(απ) −1}(α−in))/(α^2 +n^2 )) −(({(−1)^n e^(−απ) −1}(α+in})/(α^2 +n^2 ))$
$f (x) = \frac{a_{0}}{2} + \sum_{n = 1}^{\infty} a_{n} c o s (n x) w i t h$ $a_{n} = \frac{2}{T} \int_{[T]} f (x) c o s (n x) d x$ $= \frac{2}{2 π} \int_{- π}^{π} c h (α x) c o s (n x) d x$ $= \frac{2}{π} \int_{0}^{π} c h (α x) c o s (n x) d x \Rightarrow$ $\frac{π}{2} a_{n} = R e (\int_{0}^{π} \frac{e^{α x} + e^{- α x}}{2} e^{i n x} d x) = R e (A_{n})$ $A_{n} = \frac{1}{2} \int_{0}^{π} e^{(α + i n) x} d x + \frac{1}{2} \int_{0}^{π} e^{(- α + i n) x} d x$ $2 A_{n} = {[\frac{1}{α + i n} e^{(α + i n) x}]}_{0}^{π} + {[\frac{1}{- α + i n} e^{(- α + i n) x}]}_{0}^{π}$ $= \frac{e^{α π} {(- 1)}^{n} - 1}{α + i n} + \frac{1}{- α + i n} \frac{e^{- α π} {(- 1)}^{n} - 1}{1}$ $= \frac{{{(- 1)}^{n} e^{α π} - 1} (α - i n)}{α^{2} + n^{2}} - \frac{{{(- 1)}^{n} e^{- α π} - 1} (α + i n}}{α^{2} + n^{2}}$
Commented by prof Abdo imad last updated on 25/Jun/18
$2A_n = ((α{ (−1)^n e^(απ) −1}−in{(−1)^n e^(απ) −1}−α{(−1)^n e^(−απ) −1}−in{(−1)^n e^(−απ) −1))/(α^2 +n^2 )) Re( 2A_n ) = ((α(−1)^n e^(απ) −α −α(−1)^n e^(−απ) +α)/(α^2 +n^2 )) =((α(−1)^n ( e^(απ) −e^(−απ) ))/(α^2 +n^2 )) = ((2α(−1)^n sh(απ))/(α^2 +n^2 )) ⇒ (π/2)a_n = ((α(−1)^n sh(απ))/(α^2 +n^2 )) ⇒ a_n =((2α (−1)^n sh(απ))/(π( α^2 +n^2 ))) ⇒ a_0 = ((2αsh(απ))/(πα^2 )) ⇒ (a_0 /2) = ((sh(απ))/(απ)) ⇒ ch(αx)= ((sh(πα))/(πα)) +((sh(πα))/π)Σ_(n=1) ^∞ ((2α(−1)^n )/(α^2 +n^2 ))$
$2 A_{n} = \frac{α {{(- 1)}^{n} e^{α π} - 1} - i n {{(- 1)}^{n} e^{α π} - 1} - α {{(- 1)}^{n} e^{- α π} - 1} - i n {{(- 1)}^{n} e^{- α π} - 1)}{α^{2} + n^{2}}$ $R e (2 A_{n}) = \frac{α {(- 1)}^{n} e^{α π} - α - α {(- 1)}^{n} e^{- α π} + α}{α^{2} + n^{2}}$ $= \frac{α {(- 1)}^{n} (e^{α π} - e^{- α π})}{α^{2} + n^{2}} = \frac{2 α {(- 1)}^{n} s h (α π)}{α^{2} + n^{2}} \Rightarrow$ $\frac{π}{2} a_{n} = \frac{α {(- 1)}^{n} s h (α π)}{α^{2} + n^{2}} \Rightarrow$ $a_{n} = \frac{2 α {(- 1)}^{n} s h (α π)}{π (α^{2} + n^{2})} \Rightarrow$ $a_{0} = \frac{2 α s h (α π)}{π α^{2}} \Rightarrow \frac{a_{0}}{2} = \frac{s h (α π)}{α π} \Rightarrow$ $c h (α x) = \frac{s h (π α)}{π α} + \frac{s h (π α)}{π} \sum_{n = 1}^{\infty} \frac{2 α {(- 1)}^{n}}{α^{2} + n^{2}}$
Commented by prof Abdo imad last updated on 25/Jun/18

$c h (α x) = \frac{s h (π α)}{π α} + 2 α \frac{s h (π α)}{π} \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{α^{2} + n^{2}} c o s (n x)$
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