let f(x) =(1/(1+∣sinx∣)) (2π periodic even) developp f at fourier serie .


Question and Answers Forum

All Questions Topic List
Integration Questions
Previous in All Question Next in All Question
Previous in Integration Next in Integration
Question Number 39039 by maxmathsup by imad last updated on 01/Jul/18

$l e t f (x) = \frac{1}{1 + ∣ s i n x ∣} (2 π p e r i o d i c e v e n)$ $d e v e l o p p f a t f o u r i e r s e r i e .$
Commented by abdo mathsup 649 cc last updated on 05/Jul/18
$f(x)= (a_0 /2) +Σ_(n=1) ^∞ a_n cos(nx) with a_n = (2/T) ∫_([T]) f(x)cos(nx)dx =(2/(2π)) ∫_(−π) ^π (1/(1 +∣sinx∣)) cos(nx)dx =(2/π) ∫_0 ^π ((cos(nx))/(1+sinx))dx ⇒(π/2)a_n = ∫_0 ^π ((cos(nx))/(1+sinx))dx but ∫_0 ^π ((cos(nx))/(1+sinx))dx =Re( ∫_0 ^π (e^(inx) /(1+sinx))dx) changement x=2t give ∫_0 ^π (e^(inx) /(1+sinx))dx = ∫_0 ^(2π) ((2e^(i2nt) )/(1+sin(2t)))dt also chang. e^(it) =z give ∫_0 ^(2π) ((2 e^(i2nt) )/(1+sin(2t)))dt = ∫_(∣z∣=1) ((2 z^(2n) )/(1+((z^2 −z^(−2) )/(2i)))) (dz/(iz)) = ∫_(∣z∣=1) ((4i z^(2n) )/(iz( 2i+z^2 −z^(−2) )))dz = ∫_(∣z∣=1) ((4z^(2n−1) )/(2i +z^2 −(1/z^2 ))) dz = ∫_(∣z∣=1) ((4 z^(2n+1) )/(2iz^2 +z^4 −1))dz let ϕ(z) = ((4z^(2n+1) )/(z^4 +2iz^2 −1)) ϕ(z) = ((4z^(2n+1) )/((z^2 +i)^2 )) = ((4 z^(2n+1) )/((z−(√(−i)))^2 (2+(√(−i)))^2 )) = ((4 z^(2n+1) )/((z −e^(−((iπ)/4)) )^2 (z +e^(−((iπ)/4)) )^2 )) so ϕ have a double poles +^(− ) e^(−((iπ)/4)) ∫_(∣z∣=1) ϕ(2)dz =2iπ{ Res(ϕ,e^(−((iπ)/4)) ) +Res(ϕ,−e^(−((iπ)/4)) )}$
$f (x) = \frac{a_{0}}{2} + \sum_{n = 1}^{\infty} a_{n} c o s (n x) w i t h$ $a_{n} = \frac{2}{T} \int_{[T]} f (x) c o s (n x) d x$ $= \frac{2}{2 π} \int_{- π}^{π} \frac{1}{1 + ∣ s i n x ∣} c o s (n x) d x$ $= \frac{2}{π} \int_{0}^{π} \frac{c o s (n x)}{1 + s i n x} d x \Rightarrow \frac{π}{2} a_{n} = \int_{0}^{π} \frac{c o s (n x)}{1 + s i n x} d x$ $b u t \int_{0}^{π} \frac{c o s (n x)}{1 + s i n x} d x = R e (\int_{0}^{π} \frac{e^{i n x}}{1 + s i n x} d x)$ $c h a n g e m e n t x = 2 t g i v e$ $\int_{0}^{π} \frac{e^{i n x}}{1 + s i n x} d x = \int_{0}^{2 π} \frac{2 e^{i 2 n t}}{1 + s i n (2 t)} d t a l s o$ $c h a n g . e^{i t} = z g i v e$ $\int_{0}^{2 π} \frac{2 e^{i 2 n t}}{1 + s i n (2 t)} d t = \int_{∣ z ∣= 1} \frac{2 z^{2 n}}{1 + \frac{z^{2} - z^{- 2}}{2 i}} \frac{d z}{i z}$ $= \int_{∣ z ∣= 1} \frac{4 i z^{2 n}}{i z (2 i + z^{2} - z^{- 2})} d z$ $= \int_{∣ z ∣= 1} \frac{4 z^{2 n - 1}}{2 i + z^{2} - \frac{1}{z^{2}}} d z$ $= \int_{∣ z ∣= 1} \frac{4 z^{2 n + 1}}{2 {i z}^{2} + z^{4} - 1} d z l e t$ $φ (z) = \frac{4 z^{2 n + 1}}{z^{4} + 2 {i z}^{2} - 1}$ $φ (z) = \frac{4 z^{2 n + 1}}{{(z^{2} + i)}^{2}} = \frac{4 z^{2 n + 1}}{{(z - \sqrt{- i})}^{2} {(2 + \sqrt{- i})}^{2}}$ $= \frac{4 z^{2 n + 1}}{{(z - e^{- \frac{i π}{4}})}^{2} {(z + e^{- \frac{i π}{4}})}^{2}} s o φ h a v e a d o u b l e$ $p o l e s \bar{+} e^{- \frac{i π}{4}}$ $\int_{∣ z ∣= 1} φ (2) d z = 2 i π {R e s (φ, e^{- \frac{i π}{4}}) + R e s (φ, - e^{- \frac{i π}{4}})}$
Commented by abdo mathsup 649 cc last updated on 05/Jul/18
$Res(ϕ, e^(−((iπ)/4)) )=lim_(z→ e^(−((iπ)/4)) ) { (z −e^(−((iπ)/4)) )^2 ϕ(z)}^((1)) =lim_(z→ e^(−((iπ)/4)) ) { ((4 z^(2n+1) )/((z+e^(−((iπ)/4)) )^2 ))}^((1)) =lim_(z→e^(−((iπ)/4)) ) 4 (((2n+1)z^(2n) (z +e^(−((iπ)/4)) )^2 −2(z+e^(−((iπ)/4)) )z^(2n+1) )/((z +e^(−((iπ)/4)) )^4 )) =lim_(z→e^(−((iπ)/4)) ) 4 (((2n+1)z^(2n) (z +e^(−((iπ)/4)) )−2 z^(2n+1) )/((z +e^(−((iπ)/4)) )^3 )) =4(((2n+1) e^(−((inπ)/2)) (2 e^(−((iπ)/4)) ) −2 e^(−i(((2n+1)π)/4)) )/((2e^(−((iπ)/4)) )^3 )) =8(((2n+1) e^(−i((n/2) +(1/4))π) −e^(−i(((2n+1)π)/4)) )/(8 e^(−i((3π)/4)) )) = ((2n e^(−i(((2n+1)π)/4)) )/e^(−i((3π)/4)) ) =2n e^(−i{((2n+1)/4)+(3/4)}π) =2n e^(−i (((n+2)π)/2)) =−2n e^(−((inπ)/2)) =−2n{cos(((nπ)/2))−i sin(((nπ)/2))}$
$R e s (φ, e^{- \frac{i π}{4}}) = {l i m}_{z \to e^{- \frac{i π}{4}}} {{(z - e^{- \frac{i π}{4}})}^{2} φ (z)}^{(1)}$ $= {l i m}_{z \to e^{- \frac{i π}{4}}} {\frac{4 z^{2 n + 1}}{{(z + e^{- \frac{i π}{4}})}^{2}}}^{(1)}$ $= {l i m}_{z \to e^{- \frac{i π}{4}}} 4 \frac{(2 n + 1) z^{2 n} {(z + e^{- \frac{i π}{4}})}^{2} - 2 (z + e^{- \frac{i π}{4}}) z^{2 n + 1}}{{(z + e^{- \frac{i π}{4}})}^{4}}$ $= {l i m}_{z \to e^{- \frac{i π}{4}}} 4 \frac{(2 n + 1) z^{2 n} (z + e^{- \frac{i π}{4}}) - 2 z^{2 n + 1}}{{(z + e^{- \frac{i π}{4}})}^{3}}$ $= 4 \frac{(2 n + 1) e^{- \frac{i n π}{2}} (2 e^{- \frac{i π}{4}}) - 2 e^{- i \frac{(2 n + 1) π}{4}}}{{(2 e^{- \frac{i π}{4}})}^{3}}$ $= 8 \frac{(2 n + 1) e^{- i (\frac{n}{2} + \frac{1}{4}) π} - e^{- i \frac{(2 n + 1) π}{4}}}{8 e^{- i \frac{3 π}{4}}}$ $= \frac{2 n e^{- i \frac{(2 n + 1) π}{4}}}{e^{- i \frac{3 π}{4}}} = 2 n e^{- i {\frac{2 n + 1}{4} + \frac{3}{4}} π}$ $= 2 n e^{- i \frac{(n + 2) π}{2}} = - 2 n e^{- \frac{i n π}{2}}$ $= - 2 n {c o s (\frac{n π}{2}) - i s i n (\frac{n π}{2})}$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum