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UnkleRhaukus

  • 2 years ago

\(f(x)\) as a sine series

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  1. UnkleRhaukus
    • 2 years ago
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    \[ \qquad\text{\(f(x)\) as a sine series} \begin{equation*} % f(x) f(x)=1,\qquad0<x<\pi \end{equation*}\]

  2. UnkleRhaukus
    • 2 years ago
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    \[\begin{equation*} % g(x) g(x)= \begin{cases} 1,&0<x<\pi\\ -1,&-\pi<x<0 \end{cases}\qquad\text{odd extension of \(f(x)\)} \end{equation*}\] \begin{equation*} % a_0, a_n a_0=a_n=0\qquad\text{odd function} \end{equation*} \begin{align*} % b_n b_n&=\frac1\pi\int\limits_{-\pi}^{\pi} g(x)\sin(n x)\,\text dx\\ &=\frac2\pi\int\limits_0^\pi \sin(n x)\,\text dx\qquad\text{even integrand}\\ &=\frac2\pi\left(\frac{-\cos(n x)}{n}\right)\Big|_0^\pi\\ &=-\frac2\pi\left(\frac{\cos(0)-\cos(n\pi )}{n}\right)\\ &=-\frac2{\pi}\left(\frac{1-(-1)^n}{n}\right)\\ \end{align*} \begin{align*} S(x)&=-\frac2\pi\sum\limits_{n=1}^\infty\left(\frac{1-(-1)^n}{n}\right)\sin(nx)\\ &=-\frac2\pi\sum\limits_{n=1,3,5,\dots}^\infty\left(\frac{2}{n}\right)\sin(nx)\\ &=-\frac4\pi\sum\limits_{r=1}^\infty\frac{\sin\big((2r-1)x\big)}{2r-1} \end{align*}

  3. UnkleRhaukus
    • 2 years ago
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  4. experimentX
    • 2 years ago
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    just change L to pi http://mathworld.wolfram.com/FourierSeriesSquareWave.html

  5. UnkleRhaukus
    • 2 years ago
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  6. experimentX
    • 2 years ago
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    you got the opposite one .. shift the period by +pi

  7. UnkleRhaukus
    • 2 years ago
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    where did i go wrong?

  8. experimentX
    • 2 years ago
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    don't know .. don't have much time right now!! but i often use that link as reference!!

  9. experimentX
    • 2 years ago
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    sin(nx) is not even .. get rid of that minus

  10. experimentX
    • 2 years ago
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    i'll see later!!

  11. UnkleRhaukus
    • 2 years ago
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    why does (4) on that link have a sin^2 term?

  12. UnkleRhaukus
    • 2 years ago
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    ah i found my mistake

  13. UnkleRhaukus
    • 2 years ago
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    \[\begin{align*} % b_n b_n&=\frac1\pi\int\limits_{-\pi}^{\pi} g(x)\sin(n x)\,\text dx\\ &=\frac2\pi\int\limits_0^\pi \sin(n x)\,\text dx\qquad\text{even integrand}\\ &=\frac2\pi\left(\frac{-\cos(n x)}{n}\right)\Big|_0^\pi\\ &=\frac2\pi\left(\frac{\cos(0)-\cos(n\pi )}{n}\right)\\ &=\frac2{\pi}\left(\frac{1-(-1)^n}{n}\right)\\ \end{align*}\]

  14. UnkleRhaukus
    • 2 years ago
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    \begin{align*} S(x)&=\frac2\pi\sum\limits_{n=1}^\infty\left(\frac{1-(-1)^n}{n}\right)\sin(nx)\\ &=\frac4\pi\sum\limits_{n=1,3,5,\dots}^\infty\frac{\sin(nx)}{n}\\ &=\frac4\pi\sum\limits_{r=1}^\infty\frac{\sin\big((2r-1)x\big)}{2r-1} \end{align*}

  15. UnkleRhaukus
    • 2 years ago
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    just a minus sign

  16. UnkleRhaukus
    • 2 years ago
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  17. Echdip
    • 2 years ago
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  18. Echdip
    • 2 years ago
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    guy please help me with this. This is impossible tough for me

  19. UnkleRhaukus
    • 2 years ago
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    i havent studies solving PDEs yet sorry @Echdip

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