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UnkleRhaukus

  • 2 years ago

\(F\)ourier Series

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  1. UnkleRhaukus
    • 2 years ago
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  2. UnkleRhaukus
    • 2 years ago
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    please check my work for; •mathematical •typographical •formatting errors

  3. UnkleRhaukus
    • 2 years ago
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  4. richyw
    • 2 years ago
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    man I can't wait to learn this stuff

  5. Outkast3r09
    • 2 years ago
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    i would but i'm studying for my physics final =/

  6. mahmit2012
    • 2 years ago
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    that was Fourier series. And also where is your works? That was you typed?

  7. UnkleRhaukus
    • 2 years ago
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    my work is in 1.pdf,

  8. Dido525
    • 2 years ago
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    Can't wait till I learn tall this stuff.

  9. oldrin.bataku
    • 2 years ago
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    Why wait? Basic Fourier series are rather intuitive and simple to figure out!

  10. Aylin
    • 2 years ago
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    I didn't see any mistakes, though I didn't have time to check all of the integrations. I think it's good.

  11. UnkleRhaukus
    • 2 years ago
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    thanks !

  12. Mathmuse
    • 2 years ago
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    1 a) ii bounds on bn remain -pi to pi after you pull out f(x)

  13. Mathmuse
    • 2 years ago
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    didn't see anything else wrong with the integrals. this is a great primer for fourier. i like the complex form once i saw it

  14. UnkleRhaukus
    • 2 years ago
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    Your ,... Right , thank-you so much @Mathmuse \[\begin{align*} b_n&=\frac1\pi\int\limits_{-\pi}^\pi f(x)\sin(nx)\text dx\\ &=\frac1\pi\int\limits_{\color{red}{\cancel{-\pi}}0}^\pi\sin(nx)\text dx\\ &=\dots \end{align*}\]

  15. UnkleRhaukus
    • 2 years ago
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    \[\textbf 2 (b) \]\begin{align*} f(x)&=\cos(x),\qquad0<x<3 ,\qquad=f(x+3) \end{align*} \begin{align*} % a_0 a_0&=\frac23\int\limits_{0}^{3}\cos(x)\text dx\\ &=-\frac23\sin(x)\big|_0^3\\ &=-\frac23\sin(3) \end{align*} \begin{align*} % a_n a_n&=\frac23\int\limits_{0}^{3}\cos(x)\cos\left(\frac{\pi n x}{3/2}\right)\text dx\\ &=\frac13\int\limits_{0}^{3}\cos\left((\tfrac23\pi n -1)x\right)+\cos\left((\tfrac23\pi n +1)x\right)\text dx\\ &=-\frac13\left(\frac{\sin\left((\tfrac23\pi n -1)x\right)}{\tfrac23\pi n -1}-\frac{\sin\left((\tfrac23\pi n +1)x\right)}{\tfrac23\pi n +1}\right)\Big|_0^3\\ &=\frac{\sin\left(2\pi n -3\right)}{3-2\pi n }+\frac{\sin\left(2\pi n +3\right)}{2\pi n +3}\\ &=\frac{\sin\left(3 \right)}{3-2\pi n }+\frac{\sin\left(3\right)}{2\pi n +3}\\ &=\sin\left(3 \right)\left(\frac{1}{3-2\pi n }+\frac{1}{2\pi n +3}\right)\\ &=\sin\left(3 \right)\left(\frac{3+2\pi n }{(3-2\pi n)(3+2\pi n) }+\frac{3-2\pi n}{(3+2\pi n)(3-2\pi n)}\right)\\ &=\sin\left(3 \right)\left(\frac{6}{9-4\pi^2n^2}\right) \end{align*} \begin{align*} % b_n b_n&=\frac23\int\limits_{0}^{3}\cos(x)\sin\left(\frac{\pi n x}{3/2}\right)\text dx\\ &=\frac13\int\limits_{0}^{3}\sin\left((\tfrac23\pi n+1)x\right)-\sin\left((\tfrac23\pi n-1)x\right)\text dx\\\ &=-\frac13\left(\frac{\cos\left((\tfrac23\pi n+1)x\right)}{\tfrac23\pi n+1}-\frac{\cos\left((\tfrac23\pi n-1)x\right)}{\tfrac23\pi n-1}\right)\Big|_0^3\\ &=\frac{1-\cos\left((2\pi n+3)\right)}{2\pi n+3}-\frac{1-\cos\left((2\pi n-3)\right)}{2\pi n-3}\\ &=\frac{1-\cos\left(3\right)}{2\pi n+3}-\frac{1-\cos\left(3\right)}{2\pi n-3}\\ &=\left(1-\cos\left(3\right)\right)\left(\frac{1}{2\pi n+3}-\frac{1}{2\pi n-3}\right)\\ &=\left(1-\cos\left(3\right)\right)\left(\frac{2\pi n-3}{(2\pi n+3)(2\pi n-3)}-\frac{2\pi n+3}{(2\pi n-3)(2\pi n+3)}\right)\\ &=\left(1-\cos\left(3\right)\right)\left(\frac{6}{4\pi^2n^2-9}\right)\\ %&=\left(1-\cos\left(3\right)\right)\left(\frac{4\pi n}{4\pi^2n^2-9}\right)\\ \end{align*} \begin{align*} % S(x)\\ S(x)&=-\frac{\sin(3)}3+\sin\left(3 \right)\sum\limits_{n=1}^\infty\left(\frac{6}{9-4\pi^2n^2}\right)\cos\left(\tfrac23\pi n x\right)\\ &\qquad\qquad\qquad+\left(1-\cos\left(3\right)\right)\sum\limits_{n=1}^\infty\left(\frac{4\pi n}{4\pi^2n^2-9}\right)\sin\left(\tfrac23\pi n x\right)\\ &=\\ &=? \end{align*}

  16. UnkleRhaukus
    • 2 years ago
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    im going wrong somewhere here, because the correct Series should be \[S(x)=\frac{\sin(3)}3+6\sin(3)\sum\limits_{n=1}^\infty\frac{\cos(\tfrac23n\pi x)}{9-16n^2\pi^2}+4\pi(1-\cos(3))\sum\limits_{n=1}^\infty\frac{n\sin(\tfrac23n\pi x)}{4\pi^2n^2-9}\]

  17. malevolence19
    • 2 years ago
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    Let me look over it.

  18. malevolence19
    • 2 years ago
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    Okay \[=-\frac13\left(\frac{\cos\left((\tfrac23\pi n+1)x\right)}{\tfrac23\pi n+1}-\frac{\cos\left((\tfrac23\pi n-1)x\right)}{\tfrac23\pi n-1}\right)\Big|_0^3\] Since you integrate both sines they both acquire a negative so when you pull out the negative shouldn't that middle one be positive?

  19. malevolence19
    • 2 years ago
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    because when you redistribute it the signs are different.

  20. UnkleRhaukus
    • 2 years ago
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    it turns out i didn't use the trig formula correctly for a_n this somehow didnt matter , but for b_n it did should be like this \[\begin{align*} % a_n a_n&=\frac23\int\limits_{0}^{3}\cos(x)\cos\left(\frac{\pi n x}{3/2}\right)\text dx\\ &=\frac13\int\limits_{0}^{3}\cos\left((1-\tfrac23\pi n )x\right)+\cos\left((1+\tfrac23\pi n)x\right)\text dx\\ &=-\frac13\left.\left(\frac{\sin\left((1-\tfrac23\pi n )x\right)}{1-\tfrac23\pi n }-\frac{\sin\left((1+\tfrac23\pi n )x\right)}{1+\tfrac23\pi n}\right)\right|_0^3\\ &=\frac{\sin\left(3-2\pi n \right)}{3-2\pi n }+\frac{\sin\left(3+2\pi n \right)}{3+2\pi n }\\ &=\frac{\sin\left(3 \right)}{3-2\pi n }+\frac{\sin\left(3\right)}{2\pi n +3}\\ &=\sin\left(3 \right)\left(\frac{1}{3-2\pi n }+\frac{1}{3+2\pi n}\right)\\ &=\sin\left(3 \right)\left(\frac{3+2\pi n }{(3-2\pi n)(3+2\pi n) }+\frac{3-2\pi n}{(3+2\pi n)(3-2\pi n)}\right)\\ &=\left(\frac{6\sin\left(3 \right)}{9-4\pi^2n^2}\right)\\ &=\left(\frac{6\sin\left(3 \right)}{9-4\pi^2n^2}\right) \end{align*}\] \[\begin{align*} % b_n b_n&=\frac23\int\limits_{0}^{3}\cos(x)\sin\left(\frac{\pi n x}{3/2}\right)\text dx\\ &=\frac13\int\limits_{0}^{3}\sin\left((1+\tfrac23\pi n)x\right)-\sin\left((1-\tfrac23\pi n)x\right)\text dx\\\ &=-\frac13\left.\left(\frac{\cos\left((1+\tfrac23\pi n)x\right)}{1+\tfrac23\pi n}-\frac{\cos\left((1-\tfrac23\pi n)x\right)}{1-\tfrac23\pi n}\right)\right|_0^3\\ &=\frac{1-\cos\left(3+2\pi n\right)}{2\pi n+3}-\frac{1-\cos\left(3-2\pi n\right)}{2\pi n-3}\\ &=\frac{1-\cos\left(3\right)}{3+2\pi n}-\frac{1-\cos\left(3\right)}{3-2\pi n}\\ &=\left(1-\cos\left(3\right)\right)\left(\frac{1}{3+2\pi n}-\frac{1}{3-2\pi n}\right)\\ &=\left(1-\cos\left(3\right)\right)\left(\frac{3-2\pi n}{(3+2\pi n)(3-2\pi n)}-\frac{3+2\pi n}{(3-2\pi n)(3+2\pi n)}\right)\\ &=\left(1-\cos\left(3\right)\right)\left(\frac{-4\pi n}{9-4\pi^2n^2}\right)\\ &=\left(\frac{4\pi\left(1-\cos\left(3\right)\right)n}{4\pi^2n^2-9}\right)\\ \end{align*}\]

  21. UnkleRhaukus
    • 2 years ago
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    i think

  22. UnkleRhaukus
    • 2 years ago
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    how did i get a minus for a_0? lolsz

  23. UnkleRhaukus
    • 2 years ago
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    so\[S(x)=\frac{\sin(3)}3+6\sin(3)\sum\limits_{n=1}^\infty\frac{\cos(\tfrac23n\pi x)}{9-4n^2\pi^2}+4\pi(1-\cos(3))\sum\limits_{n=1}^\infty\frac{n\sin(\tfrac23n\pi x)}{4\pi^2n^2-9}\checkmark\]

  24. UnkleRhaukus
    • 2 years ago
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