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UnkleRhaukus

  • 2 years ago

Find the Fourier series for each of the following periodic function\[ f(x)=\cos(x),\qquad0<x<3,\qquad=f(x+3)\]

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  1. UnkleRhaukus
    • 2 years ago
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    \[\begin{align*} a_0&=\frac23\int\limits_{0}^{3}\cos(x)\text dx\\ &=-\frac23\sin(x)\big|_0^3\\ &=-\frac23\sin(3) \end{align*}\]

  2. UnkleRhaukus
    • 2 years ago
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    \[\begin{align*} a_n&=\frac23\int\limits_{0}^{3}\cos(x)\cos\left(\frac{\pi n x}{3/2}\right)\text dx\\ &=\frac13\int\limits_{0}^{3}\cos\left((\tfrac23\pi n -1)x\right)+\cos\left((\tfrac23\pi n +1)x\right)\text dx\\ &=-\frac13\left(\frac{\sin\left((\tfrac23\pi n -1)x\right)}{\tfrac23\pi n -1}-\frac{\sin\left((\tfrac23\pi n +1)x\right)}{\tfrac23\pi n +1}\right)\Big|_0^3\\ &=\frac{\sin\left(2\pi n -3\right)}{3-2\pi n }+\frac{\sin\left(2\pi n +3\right)}{2\pi n +3}\\ &=\frac{\sin\left(-3 \right)}{3-2\pi n }+\frac{\sin\left(3\right)}{2\pi n +3}\\ &=\sin\left(3 \right)\left(\frac{-1}{3-2\pi n }+\frac{1}{2\pi n +3}\right)\\ &=\sin\left(3 \right)\left(\frac{-3-2\pi n }{(3-2\pi n)(3+2\pi n) }+\frac{3-2\pi n}{(3+2\pi n)(3-2\pi n)}\right)\\ &=\sin\left(3 \right)\left(\frac{-4\pi n}{-4\pi^2n^2}\right) \end{align*}\]

  3. UnkleRhaukus
    • 2 years ago
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    \[\begin{align*} b_n&=\frac23\int\limits_{0}^{3}\cos(x)\sin\left(\frac{\pi n x}{3/2}\right)\text dx\\ &=\frac13\int\limits_{0}^{3}\sin\left((\tfrac23\pi n+1)x\right)-\sin\left((\tfrac23\pi n-1)x\right)\text dx\\\ &=-\frac13\left(\frac{\cos\left((\tfrac23\pi n+1)x\right)}{\tfrac23\pi n+1}-\frac{\cos\left((\tfrac23\pi n-1)x\right)}{\tfrac23\pi n-1}\right)\Big|_0^3\\ &=\frac{1-\cos\left((2\pi n+3)\right)}{2\pi n+3}-\frac{1-\cos\left((2\pi n-3)\right)}{2\pi n-3}\\ &=\frac{1-\cos\left(3\right)}{2\pi n+3}-\frac{1-\cos\left(3\right)}{2\pi n-3}\\ &=\left(1-\cos\left(3\right)\right)\left(\frac{1}{2\pi n+3}-\frac{1}{2\pi n-3}\right)\\ &=\left(1-\cos\left(3\right)\right)\left(\frac{2\pi n-3}{(2\pi n+3)(2\pi n-3)}-\frac{2\pi n+3}{(2\pi n-3)(2\pi n+3)}\right)\\ &=\left(1-\cos\left(3\right)\right)\left(\frac{-6}{4\pi^2n^2-9}\right)\\ \end{align*}\]

  4. UnkleRhaukus
    • 2 years ago
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    i can see mistakes already

  5. UnkleRhaukus
    • 2 years ago
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    got there in the end, (with help)

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