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UnkleRhaukus

  • 3 years ago

Check my work/? \[m,n\in \mathbb N,\qquad m-n=p,\quad m+n=q\] \[\delta_{a,b}=\begin{cases} 0&a\neq b\\1&a=b\end{cases}\]

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  1. UnkleRhaukus
    • 3 years ago
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    (a) \[\begin{align*} \int\limits_{-\pi}^\pi\sin(mx)\text dx &=0\qquad (\text{odd integrand})\\ \end{align*}\] (b) \[\begin{align*} \int\limits_{-\pi}^\pi\cos(mx)\text dx &=2\int\limits_0^\pi\cos(mx)\text dx\qquad(\text{even integrand})\\ &=2\frac{\sin(mx)}{m}\Big|_0^\pi\\ &=2\frac{\sin(\pi m)}m\\ \\ \text{I for }m=0\qquad&=2\lim\limits_{m\rightarrow 0}\frac{\sin(\pi m)}m\\ &\stackrel{\text{l'H}}=2\lim\limits_{m\rightarrow 0}\pi\frac{\cos(m)}1\\ &=2\pi\\\ \\ \text{II for }m\neq0\qquad&=\frac{2\sin(m\pi)}m\\ &=\frac0m\\ &=0\\ \\&=2\pi\delta_{m,0} \end{align*}\]

  2. UnkleRhaukus
    • 3 years ago
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    (c) \[\begin{align*} \int\limits_{-\pi}^\pi\cos(mx)\sin(nx)\text dx=0\qquad (\text{odd integrand}) \end{align*}\] (d) \[\begin{align*} \int\limits_{-\pi}^\pi\cos(mx)\cos(nx)\text dx&= \int\limits_0^\pi\cos\left((m-n)x\right)+\cos\left((m+n)x\right)\text dx\qquad (\text{even integrand})\\ &=\frac{\sin\left((m-n)x\right)}{m-n}+\frac{\sin((m+n)x)}{m+n}\Big|_0^\pi\\ &=\frac{\sin\left((m-n)\pi\right)}{m-n}+\frac{\sin((m+n)\pi)}{m+n}\\ \\ \text{I for }m\neq n\qquad &=\frac{\sin\left(p\pi\right)}{p}+\frac{\sin(q\pi)}{q}\\ &=\frac0p+\frac0q\\ &=0\\ \\ \text{II for }m=n\neq0\qquad &=\lim\limits_{p\rightarrow 0}\frac{\sin\left(p\pi\right)}{p}+\frac{\sin(2m\pi)}{2m}\\ &\stackrel{\text{l'H}}=\pi\lim\limits_{p\rightarrow 0}\frac{\cos(p)}1+\frac{0}{2m}\\ &=\pi+0\\ &=\pi\\ \\ \text{III for }m=n=0\qquad&=\lim\limits_{p\rightarrow 0}\frac{\sin\left(p\pi\right)}{p}+\frac{\sin(2m\pi)}{2m}\\ &\stackrel{\text{l'H}}=\pi\lim\limits_{p\rightarrow 0}\frac{\cos\left(p\pi\right)}1+2\pi\frac{\cos(2m\pi)}{2}\\ &=\pi+\pi\\ &=2\pi\\ \\ &=\pi\delta_{m,n}(1+\delta_{m,0}) \end{align*}\]

  3. UnkleRhaukus
    • 3 years ago
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    (e) \[\begin{align*} \int\limits_{-\pi}^\pi\sin(mx)\sin(nx)\text dx &=\frac12\int\limits_{-\pi}^\pi\cos((m-n)x)-\cos((m+n)x)\text dx\\ &=\int\limits_{0}^\pi\cos((m-n)x)-\cos((m+n)x)\text dx\qquad (\text{even integrand})\\\\ &=\frac{\sin((m-n)x)}{m-n}-\frac{\sin((m+n)x)}{m+n}\Big|_0^\pi\\ &=\frac{\sin((m-n)\pi)}{m-n}-\frac{\sin((m+n)\pi)}{m+n}\\ \\ \text{I for }m\neq n\qquad&=\frac{\sin(p\pi)}{p}-\frac{\sin(q\pi)}{q}\\ &=\frac 0p-\frac0q\\ &=0\\ \\ \text{II for }m=n=0\qquad&=\lim\limits_{p\rightarrow 0}\frac{\sin(p\pi)}{p}-\frac{\sin(2m\pi)}{2m}\\ &\stackrel{l'H}=\pi\frac{\cos(p\pi)}{1}-2\pi\frac{\cos(2m\pi)}{2}\\ &=\pi-\frac{2\pi}2\\ &=0\\ \\ \text{III for }m=n\neq0\qquad&=\lim\limits_{p\rightarrow 0}\frac{\sin(p\pi)}{p}-\frac{\sin(q\pi)}{q}\\ &\stackrel{l'H}=\pi\lim\limits_{p\rightarrow 0}\frac{\cos(p\pi)}{1}-\frac 0q\\ &=\pi-0\\ &=\pi\\ \\ &=\pi\delta_{m,n}(1-\delta_{m,0}) \end{align*}\] (f) \[ \begin{align*} \int\limits_{-\pi}^\pi\cos^2(mx)\text dx\\ \text{I for }m\neq 0\qquad&=2\int\limits_0^\pi\cos^2(mx)\text dx\qquad (\text{even integrand})\\ &=\int\limits_0^\pi1+\cos(2mx)\text dx\\ &=x+2m\sin(2mx)\Big|_0^\pi\\ &=\pi+2m\sin(2m\pi)-0-2m\sin(0)\\ &=\pi\\ \\ \text{II for }m= 0\qquad&=0\qquad (\text{odd integrand})\\ \\ &=\pi\delta_{m,0} \end{align*}\] \begin{align*} \int\limits_{-\pi}^\pi\sin^2(mx)\text dx\\ \text{I for }m\neq 0\qquad&=2\int\limits_0^\pi\sin^2(mx)\text dx\qquad (\text{even integrand})\\ &=\int\limits_0^\pi1-\cos(2mx)\text dx\\ &=x-2m\cos(2mx)\Big|_0^\pi\\ &=\pi-2m\cos(2m\pi)-0+2m\\ &=\pi\\ \\ \text{II for }m= 0\qquad&=0\qquad (\text{odd integrand})\\ \\ &=\pi\delta_{m,0} \end{align*}

  4. UnkleRhaukus
    • 3 years ago
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    @mahmit2012

  5. mahmit2012
    • 3 years ago
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    |dw:1354631084039:dw|

  6. mahmit2012
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  7. mahmit2012
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  8. mahmit2012
    • 3 years ago
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    |dw:1354632135095:dw|

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