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Math Help - laplace

  1. #1
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    Lightbulb laplace

    how you get the laplace of 1/tsin(wt)
    pls show me the solution
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  2. #2
    Behold, the power of SARDINES!
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    Re: laplace

    Quote Originally Posted by jomex42 View Post
    how you get the laplace of 1/tsin(wt)
    pls show me the solution
    I will leave the details of why to you.

    Given f(w,t)=\frac{1}{t}\sin(wt) \implies \frac{\partial f}{\partial w}=\cos(wt)

    \mathcal{L}\{ \frac{\partial f}{\partial x}\}=\frac{\partial }{\partial w}\mathcal{L}\{ f(w,t)\}

    \frac{s}{s^2+w^2}=\frac{\partial }{\partial w}\mathcal{L}\{ f(w,t)\}

    Now integrate both sides with respect to w to solve for the Laplace transform you want

    \int \frac{s}{s^2+w^2}dw=\int \frac{\partial }{\partial w}\mathcal{L}\{ f(w,t)\}dw

    \tan^{-1}\left( \frac{w}{s}\right)=\mathcal{L}\{ f(w,t)\}
    Thanks from jomex42
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  3. #3
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    Re: laplace

    could you show me the solution of laplace of 1/tsinwt using the formula
    intergral of limits 0 to infinity e^st(ft)dt
    then you will get the arctanw/s
    plss the exact solution my teacher want it
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  4. #4
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    Re: laplace

    laplace-untitled.jpg

    can you solve this and you will get the result of
    arctan(α/s)
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  5. #5
    Behold, the power of SARDINES!
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    Re: laplace

    Quote Originally Posted by jomex42 View Post
    Click image for larger version. 

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    can you solve this and you will get the result of
    arctan(α/s)
    Another method is to note that the integral converges uniformly so limits and integrals can be interchanged.

    Note that

    \sin(z)=\sum_{n=0}^{\infty}\frac{(-1)^nz^{2n+1}}{(2n+1)!} \quad \tan^{-1}(z)=\sum_{n=0}^{\infty}\frac{(-1)^nz^{2n+1}}{(2n+1)}

    This gives that

    \int_{0}^{\infty}\frac{e^{-st}}{t}\left( \sum_{n=0}^{\infty}\frac{(-1)^n(\alpha t)^{2n+1}}{(2n+1)!}\right)dt = \sum_{n=0}^{\infty}\frac{(-1)^n\alpha^{2n+1}}{(2n+1)!}\int_{0}^{\infty} e^{-st} t^{2n}dt = \sum_{n=0}^{\infty}\frac{(-1)^n\alpha^{2n+1}}{(2n+1)!} \mathcal{L}\{t^{2n}\}

    Now integrate or use the definition of the Laplace transform to get

    \sum_{n=0}^{\infty}\frac{(-1)^n\alpha^{2n+1}}{(2n+1)!}\frac{(2n)!}{s^{2n+1}}=  \sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)}\left( \frac{\alpha}{s}\right)^{2n+1}=\tan^{-1}\left( \frac{\alpha}{s} \right)
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  6. #6
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    Re: laplace

    What do you called that big letter e in your equation
    what method did you use?

    Because im not familiar in that equation i have not encounter that kind of function could you explain it
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  7. #7
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    Re: laplace

    what method did you use so i can explain it to my teacher how it been solved
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  8. #8
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    Re: laplace

    how the t in your first function gone in the next function

    and the summation get out of the integral sign

    What did you do in the first function to comeup with the second function
    laplace-png.png
    Last edited by jomex42; March 10th 2013 at 06:08 PM.
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