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anonymous

  • one year ago

Laplace transform of f(t)=t^n inductive proof?

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  1. anonymous
    • one year ago
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    So we know the laplace transform of this, but could anyone explain how to use an inductive method. Would we suggest letting n=1, solving for that, n=2 solving for that, n=3 solving for that and then assume it for all cases?

  2. ganeshie8
    • one year ago
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    setup the integral and see if you can get a recurrence relation

  3. anonymous
    • one year ago
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    \[F(s)=\int\limits_{0}^{\infty}t ^{n}e ^{-st}dt\]

  4. ganeshie8
    • one year ago
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    \[\large \mathcal{L}\{t^n\} ~~=~~ \int\limits_0^{\infty} t^ne^{-st}\,dt\] try by parts

  5. anonymous
    • one year ago
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    so for \[\int\limits_{}^{}fdg=fg-\int\limits_{}^{}gdf\]

  6. anonymous
    • one year ago
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    let f=t^n dg=e^(-st)

  7. ganeshie8
    • one year ago
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    looks good, keep going..

  8. anonymous
    • one year ago
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    i'll be using equation mode so bare with me : )

  9. anonymous
    • one year ago
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    \[F(s)=t ^{n}\frac{ -e ^{-st} }{ s }-\int\limits_{}^{}\frac{ -e ^{-st} }{ s }nt ^{n-1}dt\]

  10. anonymous
    • one year ago
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    you would have to do by parts again right

  11. ganeshie8
    • one year ago
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    nope, we're done

  12. anonymous
    • one year ago
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    you just keep going on forever

  13. ganeshie8
    • one year ago
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    \[\begin{align}\mathcal{L}\{t^n\} ~&=~t ^{n}\frac{ -e ^{-st} }{ s }\Bigg|_{0}^{\infty}-\int\limits_{0}^{\infty}\frac{ -e ^{-st} }{ s }nt ^{n-1}dt\\~\\ ~&=~t ^{n}\frac{ -e ^{-st} }{ s }\Bigg|_{0}^{\infty}+\dfrac{n}{s}\int\limits_{0}^{\infty} t^{n-1}e ^{-st}dt\\~\\ &=0+\dfrac{n}{s}\mathcal{L}\{t^{n-1}\} \end{align}\]

  14. ganeshie8
    • one year ago
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    so we get the recurrence relation : \[\large \mathcal{L}\{t^{n}\}~~=~~\dfrac{n}{s}\mathcal{L}\{t^{n-1}\}\]

  15. anonymous
    • one year ago
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    isn't the first part (0-(-1))?

  16. anonymous
    • one year ago
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    oh wait, the t out the front

  17. anonymous
    • one year ago
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    makes it 0

  18. anonymous
    • one year ago
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    yep, got it.

  19. ganeshie8
    • one year ago
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    are you saying \[t ^{n}\frac{ -e ^{-st} }{ s }\Bigg|_{0}^{\infty} = 0-(-1)\] ?

  20. anonymous
    • one year ago
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    my mistake

  21. anonymous
    • one year ago
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    so how is this a recurrence relationship? Its been a while since ive done this

  22. anonymous
    • one year ago
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    so the integral is actually the laplace of t^(n-1) right?

  23. ganeshie8
    • one year ago
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    so we get the recurrence relation : \[\begin{align}\mathcal{L}\{t^{n}\}~~&=~~\dfrac{n}{s}\mathcal{L}\{t^{n-1}\}\\~\\ &=~~\dfrac{n}{s}*\dfrac{n-1}{s}\mathcal{L}\{t^{n-2}\}\\~\\ &=~~\dfrac{n}{s}*\dfrac{n-1}{s}*\dfrac{n-2}{s}\mathcal{L}\{t^{n-3}\}\\~\\ &=\cdots\\~\\ &=~~\dfrac{n}{s}*\dfrac{n-1}{s}*\dfrac{n-2}{s}\cdots*\dfrac{2}{s}*\dfrac{1}{s}*\mathcal{L}\{t^{0}\}\\~\\ &=~~\dfrac{n!}{s^n}*\mathcal{L}\{t^{0}\}\\~\\ \end{align}\]

  24. anonymous
    • one year ago
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    could you simplify that more by saying that the laplace transform of t^0 is just the step change of the process; 1/s?

  25. anonymous
    • one year ago
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    \[\frac{ n! }{ s ^{n+1} }\]

  26. ganeshie8
    • one year ago
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    I think so, but im not so good with control theory...

  27. anonymous
    • one year ago
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    yeah that would make sense because that is the final laplace transform of t^n

  28. anonymous
    • one year ago
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    so im confused why you have: 2/s * 1/s *LT(t^0) why is there a 1/s factor

  29. ganeshie8
    • one year ago
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    if psble send me the meeting id, il explain it quick

  30. anonymous
    • one year ago
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    can you only run this via pc? i have a mac

  31. ganeshie8
    • one year ago
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    oh wait then

  32. anonymous
    • one year ago
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    .exe is pc only i think

  33. ganeshie8
    • one year ago
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    try this http://download.teamviewer.com/download/TeamViewer.dmg

  34. anonymous
    • one year ago
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    what do i do?

  35. anonymous
    • one year ago
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    553 005 551

  36. anonymous
    • one year ago
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    is the id

  37. ganeshie8
    • one year ago
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    message me the password

  38. anonymous
    • one year ago
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    thanks for that! i'll keep working on it. cheers for the interactive help

  39. ganeshie8
    • one year ago
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    np :) check this video when free http://ocw.mit.edu/courses/mathematics/18-03-differential-equations-spring-2010/video-lectures/lecture-19-introduction-to-the-laplace-transform/

  40. anonymous
    • one year ago
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    i'm doing a course in process control, and unfortunately we never came across laplace transforms in engineering math (my degree only uses 1st semester engineering math not second) but more learning the better!

  41. IrishBoy123
    • one year ago
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    you can also take an easy one \(\mathcal L \{1\} = \int_{0}^{\infty} e^{-st} dt = \frac{1}{s}\) and diffeentiate it \(\large \frac{d}{ds} \int_{0}^{\infty} e^{-st} dt = \int_{0}^{\infty} \frac{d}{ds}e^{-st} dt = \frac{d}{ds} \frac{1}{s}\) over and over

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