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Math Help - Inverse Laplace Transformation

  1. #1
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    Inverse Laplace Transformation

    Hello MHF members,

    If anyone can, help me sort this out... I have worked on Laplace and inverse Laplace transformations before through the table and that was fine.

    Problem is that now I am working through definitions and while Laplace worked fine, got stuck on inverse Laplace.

    Here follows the definition-formula introduced, but has no applicable example... So if anyone can help on the easiest exercise supplied, to get it...

    Many thanks in advance.
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  2. #2
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    Word documents are generally frowned on in the MHF, because they can contain viruses. Please post a pdf or a jpg, or better yet, learn LaTeX and post it that way.
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  3. #3
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    Apologies, I upload again the same attachment in pdf format.
    Attached Thumbnails Attached Thumbnails Inverse Laplace Transformation-inverse_laplace-mhf.pdf  
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  4. #4
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    Hmm. I'm not a fan of Greek letters where they're not needed or even standard. Let me rephrase the question as best as I can, and ask you a question in turn:

    Define

    \displaystyle F(t)=\lim_{s\to\infty}\sum_{v\le st}\left[(-1)^{v}\frac{g^{(v)}(s)}{v!}\,s^{v}\right],

    where g(s)=\mathcal{L}[f(t)].

    How can I use the above definition to compute the inverse Laplace transform of

    g(s)=\dfrac{\alpha}{s}+\dfrac{\beta}{s^{2}}?

    The answer is f(t)=\alpha+\beta t.

    Is that correct?

    If so, I have two questions, actually:

    1. It seems pretty clear that v\in\mathbb{Z}. What is the lower limit on the sum?

    2. What is the relationship between F(t) and f(t). Is that just a typo? Should it be

    \displaystyle f(t)=\lim_{s\to\infty}\sum_{v\le st}\left[(-1)^{v}\frac{g^{(v)}(s)}{v!}\,s^{v}\right]?
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  5. #5
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    Hello Adrian,

    Rephrase is correct (no worries with Greek or Latins, whatever suits you as long as I get how it works).

    No typo, definition on script is a bit confusing....
    says that f(t) is some density probability and F(t) its distribution that both have φ(t) as Laplace transformation....
    anyway, issue here is inverse Laplace, so let us assume that f(t) is equal to the limes...
    (guess it is indifferent in this case... as so far on script I met no Probabilities... and is not in the main subject, don't know why involved this here...)

    on s ...is supposed to be continuous interest rate... that is defined as s=ln(1+i), where i belongs to Z
    we can just use it as a continuous variable with positive values.
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  6. #6
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    Hmm. It's almost time for me to hit the sack, but I did want to say this: on page 139 of Wackerly, Mendenhall, and Scheaffer's Mathematical Statistics with Applications, 5th Ed., they provide the following Definition 4.3:

    Let F(y) be the distribution function for a continuous random variable Y. Then f(y), given by

    f(y)=\dfrac{d\,F(y)}{dy}=F'(y)

    wherever the derivative exists, is called the probability density function for the random variable Y.
    So there you have the relationship between F and f. I think that will be important in solving your problem.
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  7. #7
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    I am aware of the later, that indeed is their relationship (didn't thought of presenting it that way...brilliant) but not sure if that is of much help....

    I think (intuitively and given previous techniques) that is something more technical, more likely to do with substitutions... but this formula, totally blocks me....

    Perhaps I should also clarify, that the g^(u) inside the sum, is not a power but the u-derivative of g(s) ...rest are powers...
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  8. #8
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    I'm back, after a very long night (my almost-four-month-old daughter was a bit colicky). So it seems to me that the necessary procedure here is a two-step process:

    1. Evaluate the limit-sum expression with the given g(s). Result is F(t).
    2. Take the derivative F'(t) to obtain f(t).

    So, on #1 there. We need to find an expression for the vth derivative of g(s). We have

    g(s)=\alpha s^{-1}+\beta s^{-2}, and hence we get

    g'(s)=-\alpha s^{-2}-2\beta s^{-3}, and

    g''(s)=2\alpha s^{-3}+6\beta s^{-4}, and

    g'''(s)=-6\alpha s^{-4}-24\beta s^{-5}.

    I think you'll find the following:

    g^{(v)}(s)=(-1)^{v}(v!)\alpha s^{-(v+1)}+(-1)^{v}(v+1)!\beta s^{-(v+2)}.

    So, plugging this into your limit-sum expression yields the following:

    \displaystyle F(t)=\lim_{s\to\infty}\sum_{v=0}^{\lfloor st\rfloor}\left[(-1)^{v}\frac{1}{v!}\left[(-1)^{v}(v!)\alpha s^{-(v+1)}+(-1)^{v}(v+1)!\beta s^{-(v+2)}\right]\,s^{v}\right].

    Note: the notation \lfloor st\rfloor means the greatest integer less than or equal to st.

    \displaystyle =\lim_{s\to\infty}\sum_{v=0}^{\lfloor st\rfloor}\left[\alpha s^{-(v+1)}+(v+1)\beta s^{-(v+2)}\right]\,s^{v}

    \displaystyle =\lim_{s\to\infty}\sum_{v=0}^{\lfloor st\rfloor}\left[\alpha s^{-1}+(v+1)\beta s^{-2}\right]

    \displaystyle =\lim_{s\to\infty}\sum_{v=0}^{\lfloor st\rfloor}\left[\frac{\alpha s+\beta(v+1)}{s^{2}}\right]

    \displaystyle =\lim_{s\to\infty}\left[\frac{\alpha s+\beta}{s^{2}}\sum_{v=0}^{\lfloor st\rfloor}1+\frac{\beta}{s^{2}}\sum_{v=0}^{\lfloor st\rfloor}v\right]

    \displaystyle =\lim_{s\to\infty}\left[\frac{\alpha s+\beta}{s^{2}}\lfloor st\rfloor}+\frac{\beta}{s^{2}}\frac{\lfloor st\rfloor(\lfloor st\rfloor+1)}{2}\right].

    Again, I have to go now, but hopefully, this will stimulate your thinking a bit. The next question, of course, is how to evaluate that limit.
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  9. #9
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    Thank you a lot Adrian,

    I had hard day myself too... so I will check it in detail tomorrow,
    seems that I did some calculus mistake that made things more complex... 8)

    Beautiful thought, to specify integer value of st, on sum... makes limes much easier to solve!!
    Roughly speaking problem seems solved, will let you know tomorrow.

    (give a hug to daughter from me, )) and if baby is colicky some times helps warm (but no hot!) chamomile...
    is allowed on babies as it is also a natural antiseptic.)
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  10. #10
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    Hmm. We have chamomile; we might try that. I think the underlying reason for all this is a flu that's making the rounds in my family. I've got it now, which is no fun.

    Let me know how your calculations work out.
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