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Math Help - Convergence (Hall and Jorgenson)

  1. #1
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    Convergence (Hall and Jorgenson)

    Hall and Jorgenson uses the integral to find the present value...

    z = [integral from 0 to infinite] exp(-r s) D(s) ds

    Find z, so it's a function of t, when...

    D(s) = 2(t - s) / t^2 for 0 <= s <= t, D(s) for s > t. (Linear decrease)

    It would be appreciated a whole lot, if anyone could take a look at the above...


    Good evening,
    Simon DK
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  2. #2
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    Quote Originally Posted by sh01by View Post
    Hall and Jorgenson uses the integral to find the present value...

    z = [integral from 0 to infinite] exp(-r s) D(s) ds

    Find z, so it's a function of t, when...

    D(s) = 2(t - s) / t^2 for 0 <= s <= t, D(s) for s > t. (Linear decrease)
    Maybe that should be a "t" and not a "r", you probably made a mistake based on the juxtaposition of the buttons.

    z=\int_0^{\infty} e^{-st}D(s)ds
    ----
    But I am thinking what the function can be. It seems like a Laplace transform but I never seen anything like this before.
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    Post Thanks, but...

    Hey, thanks for taking time to look at it...

    But - I'm quite sure it's supposed to be ' r '...

    Notes on the assignment says...

    "Hall and Jorgenson uses the integral

    z = [integral from 0 to infinite] exp(-r*s)*D(s) ds

    to find the discounted present value, to rate r, by a time-dependent flow of depreciations."

    I actually have the answer, but it's getting there I'm uncertain of...


    z = (2 / (r*t)) * (1 - (1/(r*t))*(1-exp(-r*t)))

    Maybe this helps? But - how do I get there?

    Any help is still appreciated...
    Thanks though...
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  4. #4
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    Quote Originally Posted by sh01by View Post
    Hall and Jorgenson uses the integral to find the present value...

    z = [integral from 0 to infinite] exp(-r s) D(s) ds

    Find z, so it's a function of t, when...

    D(s) = 2(t - s) / t^2 for 0 <= s <= t, D(s) for s > t. (Linear decrease)

    It would be appreciated a whole lot, if anyone could take a look at the above...


    Good evening,
    Simon DK
    Well, it wouldn't completely surprise me that I goofed up somewhere, but I don't like the form of your answer and it doesn't match my work. Even if I made a mistake somewhere I don't see how you could have a factor that includes e^{-rt}.

    z = \int_0^{\infty}ds \left ( \frac{2(t-s)}{t^2} \right ) e^{-rs}

    Since the integration is independent of t:
    z = \frac{2}{t} \int_0^{\infty}ds e^{-rs} - \frac{2}{t^2} \int_0^{\infty}ds \, se^{-rs}

    The first integral is trivial:
    \int_0^{\infty}ds e^{-rs} = -\frac{1}{r}e^{-rs}|_0^{\infty} = \frac{1}{r}

    We may do the second integral using integration by parts:
    \int_0^{\infty}ds \, se^{-rs} = \left [ -\frac{se^{-rs}}{r} \right ]_0^{\infty} - \int_0^{\infty}ds \frac{-1}{r}e^{-rs}

    = \left [ -\frac{se^{-rs}}{r} \right ]_0^{\infty} - \frac{1}{r^2}e^{-rs}|_0^{\infty}

    =  0 + \frac{1}{r^2} = \frac{1}{r^2}

    So we get:
    z = \frac{2}{t} \cdot \frac{1}{r} - \frac{2}{t^2} \cdot \frac{1}{r^2}

    z = \frac{2}{rt} \left ( 1 - \frac{1}{rt} \right )

    -Dan
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  5. #5
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    Smile Hi (Lecturer has to be questioned...)

    Hi,

    Thanks for all the help... It has been a great journey, but I think I have to give in... I'll have to talk to my math teacher later today... I have a lecture in 2 hours, so - maybe he can clarify this mystery and ease the pain?

    I hope I'll come back to this forum with answers...

    Simon DK
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  6. #6
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    Question Lecturer was little help...

    Hi there...

    I talked with my lecturer and he left me with a little idea of how to proceed...

    It's for a math project, which is due till next Friday, so I have a little time to figure it out... I'll be attending another class tomorrow with a different teacher, so I'm hoping to solve the mystery tomorrow...

    But - I'll continue with other parts of the project for now...

    I'll be back with news. : o )

    Simon DK
    Last edited by sh01by; November 30th 2006 at 03:06 AM. Reason: I keep forgetting the smileys are changed from text to graphic...
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