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Math Help - Finance/Actuarial question. Please help!

  1. #1
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    Finance/Actuarial question. Please help!



    Any help on this is much appreciated!
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  2. #2
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by cirrus74 View Post


    Any help on this is much appreciated!
    i'll use s_n to denote the accumulated value of the annuity since i dont know the code in LaTex..

    so, s_{10} = \frac{(1+i)^{10} - 1}{i}

    and assuming that this is a compound interest, then we can use

    a(t) = \underbrace{e^{\int_0^t \delta_s \, ds}}_{in \, terms \, of \, force \, of \, interest} = \underbrace{(1+i)^t}_{basic \, formula}

    therefore, the middle part can be simplified (evaluated at t = 10) as \ln 2 (verify) and together with the right hand side, we have (1+i)^{10} = \ln 2 or i = \sqrt[10]{\ln 2} - 1

    therefore, s_{10} = \frac{(1+i)^{10} - 1}{i} = \frac{\ln 2 - 1}{\sqrt[10]{\ln 2} - 1}
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  3. #3
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    Please do not double post. You question was first answered in the Business Math forum: http://www.mathhelpforum.com/math-he...-function.html
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  4. #4
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    You should have these:

    a(t) = e^{\int_{0}^{t}\delta_{r}dr}

    a_{angle-n} = \sum_{t=1}^{n}\left(\frac{1}{a(t)}\right)

    s_{angle-n} = a(n)*a_{angle-n}

    Let's see what you get.

    Note: I'm not sure what Kalagota is missing. It doesn't make sense for 10 payments to be accumulated to only 8Ĺ, so you get to tell us what you manage to calulate.
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  5. #5
    MHF Contributor kalagota's Avatar
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    now i wonder what i missed?

    i think, my mistake was that \ln 2 thing.. i forgot that the integral was a power of e and thus, a(t) should be equal to 2.. this should make sense...

    now, let me correct my mistake..

    (1+i)^{10} = 2 or i = \sqrt[10]{2} - 1

    thus, s_{10} = \frac{(1+i)^{10} - 1}{i} = \frac{1}{\sqrt[10]{2} - 1} \approx 13.9327262
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  6. #6
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    Try again. I'm holding out for 14.5.
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  7. #7
    MHF Contributor kalagota's Avatar
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    how you did yours?

    is  a(t) = \frac{1}{2} ? it can't be i think, and i shall stick with a(t)=2..
    Last edited by kalagota; April 9th 2008 at 08:11 AM.
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  8. #8
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    Formulas shown above. I think yours is missing some moving parts.

    Start from s_{10} = (1+i)^{10} + (1+i)^{9} + ... + (1+i), rather than the summed version that assumes regularity of interest for each payment.

    With variable interest, a(10) \neq \left(a(5)\right)^{2}.
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  9. #9
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    Quote Originally Posted by TKHunny View Post
    Try again. I'm holding out for 14.5.
    You've got the right answer. How did you do it? And thank you both so much for helping.

    I did as was suggested in the business post and integrated exp(delta(t)dt) with limits of 1 to 20. And got 1.9... which is definitely not the answer.
    Last edited by cirrus74; April 9th 2008 at 03:02 PM.
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  10. #10
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    The integral is good. You will need that.

    Look at my first post very carefully. There is a summation in there. If you are trying to do it in one piece, you do not yet have the right idea.
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  11. #11
    MHF Contributor kalagota's Avatar
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    did you notice notice that your summation is just the sum of the first n geometric sequence.. i think, you must be the one to check your solution.. or the best way is to post your whole solution in here..

    besides, if you will push your 14.5, try to work it backward, and you will derive at a(t) = 1.71... and not 2..
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  12. #12
    MHF Contributor kalagota's Avatar
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    maybe i'll put the derivation here..

    a(t) = (1+i)^t

    thus, a_{angle-n} = \sum_{t=1}^n \frac{1}{a(t)}|_{n=10} = \frac{1}{a(1)} + \frac{1}{a(2)} + ... + \frac{1}{a(10)}

    or a_{angle-10} = \frac{1}{(1+i)} + \frac{1}{(1+i)^2} + ... + \frac{1}{(1+i)^{10}}

    and s_{angle-10} = a(10) \, a_{angle-10} = (1+i)^{10}\, a_{angle-10} = (1+i)^9 + (1+i)^8 + ... + (1+i) + 1

     = \frac{(1+i)^{10} - 1}{(1+i) - 1} and just do your substitution..
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  13. #13
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    Thanks, I've got the answer. Do you mind if I ask a question though?

    Why can I not just use this? Or at least, the answer doesn't turn out right, but I don't understand why.

    <br />
\int _1 ^{10} e^{\delta t} dt<br />

    exp(-ln(20 - 10)+ ln(20-1)) = exp (ln(19/10)) = 19/10

    Did I make a mistake somewhere?
    Last edited by cirrus74; April 10th 2008 at 12:50 AM.
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  14. #14
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    Quote Originally Posted by kalagota View Post
    thus, a_{angle-n} = \sum_{t=1}^n \frac{1}{a(t)}|_{n=10} = \frac{1}{a(1)} + \frac{1}{a(2)} + ... + \frac{1}{a(10)}
    Perfect.

    or a_{angle-10} = \frac{1}{(1+i)} + \frac{1}{(1+i)^2} + ... + \frac{1}{(1+i)^{10}}
    This is where it falls apart. Your interest is static.

    Calculate from the integrals:

    a(1) = 20/19

    a(2) = 10/9

    a(3) = 20/17

    a(4) = 5/4

    It is not a Geometric Sequence.
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  15. #15
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    Another Possible Answer:

    As it stands, we have 't' defined as the time from EACH deposit. That may not seem quite right. If we use 't' to mean the time from the beginning of the financial structure, we get:

    s(t) = e^{\int_{t}^{10}\delta^{t}\;dr}

    Then:

    s_{angle-10} = \sum_{0}^{9}s(t) = 15.5

    Very interestingly, we get:

    s(0) = 2.00

    s(1) = 1.90

    s(2) = 1.80

    etc...

    It IS an arithmetic sequence.
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