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Continuous Compound Interest with changing capital

  1. Dec 18, 2008 #1
    Is there any equation/formula for continuous compound interest to which money is added (or substracted from) periodically? Or can one be derived?


    i.e. monthly interest rate is 50% and we add 1$ every month (:bugeye:)
    Initially: 1$
    1st Month: 1.6 + 1 = 2.6$
    2nd month: 4.3 + 1 =5.28
  2. jcsd
  3. Dec 18, 2008 #2
    How are you getting your 4.3?
  4. Dec 18, 2008 #3
    from P*e^(rt)
    P for the second month is 2.6, r=0.5 t=1
  5. Dec 18, 2008 #4
    OK, so write out the first few payments using that formula.
  6. Dec 18, 2008 #5
    Initially: P
    1st Month: Pe^r + k
    2nd month: (Pe^r + k)e^r + k
    3rd month: ((Pe^r + k)e^r + k)e^r +k
    4th month: (((Pe^r + k)e^r + k)e^r +k)e^r + k
  7. Dec 18, 2008 #6
    Hmm I think I figured something out:

    The total amount of money at month m is:

    Pfinal=P0emr+((from t=0 to m-1)[tex]\Sigma[/tex])ket)+k

    Can you please check if this statement is true
    Last edited: Dec 18, 2008
  8. Dec 18, 2008 #7
    Well distributing would be my first thought to get

    [tex]1st = Pe^r + k[/tex]
    [tex]2nd = Pe^{2r} + ke^{r} + k[/tex]
    [tex]3rd = Pe^{3r} + ke^{2r} + ke^{r} + k = Pe^{3r} + k(e^{2r} + e^{r} + 1)[/tex]
    [tex]4th = Pe^{4r} + ke^{3r} + ke^{2r} + ke^{r} + k = Pe^{4r} + k(e^{3r} + e^{2r} + e^{r} + 1)[/tex]

    So the first term for n-th month is [tex]Pe^{nr}[/tex], then you have [tex]k\sum_{i=0}^{n-1}e^{ir} [/tex], which looks a lot like a geometric series...
  9. Dec 18, 2008 #8
    Your sum should go from 1 not 0 I think.
  10. Dec 18, 2008 #9
    You're right about the summation. I must have added the +k at the end by mistake. So is this a mathematically correct approach? Thank you very much by the way...
  11. Dec 18, 2008 #10
    I don't see why it wouldn't be a correct mathematical approach... I didn't check your work so I'm hoping you expressed each month correctly, I just tried to show you how to rewrite it using sums. You should be able to rewrite that sum since it converges (look up geometric sums).
  12. Dec 18, 2008 #11
    Thank you very much for your help.
    If I'm not mistaken the expression becomes:

    [tex]P_{f}=P_{0} e^{kt}+ \frac{a(1-e^{kt})}{1-e^{k}}[/tex]

    where [tex]P_{f}[/tex] denotes future value of the money, [tex]P_{0}[/tex] denotes the initial amount of the money, r denotes the annual percent interest and t denotes the total number of years and a is the amount added (or subtracted) each year.

    The statement might still be faulty, comments and critics are welcome and encouraged.
  13. Dec 18, 2008 #12
    I think the only comment I would make is that you initially said your interest rate is monthly and now you are saying it's annual.
  14. Dec 18, 2008 #13
    I tried to generalize it. It can be any period.

    Also this expression can be used for population growth where a certain amaunt of individuals die periodically (i.e. anti-viral in virus growth) or where there is periodical immigration.
  15. Dec 18, 2008 #14
    Correct it can be any period. If you are really interested in all the subtle differences and how all these formulas are derived you should check out Theory of Interest by Kellison. It's a book used by actuaries to prepare for one of their exams so the notation used might take a little while to get used to. But yes similar process applies to things of this nature, you will see it pop up in intro diff. eq. quite a bit.
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