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How to create a mathematical model of investment?

  1. Feb 1, 2017 #1
    1. The problem statement, all variables and given/known data
    "Someone with no initial capital ##S_0=0## invests ##k \frac{dollar}{year}## at an annual rate of return ##r##. Assume that investments are made continuously and that the return is compounded continuously."

    Find the solution ##S## that solves the differential equation modeling this scenario.

    2. Relevant equations
    Let ##S## be the amount of capital at any time.

    3. The attempt at a solution
    I'm trying to make sure that the units of all my quantities are the same, but the main problem I'm having is the discontinuity at ##t=0## when I divide ##S## by ##t##.

    ##\frac{dS}{dt}=\frac{S}{t}+k(r+1)##
    ##\frac{dS}{dt}-\frac{S}{t}=k(r+1)##
    ##\frac{d}{dt}(t^{-1}S)=\frac{k(r+1)}{t}##
    ##S(t)=(tlnt)(k(r+1))##

    I frankly do not know how to model this problem. Can anyone help me understand what is happening in this problem? It's hard to insert ##S## into the equation because the ##\frac{dS}{dt}## does not rely on ##S##; it relies on the rate of return ##r## which only applies to the yearly investments ##k##. Then it's compounded continually...? I do not understand this.
     
  2. jcsd
  3. Feb 1, 2017 #2

    Ray Vickson

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    For a very small time increment of ##\Delta t## years, the amount invested between times ##t## and ##t + \Delta t## is ##k \Delta t ## dollars; that is what is meant by a continuous investment at rate ##k##.

    If S(t) is the amount in the bank at time ##t## (years) we have
    $$S(t+\Delta t) = S(t) (1 + r \Delta t) + k \Delta t \hspace{2cm}(1)$$
    because the balance ##\$S(t)## at ##t## would grow by the interest-rate factor factor ##1 + r \Delta t## in the short time interval of length ##\Delta t##, even if we did not invest any more---but additional investment makes it grow more. That is what continuous compounding means: a dollar in the account at time zero grows to ##\$e^{rt}## by time ##t > 0##, and so the growth in value from ##t## to ##t + \Delta t## is
    $$e^{r(t +\Delta t)} - e^{rt} = e^{rt} \left( e^{r \Delta t}-1 \right) \doteq e^{rt} (1 + \Delta t),$$
    which is a growth rate of ##1 + r \Delta t.##

    We get the differential equation for ##S(t)## from eq.(1) above.

    For more on continuous compounding, see, eg.,
    http://www-stat.wharton.upenn.edu/~waterman/Teaching/IntroMath99/Class04/Notes/node11.htm
    http://people.duke.edu/~charvey/Classes/ba350_1997/preassignment/proof1.htm
    http://math.stackexchange.com/questions/539115/proof-of-continuous-compounding-formula
     
    Last edited: Feb 1, 2017
  4. Feb 1, 2017 #3
    Thanks. Should I worry about the equation being continuous at ##t=0##? Yes, right? So I can't have any of the variables be divided by ##t## for the ##\frac{dollar}{year}## unit.
     
  5. Feb 1, 2017 #4
    Can anyone tell me what is wrong with my current model?

    ##\frac{dS}{dt}=r(S+kt)=rS+rkt##
    ##\frac{dS}{dt}-rS=rkt##
    ##\frac{d}{dt}(Se^{-rt})=(rkt)(e^{-rt})##

    Basically, I get an equation with an extra linear term: ##S(t)=-kt+\frac{k}{r}(e^{rt}-1)##. It's another equation that satisfies the IVP at ##S(0)=0##. I'm not sure what to make of that, but at least I got something done tonight.
     
    Last edited: Feb 1, 2017
  6. Feb 1, 2017 #5

    Ray Vickson

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    *************************************

    The model above assumes that the person invests at rate ##kt## at time t, so is a variable investment rate that increases over time. The problem told you the investment rate is constant, not variable.

    ************************************

    I have no idea what you are talking about. There is no issue of discontinuity anywhere, even at ##t = 0##. You do not divide by ##t##---that is not what rates are about. I explained exactly what was meant by rates, so all you need to do is re-read the post.
     
    Last edited: Feb 1, 2017
  7. Feb 1, 2017 #6
    Oh, I got it: ##\frac{dS}{dt}=r(S+\frac{k}{r})##.

    Thanks.
     
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