# How to create a mathematical model of investment?

1. Feb 1, 2017

### Eclair_de_XII

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. Feb 1, 2017

### Ray Vickson

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
3. Feb 1, 2017

### Eclair_de_XII

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.

4. Feb 1, 2017

### Eclair_de_XII

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
5. Feb 1, 2017

### 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.

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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
6. Feb 1, 2017

### Eclair_de_XII

Oh, I got it: $\frac{dS}{dt}=r(S+\frac{k}{r})$.

Thanks.