Continuous Compounding the Hard Way
Table of Contents
1 Differential Equation for Continuous Compounding
Differential equations are equations which include variables and their derivatives. Let V be our present amount of money, and let it be compounded continuously at Q. Then we can write:
\[\frac{dV}{V} = Qdt\]
This says the percent change in value over the next instant is the quoted interest rate times and instantaneous length of time.
1.1 Future Value
We can use this differential equation to find the future value of some amount. For example, say we have $100 today. How much will this $100 be in 3 years if it is compounded continuously with an annual 5% rate? Let FV denote the future value.
\[\int_{100}^{FV}\frac{dV}{V} = \int_{t=0}^3 0.05dt\] \[ln(FV) - ln(100) = 0.05(3) \Rightarrow e^{ln(FV)} = e^{ln(100) + 0.05(3)} \Rightarrow FV = 100e^{0.05(3)} = 116.18\]
The future value is $116.18.
Note, if in the above we let the present value, quoted rate, and number of years be denoted by PV, Q, and t respectively, we have:
\[FV = PVe^{Qt}\]
1.2 Present Value
Say instead we know the future value, $116.18, and want to know the present value.
\[\int_{PV}^{116.18}\frac{dV}{V} = \int_{t=0}^3 0.05dt\] \[ln(116.18) - ln(PV) = 0.05(3) \Rightarrow PV = 116.18e^{-0.05(3)} = 100\]
1.3 Time to Reach a Given Amount
How many years will it take for $100 to become $140?
\[\int_{100}^{140}\frac{dV}{V} = \int_{t=0}^T 0.05dt\] \[ln(140) - ln(100) = 0.05(3) \Rightarrow T = \frac{ln(140) - ln(100)}{0.05} = 6.73\]
It will take 6.73 years.
1.3.1 Time to Double
Let's say we want to know how many years it will take to double our money to $200.
\[\int_{100}^{200}\frac{dV}{V} = \int_{t=0}^T 0.05dt\] \[ln(200) - ln(100) = 0.05(T) \Rightarrow T = \frac{ln(200) - ln(100)}{0.05} = 13.86\]
It will take 13.86 years.
1.4 Rate Required
Let's say we want to know what rate of return (Q) we will have to earn per year to have $175 in 10 years.
\[\int_{100}^{175}\frac{dV}{V} = \int_{t=0}^10 Qdt\] \[ln(175) - ln(100) = Q(10) \Rightarrow Q = \frac{ln(175) - ln(100)}{10} = 0.05596\]
We will have to earn about 5.60% per year.
2 FV Equation The Easy Way
You might say it is easier to just use the formula for the Future Value you get from your textbook:
\[FV = e^{rT}\]
You can rearrange this formula to answer any of the above questions. However how did your text get this formula? It usually does the foloowing.
Let Q be the annual stated rate, and let there be m compounding periods. Then the Effective Annual Rate is:
\[EAR = (1 + \frac{Q}{m})^m - 1\]
Letting the number of compounding periods increase to infinity we have:
\[lim_{n\to\infty} = (1 + \frac{Q}{m})^m - 1 = e^{Q} - 1\]
Using this rate we have:
\[FV = PV(1 + r)^t = PV(1 + e^{Q} - 1)^t = PVe^{Qt}\]
The only issue is your text probably didn't explain how \((1 + \frac{Q}{m})^m - 1 = e^{Q} - 1\) in the limit—it just says it does.
3 Alternative Differential Equation Derivation of \(FV=PVe^{Qt}\)
We can set up the following differential equation. This is just a modification of our above approach. Here \(\frac{dV}{dt}\) is the rate of change of the value of our investment, and \(QV\) is the rate at which interest increases.
\[\frac{dV}{dt} = QV\]
Also say we know our present value, \(V_0\) or more commonly PV. We can solve this equation as follows:
Using the chain rule, we can rewrite the equation as:
\[\frac{\frac{dV}{dt}}{V} = Q \Rightarrow \frac{d}{dt}ln|V| = Q\]
and integrating both sides:
\[ln|V| = Qt + C\]
where C is an arbitrary constant of integration, however given the PV, and taking the exponential of both sides we have the value at time t (defining it to be FV):
\[V \equiv FV = PVe^Qt\]
4 Reading Ease Score
"Flesch-Kincaid reading ease score: 116.47 Very easy (5th grade)"
"Flesch-Kincaid grade level score: 0.16"