Problem 8.7.5.
The function \(A(t)\) has the same form as our solution of IVP (8.18) so it must satisfy the differential equation: \(\dfdx{A}{t}= 0.05A\text{.}\) Confirm that it does.
Section 8.7 Exponential Functions and Compound Interest
The natural exponential function also arises naturally in financial mathematics. If we invest money in a bond which pays a nominal interest rate of \(5\%\) annually, and the interest is compounded quarterly then the effective yield will be \(5.09\%\text{.}\)
There is a lot of financial jargon in the previous paragraph which makes it hard to understand if we don’t speak financial-ese. So we will take a few pages to learn the vocabulary. After that things will clear up pretty quickly.
To start, the effective yield represents the actual amount of money earned at the end of one year — how much money the investment actually pays out. For example, if the investment was not compounded at all, then at the end of one year, every dollar invested would yield \(\$1.05\) in return, so the effective yield is the same as the nominal yield: \(5\) dollars.
If the investment is compounded semiannually (twice per year), then half of interest earned is paid out midway through the year. This money is then re–invested for the next half year. This is summarized in the table below. The effective yield in this case is approximately \(5.06\%\text{.}\)
| Time \(t\) in years | Amount investment is worth in dollars |
| \(0\) | \(\$ 1\) |
| \(1/2\) | \(\$1+\$1\left(\frac{0.05}{2}\right)=\$1\left(1+\frac{0.05}{2}\right)=\$1.025\) |
| \(1\) | \(\$1\left(1+\frac{0.05}{2}\right)+\$1\left(1+\frac{0.05}{2}\right)\left(\frac{0.05}{2}\right) =\$1\left(1+\frac{0.05}{2}\right)^2 =\$1.050625 \) |
The table below shows, in abbreviated form, the computations needed if the investment is compounded three times in a year.
| Time \(t\) in years | Amount investment is worth in dollars |
| \(0\) | \(\$ 1\) |
| \(1/3\) | \(\$1\left(1+\frac{0.05}{3}\right)^1\) |
| \(2/3\) | \(\$1\left(1+\frac{0.05}{3}\right)^2\) |
| \(1\) | \(\$1\left(1+\frac{0.05}{3}\right)^3\approx\$1.0508\) |
Quarterly compounding means that the interest is re–invested four times and is shown in the next table.
| Time \(t\) in years | Amount investment is worth in dollars |
| \(0\) | \(\$ 1\) |
| \(1/4\) | \(\$1\left(1+\frac{0.05}{4}\right)^1\) |
| \(1/2\) | \(\$1\left(1+\frac{0.05}{4}\right)^2\) |
| \(3/4\) | \(\$1\left(1+\frac{0.05}{4}\right)^3\) |
| \(1\) | \(\$\left(1+\frac{0.05}{4}\right)^4\approx\$1.0509\) |
Notice how these three tables differ, and how they are similar to each other. The effective yield comes from the last entry in each table. It is the difference between the amount the investment is worth at the end of the year and what it is worth at the beginning of the year. So if we compound quarterly the effective yield is \(1.0509-1=0.0509=5.09\%. \)
Similarly, if the investment is compounded daily, then the effective yield is \(\left(1+\frac{0.05}{365}\right)^{365}-1\approx0.0513 = 5.13\%\text{.}\) If we compound \(5\%\) nominal interest, \(n\) times in a year, then the return on an investment of one dollar at the end of one year would be \(\left(1+\frac{0.05}{n}\right)^{n}\) dollars, after \(2\) years \(\left(1+\frac{0.05}{n}\right)^{2n}\) dollars, after \(3\) years \(\left(1+\frac{0.05}{n}\right)^{3n}\) dollars.
In general if we let \(A(t)\) denote the value of the investment of one dollar earning \(5\%\) annually, compounded \(n\) times per year, after \(t\) years, we have
\begin{equation*}
A(t)=1\cdot\left(1+\frac{0.05}{n}\right)^{nt},\ t=0,1,2,\ldots
\end{equation*}
What if the investment was compounded continuously? That is, suppose the interest is being continuously paid out and simultaneously reinvested? Obviously, this can’t actually be done, but it comes to the same thing if at the end of the year we get an effective yield that is equal to the effective yield that would come from continuous compounding. All we have to do is figure out what this is.
To do this, it will be convenient to write \(A(t)\) as
\begin{align*}
A(t)\amp =1\cdot\left(1+\frac{0.05}{n}\right)^{nt}\\
\amp =1\cdot\left(\left(1+\frac{0.05}{n}\right)^{\frac{n}{0.05}}\right)^{0.05t}\\
\amp =1\cdot\left(\left(1+\frac{1}{\frac{n}{0.05}}\right)^{\frac{n}{0.05}}\right)^{0.05t}
\end{align*}
Setting \(m=\frac{n}{0.05}\) we see that
\begin{equation*}
A(t)=1\cdot\left(\left(1+\frac{1}{m}\right)^{m}\right)^{0.05t}.
\end{equation*}
We are particularly interested in the quantity \(\left(1+\frac{1}{m}\right)^{m}\) as we compound the interest more frequently (as \(m\) becomes very large). The following table shows some values of \(\left(1+\frac{1}{m}\right)^m\text{,}\) for large values of \(m\text{.}\) Does the numbers in the right column look familiar to you?
| \(m\) | \(\left(1+\frac{1}{m}\right)^m\) |
| \(100\) | 2.70481382942 |
| 1000 | 2.71692393224 |
| \(10000\) | 2.71814592683 |
| 100000 | 2.71826823717 |
| 1000000 | 2.71828046932 |
| 10000000 | 2.71828169255 |
In Section 8.2 we saw that \(e\approx2.7182\text{.}\) From this table it would appear that \(\left(1+\frac{1}{m}\right)^m\) approaches the number \(e\) as \(m\) grows larger — that is, as we compound the interest more frequently. We will show that this is in fact the case in Example 12.4.31.
Thus if we compound the interest continuously we have
\begin{equation}
A(t)=1\cdot\left(e\right)^{0.05t}.\tag{8.21}
\end{equation}
This equation was derived under the assumption that our initial investment was only \(1\text{.}\) In general if the initial investment is \(A_0\) we have
\begin{equation}
A(t)=A_0e^{0.05t}.\tag{8.22}
\end{equation}
Does this look familiar?
This is interesting. It appears that IVP (8.20) can be a model of continuously compounded investment return, as well as population growth. Does this make sense to you?
Think about it for a moment. When money invested the rate of return (rate of growth) is always proportional to the initial investment, in the same way that the rate of growth of a population (the rate of return) is. Speaking very loosely, the amount of baby dollars in the next generation will be proportional to the amount of mama dollars and papa dollars in the current generation in the same way that the number of baby bacteria in the next generation is proportional to the number of parent bacteria now. So it makes sense that these two very different real world phenomena are modeled by the same IVP.
7
https://www.youtube.com/watch?v=wlbuGdoAFekFrom the point of view of a banker or a biologist, these are very different problems so we assign different meanings to the variables and parameters. But since relationship between the variables and parameters is the same whatever we call them, to a mathematician these are the same problem.
This sort of abstraction is one of the strengths of mathematics. It is not at all obvious that population growth, investment income, or nuclear decay (which we will study shortly) are all essentially the same problem. It is only when we have abstracted out the critical features of each that we can see this.
Problem 8.7.6.
(a)
What would the effective yield be for a bond nominally rated at 5% annually, compounded continuously? How does this compare to the effective yield of an investment compounded daily?
(b)
Suppose we had two investments growing continuously with nominal rates of 5% and 10% annually? After one year would the effective yield of the second investment be twice that of the first? Justify your answer.
A natural question to ask is, “How long will it take for my money to double?” Would an investment compounding continuously at a nominal rate of \(10\%\) double in half the time as one growing at a nominal rate of \(5\%\text{.}\) Take a guess and write it down for later reference.
To answer this question we’d need solve \(e^{0.05t}=2\text{,}\) for \(t\) and compare this to the solution of \(e^{0.1t}=2 \text{.}\) To solve either equation we need a way to “undo” the natural exponential function. That is, we need the inverse of the natural exponential function. This leads us to a discussion of the natural logarithm function in the next section.
