4.2: Compound Interest
- Page ID
- 207994
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Applying the Compound-Interest Formula
Savings instruments, such as mutual funds and retirement accounts, that continually reinvest earnings use compound interest. The term compounding refers to interest earned not only on the original value, but on the accumulated value of the account.
The annual percentage rate (APR) of an account, also called the nominal rate, is the yearly interest rate earned by an investment account. The term nominal is used when the compounding occurs a number of times other than once per year. In fact, when interest is compounded more than once a year, the effective interest rate ends up being greater than the nominal rate! This is a powerful tool for investing.
We can calculate the compound interest using the compound interest formula, which is an exponential function of the variables time \(t\), principal \(P\), \(APR\) \(r\), and number of compounding periods in a year \(n\):
\[A(t)=P{\left (1+\dfrac{r}{n} \right )}^{nt} \nonumber\]
For example, observe Table \(\PageIndex{4}\), which shows the result of investing \($1,000\) at \(10\%\) for one year. Notice how the value of the account increases as the compounding frequency increases.
| Frequency | Value after \(1\) year |
|---|---|
| Annually | \($1100\) |
| Semiannually | \($1102.50\) |
| Quarterly | \($1103.81\) |
| Monthly | \($1104.71\) |
| Daily | \($1105.16\) |
If we invest \($3,000\) in an investment account paying \(3\%\) interest compounded quarterly, how much will the account be worth in \(10\) years?
Solution
Because we are starting with \($3,000\), \(P=3000\). Our interest rate is \(3\%\), so \(r = 0.03\). Because we are compounding quarterly, we are compounding \(4\) times per year, so \(n=4\). We want to know the value of the account in \(10\) years, so we are looking for \(A(10)\),the value when \(t = 10\).
\[\begin{align*} A(t)&= P{\left (1+\dfrac{r}{n} \right )}^{nt} \qquad \text{Use the compound interest formula}\\ A(10)&= 3000{\left (1+\dfrac{0.03}{4} \right )}^{(4)\cdot (10)} \qquad \text{Substitute using given values}\\ &\approx \$4045.05 \qquad \text{Round to two decimal places} \end{align*}\]
The account will be worth about \($4,045.05\) in \(10\) years.
A 529 Plan is a college-savings plan that allows relatives to invest money to pay for a child’s future college tuition; the account grows tax-free. Lily wants to set up a 529 account for her new granddaughter and wants the account to grow to \($40,000\) over \(18\) years. She believes the account will earn \(6\%\) compounded semi-annually (twice a year). To the nearest dollar, how much will Lily need to invest in the account now?
Solution
The nominal interest rate is \(6\%\), so \(r=0.06\). Interest is compounded twice a year, so \(k=2\).
We want to find the initial investment, \(P\), needed so that the value of the account will be worth \($40,000\) in \(18\) years. Substitute the given values into the compound interest formula, and solve for \(P\).
\[\begin{align*} A(t)&= P{\left (1+\dfrac{r}{n} \right )}^{nt} \qquad \text{Use the compound interest formula}\\ 40,000&= P{\left (1+\dfrac{0.06}{2} \right )}^{2(18)} \qquad \text{Substitute using given values } A, r, n, t\\ 40,000&= P{(1.03)}^{36} \qquad \text{Simplify}\\ \dfrac{40,000}{ {(1.03)}^{36} }&= P \qquad \text{Isolate } P\\ P&\approx \$13,801 \qquad \text{Divide and round to the nearest dollar} \end{align*}\]
Lily will need to invest \($13,801\) to have \($40,000\) in \(18\) years.
Evaluating Functions with Base \(e\)
As we saw earlier, the amount earned on an account increases as the compounding frequency increases. Table \(\PageIndex{5}\) shows that the increase from annual to semi-annual compounding is larger than the increase from monthly to daily compounding. This might lead us to ask whether this pattern will continue.
Examine the value of \($1\) invested at \(100\%\) interest for \(1\) year, compounded at various frequencies, listed in Table \(\PageIndex{5}\).
| Frequency | \(A(t)={\left (1+\dfrac{1}{n} \right )}^n\) | Value |
|---|---|---|
| Annually | \({\left (1+\dfrac{1}{1} \right )}^1\) | \($2\) |
| Semiannually | \({\left (1+\dfrac{1}{2} \right )}^2\) | \($2.25\) |
| Quarterly | \({\left (1+\dfrac{1}{4} \right )}^4\) | \($2.441406\) |
| Monthly | \({\left (1+\dfrac{1}{12} \right )}^{12}\) | \($2.613035\) |
| Daily | \({\left (1+\dfrac{1}{365} \right )}^{365}\) | \($2.714567\) |
| Hourly | \({\left (1+\dfrac{1}{8760} \right )}^{8760}\) | \($2.718127\) |
| Once per minute | \({\left (1+\dfrac{1}{525600} \right )}^{525600}\) | \($2.718279\) |
| Once per second | \({\left (1+\dfrac{1}{31536000} \right )}^{31536000}\) | \($2.718282\) |
These values appear to be approaching a limit as \(n\) increases without bound. In fact, as \(n\) gets larger and larger, the expression \({\left (1+\dfrac{1}{n} \right )}^n\) approaches a number used so frequently in mathematics that it has its own name: the letter \(e\). This value is an irrational number, which means that its decimal expansion goes on forever without repeating. The letter \(e\) is used as a base for many real-world exponential models. To work with base \(e\), we use the approximation, \(e≈2.718282\). The constant was named by the Swiss mathematician Leonhard Euler (1707–1783) who first investigated and discovered many of its properties. Its approximation to six decimal places is shown below.
Investigating Continuous Growth
So far we have worked with rational bases for exponential functions. For most real-world phenomena, however, \(e\) is used as the base for exponential functions. Exponential models that use \(e\) as the base are called continuous growth or decay models. We see these models in finance, computer science, and most of the sciences, such as physics, toxicology, and fluid dynamics.
For all real numbers \(t\),and all positive numbers \(a\) and \(r\), continuous growth or decay is represented by the formula
\[A(t)=ae^{rt}\]
where
- \(a\) is the initial value,
- \(r\) is the continuous growth rate per unit time,
- \(t\) is the elapsed time.
If \(r>0\), then the formula represents continuous growth. If \(r<0\), then the formula represents continuous decay.
For business applications, the continuous growth formula is called the continuous compounding formula and takes the form
\[A(t)=Pe^{rt}\]
where
- \(P\) is the principal or the initial invested,
- \(r\) is the growth or interest rate per unit time,
- \(t\) is the period or term of the investment.
A person invested \($1,000\) in an account earning a nominal \(10\%\) per year compounded continuously. How much was in the account at the end of one year?
Solution
Since the account is growing in value, this is a continuous compounding problem with growth rate \(r=0.10\). The initial investment was \($1,000\), so \(P=1000\). We use the continuous compounding formula to find the value after \(t=1\) year:
\[\begin{align*} A(t)&= Pe^{rt} \qquad \text{Use the continuous compounding formula}\\ &= 1000{(e)}^{0.1} \qquad \text{Substitute known values for } P, r, t\\ &\approx 1105.17 \qquad \text{Use a calculator to approximate} \end{align*}\]
The account is worth \($1,105.17\) after one year.
\(Radon-222\) decays at a continuous rate of \(17.3\%\) per day. How much will \(100 mg\) of \(Radon-222\) decay to in \(3\) days?
Solution
Since the substance is decaying, the rate, \(17.3\%\), is negative. So, \(r = −0.173\). The initial amount of \(Radon-222\) was \(100\) mg, so \(a=100\). We use the continuous decay formula to find the value after \(t=3\) days:
\[\begin{align*} A(t)&= ae^{rt} \qquad \text{Use the continuous growth formula}\\ &= 100e6{-0.173(3)} \qquad \text{Substitute known values for } a, r, t\\ &\approx 59.5115 \qquad \text{Use a calculator to approximate} \end{align*}\]
So \(59.5115\) mg of \(Radon-222\) will remain.
Key Equations
| compound interest formula |
\(A(t)=P{(1+\dfrac{r}{n})}^{nt}\), where \(A(t)\) is the account value at time \(t\) \(t\) is the number of years \(P\) is the initial investment, often called the principal \(r\) is the annual percentage rate (APR), or nominal rate \(n\) is the number of compounding periods in one year |
| continuous growth formula | \(A(t)=ae^{rt}\), where \(t\) is the number of unit time periods of growth \(a\) is the starting amount (in the continuous compounding formula a is replaced with \(P\), the principal) \(e\) is the mathematical constant, \(e≈2.718282\) |
Key Concepts
- An exponential function is defined as a function with a positive constant other than \(1\) raised to a variable exponent.
- The value of an account at any time \(t\) can be calculated using the compound interest formula when the principal, annual interest rate, and compounding periods are known.
- The initial investment of an account can be found using the compound interest formula when the value of the account, annual interest rate, compounding periods, and life span of the account are known.
- The number \(e\) is a mathematical constant often used as the base of real-world exponential growth and decay models. Its decimal approximation is \(e≈2.718282\).
- Continuous growth or decay models are exponential models that use \(e\) as the base. Continuous growth and decay models can be found when the initial value and growth or decay rate are known.