Compound interest and continuous growth are exponential models. They describe situations where the amount
grows by a percentage of the current amount, not by a fixed amount.
1. Compound interest formula
The standard compound interest formula is:
\[
A=P\left(1+\frac{r}{n}\right)^{nt}
\]
Here:
- \(A\) is the final amount.
- \(P\) is the initial amount, also called the principal.
- \(r\) is the annual rate written as a decimal.
- \(n\) is the number of compounding periods per year.
- \(t\) is the time in years.
For example, \(6\%\) is written as:
\[
r=0.06
\]
2. Meaning of compounding frequency
The value \(n\) tells how many times interest is added each year.
- Annually: \(n=1\)
- Semiannually: \(n=2\)
- Quarterly: \(n=4\)
- Monthly: \(n=12\)
- Weekly: \(n=52\)
- Daily: \(n=365\)
A larger \(n\) usually gives a slightly larger final amount when the rate is positive.
3. Continuous growth formula
Continuous compounding is the limiting case where compounding happens all the time.
The formula is:
\[
A=Pe^{rt}
\]
Continuous growth uses the number \(e\), which appears naturally in exponential growth and calculus.
4. Effective annual rate
The effective annual rate tells the actual one-year growth after compounding.
For discrete compounding:
\[
\text{EAR}=\left(1+\frac{r}{n}\right)^n-1
\]
For continuous growth:
\[
\text{EAR}=e^r-1
\]
The effective annual rate is usually larger than the nominal annual rate when compounding happens more
than once per year.
5. Solving for the initial amount
If the final amount is known, the compound interest formula can be rearranged to solve for \(P\).
\[
P=\frac{A}{\left(1+\frac{r}{n}\right)^{nt}}
\]
For continuous growth:
\[
P=Ae^{-rt}
\]
6. Solving for the rate
For discrete compounding, solve for \(r\) using:
\[
r=n\left[\left(\frac{A}{P}\right)^{1/(nt)}-1\right]
\]
For continuous growth, solve for \(r\) using:
\[
r=\frac{\ln(A/P)}{t}
\]
7. Solving for time
For discrete compounding, solve for \(t\) using:
\[
t=\frac{\ln(A/P)}{n\ln(1+r/n)}
\]
For continuous growth, solve for \(t\) using:
\[
t=\frac{\ln(A/P)}{r}
\]
Logarithms are needed because the unknown time is in the exponent.
8. Doubling and halving time
If the rate is positive, the doubling time tells how long it takes for the amount to become twice as large.
For continuous growth:
\[
T_{\text{double}}=\frac{\ln(2)}{r}
\]
If the rate is negative, the halving time tells how long it takes for the amount to become half as large.
\[
T_{\text{half}}=\frac{\ln(0.5)}{r}
\]
9. Worked example
Suppose \(P=5000\), \(r=0.06\), \(n=12\), and \(t=8\).
\[
A=5000\left(1+\frac{0.06}{12}\right)^{12\cdot8}
\]
Simplify inside the parentheses:
\[
1+\frac{0.06}{12}=1.005
\]
Then:
\[
A=5000(1.005)^{96}
\]
So the final amount is approximately:
\[
A\approx 8070.71
\]
10. Continuous comparison
With the same \(P=5000\), \(r=0.06\), and \(t=8\), continuous growth gives:
\[
A=5000e^{0.06\cdot8}
\]
\[
A=5000e^{0.48}
\]
\[
A\approx 8080.85
\]
Continuous compounding gives a slightly larger amount than monthly compounding for the same nominal rate.
11. Summary table
| Concept |
Formula |
Meaning |
| Compound interest |
\(A=P(1+r/n)^{nt}\) |
Interest is added \(n\) times per year. |
| Continuous growth |
\(A=Pe^{rt}\) |
Compounding happens continuously. |
| Discrete effective annual rate |
\(\text{EAR}=(1+r/n)^n-1\) |
Actual annual growth after compounding. |
| Continuous effective annual rate |
\(\text{EAR}=e^r-1\) |
Actual annual growth with continuous compounding. |
| Initial amount |
\(P=A/(1+r/n)^{nt}\) |
Starting amount needed to reach \(A\). |
| Continuous rate |
\(r=\ln(A/P)/t\) |
Rate needed for continuous growth. |
| Continuous time |
\(t=\ln(A/P)/r\) |
Time needed to grow from \(P\) to \(A\). |
