Understanding Compounding
Compound interest is the addition of interest to the principal sum of a loan or deposit, or in other words, interest on interest. It is the result of reinvesting interest, rather than paying it out, so that interest in the next period is then earned on the principal sum plus previously accumulated interest.
Periodic Compounding Formula
The standard formula for compound interest compounded at set periodic intervals (such as annually, monthly, or daily) is:
A = P (1 + r/n)nt
Where:
- A: The final amount of money accumulated after n years, including interest.
- P: The principal investment amount (the initial deposit).
- r: The annual nominal interest rate (decimal representation).
- n: The number of times that interest is compounded per year.
- t: The number of years the money is invested.
If you compound interest monthly ($n = 12$) or daily ($n = 365$), the frequency increases, yielding higher total returns for the same nominal interest rate. This is because interest is earned and added to the principal balance sooner, allowing subsequent interest calculations to be based on larger amounts.
Continuous Compounding and Euler's Number (e)
What happens if we compound interest more and more frequently? What if the compounding intervals become infinitely small ($n \to \infty$)? This is the concept of **Continuous Compounding**.
To derive the continuous compounding formula, we analyze the limit of the periodic formula as $n$ approaches infinity:
limn → ∞ P (1 + r/n)nt
Let's define a new variable $m = n/r$. As $n$ approaches infinity, $m$ also approaches infinity. We can rewrite the expression as:
P [ limm → ∞ (1 + 1/m)m ] rt
By mathematical definition, the limit inside the brackets is equal to **Euler's Number** ($e \approx 2.71828$):
e = limm → ∞ (1 + 1/m)m
Substituting $e$ back into the formula yields the famous continuous compounding equation:
A = P ert
Visualizing Compounding Frequency Effects
Assume an initial principal of $P = \$10,000$ at a nominal rate of $r = 10\%$ ($0.10$) over a period of $t = 1$ year. Let us look at the final amount accumulated ($A$) based on different compounding frequencies ($n$):
- Annually (n=1): $10,000 \times (1 + 0.10/1)^1 = \$11,000.00$
- Quarterly (n=4): $10,000 \times (1 + 0.10/4)^4 = \$11,038.13$
- Monthly (n=12): $10,000 \times (1 + 0.10/12)^{12} = \$11,047.13$
- Daily (n=365): $10,000 \times (1 + 0.10/365)^{365} = \$11,051.56$
- Continuously: $10,000 \times e^{0.10 \times 1} = \$11,051.71$
The difference between daily compounding and continuous compounding is very small (only $\$0.15$ on a $\$10,000$ investment), showing that the compounding curve flattens significantly as $n$ grows large.