Questions for Technical Minds

How is the compound interest formula derived

Last updated: 2026-11-10Author: WealthSyntax Editorial Team

#compounding

Short answer

The compound interest formula is derived by applying an interest rate r repeatedly to an accumulating balance. One period yields A_1 = P(1 + r); n periods produce the recurrence A_n = A_(n−1)(1 + r), resolving to A_n = P(1 + r)^n. Dividing annual rate r across k periods yields P(1 + r/k)^(kn), approaching Pe^(rn) under continuous compounding.

Educational content only, not financial advice. Examples use constant, hypothetical returns.

Mathematical Foundation

Formula and recurrence relations

Single Period: A_1 = P(1 + r)
Recurrence: A_n = A_(n−1) · (1 + r)
Closed Form: A_n = P(1 + r)^n
k Periods per Year: A_n = P(1 + r/k)^(kn)
Continuous Compounding: A_n = P · e^(rn)
With Regular Contribution: A_n = P(1 + i)^n + C · ((1 + i)^n − 1) / i

Symbol Legend:

P:Principal starting capital base

r:Nominal annual interest rate (as a decimal fraction)

k:Compounding frequency per year (e.g. 1=annual, 4=quarterly, 12=monthly)

n:Total elapsed time duration in years

A_n:Accumulated terminal balance at year n

C:Periodic cash contribution deposited at the end of each interval

i:Periodic interest rate applied each interval: i = r / k

1. One Period and the Recurrence Relation

Consider an initial principal P earning interest at an annual rate r over a single year. The interest generated at the end of the year is P * r. Adding this earned interest back to the starting principal yields the balance at the end of period 1:

A_1 = P + P*r = P(1 + r)

In the second period, interest is earned not merely on original principal P, but on the entire accumulated balance A_1. Thus A_2 = A_1(1 + r) = [P(1 + r)](1 + r) = P(1 + r)^2. For any period n, the recurrence relation is A_n = A_(n-1)(1 + r).

2. Closed Form and Proof by Mathematical Induction

Expanding the recurrence leads to the conjecture: A_n = P(1 + r)^n for all positive integers n. We prove this formally by mathematical induction.

Base Step: For n = 1, A_1 = P(1 + r)^1 = P(1 + r), which is true by definition.

Inductive Step: Assume the proposition holds for n = m, so A_m = P(1 + r)^m. By the recurrence relation, A_(m+1) = A_m(1 + r). Substituting the inductive hypothesis yields A_(m+1) = [P(1 + r)^m](1 + r) = P(1 + r)^(m+1).

Since the base case holds and the truth for m implies the truth for m + 1, the closed-form equation A_n = P(1 + r)^n holds for all positive integers n.

3. Compounding Frequency k and the Continuous Limit e^(rn)

When interest compounds k times per year, the annual rate r is divided into k equal periodic rates r/k, and the total number of compounding intervals becomes kn. The formula becomes A_n = P(1 + r/k)^(kn).

As compounding frequency approaches infinity (k -> ∞), interest compounds continuously. Let m = k / r, so k = mr. As k -> ∞, m -> ∞. Substituting m yields:

lim_(k->∞) P [ (1 + 1/m)^m ]^(rn) = P · e^(rn)

This proves why continuous compounding reaches Euler’s constant e as its theoretical mathematical ceiling.

4. Adding a Regular Periodic Contribution (Annuity Series)

When adding a regular contribution C at the end of each period, each deposit compounds for the remaining time. Deposit 1 compounds for (n-1) periods, while deposit n compounds for 0 periods:

Total Contributions Future Value = C(1 + i)^(n-1) + C(1 + i)^(n-2) + ... + C(1 + i)^0 = C * sum_(t=0)^(n-1) (1 + i)^t.

This is a finite geometric series with first term a = 1, common ratio q = (1 + i), and n terms. Applying the geometric series summation formula S_n = (q^n - 1) / (q - 1) gives C * ((1 + i)^n - 1) / i.

Combining the principal growth and contribution annuity yields the complete master formula: A_n = P(1 + i)^n + C * ((1 + i)^n - 1) / i.

Verified Worked Example

Compounding Frequency Scaling: $10,000 at 7% Over 10 Years

Type: frequency-scaling-schedule

We evaluate how terminal wealth changes as compounding frequency k shifts from annual to quarterly, monthly, daily, and continuous.

Terminal Wealth Comparison by Compounding Frequency on $10,000 at 7% for 10 Years

Compounding Frequency	Frequency k	Mathematical Formula	Terminal Balance	Gain vs Annual
Annual	1	P(1 + r)^n	$19,671.51	Baseline
Quarterly	4	P(1 + r/4)^(4n)	$20,015.97	+$344.46
Monthly	12	P(1 + r/12)^(12n)	$20,096.61	+$425.10
Daily (365)	365	P(1 + r/365)^(365n)	$20,136.18	+$464.66
Continuous	continuous	P · e^(rn)	$20,137.53	+$466.01

Numbered Step-by-Step Procedure:

1Step 1: Annual Compounding (k = 1)

$10,000 * (1 + 0.07)^10 = $10,000 * 1.967151 = $19,671.51.

$19,671.51

2Step 2: Monthly Compounding (k = 12)

$10,000 * (1 + 0.07/12)^120 = $10,000 * 2.009661 = $20,096.61 (+$425.10 vs annual).

$20,096.61

3Step 3: Continuous Compounding (k -> ∞)

$10,000 * e^(0.07 * 10) = $10,000 * e^0.70 = $10,000 * 2.013753 = $20,137.53 (+$466.02 vs annual).

$20,137.53

Observation: Increasing compounding frequency yields diminishing returns: moving from annual to monthly adds $425.10, but moving from monthly to continuous adds only $40.92.

Analytical Pitfalls

Common mistakes

✕Confusing compounding frequency k with horizon duration n (e.g. entering monthly frequency as elapsed months).
✕Omitting parentheses around exponents in code: evaluating P*(1+r)^k*n evaluates as [P*(1+r)^k] * n rather than ^(kn).
✕Mixing up ordinary annuity (deposits at month-end) with annuity due (deposits at month-start), which changes the exponent of each deposit by +1 period.

Try it interactively