Difficulty: Intermediate | Prerequisites: Exponent rules (Ch. 4 part 1), basic algebra
This section takes the exponent rules from the first half of the chapter and puts them to work in finance. You will learn how money grows when interest is compounded at different frequencies, how continuous compounding leads to Euler's number e, and how to reverse the process to find what a future sum is worth today. These formulas are the foundation of time-value-of-money reasoning, which runs through the rest of the course and through most of business finance.
Compound interest means earning interest on your interest, and the more frequently it compounds, the more your money grows. The future-value formula B(t) = P(1 + r/k)^(kt) tells you what an investment is worth after t years. Present value reverses the question: how much must you invest today to reach a target amount later.
Principal (P)
The initial amount of money invested or borrowed before any interest is applied.
Think of it as your starting balance.
Annual interest rate (r)
The percentage of the principal earned (or owed) per year, expressed as a decimal (e.g. 7% = 0.07).
In simple terms, this is the yearly "price" of borrowing money, or the yearly "reward" for lending it.
Compounding periods per year (k)
The number of times interest is calculated and added to the balance each year. Quarterly = 4, monthly = 12, daily = 365.
Think of it as how often the bank stops to add interest to your running total.
Future value, B(t)
The total amount an investment is worth after t years of compounding.
In simple terms, this is what your money grows into.
Present value, P(t)
The amount you would need to invest today, at a given rate, to reach a specific future value.
Think of it as the "today price" of a future sum of money.
Continuous compounding
The theoretical limit of compounding infinitely many times per year. Uses the formula B(t) = Pe^(rt).
In simple terms, interest is being added at every conceivable instant, not just monthly or daily.
Euler's number (e)
A mathematical constant approximately equal to 2.71828, defined as the limit of (1 + 1/n)^n as n approaches infinity.
Think of it as the number that naturally emerges when you compound interest as frequently as possible.
Simple interest on a principal P at rate r for one period gives P + rP = P(1 + r). That is one multiplication by (1 + r).
When interest is compounded k times per year, each period uses rate r/k instead of r. After one compounding period, the balance is P(1 + r/k). After two periods, the first period's balance earns interest again:
After 1 period: P₁ = P(1 + r/k)
After 2 periods: P₂ = P₁(1 + r/k) = P(1 + r/k)²
After kt periods (t years): B(t) = P(1 + r/k)^(kt)
This is the compound interest formula. Each period, the base grows because the previous interest is included in the next calculation.
The lecture demonstrates the effect of compounding frequency by investing $1,000 at 6% for 10 years:
Quarterly (k = 4): 1000(1 + 0.06/4)^(4·10) = $1,814.02
Monthly (k = 12): 1000(1 + 0.06/12)^(12·10) = $1,819.40
Daily (k = 365): 1000(1 + 0.06/365)^(365·10) = $1,822.03
Continuously: 1000e^(0.06·10) = $1,822.12
Notice that the returns increase as k grows, but with diminishing gains. The jump from quarterly to monthly is about $5; from daily to continuous is only $0.09.
As you increase k toward infinity, (1 + 1/n)^n approaches a limit: e ≈ 2.71828.
The derivation links the general formula to the continuous formula. Substitute n = k/r so that k = nr:
P(1 + r/k)^(kt) = P((1 + 1/n)^n)^(rt)
As k (and therefore n) approaches infinity, (1 + 1/n)^n approaches e.
So the limit is Pe^(rt).
This gives the continuous compounding formula: B(t) = Pe^(rt).
Present value reverses the future-value formula. If you know you need B dollars in t years, and the annual rate is r compounded k times per year:
Compound: P(t) = B(1 + r/k)^(-kt)
Continuous: P(t) = Be^(-rt)
You are solving for the principal P that would grow to B.
You want $5,000 in 4 years at 7% interest. How much must you invest now?
Compounded quarterly: 5000(1 + 0.07/4)^(-4·4) = $3,788.08
Compounded continuously: 5000e^(-0.07·4) = $3,778.92
The continuously compounded present value is slightly lower because continuous compounding grows money marginally faster, so you need a slightly smaller starting amount to reach the same target.
Compound interest (future value):
B(t) = P\left(1 + \frac{r}{k}\right)^{kt}where P = principal, r = annual rate, k = compounding periods per year, t = years.
Continuous compounding (future value):
B(t) = Pe^{rt}Present value (compound):
P(t) = B\left(1 + \frac{r}{k}\right)^{-kt}Present value (continuous):
P(t) = Be^{-rt}Definition of e:
e = \lim_{n \to \infty}\left(1 + \frac{1}{n}\right)^n \approx 2.71828Every savings account, mortgage, car loan, and bond uses these formulas. When a bank advertises an APY (annual percentage yield), it is the effective annual rate that accounts for compounding. Present value is how businesses decide whether an investment is worth making: if the present value of future cash flows exceeds the cost, the project is profitable.
Students often plug the annual rate r directly into the formula without dividing by k. If interest is compounded quarterly at 6%, the per-period rate is 0.06/4 = 0.015, not 0.06.
Students sometimes forget to multiply the exponent by t. The exponent is kt (periods per year times years), not just k.
When calculating present value, students occasionally use a positive exponent instead of a negative one. Present value discounts backward, so the exponent is negative: (1 + r/k)^(-kt).
Students sometimes assume continuous compounding gives dramatically more than daily compounding. As the $1,000 example shows, the difference is typically pennies.
⚠️ Future value and present value problems are near-certain exam questions. Know both the compound and continuous versions.
⚠️ Be prepared for problems that give you the future value and ask for the present value (or vice versa). Recognise which formula to use from the problem's phrasing: "how much will you have" = future value; "how much should you invest now" = present value.
⚠️ The derivation of e from the compound-interest formula (showing how (1 + 1/n)^n approaches e) is a conceptual question professors like to test. Understand it, do not just memorise it.
True or false: if interest is compounded monthly, k = 4.
False. Monthly means k = 12. Quarterly is k = 4.
Fill in the blank: the continuous compounding formula is B(t) = P · e^(___).
rt.
True or false: present value will always be less than future value (assuming a positive interest rate).
True. Money today grows, so you need less today to reach a larger amount later.
Fill in the blank: e ≈ ___.
2.71828.
True or false: compounding daily gives substantially more than compounding monthly.
False. The difference is small, as the lecture example shows ($1,822.03 vs $1,819.40 on $1,000 over 10 years).
Q: You invest $2,000 at 5% compounded quarterly for 6 years. What is the future value?
A: B(6) = 2000(1 + 0.05/4)^(4·6) = 2000(1.0125)^24 ≈ $2,693.78.
Q: You invest $1,000 at 4% compounded continuously for 10 years. What is the future value?
A: B(10) = 1000e^(0.04·10) = 1000e^0.4 ≈ $1,491.82.
Q: You need $10,000 in 5 years. The annual rate is 6% compounded monthly. How much should you invest today?
A: P = 10000(1 + 0.06/12)^(-12·5) = 10000(1.005)^(-60) ≈ $7,413.72.
Q: Explain why the continuous compounding formula uses e.
A: As the number of compounding periods k grows without bound, (1 + r/k)^(kt) approaches Pe^(rt). This is because (1 + 1/n)^n converges to e as n approaches infinity, and the compound-interest formula can be rewritten in terms of this limit by substituting n = k/r.
Q: A trip will cost $5,000 in 4 years. At 7% compounded quarterly, how much should you invest now?
A: P = 5000(1 + 0.07/4)^(-4·4) = 5000(1.0175)^(-16) ≈ $3,788.08.
This connects to derivatives and integration because continuous growth models (Pe^(rt)) are differentiated and integrated in later chapters when you study marginal analysis and accumulated change. Present value is the basis for net present value (NPV), which you will use in business decision-making. The exponential function e^x, introduced here through the interest context, becomes a central function in differential calculus.
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