📈 Compound & Continuous Interest

01

Compound Interest · Vocabulary

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PRINCIPAL = starting money | RATE = r | TIME = t | n = compounding times/year

🔑 Core Formula

\[ A = P\!\left(1 + \frac{r}{n}\right)^{nt} \]

A = final amount, P = principal, r = annual rate (decimal), n = times compounded/year, t = years

Q1. You invest $1,000 at an annual interest rate of 6%, compounded annually (n = 1) for 1 year. What is the final amount?

✏️ Solution

Use $ A = P\left(1+\frac{r}{n}\right)^{nt} $.
$ A = 1000\left(1+\frac{0.06}{1}\right)^{1\cdot1} = 1000 \times 1.06 = \mathbf{\$1,060.00} $

Watch out! r must be a decimal: 6% → 0.06. Not 6!

02

Compound Interest · Identifying n

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n VALUES to memorize:
annually = 1 · semi-annually = 2 · quarterly = 4 · monthly = 12 · daily = 365

Q2. A bank compounds interest "quarterly." What is the value of n?

✏️ Solution

Quarterly = 4 times per year → n = 4.
Think: "quarter" = 1/4 of a year, so 4 quarters in a year.

03

Compound Interest · Interest Earned

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INTEREST EARNED = A − P (Final Amount minus Principal)

Q3. $2,000 is invested at 5% annual interest, compounded annually for 3 years. How much interest is earned?
(Round to the nearest cent)

✏️ Solution

$ A = 2000(1.05)^3 = 2000 \times 1.157625 = \$2315.25 $
Interest = $ A - P = 2315.25 - 2000 = \mathbf{\$315.25} $

Trap! Simple interest would be $2000 \times 0.05 \times 3 = \$300$. Compound interest earns MORE because you earn interest on interest!

04

Compound Interest · Monthly Compounding

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Q4. $500 is deposited at 12% annual interest, compounded monthly. What is the amount after 1 year?
(Round to the nearest cent)

💡 Hint: monthly → n = 12, so each period rate = 12%/12 = 1%

✏️ Solution

$ A = 500\!\left(1+\frac{0.12}{12}\right)^{12\cdot1} = 500(1.01)^{12} $
$ (1.01)^{12} \approx 1.126825 $
$ A \approx 500 \times 1.126825 = \mathbf{\$563.41} $

Note: Simple interest would give only $560. Monthly compounding adds extra $3.41!

05

Continuous Compounding · Introduction

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CONTINUOUS COMPOUNDING = interest is added every instant
Formula: $ A = Pe^{rt} $
e ≈ 2.71828 (Euler's number)

🔑 Continuous Formula

\[ A = Pe^{rt} \]

When n → ∞, the formula becomes this beautiful result!

Q5. $1,000 is invested at 5% continuously compounded for 2 years. What is A?
(Use $e^{0.1} \approx 1.10517$)

✏️ Solution

$ A = 1000 \cdot e^{0.05 \times 2} = 1000 \cdot e^{0.10} \approx 1000 \times 1.10517 = \mathbf{\$1105.17} $

Remember: rt goes in the exponent together. Here r=0.05, t=2, so rt = 0.10.

06

Comparing Compounding Frequencies

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MORE FREQUENT compounding → MORE MONEY
daily > monthly > quarterly > semi-annually > annually

Q6. Which option gives the most money after 5 years for the same P and r = 8%?

✏️ Solution

Continuous compounding is the mathematical limit as n → ∞, giving the maximum possible return.
Order: Continuous > Daily > Monthly > Quarterly > Annually
The difference is small, but continuous always wins!

07

Effective Annual Rate (EAR)

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EAR (Effective Annual Rate) = actual yearly return
$ \text{EAR} = \left(1+\frac{r}{n}\right)^n - 1 $ for continuous: $ \text{EAR} = e^r - 1 $

Q7. A nominal rate is 12% compounded monthly. What is the Effective Annual Rate?
(Round to 4 decimal places)

✏️ Solution

$ \text{EAR} = \left(1+\frac{0.12}{12}\right)^{12} - 1 = (1.01)^{12} - 1 $
$ = 1.126825 - 1 = 0.126825 = \mathbf{12.6825\%} $

The stated rate (12%) is the nominal rate. The EAR (12.6825%) is what you actually earn!

08

Solving for Principal (P)

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PRESENT VALUE: solve for P by dividing both sides by the growth factor.
$ P = \frac{A}{\left(1+\frac{r}{n}\right)^{nt}} $ or $ P = Ae^{-rt} $ (continuous)

Q8. How much must you invest today at 8% compounded annually to have $5,000 in 3 years?
(Round to nearest cent)

✏️ Solution

$ P = \frac{A}{\left(1+r\right)^t} = \frac{5000}{(1.08)^3} = \frac{5000}{1.259712} \approx \mathbf{\$3969.16} $

This is called discounting — finding today's value of a future amount. Key skill in finance!

09

Continuous Interest · Solving for Time

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To solve for t in $A = Pe^{rt}$:
Take ln of both sides → $ t = \frac{\ln(A/P)}{r} $

Q9. $1,000 grows to $2,000 with continuous compounding at 7% annual rate. How many years does this take?
Use $\ln 2 \approx 0.6931$

✏️ Solution

$ 2000 = 1000 e^{0.07t} \Rightarrow 2 = e^{0.07t} $
$ \ln 2 = 0.07t \Rightarrow t = \frac{0.6931}{0.07} \approx \mathbf{9.90 \text{ years}} $

🌟 Rule of 70: doubling time ≈ 70/rate% = 70/7 = 10 years. Great quick estimate!

10

Solving for Rate r

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To solve for r in $A=Pe^{rt}$:
$ r = \frac{\ln(A/P)}{t} $

Q10. $3,000 grew to $4,500 in 5 years with continuous compounding. What was the annual rate?
Use $\ln(1.5) \approx 0.4055$

✏️ Solution

$ \frac{4500}{3000} = e^{5r} \Rightarrow 1.5 = e^{5r} $
$ \ln(1.5) = 5r \Rightarrow r = \frac{0.4055}{5} \approx 0.0811 = \mathbf{8.11\%} $

11

APR vs APY — Common Confusion!

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APR = Annual Percentage Rate = nominal rate (stated)
APY = Annual Percentage Yield = effective rate (actual earned)
APY > APR (always, unless n=1)

Q11. Bank A offers 6% APR compounded monthly. Bank B offers 6.1% APR compounded annually. Which gives a higher APY?

✏️ Solution

Bank A: $ \text{APY} = (1+\frac{0.06}{12})^{12}-1 = (1.005)^{12}-1 \approx 6.168\% $
Bank B: APR = APY = 6.1% (compounded annually, n=1)
6.168% > 6.1% → Bank A wins!

Don't be tricked by a higher stated rate — always compare APY!

12

Rule of 72

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RULE OF 72: Doubling time ≈ $\frac{72}{\text{rate \%}}$
Quick mental math trick for compound interest!

Q12. At 9% annual compound interest, approximately how many years to double your money?

✏️ Solution

$ \text{Doubling time} \approx \frac{72}{9} = 8 \text{ years} $

Exact answer: $ t = \frac{\ln 2}{\ln 1.09} = \frac{0.6931}{0.0862} \approx 8.04 $ years
The Rule of 72 is a great approximation!

13

Semi-Annual Compounding + Solving for t

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Solving for t in compound formula:
$ t = \frac{\ln(A/P)}{n \cdot \ln(1+r/n)} $

Q13. $4,000 is invested at 6% compounded semi-annually. How long until it reaches $8,000?
Use $\ln 2 \approx 0.6931$, $\ln(1.03) \approx 0.02956$

✏️ Solution

$ 8000 = 4000(1.03)^{2t} \Rightarrow 2 = (1.03)^{2t} $
$ \ln 2 = 2t \cdot \ln(1.03) \Rightarrow t = \frac{0.6931}{2 \times 0.02956} = \frac{0.6931}{0.05912} \approx \mathbf{11.72 \text{ yrs}} $

14

Continuous Compounding — Present Value

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Q14. How much do you need to invest today with continuous compounding at 5% to have $10,000 in 8 years?
Use $e^{0.4} \approx 1.4918$

✏️ Solution

$ P = Ae^{-rt} = 10000 \cdot e^{-0.05 \times 8} = 10000 \cdot e^{-0.4} $
$ = \frac{10000}{1.4918} \approx \mathbf{\$6703.20} $

The negative exponent means we're going backwards in time — discounting!

15

Continuous vs Monthly — Numerical Comparison

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Q15. $2,000 at 10% for 5 years. What's the difference between continuous and monthly compounding?
$e^{0.5} \approx 1.6487$, $(1+\frac{0.10}{12})^{60} \approx 1.6453$

✏️ Solution

Continuous: $ 2000 \times 1.6487 = \$3297.44 $
Monthly: $ 2000 \times 1.6453 = \$3290.62 $
Difference: $ 3297.44 - 3290.62 = \mathbf{\$6.82} $

Continuous compounding sounds amazing but in practice, the difference is tiny!

16

Tripling Time — Continuous

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TRIPLING: set $A/P = 3$, then $t = \frac{\ln 3}{r}$
$\ln 3 \approx 1.0986$

Q16. At what annual rate (continuous compounding) does money triple in 12 years?
$\ln 3 \approx 1.0986$

✏️ Solution

$ 3 = e^{12r} \Rightarrow \ln 3 = 12r \Rightarrow r = \frac{1.0986}{12} \approx 0.09155 \approx \mathbf{9.16\%} $

17

Inflation — Real Value

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REAL RATE ≈ nominal rate − inflation rate
Exact: $ r_{real} = \frac{1+r_{nominal}}{1+r_{inflation}} - 1 $

Q17. An account earns 8% compounded annually. Inflation is 3% per year. What is the exact real annual rate of return?

✏️ Solution

$ r_{real} = \frac{1.08}{1.03} - 1 = 1.04854 - 1 = \mathbf{4.854\%} $

Common mistake: subtracting 8% - 3% = 5% is an approximation. The exact answer is 4.854%. The approximation only works for small rates!

18

Two Different Investments — Which Wins?

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Q18. Investment A: $5,000 at 7% compounded quarterly for 10 years.
Investment B: $4,500 at 8% compounded continuously for 10 years.
Which has a higher final value?
$(1.0175)^{40} \approx 2.0016$ · $e^{0.8} \approx 2.2255$

✏️ Solution

A: $ 5000 \times 2.0016 = \$10,008 $
B: $ 4500 \times e^{0.8} = 4500 \times 2.2255 = \$10,015 $
Investment B wins by ~$7 despite a smaller principal, because the higher rate + continuous compounding overcomes the $500 difference!

19

Deriving the Continuous Formula

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KEY LIMIT: $\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n = e$
This is how $Pe^{rt}$ comes from $P\!\left(1+\frac{r}{n}\right)^{nt}$ as $n \to \infty$

Q19. Which expression correctly shows the step connecting compound to continuous interest?

✏️ Solution

Let $ m = n/r $, so $ n = mr $. Then:
$ \left(1+\frac{r}{n}\right)^{nt} = \left(1+\frac{1}{m}\right)^{mrt} = \left[\left(1+\frac{1}{m}\right)^m\right]^{rt} \to e^{rt} $

As $n\to\infty$, $m\to\infty$ too. The key is that both n and t appear in the exponent as nt — not just one of them!

20

⚡ BOSS LEVEL — Multi-Step

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STRATEGY: complex problems → break into steps → apply formulas one at a time

Q20. Alice invests $10,000 at 6% compounded monthly. Bob invests $X at 6% continuous compounding. Both invest for 20 years. For Bob to end up with the same amount as Alice, what should X be?
$(1.005)^{240} \approx 3.3102$ · $e^{1.2} \approx 3.3201$

✏️ Solution

Step 1 — Alice's final amount:
$ A_{Alice} = 10000(1.005)^{240} = 10000 \times 3.3102 = \$33,102 $

Step 2 — Set Bob equal to Alice:
$ X \cdot e^{0.06 \times 20} = 33102 \Rightarrow X \cdot e^{1.2} = 33102 $
$ X = \frac{33102}{3.3201} \approx \mathbf{\$9,970} $

Bob needs slightly less principal because continuous compounding is marginally more efficient!