First — talk for 2 minHas anyone in your family had a savings account or invested money? What grew the most? What didn't grow at all? If no one, Mr. Bennett will share.

How fast can money grow?

In 1683, a Swiss mathematician named Jakob Bernoulli asked a strange question. Imagine a bank that pays you 100% interest each year on a $1 deposit. At year's end, you would have $2. But what if the bank added the interest twice a year — 50% halfway, then another 50% on the new total? You would do a little better. Every month? Better still. Every second?

Bernoulli could not write the answer down as a fraction or a square root. Fifty years later, the mathematician Euler gave the limit a name: the letter e. Like π, it is irrational — its decimal never repeats or ends. And like π, it shows up everywhere — in finance, in biology, even in how a hot drink cools.

Today we will use e to look at two families. Both invest at the same rate. Both wait the same time. Family A starts with $10,000; Family B starts with $1,000,000. The dollar gap between them at the start is $990,000. What happens when interest compounds not annually, not monthly, not daily, but continuously — at every instant — and what does that limit do to the gap after forty years?

Worked Example #1 — Bernoulli's compounding question (build the limit)
ProblemA bank pays 100% interest per year on $1. If it compounds n times per year, after 1 year you have \(P\!\left(1 + \dfrac{r}{n}\right)^{n t}\). With \(P=1\), \(r=1\), \(t=1\), this becomes \(\left(1 + \tfrac{1}{n}\right)^{n}\). What does the value approach as n grows very large?
DataCompute the value for several n. As n grows, the base \(1 + \tfrac{1}{n}\) shrinks toward 1, but the exponent n grows without bound.
Word Bank exponent base level off irrational limit infinity round up subtract
Number Bank 2.0000 2.2500 2.7146 2.71828… 3.1416 2.5000
#Reasoning (fill in)Math
1 For \(n = 1\) (annually), the is \(1 + 1 = 2\). Raise to the power 1. \(\left(1+\tfrac{1}{1}\right)^{1} = \)
2 For \(n = 2\) (semi-annually), the base is \(1.5\). Raise to the 2. \(\left(1+\tfrac{1}{2}\right)^{2} = 1.5^{2} = \)
3 For \(n = 365\) (daily), use a calculator. The value is just over 2.7. \(\left(1+\tfrac{1}{365}\right)^{365} \approx \)
4 The values — they do not blow up to infinity. The ceiling is a special number. \(\displaystyle \lim_{n \to \infty}\left(1+\tfrac{1}{n}\right)^{n} = \) \(e\)
5 The number e is — its decimal never repeats. Write its first few digits. \(e \approx \)
Worked Example #2 — Family A: continuous compounding for 40 years
ProblemFamily A invests a starting principal at a continuous rate. Use the formula \(A = P \cdot e^{r t}\) to find the amount after 40 years.
DataFamily A starts with P = $10,000. The S&P 500 has returned about r = 0.07 per year (7%, after inflation) over the last 50 years. The horizon is t = 40 years (one career to retirement).
Word Bank multiply exponent principal continuous substitute divide add up
Number Bank 2.8 16.4446 164,446 280,000 10,000 2.7183 47
#Reasoning (fill in)Math
1 Identify the P, the rate r, and the time t. \(P = \) , \(r = 0.07\), \(t = 40\)
2 Compute the \(r \cdot t\) first. \(r \cdot t = 0.07 \cdot 40 = \)
3 Use the calculator's \(e^{x}\) key to find \(e^{x}\) at \(x = 2.8\). \(e^{2.8} \approx \)
4 the principal P by the growth factor \(e^{2.8}\). \(A = 10{,}000 \cdot 16.4446\)
5 Round to the nearest dollar. Family A's $10,000 grows to about $. \(A \approx \$164{,}446\)
Activity 2 — Family B & the gap (fill in from your work)
Family B: P = $1,000,000, r = 0.07, t = 40. Same multiplier \(e^{2.8} \approx\) . So \(A_B = 1{,}000{,}000 \cdot e^{2.8} \approx\) $.
Dollar gap (B − A): at year 0 the gap is $. At year 40 the gap is about $.
Ratio B/A: at year 0 it is ×. At year 40 it is ×. The ratio is (the same / different) — circle one.
Conclude

The math is "fair." The dollars are not. What lever would you pull?

Same rate. Same time. Same \(e^{2.8}\). The ratio is preserved at 100× — that is just algebra: the factor cancels. But the dollar gap grew by the same multiplier as everyone's wealth: from $990,000 at year 0 to about $16.28 million at year 40.

Look at \(A = P \cdot e^{r t}\). Three letters can move: P (who has a starting principal?), r (fees, redlining, predatory lending bend it), t (an earlier start = more years of compounding). Which lever does each policy pull?

Reflect: Pick one intervention from the slide deck (Baby Bonds, expanded CTC, estate-tax restoration, or your own). Which letter in the formula does it change — and why does that move the outcome more than changing the rate?

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