Compounding Shortcuts for Repeated Growth

An interviewer asks: "$100 grows at 8% a year, compounded, for 9 years. Roughly what's the final value, and please don't reach for a calculator." Multiplying $1.08$ by itself nine times in your head is a non-starter. But you already know, roughly, that repeated multiplicative growth behaves almost like repeated addition when the rate is small — and "almost" is close enough for an interview, if you know how close.

Try estimating $1.06^{12}$ in your head before reading on. Then check it against a calculator and see how far off you were.

The trap: trying to multiply it out exactly

Compounding is inherently multiplicative — $(1+r)$ multiplied by itself $n$ times — and multiplying decimals repeatedly in your head is exactly the kind of task that eats your working memory and produces silently wrong answers. The fix is to notice that for small $r$, each multiplication by $(1+r)$ adds approximately $r$ to the running total, because the cross-terms from compounding are second order and small.

Where the approximation comes from

Expand $(1+r)^n$ using the binomial theorem:

$$(1+r)^n = 1 + nr + \binom{n}{2}r^2 + \binom{n}{2}r^2\cdots$$

In words: the first two terms are "1" and "$n$ copies of $r$ added up" — that's the linear approximation. Every term after that involves $r^2$ or higher, and if $r$ is a small number like 0.08, $r^2 = 0.0064$ is tiny, so those terms barely move the answer unless $n$ is also large. The whole approximation is just "keep the first two terms of the expansion, throw away the rest."

Worked example: 8% for 9 years

Linear approximation: $1 + 9\times0.08 = 1.72$, so $100 becomes roughly $172.

The true value: $1.08^9 = 1.999\ldots \approx 1.999$, so $100 becomes about $199.90.

That's a 16% miss — too large to call "close enough" for a real answer, which is the whole point of showing it: the plain linear rule quietly breaks down once $nr$ gets large (here $nr = 0.72$, not small at all). You need the standard correction.

The correction: add half of $nr$ squared-ish, or just use $e^{nr}$

A much better approximation for moderate $nr$ comes from remembering that $(1+r)^n \approx e^{nr}$ when $r$ is small (since $\ln(1+r)\approx r$). So instead of $1+nr$, estimate $e^{nr}$.

$$(1+r)^n \approx e^{nr}.$$

In words: compounding at rate $r$ for $n$ periods is close to continuous growth at total rate $nr$. Here $nr = 0.72$. You likely have $e^{0.7}\approx 2.01$ memorised or close to it (from the rule that $e^{0.7}\approx 2$, tied to the rule of 72's $\ln 2\approx 0.693$). So the estimate is $100 times 2.01, about $201 — within half a percent of the true $199.90, versus the linear rule's 16% miss.

A second worked example, going the other way: -5% for 14 years

A portfolio loses 5% a year for 14 years. Linear estimate: $1 - 14\times0.05 = 1-0.7=0.3$, i.e. 70% of value lost — but this rule can't even go negative gracefully once $nr$ passes 1, a sign you've left its comfort zone. Exponential estimate: $e^{-0.7}\approx 0.497$, so about half the value remains, a 50% loss. The true value is $0.95^{14}\approx 0.488$ — the exponential estimate is off by about two percentage points, dramatically better than the linear rule's answer, which is off by twenty.

The transferable technique

Any repeated multiplicative process — compounding returns, population growth, radioactive decay, volatility scaling with $\sqrt{t}$ — has the same two-tier mental toolkit: a cheap linear approximation for small total change, and the exponential form $e^{nr}$ (or $e^{-nr}$ for decay) for anything bigger, anchored to memorised values like $e^{0.7}\approx2$ and $e\approx2.718$. Recognising which regime $nr$ puts you in, before you start calculating, is the actual skill being tested.