Exponentials and growth

Money in an account earning five per cent does not grow by a fixed amount each year; it grows by a fixed factor. No polynomial does that, however high its degree, so the family of functions built in the previous lessons cannot describe interest, populations, or radioactive decay.

This lesson builds the family that can. It starts from repeated multiplication, which is arithmetic, and ends with a function defined for every real exponent, a specific irrational number that falls out of compounding, and a comparison that settles which kind of growth wins in the long run.

Counting the factors

Write $b^{n}$ for $b$ multiplied by itself $n$ times. With that reading the index laws are not rules to memorise but observations about counting. Multiplying $b^{m}$ by $b^{n}$ writes down $m$ factors and then $n$ more, so

b^{m} \times b^{n} = b^{m + n}

Raising a power to a power writes down $n$ copies of a block of $m$ factors, giving $(b^{m})^{n} = b^{m n}$ . Distributing across a product, $(a b)^{n} = a^{n} b^{n}$ , because multiplication is commutative and the factors can be sorted. That is the whole of the theory for whole-number exponents, and every one of the three laws is proved by counting.

Division follows from the first law: $b^{m} / b^{n} = b^{m - n}$ whenever $m$ exceeds $n$ , since cancelling $n$ factors from $m$ leaves $m - n$ . So $2^{5} \times 2^{3} / 2^{6} = 2^{2} = 4$ , which is worth checking directly: $32 \times 8 / 64 = 4$ .

The extensions are forced, not chosen

Nothing in the counting picture makes sense of $b^{0}$ , or $b^{- 3}$ , or $b^{1 / 2}$ : multiplying something by itself zero times, or minus three times, is not an operation. Yet all three have standard values, and the reason is a principle used repeatedly in the first lesson. Extend the notation in whatever way keeps the existing laws true, then check that nothing already working breaks.

Apply the first law with $n = 0$ : $b^{m} \times b^{0} = b^{m + 0} = b^{m}$ . Dividing by $b^{m}$ , which is legal when $b$ is not zero, gives $b^{0} = 1$ . That is not a convention adopted for tidiness, it is the only value compatible with the law. Now apply the law with $n = - m$ : $b^{m} \times b^{- m} = b^{0} = 1$ , so $b^{- m}$ must be $1 / b^{m}$ . A negative exponent is a reciprocal, forced.

Fractional exponents come from the second law. Whatever $b^{1 / 2}$ is, squaring it gives $(b^{1 / 2})^{2} = b^{1} = b$ , so it is a square root of $b$ , and the positive root is chosen to keep the function single valued for positive $b$ . In general $b^{p / q}$ is the $q$ -th root of $b^{p}$ . So $8^{2 / 3}$ is the cube root of $64$ , which is $4$ , and equally the square of the cube root of $8$ , which is $2^{2} = 4$ . Both routes must agree, and they do.

Two restrictions come with this. The base is kept positive, because $(- 8)^{1 / 2}$ has no real value while $(- 8)^{1 / 3} = - 2$ does, and a function that exists at some fractional exponents and not others is unusable. And $0^{0}$ is left undefined, since the pattern $b^{0} = 1$ and the pattern $0^{n} = 0$ give different answers and neither has priority.

Example. Evaluate $16^{- 3 / 4}$ and simplify $(2 a^{3})^{4}$ .

For the first, the negative sign inverts and the fraction takes a root: $16^{- 3 / 4} = 1 / 16^{3 / 4}$ , and $16^{3 / 4}$ is the cube of the fourth root of $16$ , which is $2^{3} = 8$ . So the value is $1 / 8 = 0.125$ . For the second, distribute the exponent across the product: $2^{4} (a^{3})^{4} = 16 a^{12}$ . Check at $a = 2$ : the left is $(2 \times 8)^{4} = 16^{4} = 65 {,} 536$ and the right is $16 \times 4096 = 65 {,} 536$ .

Now you. Evaluate $27^{4 / 3}$ and simplify $(3 x^{2})^{3} x^{- 4}$ .

Answer

$27^{4 / 3}$ is the fourth power of the cube root of $27$ , which is $3^{4} = 81$ . And $(3 x^{2})^{3} x^{- 4} = 27 x^{6} x^{- 4} = 27 x^{2}$ . Check at $x = 2$ : $27 \times 64 \times (1 / 16) = 108$ , and $27 \times 4 = 108$ .

Irrational exponents, and where the real line earns its keep

Every exponent so far has been rational. What is $2^{\sqrt{2}}$ ?

The honest answer needs the completeness of the real line from the first lesson. The rationals approach $\sqrt{2}$ as closely as you like, and the corresponding powers approach a single value: $2^{1.4} = 2.639016$ , $2^{1.41} = 2.657372$ , $2^{1.414} = 2.664750$ , and the sequence closes in on $2.665144$ . Completeness guarantees that a real number sits exactly where the sequence is heading, and that number is defined to be $2^{\sqrt{2}}$ .

The definition is worth pausing on, because it is the first place in this course where a value is defined by a limiting process rather than computed by an operation. The gap the Pythagoreans found is what makes the definition necessary, and the completeness Dedekind and Cantor supplied is what makes it work. With that step taken, $b^{x}$ is defined for every real $x$ and every positive base $b$ , the index laws still hold, and the result is a genuine function with domain all of $R$ .

Its range is only the positive numbers. Raising a positive base to any real power never produces zero or a negative, which means the graph of $y = 2^{x}$ lies entirely above the horizontal axis, approaching it as $x$ falls without ever touching. For $b > 1$ the function increases and models growth; for $0 < b < 1$ it decreases and models decay; and $b = 1$ gives the constant $1$ , which is why that base is excluded from anything interesting.

Compound interest, and the number it converges on

Put $P$ in an account paying an annual rate $r$ , compounded $n$ times a year. Each period multiplies the balance by $1 + r / n$ , and after $t$ years there have been $n t$ periods, so

A = P {(1 + \frac{r}{n})}^{n t}

Example. Put $1000$ pounds at $6$ per cent for five years, compounded monthly. What is the balance?

Here $r / n = 0.06 / 12 = 0.005$ and $n t = 60$ , so $A = 1000 \times {1.005}^{60} = 1000 \times 1.34885 = 1348.85$ pounds. Compounding annually instead gives $1000 \times {1.06}^{5} = 1338.23$ , and daily gives $1349.83$ . More frequent compounding pays more, and the increments are shrinking.

Now you. Put $2500$ pounds at $4.5$ per cent for eight years, compounded monthly.

Answer

$r / n = 0.00375$ and $n t = 96$ , so $A = 2500 \times {1.00375}^{96} = 2500 \times 1.432365 = 3580.91$ pounds.

The shrinking increments raise the obvious question: what happens as the compounding gets infinitely frequent? Take one pound at one hundred per cent for one year, so the balance is $(1 + 1 / n)^{n}$ , and push $n$ up. Annually it gives $2$ . Semi-annually, $2.25$ . Monthly, $2.613035$ . Daily, $2.714567$ . Hourly, $2.718127$ . At a million periods, $2.718280$ .

The sequence does not run away; it converges, and its limit is

e = 2.718281828459045 dots

named by Euler, who computed it to eighteen places in 1748 and proved it irrational. Continuous compounding therefore gives $A = P e^{r t}$ , and for the example above that is $1000 \times e^{0.3} = 1349.86$ pounds, one penny more than daily compounding. The interesting part is not the penny. It is that a limit of purely financial bookkeeping produces a constant that turns out to be the natural base for every growth process, for reasons that only calculus makes fully clear: $e^{x}$ is the unique exponential whose rate of increase at each point equals its own value there.

Decay, and half-lives

When the base is less than one the same function runs downhill. Radioactive decay is the cleanest case, because the fraction lost per unit time is fixed by physics and unaffected by temperature, pressure or chemistry. It is usual to write the law using the half-life $T$ , the time for half of any sample to decay:

N (t) = N_{0} {(\frac{1}{2})}^{t / T}

Carbon-14 has a half-life of $5730$ years. After three half-lives, that is $17 {,} 190$ years, an eighth is left, which needs no calculation. For a time that is not a whole number of half-lives, the fractional exponent does the work.

Example. What fraction of the carbon-14 in a sample remains after $15 {,} 000$ years?

The exponent is $15000 / 5730 = 2.6178$ , so the fraction is ${0.5}^{2.6178} = 0.1629$ , about $16$ per cent. As a sanity check, this sits between the eighth left after three half-lives and the quarter left after two, nearer the eighth, which matches an exponent nearer three than two.

Now you. Caesium-137, the isotope of most concern after the Chernobyl accident, has a half-life of $30.08$ years. What fraction remains after a century?

Answer

The exponent is $100 / 30.08 = 3.324$ , so the fraction is ${0.5}^{3.324} = 0.0998$ , just under a tenth.

Why exponentials always win

Set an exponential against a polynomial and the exponential eventually wins, no matter how modest its base or how high the polynomial's degree. The comparison is worth doing with real numbers, because the crossover can be a long way out and the intermediate behaviour is genuinely misleading.

Compare $2^{x}$ with $x^{3}$ . At $x = 5$ the cube is well ahead, $125$ against $32$ . At $x = 9$ it is still ahead, $729$ against $512$ . At $x = 10$ the exponential has taken the lead, $1024$ against $1000$ , and it never loses it again: at $x = 20$ the figures are $1 {,} 048 {,} 576$ against $8000$ .

Raise the stakes to $x^{10}$ and the exponential looks hopeless for a long time, trailing by many orders of magnitude through the whole of the range anyone would plot. It overtakes just before $x = 59$ , where $2^{59} = 5.76 \times 10^{17}$ against $59^{10} = 5.11 \times 10^{17}$ , and thereafter the gap widens without limit. The pattern is general: any exponential with base above one eventually exceeds any polynomial, permanently. This single fact is why an algorithm whose cost grows exponentially with the size of its input is useless at scale while a cubic one is merely slow.

What nothing grows exponentially forever

Exponential models describe the early part of many processes and the whole of almost none, and saying so is not a hedge. A single bacterium dividing every twenty minutes would after eight hours be $2^{24}$ cells, about seventeen million, which is realistic in fresh broth. Continue for two days and the same model predicts a mass exceeding that of the Earth. What stops it is not the mathematics but food, space and waste, and the curve bends over into a shape called logistic that this course does not cover.

The same warning applies to money and to populations. Compound interest is exact as long as the rate holds, and rates do not hold for centuries. World population grew at over two per cent a year in the late 1960s and grows at under one per cent now, so any exponential fitted then would badly overshoot today.

Use exponentials for what they are: the correct description of a fixed proportional rate, valid while that rate is fixed. Within that scope they are exceptionally accurate, which raises the practical question this lesson cannot answer. Computing $2^{x}$ is easy for any $x$ . Going backwards, finding the $x$ for which $2^{x} = 10$ , has no method here at all. That inverse function is the next lesson.