The Number E

The gradient of $2^x$2ˣ is its height times a fixed number

Take $y=2^x$y = 2ˣ and a chord from x to x + h. Its gradient is $\frac{2^{x+h}-2^x}{h}$(2^(x + h) − 2ˣ)/h, and since $2^{x+h}=2^x\times 2^h$2^(x + h) = 2ˣ × 2^h, that is $2^x\times \frac{2^h-1}{h}$2ˣ × (2^h − 1)/h.

The second factor, $\frac{2^h-1}{h}$(2^h − 1)/h, does not depend on x at all. So as $h\to 0$h → 0, the gradient of $2^x$2ˣ at every point is its height $2^x$2ˣ times one fixed number: the limit of $\frac{2^h-1}{h}$(2^h − 1)/h, which is also the gradient at x = 0, where the height is 1.

With h = 0.000001, $\frac{2^h-1}{h}=0.6931$(2^h − 1)/h = 0.6931. So $2^x$2ˣ climbs at 0.6931 times its own height. At x = 3 the height is 8 and the gradient is 8 × 0.6931 = 5.545. At x = 0 the height is 1 and the gradient is 0.6931, a little less than the height.

The gold curve $y=2^x$y = 2ˣ and its tangent at (0, 1), the gold line with gradient 0.6931. The dashed line through (0, 1) has gradient 1, equal to the height there; the tangent is less steep.

$3^x$3ˣ overshoots

The same argument works for any base b > 0: the gradient of $b^x$bˣ is $b^x$bˣ times the limit of $\frac{b^h-1}{h}$(b^h − 1)/h. For b = 3, with h = 0.000001, that number is 1.0986. So $3^x$3ˣ climbs faster than its own height: at x = 0 its height is 1 and its gradient is 1.0986.

The multiplier grows with the base. With h = 0.000001 it is 0.6931 for b = 2, 0.9163 for b = 2.5, 0.9933 for b = 2.7, 1.0006 for b = 2.72 and 1.0986 for b = 3. It crosses 1 between b = 2.7 and b = 2.72.

The curve $y=b^x$y = bˣ with its tangent at (0, 1). The plain bar is the height 1 at x = 0; the gold bar at x = 1 is the tangent’s rise over a run of 1, which is its gradient. At b = 2 the gold bar is 0.69, shorter than the plain one; drag b up until the two bars match.

The base where they match

Narrowing the search, the multiplier is 0.99695 for b = 2.71 and 1.00063 for b = 2.72, so the base where it is exactly 1 lies between them. Halving the interval again and again pins it down to 2.71828 to five decimal places. That number is called e.

So e is the base for which the gradient of $b^x$bˣ at x = 0 is exactly 1. For that base the fixed multiplier is 1, so the gradient of $e^x$eˣ equals its height at every point, not only at x = 0. Like $\pi$π, e is irrational: its decimals never end and never repeat.

The gold curve $y=e^x$y = eˣ between the dashed curves $y=2^x$y = 2ˣ, below it for x > 0, and $y=3^x$y = 3ˣ, above it. All three pass through (0, 1). The gold line y = x + 1 is the tangent to $e^x$eˣ there, with gradient 1. At x = 2 the three heights are 4, 7.389 and 9.

e as a limit

Near x = 0 the curve $y=e^x$y = eˣ runs close to its tangent y = 1 + x, so for a small h, $e^h\approx 1+h$e^h ≈ 1 + h. At h = 0.001, $e^{0.001}=1.0010005$e^0.001 = 1.0010005, against 1.001.

Put $h=\frac{1}{n}$h = 1/n for a large whole number n. Then $e^{\frac{1}{n}}\approx 1+\frac{1}{n}$e^(1/n) ≈ 1 + 1/n, and raising both sides to the power n gives $e\approx {\left(1+\frac{1}{n}\right)}^n$e ≈ (1 + 1/n)ⁿ, more closely as n grows. ${\left(1+\frac{1}{n}\right)}^n$(1 + 1/n)ⁿ is 2 for n = 1, 2.59374 for n = 10, 2.70481 for n = 100, 2.71692 for n = 1000 and 2.71828 for n = 1,000,000.

So e is also the limit of ${\left(1+\frac{1}{n}\right)}^n$(1 + 1/n)ⁿ as $n\to \infty$n → ∞, which is the other common way to define it. In money, it is what $1 grows to in a year at 100% interest, compounded more and more often.

The usual mistakes

Taking e to be 3.142, which is $\pi$π, or 1.618, the golden ratio. e is about 2.718, between 2 and 3.

Saying the gradient of $e^x$eˣ is always 1. It is 1 only at x = 0, where the height is 1; everywhere the gradient equals the height $e^x$eˣ.

Differentiating $e^x$eˣ by the power rule, as $x{e}^{x-1}$x eˣ⁻¹. The power rule needs the variable in the base; here x is in the exponent.

Thinking any exponential has a gradient equal to its height. For $2^x$2ˣ the gradient is 0.6931 times the height, and for $3^x$3ˣ it is 1.0986 times; e is the one base where the multiplier is 1.