Maclaurin Series

Start with the tangent

Near x = 0, the curve $y=e^x$y = eˣ is close to its tangent there. The curve passes through (0, 1), because $e^0=1$e⁰ = 1, and its gradient there is also $e^0=1$e⁰ = 1, so the tangent is y = 1 + x.

At x = 0.1 the tangent gives 1.1, and $e^{0.1}=1.105171$e^0.1 = 1.105171. At x = 0.5 it gives 1.5, against $e^{0.5}=1.648721$e^0.5 = 1.648721. The tangent matches the curve's height and gradient at 0, and nothing more, so it drifts away as x moves off 0.

Match the bend as well

The tangent is straight, but $e^x$eˣ bends upward. To follow the bend, add a term in $x^2$x²: try 1 + x + cx², and choose c so that this polynomial has the same second derivative at 0 as $e^x$eˣ.

The second derivative of 1 + x + cx² is 2c, and the second derivative of $e^x$eˣ is $e^x$eˣ, which is 1 at 0. So 2c = 1 and $c=\frac{1}{2}$c = 1/2, giving $1+x+\frac{x^2}{2}$1 + x + x²/2.

At x = 0.5 this gives 1 + 0.5 + 0.125 = 1.625, against the true 1.648721: out by 0.023721, where the tangent alone was out by 0.148721.

The gold curve $y=e^x$y = eˣ, with the plain straight line y = 1 + x and the plain parabola $y=1+x+\frac{x^2}{2}$y = 1 + x + x²/2. All three pass through (0, 1). The parabola stays with the curve further from 0 than the line does.

The general term

Carry on, one term per derivative. Term n is $\frac{f^{(n)}\left(0\right)x^n}{n!}$f⁽ⁿ⁾(0)xⁿ/n!: the nth derivative of the function at 0, times $x^n$xⁿ, divided by n factorial. Written out, $f\left(x\right)=f\left(0\right)+f^{\prime }\left(0\right)x+\frac{f^{(2)}\left(0\right)x^2}{2!}+\frac{f^{(3)}\left(0\right)x^3}{3!}+\dots$f(x) = f(0) + f'(0)x + f⁽²⁾(0)x²/2! + f⁽³⁾(0)x³/3! + …, and this is the Maclaurin series of f.

The factorial is there to make the derivatives match. Differentiate $x^{n}a$xⁿ a total of n times and it becomes n × (n − 1) × … × 1 = n!. So the nth derivative of the term $\frac{f^{(n)}\left(0\right)x^n}{n!}$f⁽ⁿ⁾(0)xⁿ/n! is exactly $f^{(n)}\left(0\right)$f⁽ⁿ⁾(0). Every other term has either vanished by then or still carries a power of x, which is 0 at x = 0. So at 0 the polynomial and the function agree in height, gradient, second derivative, and every derivative the polynomial includes.

Everything in the series is measured at zero. The function's values anywhere else are not used.

The series for $e^x$eˣ

Every derivative of $e^x$eˣ is $e^x$eˣ, and at 0 each of them equals 1. So every coefficient is $\frac{1}{n!}$1/n!, and $e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\dots$eˣ = 1 + x + x²/2! + x³/3! + …, which is $1+x+\frac{x^2}{2}+\frac{x^3}{6}+\dots$1 + x + x²/2 + x³/6 + … with the factorials worked out.

Put in x = 1 and add one term at a time. The sums are 1, then 2, then 2.5, then 2.666667 after four terms, 2.708333 after five, 2.716667 after six, and 2.718056 after seven. The true value is e = 2.718282, so seven terms are out by 0.000226.

At x = 0.5 the same four terms, 1 + 0.5 + 0.125 + 0.020833, give 1.645833, out by 0.002888 from 1.648721. Closer to 0, fewer terms are needed for the same accuracy.

The series for sin x

Differentiate sin x again and again: cos x, then −sin x, then −cos x, then sin x, and round again. At 0 these are 1, 0, −1, 0, and sin 0 = 0 itself. So the terms in even powers vanish, and the odd ones alternate in sign: $\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\dots$sin x = x − x³/3! + x⁵/5! − …, which is $x-\frac{x^3}{6}+\frac{x^5}{120}-\dots$x − x³/6 + x⁵/120 − …

The first term alone says $\sin x\approx x$sin x ≈ x for small x, measured in radians. That is the small-angle approximation. At x = 0.5 it gives 0.5; two terms give 0.5 − 0.020833 = 0.479167; three give 0.479427. The true value is sin 0.5 = 0.479426.

At x = 1 the sums are 1, then 0.833333, then 0.841667, then 0.841468 with a fourth term, against sin 1 = 0.841471.

The same series settles a limit. Divide by x: $\sin x/x=1-\frac{x^2}{6}+\dots$sin x / x = 1 − x²/6 + …, and every term after the 1 shrinks to 0 as x shrinks, so $\sin x/x$sin x / x tends to 1 as x tends to 0.

The plain curve is y = sin x and the gold one is the series stopped at the $x^3$x³ term, $x-\frac{x^3}{3!}$x − x³/3!. The shaded band is where the two differ by less than 0.01: out to about |x| = 1.04. Drag N to add terms: with the $x^5$x⁵ term the band reaches about 1.76, and with the $x^7$x⁷ term about 2.50.

The usual mistakes

Leaving out the factorial. The $x^2$x² term of $e^x$eˣ is $\frac{x^2}{2}$x²/2, not $x^2$x²: its coefficient is $\frac{f^{(2)}\left(0\right)}{2!}=\frac{1}{2}$f⁽²⁾(0)/2! = 1/2.

Putting the wrong factorial under a term. 3! = 6 belongs to the $x^3$x³ term; the $x^2$x² term divides by 2! = 2.

Getting the sign of the $x^3$x³ term of sin x wrong. It is $x-\frac{x^3}{6}$x − x³/6; the minus bends the line y = x back down toward the curve.

Using degrees. The derivatives of sin x and cos x are the ones above only when x is in radians, so the series needs x in radians.

The cosine, and a pendulum

Differentiating cos x gives −sin x, −cos x, sin x, cos x, which at 0 are 0, −1, 0, 1, and cos 0 = 1. So $\cos x=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\dots$cos x = 1 − x²/2! + x⁴/4! − …, which is $1-\frac{x^2}{2}+\frac{x^4}{24}-\dots$1 − x²/2 + x⁴/24 − …

In the application below, a pendulum 2 meters long is drawn aside through 0.2 radians, and the series for cos x up to the $x^4$x⁴ term gives how far the bob rises.