Taylor Series About a Point

When 0 will not do

A Maclaurin series is built from a function’s value and derivatives at x = 0. ln x has no value there: as x shrinks toward 0, ln x falls without limit. Its derivatives, $\frac{1}{x}$1/x, $-\frac{1}{x^2}$−1/x², $\frac{2}{x^3}$2/x³, …, have no value at 0 either. So ln x has no Maclaurin series.

The way out is to measure somewhere else. At x = 1 everything is known exactly: ln 1 = 0, and each derivative is a whole number. A series built at x = 1 is called the Taylor series of ln x about 1, and 1 is its center.

The same formula, a new center

The Taylor series of f about a is $f\left(x\right)=f\left(a\right)+f^{\prime }\left(a\right)\left(x-a\right)+\frac{f^{(2)}\left(a\right){\left(x-a\right)}^2}{2!}+\frac{f^{(3)}\left(a\right){\left(x-a\right)}^3}{3!}+\dots$f(x) = f(a) + f'(a)(x − a) + f⁽²⁾(a)(x − a)²/2! + f⁽³⁾(a)(x − a)³/3! + …. Term n is $\frac{f^{(n)}\left(a\right){\left(x-a\right)}^n}{n!}$f⁽ⁿ⁾(a)(x − a)ⁿ/n!: the nth derivative at a, times the nth power of x − a, divided by n factorial.

It works for the same reason as at 0. Every power of x − a is 0 at x = a, so at x = a the polynomial equals f(a). Take the term with ${\left(x-a\right)}^n$(x − a)ⁿ and differentiate it n times: it becomes $f^{(n)}\left(a\right)$f⁽ⁿ⁾(a): the n! from the power cancels the n! underneath. The terms with lower powers have been differentiated away by then, and the higher ones still carry a factor of x − a, which is 0 at a. So at a the polynomial and the function have the same value, the same gradient, and the same derivatives as far as the polynomial goes.

Everything in the series is measured at a. The powers are powers of x − a, the distance from the center, and not of x.

The series of ln x about 1

Differentiate ln x four times: $f^{\prime }\left(x\right)=\frac{1}{x}$f'(x) = 1/x, $f^{(2)}\left(x\right)=-\frac{1}{x^2}$f⁽²⁾(x) = −1/x², $f^{(3)}\left(x\right)=\frac{2}{x^3}$f⁽³⁾(x) = 2/x³ and $f^{(4)}\left(x\right)=-\frac{6}{x^4}$f⁽⁴⁾(x) = −6/x⁴. At x = 1 these are 1, −1, 2 and −6, and ln 1 = 0.

Divide each by its factorial: $\frac{1}{1!}=1$1/1! = 1, $-\frac{1}{2!}=-\frac{1}{2}$−1/2! = −1/2, $\frac{2}{3!}=\frac{1}{3}$2/3! = 1/3 and $-\frac{6}{4!}=-\frac{1}{4}$−6/4! = −1/4. So $\ln x=\left(x-1\right)-\frac{{\left(x-1\right)}^2}{2}+\frac{{\left(x-1\right)}^3}{3}-\frac{{\left(x-1\right)}^4}{4}+\dots$ln x = (x − 1) − (x − 1)²/2 + (x − 1)³/3 − (x − 1)⁴/4 + …. Write u for x − 1 and this is $u-\frac{u^2}{2}+\frac{u^3}{3}-\dots$u − u²/2 + u³/3 − …, the series for ln(1 + u), as it must be, since x = 1 + u.

Check at x = 1.1, where x − 1 = 0.1. The running totals are 0.1, 0.095, 0.095333 and 0.095308, and ln 1.1 = 0.095310. At x = 1.5 they are 0.5, 0.375, 0.416667 and 0.401042, against ln 1.5 = 0.405465: further from the center, more terms are needed for the same accuracy.

The gold curve is y = ln x, and the plain one is the first two terms, $\left(x-1\right)-\frac{{\left(x-1\right)}^2}{2}$(x − 1) − (x − 1)²/2. They touch at (1, 0) and bend the same way there. At x = 1.5 the two terms give 0.375 against 0.405465, at x = 2 they give 0.5 against 0.693147, and at x = 3 they give 0 against 1.098612.

Near the center and far from it

Two terms match the value, the gradient and the bend of ln x at 1, and nothing more. Close to 1 that is enough; further off the curve and the parabola drift apart, on both sides: at x = 0.5 the two terms give −0.625, and ln 0.5 = −0.693147.

More terms carry the agreement further, but only so far. The series is the ln(1 + u) series with u = x − 1, which converges for $-1<u\le 1$−1 < u ≤ 1, so the Taylor series of ln x about 1 converges for $0<x\le 2$0 < x ≤ 2. The stretch reaches from the center 1 to the point 0 where ln x breaks down, and the same distance on the other side.

Choosing the center

Put a = 0 and every power of x − a becomes a power of x, every derivative is taken at 0, and the Taylor series is the Maclaurin series. The Maclaurin series is the Taylor series about 0.

Any center will do where the function and its derivatives are known. The best center is one close to the x you need, so that x − a is small and few terms are enough.

For example, $\sqrt{x}$√x has no gradient at 0, since its derivative $\frac{1}{2\sqrt{x}}$1/(2√x) has no value there, but at 4 everything is easy. f(4) = 2, $f^{\prime }\left(4\right)=\frac{1}{2\times 2}=\frac{1}{4}$f'(4) = 1/(2 × 2) = 1/4, and the second derivative is $-\frac{1}{4x^{\frac{3}{2}}}$−1/(4x^(3/2)), which is $-\frac{1}{32}$−1/32 at x = 4. Dividing that by 2! gives $-\frac{1}{64}$−1/64, so $\sqrt{x}\approx 2+\frac{x-4}{4}-\frac{{\left(x-4\right)}^2}{64}$√x ≈ 2 + (x − 4)/4 − (x − 4)²/64 near 4.

At x = 4.2, x − 4 = 0.2, and the polynomial gives 2 + 0.05 − 0.000625 = 2.049375, against $\sqrt{4.2}=2.049390$√4.2 = 2.049390. At x = 5, further from the center, it gives 2.234375 against $\sqrt{5}=2.236068$√5 = 2.236068.

The usual mistakes

Using powers of x instead of powers of x − a. About 1, the second term of ln x is $-\frac{{\left(x-1\right)}^2}{2}$−(x − 1)²/2, not $-\frac{x^2}{2}$−x²/2.

Taking the derivatives at 0. Every derivative is measured at the center a; derivatives at 0 build the Maclaurin series, and for ln x they do not exist.

Leaving out the factorials. The third derivative of ln x at 1 is 2, but the coefficient of ${\left(x-1\right)}^3$(x − 1)³ is $\frac{2}{3!}=\frac{1}{3}$2/3! = 1/3.

Starting with a constant of 1. The constant term is f(a), and for ln x about 1 that is ln 1 = 0.

A controller calibrated at 20 degrees

In the application below, a controller knows a setting time and its first two derivatives only at 20 degrees Celsius. So the polynomial is built about 20, in powers of T − 20, and then read at 23 degrees.