4.5 Taylor Series: A Function from Its Derivatives at One Point

The problem

A first-order decay is [\mathrm{A}]/[\mathrm{A}]_0 = e^{-kt}. Every kinetics student learns a shortcut: for a short time, e^{-kt}\approx 1-kt. With k=0.050\ \mathrm{s^{-1}} and t=2.0 s the shortcut gives 0.900, the true value is 0.9048, and the shortcut is off by half a percent. Wait twenty seconds (kt=1) and the shortcut says the reactant is gone (1-1=0) while e^{-1}=0.368 of it remains. Why does the shortcut work, how good is it, and can it be improved without computing an exponential?

The question matters because chemistry runs on functions that cannot be evaluated by hand (e^{-\varepsilon/k_BT}, \ln K, \sin\theta, 1/T), yet near a state we understand they behave simply. We want a systematic way to replace a hard function by a short polynomial, with an honest estimate of how far to trust it.

What a good local copy must do

Suppose we know a function only at one point a, but we know its derivatives there. A copy of f built from that information should agree with f as far as the information goes. The simplest copy is a constant that matches the value, f(a). A better one also has the same slope: f(a)+f'(a)(x-a) — the tangent line from §4.1. Better still, it should bend the same way, so its second derivative should match f''(a). Each extra derivative we match buys a better copy near a.

Take f(x)=e^{x} at a=0, where every derivative equals 1. Try the polynomial p(x)=c_0+c_1x+c_2x^{2}+c_3x^{3} and demand p(0)=p'(0)=p''(0)=p'''(0)=1. Differentiating gives p(0)=c_0, p'(0)=c_1, p''(0)=2c_2, p'''(0)=6c_3. Setting each to 1: c_0=1,\quad c_1=1,\quad c_2=\tfrac12,\quad c_3=\tfrac16. The factorials are not decoration. They are exactly what cancels the factor that the power rule produces when a term is differentiated k times: \dfrac{d^{k}}{dx^{k}}\,x^{k} = k!.

Definition 4.5.1 (Taylor and Maclaurin series). For a function f with derivatives at a, f(x)\;\approx\; f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}(x-a)^{2}+\frac{f'''(a)}{3!}(x-a)^{3}+\cdots is the Taylor series of f about a. When a=0 it is called the Maclaurin series. Stopping after the (x-a)^{N} term gives the polynomial of order N.

Examples, each from a derivative table

Example 4.5.1 — the decay shortcut, improved. Since \dfrac{d^{k}}{dx^{k}}e^{-x}=(-1)^{k}e^{-x}, the derivatives at 0 are 1,-1,1,-1,\dots and e^{-x}\approx 1-x+\frac{x^{2}}{2}-\frac{x^{3}}{6}. Set x=kt=0.10: 1-0.1+0.005-0.000167=0.904833, against the exact 0.904837. The textbook shortcut 1-kt is just the first two terms; the next term, +(kt)^{2}/2, is exactly the size of the shortcut’s error (0.005).

Check. At x=0 the series gives 1 ✓. At x=1 (kt=1) it gives 1-1+0.5-0.1667=0.333, not 0.368 — the series is losing accuracy as x grows, which is the point of the next section.

Example 4.5.2 — expanding about a point other than zero. Take f(x)=\sqrt{x} about a=4. The derivatives are f'=\dfrac{1}{2\sqrt{x}}, f''=-\dfrac{1}{4x^{3/2}}, which at 4 give f(4)=2, f'(4)=\tfrac14, f''(4)=-\tfrac{1}{32}. Hence \sqrt{x}\approx 2+\tfrac14(x-4)-\tfrac1{64}(x-4)^{2}. At x=4.4: 2+0.1-0.0025=2.0975 against the exact 2.0976. Why a=4 and not a=0? Because we know \sqrt4 exactly, and 0.4 is a small step away. The expansion point should be a place where you know the answer, near where you want it.

See it grow

For e^{x}, how many terms do you think you need before the polynomial and the curve are indistinguishable on -1\le x\le 1? On -3\le x\le 3? Then repeat with \sin x: its polynomial uses only odd powers. Predict first, then raise the order.

function e^xsin xln(1+x) order N test point x

Near x=0 the polynomial hugs the curve after only two or three terms, and the agreement spreads outward as you raise the order. For e^{x} and \sin x it eventually covers the whole window. For \ln(1+x) it never does: past x=1 the polynomial wanders off no matter how many terms are kept, and at x\le-1 the function does not exist at all. A series has a radius beyond which it fails.

When can I trust it?

The terms of a series for e^{x} are x^{k}/k!. For small x they shrink fast, so dropping the later ones barely matters; the first dropped term is a good estimate of the error. For x=5 the terms are 1,5,12.5,20.8,26.0,\dots — still growing — and four terms give 39.3 against 148.4. The rule is therefore practical rather than mysterious: the polynomial is trustworthy when |x-a| is small enough that successive terms are shrinking quickly. Some series (ours for \ln(1+x)) fail beyond a fixed radius; others (e^{x}, \sin x) never do, but may need many terms for large x.

What chemists do with it

Example 4.5.3 — the harmonic approximation. A diatomic bond is modelled by the Morse potential V(x)=D\,(1-e^{-ax})^{2}, with x the stretch from equilibrium. Using e^{-ax}\approx 1-ax+\tfrac{(ax)^{2}}{2}, we get 1-e^{-ax}\approx ax-\tfrac{(ax)^{2}}{2}, and to lowest order V(x)\approx D\,(ax)^{2}=\tfrac12\,(2Da^{2})\,x^{2}. This is Hooke’s law with force constant k=2Da^{2}. For D=5.0 eV and a=1.9\ \text{Å}^{-1}, k=2(5.0)(1.9)^{2}=36\ \mathrm{eV/Å^{2}}. Every spring-like picture of a bond — vibrational spectroscopy included — is the first nonzero term of a Taylor series.

Check. Units: D\,a^{2} is energy per length squared ✓. And the approximation is only good for small stretch; it cannot describe dissociation, where V\to D rather than growing without bound.

The same device linearises temperature dependence. Expanding 1/T about T_0 gives 1/T\approx 1/T_0-(T-T_0)/T_0^{2}, so a 10 K rise from 300 K changes 1/T by only about 3\% — and Arrhenius plots, which are linear in 1/T, are nearly straight over short ranges.

Cautions

Stay near the expansion point. 1-kt for kt=1 predicts zero reactant. A polynomial is a local statement; use it only where the next term is much smaller than the last.

Radians, and expansion point. \sin\theta\approx\theta holds only with \theta in radians (the derivative \cos\theta equals 1 at 0 only then). And \ln x has no Maclaurin series: it is undefined at 0, so expand \ln(1+x), or expand \ln x about a=1.

Summary

A function known only at one point can be copied by a polynomial whose coefficients are the derivatives there divided by factorials, so that value, slope, curvature and higher derivatives all agree at that point. The first two terms recover the tangent line, the next term measures the curvature, and the size of the first dropped term estimates the error. The copy is local: it is excellent near the expansion point and degrades with distance, and some series stop working beyond a fixed radius. In chemistry the pattern appears as e^{-kt}\approx 1-kt, \sin\theta\approx\theta, the linearisation of 1/T, and the harmonic approximation of a bond.

Practice until it sticks

Problem Set 9 builds each of these series from a derivative table and compares every polynomial with the exact value.