Appendix K — Problem Set 9 | Maclaurin and Taylor Series

Companion to §4.5 — replacing a hard function by a short polynomial

Purpose

Read §4.5 first; its figure shows the polynomial growing toward the curve.

Chemistry is full of functions we cannot evaluate in our heads: e^{-kt}, \ln K, \sin\theta, 1/T. Near a point we know well, each of them behaves like a short polynomial, and the polynomial is built from derivatives at that point. This set is simple practice in building that polynomial and checking it against the real value.

The recipe has three steps:

Differentiate f as many times as you need (usually 2 or 3).
Evaluate f, f', f'', \dots at the expansion point a.
Assemble the polynomial and test it against the exact value.

f(x) \approx f(a) + f'(a)\,(x-a) + \frac{f''(a)}{2!}\,(x-a)^{2} + \frac{f'''(a)}{3!}\,(x-a)^{3} + \cdots

When a = 0 this is called the Maclaurin series; for any other a it is the Taylor series about a. The factorials are there so that the polynomial and f share the same value, slope, curvature, and so on, at a. The series is good near a and gets worse as |x-a| grows.

Worked example 1 — a Maclaurin series

Expand f(x)=e^{x} about a=0 through x^{3}, then estimate e^{0.1}.

Step 1 — differentiate. Every derivative of e^{x} is e^{x}: f' = f'' = f''' = e^{x}.

Step 2 — evaluate at 0. f(0)=f'(0)=f''(0)=f'''(0)=1.

Step 3 — assemble. e^{x} \approx 1 + x + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} = 1 + x + \frac{x^{2}}{2} + \frac{x^{3}}{6}.

Test. At x = 0.1: 1 + 0.1 + 0.005 + 0.000167 = 1.105167. A calculator gives e^{0.1} = 1.105171, so the error is 4\times10^{-6}. Dropping to two terms (1 + x = 1.100) is off by 0.005, so each extra term helps. Check that the polynomial gets x=0 exactly right: it gives 1 = e^{0} ✓.

Worked example 2 — a Taylor series about a \ne 0

Expand f(x)=\sqrt{x} about a=4 through (x-4)^{2}, then estimate \sqrt{4.4}.

Why not Maclaurin? \sqrt{x} has infinite slope at 0 and we know nothing useful there. But \sqrt{4}=2 is known exactly, so expand about a=4.

Step 1 — differentiate. f' = \dfrac{1}{2\sqrt{x}}, \quad f'' = -\dfrac{1}{4\,x^{3/2}}.

Step 2 — evaluate at 4. f(4)=2, \;f'(4)=\dfrac{1}{4}, \;f''(4)=-\dfrac{1}{4\cdot 8} = -\dfrac{1}{32}.

Step 3 — assemble. \sqrt{x} \approx 2 + \frac{1}{4}(x-4) - \frac{1}{64}(x-4)^{2}. (the 64 is 2!\times 32.)

Test. At x = 4.4, x-4 = 0.4: 2 + 0.1 - \dfrac{0.16}{64} = 2 + 0.1 - 0.0025 = 2.0975. A calculator gives \sqrt{4.4}=2.09762, so the error is 1\times 10^{-4}.

Both examples used the same three steps. Do the same in every problem below.

A. Maclaurin series (a=0)

1. Four standard series

Build each Maclaurin polynomial from derivatives at 0 (show the derivative table):

e^{-x} through x^{3}.
\sin x through x^{3}.
\cos x through x^{4}.
\ln(1+x) through x^{3}.

Hint for (d): f' = 1/(1+x); keep differentiating.

2. First-order decay

A first-order reaction has [\mathrm{A}]/[\mathrm{A}]_0 = e^{-kt} with k = 0.050\ \mathrm{s^{-1}}.

Use your series from 1a with x = kt (so e^{-kt}\approx 1 - kt + (kt)^{2}/2 - (kt)^{3}/6) to estimate [\mathrm{A}]/[\mathrm{A}]_0 at t = 2.0 s.
Compare with the exact value e^{-0.10}=0.904837. What is the error?
The one-term shortcut is 1 - kt. What does it give, and what is its error?
One sentence: for what kind of kt is the shortcut 1 - kt safe?

3. Small angles

Use \sin\theta \approx \theta - \theta^{3}/6 and \cos\theta\approx 1-\theta^{2}/2 to estimate \sin 0.20 and \cos 0.20 (radians).
Compare with \sin 0.20=0.198669 and \cos 0.20=0.980067.
What does the simplest approximation \sin\theta\approx\theta give, and by what percent is it off?
One sentence: why must \theta be in radians?

4. A Boltzmann factor

For an energy gap that is small compared with k_BT, let x = \varepsilon/k_BT = 0.20.

Estimate e^{-x} with the series from 1a through x^{3}.
Compare with the exact e^{-0.20}=0.818731.
Estimate the fraction e^{-x} for x = 2.0 with the same three-term series and compare with the exact 0.135335. What happened, and why?

B. Taylor series about a \neq 0

5. The logarithm about x=1

Show that f(x)=\ln x has f(1)=0, f'(1)=1, f''(1)=-1, f'''(1)=2.
Write the Taylor polynomial through (x-1)^{3}.
Estimate \ln 1.1 and \ln 1.2 and compare with 0.095310 and 0.182322.
Why can there be no Maclaurin series for \ln x about x=0?

6. A square root about a=9

Expand f(x)=\sqrt{x} about a=9 through (x-9)^{2}.
Estimate \sqrt{9.5} and compare with 3.082207.
One sentence: why is a=9 a better choice than a=4 for this number?

7. The exponential about a=2

Show that every derivative of e^{x} at a=2 equals e^{2}=7.389.
Write the Taylor polynomial through (x-2)^{2}.
Estimate e^{2.1} and compare with 8.1662.

8. Temperature dependence of 1/T

Rate constants depend on 1/T (Arrhenius). Expand f(T)=1/T about T_0=300 K.

Find f'(T) and f''(T) and evaluate them at 300 K, with units.
Write the Taylor polynomial through (T-300)^{2}.
Estimate 1/310 with two terms (through (T-300)) and with three terms. Compare with 1/310 = 0.0032258\ \mathrm{K^{-1}}.
One sentence of chemistry: the one-term result is 1/T_0 - \Delta T/T_0^{2}. Why does a 10 K rise change 1/T by only about 3\%?

C. Reading the series

9. How many terms do I need?

For x = 0.5, compute 1+x, 1 + x + x^{2}/2, and 1+x+x^{2}/2+x^{3}/6 and compare with e^{0.5}=1.64872.
Does each added term move you closer to the exact value?
Repeat for x=0.1 (use Worked example 1). Which x needs more terms to reach 0.1\% accuracy, and why?

10. The tempting error

A classmate writes “e^{x} \approx 1 + x + x^{2}/2 + x^{3}/6, so e^{5} \approx 1+5+12.5+20.8 = 39.3.”

Compare with the exact e^{5}=148.41.
Explain in one sentence why the series failed here although it worked at x=0.1.
State the rule of thumb: when is a Maclaurin polynomial trustworthy?

Expectations

For every series, show the derivative table: f, f', f'', … evaluated at a.
Write the polynomial before substituting numbers.
State the expansion point a and keep (x-a) together.
Always compare with the exact value and report the error. Show units where there are any.
Submit on the course submission page as Set 9, or on paper.