Mathematics · Complete guide

Taylor series, and the simple trick for writing one

A Taylor series rebuilds a complicated function as a polynomial, the easiest thing in maths to work with, by copying its value, slope and bend at one point. Exams ask for a second-order expansion, a limit, an approximation or a growth rate, and all of them come from one table. This guide teaches that table, the shortcuts that skip it, and shows the polynomial hugging the curve as you add terms.

Start here

The idea: copy a curve with a polynomial

Stand at one point on a curve, x = a. How would you build a polynomial that looks as much like the curve as possible near that point? Match it one feature at a time:

Degree 0

T_{0} = f (a)

Same height. A flat line: right at a, useless anywhere else.

Degree 1

T_{1} = f (a) + f^{'} (a) (x - a)

Same height and slope: the tangent line, or linear approximation.

Degree 2

T_{2} = T_{1} + \frac{f ^{''} ( a )}{2 !} (x - a)^{2}

Same bend too. A parabola that curves the way the function curves.

Keep going

T_{n} = T_{n - 1} + \frac{f ^{(n)} ( a )}{n !} (x - a)^{n}

Each new term matches one more derivative, and the fit spreads further from a.

Key idea

A Taylor polynomial of degree n is the one polynomial that agrees with f at a in value and in its first n derivatives. Let n run to infinity and you get the Taylor series; about a = 0 it is called the Maclaurin series.

The formula

The formula, and where the k! comes from

Taylor series of f about x = a

f (x) = k = 0 \sum \infty \frac{f ^{(k)} ( a )}{k !} (x - a)^{k} = f (a) + f^{'} (a) (x - a) + \frac{f ^{''} ( a )}{2 !} (x - a)^{2} + \frac{f ^{'''} ( a )}{3 !} (x - a)^{3} + \dots

Put a = 0 for the Maclaurin series: f(x) = f(0) + f′(0)x + f″(0)x²/2! + …

Why divide by k!? Differentiate $(x - a)^{k}$ k times and you get k! (the power comes down as k, then k − 1, and so on). So the k-th derivative of the term $c_{k} (x - a)^{k}$ at a is $k! c_{k}$ , and every other term contributes 0 there. For that to equal $f^{(k)} (a)$ we need $c_{k} = f^{(k)} (a) / k!$ .

The same fact, read backwards

If a series is $\sum c_{k} x^{k}$ , then $f^{(k)} (0) = k! c_{k}$ . That turns “find the 6th derivative at 0” into “find one coefficient”, often with no differentiating at all (chapter 6).

The method

The simple trick: the table method

Never write the formula out in one go. Build a table with four columns and fill it row by row:

Differentiate: list f, f′, f″, f‴ down the first column, as many as the degree you need.
Plug in a: evaluate each at the centre.
Divide by k!: 1, 1, 2, 6, 24, 120… That gives the coefficient.
Attach (x − a)^k and add the rows.

Worked question · the table method

Write the third-order Taylor polynomial of

f (x) = ln x

about

a = 1

, and use it to estimate ln 1.1.

Fill the table.
k f⁽ᵏ⁾(x) at x = 1 ÷ k!
0 $ln x$ 0 0
1 $1/ x$ 1 1
2 $- 1/ x^{2}$ −1 −1/2
3 $2/ x^{3}$ 2 1/3
Read off the polynomial.
$T_{3} (x) = (x - 1) - \frac{1}{2} (x - 1)^{2} + \frac{1}{3} (x - 1)^{3}$
Estimate ln 1.1 with x − 1 = 0.1.
$ln 1.1 \approx 0.1 - 0.005 + 0.000333 = 0.095333 (true value 0.095310)$

k	f⁽ᵏ⁾(x)	at x = 1	÷ k!
0	$ln x$	0	0
1	$1/ x$	1	1
2	$- 1/ x^{2}$	−1	−1/2
3	$2/ x^{3}$	2	1/3

Answer

T_{3} (x) = (x - 1) - \frac{1}{2} (x - 1)^{2} + \frac{1}{3} (x - 1)^{3}

; ln 1.1 ≈ 0.0953, correct to four decimals from three terms.

Interactive

The Taylor lab

Pick a function and press play: terms are added one at a time and the blue polynomial wraps itself around the black curve. Move the centre to get a Taylor series about a different point. The green band is where the series converges; outside it, extra terms make things worse. The table under the graph is the table method, filled in live.

T_{3} (x) = 1 + x + \frac{1}{2} x^{2} + \frac{1}{6} x^{3}

eˣTaylor polynomial, degree 3

At x = 1

True value 2.7183, polynomial gives 2.6667: an error of 0.0516. Add terms and the error shrinks: each new term fixes one more derivative at the centre.

The table method, filled in for you

k	k-th derivative	Value at a = 0	÷ k!	Term
0	$f (x) = e^{x}$	1	÷ 1	1
1	$f^{(1)} (x) = e^{x}$	1	÷ 1	1 $x$
2	$f^{(2)} (x) = e^{x}$	1	÷ 2	0.5 $x^{2}$
3	$f^{(3)} (x) = e^{x}$	1	÷ 6	0.1667 $x^{3}$

Try eˣ, sin x, cos x

These converge everywhere. However far out you go, enough terms will catch the curve.

Try ln(1 + x) and 1/(1 − x)

These only converge for |x| < 1 about 0. Past x = 1 the polynomial flaps wildly however many terms you add.

Try sin x with n = 1, 2

No change from 1 to 2: sin is odd, so its even-power terms are all zero. The same happens to cos with odd powers.

Memorise

Five series to know by heart

Function	Maclaurin series	Valid for
$e^{x}$	$1 + x + \frac{x ^{2}}{2 !} + \frac{x ^{3}}{3 !} + \dots$	all x
$sin x$	$x - \frac{x ^{3}}{3 !} + \frac{x ^{5}}{5 !} - \dots$	all x
$cos x$	$1 - \frac{x ^{2}}{2 !} + \frac{x ^{4}}{4 !} - \dots$	all x
$ln (1 + x)$	$x - \frac{x ^{2}}{2} + \frac{x ^{3}}{3} - \frac{x ^{4}}{4} + \dots$	$- 1 < x \leq 1$
$\frac{1}{1 - x}$	$1 + x + x^{2} + x^{3} + \dots$	$∣ x ∣ < 1$
$(1 + x)^{m}$	$1 + m x + \frac{m ( m - 1 )}{2 !} x^{2} + \dots$	$∣ x ∣ < 1$ (any real m)

Patterns that make them easy to remember

eˣ has every power with k! underneath. sin takes the odd powers and cos the even ones, both with alternating signs. ln(1 + x) has no factorials, just 1/k with alternating signs. The last row is the binomial theorem, which also works for fractional and negative m: put m = −1 and replace x by −x to get 1/(1 − x).

Faster

Shortcuts: substitute, don't differentiate

Differentiating $e^{- x^{2}}$ four times is miserable. You don't need to: a Maclaurin series stays valid when you substitute something else for x, multiply two series together, or differentiate and integrate term by term.

Worked question · substitution

Find the Maclaurin series of

f (x) = e^{- x^{2}}

up to

x^{6}

, and hence

f^{(4)} (0)

and

f^{(3)} (0)

Replace x by −x² in the series for eˣ.
$e^{u} = 1 + u + \frac{u ^{2}}{2} + \frac{u ^{3}}{6} + \dots \Rightarrow e^{- x^{2}} = 1 - x^{2} + \frac{x ^{4}}{2} - \frac{x ^{6}}{6} + \dots$
Read derivatives off the coefficients.
$f^{(4)} (0) = 4! \times \frac{1}{2} = 12$ . There is no $x^{3}$ term, so $f^{(3)} (0) = 3! \times 0 = 0$ .

Answer

1 - x^{2} + \frac{1}{2} x^{4} - \frac{1}{6} x^{6}

;

f^{(4)} (0) = 12

f^{(3)} (0) = 0

Substitute

$ln (1 - x) = - x - \frac{x ^{2}}{2} - \frac{x ^{3}}{3} - \dots$ (x → −x). $e^{2 x} = 1 + 2 x + 2 x^{2} + \frac{4}{3} x^{3} + \dots$ (x → 2x).

Multiply

$x e^{x} = x + x^{2} + \frac{x ^{3}}{2} + \dots$ . Multiply each term of the series by x.

Differentiate term by term

$\frac{d}{d x} \frac{1}{1 - x} = \frac{1}{( 1 - x ) ^{2}} = 1 + 2 x + 3 x^{2} + \dots$

Applications

What it's for: limits, approximations, growth rates

Limits

Find

x \to 0 lim \frac{1 - cos x}{x ^{2}}

and

x \to 0 lim \frac{sin x - x}{x ^{3}}

Expand the top.
$1 - cos x = \frac{x ^{2}}{2} - \frac{x ^{4}}{24} + \dots, sin x - x = - \frac{x ^{3}}{6} + \frac{x ^{5}}{120} - \dots$
Divide and let x → 0.
Every term after the first still has a power of x, so it vanishes.

Answer

1/2 and −1/6. No L'Hôpital needed.

Approximations

Estimate

4.1

and

e^{0.1}

without a calculator.

Write it as (1 + small).
$4.1 = 2 1 + 0.025 \approx 2 (1 + \frac{0.025}{2}) = 2.025$
Use the series.
$e^{0.1} \approx 1 + 0.1 + \frac{0.01}{2} + \frac{0.001}{6} = 1.10517$

Answer

2.025 (true 2.0248) and 1.10517 (true 1.10517).

Economics: why growth rates are log differences

The first-order series $ln (1 + g) \approx g$ for small g is used all the time. It is why a growth rate can be computed as $ln Y_{t} - ln Y_{t - 1}$ , why the growth rate of a product is the sum of growth rates, and where the rule of 70 comes from:

Doubling time

(1 + g)^{T} = 2 \Rightarrow T = \frac{ln 2}{ln ( 1 + g )} \approx \frac{0.693}{g} \approx \frac{70}{growth rate in %}

At 5% growth: 70 / 5 = 14 years (exact answer ln 2 / ln 1.05 = 14.2).

Why +10% then −10% loses money

1.1 × 0.9 = 0.99. In logs: $ln 1.1 + ln 0.9 \approx (0.1 - 0.005) + (- 0.1 - 0.005) = - 0.01$ . The first-order terms cancel; the second-order terms $- \frac{g ^{2}}{2}$ don't. Volatility drags down compound growth, here by about 0.5% a year.

Risk premium (second order)

Expanding a utility function to second order around expected wealth gives the Arrow–Pratt result: the risk premium is about $\frac{1}{2} σ^{2} \cdot (- \frac{u ^{''}}{u ^{'}})$ . Curvature (u″) is exactly what the second Taylor term measures.

Accuracy

How good is the approximation?

Lagrange form of the remainder

f (x) = T_{n} (x) + R_{n} (x), R_{n} (x) = \frac{f ^{(n + 1)} ( ξ )}{( n + 1 )!} (x - a)^{n + 1}

ξ is some unknown point between a and x. Bound |f⁽ⁿ⁺¹⁾| on that interval to bound the error.

Close to a

The error behaves like $(x - a)^{n + 1}$ . Halve the distance from the centre and a second-order approximation becomes about 8 times more accurate.

More terms

The $(n + 1)!$ underneath grows very fast, which is why eˣ, sin and cos converge everywhere.

Radius of convergence

Some series only converge within a distance R of the centre. For ln(1 + x) and 1/(1 − x) about 0, R = 1: the distance to the nearest point where the function breaks.

Practice

Exercises, with answers

0 of 13 correct so far

A. Coefficients and derivatives

Decimals or fractions (like 4/3) are both fine.

Problem 1 · a coefficient

What is the coefficient of

x^{3}

in the Maclaurin series of

e^{2 x}

Coefficient

Problem 2 · a derivative from a series

For

f (x) = e^{x^{2}}

, find

f^{(4)} (0)

f⁽⁴⁾(0)

Problem 3 · a coefficient of ln

What is the coefficient of

x^{4}

in the Maclaurin series of

ln (1 + x)

Coefficient

B. Limits and approximations

Problem 4 · a limit

x \to 0 lim \frac{e ^{x} - 1 - x}{x ^{2}}

Limit

Problem 5 · first and second order

Approximate

9.18

using the Taylor polynomial of

x

about 9, first to first order and then to second order (4 decimals).

First orderSecond order

Problem 6 · rule of 70

A country's GDP grows at 3.5% a year. Roughly how many years does it take to double?

Years

C. Concept checks

One attempt each. Read the explanation even when you get it right.

1The second-order Maclaurin approximation of e⁻ˣ is:

2The coefficient of (x − a)ᵏ in the Taylor series of f about a is:

3The Maclaurin series of an odd function such as sin x contains:

4The series x − x²/2 + x³/3 − … converges to ln(1 + x) for:

5The first-order Taylor polynomial of f about a is:

6If f(x) = 1 + 3x + 5x² + 7x³ + …, then f″(0) is:

Revise

One-page revision sheet

The formula

$\sum \frac{f ^{(k)} ( a )}{k !} (x - a)^{k}$ . About a = 0 it is the Maclaurin series.

The table method

Differentiate → plug in a → divide by k! → attach (x − a)^k → add.

Backwards

$f^{(k)} (0) = k! \times$ (coefficient of $x^{k}$ ). A missing power means that derivative is 0.

Know by heart

eˣ, sin x, cos x, ln(1 + x) (−1 < x ≤ 1), 1/(1 − x) and (1 + x)ᵐ (|x| < 1).

Shortcuts

Substitute (x → −x², 2x, …), multiply by powers of x, differentiate or integrate term by term.

Uses

Limits (expand the top), approximations (write as 1 + small), ln(1 + g) ≈ g, rule of 70, second-order risk.

Watch out for

The mistakes examiners are fishing for

Forgetting the k!. The coefficient of (x − a)² is f″(a)/2, not f″(a).
Evaluating the derivatives at x instead of at a. The coefficients are numbers: derivatives at the centre.
Expanding powers of x when the centre isn't 0. About a = 1 the terms are powers of (x − 1).
Using a series outside its range. ln(1 + x) = x − x²/2 + … fails for x > 1, however many terms you take.
Mixing up the signs of ln(1 + x) and ln(1 − x). The second has every term negative.

Try it on real questions

Eight past-paper questions this unlocks

From CUET-PG, JNU, IIT JAM and the University of Hyderabad: second-order Taylor expansions, limits that fall straight out of a series, e as a limit, and a growth-rate approximation.

These are questions from the TestStop bank. A free account answers 10 a day, each with its worked solution; All-Access opens every set with no daily limit.

Practice these →All questions on this topic →More resources →

Taylor series, and the simple trick for writing one

The idea: copy a curve with a polynomial

The formula, and where the k! comes from

The simple trick: the table method

The Taylor lab

Five series to know by heart

Shortcuts: substitute, don't differentiate

What it's for: limits, approximations, growth rates

Economics: why growth rates are log differences

How good is the approximation?

Exercises, with answers

A. Coefficients and derivatives

B. Limits and approximations

C. Concept checks

One-page revision sheet

The formula

The table method

Backwards

Know by heart

Shortcuts

Uses

The mistakes examiners are fishing for

Eight past-paper questions this unlocks

Read next

Taylor series, and the simple trick for writing one

The idea: copy a curve with a polynomial

The formula, and where the k! comes from

The simple trick: the table method

The Taylor lab

Five series to know by heart

Shortcuts: substitute, don't differentiate

What it's for: limits, approximations, growth rates

Economics: why growth rates are log differences

How good is the approximation?

Exercises, with answers

A. Coefficients and derivatives

B. Limits and approximations

C. Concept checks

One-page revision sheet

The formula

The table method

Backwards

Know by heart

Shortcuts

Uses

The mistakes examiners are fishing for

Eight past-paper questions this unlocks

Read next