2.7.1 Taylor polynomials

The tangent is the degree-1 copy

A Taylor polynomial copies a function near one point $a$ with a polynomial, which needs only adding and multiplying. The simplest copy is the tangent line of lesson 2.2.1.

Rule: Degree-1 copy: the tangent

p_1(x) = f(a) + f'(a)\,(x - a)

It copies two facts about $f$ at $a$ : the value, since $p_1(a) = f(a)$ , and the slope, since the slope of $p_1$ is $f'(a)$ everywhere.

For $f(x) = e^x$ at $a = 0$ , both the value and the slope are $e^0 = 1$ (lesson 2.3.4), so $p_1(x) = 1 + x$ . Lesson 2.4.2 measured the gaps: 0.3679 at $x = -1$ , 0.1487 at 0.5 and 0.7183 at 1. The curve bends away from the line.

Case studies 2.2.4 and 2.3.7 used this line for a bond price: minus its slope, per unit of price, is the duration.

The quadratic copy, from $f''$

To follow the bend, add a term $c\,(x - a)^2$ . At $a$ it is 0, and so is its slope $2c\,(x - a)$ . So it leaves the value and the slope of the copy alone.

Its second derivative is $2c$ everywhere. To match the bend of $f$ at $a$ , set $2c = f''(a)$ , so $c = \frac{f''(a)}{2}$ . That is where the $\frac{1}{2}$ comes from.

Rule: Degree-2 copy

\begin{gathered} p_2(x) = f(a) + f'(a)\,(x - a) \\ + \frac{f''(a)}{2}\,(x - a)^2 \end{gathered}

The best quadratic copy of $f$ near $a$ : it matches the value, the slope and the second derivative. Without the $\frac{1}{2}$ its second derivative would be $2f''(a)$ , twice the bend.

This is the copy lesson 2.4.2 promised, built from $f''$ . Case study 2.4.5's duration–convexity estimate has the same $\frac{1}{2}$ , because it is this copy of a bond price (case study 2.7.4).

Example 1: $e^x$ at 0. Every derivative of $e^x$ is $e^x$ , so $f''(0) = 1$ and $p_2(x) = 1 + x + \frac{x^2}{2}$ .

$x$	$-1$	$-0.5$	0.5	1
$e^x$	0.3679	0.6065	1.6487	2.7183
gap to $p_1$	0.3679	0.1065	0.1487	0.7183
gap to $p_2$	$-0.1321$	$-0.0185$	0.0237	0.2183

The bend term cuts the gap at 0.5 from 0.1487 to 0.0237. Left of 0 the gap turns negative: there $p_2$ is slightly above the curve. A later section explains the sign.

Example 2: $\ln x$ at 1. Here $f(1) = 0$ , $f'(1) = \frac{1}{1} = 1$ and $f''(1) = -\frac{1}{1^2} = -1$ (lessons 2.3.4 and 2.4.2). So

p_2(x) = (x - 1) - \frac{(x - 1)^2}{2}

The bend term is negative, because $\ln x$ is concave down. At 1.1, $\ln 1.1 = 0.0953$ against $p_2 = 0.095$ : a gap of 0.0003, where the tangent's gap is $-0.0047$ . At 1.5 the gap is 0.0305, against the tangent's $-0.0945$ .

a

import math import sympy as sp x = sp.symbols("x") def quadratic_copy(f, a): d0, d1, d2 = [sp.diff(f, x, k).subs(x, a) for k in range(3)] return d0 + d1 * (x - a) + d2 / 2 * (x - a)**2 print(quadratic_copy(sp.exp(x), 0)) print(quadratic_copy(sp.log(x), 1)) for t in [-1, -0.5, 0.5, 1]: print(t, round(math.exp(t) - (1 + t), 4), round(math.exp(t) - (1 + t + t**2 / 2), 4)) for t in [1.1, 1.5]: print(t, round(math.log(t) - (t - 1), 4), round(math.log(t) - ((t - 1) - (t - 1)**2 / 2), 4))

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Where Newton's constant comes from

Lesson 2.4.3 found that near a root $r$ , each Newton error is about $\frac{f''(r)}{2f'(r)}$ times the square of the last. Newton's step follows the tangent at the guess $x_n$ . The quadratic copy about $x_n$ shows what that leaves out.

At the root, $f(r) = 0$ . Write the quadratic copy about $x_n$ and evaluate it at $r$ :

\begin{gathered} 0 \approx f(x_n) + f'(x_n)\,(r - x_n) \\ + \frac{f''(x_n)}{2}\,(r - x_n)^2 \end{gathered}

Divide by $f'(x_n)$ . Newton's step is $x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$ , so the first two terms become $r - x_{n+1}$ . With errors $e_n = x_n - r$ , this leaves

e_{n+1} \approx \frac{f''(x_n)}{2f'(x_n)}\, e_n^2

As $x_n$ closes in on $r$ , the constant becomes $\frac{f''(r)}{2f'(r)}$ : 0.819 for $xe^x - 1$ and 0.354 for $x^2 - 2$ , as in lesson 2.4.3. For $x^2 - 2$ the quadratic copy is exact, since $f$ is a quadratic. So $e_{n+1} = \frac{e_n^2}{2x_n}$ holds exactly: from $x_0 = 1$ , $e_0 = -0.4142$ and $e_1 = 0.0858$ .

\frac{e_{n+1}}{e_n^2}

import math def errors(f, slope, x0): xs = [x0] for _ in range(12): xs.append(xs[-1] - f(xs[-1]) / slope(xs[-1])) return [v - xs[-1] for v in xs[:5]], xs[-1] e, r = errors(lambda t: t * math.exp(t) - 1, lambda t: math.exp(t) * (1 + t), 1.0) print([round(e[n + 1] / e[n]**2, 3) for n in range(4)], round((2 + r) / (2 * (1 + r)), 3)) e, r = errors(lambda t: t * t - 2, lambda t: 2 * t, 1.0) print([round(e[n + 1] / e[n]**2, 3) for n in range(4)], round(2 / (2 * 2 * r), 3)) print(round(e[0], 4), round(e[1], 4), round(e[0]**2 / 2, 4))

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Degree $n$ about any point

Keep going: add a cube term $c_3(x - a)^3$ , then a fourth power, and so on. Each new term is chosen to match one more derivative of $f$ at $a$ . Write $f^{(k)}$ for the $k$ -th derivative, so $f^{(1)} = f'$ , $f^{(2)} = f''$ , and $f^{(0)} = f$ itself.

Why each term has a $k!$

Differentiate $(x - a)^3$ three times with the power rule (lesson 2.3.1):

(x - a)^3 \;\to\; 3(x - a)^2 \;\to\; 6(x - a) \;\to\; 6

The power rule brings down 3, then 2, then 1, so the third derivative is $3 \times 2 \times 1 = 6$ . At $a$ , the lower derivatives are 0, because a factor $x - a$ is left. Higher derivatives are 0, because the 6 is a constant.

So the $k$ -th derivative of $c_k(x - a)^k$ at $a$ is $c_k \times k!$ , read " $k$ factorial": $k! = 1 \times 2 \times \cdots \times k$ . To match $f^{(k)}(a)$ , divide by it: $c_k = \frac{f^{(k)}(a)}{k!}$ .

$k$	0	1	2	3	4	5
$k!$	1	1	2	6	24	120

By convention $0! = 1$ , so the same formula gives the constant term $f(a)$ . The factorials grow fast, which is why later terms are often small.

Rule: Taylor polynomial of degree n about a

\begin{gathered} p_n(x) = \sum_{k=0}^{n} \frac{f^{(k)}(a)}{k!}\,(x - a)^k \\[4pt] = f(a) + f'(a)(x - a) \\ + \frac{f''(a)}{2!}(x - a)^2 + \cdots \end{gathered}

It matches $f$ and its first $n$ derivatives at $a$ . Each term has the units of $f$ . With a step $h = x - a$ , the same copy reads $f(a + h) \approx f(a) + f'(a)h + \frac{f''(a)}{2}h^2 + \cdots$ .

The coefficients are numbers: the derivatives are taken at $a$ , not at $x$ . The step form is the one case study 2.7.4 and lesson 2.8.1 use.

Example 1: $e^x$ about 0. Every derivative is $e^x$ , which is 1 at 0, so

p_n(x) = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots + \frac{x^n}{n!}

At $x = 0.5$ , $p_3 = 1.6458$ against $e^{0.5} = 1.6487$ : a gap of 0.0029. The figure shows each degree hugging the curve over a wider stretch.

$e^x$ and its Taylor polynomials of degree 1, 2 and 3 about 0. Each extra term matches one more derivative at 0; at $x = 1$ the gap falls from 0.72 to 0.22 to 0.05.

Example 2: $x^3$ about 2. The derivatives are $3x^2$ , $6x$ and 6, then 0. At 2 they are 12, 12 and 6.

$k$	0	1	2	3	4
$f^{(k)}(2)$	8	12	12	6	0
$k!$	1	1	2	6	24
coefficient	8	12	6	1	0

So $p_3(x) = 8 + 12(x - 2) + 6(x - 2)^2 + (x - 2)^3$ . With $h = x - 2$ this is $8 + 12h + 6h^2 + h^3$ , the expansion of $(2 + h)^3$ in lesson 2.2.1. A polynomial of degree 3 is its own Taylor polynomial of degree 3, with no error at all.

Try it below. Raise the degree $n$ from 1 to 5 and watch the copy of $e^x$ hug the curve further from the centre. Then move the centre $a$ to 1 and slide $x$ towards it: the error shrinks fast as $x$ nears $a$ .

Explore

Raise the degree and watch the polynomial hug the curve further from its centre. Move the centre, then the point, and compare the error with the first term left out.

y = eˣits Taylor polynomial p₁The large dot is the centre; the dashed line at x is the error.

Degree n = 1

Centre a = 0

Point x = 1

Polynomial: $p_{1}(x) = 1 + x$
At x = 1: $e^x$ = 2.71828, $p_{1}(x)$ = 2
Error: 0.71828
First term left out: 0.5
Reach: The series reaches every x: with enough terms the error shrinks to zero.

p_n

import math import sympy as sp x = sp.symbols("x") def taylor(f, a, n): return sum(sp.diff(f, x, k).subs(x, a) / sp.factorial(k) * (x - a)**k for k in range(n + 1)) print([math.factorial(k) for k in range(6)]) p3 = taylor(sp.exp(x), 0, 3) print(p3, "|", round(float(p3.subs(x, 0.5)), 4), round(math.exp(0.5), 4)) q = taylor(x**3, 2, 3) print(q, "|", sp.expand(q - x**3))

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How fast the error falls

Near $a$ , the error $f(x) - p_n(x)$ behaves like a constant times $(x - a)^{n+1}$ : the first power the copy leaves out. Halve the distance to $a$ and the error shrinks about $2^{n+1}$ times.

$x$	0.4	0.2	0.1	shrinks by
error of $p_1$	0.0918	0.0214	0.00517	about 4
error of $p_2$	0.0118	0.0014	0.000171	about 8
error of $p_3$	0.00116	0.0000694	0.00000425	about 16

These are the errors for $e^x$ about 0. The tangent's error falls like $x^2$ , as in lesson 2.2.1; one more term makes it fall like $x^3$ , then $x^4$ .

Rule: Error near a

f(x) - p_n(x) \approx C\,(x - a)^{n+1}

For $x$ near $a$ , with a constant $C$ that depends on $f$ and $a$ . Lesson 2.7.2 shows how to find $C$ , and how big the error can be further out.

On log scales a power of $x$ is a straight line whose slope is the power. So each error is a line of slope $n + 1$ .

The error of $p_1$ , $p_2$ and $p_3$ for $e^x$ against the distance from 0, on log scales. The slopes are 2, 3 and 4: one more term adds one to the power.

Both sides of $a$

The sign of the error follows $(x - a)^{n+1}$ . An even power is positive on both sides of $a$ ; an odd power changes sign at $a$ .

For $e^x$ the tangent's error, like $x^2$ , is positive both ways: $e^x$ lies above its tangent (lesson 2.4.2). The quadratic copy's error, like $x^3$ , is negative left of 0.

That is why $p_2$ overshoots at $-0.5$ : 0.625 against 0.6065. Its size still shrinks about 8 times per halving: 0.00968, 0.00127 and 0.000163 at $-0.4$ , $-0.2$ and $-0.1$ .

When a term is missing

If a coefficient is 0, the copy gains nothing from that degree. At an inflection point $f''(a) = 0$ (lesson 2.4.2), so the quadratic copy is just the tangent.

Take $x^3 - 3x$ at 0. Its derivatives there are 0, $-3$ , 0 and 6, so $p_1 = p_2 = -3x$ . The error is $x^3$ exactly: it falls like the next power that is actually present.

Far from $a$ , and where there is no copy

Far from $a$ the rule says nothing. At $x = 3$ , $p_3$ for $e^x$ gives $1 + 3 + 4.5 + 4.5 = 13$ , against $e^3 = 20.0855$ . Lesson 2.7.2 asks how many terms fix that, and finds functions where no number of terms will.

A Taylor polynomial needs the derivatives at $a$ . A corner, such as $|x|$ at 0, has no slope (lesson 2.2.2), and neither has a vertical tangent, such as $\sqrt[3]{x}$ at 0. At such a point there is no copy of degree 1 or more.

p_1

import math def p(n, t): return sum(t**k / math.factorial(k) for k in range(n + 1)) for n in [1, 2, 3]: errs = [math.exp(t) - p(n, t) for t in [0.4, 0.2, 0.1]] print(n, [f"{e:.3g}" for e in errs], [round(errs[i] / errs[i + 1], 1) for i in range(2)]) print([f"{math.exp(t) - p(2, t):.3g}" for t in [-0.4, -0.2, -0.1]], p(2, -0.5)) print(p(3, 3), round(math.exp(3), 4))

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Common mistakes

Keep the $k!$ . The cube term of $e^x$ about 0 is $\frac{x^3}{6}$ , not $x^3$ ; the quadratic term is $\frac{x^2}{2}$ , not $x^2$ .
Take the derivatives at $a$ . The coefficients are numbers such as $f''(a)$ , not functions of $x$ .
Use powers of $x - a$ . About $a = 1$ , the copy of $\ln x$ is $(x - 1) - \frac{(x - 1)^2}{2}$ , not $x - \frac{x^2}{2}$ .
The error rule holds near $a$ . Far away, more degree may still leave a large error.
Mind the sign on the left. An odd power $(x - a)^{n+1}$ makes the error change sign at $a$ .
A zero coefficient adds nothing. At an inflection point $p_2 = p_1$ .

Taylor polynomials

0 of 12 correct

1.
Find the degree-1 Taylor polynomial of $e^x$ about 0, and use it to estimate $e^{0.1}$
2.
$f(2) = 5$ , $f'(2) = 3$ and $f''(2) = 4$ . Evaluate the degree-2 copy about 2 at $x = 2.1$
3.
Evaluate $5!$
4.
What is the coefficient of $x^4$ in the Taylor polynomial of $e^x$ about 0?
5.
A function has $f'''(0) = 12$ . What is the coefficient of $x^3$ in its Taylor polynomial about 0?
6.
Use the tangent of $\sqrt{x}$ at 4, with slope $\frac{1}{4}$ , to estimate $\sqrt{4.4}$
7.
Evaluate the degree-2 copy of $\ln x$ about 1 at $x = 1.2$
8.
$x^3$ about 2 is $8 + 12(x - 2) + c\,(x - 2)^2 + (x - 2)^3$ . Find $c$
9.
Evaluate the degree-3 Taylor polynomial of $e^x$ about 0 at $x = -1$
10.
Near $a$ , the error of a degree-2 copy is 0.0016 at a distance 0.2 from $a$ . Estimate it at a distance 0.1
11.
Write the degree-2 Taylor polynomial of $x^3 - 3x$ about 0
12.
Near a root, $f'(r) = 2$ and $f''(r) = 8$ . A Newton error is 0.01. Estimate the next one

How to type answers

Powers with ^: x^7, 2^(1/3), a^-2.
Multiply with * or by writing side by side: 3x^2, 2(x+1).
ln(x) is the natural log, log(x, 2) is log base 2, e is Euler's number, sqrt(x) is a square root.
sin(x), cos(x) and tan(x) take radians; write pi for π and sin(x)^2 for sin²x.
Your first answer to each question is saved: it counts towards your score and brings the question back for review.
Exact answers like ln(20)/ln(3) are best. Decimals need about 4 significant figures.

p_1(x) = f(a) + f'(a)(x - a)

Taylor polynomials

The tangent is the degree-1 copy

Rule: Degree-1 copy: the tangent

The quadratic copy, from $f''$

Rule: Degree-2 copy

Where Newton's constant comes from

Degree $n$ about any point

Why each term has a $k!$

Rule: Taylor polynomial of degree n about a

How fast the error falls

Rule: Error near a

Both sides of $a$

When a term is missing

Far from $a$ , and where there is no copy

Common mistakes

Taylor polynomials

Summary card

Free further reading

The tangent is the degree-1 copy

Rule: Degree-1 copy: the tangent

The quadratic copy, from f′′f''f′′

Rule: Degree-2 copy

Where Newton's constant comes from

Degree nnn about any point

Why each term has a k!k!k!

Rule: Taylor polynomial of degree n about a

How fast the error falls

Rule: Error near a

Both sides of aaa

When a term is missing

Far from aaa, and where there is no copy

Common mistakes

Summary card

Free further reading

The quadratic copy, from $f''$

Degree $n$ about any point

Why each term has a $k!$

Both sides of $a$

Far from $a$ , and where there is no copy