2.7.2 Standard series and the size of the error

In this lesson

From Taylor polynomials to Taylor series: infinitely many terms, as in lesson 1.5.3
The standard series: $e^x$ , $\sin x$ , $\cos x$ , $\ln(1 + x)$ and $\frac{1}{1 - x}$
How big the error is: the first left-out term as an estimate, and a bound you can trust
How many terms you need, and why it depends on $x$
Where a series stops working: $\ln(1 + x)$ and $\frac{1}{1 - x}$ beyond $|x| = 1$
Three uses: $\ln(1 + x) \approx x - \frac{x^2}{2}$ , the gap between $\left(1 + \frac{r}{n}\right)^n$ and $e^r$ , and Euler's formula

You need Taylor polynomials (lesson 2.7.1), infinite series (lesson 1.5.3), and the derivatives of $e^x$ , $\ln x$ , $\sin x$ and $\cos x$ (lessons 2.3.4 and 2.3.5). The last section uses complex numbers from lessons 1.7.2 and 1.7.3.

From polynomials to series

Lesson 2.7.1 built the Taylor polynomial $p_n$ from the first $n$ derivatives at $a$ . Let $n$ grow without end and you get an infinite series, the Taylor series of $f$ about $a$ .

As in lesson 1.5.3, an infinite series means the limit of its partial sums. Here the partial sums are the Taylor polynomials $p_0, p_1, p_2, \ldots$ , so the question is whether $p_n(x)$ settles on $f(x)$ .

Rule: Taylor series

f(x) = \sum_{k=0}^{\infty} \frac{f^{(k)}(a)}{k!}\,(x - a)^k

This holds at every $x$ where the partial sums $p_n(x)$ settle on $f(x)$ . This lesson finds where that is for five standard series, all about $a = 0$ .

The series for $e^x$

Every derivative of $e^x$ is $e^x$ (lesson 2.3.4), which is 1 at 0. So every coefficient is $\frac{1}{k!}$ .

Rule: Series for the exponential

\begin{gathered} e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots \\ \text{for every } x \end{gathered}

The terms shrink fast because $k!$ eventually outgrows any power $x^k$ . The Proof Gym (stage PG5) proves that the series settles on $e^x$ for every $x$ .

At $x = 1$ the series gives $e = 1 + 1 + \frac{1}{2} + \frac{1}{6} + \cdots$ . The partial sums are 2.5, 2.6667, 2.7083, 2.7167 and 2.7181 after degrees 2 to 6. After degree 10 the gap to $e$ is only $2.7 \times 10^{-8}$ .

Sine and cosine

The derivatives of $\sin x$ go round the cycle of four from lesson 2.3.5: $\sin$ , $\cos$ , $-\sin$ , $-\cos$ . At 0 their values are 0, 1, 0, $-1$ , then the cycle repeats.

So the even powers drop out, and the odd ones alternate in sign. Cosine starts one step along the cycle, at 1, 0, $-1$ , 0, so it keeps only the even powers.

Rule: Series for sine and cosine

\begin{gathered} \sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots \\[4pt] \cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots \\[4pt] \text{for every } x, \text{ in radians} \end{gathered}

Sine has only odd powers, like its symmetry $\sin(-x) = -\sin x$ ; cosine has only even powers, like $\cos(-x) = \cos x$ (lesson 1.7.1).

The first terms are old friends. $\sin x \approx x$ is the tangent at 0 from lesson 2.3.5, and $\cos x \approx 1 - \frac{x^2}{2}$ is the quadratic copy at the top of the wave.

Far from 0 you need many terms. Here are the partial sums for $\sin 3$ , whose true value is 0.14112.

degree	1	3	5	7	9	11	13
partial sum	3	$-1.5$	0.525	0.091071	0.145312	0.140875	0.141131

The terms $\frac{3^k}{k!}$ first grow, to 4.5 at $k = 3$ and $k = 4$ , and only then shrink as $k!$ wins. The partial sums swing wildly before they settle.

$\ln(1 + x)$ and the geometric series

The derivatives of $\ln(1 + x)$ are $\frac{1}{1 + x}$ , $-\frac{1}{(1 + x)^2}$ , $\frac{2}{(1 + x)^3}$ , $-\frac{6}{(1 + x)^4}$ , by the rules of lessons 2.3.1 and 2.3.4. At 0 they are 1, $-1$ , 2, $-6$ . Divided by $1!, 2!, 3!, 4!$ , they give the coefficients 1, $-\frac{1}{2}$ , $\frac{1}{3}$ , $-\frac{1}{4}$ .

For $\frac{1}{1 - x}$ the $k$ -th derivative at 0 is $k!$ , so every coefficient is 1. Its Taylor series is the geometric series of lesson 1.5.3, with first term 1 and ratio $x$ .

Rule: Series for ln(1 + x) and 1/(1 − x)

\begin{gathered} \ln(1 + x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots \\ \text{for } -1 < x \le 1 \\[6pt] \frac{1}{1 - x} = 1 + x + x^2 + \cdots \\ \text{for } -1 < x < 1 \end{gathered}

The bottoms are $k$ , not $k!$ , for $\ln(1 + x)$ , so its terms shrink slowly. Neither series works beyond $|x| = 1$ ; a later section shows what happens there.

Case study 1.4.7 used $\ln(1 + R) \approx R - \frac{R^2}{2}$ for small returns $R$ : these are the first two terms. At $R = 0.05$ , $\ln 1.05 = 0.04879$ against $0.04875$ .

x^6

import math import sympy as sp x = sp.symbols("x") for f in [sp.exp(x), sp.sin(x), sp.cos(x), sp.log(1 + x), 1 / (1 - x)]: print(sp.series(f, x, 0, 7)) for n in [2, 3, 4, 5, 6, 10]: partial = sum(1 / math.factorial(k) for k in range(n + 1)) print(n, round(partial, 4), f"{math.e - partial:.1e}") print(round(math.log(1.05), 6), round(0.05 - 0.05**2 / 2, 6))

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How big is the error?

The first left-out term

Lesson 2.7.1 found that near $a$ the error of $p_n$ is about $C(x - a)^{n+1}$ . The constant is the next coefficient, $\frac{f^{(n+1)}(a)}{(n+1)!}$ .

So the first term you leave out is an estimate of the error. If its coefficient is 0, as half of sine's are, use the next term that is not 0.

approximation	error	first left-out term
$e^{0.5} \approx 1 + 0.5 + \frac{0.5^2}{2} + \frac{0.5^3}{6}$	0.00289	$\frac{0.5^4}{24} = 0.0026$
$\sin 0.5 \approx 0.5 - \frac{0.5^3}{6}$	0.000259	$\frac{0.5^5}{120} = 0.00026$
$\ln 1.05 \approx 0.05 - \frac{0.05^2}{2}$	0.0000402	$\frac{0.05^3}{3} = 0.0000417$

The estimate is close, but it is not a promise: for $e^{0.5}$ the true error is larger than the estimate.

A bound you can trust

When you need a guarantee, use the remainder bound. This lesson states it without proof.

Rule: Remainder bound

\lvert f(x) - p_n(x) \rvert \le \frac{M\,\lvert x - a \rvert^{n+1}}{(n+1)!}

$M$ is the largest size of $f^{(n+1)}$ anywhere between $a$ and $x$ . It is the first left-out term with the derivative at its worst, instead of at $a$ .

For $e^{0.5}$ with $p_3$ , the fourth derivative $e^x$ is at most $e^{0.5}$ between 0 and 0.5. So the error is at most $e^{0.5} \times \frac{0.5^4}{24} = 0.0043$ , and the true error, 0.00289, is below that.

For sine and cosine every derivative is a sine or cosine, at most 1 in size. So $M = 1$ , and the error is at most $\frac{|x|^{n+1}}{(n+1)!}$ .

How many terms

To get $e$ within $10^{-6}$ , use the bound with $M = e < 3$ : the error after degree $n$ is below $\frac{3}{(n + 1)!}$ . That first drops below $10^{-6}$ at degree 9, where it is $8.3 \times 10^{-7}$ , so 10 terms are enough. The true error is $3.0 \times 10^{-7}$ .

The answer depends on $x$ . For $e^x$ within a relative error of $10^{-6}$ , you need 10 terms at $x = 1$ , 20 at $x = 5$ , 29 at $x = 10$ and 46 at $x = 20$ .

The slow series is $\ln(1 + x)$ at $x = 1$ : $\ln 2 = 1 - \frac{1}{2} + \frac{1}{3} - \cdots$ . After 10, 100 and 1,000 terms the error is 0.0475, 0.00498 and 0.0005, about $\frac{1}{2n}$ . Six correct decimals would take about a million terms.

A smaller $x$ fixes that. Since $\ln 2 = -\ln(1 - 0.5)$ , use the series at $x = -0.5$ : its terms shrink like $0.5^k$ , and 20 of them give $\ln 2$ within $4.4 \times 10^{-8}$ .

e^{0.5}

import math p3 = 1 + 0.5 + 0.5**2 / 2 + 0.5**3 / 6 print(round(math.exp(0.5) - p3, 5), round(0.5**4 / 24, 5), round(math.exp(0.5) * 0.5**4 / 24, 4)) print(round(math.sin(0.5) - (0.5 - 0.5**3 / 6), 6), round(0.5**5 / 120, 6)) print(round(math.log(1.05) - (0.05 - 0.05**2 / 2), 7), round(0.05**3 / 3, 7)) def terms_for(x, tol=1e-6): total, k = 0.0, 0 while abs(math.exp(x) - total) >= tol * math.exp(x): total += x**k / math.factorial(k) k += 1 return k print(f"{3 / math.factorial(10):.1e}", f"{math.e - sum(1 / math.factorial(k) for k in range(10)):.1e}") print([terms_for(x) for x in [1, 5, 10, 20]]) def ln1p_series(x, n): return sum((-1)**(k + 1) * x**k / k for k in range(1, n + 1)) print([f"{math.log(2) - ln1p_series(1, n):.3g}" for n in [10, 100, 1000]]) print(f"{math.log(2) + ln1p_series(-0.5, 20):.1e}")

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Where a series stops working

The series for $e^x$ , $\sin x$ and $\cos x$ work for every $x$ . The other two do not.

$\ln(1 + x)$ at $x = 1.5$ . The function is fine there: $\ln 2.5 = 0.9163$ . But the terms $\frac{1.5^k}{k}$ grow, so the partial sums swing ever wider.

terms	1	2	3	4	5	10	20	30
partial sum	1.5	0.375	1.5	0.2344	1.7531	$-2.4031$	$-96.8286$	$-3782.6561$

Terms that do not shrink to 0 mean the series diverges (lesson 1.5.3). At $x = -1$ it is $-\left(1 + \frac{1}{2} + \frac{1}{3} + \cdots\right)$ , minus the harmonic series of lesson 1.5.3, which has no sum either. That matches $\ln 0$ , which does not exist.

$\frac{1}{1 - x}$ at $x = 2$ . The series is $1 + 2 + 4 + \cdots$ , whose partial sums grow without end, while $\frac{1}{1 - 2} = -1$ . At $x = -1$ it is $1 - 1 + 1 - \cdots$ , which bounces, while $\frac{1}{1 - (-1)} = \frac{1}{2}$ . Lesson 1.5.3 met both.

Rule: Radius of convergence

\begin{gathered} \text{converges for } \lvert x - a \rvert < R \\ \text{diverges for } \lvert x - a \rvert > R \end{gathered}

Every Taylor series has such a distance $R$ . For $\ln(1 + x)$ and $\frac{1}{1 - x}$ , $R = 1$ ; for $e^x$ , $\sin x$ and $\cos x$ there is no limit. At $|x - a| = R$ itself, check each end.

The trouble at one end limits the other. $\ln(1 + x)$ breaks at $x = -1$ , one unit from 0, so its series fails beyond 1 too, where the function is fine.

Near $|x| = R$ the series still converges, but slowly: $\ln 2$ took a million terms, and lesson 1.5.3's $1 + 0.9 + 0.81 + \cdots$ needed 66 terms to come within 0.01 of its sum. A Taylor polynomial is still a good copy near 0; it is adding more terms far out that fails.

More terms carry the copy of $\sin x$ further out, without limit. The copies of $\ln(1 + x)$ hug the curve only for $|x| \le 1$ ; beyond 1, more terms make them worse.

Try both sides of the edge. With $x = 0.5$ , raise the degree $n$ and watch the error shrink. Then set $x = 1.5$ : every extra term makes the error larger.

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 = ln(1 + x)its Taylor polynomial p₅The large dot is the centre; the dashed line at x is the error.The dotted line marks x = −1, where the function breaks.

Degree n = 5

Centre a = 0

Point x = 0.5

Polynomial: $p_{5}(x) = x - \dfrac{x^{2}}{2} + \dfrac{x^{3}}{3} - \dfrac{x^{4}}{4} + \dfrac{x^{5}}{5}$
At x = 0.5: $\ln(1 + x)$ = 0.40547, $p_{5}(x)$ = 0.40729
Error: −0.00183
First term left out: −0.0026
Reach: x is within 1 of the centre, the distance to the break at x = −1: more terms shrink the error.

\ln(1 + x)

import math def ln1p_series(x, n): return sum((-1)**(k + 1) * x**k / k for k in range(1, n + 1)) print([round(ln1p_series(1.5, n), 4) for n in [1, 2, 3, 4, 5, 10, 20, 30]], round(math.log(2.5), 4)) print([round(ln1p_series(-1, n), 4) for n in [1, 10, 100, 1000]]) print(sum(2**k for k in range(20)), 1 / (1 - 2)) print(sum((-1)**k for k in range(20)), sum((-1)**k for k in range(21)), 1 / (1 + 1))

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Two series at once: $\left(1 + \frac{r}{n}\right)^n$ against $e^r$

Case study 2.1.5 found that $e^r - \left(1 + \frac{r}{n}\right)^n$ is about $\frac{r^2 e^r}{2n}$ , which is $\frac{0.0013141}{n}$ at $r = 0.05$ . The series for $\ln(1 + x)$ and $e^x$ show where that comes from.

1
$\left(1 + \frac{r}{n}\right)^n = e^{\,n \ln(1 + r/n)}$
2
$\begin{gathered} n \ln\left(1 + \frac{r}{n}\right) \\ = r - \frac{r^2}{2n} + \frac{r^3}{3n^2} - \cdots \end{gathered}$
3
$\begin{gathered} \left(1 + \frac{r}{n}\right)^n = e^r e^u \\ \approx e^r\left(1 - \frac{r^2}{2n}\right) \end{gathered}$

So the gap $e^r - \left(1 + \frac{r}{n}\right)^n$ is about $e^r \times \frac{r^2}{2n}$ . At $r = 0.05$ , $\frac{r^2 e^r}{2} = 0.0013141$ , the number the table of case study 2.1.5 settled on.

n = 12

import math r, n = 0.05, 12 print(round(n * math.log1p(r / n), 7), round(r - r**2 / (2 * n), 7)) for n in [12, 365, 8760]: gap = math.exp(r) - (1 + r / n)**n print(n, f"{gap * n:.7f}", f"{r**2 * math.exp(r) / 2:.7f}")

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Euler's formula from the series

Lesson 1.7.3 checked Euler's formula $e^{i\theta} = \cos\theta + i\sin\theta$ with numbers. The series give one way to see why it holds for every angle $\theta$ , in radians.

Put $x = i\theta$ into the series for $e^x$ . The powers of $i$ go round a cycle of four, $1, i, -1, -i$ (lesson 1.7.2):

\begin{gathered} e^{i\theta} = 1 + i\theta - \frac{\theta^2}{2!} - \frac{i\theta^3}{3!} \\ + \frac{\theta^4}{4!} + \frac{i\theta^5}{5!} - \cdots \end{gathered}

Sort the terms. The real ones, $1 - \frac{\theta^2}{2!} + \frac{\theta^4}{4!} - \cdots$ , are the series for $\cos\theta$ . The others are $i$ times $\theta - \frac{\theta^3}{3!} + \frac{\theta^5}{5!} - \cdots$ , the series for $\sin\theta$ .

So $e^{i\theta} = \cos\theta + i\sin\theta$ . This takes on trust that the series for $e^x$ works for complex $x$ as well; all three series work for every input.

\sum \frac{(i\theta)^k}{k!}

import cmath import math def tidy(z): return complex(round(z.real, 6) + 0.0, round(z.imag, 6) + 0.0) for theta in [1, math.pi]: series = sum((1j * theta)**k / math.factorial(k) for k in range(25)) print(tidy(series), tidy(cmath.exp(1j * theta)), tidy(complex(math.cos(theta), math.sin(theta))))

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Common mistakes

Radians for sine and cosine. The series $x - \frac{x^3}{6} + \cdots$ needs $x$ in radians, as the derivatives of lesson 2.3.5 do.
Which signs alternate. Sine, cosine and $\ln(1 + x)$ alternate; $e^x$ and $\frac{1}{1 - x}$ do not.
$k$ , not $k!$ , under $\ln(1 + x)$ . Its terms are $\frac{x^k}{k}$ , with a minus sign on the even ones.
A series outside its range gives nonsense. The series for $\ln(1 + x)$ at $x = 1.5$ and for $\frac{1}{1 - x}$ at $x = 2$ diverge, though both functions have values there.
The first left-out term is an estimate, not a guarantee. For a guarantee, use the remainder bound.
More terms for larger $|x|$ . Ten terms give $e$ within $10^{-6}$ ; $e^{20}$ needs 46 for the same relative accuracy.

Standard series and the size of the error

0 of 12 correct

1.
Use the series for $e^x$ up to the $x^2$ term to estimate $e^{0.2}$
2.
What is the coefficient of $x^5$ in the series for $\sin x$ ?
3.
What is the coefficient of $x^4$ in the series for $\cos x$ ?
4.
What is the coefficient of $x^4$ in the series for $\ln(1 + x)$ ?
5.
Sum $1 + x + x^2 + \cdots$ at $x = -0.5$
6.
$\ln 1.3 \approx 0.3 - \frac{0.3^2}{2} = 0.255$ . Estimate its error with the first left-out term
7.
Every derivative of $\sin x$ is at most 1 in size. Bound $\lvert \sin 0.6 - 0.6 \rvert$ with the remainder bound for degree 2
8.
After degree $n$ , the series for $e$ is within $\frac{3}{(n + 1)!}$ of $e$ . What is the smallest $n$ that makes this less than 0.001?
9.
The series for $\ln(1 + x)$ fails at $x = 2$ , though $\ln 3$ exists. What is its radius of convergence about 0?
10.
$\ln(1 + R) \approx R - \frac{R^2}{2}$ . Use it for $R = 0.1$
11.
In the series for $e^{i\theta}$ , the term $\frac{(i\theta)^2}{2!}$ equals $c\,\theta^2$ . Find $c$
12.
Estimate $e^r - \left(1 + \frac{r}{n}\right)^n$ for $r = 0.1$ and $n = 100$ with $\frac{r^2 e^r}{2n}$ . Type e for $e$

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.

n \to \infty

Standard series and the size of the error

From polynomials to series

Rule: Taylor series

The series for $e^x$

Rule: Series for the exponential

Sine and cosine

Rule: Series for sine and cosine

$\ln(1 + x)$ and the geometric series

Rule: Series for ln(1 + x) and 1/(1 − x)

How big is the error?

The first left-out term

A bound you can trust

Rule: Remainder bound

How many terms

Where a series stops working

Rule: Radius of convergence

Two series at once: $\left(1 + \frac{r}{n}\right)^n$ against $e^r$

Euler's formula from the series

Common mistakes

Standard series and the size of the error

Summary card

Free further reading

From polynomials to series

Rule: Taylor series

The series for exe^xex

Rule: Series for the exponential

Sine and cosine

Rule: Series for sine and cosine

ln⁡(1+x)\ln(1 + x)ln(1+x) and the geometric series

Rule: Series for ln(1 + x) and 1/(1 − x)

How big is the error?

The first left-out term

A bound you can trust

Rule: Remainder bound

How many terms

Where a series stops working

Rule: Radius of convergence

Two series at once: (1+rn)n\left(1 + \frac{r}{n}\right)^n(1+nr​)n against ere^rer

Euler's formula from the series

Common mistakes

Summary card

Free further reading

The series for $e^x$

$\ln(1 + x)$ and the geometric series

Two series at once: $\left(1 + \frac{r}{n}\right)^n$ against $e^r$