MATH 141 Calculus 2 • McGill University, Montreal

Revision sheet: Taylor polynomials and error bounds (MATH 141)

Q: How do I choose M in Taylor's inequality?

M must be at least the absolute value of the next derivative at every point between the centre and the point where you approximate. Compute that derivative, decide whether it increases or decreases on the interval, and take its value at the worst end. For the exponential that is the right end, for logarithms and powers of one plus x usually the left end, and for sine and cosine the number one always works.

Q: When can I use the alternating series estimate instead of Taylor's inequality?

When the Taylor series, evaluated at your point, has terms that alternate in sign, decrease in size and tend to zero. Then the error is smaller than the first term you left out and has the same sign. It works for sine, cosine and arctangent at small values, for e to a negative power and for the logarithm of one plus a positive number, but never for a series whose terms all have the same sign.

Q: Why is the degree 3 Taylor polynomial of sin x also the degree 4 one?

Because the fourth derivative of sine is sine itself, which is zero at zero, so the term of degree four is zero. The same polynomial can therefore be treated as the degree four polynomial, and Taylor's inequality with n equal to four uses the fifth power of x instead of the fourth, a much smaller bound near zero. The same happens for cosine with odd degrees.

Q: How do I find how many terms of a Taylor series I need for a given accuracy?

Write the error bound as a function of n, either the Taylor inequality bound with its factorial or the first omitted term if the series alternates, then test n equal to one, two, three and so on until the bound falls below the tolerance. The first n that works is the answer. Do not stop because the next term is small when the terms are all positive: the omitted terms add up.

This sheet is not a summary of sections 11.10 and 11.11 of Stewart: you already have the course notes. It answers one question only, what makes students lose marks on Taylor polynomials and error bounds in MATH 141 at McGill University, and which precise gesture avoids each loss.

Every number below is done by hand, as on the exam, and every value quoted has been checked by an independent computation. The rule that runs through the whole sheet: an approximation without a guaranteed bound answers half the question.

The thread of the chapter

An approximation is a value AND a guaranteed bound: M bounds the next derivative at its WORST point between the centre and x, never at the centre; the first omitted term is a bound only when the series alternates; and when a zero term makes T_3 equal T_4, the inequality is used with the larger degree.

This chapter is part of MATH 141, Calculus 2 (McGill)

What is on this sheet

1. The essentials
2. The rules in table form
3. The mistakes that cost marks
4. Which method to choose
5. How the answer is expected to be written
6. Check before you hand in
7. The typical problem, taken apart
8. Learn by heart
9. Frequently asked questions

Topics covered

Taylor polynomial
Remainder of a Taylor series
Taylor's inequality
Lagrange form of the remainder
Alternating series estimation theorem
Error bound
Linear approximation
Centre of a Taylor polynomial
Small-angle approximation
Binomial series

The essentials

Taylor's inequality: M lives at the worst point

• $T_n(x) = \sum_{k=0}^{n}\frac{f^{(k)}(a)}{k!}(x - a)^k$ and the remainder is $R_n(x) = f(x) - T_n(x)$ . Keep the powers of $x - a$ ; never expand them.
• If $\lvert f^{(n+1)}(x)\rvert \le M$ for EVERY $x$ with $\lvert x - a\rvert \le d$ , then $\lvert R_n(x)\rvert \le \frac{M}{(n + 1)!}\lvert x - a\rvert^{n+1}$ on that interval.
• $\lvert f^{(n+1)}\rvert$ is almost always monotone on the interval, so its maximum sits at an END: evaluate both ends and take the larger. Any larger $M$ is valid, any smaller one is false.
• $e^x$ : right end. Powers of $\frac{1}{1 + x}$ , from $\ln(1 + x)$ or $(1 + x)^k$ : left end. $\frac{1}{1 - x}$ : right end. $\sin$ , $\cos$ : $M = 1$ .
• When $T_n = T_{n+1}$ ( $\sin$ and $\cos$ at $0$ ), use $n + 1$ : the bound gains a full power of $\lvert x - a\rvert$ .
• Lagrange form $R_n(x) = \frac{f^{(n+1)}(z)}{(n + 1)!}(x - a)^{n+1}$ : if $f^{(n+1)}$ keeps one sign on the interval, the error has a known sign.

For

f(x) = \frac{1}{1 - x}

[0, 0.5]

f'''

6

at the centre and

96

at the right end: the centre underestimates the worst derivative sixteen times.

Write every $M$ with its interval: $\lvert f'''(x)\rvert \le 16$ for $-0.5 \le x \le 0.5$ . A bound without its interval cannot be checked, and markers do not check it for you.

The two other tools: the first omitted term and the geometric tail

• If the Taylor series AT THE POINT alternates with terms decreasing to $0$ , the error is smaller than the first omitted term and has its sign. No $M$ to find.
• It is often sharper: $\cos 0.5 \approx \frac{7}{8}$ with error below $\frac{1}{384} \approx 0.0026$ , and the true error is $0.0026$ .
• Alternation is checked on the series at the point, not on the function: $\ln(1 + x)$ alternates at $x = 0.2$ and not at $x = -0.2$ ; $e^{x}$ alternates at $x = -0.2$ only.
• Terms all positive and derivatives unwieldy: bound the tail by a geometric series, as in $2\sum_{k \ge 3}\frac{x^{2k+1}}{2k + 1} \le \frac{2}{7} \cdot \frac{x^7}{1 - x^2}$ .
• Bounds of a sum add: approximating $\ln 3 = \ln 2 + \ln\frac{3}{2}$ costs the sum of the two bounds.

The rules in table form

Each row reads from left to right: the assumptions, then the result. A red cell is not an answer, it is the finding that the form settles nothing and the instruction to rewrite it. Every case is followed by a worked example.

Where the worst derivative sits, and which M to take

The left column is the function, the middle one how $\lvert f^{(n+1)}\rvert$ behaves on the interval, the right one the $M$ to write. The red line is the one that appears on half the papers.

Function	Behaviour of the derivative on the interval	M to take
$e^x$ on $[a, b]$	increasing	$e^b$ , the right end
Example: $e^{0.2}$ with $T_3$ : $M = e^{0.2} < 1.25$ since $1.25^5 > 3$ , bound $\frac{1.25(0.2)^4}{24} = \frac{1}{12000}$ .
$\sin x$ , $\cos x$	bounded by $1$ everywhere	$1$ , on any interval
Example: $\cos 31^\circ$ with $T_2$ at $\frac{\pi}{6}$ : $\lvert R_2\rvert \le \frac{1}{6}\left(\frac{\pi}{180}\right)^3 < \frac{1}{750\,000}$ .
$\ln(1 + x)$ , $(1 + x)^k$	powers of $\frac{1}{1 + x}$ , decreasing	the value at the LEFT end
Example: $\ln(1 + x)$ , $T_2$ on $[-0.5, 0.5]$ : $f''' = \frac{2}{(1 + x)^3}$ , $M = 16$ at $x = -0.5$ , bound $\frac{1}{3}$ .
$\frac{1}{1 - x}$ on $[0, b]$	increasing	the value at the RIGHT end
Example: $T_2$ on $[0, 0.5]$ : $f''' = \frac{6}{(1 - x)^4}$ , $M = 96$ , bound $\frac{96}{6}(0.5)^3 = 2$ .
any $f$	not checked	$\lvert f^{(n+1)}(a)\rvert$ , the value at the centre not a bound
Example: $\frac{1}{1 - x}$ on $[0, 0.5]$ : $M = 6$ claims $\lvert R_2\rvert \le \frac{1}{8}$ , and the true error at $0.5$ is $2 - 1.75 = \frac{1}{4}$ . What to do: Evaluate $\lvert f^{(n+1)}\rvert$ at both ends of the interval and take the larger. The centre is right only when it is an end and the worst point, and that must be said.

The red line is not a weaker bound, it is no bound at all: the function itself breaks it. A larger $M$ only loses sharpness; a smaller one loses the question.

The mistakes that cost marks

These are the errors I correct most often in session. Each one costs marks on a paper, even when the reasoning behind it is right.

1. Taking M at the centre of the interval

the whole question: a bound the function violates is worth zero

What not to write

“For $\ln(1 + x)$ on $[-0.5, 0.5]$ : $\lvert f'''(x)\rvert \le f'''(0) = 2$ , so $\lvert R_2(x)\rvert \le \frac{2}{6}(0.5)^3 = \frac{1}{24}$ .”

What to write

“ $f'''(x) = \frac{2}{(1 + x)^3}$ is decreasing, largest at $x = -0.5$ : $M = 16$ , so $\lvert R_2(x)\rvert \le \frac{16}{6}(0.5)^3 = \frac{1}{3}$ for $\lvert x\rvert \le 0.5$ .”

Why: At $x = -0.5$ the true error is $\ln 2 - \frac{5}{8} \approx 0.068$ , above $\frac{1}{24} \approx 0.042$ . The inequality asks for a bound on the WHOLE interval, and the centre is rarely the worst point.

2. Using the first omitted term on a series of positive terms

2 to 3 marks, and the stated bound is false

What not to write

“The terms of $\sum\frac{1}{k!}$ decrease, so $e - T_3(1) \le \frac{1}{24}$ .”

What to write

“The series of $e$ does not alternate. On $[0, 1]$ , $f^{(4)} = e^x \le e < 3$ , so $0 < e - T_3(1) \le \frac{3}{24} = \frac{1}{8}$ .”

The sums of

e

climb from below, so the gap is the WHOLE tail; the sums of

\frac{1}{e}

jump from side to side, so the gap is less than the next jump.

Why: The first-omitted-term rule is the alternating series estimate. With positive terms the omitted terms all push the same way and add up: $e - \frac{8}{3} \approx 0.052 > \frac{1}{24}$ .

3. Not using T_3 as T_4 for the sine

1 to 2 marks, the whole question if the conclusion is wrong

What not to write

“ $\lvert\sin x - (x - \frac{x^3}{6})\rvert \le \frac{x^4}{24}$ , so $T_3$ cannot give $\sin 0.1$ to within $10^{-6}$ .”

What to write

“The sine has no $x^4$ term, so $T_3 = T_4$ and $\lvert R_4(0.1)\rvert \le \frac{(0.1)^5}{120} = \frac{1}{12\,000\,000}$ .”

Why: $\frac{x^4}{24}$ is true but fifty times too large at $x = 0.1$ . A large bound never proves a large error: it only proves you used a weak bound.

4. Plugging degrees into a formula written in radians

the whole question

What not to write

“ $\cos 31^\circ \approx \frac{\sqrt{3}}{2} - \frac{1}{2}(1) - \frac{\sqrt{3}}{4}(1)^2$ .”

What to write

“ $31^\circ = \frac{\pi}{6} + \frac{\pi}{180}$ , so $h = \frac{\pi}{180}$ and $\cos 31^\circ \approx \frac{\sqrt{3}}{2} - \frac{\pi}{360} - \frac{\sqrt{3}\,\pi^2}{129\,600}$ .”

Why: The faux line gives a negative cosine for an acute angle. Every derivative of $\sin$ and $\cos$ , hence every Taylor coefficient, assumes radians.

5. Reading an error bound as correct decimals

1 mark

What not to write

“ $e^{0.2} \approx 1.22133$ with an error below $10^{-4}$ , so $e^{0.2} = 1.2213\ldots$ ”

What to write

“ $1.22133 < e^{0.2} < 1.22142$ : the digits $1.221$ are certain, the fourth decimal is not decided by this bound.”

Why: A bound gives an interval, and an interval can straddle a change of digit. Here it does: $e^{0.2} = 1.22140\ldots$ , so the fourth decimal of the approximation is wrong although the error is below $10^{-4}$ .

6. Centring at 0 by habit

the whole question: the tolerance is not guaranteed

What not to write

“For $\cos 31^\circ$ within $10^{-5}$ : $T_2$ at $0$ with $x = \frac{31\pi}{180}$ , error at most $\frac{x^4}{24}$ .”

What to write

“Centre at $\frac{\pi}{6}$ , $h = \frac{\pi}{180}$ : $\lvert R_2\rvert \le \frac{h^3}{6} < \frac{1}{750\,000} < 10^{-5}$ .”

31^\circ

, the dashed line, the bound centred at

\frac{\pi}{6}

is practically zero while the one centred at

0

is already

3.6 \times 10^{-3}

: each bound is small only near its own centre.

Why: At $0$ the bound is $\frac{x^4}{24} \approx 0.0036$ , some three hundred times the tolerance. The distance to the centre is raised to a power, so the nearest point with exact values wins by orders of magnitude.

7. Giving the value without the bound or its sign

half the question

What not to write

“ $\sqrt{10} \approx \frac{683}{216}$ .”

What to write

“ $\frac{683}{216} < \sqrt{10} < \frac{683}{216} + \frac{1}{3888}$ , because $f''' > 0$ on $[9, 10]$ makes $R_2$ positive.”

Why: Approximate, on a MATH 141 paper, means value AND guaranteed bound. The sign costs one line and halves the width of the answer: the approximation is one end of the interval.

Which method to choose

Which bound, by the FORM of the question

Write the series at the point, look at the signs of its terms and at the derivatives, then choose

If the series at the point alternates, with terms decreasing to 0 → the first omitted term bounds the error and gives its sign

Example: $\cos 0.5 \approx \frac{7}{8}$ , error below $\frac{1}{384}$ , $\frac{7}{8}$ too small
If the terms have one sign and the derivatives are easy → Taylor's inequality, $M$ at the worst point, sign from $f^{(n+1)}$

Example: $e^{0.2} \approx \frac{458}{375}$ , error below $\frac{1}{12000}$ , too small
If the terms have one sign and look like $\frac{x^k}{k}$ → compare the tail with a geometric series

Example: $\ln 2 \approx \frac{842}{1215}$ from $x = \frac{1}{3}$ , tail below $\frac{1}{6804}$
If the point is far from 0 but close to a point with exact values → centre there, and convert angles to radians

Example: $\cos 31^\circ$ at $\frac{\pi}{6}$ with $h = \frac{\pi}{180}$ , error below $\frac{1}{750\,000}$
If the accuracy is given and $n$ is the unknown → write the bound as a function of $n$ , find the smallest $n$

Example: $e$ within $10^{-3}$ : $\frac{3}{(n + 1)!} < 10^{-3}$ gives $n = 6$
If the approximation must hold on a whole interval → bound at the worst $x$ of the interval, for $M$ and for $\lvert x - a\rvert^{n+1}$ alike

Example: $e^x$ on $[0, 2]$ within $0.01$ : $\frac{9 \cdot 2^{n+1}}{(n + 1)!}$ gives $n = 9$

Several tools may apply; choose the one whose hypotheses you can CHECK in one line. For $\arctan\frac{1}{2}$ , the alternating estimate is one line and Taylor's inequality is a page.

How the answer is expected to be written

A marker ticks steps. Here they are in order, with the concluding sentence expected word for word.

Writing an approximation with a guaranteed bound

When to use it: Any question that says approximate, estimate or with an error less than, on a value such as $e^{0.2}$ , $\sqrt[3]{1.1}$ , $\sin 0.1$ or $\cos 31^\circ$

1 Name $f$ , the centre $a$ and the degree $n$ , and write $T_n$ in powers of $x - a$ .
2 Evaluate $T_n$ at the point in fractions, the decimal only at the end.
3 Name the tool and check its hypotheses: for Taylor's inequality, the interval and $M$ at its worst point; for the alternating estimate, the alternation and the decrease.
4 Compute the bound as a simple fraction.
5 Give the sign of the error and conclude with an interval.

Concluding sentence

“ $T_3(0.2) = \frac{458}{375}$ . For $0 \le x \le 0.2$ , $\lvert f^{(4)}(x)\rvert = e^x \le e^{0.2} < 1.25$ , so $0 < e^{0.2} - \frac{458}{375} \le \frac{1.25(0.2)^4}{24} = \frac{1}{12000}$ .”

The trap: Writing $M = e^{0.2}$ and stopping there: $M$ must be a number you can compute by hand, so bound it ( $e^{0.2} < e < 3$ , or $< 1.25$ ).

Marking: Typically 1 mark for the polynomial, 1 for its value, 1 to 2 for M with its interval or for the hypotheses of the alternating estimate, 1 for the bound, and the sign or interval is what separates a full answer from a partial one.

Check before you hand in

Five minutes of checking recover more marks than one more problem started in a hurry.

The worst-point test

Evaluate $\lvert f^{(n+1)}\rvert$ at BOTH ends of the interval before writing $M$ . Your $M$ must be at least the larger value.

$\frac{2}{(1 + x)^3}$ on $[-0.5, 0.5]$ : $16$ at the left end, $\frac{16}{27}$ at the right end, so $M = 16$ , not $2$ .
The first-term test

A good bound is of the order of the first omitted term. If yours is much larger, look for a zero term that makes the polynomial of degree n also the one of degree n + 1.

$\sin 0.1$ : the bound $\frac{1}{240\,000}$ is fifty times the first omitted term $\frac{1}{12\,000\,000}$ , the sign that $T_3 = T_4$ was missed.
The sign test

If $f^{(n+1)}$ keeps one sign on the interval, the error has that sign times the sign of $(x - a)^{n+1}$ . Your interval must be one-sided accordingly.

$\sqrt[3]{1 + x}$ : $f''' > 0$ on $[0, 0.1]$ , so $\frac{929}{900}$ is too small and $\sqrt[3]{1.1} \in \left(\frac{929}{900}, \frac{929}{900} + \frac{1}{16200}\right)$ .
The radian test

An increment of a few degrees must become a small number of radians. If h is 1, 5 or 30, the degrees were not converted.

$1^\circ = \frac{\pi}{180} \approx 0.017$ rad, so the $h^3$ of the remainder is about $5 \times 10^{-6}$ .

The typical problem, taken apart

The square root of 10 by hand, with a guaranteed bound

Approximate $\sqrt{10}$ with a Taylor polynomial of degree $2$ of $f(x) = \sqrt{x}$ at a well chosen centre. Give a guaranteed bound on the error and its sign.

No calculator. Every step must be justified as on a MATH 141 final.

Step 1

Centre: $a = 9$ , the perfect square nearest to $10$ , so $x - a = 1$ .

Why

The bound grows like $\lvert x - a\rvert^3$ . The centre must be a point where $f$ and its derivatives are exact: $a = 0$ is impossible, $\sqrt{x}$ is not differentiable there.

Step 2

$f(9) = 3$ , $f'(x) = \frac{1}{2}x^{-1/2}$ so $f'(9) = \frac{1}{6}$ , $f''(x) = -\frac{1}{4}x^{-3/2}$ so $f''(9) = -\frac{1}{108}$ . $T_2(10) = 3 + \frac{1}{6} - \frac{1}{2 \cdot 108} = \frac{648 + 36 - 1}{216} = \frac{683}{216} \approx 3.16204$ .

Why

The $\frac{1}{2!}$ of the last coefficient is the slip of the day: without it, $-\frac{1}{108}$ gives $\frac{682}{216}$ .

Step 3

$f'''(x) = \frac{3}{8}x^{-5/2}$ is positive and DECREASING on $[9, 10]$ : its maximum is at $x = 9$ , $M = \frac{3}{8} \cdot \frac{1}{243} = \frac{1}{648}$ .

Why

Here the centre happens to be the worst point, and that is shown by the monotonicity, not assumed. Taking $M$ at $x = 10$ would give a smaller number that is not a bound.

Step 4

Taylor's inequality: $\lvert R_2(10)\rvert \le \frac{1/648}{3!} \cdot 1^3 = \frac{1}{3888} \approx 0.00026$ .

Why

A simple fraction, computed by hand. The marker checks the factorial and the power, so write them.

Step 5

Sign: $R_2(10) = \frac{f'''(z)}{6} \cdot 1^3$ with $f'''(z) > 0$ , so the error is positive and $T_2(10)$ is an underestimate.

Why

The sign costs one line and turns a symmetric band into a one-sided interval, half as wide.

The conclusion, written out

“ $\frac{683}{216} < \sqrt{10} < \frac{683}{216} + \frac{1}{3888}$ , that is $3.16203 < \sqrt{10} < 3.16230$ .”

The classic mistake on this problem: Centring at $a = 16$ (the bound grows like $6^3$ instead of $1^3$ ), or writing $\frac{1}{108}$ for the coefficient of $(x - 9)^2$ and missing it because the value still looks plausible: only the derivative check of $T_2$ catches it.

Learn by heart

• Approximate means value AND guaranteed bound, with its sign when available.
• Taylor's inequality: $\lvert R_n(x)\rvert \le \frac{M}{(n + 1)!}\lvert x - a\rvert^{n+1}$ , with $M \ge \lvert f^{(n+1)}\rvert$ on the WHOLE interval.
• $M$ at the worst point: an end of the interval, checked; never the centre by default.
• $\sin$ and $\cos$ at $0$ : $T_{2k+1} = T_{2k+2}$ and $T_{2k} = T_{2k+1}$ , use the larger $n$ .
• First omitted term: a bound only if the series at the point ALTERNATES with decreasing terms.
• Radians always: $1^\circ = \frac{\pi}{180}$ .
• Centre at the nearest point with exact values: the bound grows like $\lvert x - a\rvert^{n+1}$ .

Frequently asked questions

How do I choose M in Taylor's inequality?

M must be at least the absolute value of the next derivative at every point between the centre and the point where you approximate. Compute that derivative, decide whether it increases or decreases on the interval, and take its value at the worst end. For the exponential that is the right end, for logarithms and powers of one plus x usually the left end, and for sine and cosine the number one always works.

When can I use the alternating series estimate instead of Taylor's inequality?

When the Taylor series, evaluated at your point, has terms that alternate in sign, decrease in size and tend to zero. Then the error is smaller than the first term you left out and has the same sign. It works for sine, cosine and arctangent at small values, for e to a negative power and for the logarithm of one plus a positive number, but never for a series whose terms all have the same sign.

Why is the degree 3 Taylor polynomial of sin x also the degree 4 one?

Because the fourth derivative of sine is sine itself, which is zero at zero, so the term of degree four is zero. The same polynomial can therefore be treated as the degree four polynomial, and Taylor's inequality with n equal to four uses the fifth power of x instead of the fourth, a much smaller bound near zero. The same happens for cosine with odd degrees.

How do I find how many terms of a Taylor series I need for a given accuracy?

Write the error bound as a function of n, either the Taylor inequality bound with its factorial or the first omitted term if the series alternates, then test n equal to one, two, three and so on until the bound falls below the tolerance. The first n that works is the answer. Do not stop because the next term is small when the terms are all positive: the omitted terms add up.

Practise it

Corrected exercises: Taylor polynomials and error bounds, MATH 141 at McGill

A method is proved on a paper, not on a sheet. The set for the same chapter takes each of these traps into a problem, with the solution written out step by step.

10 corrected exercises
100 points
150 minutes

Do the exercises

Previous sheet Taylor and Maclaurin series

Looking for a MATH 141 tutor in Montreal?

Get in touch for a first session. Error bounds close the course and open every final: the marks are in the choice of M and in the sign of the error, not in the arithmetic.

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Revision sheet: Taylor polynomials and error bounds (MATH 141)

The essentials

Taylor's inequality: M lives at the worst point

The two other tools: the first omitted term and the geometric tail

The rules in table form

Where the worst derivative sits, and which M to take

The mistakes that cost marks

1. Taking M at the centre of the interval

2. Using the first omitted term on a series of positive terms

3. Not using T_3 as T_4 for the sine

4. Plugging degrees into a formula written in radians

5. Reading an error bound as correct decimals

6. Centring at 0 by habit

7. Giving the value without the bound or its sign

Which method to choose

Which bound, by the FORM of the question

How the answer is expected to be written

Writing an approximation with a guaranteed bound

Check before you hand in

The typical problem, taken apart

The square root of 10 by hand, with a guaranteed bound

Learn by heart

Frequently asked questions

How do I choose M in Taylor's inequality?

When can I use the alternating series estimate instead of Taylor's inequality?

Why is the degree 3 Taylor polynomial of sin x also the degree 4 one?

How do I find how many terms of a Taylor series I need for a given accuracy?

Corrected exercises: Taylor polynomials and error bounds, MATH 141 at McGill

See also

Looking for a MATH 141 tutor in Montreal?