MATH 141 Calculus 2 • McGill University, Montreal

Revision sheet: Taylor and Maclaurin series (MATH 141)

Q: What is the difference between a Taylor series and a Maclaurin series?

A Maclaurin series is simply a Taylor series centred at zero. The Taylor series of a function at a point a is written in powers of x minus a, with the n-th derivative at a divided by n factorial as coefficient. Choose the centre where the function and its derivatives are easy to evaluate, and where the function is defined.

Q: How do I find a Maclaurin series without computing derivatives?

Start from the table of known series, those of e to the x, sine, cosine, the natural log of one plus x, arctangent and the binomial series, then substitute, multiply by a power of x, multiply or divide two series, or integrate term by term. By uniqueness the result is the Maclaurin series. Always restate the interval where the new series is valid.

Q: How do I find the n-th derivative at zero from a power series?

Find the coefficient of x to the n in the series, then multiply it by n factorial. For e to the x cubed, the x to the ninth term is x to the ninth over 3 factorial, so the ninth derivative at zero is 9 factorial over 3 factorial, 60480. If x to the n never appears, that derivative is zero.

Q: How do I use Taylor series to compute a limit?

Replace each function by its series around the point, keeping enough terms that something nonzero survives after the cancellations, usually up to the degree of the denominator. Simplify, divide, and the limit is the coefficient of the lowest surviving power. Stopping too early, for instance replacing sine of x by x, is the most common way to get zero by mistake.

This sheet is not a summary of section 11.10 of Stewart: you already have the course notes. It answers one question only, what makes students lose marks on Taylor and Maclaurin series 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: build from the table, read back with $n!$ , and stay inside the interval.

The thread of the chapter

The coefficient of the n-th power of (x - a) is the n-th derivative at a over n factorial, read in both directions: build a series from the table instead of differentiating, read derivatives back as n factorial times the coefficient, and never use a series outside the interval on which it equals its function.

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 series
Maclaurin series
Taylor coefficients
Uniqueness of power series
Binomial series
Binomial coefficients
Interval of convergence
Product of power series
Composition of power series
Limits by series expansion
Term-by-term integration

The essentials

One formula, read in both directions

• Taylor series of $f$ at $a$ : $\sum_{n=0}^{\infty}\frac{f^{(n)}(a)}{n!}(x-a)^n$ . Maclaurin series: the case $a = 0$ .
• Uniqueness: if $f(x) = \sum c_n(x-a)^n$ on an interval around $a$ , then $c_n = \frac{f^{(n)}(a)}{n!}$ , whatever route produced the series.
• Forwards, from the definition: differentiate, find and prove the pattern, evaluate AT $a$ , divide by $n!$ , keep $(x - a)^n$ . Only when no table route exists.
• Backwards: $f^{(n)}(a) = n!\,c_n$ . From $e^{x^3} = \sum\frac{x^{3n}}{n!}$ : $f^{(9)}(0) = \frac{9!}{3!} = 60480$ and $f^{(10)}(0) = 0$ .
• A convergent Taylor series may still not equal $f$ : the table's equalities are proved theorems, and a new function is best BUILT from them.

On a copy, say which direction you use. A derivative read from a series is worth full marks only if the $n!$ appears on the page.

The table, with the interval of each equality

• Every $x$ : $e^x = \sum\frac{x^n}{n!}$ , $\sin x = \sum\frac{(-1)^n x^{2n+1}}{(2n+1)!}$ , $\cos x = \sum\frac{(-1)^n x^{2n}}{(2n)!}$ .
• $\frac{1}{1-x} = \sum x^n$ on $(-1, 1)$ ; $\ln(1+x) = \sum_{n\ge1}\frac{(-1)^{n-1}x^n}{n}$ on $(-1, 1]$ ; $\arctan x = \sum\frac{(-1)^n x^{2n+1}}{2n+1}$ on $[-1, 1]$ , both from the power series chapter.
• $(1+x)^k = \sum\binom{k}{n}x^n$ for $|x| < 1$ , with $\binom{k}{n} = \frac{k(k-1)\cdots(k-n+1)}{n!}$ , exactly $n$ factors on top. The constant inside must be $1$ .
• Recognize a table series by its DENOMINATOR: $n!$ is the exponential, $(2n)!$ or $(2n+1)!$ with alternating signs is the cosine or the sine, $n$ is the logarithm, $2n + 1$ is the arctangent.

Red dots mark the endpoints: filled means the equality holds there, hollow means it does not. Only

e^x

\sin

and

\cos

are valid everywhere; everything else stops at

\pm1

before any substitution.

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 does the new series live?

Every construction moves the interval in its own way. The last column is the interval on which the new series equals its function; the red line is the series that students write and that does not exist.

Construction	Example	Interval of the result
substitute $u = cx$	$\ln(1 + 2x)$ from $\ln(1+u)$	$\left(-\frac{1}{2}, \frac{1}{2}\right]$
Example: $-1 < 2x \le 1$ ; at $x = \frac{1}{2}$ the series is $\sum\frac{(-1)^{n-1}}{n}$ , it converges; at $x = -\frac{1}{2}$ it is $-\sum\frac{1}{n}$ .
substitute $u = x^m$	$\arctan(x^3)$	$[-1, 1]$
Example: $\|x^3\| \le 1 \iff \|x\| \le 1$ ; the powers become $x^{6n+3}$ , exponents multiply.
multiply by $x^k$	$x^2e^{-x}$	unchanged: every $x$
Example: $x^2e^{-x} = \sum\frac{(-1)^m x^{m+2}}{m!}$ , so $f^{(21)}(0) = -\frac{21!}{19!} = -420$ .
factor to reach $(1 + u)^k$	$\sqrt[3]{8 + x} = 2\left(1 + \frac{x}{8}\right)^{1/3}$	$\|x\| < 8$
Example: $2 + \frac{x}{12} - \frac{x^2}{288} + \dots$ , and $f'(0) = \frac{1}{3} \cdot 8^{-2/3} = \frac{1}{12}$ confirms $c_1$ .
integrate term by term	$\int_0^x e^{-t^2}\,dt$	same radius: every $x$
Example: $\sum\frac{(-1)^n x^{2n+1}}{n!\,(2n+1)}$ ; at $x = 1$ the exact value $1 - \frac{1}{3} + \frac{1}{10} - \frac{1}{42} + \dots$
Maclaurin series of $\ln x$	$\ln x$ at $a = 0$	no such series no such series
Example: $\ln x$ is not even defined at $0$ , so no $f^{(n)}(0)$ exists and no series in powers of $x$ can equal it. Same form, other result: $\frac{1}{x}$ has no Maclaurin series either, yet $\frac{1}{x} = \sum(-1)^n(x-1)^n$ on $(0, 2)$ : centred at $1$ it works. What to do: Change the centre: $\ln x = \ln 2 + \ln\left(1 + \frac{x-2}{2}\right) = \ln 2 + \sum_{n\ge1}\frac{(-1)^{n-1}(x-2)^n}{n\,2^n}$ on $(0, 4]$ , or work with $\ln(1 + x)$ .

Endpoints never follow a rule of thumb: after a substitution, each endpoint is the image of an endpoint of the table, and it is included exactly when the table includes it.

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. Reading a derivative off a coefficient without the factorial

the whole question

What not to write

“ $\sin(x^2) = x^2 - \frac{x^6}{6} + \dots$ , so the sixth derivative of $\sin(x^2)$ at $0$ is $-\frac{1}{6}$ .”

What to write

“ $c_6 = -\frac{1}{6}$ , so $f^{(6)}(0) = 6!\,c_6 = 720 \cdot \left(-\frac{1}{6}\right) = -120$ .”

Why: The coefficient is $\frac{f^{(n)}(a)}{n!}$ , not $f^{(n)}(a)$ . The factorial is the only computation of the question, so forgetting it leaves nothing to mark.

2. Losing the centre

2 to 3 marks, and every later use of the series

What not to write

“The Taylor series of $\frac{1}{x}$ at $a = 1$ is $\sum(-1)^n x^n$ .”

What to write

“ $\frac{1}{x} = \frac{1}{1 + (x-1)} = \sum_{n\ge0}(-1)^n(x-1)^n$ for $0 < x < 2$ .”

Why: A series centred at $a$ is written in powers of $(x - a)$ . The faux line is the series of $\frac{1}{1+x}$ at $0$ , a different function. The value-at-the-centre test catches it: at $x = 1$ it gives $\sum(-1)^n$ , which does not even converge.

3. Evaluating a series outside its interval

the whole question

What not to write

“ $\frac{\pi}{3} = \arctan\sqrt{3} = \sum_{n\ge0}\frac{(-1)^n(\sqrt{3})^{2n+1}}{2n+1}$ .”

What to write

“ $\frac{1}{\sqrt{3}}$ is in $[-1, 1]$ : $\frac{\pi}{6} = \arctan\frac{1}{\sqrt{3}}$ , so $\pi = 2\sqrt{3}\sum_{n\ge0}\frac{(-1)^n}{(2n+1)3^n}$ .”

Why: The arctangent series equals $\arctan x$ only on $[-1, 1]$ . At $\sqrt{3}$ the terms $\frac{3^n\sqrt{3}}{2n+1}$ tend to infinity: the series diverges and equals nothing. Rewrite the number so that the argument falls inside the interval.

4. Copying the interval of the table after a substitution

1 to 2 marks

What not to write

“ $\ln(1 + 2x) = \sum_{n\ge1}\frac{(-1)^{n-1}2^n x^n}{n}$ for $-1 < x \le 1$ .”

What to write

“The table needs $-1 < u \le 1$ with $u = 2x$ , so the series is valid for $-\frac{1}{2} < x \le \frac{1}{2}$ .”

Why: The interval belongs to the variable of the table, $u$ , not to $x$ . At $x = 1$ the faux line would add terms $\frac{2^n}{n}$ that grow without bound.

5. Applying the binomial series before factoring

2 to 3 marks

What not to write

“ $\sqrt[3]{8 + x} = 2 + \frac{1}{3}x - \frac{1}{9}x^2 + \dots$ ”

What to write

“ $\sqrt[3]{8 + x} = 2\left(1 + \frac{x}{8}\right)^{1/3} = 2 + \frac{x}{12} - \frac{x^2}{288} + \dots$ for $|x| < 8$ .”

Why: The binomial series is the series of $(1 + u)^k$ : the constant must be $1$ . The first-derivative test settles it in five seconds: $f'(0) = \frac{1}{3} \cdot 8^{-2/3} = \frac{1}{12}$ , not $\frac{1}{3}$ .

6. Multiplying two series coefficient by coefficient

the whole question

What not to write

“ $e^x\cos x$ : multiply the coefficients of equal powers, $1 + 0 \cdot x - \frac{1}{4}x^2 + \dots$ ”

What to write

“The coefficient of $x^2$ collects $1 \cdot \left(-\frac{1}{2}\right) + 1 \cdot 0 + \frac{1}{2} \cdot 1 = 0$ : $e^x\cos x = 1 + x - \frac{x^3}{3} - \frac{x^4}{6} + \dots$ ”

Why: A product of series is a product of polynomials that never ends: the coefficient of $x^n$ gathers every pair of degrees adding to $n$ . The faux line even fails $c_1 = f'(0) = 1$ .

7. Stopping the expansion before anything survives

the whole question

What not to write

“Near $0$ , $\cos x \approx 1$ , so $\frac{1 - \cos x}{x^2} \approx \frac{0}{x^2} = 0$ and the limit is $0$ .”

What to write

“ $1 - \cos x = \frac{x^2}{2} - \frac{x^4}{24} + \dots$ , so $\frac{1 - \cos x}{x^2} = \frac{1}{2} - \frac{x^2}{24} + \dots \to \frac{1}{2}$ .”

The curve does not head for

0

: it climbs to the hollow point at height

\frac{1}{2}

, the coefficient of the first term that survives,

\frac{x^2}{2}

Why: The approximation $\cos x \approx 1$ throws away exactly the term that decides. Expand until a nonzero term survives the cancellation, to the degree of the denominator.

8. Taking a convergent Taylor series for its function

the whole question

What not to write

“Every derivative of $f(x) = e^{-1/x^2}$ (with $f(0) = 0$ ) is $0$ at $0$ , so its Maclaurin series is $0$ and $f(x) = 0$ for all $x$ .”

What to write

“The Maclaurin series of $f$ is $0$ and converges everywhere, but $f(1) = \frac{1}{e} \ne 0$ : the series equals $f$ only at $x = 0$ .”

The blue curve is so flat at

0

that every derivative vanishes there, so its Maclaurin series is the red line

y = 0

; the two agree only at the origin.

Why: Convergence of the Taylor series says nothing about its limit. The equalities of the table are theorems proved with the remainder; a function outside the table is either built from it or checked the same way.

Which method to choose

Which route to the series, by the FORM of the function

Before differentiating anything, look for a table function hidden in the formula

If a table function of a simple expression: $e^{-x^2}$ , $\cos\sqrt{x}$ , $\ln(1 + 2x)$ → substitute in parentheses, then rescale the interval

Example: $e^{-x^2} = \sum\frac{(-1)^n x^{2n}}{n!}$ , every $x$
If a power of $x$ times a table function → multiply, and shift the exponents

Example: $x^2\arctan(x^3) = \sum\frac{(-1)^n x^{6n+5}}{2n+1}$ on $[-1, 1]$
If a product or a quotient of two table functions → collect every pair of degrees, or write the unknown series and match coefficients

Example: $\tan x = x + \frac{x^3}{3} + \frac{2x^5}{15} + \dots$ from $\sin x = \tan x \cdot \cos x$
If a composition $f(g(x))$ → make the inner quantity tend to $0$ , then expand each power of it to the degree asked

Example: $\ln(\cos x)$ with $\cos x = 1 + u$ : $-\frac{x^2}{2} - \frac{x^4}{12} + \dots$
If a constant other than $1$ inside a power, or a centre $a \ne 0$ → factor to reach $(1 + u)^k$ or $\ln(1 + u)$

Example: $\ln x = \ln 2 + \ln\left(1 + \frac{x-2}{2}\right)$ on $(0, 4]$
If an integral with no elementary antiderivative, or a function defined by one → expand the integrand, integrate term by term on a bounded interval

Example: $\text{Si}(x) = x - \frac{x^3}{18} + \frac{x^5}{600} - \dots$
If none of the above, and the question says from the definition → differentiate, find the pattern, prove it, evaluate at the centre, divide by n!

Example: $\ln x$ at $2$ : $f^{(n)}(2) = \frac{(-1)^{n-1}(n-1)!}{2^n}$ , $c_n = \frac{(-1)^{n-1}}{n\,2^n}$

A table route and the definition give the SAME series, by uniqueness. When time allows, the second route is the best check of the first.

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 a series built from the table

When to use it: Any question that says find the Taylor or Maclaurin series and state where it is valid

1 Name the table series you start from, with its interval, in the variable $u$ .
2 Say what $u$ is, substitute in parentheses, and multiply or factor as needed.
3 Simplify the general term: powers of powers multiply, signs come out as $(-1)^n$ .
4 Translate the interval of $u$ into an interval of $x$ , and give each endpoint's verdict from the table.
5 Check the first term, or $c_1 = f'(a)$ , before moving on.

Concluding sentence

“Since $\ln(1+u) = \sum_{n\ge1}\frac{(-1)^{n-1}u^n}{n}$ for $-1 < u \le 1$ , with $u = 2x$ : $\ln(1+2x) = \sum_{n\ge1}\frac{(-1)^{n-1}2^n x^n}{n}$ for $-\frac{1}{2} < x \le \frac{1}{2}$ .”

The trap: Writing the table series and the final series without saying what $u$ is: the marker cannot tell a substitution from a guess, and the interval has no justification.

Marking: Typically 1 mark for the table series quoted with its interval, 1 to 2 for the substitution and the general term, 1 for the interval of the result with its endpoints.

Check before you hand in

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

The value at the centre

Put x = a in the series: every term but the first vanishes, and what is left must be f(a).

The series of $\ln x$ at $2$ gives $\ln 2$ at $x = 2$ ; the faux series $\sum(-1)^n x^n$ for $\frac{1}{x}$ at $1$ gives a divergent $\sum(-1)^n$ .
The first-derivative test

Compute f'(a) directly, in one line, and compare with the coefficient c_1.

For $\sqrt[3]{8 + x}$ : $f'(0) = \frac{1}{12}$ , which rules out the unfactored $\frac{1}{3}$ at once.
The parity test

An even function has only even powers in its Maclaurin series, an odd function only odd powers.

$\ln(\cos x)$ and $\frac{\sin x}{x}$ are even, $\tan x$ , $\arctan x$ and Si are odd: an $x^3$ in the series of $\ln(\cos x)$ is a slip.
The square test

For a root, square the series and check that the result is the function under the root.

$\left(1 + \frac{x}{2} - \frac{x^2}{8} + \frac{x^3}{16}\right)^2 = 1 + x + 0x^2 + 0x^3 + \dots$

The typical problem, taken apart

A limit by series, composition included

Compute $\lim_{x\to0}\frac{e^{x^2} - \cos x - \frac{3}{2}x^2}{x^4}$ .

No calculator, no L'Hôpital's rule: the answer is expected by series, as on a MATH 141 final.

Step 1

Form $\frac{0}{0}$ , denominator $x^4$ : every piece of the numerator will be expanded up to $x^4$ , not further.

Why

The degree of the denominator fixes the depth of every expansion. Going further costs time; stopping earlier loses the answer.

Step 2

$e^u = 1 + u + \frac{u^2}{2} + \dots$ with $u = x^2$ : $e^{x^2} = 1 + x^2 + \frac{x^4}{2} + \dots$ , the next term being of degree $6$ .

Why

A substitution into the table, no derivative. Stopping at $1 + x^2$ because $x^4$ looks small is exactly the error that ruins the question.

Step 3

$\cos x = 1 - \frac{x^2}{2} + \frac{x^4}{24} + \dots$ , the next term being of degree $6$ .

Why

Straight from the table. Both expansions now stop at the same degree, $4$ , which is the condition for the subtraction to be exact up to $x^4$ .

Step 4

Numerator: $(1 - 1) + \left(1 + \frac{1}{2} - \frac{3}{2}\right)x^2 + \left(\frac{1}{2} - \frac{1}{24}\right)x^4 + \dots = \frac{11}{24}x^4 + \dots$

Why

Collect degree by degree and write the cancellations on the page: the constants and the $x^2$ terms vanish, which is why the limit is decided at $x^4$ .

Step 5

Divide by $x^4$ : $\frac{11}{24} + (\text{terms of degree} \ge 2) \to \frac{11}{24}$ .

Why

The limit is the coefficient of the first surviving power, once it has the degree of the denominator.

The conclusion, written out

“ $\lim_{x\to0}\frac{e^{x^2} - \cos x - \frac{3}{2}x^2}{x^4} = \lim_{x\to0}\left(\frac{11}{24} + \dots\right) = \frac{11}{24}$ .”

The classic mistake on this problem: Stopping $e^{x^2}$ at $1 + x^2$ turns the numerator into $-\frac{x^4}{24}$ and the limit into $-\frac{1}{24}$ : a wrong sign and a wrong size, from one term left out.

Learn by heart

• $c_n = \frac{f^{(n)}(a)}{n!}$ , and backwards $f^{(n)}(a) = n!\,c_n$ . Never forget the factorial.
• Centre $a$ : powers of $(x - a)$ . The series of $\ln x$ and $\frac{1}{x}$ exist only around $a \ne 0$ .
• Only $e^x$ , $\sin x$ , $\cos x$ are valid for every $x$ . The others stop at $\pm1$ in their own variable.
• Substitute in parentheses, then rescale the interval and recheck its endpoints.
• Binomial series: make the constant $1$ first; $\binom{k}{n}$ has exactly $n$ factors on top.
• Products collect all pairs of degrees; compositions need an inner quantity tending to $0$ .
• Limits: expand until something survives, to the degree of the denominator.
• A convergent Taylor series may converge to something else: $e^{-1/x^2}$ .

Frequently asked questions

What is the difference between a Taylor series and a Maclaurin series?

A Maclaurin series is simply a Taylor series centred at zero. The Taylor series of a function at a point a is written in powers of x minus a, with the n-th derivative at a divided by n factorial as coefficient. Choose the centre where the function and its derivatives are easy to evaluate, and where the function is defined.

How do I find a Maclaurin series without computing derivatives?

Start from the table of known series, those of e to the x, sine, cosine, the natural log of one plus x, arctangent and the binomial series, then substitute, multiply by a power of x, multiply or divide two series, or integrate term by term. By uniqueness the result is the Maclaurin series. Always restate the interval where the new series is valid.

How do I find the n-th derivative at zero from a power series?

Find the coefficient of x to the n in the series, then multiply it by n factorial. For e to the x cubed, the x to the ninth term is x to the ninth over 3 factorial, so the ninth derivative at zero is 9 factorial over 3 factorial, 60480. If x to the n never appears, that derivative is zero.

How do I use Taylor series to compute a limit?

Replace each function by its series around the point, keeping enough terms that something nonzero survives after the cancellations, usually up to the degree of the denominator. Simplify, divide, and the limit is the coefficient of the lowest surviving power. Stopping too early, for instance replacing sine of x by x, is the most common way to get zero by mistake.

Practise it

Corrected exercises: Taylor and Maclaurin series, 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

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Revision sheet: Taylor and Maclaurin series (MATH 141)

The essentials

One formula, read in both directions

The table, with the interval of each equality

The rules in table form

Where does the new series live?

The mistakes that cost marks

1. Reading a derivative off a coefficient without the factorial

2. Losing the centre

3. Evaluating a series outside its interval

4. Copying the interval of the table after a substitution

5. Applying the binomial series before factoring

6. Multiplying two series coefficient by coefficient

7. Stopping the expansion before anything survives

8. Taking a convergent Taylor series for its function

Which method to choose

Which route to the series, by the FORM of the function

How the answer is expected to be written

Writing a series built from the table

Check before you hand in

The typical problem, taken apart

A limit by series, composition included

Learn by heart

Frequently asked questions

What is the difference between a Taylor series and a Maclaurin series?

How do I find a Maclaurin series without computing derivatives?

How do I find the n-th derivative at zero from a power series?

How do I use Taylor series to compute a limit?

Corrected exercises: Taylor and Maclaurin series, MATH 141 at McGill

See also

Looking for a MATH 141 tutor in Montreal?