Integral Calculus 201-SN3-RE (201-NYB) • CEGEP, Montreal

Revision sheet: power series and Taylor series, 201-SN3-RE (201-NYB)

Q: Why do you have to give the interval of convergence?

Because a series expansion is a conditional equality. Outside the interval, the series diverges while the function remains perfectly well defined, so the equality is false. The interval is part of the answer, just like the coefficients.

Q: Why test the endpoints of the interval?

Because the Ratio Test is inconclusive when the limit equals exactly one, which is the case at both ends. Three series with the same radius can have three different intervals: open, closed or half-open. Only the direct test, substituting each endpoint, settles the question.

Q: Do you have to compute successive derivatives to get a Taylor series?

Rarely. Almost every expansion asked for is obtained by substituting into, differentiating or integrating one of the six reference series. Taylor's formula is used to establish those six series, not to rediscover them, and successive derivatives quickly become unmanageable.

Q: Does the radius of convergence change when you integrate term by term?

No, the radius is preserved. The behaviour at the endpoints, however, can change, and it often does. The series of one over one plus x squared diverges at both ends, whereas that of the arctangent, obtained by integrating it, converges there.

Q: How do you know how many terms to keep for a given precision?

You bound the remainder. For an alternating series whose terms decrease, the error is less than the first neglected term, which gives the order in one line. Otherwise you use the Lagrange remainder, bounding the next derivative on the interval concerned.

This sheet does not re-derive Taylor's formula: it is in your notes. It deals with what decides the mark on the 201-SN3-RE (formerly 201-NYB) final exam, namely the interval of convergence with both endpoints tested, and substitution into a known series, which nine times out of ten replaces a computation of successive derivatives.

It is written for CEGEP students in the Science program (Sciences de la nature). Once the sheet is read, the corrected exercise set on the same chapter puts every reflex to the test.

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The thread of the chapter

A Taylor series IS not the function: it represents it only on its INTERVAL OF CONVERGENCE. Giving an expansion without giving that interval is asserting an equality without knowing where it holds.

This chapter is part of 201-SN3-RE (201-NYB), Integral Calculus

The essentials

Radius, interval, and what must be tested

• A power series $\sum a_{n}\left(x-c\right)^{n}$ converges on an interval CENTRED at $c$ , of radius $R$ .
• $R$ is obtained with the Ratio Test: $\lim\left|\frac{a_{n+1}}{a_{n}}\right|\left|x-c\right|<1$ .
• The test gives the OPEN interval $\left(c-R,c+R\right)$ ; the two ENDPOINTS must be tested separately.
• Three cases: $R=0$ , the series converges only at $c$ ; $R=+\infty$ , it converges everywhere; otherwise a bounded interval.
• On $\left(c-R,c+R\right)$ , the series converges ABSOLUTELY; at the endpoints, anything can happen.

The partial sums of

1+x+x^{2}+\cdots

stick to

\frac{1}{1-x}

as long as

\left|x\right|<1

and pull away sharply beyond: the equality holds only on the interval.

An interval of convergence can be open, closed or half-open: the endpoint test, and it alone, tells you which.

The expansions to know by heart

• $\frac{1}{1-x}=\sum_{n=0}^{\infty}x^{n}$ for $\left|x\right|<1$ .
• $e^{x}=\sum_{n=0}^{\infty}\frac{x^{n}}{n!}$ for every $x$ .
• $\sin x=\sum_{n=0}^{\infty}\frac{(-1)^{n}x^{2n+1}}{(2n+1)!}$ and $\cos x=\sum_{n=0}^{\infty}\frac{(-1)^{n}x^{2n}}{(2n)!}$ , for every $x$ .
• $\ln(1+x)=\sum_{n=1}^{\infty}\frac{(-1)^{n+1}x^{n}}{n}$ for $-1<x\le 1$ .
• $\arctan x=\sum_{n=0}^{\infty}\frac{(-1)^{n}x^{2n+1}}{2n+1}$ for $-1\le x\le 1$ .

Near

0

1+x

then

1+x+\frac{x^{2}}{2}

stick better and better to

e^{x}

: each term

\frac{x^{n}}{n!}

adds one degree of contact.

These six expansions generate almost all the others by SUBSTITUTION, differentiation or term-by-term integration: recomputing them with Taylor's formula is a waste of time on an exam.

Taylor's formula and its remainder

• $f(x)=\sum_{n=0}^{\infty}\frac{f^{(n)}(c)}{n!}\left(x-c\right)^{n}$ , the TAYLOR series centred at $c$ .
• Centred at $c=0$ , it is called a MACLAURIN series.
• Taylor polynomial of order $N$ : the sum up to $n=N$ . It approximates $f$ near $c$ .
• Lagrange remainder: $\left|R_{N}(x)\right|\le\frac{M\left|x-c\right|^{N+1}}{(N+1)!}$ , where $M$ bounds $\left|f^{(N+1)}\right|$ between $c$ and $x$ .
• The factorial in the denominator crushes everything: that is why a few terms are enough near the centre.

x=2

, the red segment between

e^{x}

and

P_{3}

is the remainder: it is what the Lagrange bound controls.

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. Giving an expansion without its interval of convergence

1 to 2 marks, every time

What not to write

“ $\frac{1}{1-x}=1+x+x^{2}+x^{3}+\cdots$ ”

What to write

“ $\frac{1}{1-x}=\sum_{n=0}^{\infty}x^{n}$ for $\left|x\right|<1$ .”

Why: At $x=2$ , the left-hand side equals $-1$ and the right-hand side diverges: the equality is false. An expansion is a CONDITIONAL equality, and the condition is part of the answer, not an optional remark.

2. Not testing the endpoints of the interval

1 to 2 marks, and the interval is wrong half the time

What not to write

“The Ratio Test gives $\left|x\right|<1$ , so the interval is $\left(-1, 1\right)$ .”

What to write

“The test gives the open interval; I test $x=-1$ and $x=1$ SEPARATELY, which may close one end, the other, or both.”

For

\ln(1+x)

, the Ratio Test gives

\left(-1, 1\right)

; testing the endpoints closes it at

1

(alternating series) and leaves it open at

-1

(harmonic series):

\left(-1, 1\right]

Why: $\sum x^{n}$ diverges at both ends, $\sum\frac{x^{n}}{n}$ converges at $-1$ and diverges at $1$ , $\sum\frac{x^{n}}{n^{2}}$ converges at both. Same radius, three different intervals: only the endpoint test tells them apart.

3. Forgetting the factorial in the denominator

the whole question, and a wrong radius of convergence

What not to write

$e^{x}=1+x+x^{2}+x^{3}+\cdots$

What to write

$e^{x}=1+x+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+\cdots$

Without the factorials,

1+x+x^{2}+\cdots

\frac{1}{1-x}

: it blows up at

1

, whereas

e^{x}

(green) exists everywhere.

Why: Without the factorials, the series is geometric and its radius is $1$ , whereas the radius of the exponential is INFINITE. Check at $x=1$ : the true series gives $2.718$ , the wrong one diverges. The factorials are what make the exponential representable everywhere.

4. Recomputing an expansion through successive derivatives

twenty minutes of exam time, and almost always a derivative error

What not to write

“ $f(x)=e^{-x^{2}}$ : I compute $f'$ , $f''$ , $f'''$ , $f^{(4)}$ ...”

What to write

“I substitute $-x^{2}$ into the series of $e^{u}$ : $e^{-x^{2}}=\sum_{n=0}^{\infty}\frac{(-1)^{n}x^{2n}}{n!}$ , for every $x$ .”

Why: The successive derivatives of $e^{-x^{2}}$ become monstrous from the third order on, whereas substitution gives the complete expansion in one line. Taylor's formula is there to ESTABLISH the six reference series, not to rediscover them at every question.

5. Forgetting to translate the radius after a substitution

the whole radius question

What not to write

“ $\frac{1}{1-x}$ converges for $\left|x\right|<1$ , so $\frac{1}{1-3x}$ does too.”

What to write

“Substitute $u=3x$ : the condition $\left|u\right|<1$ becomes $\left|3x\right|<1$ , so $\left|x\right|<\frac{1}{3}$ and $R=\frac{1}{3}$ .”

With

u=3x

, the condition

\left|u\right|<1

gives

\left|x\right|<\frac{1}{3}

: the interval shrinks, it does not stay

\left(-1, 1\right)

Why: The condition is on the SUBSTITUTED variable, and it must be translated back into $x$ . The same reasoning gives $R=+\infty$ for $e^{-x^{2}}$ , because the series of $e^{u}$ converges for every $u$ , hence for every $x$ . Every substitution calls for this explicit translation.

6. Confusing the centre of the expansion with zero

the whole question

What not to write

“Taylor series of $\ln x$ centred at $1$ : $\ln x=x-\frac{x^{2}}{2}+\cdots$ ”

What to write

“The powers are those of $(x-1)$ : $\ln x=(x-1)-\frac{(x-1)^{2}}{2}+\frac{(x-1)^{3}}{3}-\cdots$ ”

Centred at

1

, the expansion in powers of

(x-1)

sticks to

\ln x

1

(green);

x-\frac{x^{2}}{2}

(red) sticks nowhere.

Why: A series centred at $c$ is written in powers of $(x-c)$ , never of $x$ . Immediate check: at $x=c$ , every term except the first must vanish, and the sum must equal $f(c)$ . Here $\ln 1=0$ ✓ with the right form, and $1-\frac{1}{2}+\cdots$ with the wrong one.

7. Using a Taylor polynomial far from its centre

the whole question, with an answer outside

\left[-1, 1\right]

What not to write

“ $\sin x\approx x-\frac{x^{3}}{6}$ , so $\sin 5\approx 5-\frac{125}{6}=-15.83$ .”

What to write

“The polynomial of order $3$ is reliable only near $0$ ; at $x=5$ , the error exceeds $14$ , and $\sin 5=-0.959$ .”

Each Taylor polynomial sticks to the sine on a wider interval than the previous one, then breaks away sharply: the quality of the approximation depends on the DISTANCE from the centre.

Why: A Taylor polynomial approximates the function NEAR the centre, and the error grows like $\left|x-c\right|^{N+1}$ . The check that saves you: a sine greater than $1$ in absolute value is impossible, whatever computation produced it.

8. Not bounding the remainder when a precision is asked for

half the question

What not to write

“I take four terms, that should be enough.”

What to write

“I bound $\left|R_{N}\right|\le\frac{M\left|x\right|^{N+1}}{(N+1)!}$ and solve for the $N$ that guarantees the precision.”

Why: “That should be enough” is not a justification. For an ALTERNATING series, there is something simpler still: the error is bounded by the first NEGLECTED term, which gives the required order in one line, without computing any derivative.

Which method to choose

How to get an expansion without differentiating

Always start from one of the six reference series and transform it. Taylor's formula is the last resort, not the first.

If the function is a known series with a modified argument → SUBSTITUTION, then translate the convergence condition

Example: $e^{-x^{2}}$ is obtained by setting $u=-x^{2}$ in $e^{u}$
If the function is the DERIVATIVE of a known series → differentiate term by term; the radius does not change

Example: $\frac{1}{(1-x)^{2}}$ is the derivative of $\frac{1}{1-x}$
If the function is an ANTIDERIVATIVE of a known series → integrate term by term; the radius does not change, but the endpoints may change their behaviour

Example: $\arctan x$ is obtained by integrating $\frac{1}{1+x^{2}}$
If the function is a PRODUCT or a sum of known series → multiply or add term by term, keeping the smaller of the two radii

Example: $e^{x}\sin x$ is expanded by multiplying the two series
If no known series appears → Taylor's formula, with the successive derivatives at $c$

Example: a piecewise function, or one defined by an integral

Differentiating or integrating term by term preserves the RADIUS but not always the interval: the endpoints must be tested again every time.

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.

An interval of convergence, written up for the marker

When to use it: Any question asking for the radius or the interval of convergence of a power series.

1 Apply the Ratio Test to the general term, keeping $\left|x-c\right|$ as a factor.
2 Write the condition $L\left|x-c\right|<1$ and deduce $R$ .
3 State the OPEN interval obtained.
4 Test the LEFT endpoint by substituting, which gives a numerical series whose behaviour you determine.
5 Test the RIGHT endpoint the same way, then conclude with the right brackets.

Concluding sentence

“The series converges for $-1\le x<1$ : the radius is $1$ , it converges at $x=-1$ by the Alternating Series Test and diverges at $x=1$ , where it becomes the harmonic series.”

The trap: Stopping at the open interval. The two endpoint tests are often worth one mark each, and they are independent: one can converge and the other not.

Marking: Typically 2 marks for the Ratio Test and the radius, 1 mark per endpoint tested, 1 mark for the conclusion with the right brackets.

Check before you hand in

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

Test the expansion at a simple point

Substitute a value inside the interval into the truncated series and into the function: the two must agree to the first few decimals.

$e^{x}$ at $x=1$ : four terms give $2.667$ , seven terms $2.718$ , and $e=2.718$ ✓.
Check the value at the centre

At $x=c$ , every term except the first vanishes: the sum must equal exactly $f(c)$ . This is the check that catches a misplaced centre.

The series of $\ln x$ centred at $1$ must give $0$ at $x=1$ ✓; written in powers of $x$ , it would give something else.
Check that the result lies in the range of the function

A sine outside $\left[-1, 1\right]$ , a negative exponential, the logarithm of a negative number: each signals a Taylor polynomial used too far from its centre.

$-15.83$ for $\sin 5$ is impossible: order $3$ is worthless at that distance.
Bound the error by the first neglected term

For an ALTERNATING series whose terms decrease, the truncation error is less than the first term left out. It is the fastest bound in the chapter.

$\arctan\frac{1}{2}$ truncated after $\frac{x^{5}}{5}$ has an error less than $\frac{(1/2)^{7}}{7}=0.0011$ .

The typical problem, taken apart

An expansion by substitution, then by integration

Find the Maclaurin series of $\dfrac{1}{1+x^{2}}$ and its interval of convergence, then deduce that of $\arctan x$ .

The series of

\arctan x

hugs the curve between

-1

and

1

and pulls away right after: term-by-term integration preserves the radius, not global validity.

Step 1

Start from $\frac{1}{1-u}=\sum_{n=0}^{\infty}u^{n}$ , valid for $\left|u\right|<1$ , and substitute $u=-x^{2}$ .

Why

Substitution is the first reflex: it avoids four successive differentiations, the third of which is already painful. Write carefully $u=-x^{2}$ and not $x^{2}$ : the sign is what will produce the alternation.

Step 2

$\frac{1}{1+x^{2}}=\frac{1}{1-\left(-x^{2}\right)}=\sum_{n=0}^{\infty}\left(-x^{2}\right)^{n}=\sum_{n=0}^{\infty}(-1)^{n}x^{2n}=1-x^{2}+x^{4}-x^{6}+\cdots$

Why

$\left(-x^{2}\right)^{n}$ splits into $(-1)^{n}x^{2n}$ : that is where the alternating signs and the even powers come from. Writing the first four terms lets you check the alternation at a glance.

Step 3

Condition: $\left|u\right|<1$ becomes $\left|-x^{2}\right|<1$ , that is $x^{2}<1$ , that is $\left|x\right|<1$ . The radius is $R=1$ . At the endpoints $x=\pm 1$ , the general term equals $(-1)^{n}$ , which does not tend to $0$ : the series DIVERGES at both ends.

Why

Translating the condition is compulsory, and the endpoints are tested here with the Divergence Test, the fastest of the five. The interval is therefore $\left(-1, 1\right)$ , open on both sides.

Step 4

Since $\arctan x=\int_{0}^{x}\frac{dt}{1+t^{2}}$ , integrate term by term: $\arctan x=\sum_{n=0}^{\infty}\frac{(-1)^{n}x^{2n+1}}{2n+1}=x-\frac{x^{3}}{3}+\frac{x^{5}}{5}-\cdots$

Why

Term-by-term integration is legitimate inside the radius of convergence, and the constant of integration is zero because $\arctan 0=0$ . Checking this value at the centre is the fastest check of the computation.

Step 5

The radius stays $R=1$ , but the endpoints CHANGE their behaviour: at $x=1$ , the series becomes $1-\frac{1}{3}+\frac{1}{5}-\cdots$ , alternating with decreasing terms tending to $0$ , hence CONVERGENT; likewise at $x=-1$ . The interval is $\left[-1, 1\right]$ , CLOSED.

Why

This is the point that sets a good paper apart: integration divided each term by $2n+1$ , which is enough to make the endpoints converge where they diverged. The radius is preserved, the interval is not.

Step 6

Numerical check: at $x=1$ , the series gives $\frac{\pi}{4}=0.7854$ ; three terms give $0.8667$ , ten terms $0.7605$ , a hundred terms $0.7829$ : the convergence is slow but real ✓.

Why

The slowness is the sign of a conditionally convergent series: it converges, but not absolutely, exactly like the alternating series of reciprocals. The check confirms the closed interval.

The conclusion, written out

$\frac{1}{1+x^{2}}=\sum_{n=0}^{\infty}(-1)^{n}x^{2n}$ on $\left(-1, 1\right)$ , and by term-by-term integration $\arctan x=\sum_{n=0}^{\infty}\frac{(-1)^{n}x^{2n+1}}{2n+1}$ on $\left[-1, 1\right]$ , a CLOSED interval this time.

The classic mistake on this problem: Copying the interval of the first series for the second. Term-by-term integration preserves the radius, never automatically the endpoints: they must be tested again, and here both of them change their behaviour.

Learn by heart

• A power series converges on an interval CENTRED at $c$ , of radius $R$ .
• The Ratio Test gives the OPEN interval: the two endpoints are tested separately.
• Six reference series: $\frac{1}{1-x}$ , $e^{x}$ , $\sin x$ , $\cos x$ , $\ln(1+x)$ and $\arctan x$ .
• Substituting into, differentiating or integrating a known series is almost always faster than Taylor's formula.
• After a substitution, TRANSLATE the convergence condition back into the original variable.
• Series centred at $c$ : powers of $(x-c)$ , never of $x$ .
• Differentiating or integrating term by term preserves the radius, not the behaviour at the endpoints.
• Alternating series: the truncation error is bounded by the first NEGLECTED term.

Frequently asked questions

Why do you have to give the interval of convergence?

Because a series expansion is a conditional equality. Outside the interval, the series diverges while the function remains perfectly well defined, so the equality is false. The interval is part of the answer, just like the coefficients.

Why test the endpoints of the interval?

Because the Ratio Test is inconclusive when the limit equals exactly one, which is the case at both ends. Three series with the same radius can have three different intervals: open, closed or half-open. Only the direct test, substituting each endpoint, settles the question.

Do you have to compute successive derivatives to get a Taylor series?

Rarely. Almost every expansion asked for is obtained by substituting into, differentiating or integrating one of the six reference series. Taylor's formula is used to establish those six series, not to rediscover them, and successive derivatives quickly become unmanageable.

Does the radius of convergence change when you integrate term by term?

No, the radius is preserved. The behaviour at the endpoints, however, can change, and it often does. The series of one over one plus x squared diverges at both ends, whereas that of the arctangent, obtained by integrating it, converges there.

How do you know how many terms to keep for a given precision?

You bound the remainder. For an alternating series whose terms decrease, the error is less than the first neglected term, which gives the order in one line. Otherwise you use the Lagrange remainder, bounding the next derivative on the interval concerned.

Practise it

Corrected exercises: Power series and Taylor series, 201-SN3-RE (201-NYB)

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.

15 corrected exercises
150 points
225 minutes

Do the exercises

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Revision sheet: power series and Taylor series, 201-SN3-RE (201-NYB)

The essentials

Radius, interval, and what must be tested

The expansions to know by heart

Taylor's formula and its remainder

The mistakes that cost marks

1. Giving an expansion without its interval of convergence

2. Not testing the endpoints of the interval

3. Forgetting the factorial in the denominator

4. Recomputing an expansion through successive derivatives

5. Forgetting to translate the radius after a substitution

6. Confusing the centre of the expansion with zero

7. Using a Taylor polynomial far from its centre

8. Not bounding the remainder when a precision is asked for

Which method to choose

How to get an expansion without differentiating

How the answer is expected to be written

An interval of convergence, written up for the marker

Check before you hand in

The typical problem, taken apart

An expansion by substitution, then by integration

Learn by heart

Frequently asked questions

Why do you have to give the interval of convergence?

Why test the endpoints of the interval?

Do you have to compute successive derivatives to get a Taylor series?

Does the radius of convergence change when you integrate term by term?

How do you know how many terms to keep for a given precision?

Corrected exercises: Power series and Taylor series, 201-SN3-RE (201-NYB)

See also

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