MA225 - Power Series

Pedro Vilanova

February 19, 2025

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 Contents

• Introduction
• Power series
• Convergence of a power series
• Interval and radius of convergence
• Representing functions as power series
• Combining power series
• Multiplication of power series
• Differentiating and integrating power series
• Taylor/Maclaurin series

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 Introduction

A power series is a type of series with terms involving a variable. If the

variable is x, then all the terms of the series involve powers of x.
In other words, a power series is a function of x. A power series can be
thought of as an infinite polynomial.
Power series are used to represent common functions and also to define
new functions. In this section we define power series and show how to
determine when a power series converges and when it diverges. We also
show how to represent certain functions using power series.

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 Power series

A series of the form

∞
X
cn (x − a)n = c0 + c1 (x − a) + c2 (x − a)2 + . . .
n=0

is a power series centered at x = a. A particular case is a power series of

the form
X∞
cn x n = c0 + c1 x + c2 x 2 + . . .
n=0

which is centered at x = 0.
Recall x 0 = 1 and (x − a)0 = 1.

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 Power series

The series
∞
X xn x2 x3
=1+x + + + ... do you recognize this series?
n=0
n! 2! 3!

and
∞
X
n!x n = 1 + x + 2!x 2 + 3!x 3 + . . .
n=0

are both power series centered at x = 0. The series

∞
X (x − 2)n x − 2 (x − 2)2 (x − 2)3
= 1 + + + + ...
n=0
(n + 1)3n 2·3 3 · 32 4 · 33

is a power series centered at x = 2.

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 Convergence of power series

Important note: Since the terms in a power series involve a variable x,

the series may converge for certain values ofP x and diverge for other
∞
values of x. Example: The geometric series n=0 x n converges for all x
in the interval (−1, 1), but diverges for all x outside that interval.

For a power series centered at x = a, the value of the series at x = a is

given by c0 . Therefore, a power series always converges at its center.

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 Convergence of power series

Proposition
P∞
Consider the power series n=0 cn (x − a)n . The series satisfies exactly
one of the following properties:
1 The series converges at x = a and diverges for all x 6= a.
2 The series converges for all real numbers x.
3 There exists a real number R > 0 such that the series converges if
|x − a| < R and diverges if |x − a| > R. At the values x where
|x − a| = R, the series may converge or diverge.

Note (important): To determine the interval of convergence for a

power series, we typically apply the ratio test.

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 Convergence of power series
Converges
Diverges Diverges

a x
Converges

a x
Diverges Converges Diverges
? ?

a x
a−R a+R
P∞
Let n=0 cn (x − a)n :
• Top: interval of convergence is {a}, the radius is R = 0,
• Middle: interval of convergence is R, the radius is R = ∞,
• Bottom: interval of convergence may be (a−R, a+R), or [a−R, a+R) or
(a−R, a+R], [a−R, a+R], the radius is R.
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 Interval and radius of convergence

Definition (Interval of convergence)

P∞
The set of values x for which the series n=0 cn (x − a)n converges is
known as the interval of convergence.
P∞
Example: The series n=0 x n (geometric series) converges for all values
x in the interval (−1, 1) and diverges for all values x such that |x| ≥ 1,
the interval of convergence of this series is (−1, 1). Since the length of
the interval is 2, the radius of convergence is 1 (see next slide).

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 Interval and radius of convergence

Definition (Radius of convergence)

1 If the series converges only at x = a, then the radius of convergence
is R = 0,
2 If the series converges for all real numbers x, we say the radius of
convergence is R = ∞.
3 If there exists a real number R > 0 such that the series converges
for |x − a| < R and diverges for |x − a| > R, then R is the radius of
convergence.

Observation: In the context of power series, in this course convergence

of a power series on a given interval means absolute convergence except
possibly at the endpoints.

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 Convergence of power series
Example: For each of the following series, find the interval and radius of
convergence. Recall interval is always at least the center point.
P∞ x n
1 n=0 n! . Hint: x ∈ R, including x = 0.
P∞ n
2 n=0 n!x .
P∞ (x−2)n
3 n=0 (n+1)3n .

Exercise: Find the interval and radius of convergence for the series
∞
X xn
√ .
n=1
n

Hint: Apply the ratio test to check for absolute convergence.

Answer: The interval of convergence is [−1, 1). The radius of
convergence is R = 1.

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 Summary

Series Interval of convergence Radius

∞
xn
P
(−1, 1) R=1
n=0
∞
xn
P
n! R R=∞
n=0
∞
n!x n
P
{0} R=0
n=0
∞
P (x−3)n
n [2, 4) R=1
n=1

Table: Examples of series with their radii and intervals of convergence.

Exercise: Check last one.

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 Summary

What is more interesting is that it can be shown (later) that

∞
• 1
xn =
P
1−x , for |x| < 1;
n=0
∞
• xn
= e x , for x ∈ R;
P
n!
n=0
∞
(x−3)n
•
P
n = − ln(4 − x), for x ∈ [2, 4)
n=1

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 Representing functions as power series

Being able to represent a function by an “infinite polynomial” is a

powerful tool.
Why? Polynomial functions are “easy” functions to work with. If we can
represent a complicated function by an infinite polynomial, we can use
the polynomial representation to differentiate or integrate it.
A polynomial in a single variable x can be written

a0 x 0 + a1 x 1 + a2 x 2 + ... + an−1 x n−1 + an x n ,

where a0 , a1 , .., an ∈ R.
The degree of a polynomial is the largest degree of any term with
nonzero coefficient.

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 Representing functions as power series

The question is, when can we represent a function by a power series?

1
Consider
∞
X
1 + x + x2 + x3 + . . . = xn .
n=0
1
If |x| < 1, converges to 1−x and we write

1
1 + x + x2 + x3 + . . . = , for |x| < 1.
1−x

1
Recall the geometric series
a
a + ar + ar 2 + ar 3 + . . . = , for |r | < 1 .
1−r

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 Representing functions as power series

1
As a result, we are able to represent the function f (x) = 1−x by the
power series
1 + x + x 2 + x 3 + . . . when |x| < 1.

The immediate conclusion is that the partial sum approximates the

function f .

1
Example 1: Sketch the graph of f (x) = 1−x and the graphs of the
PN n
corresponding partial sums SN (x) = n=0 x on the interval (−1, 1).
https://www.wolframcloud.com/env/pguerra0/PowerSeries.nb

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 Representing functions as power series

1
Example 2: Sketch a graph of f (x) = 1−x 2 and the corresponding
PN
partial sums SN (x) = n=0 x 2n on the interval (−1, 1).
2(N+1)
Hint: SN (x) = 1 + x 2 + . . . + x 2N = 1−x
1−x 2 .

Examples: Use a power series (geometric) to represent each of the

following functions. Find the interval of convergence.
1
1 f (x) = 1+x 3

x2
2 f (x) = 4−x 2

Remark: To find the interval of convergence, we need to solve for x. For

example, | − x 3 | < 1 is an incomplete derivation of an interval of
convergence.

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 Representing functions as power series
Solution (1) This is again the sum of a geometric series:

1 1
3
= .
1+x 1 − (−x 3 )

So the ratio is −x 3 and if −x 3 < 1 we have

1 1
=
1 + x3 1 − (−x 3 )
∞
X
n
= −x 3
n=0
= 1 − x3 + x6 − x9 + . . . .

Interval of convergence? We will show that | − x 3 | < 1 iff |x| < 1.

Observe | − x 3 | = |x 3 | thus |x 3 | < 1.

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 Representing functions as power series

Recall that |x 3 | = |x|3 because the absolute value of a product is the

product of the absolute values. Hence,

|x|3 < 1 .

To remove the cube, consider that |x| ≥ 0 for any x, and thus we can
take cubic root in both sides of |x|3 < 1 obtaining the result since cube
function is strictly increasing.
The interval of convergence is thus (−1, 1), and we have

1
= 1 − x 3 + x 6 − x 9 + . . . for |x| < 1.
1 + x3

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 Representing functions as power series
Solution (2): A little algebraic manipulation can relate f to a geometric
series.
By factoring 4 out of the two terms in the denominator, we get

x2 x2
2
=  
4−x 4 1− x2
4

x2
=  
x 2


4 1− 2
x2
4
=
2
1 − x2
∞
X x 2  x 2n
= .
n=0
4 2

x 2


The series converges as long as 2 < 1.

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 Representing functions as power series

Solving this inequality gives the interval of convergence (−2, 2) and thus
∞
x2 X x 2n+2
2
=
4−x n=0
4n+1
x2 x4 x6
= + 2 + 3 + ...
4 4 4
for |x| < 2.

x3
Exercise: Represent the function f (x) = 2−x using a power series and
find the interval of convergence.
a(z)
Hint: Rewrite f in the form f (z) = 1−r (z) for some functions a and r and
considering the center of the series.
P∞ n+3
Solution: n=0 x2n+1 with interval of convergence (−2, 2).

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 Working with power series
We now discuss how power series can be combined, differentiated, or
integrated to create new power series.

1 Allows to find power series representations for certain elementary

functions, by writing those functions in terms of functions with
known power series.
1
For example, given the power series representation for f (x) = 1−x ,
we can find a power series representation for f 0 (x) = (1−x)
1
2.

2 Allows to define new functions that cannot be written in terms of

elementary functions.

For example, this is particularly useful for solving differential

equations for which there is no solution in terms of elementary
functions.

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 Combining power series
If we have two power series with the same interval of convergence we can
create a new power series:
1 We can add or subtract the two series to create a new power series,
also with the same interval of convergence,
2 We can multiply a power series by a power of x,
3 We can evaluate a power series at x m for a positive integer m.

For example, since we know the power series representation for

1
f (x) = 1−x , we can find power series representations for related
functions, such as
3x
y=
1 − x2
and
1
y= .
(x − 1)(x − 3)

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 Combining power series

Proposition (Combining power series)

P∞ P∞
Suppose that the two power series n=0 cn x n and n=0 dn x n converge
to the functions f and g , respectively, on a common interval I .
P∞ n n
1 The power series n=0 (cn x ± dn x ) converges to f ± g on I .

2 For any integer m ≥ 0 and any real number b, the power series
P ∞ m n m
n=0 bx cn x converges to bx f (x) on I .

3 For any integer m ≥ 0 and any real number b, the series

P ∞ m n m m
n=0 cn (bx ) converges to f (bx ) for all x such that bx is on I.

Remark: For simplicity, the property is stated for series centered at

x = 0. Similar results hold for power series centered at x = a.

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 Combining power series

P∞
Example: Suppose that n=0 an x n is a power
P∞ series whose interval of
convergence is (−1, 1), and suppose that n=0 bn x n is a power series
whose interval of convergence is (−2, 2).
P∞ n n
1 Find the interval of convergence of the series n=0 (an x + bn x ).
P∞ n n
2 Find the interval of convergence of the series n=0 an 3 x .

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 Combining power series

Solution (1): Since the interval

P∞ (−1, 1) isPa∞common interval of
convergence of the series Pn=0 an x n and n=0 bn x n , the interval of
∞
convergence of the series n=0 (an x n + bn x n ) is (−1, 1).
P∞
Solution (2): Since n=0 an x n is a power series centered at zero with
radius of convergence 1, it converges for all x in the interval (−1, 1). By
(3), the series
X∞ ∞
X
an 3n x n = an (3x)n
n=0 n=0

converges if 3x is in the interval (−1, 1). Therefore, the series converges

for all x in the interval − 31 , 13 .

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 Constructing power series from known
ones

1
Example: Use the power series representation for f (x) = 1−h(x) to
construct a power series for each of the following functions. Find the
interval of convergence of the power series.
3x
1 f (x) = 1+x 2 .

1
2 f (x) = (x−1)(x−3) .

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 Constructing power series from known
ones
Solution (1): First write f (x) as
 
1
f (x) = 3x .
1 − (−x 2 )
1
Using the power series representation f (x) = 1−(−x 2 ) and (3), we find

that a (convergent) power series representation for f is given by

∞ ∞
X
n X
3x −x 2 = 3(−1)n x 2n+1 .
n=0 n=0

1
Since the interval of convergence of the series for 1−h(x) is (−1, 1), the
interval of convergence for this new series is the set of real numbers x
such that −x 2 = x 2 < 1 iff |x| < 1. Therefore, the interval of
convergence is (−1, 1).

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 Constructing power series from known
ones
Solution (2): To find the power series representation, use partial
1
fractions to write f (x) = (x−1)(x−3) as the sum of two fractions. We have

1 −1/2 1/2 1/2 1/2 1/2 1/6

= + = − = − .
(x − 1)(x − 3) x −1 x −3 1−x 3−x 1−x 1 − x3

First, using (2), we obtain

∞
1/2 X1
= xn for |x| < 1 .
1−x n=0
2

Then, using (2) and (3), we have

∞
X 1  x n
1/6
= for |x| < 3 .
1 − x/3 n=0
6 3

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 Constructing power series from known
ones

Since we are combining these two power series, the interval of

convergence of the difference must be the smaller of these two intervals.
Using this fact and (1), we have
∞  
1 X 1 1
= − xn
(x − 1)(x − 3) n=0
2 6 · 3n

where the interval of convergence is (−1, 1).

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 Finding the function represented by a
given power series

Up to now, given a function we found the power series. We can also do

the opposite: given a power series, determine which function it represents.
P∞
Example: Consider n=1 2n x n . Find the function f represented by this
series.
Notice that we can rewrite the series as
∞
X
(2x)n ,
n=1

which is “almost” a geometric series with common ratio r = 2x.

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 Finding the function represented by a
given power series
Solution: Recall that a geometric series starting at n = 0 is given by
∞
X 1
rn = , |r | < 1.
n=0
1−r

Since our series starts at n = 1, we have

∞ ∞
!
X
n
X
n 1
r = r −1= − 1.
n=1 n=0
1−r

Substituting r = 2x, we obtain

∞
X 1
(2x)n = − 1.
n=1
1 − 2x

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 Finding the function represented by a
given power series

Now simplify the right-hand side:

1 1 − (1 − 2x) 2x
−1= = .
1 − 2x 1 − 2x 1 − 2x
Thus, the function represented by the series is

2x 1
f (x) = , |2x| < 1 iff |x| < .
1 − 2x 2

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 Multiplication of power series
We can also create new power series by multiplying power series. Being
able to multiply two power series provides another way of finding power
series representations for functions. The way we multiply them is similar
to how we multiply polynomials. For example, suppose we want to
multiply
X∞
cn x n = c0 + c1 x + c2 x 2 + . . .
n=0

and
∞
X
dn x n = d0 + d1 x + d2 x 2 + . . .
n=0

We have
∞ ∞
! !
X X
n n
= c0 + c1 x + c2 x 2 + . . . d0 + d1 x + d2 x 2 + . . .



cn x dn x
n=0 n=0
= c0 d0 + (c1 d0 + c0 d1 ) x + (c2 d0 + c1 d1 + c0 d2 ) x 2 + . . .

Note the sum of the indices are 0,1,2..

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 Multiplication of power series

Proposition (Multiplying power series)

P∞ P∞
Suppose that the power series n=0 cn x n and n=0 dn x n converge to f
and g , respectively, on a common interval I . Let
n
X
en = c0 dn + c1 dn−1 + c2 dn−2 + . . . + cn−1 d1 + cn d0 = ck dn−k .
k=0

Then
∞ ∞ ∞
! !
X X X
n n
cn x dn x = en x n
n=0 n=0 n=0

and
∞
X
en x n converges to f (x)g (x) on I .
n=0
P∞ n
The series n=0 en x is known as the Cauchy product.

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 Example multiplication
Note: Recall original algebraic rules for series lacked multiplication case.
Multipliying series (that are not necessarily power series) requires
absolute convergence.

Example: Find the power series representation for

1
f (x) = .
(1 − x) (1 − x 2 )

Multiply
∞
1 X
= xn = 1 + x + x2 + x3 + . . . for |x| < 1 ,
1−x n=0

with
∞
1 X
n
2
= x2 = 1 + x2 + x4 + x6 + . . . for |x| < 1 ,
1−x n=0

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 Example multiplication
1
to construct a power series for f (x) = (1−x)(1−x 2 ) on the interval (−1, 1).
That is

1 + x + x2 + x3 + . . . 1 + x2 + x4 + x6 + . . . .

Writing out the first several terms, we see that the product is given by

1 + x2 + x4 + x6 + . . .

+ x + x3 + x5 + x7 + . . .

+ x2 + x4 + x6 + x8 + . . .

+ x3 + x5 + x7 + x9 + . . . + . . .

= 1 + x + (1 + 1)x 2 + (1 + 1)x 3 + (1 + 1 + 1)x 4 + (1 + 1 + 1)x 5 + . . .

∞  
2 3 4 5
X n+2 n
= 1 + x + 2x + 2x + 3x + 3x + . . . = x
n=0
2
1 1
Since the series for y = 1−x and y = 1−x 2 both converge on the interval

(−1, 1), the series for the product also converges on the interval (−1, 1).
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 Example multiplication

Another way is to construct a “folding” table. First, consider the

coefficients
x0 x1 x2 x3 x4 x5 ···
1
1−x 1 1 1 1 1 1 ···
1
1−x 2 1 0 1 0 1 0 ···

In the next slide we show the table. To find the coefficient of x n , you add
up all the entries in the table that lie along the diagonal corresponding to
i + j = n.

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 Example multiplication

× 1 (x 0 ) 0 (x 1 ) 1 (x 2 ) 0 (x 3 ) 1 (x 4 ) 0 (x 5 )

1 (x 0 ) 1 0 1 0 1 0

1 (x 1 ) 1 0 1 0 1 0

1 (x 2 ) 1 0 1 0 1 0

1 (x 3 ) 1 0 1 0 1 0

.. .. .. .. .. .. ..
. . . . . . .

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 Example multiplication
Example 2: Consider the product of
∞
1 X
f (x) = = xn = 1 + x + x2 + x3 + · · · , |x| < 1,
1−x n=0

and
∞
1 X
g (x) = 2
= (n+1)x n = 1+2x+3x 2 +4x 3 +5x 4 +· · · , |x| < 1.
(1 − x) n=0

The product is
∞
1 X
f (x)g (x) = = cn x n .
(1 − x)3 n=0

It is known that (we will see later how to get this)

(n + 1)(n + 2)
cn = .
2
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 Example multiplication
Let’s see how this comes out via the Cauchy product using a table.

x0 x1 x2 x3 x4 x5 ···
1
f (x) = 1−x 1 1 1 1 1 1 ···
1
g (x) = (1−x) 2 1 2 3 4 5 6 ···

Now, we set up a table for the product where the rows come from f (x)
and the columns from g (x). Each entry is the product of the coefficient
from f (x) (on its row) with that from g (x) (on its column). Since every
coefficient in f (x) is 1, each row will be identical to the sequence from
g (x).

× 1 (x 0 ) 2 (x 1 ) 3 (x 2 ) 4 (x 3 ) 5 (x 4 ) 6 (x 5 )
0
1 (x ) 1 2 3 4 5 6
1 (x 1 ) 1 2 3 4 5 6
1 (x 2 ) 1 2 3 4 5 6
1 (x 3 ) 1 2 3 4 5 6
.. .. .. .. .. .. ..
. . . . . . .
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 Example multiplication

Collecting the diagonal sums, we get:

c0 = 1,
c1 = 3,
c2 = 6,
c3 = 10,
c4 = 15,
c5 = 21, etc.

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 Example multiplication
Example 3:
1 X
f (x) = = 2n x n = 1 + 2x + 4x 2 + 8x 3 + 16x 4 + · · · ,
1 − 2x
n≥0

1 X
g (x) = = 3n x n = 1 + 3x + 9x 2 + 27x 3 + 81x 4 + · · · ,
1 − 3x
n≥0

1 1
for |x| < and |x| <
2 3 respectively.
Their product is
1
f (x)g (x) = ,
(1 − 2x)(1 − 3x)
and its coefficient for x n is given by the Cauchy product
n n  k
X X 2
cn = 2k 3n−k = 3n .
3
k=0 k=0

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 Example multiplication
It is the partial sum of a geometric series with r = 2/3 and a = 1. Recall
the formula for a geometric series:
n
X 1 − r n+1
ri = .
1−r
i=0

Substitute r = 23 :

n  i
n+1 2 n+1  n+1 !

1 − 23


X 2 1− 3 2
= 2 = 1 =3 1− .
3 1− 3 3
3
i=0

Now plug this back into the expression for cn :

 n+1 !  n+1 !
n 2 n+1 2
cn = 3 · 3 1 − =3 1− .
3 3

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 Example multiplication
Simplify by noting that:
 n+1
2
3n+1 = 2n+1 .
3

Thus,
cn = 3n+1 − 2n+1 .
We now construct the “folding” table. We first list a few coefficients of
each series:
x0 x1 x2 x3 x4 x5
1
f (x) = 1−2x 1 2 4 8 16 32
1
g (x) = 1−3x 1 3 9 27 81 243

Now form the multiplication table where the entry in row i and column j
is
(row i coefficient) × (column j coefficient).

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 Example multiplication

1 (x 0 ) 3 (x 1 ) 9 (x 2 ) 27 (x 3 ) 81 (x 4 ) 243 (x 5 )
0
1 (x ) 1 3 9 27 81 243
2 (x 1 ) 2 6 18 54 162 486
4 (x 2 ) 4 12 36 108 324 972
8 (x 3 ) 8 24 72 216 648 1944
16 (x 4 ) 16 48 144 432 1296 3888
32 (x 5 ) 32 96 288 864 2592 7776
In other words,
Diagonal i +j = 0: 1
Diagonal i +j = 1: 3+2=5
Diagonal i +j = 2: 9 + 6 + 4 = 19
Diagonal i +j = 3: 27 + 18 + 12 + 8 = 65
Diagonal i +j = 4: 81 + 54 + 36 + 24 + 16 = 211
Diagonal i +j = 5: 243 + 162 + 108 + 72 + 48 + 32 = 665

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 Differentiating and Integrating power
series

Consider a power series

∞
X
cn x n = c0 + c1 x + c2 x 2 + . . .
n=0

that converges on some interval I , and let f be the function defined by

this series.
Here we address two questions about f .
• Is f differentiable, and if so, how do we determine the derivative f 0 ?
• How do we evaluate the indefinite integral f (x)dx?
R

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 Differentiating and Integrating power
series
For convergent power series we can differentiate and integrate
term-by-term like in a finite polynomial. That is, if
∞
X
f (x) = cn x n = c0 + c1 x + c2 x 2 + . . .
n=0

converges on some interval I , then

f 0 (x) = c1 + 2c2 x + 3c3 x 2 + . . .

and
x2 x3
Z
f (x)dx = C + c0 x + c1 + c2 + . . .
2 3
converge on the same interval I .
Note: This is not an obvious result. This
P∞ says differentiation and
integration operators commutes with i=1 .

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 Differentiating and Integrating power
series

The ability to differentiate and integrate power series term-by-term also

allows us to use known power series representations to find power series
representations for other functions.
1
For example, given the power series for f (x) = 1−x , we can differentiate
0 1
term-by-term to find the power series for f (x) = (1−x) 2 . Similarly, using
1
the power series for g (x) = 1+x , we can integrate term-by-term to find
the power series for G (x) = ln(1 + x), an antiderivative of g .

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 Differentiating and Integrating power
series
Proposition (Term-by-term differentiation)
P∞
Suppose that the power series n=0 cn (x − a)n converges on the interval
(a − R, a + R) for some R > 0. Let f be the function defined by the series
∞
X
f (x) = cn (x − a)n = c0 + c1 (x − a) + c2 (x − a)2 + c3 (x − a)3 + . . .
n=0

for |x − a| < R. Then f is differentiable on the interval (a − R, a + R)

and we can find f 0 by differentiating the series term-by-term:
∞
X
f 0 (x) = ncn (x − a)n−1 = c1 + 2c2 (x − a) + 3c3 (x − a)2 + . . .
n=1

for |x − a| < R.

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 Differentiating and Integrating power
series
Proposition (Term-by-term integration)
RLet f be a function as defined in the previous proposition. To find
f (x)dx, we can integrate the series term-by-term. The resulting series
converges on (a − R, a + R), and we have
∞
(x − a)n+1 (x − a)2 (x − a)3
Z X
f (x)dx = C + cn = C +c0 (x−a)+c1 +c2 +. . .
n=0
n+1 2 3

for |x − a| < R.
Here the integration constant C must be determined.

Important note: The previous properties say nothing about behaviour

at the endpoints. It is possible that the differentiated and integrated
power series have different behavior at the endpoints than does the
original series.

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 Differentiating power series

Example: Use the power series representation

∞
1 X
f (x) = = xn = 1 + x + x2 + x3 + . . .
1−x n=0

for |x| < 1 to find a power series representation for

1
g (x) =
(1 − x)2

on the interval (−1, 1).

Note: In this case, the question asks about the convergence on an open
interval so it is not required to investigate the behaviour at the endpoints.
1 1
Hint: g (x) = (1−x)2 is the derivative of f (x) = 1−x .

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 Differentiating power series
Solution: We find a power series representation for g by differentiating
the power series for f term-by-term. The result is
  ∞
1 d 1 X d
g (x) = = = (x n )
(1 − x)2 dx 1−x n=0
dx
d
1 + x + x2 + x3 + . . .


=
dx
= 0 + 1 + 2x + 3x 2 + 4x 3 + . . . (write the series in terms of x n )
X∞
= (n + 1)x n ,
n=0

for |x| < 1.

Observation: By using the divergence test, we find that the series
diverges at both endpoints x = ±1.

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 Differentiating power series
Another way to solve this problem is to write
1 1
2
= ,
(1 − x) 1 − (2x − x 2 )

so that a(x) = 1 and h(x) = 2x − x 2 . Provided that |2x − x 2 | < 1, we

have
∞
1 X
n
= 2x − x 2 .
1 − (2x − x 2 ) n=0

Now, while this expression looks quite different at first glance, if we

expand (2x − x 2 )n using the binomial theorem and then collect like
powers of x, one recovers exactly the same power series as before:
∞
1 X
= (n + 1)x n .
(1 − x)2 n=0

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 Differentiating power series

Exercise: P
Use the result of the previous example to evaluate the sum of
∞
the series n=0 n+1
4n .

Note from previous example we know that

∞
X 1
(n + 1)x n = .
n=0
(1 − x)2

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 Differentiating power series

Solution: We have
∞ ∞  n
X n+1 X 1
= (n + 1)
n=0
4n n=0
4
1 16
= = .
1 2 9


1− 4

So looking back to geometric series sum formula, this exercise shows that
we can apply a similar formula to a non-geometric series.
Gives a conceptual insight: It shows how interconnected different series
are and how a small transformation (like differentiation) can reveal new
relationships.

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 Differentiating power series

1
P∞ n
Exercise: Differentiate the series (1−x) 2 = n=0 (n + 1)x term-by-term
2
to find a power series representation for (1−x)3 on the interval (−1, 1).

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 Differentiating power series

Solution:
d 1 0 − 2(1 − x) · (−1) 2
= =
dx (1 − x)2 (1 − x)4 (1 − x)3
X∞
= (n + 1)nx n−1 (write the series in terms of x n )
n=0
∞
X
= (n + 1)nx n−1
n=1
= 2x + 6x 1 + 12x 2 + ...
0

X∞
= (n + 2)(n + 1)x n (written in terms of x n and starting from n=0 ∗ )
n=0

∗ whenever possible

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 Integrating power series

Example: Find a power series representation for f (x) = ln(1 + x) by

integrating the power series for f 0 and find its interval of convergence.

Observe that, for f (x) = ln(1 + x), the derivative is f 0 (x) = 1

1+x . We
know that
∞
1 1 X
= = (−x)n = 1 − x + x 2 − x 3 + . . .
1+x 1 − (−x) n=0

for |x| < 1.

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 Integrating power series

Solution: To find a power series for f (x) = ln(1 + x), we integrate the
series term-by-term.

x2 x3 x4
Z Z
f 0 (x)dx = 1 − x + x 2 − x 3 + . . . dx = C + x − + − + . . .


2 3 4

It remains to solve for the constant C . Select a value of x within the

interval of convergence where the function’s value is known and the
series simplifies. Common choices: center of the series or x = 0.

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 Integrating power series
We have
x2 x3 x4
C +x − + − + . . . = ln (1 + x) .
2 3 4
At x = 0 we have

C + 0 + 0... = ln (1 + 0) = 0 .

Therefore, a power series representation for f (x) = ln(1 + x) is

∞
x2 x3 x4 X xn
ln(1 + x) = x − + − + ... = (−1)n+1 for |x| < 1.
2 3 4 n=1
n

Checking the convergence at endpoints: x = 1 the series is the

alternating harmonic series, which converges. Also, at x = −1, the series
is the harmonic series, which diverges. So the actual interval of
convergence is (−1, 1].
https://www.wolframcloud.com/env/pguerra0/PowerSeries.nb

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 Integrating power series

n+1 x n
P∞
Exercise: Integrate the power
R series ln(1 + x) = n=1 (−1) n
term-by-term to evaluate ln(1 + x)dx.

You
R can still evaluate the integral
ln (1 + x)dx = (1 + x) ln (1 + x) − x + C . Steps: use substitution
u = 1 + x and then integration by parts. But the idea is that integrating
the power series is easier.
x n+1 xn
Hint: Use the fact that (n+1)n is an antiderivative of n .

Answer:
∞
X (−1)n+1 x n+1
.
n=2
(n + 1)n

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 Remark: Uniqueness of power series
Up to this point, we have shown several techniques for finding power
series representations for functions. However, how do we know that these
power series are unique? That is, given a function f and a power series
for f at a, is it possible that there is a different power series for f at a
that we could have found if we had used a different technique? The
answer to this question is no.

Proposition (Uniqueness of power series)

P∞ P∞
Let n=0 cn (x − a)n and n=0 dn (x − a)n be two convergent power
series such that
∞
X ∞
X
f (x) = cn (x − a)n = dn (x − a)n
n=0 n=0

for all x in an open interval containing a. Then cn = dn for all n ≥ 0

This is analogous to the uniqueness of coefficients in a polynomial: if two

polynomials agree on an interval (or on infinitely many points), then their
corresponding coefficients must be equal. The infinite series, obeys the
same uniqueness principle. 63 / 87
 Power series and differential equations

Example 1: Show that the function

∞
X (−1)n x 2n
f (x) =
n=0
(2n)!

is a solution of the differential equation

f 00 (x) + f (x) = 0 .

https://en.wikipedia.org/wiki/Harmonic_oscillator#Simple_harmonic_oscillator

https://en.wikipedia.org/wiki/Harmonic_oscillator#Damped_harmonic_oscillator

P∞ (−1)n x 2n
Observation: We will see later that f (x) = n=0 (2n)! = cos(x)

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 Power series and differential equations
Solution 1: We start with
∞
X (−1)n x 2n
f (x) = .
n=0
(2n)!

Differentiating term by term gives:

∞
X (−1)n · 2n x 2n−1
f 0 (x) = ,
n=1
(2n)!

and differentiating once more,

∞
X (−1)n · 2n(2n − 1) x 2n−2
f 00 (x) = .
n=1
(2n)!

Notice that
(2n)! = 2n(2n − 1)(2n − 2)!,

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 Power series and differential equations
so
2n(2n − 1) 1
= .
(2n)! (2n − 2)!
Thus,
∞
X (−1)n x 2n−2
f 00 (x) = .
n=1
(2n − 2)!

Let m = n − 1 (so x 2n−2 = x 2m and (2n − 2)! = (2m)!); then

∞ ∞
X (−1)m+1 x 2m X (−1)m x 2m
f 00 (x) = =− = −f (x).
m=0
(2m)! m=0
(2m)!

Thus,
f 00 (x) + f (x) = 0.
This shows that f (x) is a solution of the differential equation.

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 Power series and differential equations

Example 2: The Bessel function of order 0 is defined by

∞
X (−1)n x 2n
J0 (x) = .
n=0
22n (n!)2

(a) Show that J0 satisfies the differential equation

x 2 J000 (x) + xJ00 (x) + x 2 J0 (x) = 0 .

https://en.wikipedia.org/wiki/Vibrations_of_a_circular_membrane
R1
(b) Evaluate 0
J0 (x) dx correct to three decimal places.
Observation: For large x, Bessel functions behave like damped sine and
cosine functions.

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 Power series and differential equations
Since the series for J0 (x) converges (absolutely) on [0, 1], we can
integrate term by term:
1 ∞ Z 1 ∞
(−1)n (−1)n
Z X
2n
X 1
J0 (x) dx = 2n 2
x dx = 2n (n!)2 2n + 1
.
0 n=0
2 (n!) 0 n=0
2

Compute terms until magnitude is less than 3 places:

For n = 0:
(−1)0 1
· = 1.
20 (0!)2 1
For n = 1:
(−1)1 1 1 1 1
· =− · =− ≈ −0.08333 .
22 (1!)2 3 4 3 12

For n = 2:
(−1)2 1 1 1 1
· = · = ≈ 0.003125 .
24 (2!)2 5 16 · 4 5 320
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 Power series and differential equations

For n = 3:
(−1)3 1 1 1 1
6 2
· =− · =− ≈ −0.000062 .
2 (3!) 7 64 · 36 7 16128

Adding these:

1 − 0.08333 + 0.003125 − 0.000062 ≈ 0.91973 .

Thus, to three decimal places we have

Z 1
J0 (x) dx ≈ 0.920 .
0

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 Power series and differential equations

Exercise: The Bessel function of order 1 is defined by

∞
X (−1)n x 2n+1
J1 (x) = .
n=0
n!(n + 1)!22n+1

(a) Show that J1 satisfies the differential equation

x 2 J100 (x) + xJ10 (x) + (x 2 − 1)J1 (x) = 0 .

(b) Show that J00 (x) = −J1 (x).

70 / 87
 Taylor/Maclaurin Series

Up to here, we discussed how to find power series representations for

functions related to geometric series.
We now address the following questions:
1 What functions can be represented by power series and how do we
find such representations?
2 If we can find a power series representation for a particular function
f and the series converges on some interval, how do we prove that
the series actually converges to f ?

71 / 87
 Taylor/Maclaurin Series
Idea: smoothness implies predictability and behaviour can be captured
by examining its derivatives at that point. P-S is a polynomial object. To
make the polynomial a good approximation, we want it to match f not
just at the point a but also in how it changes.
Consider a function f that has a power series representation at x = a:
∞
X
cn (x − a)n = c0 + c1 (x − a) + c2 (x − a)2 + . . .
n=0

We now choose the coefficients cn .

First, we want the series to equal f (a) at x = a:
∞
set
X
f (a) = cn (x − a)n = c0 + c1 (a − a) + c2 (a − a)2 + · · · = c0 .
x=a
n=0

Thus, the series equals f (a) if the coefficient c0 = f (a).

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 Taylor/Maclaurin Series
In addition, we would like the first derivative of the power series to equal
f 0 (a) at x = a. Differentiating term-by-term, we see that
∞
!
d X
cn (x − a)n = c1 + 2c2 (x − a) + 3c3 (x − a)2 + . . .
dx n=0

Therefore, at x = a, the derivative is

∞
!
set d
X
0 n
f (a) = cn (x − a) = c1 +2c2 (a−a)+3c3 (a−a)2 +· · · = c1 .
dx n=0 x=a

So c1 = f 0 (a).
Continuing in this way, we look for coefficients cn such that all the
derivatives of the power series will agree with all the corresponding
derivatives of f at x = a.

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 Taylor/Maclaurin Series

Therefore, at x = a, the second and third derivatives:

∞
!
00 d2 set X
n
f (a) = cn (x − a) = 2c2 + 3 · 2c3 (a − a) + 4 · 3c4 (a − a)2 + · · · = 2c2
dx 2 n=0
x=a

∞
!
000 setd3 X
n
f (a) = cn (x − a) = 3·2c3 +4·3·2c4 (a−a)+5·4·3c5 (a−a)2 +· · · = 3·2c3
dx 3 n=0
x=a

f 00 (a) f 000 (a)

equal f 00 (a) and f 000 (a), respectively, if c2 = 2 and c3 = 3·2 .

More generally, we see that if f has a power series representation at

x = a, then the coefficients should be given by

f (n) (a)
cn = .
n!

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 Taylor/Maclaurin Series

Definition (Taylor/Maclaurin series)

If f has derivatives of all orders at a, then the Taylor series for the
function f at a is
∞
X f (n) (a) f 00 (a) f (n) (a)
(x−a)n = f (a)+f 0 (a)(x−a)+ (x−a)2 +· · ·+ (x−a)n +· · ·
n=0
n! 2! n!

The Taylor series for f at a = 0 is known as the Maclaurin series for f .

Observation: Taylor/Maclaurin series is just a power series with the

f (n) (a)
particular coefficients cn = n! .

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 Taylor/Maclaurin Series

We now state an important result: If a function f has a power series

representation at a, then it must be the Taylor series for f at a.
Proposition (Uniqueness of Taylor Series)
If a function f has a power series at a that converges to f on some open
interval containing a, then that power series is the Taylor series for f at a.

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 Taylor/Maclaurin Series

Definition (Taylor polynomial is the partial sum)

The nth partial sum of the Taylor series for a function f at a is known as
the n-th degree Taylor polynomial. That is

p0 (x) = f (a)
p1 (x) = f (a) + f 0 (a)(x − a)
f 00 (a)
p2 (x) = f (a) + f 0 (a)(x − a) + (x − a)2
2!
f 00 (a) f 000 (a)
p3 (x) = f (a) + f 0 (a)(x − a) + (x − a)2 + (x − a)3
2! 3!
respectively.

77 / 87
 Using Taylor/Maclaurin polynomial as
approximation
Definition (nth -degree Taylor polynomial)
If f has n derivatives at a, then the nth -degree Taylor polynomial of f at
a is
f 00 (a) f 000 (a) f (n) (a)
pn (x) = f (a)+f 0 (a)(x−a)+ (x−a)2 + (x−a)3 +· · ·+ (x−a)n .
2! 3! n!

Definition (nth -degree Maclaurin polynomial)

The nth -degree Taylor polynomial for f at a = 0 is known as the
nth -degree Maclaurin polynomial for f .

Observation: From the theoretical point of view, there is a significant

difference between the previous result and this one. A function that has a
power series representation is called analytic. Here we are just using the
Taylor polynomial as an approximating tool, regardless if the function is
analytic or not.
78 / 87
 Using Taylor/Maclaurin polynomial as
approximation
Example: Find the Taylor polynomials p0 , p1 , p2 and p3 for f (x) = ln(x)
at x = 1. Compare the graph of f with the graphs of p0 , p1 , p2 and p3 .
Recall

p0 (x) = f (a)
p1 (x) = f (a) + f 0 (a)(x − a)
f 00 (a)
p2 (x) = f (a) + f 0 (a)(x − a) + (x − a)2 ,
2!
and so on.
Q: What is a?

https://www.wolframcloud.com/env/pguerra0/TaylorSeries.nb

79 / 87
 Using Taylor/Maclaurin polynomial as
approximation

Solution: To find these Taylor polynomials, we need to evaluate f and

its first three derivatives at x = 1:

f (x) = ln(x), f (1) = 0

f 0 (x) = x1 f 0 (1) = 1
f 00 (x) = − x12 f 00 (1) = −1
f 000 (x) = x23 f 000 (1) = 2

80 / 87
 Using Taylor/Maclaurin polynomial as
approximation

Therefore,

p0 (x) = f (1) = 0,
p1 (x) = f (1) + f 0 (1)(x − 1) = x − 1,
f 00 (1) 1
p2 (x) = f (1) + f 0 (1)(x − 1) + (x − 1)2 = (x − 1) − (x − 1)2 ,
2 2
0 f 00 (1) 2 f 000 (1)
p3 (x) = f (1) + f (1)(x − 1) + (x − 1) + (x − 1)3
2 3!
1 1
= (x − 1) − (x − 1)2 + (x − 1)3 .
2 3

81 / 87
 Using Taylor/Maclaurin polynomial as
approximation

Example: For each of the following functions, find formulas for the
Maclaurin polynomials p0 , p1 , p2 and p3 . Find a formula for the nth -degree
Maclaurin polynomial. Compare the graphs of p0 , p1 , p2 and p3 with f .
1 f (x) = e x
2 f (x) = sin(x)
3 f (x) = cos(x)

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 Taylor/Maclaurin Series

Solution (1): First, we’ll find the required derivatives of f = e x and

evaluate them at x = 0:
f (x) = e x , f (0) = 1
f 0 (x) = e x , f 0 (0) = 1
f 00 (x) = e x , f 00 (0) = 1
f 000 (x) = e x , f 000 (0) = 1

Using these values:

p0 (x) = f (0) = 1,
p1 (x) = f (0) + f 0 (0)x = 1 + x,
f 00 (0) 2 1
p2 (x) = f (0) + f 0 (0)x + x = 1 + x + x 2,
2! 2
0 f 00 (0) 2 f 000 (0) 3 1 1
p3 (x) = f (0) + f (0)x + x + x = 1 + x + x 2 + x 3.
2! 3! 2 6

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 Taylor/Maclaurin Series

The general formula for the nth -degree Maclaurin polynomial for
f (x) = e x is:
n
X f (k) (0) k
pn (x) = x
k!
k=0

Given that f (k) (x) = e x for all k and f (k) (0) = 1 for all k, the general
formula simplifies to:
n
X xk
pn (x) =
k!
k=0

This is the Maclaurin series representation of e x .

84 / 87
 Taylor/Maclaurin Series

Solution (2): We have

f (x) = sin(x), f (0) = 0

f 0 (x) = cos(x), f 0 (0) = 1
f 00 (x) = − sin(x), f 00 (0) = 0
f 000 (x) = − cos(x), f 000 (0) = −1

From the evaluated derivatives:

p0 (x) = f (0) = 0,
p1 (x) = f (0) + f 0 (0)x = x,
f 00 (0) 2
p2 (x) = f (0) + f 0 (0)x + x = x,
2!
f 00 (0) 2 f 000 (0) 3 1
p3 (x) = f (0) + f 0 (0)x + x + x = x − x 3.
2! 3! 6

85 / 87
 Taylor/Maclaurin Series
The general formula for the nth -degree Maclaurin polynomial for
f (x) = sin(x) is
n
X f (k) (0) k
pn (x) = x .
k!
k=0

However, due to the periodic nature of sine’s derivatives, only the odd
terms will have non-zero coefficients, and even terms will vanish.
Therefore, we have:

x3 x5 x7
pn (x) = x − + − + ...
3! 5! 7!
General formula:
n/2
X (−1)k 2k+1
pn (x) = x .
(2k + 1)!
k=0

This is the Maclaurin series representation of sin(x).

86 / 87
 Taylor/Maclaurin Series
Exercise: Find formulas for the Maclaurin polynomials p0 , p1 , p2 and p3
1
for f (x) = 1+x . Find a formula for the nth -degree Maclaurin polynomial.

Hint: Evaluate the first four derivatives of f and look for a pattern.

Answer:

p0 (x) = 1
p1 (x) = 1 − x
p2 (x) = 1 − x + x 2
p3 (x) = 1 − x + x 2 − x 3
n
X
pn (x) = 1 − x + x 2 − x 3 + · · · + (−1)n x n = (−1)k x k .
k=0

87 / 87