Lesson 3

POWER SERIES: TAYLOR and MACLAURIN

POLYNOMIALS and SERIES
 DEFINITION OF POWER SERIES

If x is a variable, then an infinite series of the form

∑$ ! & ' !
!"# 𝑎! 𝑥 = 𝑎# + 𝑎% 𝑥 + 𝑎& 𝑥 + 𝑎' 𝑥 + ⋯ + 𝑎! 𝑥 + ⋯

is called a 𝒑𝒐𝒘𝒆𝒓 𝒔𝒆𝒓𝒊𝒆𝒔 𝒄𝒆𝒏𝒕𝒆𝒓𝒆𝒅 𝒂𝒕 𝟎.

More generally, an infinite series of the form

$
5 𝑎! 𝑥 − 𝑐 ! = 𝑎# + 𝑎% 𝑥 − 𝑐 + 𝑎& 𝑥 − 𝑐 & + ⋯ + 𝑎! 𝑥 − 𝑐 ! +⋯
!"#

Is called a 𝒑𝒐𝒘𝒆𝒓 𝒔𝒆𝒓𝒊𝒆𝒔 𝒄𝒆𝒏𝒕𝒆𝒓𝒆𝒅 𝒂𝒕 𝒄, where c is a constant.

 ILLUSTRATIVE EXAMPLE

1. A power series centered at 0.

% ! %" %#
∑$
!"# !! =1+𝑥+ + +⋯
' (!

2. A power series centered at −1.

$

8 (−1)! (𝑥 + 1)! = 1 − 𝑥 + 1 + 𝑥 + 1 ' − 𝑥+1 ( +⋯

!"#

3. A power series centered at 1.

) ) )
∑$
!") (𝑥 − 1)! = 𝑥 − 1 + 𝑥−1 ' + 𝑥−1 ( +⋯
! ' (
 DEFINITION
Taylor Polynomial

If 𝑓 can be differentiated 𝑛 times at 𝑥# , then we define the 𝑛𝑡ℎ 𝑇𝑎𝑦𝑙𝑜𝑟 𝑃𝑜𝑙𝑦𝑛𝑜𝑚𝑖𝑎𝑙 for 𝑓
about 𝑥 = 𝑥# to be:
)"(,! ) )((((,! ) ) " ,!
𝑝! 𝑥 = 𝑓 𝑥# + 𝑓( 𝑥# 𝑥 − 𝑥# + 𝑥 − 𝑥# & + (𝑥 − 𝑥# )' + . . . + (𝑥 − 𝑥# )!
&! '! !!

Using the Sigma Notation, this can be expressed as:

! ) # ,! /
𝑝! 𝑥 = ∑/"# 𝑥 − 𝑥# = 𝑓 𝑥# + 𝑓 ( 𝑥# 𝑥 − 𝑥# +
/!
)"(,! ) )((((,! ) ) " ,!
(𝑥 − 𝑥# )& + (𝑥 − 𝑥# )' + . . . + (𝑥 − 𝑥# )!
&! '! !!
 ILLUSTRATIVE EXAMPLE

1. Find the first four Taylor Polynomials for f(x) = ln 𝑥 about 𝑥 = 2

Solution: Let 𝑓 𝑥 = ln 𝑥. 𝑓 2 = ln 2
) )
𝑓* 𝑥 = 𝑓* 2 =
% '
+) +)
𝑓 ** 𝑥 = 𝑓 ** 2 =
%" ,
' )
𝑓 *** 𝑥 = 𝑓 *** 2 =
%# ,

𝑝# 𝑥 = ln 2
)
𝑝) 𝑥 = ln 2 + ' 𝑥 − 2
) ) %+' "
𝑝' 𝑥 = ln 2 + 𝑥−2 −
' , '!

' (
1 1 𝑥−2 1 𝑥−2
𝑝( 𝑥 = ln 2 + 𝑥 − 2 − +
2 4 2! 4 3!
) %+' " %+' #
𝑝( 𝑥 = ln 2 + 𝑥−2 − +
' - ',
 ILLUSTRATIVE EXAMPLE
)
2. Find the nth Taylor polynomial for about 𝑥 = 1 and express it in sigma notation.
%
Solution:
) Note: express the values
Let 𝑓 𝑥 = = 𝑥 +) 𝑓 1 = 1 = 0!
%
in terms of factorial
𝑓 𝑥 = −𝑥 +'
* 𝑓 * 1 = −1 = −(1!)
notation to be able to
𝑓"(𝑥) = 2𝑥 +( 𝑓"(1) = 2 = 1 (2) = 2!
recognize the pattern of
𝑓 *** 𝑥 = −6𝑥 +, 𝑓 *** 1 = −6 = − 1 2 3 = −(3!)
the terms.
𝑓 ./ 𝑥 = 24𝑥 +0 𝑓 ./ 1 = 24 = 4!

! 2 $ %% 1
𝑝! 𝑥 = ∑1"# 𝑥 − 𝑥# = 𝑓 𝑥# + 𝑓 * 𝑥# 𝑥 − 𝑥# +
1!
2"(%%) 2***(%%) 2 ! %%
'!
(𝑥 − 𝑥# )' + (!
(𝑥 − 𝑥# )( + . . . + !!
(𝑥 − 𝑥# )!

) ! %+) " (! %+) # ,! %+) & !

! !! %+) ! !! %+)
%
= ∑!1"# −1 !!
= 1 − 𝑥 − 1 + 2! '!
− (!
+ ,!
+. . . + −1 !!

) ,! %+) &
%
= ∑!1"# −1 !
𝑥−1 !
=1− 𝑥−1 + 𝑥−1 '
− 𝑥−1 (
+ ,!
+. . . + −1 !
𝑥−1 !
 DEFINITION

Maclaurin’s Polynomial
If 𝑓 can be differentiated 𝑛 times at 0, then we define the
𝑛𝑡ℎ 𝑀𝑎𝑐𝑙𝑎𝑢𝑟𝑖𝑛 𝑝𝑜𝑙𝑦𝑛𝑜𝑚𝑖𝑎𝑙 for 𝑓 to be

𝑓"(0) 𝑓′′′(0) 𝑓 D 0
𝑝D 𝑥 = 𝑓 0 + 𝑓 E 0 𝑥 + 𝑥F + 𝑥 G+ . . . + 𝑥D
2! 3! 𝑛!
Or in Sigma Notation:
!
𝑓 1 𝑥# 𝑓"(0) 𝑓′′′(0) 𝑓 ! 0
𝑝! 𝑥 = 8 = 𝑓 0 + 𝑓* 0 𝑥 + 𝑥' + 𝑥(+ . . . + 𝑥!
𝑘! 2! 3! 𝑛!
1"#
 ILLUSTRATIVE EXAMPLE
1. Find the Maclaurin polynomials 𝑝# , 𝑝% , 𝑝& , 𝑝' , and 𝑝! for 𝑒 , .

Solution: Get the first, second, third, and nth derivative of the function,
Let 𝑓 𝑥 = 𝑒 %
𝑓 * 𝑥 = 𝑒 % ; 𝑓"(𝑥) = 𝑒 % ; 𝑓′′′(𝑥) = 𝑒 % ; 𝑓 ! 𝑥 = 𝑒 %
Assign 𝑥 = 0, thus
𝑓 * 0 = 𝑓 ** 0 = 𝑓 *** 0 =. . . = 𝑓 ! 0 = 𝑒 # = 1
Therefore,
𝑝# 𝑥 = 𝑓 0 = 1
𝑝) 𝑥 = 𝑓 0 + 𝑓 * 0 𝑥 = 1 + 𝑥
2 '' # % " %"
𝑝' 𝑥 = 𝑓 0 + 𝑓 * 0 𝑥 + =1+𝑥+
'! '!

2 '' # % " 2 ''' # % # %" %#

𝑝( 𝑥 = 𝑓 0 + 𝑓* 0 𝑥+ + =1+𝑥+ +
'! (! '! (!
**
𝑓 0 𝑥' 𝑓! 0 𝑥 !
𝑝! 𝑥 = 𝑓 0 + 𝑓 * 0 𝑥 + +. . . +
2! 𝑛!
%" %!
=1+𝑥+ +. . . +
'! !!
 ILLUSTRATIVE EXAMPLE

2. Find the nth Maclaurin polynomials for 𝑠𝑖𝑛 𝑥.

Solution: let 𝑓 𝑥 = sin 𝑥 𝑓 0 = sin 0 = 0
𝑓 ( 𝑥 = cos 𝑥 𝑓 ( 0 = cos 0 = 1
𝑓 (( 𝑥 = − sin 𝑥 𝑓 (( 0 = − sin 0 = 0
𝑓 ((( 𝑥 = − cos 𝑥 𝑓 ((( 0 = − cos 0 = −1
𝑓 01 𝑥 = sin 𝑥 𝑓 01 0 = sin 0 = 0
𝑓 1 𝑥 = cos 𝑥 𝑓 1 0 = cos 0 = 1
Thus, the nth Maclaurin polynomial for sin 𝑥 based on
)"(#) & )((((#) ' )" #
𝑝! 𝑥 = 𝑓 0 + 𝑓( 0 𝑥+ 𝑥 + 𝑥 + ...+ 𝑥 ! is given by
&! '! !!

𝑥' 𝑥2 !
𝑥 &!3%
𝑝! 𝑥 = 𝑥 − + +. . . + −1 𝑛 = 0, 1, 2, . . .
3! 5! 2𝑛 + 1 !
 ILLUSTRATIVE EXAMPLE

3. Find the nth Maclaurin polynomials for cos 𝑥.

Solution: let 𝑓 𝑥 = cos 𝑥 𝑓 0 = cos 0 = 1
𝑓 ( 𝑥 = −sin 𝑥 𝑓 ( 0 = −sin 0 = 0
𝑓 (( 𝑥 = − cos 𝑥 𝑓 (( 0 = − cos 0 = −1
𝑓 ((( 𝑥 = sin 𝑥 𝑓 ((( 0 = sin 0 = 0
𝑓 01 𝑥 = co𝑠 𝑥 𝑓 01 0 = cos 0 = 1
𝑓 1 𝑥 = −sin 𝑥 𝑓 1 0 = −sin 0 = 0

Thus, the nth Maclaurin polynomial for cos 𝑥 is

𝑥& 𝑥4 𝑥5 !
𝑥 &!
𝑝! 𝑥 = 1 − + − . . . + −1 𝑛 = 0, 1, 2, . . .
2! 4! 6! 2𝑛 !
 ILLUSTRATIVE EXAMPLE
%
4. Find the nth Maclaurin polynomial for and express it in sigma notation.
%6,
Solution:
%
Let 𝑓 𝑥 = = 1 − 𝑥 6% 𝑓 0 = 1 = 0!
%6,
%
𝑓( 𝑥 = 1 − 𝑥 6& = 𝑓 ( 0 = 1 = 1!
%6, $
&
𝑓 (( 𝑥 = 2 1 − 𝑥 6' = 𝑓 (( 0 = 2 = 2!
%6, %
5
𝑓 ((( 𝑥 = 6 1 − 𝑥 64 = 𝑓 ((( 0 = 6 = 3!
%6, &
! &
1 𝑥 𝑥' 𝑥!
= 5 𝑥 / = 1 + 𝑥 + 2! + 3! +. . . +𝑛!
1−𝑥 2! 3! 𝑛!
/"#

%
= ∑!/"# 𝑥 / = 1 + 𝑥 + 𝑥 & + 𝑥 ' +. . . +𝑥 !
%6,
 EXERCISES
1. Find the Maclaurin polynomials of orders n = 0, 1, 2, 3, and 4, and then
find the nth Maclaurin polynomials for the function in sigma notation.
a. 𝑒 +%
b. sin 𝜋𝑥
c. ln 1 + 𝑥

2. Find the Taylor polynomials of orders n = 0, 1, 2, 3, and 4, about 𝑥 = 𝑥#

and then find the nth Taylor polynomials for the function in sigma notation.
a. 𝑒 % ; 𝑥# = 1
)
b. ; 𝑥# = −1
%
POWER SERIES, RADIUS AND
INTERVAL OF CONVERGENCE
 DEFINITION OF POWER SERIES

If x is a variable, then an infinite series of the form

∑$ ! & ' !
!"# 𝑎! 𝑥 = 𝑎# + 𝑎% 𝑥 + 𝑎& 𝑥 + 𝑎' 𝑥 + ⋯ + 𝑎! 𝑥 + ⋯

is called a 𝒑𝒐𝒘𝒆𝒓 𝒔𝒆𝒓𝒊𝒆𝒔 𝒄𝒆𝒏𝒕𝒆𝒓𝒆𝒅 𝒂𝒕 𝟎.

More generally, an infinite series of the form

$
5 𝑎! 𝑥 − 𝑐 ! = 𝑎# + 𝑎% 𝑥 − 𝑐 + 𝑎& 𝑥 − 𝑐 & + ⋯ + 𝑎! 𝑥 − 𝑐 ! +⋯
!"#

Is called a 𝒑𝒐𝒘𝒆𝒓 𝒔𝒆𝒓𝒊𝒆𝒔 𝒄𝒆𝒏𝒕𝒆𝒓𝒆𝒅 𝒂𝒕 𝒄, where c is a constant.

 CONVERGENCE OF A POWER SERIES

A power series in 𝑥

𝑎# +𝑎% 𝑥 + 𝑎& 𝑥 & + ⋯ + 𝑎!6% 𝑥 !6% + ⋯

Or 𝑎# +𝑎%( 𝑥 − 𝑐) + 𝑎& (𝑥 − 𝑐)& + ⋯ + 𝑎!6% (𝑥 − 𝑐)!6% + ⋯

where 𝑥 is a variable and the coefficients 𝑎# , 𝑎% , 𝑎& ,⋯ , 𝑎!6% ,⋯ are arbitrary constants

may converge for all values of 𝑥, or for no value except 𝑥 = 0 𝑜𝑟 𝑐, or for some finite number of values
of x. The totality of values of 𝑥 for which the power series converges is called the interval of
convergence of the series. This interval of convergence can be determined using the Ratio or Root test.;
the set of values of x over which the power series converges is called the Convergence set.
 RADIUS AND INTERVAL OF CONVERGENCE

The convergence of the series ∑P

DNO 𝑎D 𝑥 − 𝑐
D is described by one of the
following three cases:

1. There is a positive number R such that the series diverges for x with | x – c| >
R but converges absolutely for x with | x – c | < R . The series may or may not
converge at either of the endpoints x = c – R and x = c + R.

2. The series converges absolutely for every x 𝑅 → ∞ .

3. The series converges at x = c and diverges elsewhere ( R = 0 ).

 ILLUSTRATIVE EXAMPLE

1. Determine the radius of convergence and interval of convergence for the power series.

Using the ratio test for absolute convergence to the given series,

6% " 6% !3% ,3' " ,3' 4"

= lim
!→$ 4" 4 (6%)" ! ( ,3')"

By the ratio test, the series converges if L < 1,

thus:
the series converges if
)
,
𝑥+3 <1
𝑥 + 3 < 4 ; that is, R = 4
–4<(x+3)<4
- 4 – 3 < x < 4 -3
-7 < x < 1 ; interval of convergence
To test convergence at the endpoints , simply substitute the values into the original power series and see if the series
converges or diverges using any applicable test.

Verifying convergence at the endpoint 𝑥 = −7

NOTE:
This series diverges by the
Divergence Test since the limit of
the nth term as n increases without
This series is divergent by the Divergence Test since
bound is unbounded as well; that is,
lim 𝑛 → ∞.
!→$

At the endpoint 𝑥 = 1:
This series diverges by the Alternating Series Test since the two
conditions are not satisfied. Thus, this power series will not converge
at either endpoint.

Therefore for the given series, the interval of convergence is −7, 1 with radius of
convergence 𝑅 = 4.
 ILLUSTRATIVE EXAMPLE
2. Determine the radius of convergence and interval of convergence for the power series

Solution:
In this example the root test is more appropriate to use. So

With 𝐿 = 0 < 1 and since regardless of the value of x substituted in | x – 6 | , the limit still will be zero then this power
series will converge for every
. x, a real number. Hence , the radius of convergence is 𝑅 = ∞
and the interval of convergence is ( −∞ , ∞ .
 ILLUSTRATIVE EXAMPLE
3. Determine the radius of convergence and interval of convergence for the power series .

Solution: Using the ratio test for absolute convergence,

𝑛 + 1 𝑛! (2𝑥 + 1)! (2𝑥 + 1)
= lim
!→$ 𝑛! 2𝑥 + 1 !

𝑊ℎ𝑒𝑟𝑒 lim 𝑛 + 1 → ∞ =
!→$

%
We’ll have 𝐿 → ∞ > 1 and the limit will be so provided 𝑥 ≠ − ( this is to avoid the indeterminate form
&
%
0 ∗ ∞) ; thus, this series diverges for all 𝑥≠− . This result will further imply that this power series will
&
)
only converge if 𝑥 = − . In this case we have the radius of convergence 𝑅 = 0 and the interval of
'
)
convergence −' .
 EXERCISES
Find the radius of convergence and the interval of
convergence.
$ ,#
1. ∑/"#
&/3'

$ 6% # ,#
2. ∑/"#
/!

5 #
3. ∑$
/"# / $ 𝑥
/