9 9 9 10 Power Taylor Maclaurin Series
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Transcript 9 9 9 10 Power Taylor Maclaurin Series
9-9/9-10: Taylor & Maclaurin Series
Objectives:
1. To represent
functions with a
power, Taylor, or
Maclaurin series
Assignment:
โข P. 674-675: 5, 9, 13, 19,
21, 35, 37
โข P. 685-686: 1-7 odd, 2125 odd, 37-41 odd, 47,
55, 59
โข Homework Supplement
Warm Up 1
Write the 4th Maclaurin polynomial for ๐ ๐ฅ = ๐ ๐ฅ .
Warm Up 2
Evaluate
1 โ๐ฅ 2
๐
0
๐๐ฅ.
No closed-form antiderivative
Can be approximated with a seriesโฆ
Objective 1
You will be able to
represent a function
with a power, Taylor, or
Maclaurin series
Exercise 1
Find the interval of convergence for
๐ฅ <1
Radius of
Convergence
โ1 < ๐ฅ < 1
Interval of
Convergence
โ
๐
๐ฅ
.
๐=0
Exercise 2
๐
Treating โ
๐ฅ
as a geometric series, find
๐=0
the sum of the series in terms of ๐ฅ.
โ
๐ฅ๐
๐=๐ฅ
๐=0
๐=1
1
๐=
1โ๐ฅ
A Powerful Function
If we define a function
๐ , its
as ๐ ๐ฅ = โ
๐ฅ
๐=0
domain would be
โ1 < ๐ฅ < 1, all the
๐
values for which โ
๐ฅ
๐=0
converges.
Each output value would
be the infinite sum at
each ๐ฅ-value in the
domain.
โ
๐ฅ๐
๐ ๐ฅ =
๐=0
๐ ๐ฅ = 1 + ๐ฅ + ๐ฅ2 + ๐ฅ3 + ๐ฅ4 + โฏ
1
๐ ๐ฅ =
for โ1 < ๐ฅ < 1
1โ๐ฅ
A Powerful Function
If we define a function
๐ , its
as ๐ ๐ฅ = โ
๐ฅ
๐=0
domain would be
โ1 < ๐ฅ < 1, all the
๐
values for which โ
๐ฅ
๐=0
converges.
Each output value would
be the infinite sum at
each ๐ฅ-value in the
domain.
A Powerful Function
If we define a function
๐ , its
as ๐ ๐ฅ = โ
๐ฅ
๐=0
domain would be
โ1 < ๐ฅ < 1, all the
๐
values for which โ
๐ฅ
๐=0
converges.
Each output value would
be the infinite sum at
each ๐ฅ-value in the
domain.
Representing Functions as Power Series
In the previous example,
we found a function to
which the power series
converges over its
interval of convergence.
Usually, we want to do
the reverse; we want to
find a power series
representation for a
particular function.
This can be accomplished in a
variety of ways:
Manipulation of a geometric series
Algebraic processes
Operations on known series
Composition
Term-by-term differentiation
Term-by-term integration
Exercise 3
Find a power series for ๐ ๐ฅ =
๐ ๐ฅ =
Needs to be 1
Divide by 2
4
๐ฅ+2
4
,
๐ฅ+2
Needs to look like
centered at 0.
๐
1โ๐
4
2
2
4/2
=
=
=
2 + ๐ฅ 2/2 + ๐ฅ/2 1 + ๐ฅ/2 1 โ โ๐ฅ/2
Needs to be โ
๐=2
๐ = โ๐ฅ/2
Exercise 3
Find a power series for ๐ ๐ฅ =
4
,
๐ฅ+2
centered at 0.
4
๐ฅ2
๐ฅ3
๐ฅ4
๐ ๐ฅ =
=2โ๐ฅ +
โ
+
โโฏ
๐ฅ+2
2
4
8
โ
4
๐ ๐ฅ =
=
๐ฅ+2
๐ฅ
2
โ
๐=0
โ
๐ฅ
2
๐ฅ
2 โ
2
โ
๐
๐ฅ
2
โ
๐ฅ
2
๐ฅ
๐ = โ <1
2
๐ฅ
โ1 < < 1
2
โ2 < ๐ฅ < 2
Exercise 3
Find a power series for ๐ ๐ฅ =
๐ ๐ฅ =
4
๐ฅ+2
Implies Division
4
,
๐ฅ+2
centered at 0.
Exercise 4
Find a power series for ๐ ๐ฅ =
1
,
1โ๐ฅ
centered at โ1.
Exercise 5
Find a power series for ๐ ๐ฅ =
1
,
๐ฅ
centered at 1.
Operations with Power Series
Besides manipulating the sum of a geometric
series, we can also use algebraic means to derive
a power series for a function.
Let ๐ ๐ฅ =
โ
๐=0 ๐๐
๐ฅ ๐ and ๐ ๐ฅ =
๐ ๐ฅ =
โ
๐๐ ๐ ๐ ๐ฅ ๐
๐ ๐๐ฅ =
1
1โ๐ฅ
โ
๐=0 ๐๐
๐ฅ๐
๐ 2๐ฅ =
โ
๐=0
โ
๐ฅ๐
๐ ๐ฅ =
๐=0
1
1 โ 2๐ฅ
2๐ ๐ฅ ๐
๐ 2๐ฅ =
๐=0
Operations with Power Series
Besides manipulating the sum of a geometric
series, we can also use algebraic means to derive
a power series for a function.
Let ๐ ๐ฅ =
๐ฅ ๐ and ๐ ๐ฅ =
๐ ๐ฅ =
โ
๐ ๐ฅ๐ =
โ
๐=0 ๐๐
๐๐ ๐ฅ ๐๐
1
1โ๐ฅ
โ
๐=0 ๐๐
๐ฅ๐
๐ ๐ฅ2 =
โ
๐=0
โ
๐ฅ๐
๐ ๐ฅ =
๐=0
1
1 โ ๐ฅ2
๐ ๐ฅ2 =
๐ฅ 2๐
๐=0
Operations with Power Series
Besides manipulating the sum of a geometric
series, we can also use algebraic means to derive
a power series for a function.
Let ๐ ๐ฅ =
โ
๐=0 ๐๐
๐ฅ ๐ and ๐ ๐ฅ =
โ
๐=0 ๐๐
๐ฅ๐
โ
๐๐ ± ๐๐ ๐ฅ ๐
๐ ๐ฅ ±๐ ๐ฅ =
๐=0
See next exampleโฆ
Exercise 6
Find a power series for ๐ ๐ฅ =
3๐ฅโ1
,
๐ฅ 2 โ1
centered at 0.
M AT H
Use partial fraction decomposition
Term-By-Term
Lastly, we can derive a power series for a
function by term-by-term differentiation or
integration.
The radius of convergence of the series
obtained by differentiating or integrating a
power series is the same as the original series.
The interval of convergence, on the other hand, may
differ as a result of the behavior at the endpoints.
Exercise 7
Find a power series for ๐ ๐ฅ = ln ๐ฅ, centered at 1.
Differentiate
Integrate
๐ ๐ฅ = ln ๐ฅ
1
=
๐โฒ ๐ฅ =
๐ฅ
โ
โ1
๐
๐ฅโ1
๐
Check endpoints
๐
Use Ratio Test to
find interval of
convergence
๐=0
โ
๐โฒ ๐ฅ ๐๐ฅ =
โ1
๐
โ1
๐
๐ฅโ1
๐๐ฅ
๐=0
โ
ln ๐ฅ = ๐ถ +
๐=0
๐ฅโ1
๐+1
๐+1
Use initial condition
ln 1 = 0
Exercise 8
Use the previous power series to approximate
ln 1.1 using the 4th partial sum.
0.12 0.13 0.14
โ 0.0953083
0.1 โ
+
โ
2
3
4
Since this is an alternating series,
the error is less than (or equal to)
the first neglected term.
๐
4
0.15
โค
5
๐
4 โค 0.000002
Exercise 9
Find a power series for ๐ ๐ฅ = arctan ๐ฅ, centered at
0.
Differentiate
Integrate
๐ ๐ฅ = arctan ๐ฅ
1
1
๐โฒ ๐ฅ =
=
=
1 + ๐ฅ2
1 โ โ๐ฅ 2
โ
โ
โ๐ฅ 2
๐=0
๐
โ1 ๐ ๐ฅ 2๐
=
๐=0
โ
โ1 ๐ ๐ฅ 2๐ ๐๐ฅ
๐โฒ ๐ฅ ๐๐ฅ =
๐=0
โ
arctan ๐ฅ = ๐ถ +
๐=0
โ1 ๐ ๐ฅ 2๐+1
2๐ + 1
Use Ratio Test to find
interval of convergence
Use initial condition
arctan 0 = 0
Exercise 9
Find a power series for ๐ ๐ฅ = arctan ๐ฅ, centered at
0.
Differentiate
Integrate
๐ ๐ฅ = arctan ๐ฅ
1
1
๐โฒ ๐ฅ =
=
=
1 + ๐ฅ2
1 โ โ๐ฅ 2
โ
โ
โ๐ฅ 2
๐=0
๐
โ1 ๐ ๐ฅ 2๐
=
๐=0
โ
โ1 ๐ ๐ฅ 2๐ ๐๐ฅ
๐โฒ ๐ฅ ๐๐ฅ =
๐=0
โ
arctan ๐ฅ = ๐ถ +
๐=0
โ1 ๐ ๐ฅ 2๐+1
2๐ + 1
Check endpoints
Use initial condition
arctan 0 = 0
Convergent Power Series
If ๐ is represented by a power series
๐
๐ ๐ฅ = โ
๐
๐ฅ
โ
๐
for all ๐ฅ in an open
๐=0 ๐
interval ๐ผ containing ๐, then
๐๐ =
๐
๐
๐!
๐
Coefficients of the Taylor polynomials
and
๐โฒโฒ ๐
๐ ๐ฅ = ๐ ๐ + ๐โฒ ๐ ๐ฅ โ ๐ +
2!
Taylor series
๐ฅโ๐
2
+ โฏ+
๐
๐
๐!
๐
๐ฅโ๐
๐
+โฏ
If a power
series
converges,
then it is a
Taylor series.
Definition of Taylor Series
If a function ๐ has derivatives at all orders at
๐ฅ = ๐, then the series
โ
๐
๐
๐
๐!
๐=0
๐ฅโ๐
๐
๐โฒโฒ ๐
= ๐ ๐ + ๐โฒ ๐ ๐ฅ โ ๐ +
2!
๐ฅโ๐
2
+โฏ
is called the Taylor series for ๐ ๐ฅ at ๐. If ๐ = 0, then
โ
๐=0
๐
๐
๐!
0
๐ฅ๐
๐โฒโฒ 0 2
= ๐ 0 + ๐โฒ 0 ๐ฅ +
๐ฅ +โฏ
2!
is called the Maclaurin series for ๐ ๐ฅ .
Essentially, a
Taylor
polynomial is a
partial sum of a
Taylor series.
Making Taylor Series
Take lots
of
derivatives
of ๐ ๐ฅ
1
๐
5
0 =1
๐
4
0 =1
๐โฒโฒโฒ 0 = 1
๐โฒโฒ 0 = 1
Find a
pattern to
write the
๐th term
3
๐โฒ 0 = 1
๐ ๐ฅ = ๐๐ฅ
๐ 0 =1
๐โฒ ๐ฅ = ๐ ๐ฅ
๐โฒโฒ ๐ฅ = ๐ ๐ฅ
๐โฒโฒโฒ ๐ฅ = ๐ ๐ฅ
๐
4
๐๐ฅ
๐ฅ =
๐
5
๐ฅ = ๐๐ฅ
Evaluate
said
derivatives
at ๐
2
๐ฅ2 ๐ฅ3 ๐ฅ4
๐ ๐ฅ = 1+๐ฅ+ +
+
+โฏ
2! 3! 4!
โ
๐ ๐ฅ =
๐=0
๐ฅ๐
๐!
Determine
the interval
of
convergence
4
Ratio Test
Exercise 10
Write the Maclaurin series for ๐ ๐ฅ = sin ๐ฅ.
๐ ๐ฅ = sin ๐ฅ
๐ 0 =0
๐โฒ ๐ฅ = cos ๐ฅ
๐โฒ 0 = 1
๐โฒโฒ ๐ฅ = โ sin ๐ฅ
๐โฒโฒ 0 = 0
๐โฒโฒโฒ ๐ฅ = โcos ๐ฅ
๐ โฒโฒโฒ 0 = โ1
๐
4
๐ฅ = sin ๐ฅ
๐
4
0 =0
๐
5
๐ฅ = cos ๐ฅ
๐
5
0 =1
๐
6
๐ฅ = โsin ๐ฅ
๐
6
0 =0
โฎ
โฎ
Exercise 10
Write the Maclaurin series for ๐ ๐ฅ = sin ๐ฅ.
๐ 0 =0
๐โฒ 0 = 1
๐โฒโฒ 0 = 0
๐ โฒโฒโฒ 0 = โ1
๐
4
๐
5
0 =1
๐
6
0 =0
0 2 1 3 0 4 1 5 0 6
๐ ๐ฅ = 0 + 1๐ฅ + ๐ฅ โ ๐ฅ + ๐ฅ + ๐ฅ + ๐ฅ + โฏ
2!
3!
4!
5!
6!
1 3 1 5 1 7 1 9
๐ ๐ฅ =๐ฅโ ๐ฅ + ๐ฅ โ ๐ฅ + ๐ฅ โโฏ
7!
9!
3!
5!
0 =0
โฎ
โ
๐ ๐ฅ =
๐=0
โ1 ๐ 2๐+1
๐ฅ
2๐ + 1 !
Use the Ratio Test to
determine the interval
of convergence
Exercise 10
Write the Maclaurin series for ๐ ๐ฅ = sin ๐ฅ.
๐ 0 =0
๐โฒ 0 = 1
๐โฒโฒ 0 = 0
๐ โฒโฒโฒ 0 = โ1
๐
4
๐
5
0 =1
๐
6
0 =0
0 2 1 3 0 4 1 5 0 6
๐ ๐ฅ = 0 + 1๐ฅ + ๐ฅ โ ๐ฅ + ๐ฅ + ๐ฅ + ๐ฅ + โฏ
2!
3!
4!
5!
6!
1 3 1 5 1 7 1 9
๐ ๐ฅ =๐ฅโ ๐ฅ + ๐ฅ โ ๐ฅ + ๐ฅ โโฏ
7!
9!
3!
5!
0 =0
โฎ
โ
๐ ๐ฅ =
๐=0
โ1 ๐ 2๐+1
๐ฅ
2๐ + 1 !
โโ < ๐ฅ < โ
Exercise 10
Write the Maclaurin series for ๐ ๐ฅ = sin ๐ฅ.
Exercise 11
Find the Maclaurin series ๐ ๐ฅ = sin ๐ฅ 2 .
Exercise 12
Find the Maclaurin series ๐ ๐ฅ = cos ๐ฅ.
Exercise 13
Find the Maclaurin series ๐ ๐ฅ = ๐ ๐ฅ .
Exercise 14
Find the Maclaurin series ๐ ๐ฅ = ๐
โ๐ฅ 2
.
Exercise 15
1 โ๐ฅ 2
๐
0
Approximate
๐๐ฅ using a power series
with an error of less than 0.01.
Exercise 16
Find the first three nonzero terms in each of
the Maclaurin series.
1. ๐ ๐ฅ arctan ๐ฅ
Exercise 16
Find the first three nonzero terms in each of
the Maclaurin series.
1. ๐ ๐ฅ arctan ๐ฅ
Exercise 16
Find the first three nonzero terms in each of
the Maclaurin series.
2. tan ๐ฅ
Exercise 16
Find the first three nonzero terms in each of
the Maclaurin series.
2. tan ๐ฅ
9-9/9-10: Taylor & Maclaurin Series
Objectives:
1. To represent
functions with a
power, Taylor, or
Maclaurin series
Assignment:
โข P. 674-675: 5, 9, 13, 19,
21, 35, 37
โข P. 685-686: 1-7 odd, 2125 odd, 37-41 odd, 47,
55, 59
โข Homework Supplement