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:
• P. 674-675: 5, 9, 13, 19,
21, 35, 37
• P. 685-686: 1-7 odd, 2125 odd, 37-41 odd, 47,
55, 59
• Homework Supplement
Warm Up 1
Write the 4th Maclaurin polynomial for 𝑓 𝑥 = 𝑒 𝑥 .
Warm Up 2
Evaluate
1 −𝑥 2
𝑒
0
𝑑𝑥.
No closed-form antiderivative
Can be approximated with a series…
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
𝑥 <1
Radius of
Convergence
−1 < 𝑥 < 1
Interval of
Convergence
∞
𝑛
𝑥
.
𝑛=0
Exercise 2
𝑛
Treating ∞
𝑥
as a geometric series, find
𝑛=0
the sum of the series in terms of 𝑥.
∞
𝑥𝑛
𝑟=𝑥
𝑛=0
𝑎=1
1
𝑆=
1−𝑥
A Powerful Function
If we define a function
𝑛 , its
as 𝑓 𝑥 = ∞
𝑥
𝑛=0
domain would be
−1 < 𝑥 < 1, all the
𝑛
values for which ∞
𝑥
𝑛=0
converges.
Each output value would
be the infinite sum at
each 𝑥-value in the
domain.
∞
𝑥𝑛
𝑓 𝑥 =
𝑛=0
𝑓 𝑥 = 1 + 𝑥 + 𝑥2 + 𝑥3 + 𝑥4 + ⋯
1
𝑓 𝑥 =
for −1 < 𝑥 < 1
1−𝑥
A Powerful Function
If we define a function
𝑛 , its
as 𝑓 𝑥 = ∞
𝑥
𝑛=0
domain would be
−1 < 𝑥 < 1, all the
𝑛
values for which ∞
𝑥
𝑛=0
converges.
Each output value would
be the infinite sum at
each 𝑥-value in the
domain.
A Powerful Function
If we define a function
𝑛 , its
as 𝑓 𝑥 = ∞
𝑥
𝑛=0
domain would be
−1 < 𝑥 < 1, all the
𝑛
values for which ∞
𝑥
𝑛=0
converges.
Each output value would
be the infinite sum at
each 𝑥-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 𝑓 𝑥 =
𝑓 𝑥 =
Needs to be 1
Divide by 2
4
𝑥+2
4
,
𝑥+2
Needs to look like
centered at 0.
𝑎
1−𝑟
4
2
2
4/2
=
=
=
2 + 𝑥 2/2 + 𝑥/2 1 + 𝑥/2 1 − −𝑥/2
Needs to be −
𝑎=2
𝑟 = −𝑥/2
Exercise 3
Find a power series for 𝑓 𝑥 =
4
,
𝑥+2
centered at 0.
4
𝑥2
𝑥3
𝑥4
𝑓 𝑥 =
=2−𝑥 +
−
+
−⋯
𝑥+2
2
4
8
−
4
𝑓 𝑥 =
=
𝑥+2
𝑥
2
∞
𝑛=0
−
𝑥
2
𝑥
2 −
2
−
𝑛
𝑥
2
−
𝑥
2
𝑥
𝑟 = − <1
2
𝑥
−1 < < 1
2
−2 < 𝑥 < 2
Exercise 3
Find a power series for 𝑓 𝑥 =
𝑓 𝑥 =
4
𝑥+2
Implies Division
4
,
𝑥+2
centered at 0.
Exercise 4
Find a power series for 𝑓 𝑥 =
1
,
1−𝑥
centered at −1.
Exercise 5
Find a power series for 𝑓 𝑥 =
1
,
𝑥
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 𝑓 𝑥 =
∞
𝑛=0 𝑎𝑛
𝑥 𝑛 and 𝑔 𝑥 =
𝑓 𝑥 =
∞
𝑎𝑛 𝑘 𝑛 𝑥 𝑛
𝑓 𝑘𝑥 =
1
1−𝑥
∞
𝑛=0 𝑏𝑛
𝑥𝑛
𝑓 2𝑥 =
∞
𝑛=0
∞
𝑥𝑛
𝑓 𝑥 =
𝑛=0
1
1 − 2𝑥
2𝑛 𝑥 𝑛
𝑓 2𝑥 =
𝑛=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 𝑓 𝑥 =
𝑥 𝑛 and 𝑔 𝑥 =
𝑓 𝑥 =
∞
𝑓 𝑥𝑁 =
∞
𝑛=0 𝑎𝑛
𝑎𝑛 𝑥 𝑛𝑁
1
1−𝑥
∞
𝑛=0 𝑏𝑛
𝑥𝑛
𝑓 𝑥2 =
∞
𝑛=0
∞
𝑥𝑛
𝑓 𝑥 =
𝑛=0
1
1 − 𝑥2
𝑓 𝑥2 =
𝑥 2𝑛
𝑛=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 𝑓 𝑥 =
∞
𝑛=0 𝑎𝑛
𝑥 𝑛 and 𝑔 𝑥 =
∞
𝑛=0 𝑏𝑛
𝑥𝑛
∞
𝑎𝑛 ± 𝑏𝑛 𝑥 𝑛
𝑓 𝑥 ±𝑔 𝑥 =
𝑛=0
See next example…
Exercise 6
Find a power series for 𝑓 𝑥 =
3𝑥−1
,
𝑥 2 −1
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 𝑓 𝑥 = ln 𝑥, centered at 1.
Differentiate
Integrate
𝑓 𝑥 = ln 𝑥
1
=
𝑓′ 𝑥 =
𝑥
∞
−1
𝑛
𝑥−1
𝑛
Check endpoints
𝑛
Use Ratio Test to
find interval of
convergence
𝑛=0
∞
𝑓′ 𝑥 𝑑𝑥 =
−1
𝑛
−1
𝑛
𝑥−1
𝑑𝑥
𝑛=0
∞
ln 𝑥 = 𝐶 +
𝑛=0
𝑥−1
𝑛+1
𝑛+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
≈ 0.0953083
0.1 −
+
−
2
3
4
Since this is an alternating series,
the error is less than (or equal to)
the first neglected term.
𝑅4
0.15
≤
5
𝑅4 ≤ 0.000002
Exercise 9
Find a power series for 𝑓 𝑥 = arctan 𝑥, centered at
0.
Differentiate
Integrate
𝑓 𝑥 = arctan 𝑥
1
1
𝑓′ 𝑥 =
=
=
1 + 𝑥2
1 − −𝑥 2
∞
∞
−𝑥 2
𝑛=0
𝑛
−1 𝑛 𝑥 2𝑛
=
𝑛=0
∞
−1 𝑛 𝑥 2𝑛 𝑑𝑥
𝑓′ 𝑥 𝑑𝑥 =
𝑛=0
∞
arctan 𝑥 = 𝐶 +
𝑛=0
−1 𝑛 𝑥 2𝑛+1
2𝑛 + 1
Use Ratio Test to find
interval of convergence
Use initial condition
arctan 0 = 0
Exercise 9
Find a power series for 𝑓 𝑥 = arctan 𝑥, centered at
0.
Differentiate
Integrate
𝑓 𝑥 = arctan 𝑥
1
1
𝑓′ 𝑥 =
=
=
1 + 𝑥2
1 − −𝑥 2
∞
∞
−𝑥 2
𝑛=0
𝑛
−1 𝑛 𝑥 2𝑛
=
𝑛=0
∞
−1 𝑛 𝑥 2𝑛 𝑑𝑥
𝑓′ 𝑥 𝑑𝑥 =
𝑛=0
∞
arctan 𝑥 = 𝐶 +
𝑛=0
−1 𝑛 𝑥 2𝑛+1
2𝑛 + 1
Check endpoints
Use initial condition
arctan 0 = 0
Convergent Power Series
If 𝑓 is represented by a power series
𝑛
𝑓 𝑥 = ∞
𝑎
𝑥
−
𝑐
for all 𝑥 in an open
𝑛=0 𝑛
interval 𝐼 containing 𝑐, then
𝑎𝑛 =
𝑓
𝑛
𝑛!
𝑐
Coefficients of the Taylor polynomials
and
𝑓′′ 𝑐
𝑓 𝑥 = 𝑓 𝑐 + 𝑓′ 𝑐 𝑥 − 𝑐 +
2!
Taylor series
𝑥−𝑐
2
+ ⋯+
𝑓
𝑛
𝑛!
𝑐
𝑥−𝑐
𝑛
+⋯
If a power
series
converges,
then it is a
Taylor series.
Definition of Taylor Series
If a function 𝑓 has derivatives at all orders at
𝑥 = 𝑐, then the series
∞
𝑓
𝑛
𝑐
𝑛!
𝑛=0
𝑥−𝑐
𝑛
𝑓′′ 𝑐
= 𝑓 𝑐 + 𝑓′ 𝑐 𝑥 − 𝑐 +
2!
𝑥−𝑐
2
+⋯
is called the Taylor series for 𝑓 𝑥 at 𝑐. If 𝑐 = 0, then
∞
𝑛=0
𝑓
𝑛
𝑛!
0
𝑥𝑛
𝑓′′ 0 2
= 𝑓 0 + 𝑓′ 0 𝑥 +
𝑥 +⋯
2!
is called the Maclaurin series for 𝑓 𝑥 .
Essentially, a
Taylor
polynomial is a
partial sum of a
Taylor series.
Making Taylor Series
Take lots
of
derivatives
of 𝑓 𝑥
1
𝑓
5
0 =1
𝑓
4
0 =1
𝑓′′′ 0 = 1
𝑓′′ 0 = 1
Find a
pattern to
write the
𝑛th term
3
𝑓′ 0 = 1
𝑓 𝑥 = 𝑒𝑥
𝑓 0 =1
𝑓′ 𝑥 = 𝑒 𝑥
𝑓′′ 𝑥 = 𝑒 𝑥
𝑓′′′ 𝑥 = 𝑒 𝑥
𝑓
4
𝑒𝑥
𝑥 =
𝑓
5
𝑥 = 𝑒𝑥
Evaluate
said
derivatives
at 𝑐
2
𝑥2 𝑥3 𝑥4
𝑓 𝑥 = 1+𝑥+ +
+
+⋯
2! 3! 4!
∞
𝑓 𝑥 =
𝑛=0
𝑥𝑛
𝑛!
Determine
the interval
of
convergence
4
Ratio Test
Exercise 10
Write the Maclaurin series for 𝑓 𝑥 = sin 𝑥.
𝑓 𝑥 = sin 𝑥
𝑓 0 =0
𝑓′ 𝑥 = cos 𝑥
𝑓′ 0 = 1
𝑓′′ 𝑥 = − sin 𝑥
𝑓′′ 0 = 0
𝑓′′′ 𝑥 = −cos 𝑥
𝑓 ′′′ 0 = −1
𝑓
4
𝑥 = sin 𝑥
𝑓
4
0 =0
𝑓
5
𝑥 = cos 𝑥
𝑓
5
0 =1
𝑓
6
𝑥 = −sin 𝑥
𝑓
6
0 =0
⋮
⋮
Exercise 10
Write the Maclaurin series for 𝑓 𝑥 = sin 𝑥.
𝑓 0 =0
𝑓′ 0 = 1
𝑓′′ 0 = 0
𝑓 ′′′ 0 = −1
𝑓
4
𝑓
5
0 =1
𝑓
6
0 =0
0 2 1 3 0 4 1 5 0 6
𝑓 𝑥 = 0 + 1𝑥 + 𝑥 − 𝑥 + 𝑥 + 𝑥 + 𝑥 + ⋯
2!
3!
4!
5!
6!
1 3 1 5 1 7 1 9
𝑓 𝑥 =𝑥− 𝑥 + 𝑥 − 𝑥 + 𝑥 −⋯
7!
9!
3!
5!
0 =0
⋮
∞
𝑓 𝑥 =
𝑛=0
−1 𝑛 2𝑛+1
𝑥
2𝑛 + 1 !
Use the Ratio Test to
determine the interval
of convergence
Exercise 10
Write the Maclaurin series for 𝑓 𝑥 = sin 𝑥.
𝑓 0 =0
𝑓′ 0 = 1
𝑓′′ 0 = 0
𝑓 ′′′ 0 = −1
𝑓
4
𝑓
5
0 =1
𝑓
6
0 =0
0 2 1 3 0 4 1 5 0 6
𝑓 𝑥 = 0 + 1𝑥 + 𝑥 − 𝑥 + 𝑥 + 𝑥 + 𝑥 + ⋯
2!
3!
4!
5!
6!
1 3 1 5 1 7 1 9
𝑓 𝑥 =𝑥− 𝑥 + 𝑥 − 𝑥 + 𝑥 −⋯
7!
9!
3!
5!
0 =0
⋮
∞
𝑓 𝑥 =
𝑛=0
−1 𝑛 2𝑛+1
𝑥
2𝑛 + 1 !
−∞ < 𝑥 < ∞
Exercise 10
Write the Maclaurin series for 𝑓 𝑥 = sin 𝑥.
Exercise 11
Find the Maclaurin series 𝑓 𝑥 = sin 𝑥 2 .
Exercise 12
Find the Maclaurin series 𝑓 𝑥 = cos 𝑥.
Exercise 13
Find the Maclaurin series 𝑓 𝑥 = 𝑒 𝑥 .
Exercise 14
Find the Maclaurin series 𝑓 𝑥 = 𝑒
−𝑥 2
.
Exercise 15
1 −𝑥 2
𝑒
0
Approximate
𝑑𝑥 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. 𝑒 𝑥 arctan 𝑥
Exercise 16
Find the first three nonzero terms in each of
the Maclaurin series.
1. 𝑒 𝑥 arctan 𝑥
Exercise 16
Find the first three nonzero terms in each of
the Maclaurin series.
2. tan 𝑥
Exercise 16
Find the first three nonzero terms in each of
the Maclaurin series.
2. tan 𝑥
9-9/9-10: Taylor & Maclaurin Series
Objectives:
1. To represent
functions with a
power, Taylor, or
Maclaurin series
Assignment:
• P. 674-675: 5, 9, 13, 19,
21, 35, 37
• P. 685-686: 1-7 odd, 2125 odd, 37-41 odd, 47,
55, 59
• Homework Supplement