§8.7--Taylor and Maclaurin Seriesmath.furman.edu/~mwoodard/math151/docs/sec7.pdfMark Woodard...
Transcript of §8.7--Taylor and Maclaurin Seriesmath.furman.edu/~mwoodard/math151/docs/sec7.pdfMark Woodard...
§8.7–Taylor and Maclaurin Series
Mark Woodard
Furman University
Fall 2007
Mark Woodard (Furman University) §8.7–Taylor and Maclaurin Series Fall 2007 1 / 23
Outline
1 Taylor’s Theorem
2 When does a function equal its Taylor series?
3 Important examples: ex , sin(x), cos(x)
4 Newton’s binomial theorem
5 Some uses of Taylor series
6 Multiplication and division of series
Mark Woodard (Furman University) §8.7–Taylor and Maclaurin Series Fall 2007 2 / 23
Taylor’s Theorem
Overview
In the last section we learned that some functions can be expressed aspower series. In this section we explore a general method for expressing afunction as a power series:
f (x) =∞∑
n=0
cn(x − a)n = c0 + c1(x − a)1 + c2(x − a)2 + · · ·
Some questions:
Which functions have power series representations?
If a function can be represented by a power series, what are thecoefficients {cn}?Why does it matter? Why is it helpful for a function to berepresented by a power series?
Mark Woodard (Furman University) §8.7–Taylor and Maclaurin Series Fall 2007 3 / 23
Taylor’s Theorem
Overview
In the last section we learned that some functions can be expressed aspower series. In this section we explore a general method for expressing afunction as a power series:
f (x) =∞∑
n=0
cn(x − a)n = c0 + c1(x − a)1 + c2(x − a)2 + · · ·
Some questions:
Which functions have power series representations?
If a function can be represented by a power series, what are thecoefficients {cn}?Why does it matter? Why is it helpful for a function to berepresented by a power series?
Mark Woodard (Furman University) §8.7–Taylor and Maclaurin Series Fall 2007 3 / 23
Taylor’s Theorem
Overview
In the last section we learned that some functions can be expressed aspower series. In this section we explore a general method for expressing afunction as a power series:
f (x) =∞∑
n=0
cn(x − a)n = c0 + c1(x − a)1 + c2(x − a)2 + · · ·
Some questions:
Which functions have power series representations?
If a function can be represented by a power series, what are thecoefficients {cn}?
Why does it matter? Why is it helpful for a function to berepresented by a power series?
Mark Woodard (Furman University) §8.7–Taylor and Maclaurin Series Fall 2007 3 / 23
Taylor’s Theorem
Overview
In the last section we learned that some functions can be expressed aspower series. In this section we explore a general method for expressing afunction as a power series:
f (x) =∞∑
n=0
cn(x − a)n = c0 + c1(x − a)1 + c2(x − a)2 + · · ·
Some questions:
Which functions have power series representations?
If a function can be represented by a power series, what are thecoefficients {cn}?Why does it matter? Why is it helpful for a function to berepresented by a power series?
Mark Woodard (Furman University) §8.7–Taylor and Maclaurin Series Fall 2007 3 / 23
Taylor’s Theorem
Theorem (The form of the series)
Suppose that
f (x) =∞∑
n=0
cn(x − a)n = c0 + c1(x − a)1 + c2(x − a)2 + · · ·
in some symmetric interval about the point a, then the coefficients mustbe of the form
cn =f (n)(a)
n!, n ≥ 0.
Thus, if f is represented by a power series, it must assume the form:
f (x) =∞∑
n=0
f (n)(a)
n!(x − a)n.
Mark Woodard (Furman University) §8.7–Taylor and Maclaurin Series Fall 2007 4 / 23
Taylor’s Theorem
Proof.
Suppose f (x) = c0 + c1(x − a)1 + c2(x − a)2 + c3(x − a)3 + · · · insome interval about the point a.
Evaluating both sides at x = a shows that f (a) = c0; thus, c0 = f (a).
Differentiate both sides: f ′(x) = c1 + 2c2(x − a)1 + 3c3(x − a)2 + · · · .Evaluate both sides at x = a and note that f ′(a) = c1.
Likewise f ′′(x) = 2c2 + 3 · 2(x − a)1 + · · · . Upon evaluating bothsides at x = a, we see that f ′′(a) = 2c2 or c2 = f ′′(a)/2!.
Differentiating and evaluating once more leads to c3 = f ′′′(a)/3!.
Continue in the same manner.
Mark Woodard (Furman University) §8.7–Taylor and Maclaurin Series Fall 2007 5 / 23
Taylor’s Theorem
Proof.
Suppose f (x) = c0 + c1(x − a)1 + c2(x − a)2 + c3(x − a)3 + · · · insome interval about the point a.
Evaluating both sides at x = a shows that f (a) = c0; thus, c0 = f (a).
Differentiate both sides: f ′(x) = c1 + 2c2(x − a)1 + 3c3(x − a)2 + · · · .Evaluate both sides at x = a and note that f ′(a) = c1.
Likewise f ′′(x) = 2c2 + 3 · 2(x − a)1 + · · · . Upon evaluating bothsides at x = a, we see that f ′′(a) = 2c2 or c2 = f ′′(a)/2!.
Differentiating and evaluating once more leads to c3 = f ′′′(a)/3!.
Continue in the same manner.
Mark Woodard (Furman University) §8.7–Taylor and Maclaurin Series Fall 2007 5 / 23
Taylor’s Theorem
Proof.
Suppose f (x) = c0 + c1(x − a)1 + c2(x − a)2 + c3(x − a)3 + · · · insome interval about the point a.
Evaluating both sides at x = a shows that f (a) = c0; thus, c0 = f (a).
Differentiate both sides: f ′(x) = c1 + 2c2(x − a)1 + 3c3(x − a)2 + · · · .Evaluate both sides at x = a and note that f ′(a) = c1.
Likewise f ′′(x) = 2c2 + 3 · 2(x − a)1 + · · · . Upon evaluating bothsides at x = a, we see that f ′′(a) = 2c2 or c2 = f ′′(a)/2!.
Differentiating and evaluating once more leads to c3 = f ′′′(a)/3!.
Continue in the same manner.
Mark Woodard (Furman University) §8.7–Taylor and Maclaurin Series Fall 2007 5 / 23
Taylor’s Theorem
Proof.
Suppose f (x) = c0 + c1(x − a)1 + c2(x − a)2 + c3(x − a)3 + · · · insome interval about the point a.
Evaluating both sides at x = a shows that f (a) = c0; thus, c0 = f (a).
Differentiate both sides: f ′(x) = c1 + 2c2(x − a)1 + 3c3(x − a)2 + · · · .Evaluate both sides at x = a and note that f ′(a) = c1.
Likewise f ′′(x) = 2c2 + 3 · 2(x − a)1 + · · · . Upon evaluating bothsides at x = a, we see that f ′′(a) = 2c2 or c2 = f ′′(a)/2!.
Differentiating and evaluating once more leads to c3 = f ′′′(a)/3!.
Continue in the same manner.
Mark Woodard (Furman University) §8.7–Taylor and Maclaurin Series Fall 2007 5 / 23
Taylor’s Theorem
Proof.
Suppose f (x) = c0 + c1(x − a)1 + c2(x − a)2 + c3(x − a)3 + · · · insome interval about the point a.
Evaluating both sides at x = a shows that f (a) = c0; thus, c0 = f (a).
Differentiate both sides: f ′(x) = c1 + 2c2(x − a)1 + 3c3(x − a)2 + · · · .Evaluate both sides at x = a and note that f ′(a) = c1.
Likewise f ′′(x) = 2c2 + 3 · 2(x − a)1 + · · · . Upon evaluating bothsides at x = a, we see that f ′′(a) = 2c2 or c2 = f ′′(a)/2!.
Differentiating and evaluating once more leads to c3 = f ′′′(a)/3!.
Continue in the same manner.
Mark Woodard (Furman University) §8.7–Taylor and Maclaurin Series Fall 2007 5 / 23
Taylor’s Theorem
Proof.
Suppose f (x) = c0 + c1(x − a)1 + c2(x − a)2 + c3(x − a)3 + · · · insome interval about the point a.
Evaluating both sides at x = a shows that f (a) = c0; thus, c0 = f (a).
Differentiate both sides: f ′(x) = c1 + 2c2(x − a)1 + 3c3(x − a)2 + · · · .Evaluate both sides at x = a and note that f ′(a) = c1.
Likewise f ′′(x) = 2c2 + 3 · 2(x − a)1 + · · · . Upon evaluating bothsides at x = a, we see that f ′′(a) = 2c2 or c2 = f ′′(a)/2!.
Differentiating and evaluating once more leads to c3 = f ′′′(a)/3!.
Continue in the same manner.
Mark Woodard (Furman University) §8.7–Taylor and Maclaurin Series Fall 2007 5 / 23
Taylor’s Theorem
Proof.
Suppose f (x) = c0 + c1(x − a)1 + c2(x − a)2 + c3(x − a)3 + · · · insome interval about the point a.
Evaluating both sides at x = a shows that f (a) = c0; thus, c0 = f (a).
Differentiate both sides: f ′(x) = c1 + 2c2(x − a)1 + 3c3(x − a)2 + · · · .Evaluate both sides at x = a and note that f ′(a) = c1.
Likewise f ′′(x) = 2c2 + 3 · 2(x − a)1 + · · · . Upon evaluating bothsides at x = a, we see that f ′′(a) = 2c2 or c2 = f ′′(a)/2!.
Differentiating and evaluating once more leads to c3 = f ′′′(a)/3!.
Continue in the same manner.
Mark Woodard (Furman University) §8.7–Taylor and Maclaurin Series Fall 2007 5 / 23
Taylor’s Theorem
Definition
Given a function f and a point a, the series
f (x) =∞∑
n=0
f (n)(a)
n!(x − a)n
is called a Taylor series for f centered at a.
When a = 0, the series is called a Maclaurin series.
Mark Woodard (Furman University) §8.7–Taylor and Maclaurin Series Fall 2007 6 / 23
Taylor’s Theorem
Definition
Given a function f and a point a, the series
f (x) =∞∑
n=0
f (n)(a)
n!(x − a)n
is called a Taylor series for f centered at a.
When a = 0, the series is called a Maclaurin series.
Mark Woodard (Furman University) §8.7–Taylor and Maclaurin Series Fall 2007 6 / 23
Taylor’s Theorem
Definition
Given a function f and a point a, the series
f (x) =∞∑
n=0
f (n)(a)
n!(x − a)n
is called a Taylor series for f centered at a.
When a = 0, the series is called a Maclaurin series.
Mark Woodard (Furman University) §8.7–Taylor and Maclaurin Series Fall 2007 6 / 23
Taylor’s Theorem
Problem
Find the Maclaurin series for ex . What is its radius of convergence?
Solution
If f (x) = ex , then f (n)(x) = ex for all n ≥ 0.
Thus the coefficients for the Maclaurin series are
cn =f (n)(0)
n!=
1
n!
The Maclaurin series is thus
∞∑n=0
1
n!xn = 1 + x +
x2
2!+
x3
3!+ · · ·
It is easy to see that the radius of convergence is R = +∞.
Mark Woodard (Furman University) §8.7–Taylor and Maclaurin Series Fall 2007 7 / 23
Taylor’s Theorem
Problem
Find the Maclaurin series for ex . What is its radius of convergence?
Solution
If f (x) = ex , then f (n)(x) = ex for all n ≥ 0.
Thus the coefficients for the Maclaurin series are
cn =f (n)(0)
n!=
1
n!
The Maclaurin series is thus
∞∑n=0
1
n!xn = 1 + x +
x2
2!+
x3
3!+ · · ·
It is easy to see that the radius of convergence is R = +∞.
Mark Woodard (Furman University) §8.7–Taylor and Maclaurin Series Fall 2007 7 / 23
Taylor’s Theorem
Problem
Find the Maclaurin series for ex . What is its radius of convergence?
Solution
If f (x) = ex , then f (n)(x) = ex for all n ≥ 0.
Thus the coefficients for the Maclaurin series are
cn =f (n)(0)
n!=
1
n!
The Maclaurin series is thus
∞∑n=0
1
n!xn = 1 + x +
x2
2!+
x3
3!+ · · ·
It is easy to see that the radius of convergence is R = +∞.
Mark Woodard (Furman University) §8.7–Taylor and Maclaurin Series Fall 2007 7 / 23
Taylor’s Theorem
Problem
Find the Maclaurin series for ex . What is its radius of convergence?
Solution
If f (x) = ex , then f (n)(x) = ex for all n ≥ 0.
Thus the coefficients for the Maclaurin series are
cn =f (n)(0)
n!=
1
n!
The Maclaurin series is thus
∞∑n=0
1
n!xn = 1 + x +
x2
2!+
x3
3!+ · · ·
It is easy to see that the radius of convergence is R = +∞.
Mark Woodard (Furman University) §8.7–Taylor and Maclaurin Series Fall 2007 7 / 23
Taylor’s Theorem
Problem
Find the Maclaurin series for ex . What is its radius of convergence?
Solution
If f (x) = ex , then f (n)(x) = ex for all n ≥ 0.
Thus the coefficients for the Maclaurin series are
cn =f (n)(0)
n!=
1
n!
The Maclaurin series is thus
∞∑n=0
1
n!xn = 1 + x +
x2
2!+
x3
3!+ · · ·
It is easy to see that the radius of convergence is R = +∞.
Mark Woodard (Furman University) §8.7–Taylor and Maclaurin Series Fall 2007 7 / 23
Taylor’s Theorem
Problem
Find the Maclaurin series for ex . What is its radius of convergence?
Solution
If f (x) = ex , then f (n)(x) = ex for all n ≥ 0.
Thus the coefficients for the Maclaurin series are
cn =f (n)(0)
n!=
1
n!
The Maclaurin series is thus
∞∑n=0
1
n!xn = 1 + x +
x2
2!+
x3
3!+ · · ·
It is easy to see that the radius of convergence is R = +∞.
Mark Woodard (Furman University) §8.7–Taylor and Maclaurin Series Fall 2007 7 / 23
When does a function equal its Taylor series?
Definition
For each N ≥ 0, let
TN(x) =N∑
n=0
f (n)(a)
n!(x − a)n
This is called the Taylor polynomial of order N for f (x).
Let RN(x) = f (x)− TN(x). The function RN is called the remainder.
Theorem
If RN(x)→ 0 as N →∞, then
f (x) = limN→∞
TN(x) =∞∑
n=0
f (n)(a)
n!(x − a)n,
that is, f (x) is equal to its Taylor series.
Mark Woodard (Furman University) §8.7–Taylor and Maclaurin Series Fall 2007 8 / 23
When does a function equal its Taylor series?
Definition
For each N ≥ 0, let
TN(x) =N∑
n=0
f (n)(a)
n!(x − a)n
This is called the Taylor polynomial of order N for f (x).
Let RN(x) = f (x)− TN(x). The function RN is called the remainder.
Theorem
If RN(x)→ 0 as N →∞, then
f (x) = limN→∞
TN(x) =∞∑
n=0
f (n)(a)
n!(x − a)n,
that is, f (x) is equal to its Taylor series.
Mark Woodard (Furman University) §8.7–Taylor and Maclaurin Series Fall 2007 8 / 23
When does a function equal its Taylor series?
Definition
For each N ≥ 0, let
TN(x) =N∑
n=0
f (n)(a)
n!(x − a)n
This is called the Taylor polynomial of order N for f (x).
Let RN(x) = f (x)− TN(x). The function RN is called the remainder.
Theorem
If RN(x)→ 0 as N →∞, then
f (x) = limN→∞
TN(x) =∞∑
n=0
f (n)(a)
n!(x − a)n,
that is, f (x) is equal to its Taylor series.
Mark Woodard (Furman University) §8.7–Taylor and Maclaurin Series Fall 2007 8 / 23
When does a function equal its Taylor series?
Definition
For each N ≥ 0, let
TN(x) =N∑
n=0
f (n)(a)
n!(x − a)n
This is called the Taylor polynomial of order N for f (x).
Let RN(x) = f (x)− TN(x). The function RN is called the remainder.
Theorem
If RN(x)→ 0 as N →∞, then
f (x) = limN→∞
TN(x) =∞∑
n=0
f (n)(a)
n!(x − a)n,
that is, f (x) is equal to its Taylor series.
Mark Woodard (Furman University) §8.7–Taylor and Maclaurin Series Fall 2007 8 / 23
When does a function equal its Taylor series?
Remark
Our ability to faithfully represent a function by its Taylor series hinges onour ability to show that RN tends to 0 as N →∞. Our next theorem givesan important way to represent RN .
Theorem (Taylor’s Theorem)
If f has N + 1 derivatives in an interval I that contains the number a, thenfor x ∈ I there is a number z strictly between x and a such that theremainder term in the Taylor series can be expressed as
RN(x) =f (N+1)(z)
(N + 1)!(x − a)N+1.
Mark Woodard (Furman University) §8.7–Taylor and Maclaurin Series Fall 2007 9 / 23
When does a function equal its Taylor series?
Remark
Our ability to faithfully represent a function by its Taylor series hinges onour ability to show that RN tends to 0 as N →∞. Our next theorem givesan important way to represent RN .
Theorem (Taylor’s Theorem)
If f has N + 1 derivatives in an interval I that contains the number a, thenfor x ∈ I there is a number z strictly between x and a such that theremainder term in the Taylor series can be expressed as
RN(x) =f (N+1)(z)
(N + 1)!(x − a)N+1.
Mark Woodard (Furman University) §8.7–Taylor and Maclaurin Series Fall 2007 9 / 23
Important examples: ex , sin(x), cos(x)
Theorem
For each x ∈ R, xn/n!→ 0 as n→∞.
Problem
Show that ex , sin(x) and cos(x) are represented by their Maclaurin seriesfor all x.
Mark Woodard (Furman University) §8.7–Taylor and Maclaurin Series Fall 2007 10 / 23
Important examples: ex , sin(x), cos(x)
Theorem
For each x ∈ R, xn/n!→ 0 as n→∞.
Problem
Show that ex , sin(x) and cos(x) are represented by their Maclaurin seriesfor all x.
Mark Woodard (Furman University) §8.7–Taylor and Maclaurin Series Fall 2007 10 / 23
Important examples: ex , sin(x), cos(x)
Solution (The exponential function)
Let f (x) = ex . Then f (n)(x) = ex for all n ≥ 0. In particularf (n)(0) = 1 for all n ≥ 0.
The Maclaurin series for ex is thus
∞∑n=0
1
n!xn
Now we need to show that the series represents ex . To this end, fixx ∈ R. Then
RN(x) =ez
(N + 1)!xN+1
It is clear that this tends to 0 as N →∞. Thus the series representsex for all x.
Mark Woodard (Furman University) §8.7–Taylor and Maclaurin Series Fall 2007 11 / 23
Important examples: ex , sin(x), cos(x)
Solution (The sine function)
Let f (x) = sin(x). The function f and its derivatives will cycle asfollows: sin(x), cos(x), − sin(x), − cos(x), sin(x), etc.
The Maclaurin series for sin(x) is as follows:
x − x3
3!+
x5
5!− x7
7!+ · · · =
∞∑n=0
(−1)n
(2n + 1)!x2n+1.
Now we must show that the series represents the sine function. Fixx ∈ R and note that
RN(x) =f N+1(z)
(N + 1)!xN+1
The N + 1st derivative of f is one of the four functions listed above;thus, |RN(x)| ≤ |x |N+1/(N + 1)!. Since |RN(x)| → 0 as N →∞, wemay conclude that the series represents sine for all x ∈ R.
Mark Woodard (Furman University) §8.7–Taylor and Maclaurin Series Fall 2007 12 / 23
Important examples: ex , sin(x), cos(x)
Solution (The sine function)
Let f (x) = sin(x). The function f and its derivatives will cycle asfollows: sin(x), cos(x), − sin(x), − cos(x), sin(x), etc.
The Maclaurin series for sin(x) is as follows:
x − x3
3!+
x5
5!− x7
7!+ · · · =
∞∑n=0
(−1)n
(2n + 1)!x2n+1.
Now we must show that the series represents the sine function. Fixx ∈ R and note that
RN(x) =f N+1(z)
(N + 1)!xN+1
The N + 1st derivative of f is one of the four functions listed above;thus, |RN(x)| ≤ |x |N+1/(N + 1)!. Since |RN(x)| → 0 as N →∞, wemay conclude that the series represents sine for all x ∈ R.
Mark Woodard (Furman University) §8.7–Taylor and Maclaurin Series Fall 2007 12 / 23
Important examples: ex , sin(x), cos(x)
Solution (The sine function)
Let f (x) = sin(x). The function f and its derivatives will cycle asfollows: sin(x), cos(x), − sin(x), − cos(x), sin(x), etc.
The Maclaurin series for sin(x) is as follows:
x − x3
3!+
x5
5!− x7
7!+ · · · =
∞∑n=0
(−1)n
(2n + 1)!x2n+1.
Now we must show that the series represents the sine function. Fixx ∈ R and note that
RN(x) =f N+1(z)
(N + 1)!xN+1
The N + 1st derivative of f is one of the four functions listed above;thus, |RN(x)| ≤ |x |N+1/(N + 1)!. Since |RN(x)| → 0 as N →∞, wemay conclude that the series represents sine for all x ∈ R.
Mark Woodard (Furman University) §8.7–Taylor and Maclaurin Series Fall 2007 12 / 23
Important examples: ex , sin(x), cos(x)
Solution (The sine function)
Let f (x) = sin(x). The function f and its derivatives will cycle asfollows: sin(x), cos(x), − sin(x), − cos(x), sin(x), etc.
The Maclaurin series for sin(x) is as follows:
x − x3
3!+
x5
5!− x7
7!+ · · · =
∞∑n=0
(−1)n
(2n + 1)!x2n+1.
Now we must show that the series represents the sine function. Fixx ∈ R and note that
RN(x) =f N+1(z)
(N + 1)!xN+1
The N + 1st derivative of f is one of the four functions listed above;thus, |RN(x)| ≤ |x |N+1/(N + 1)!. Since |RN(x)| → 0 as N →∞, wemay conclude that the series represents sine for all x ∈ R.
Mark Woodard (Furman University) §8.7–Taylor and Maclaurin Series Fall 2007 12 / 23
Important examples: ex , sin(x), cos(x)
Solution (The sine function)
Let f (x) = sin(x). The function f and its derivatives will cycle asfollows: sin(x), cos(x), − sin(x), − cos(x), sin(x), etc.
The Maclaurin series for sin(x) is as follows:
x − x3
3!+
x5
5!− x7
7!+ · · · =
∞∑n=0
(−1)n
(2n + 1)!x2n+1.
Now we must show that the series represents the sine function. Fixx ∈ R and note that
RN(x) =f N+1(z)
(N + 1)!xN+1
The N + 1st derivative of f is one of the four functions listed above;thus, |RN(x)| ≤ |x |N+1/(N + 1)!. Since |RN(x)| → 0 as N →∞, wemay conclude that the series represents sine for all x ∈ R.
Mark Woodard (Furman University) §8.7–Taylor and Maclaurin Series Fall 2007 12 / 23
Important examples: ex , sin(x), cos(x)
Solution (The cosine function)
The analysis for the cosine is almost identical to that of the sinefunction. The Maclaurin series is
1− x2
2!+
x4
4!− x6
6!+ · · · =
∞∑n=0
(−1)n
(2n)!x2n.
This series represents cosine for all x ∈ R. The proof is almostidentical to that given above for sine.
Mark Woodard (Furman University) §8.7–Taylor and Maclaurin Series Fall 2007 13 / 23
Important examples: ex , sin(x), cos(x)
Solution (The cosine function)
The analysis for the cosine is almost identical to that of the sinefunction. The Maclaurin series is
1− x2
2!+
x4
4!− x6
6!+ · · · =
∞∑n=0
(−1)n
(2n)!x2n.
This series represents cosine for all x ∈ R. The proof is almostidentical to that given above for sine.
Mark Woodard (Furman University) §8.7–Taylor and Maclaurin Series Fall 2007 13 / 23
Important examples: ex , sin(x), cos(x)
Solution (The cosine function)
The analysis for the cosine is almost identical to that of the sinefunction. The Maclaurin series is
1− x2
2!+
x4
4!− x6
6!+ · · · =
∞∑n=0
(−1)n
(2n)!x2n.
This series represents cosine for all x ∈ R. The proof is almostidentical to that given above for sine.
Mark Woodard (Furman University) §8.7–Taylor and Maclaurin Series Fall 2007 13 / 23
Important examples: ex , sin(x), cos(x)
Problem
Find the sum of the series S =2
1!+
4
2!+
8
3!+
16
4!+ · · ·.
Solution
We can see that this is close to the series for e2:
e2 =∞∑
n=0
2n
n!= 1 +
2
1!+
22
2!+
23
3!+ · · · = 1 + S .
Thus S = e2 − 1.
Mark Woodard (Furman University) §8.7–Taylor and Maclaurin Series Fall 2007 14 / 23
Important examples: ex , sin(x), cos(x)
Problem
Find the sum of the series S =2
1!+
4
2!+
8
3!+
16
4!+ · · ·.
Solution
We can see that this is close to the series for e2:
e2 =∞∑
n=0
2n
n!= 1 +
2
1!+
22
2!+
23
3!+ · · · = 1 + S .
Thus S = e2 − 1.
Mark Woodard (Furman University) §8.7–Taylor and Maclaurin Series Fall 2007 14 / 23
Newton’s binomial theorem
Definition (Falling factorial)
Given x ∈ R and an integer n ≥ 1, let
(x)n = (x)(x − 1)(x − 2) · · · (x − n + 1)︸ ︷︷ ︸n terms
Let (x)0 = 1 for completeness.
Definition (Binomial coefficients)
Given x ∈ R and an integer n ≥ 0, let(x
n
)=
(x)n
n!
Mark Woodard (Furman University) §8.7–Taylor and Maclaurin Series Fall 2007 15 / 23
Newton’s binomial theorem
Definition (Falling factorial)
Given x ∈ R and an integer n ≥ 1, let
(x)n = (x)(x − 1)(x − 2) · · · (x − n + 1)︸ ︷︷ ︸n terms
Let (x)0 = 1 for completeness.
Definition (Binomial coefficients)
Given x ∈ R and an integer n ≥ 0, let(x
n
)=
(x)n
n!
Mark Woodard (Furman University) §8.7–Taylor and Maclaurin Series Fall 2007 15 / 23
Newton’s binomial theorem
Problem
Evaluate(−1/2
3
).
Solution(−1/2
3
)=
(−1/2)3
3!=
(−1/2)(−3/2)(−5/2)
6=−15
48= − 5
16
Mark Woodard (Furman University) §8.7–Taylor and Maclaurin Series Fall 2007 16 / 23
Newton’s binomial theorem
Problem
Evaluate(−1/2
3
).
Solution(−1/2
3
)=
(−1/2)3
3!=
(−1/2)(−3/2)(−5/2)
6=−15
48= − 5
16
Mark Woodard (Furman University) §8.7–Taylor and Maclaurin Series Fall 2007 16 / 23
Newton’s binomial theorem
Problem
Let k ∈ R and let f (x) = (1 + x)k . Show that the Maclaurin series for f is
∞∑n=0
(k
n
)xn
Find the radius of convergence of the series.
Theorem (Newton’s binomial theorem)
If k is any real number and |x | < 1, then
(1 + x)k =∞∑
n=0
(k
n
)xn.
Mark Woodard (Furman University) §8.7–Taylor and Maclaurin Series Fall 2007 17 / 23
Newton’s binomial theorem
Problem
Let k ∈ R and let f (x) = (1 + x)k . Show that the Maclaurin series for f is
∞∑n=0
(k
n
)xn
Find the radius of convergence of the series.
Theorem (Newton’s binomial theorem)
If k is any real number and |x | < 1, then
(1 + x)k =∞∑
n=0
(k
n
)xn.
Mark Woodard (Furman University) §8.7–Taylor and Maclaurin Series Fall 2007 17 / 23
Newton’s binomial theorem
Problem
Find the Maclaurin series for√
1− x. For which x does the seriesrepresent the function?
Solution
We have
√1− x = (1 + (−x))1/2 =
∞∑n=0
(1/2
n
)(−x)n =
∞∑n=0
(−1)n
(1/2
n
)xn.
This is true for all | − x | < 1 or for |x | < 1.
Mark Woodard (Furman University) §8.7–Taylor and Maclaurin Series Fall 2007 18 / 23
Newton’s binomial theorem
Problem
Find the Maclaurin series for√
1− x. For which x does the seriesrepresent the function?
Solution
We have
√1− x = (1 + (−x))1/2 =
∞∑n=0
(1/2
n
)(−x)n =
∞∑n=0
(−1)n
(1/2
n
)xn.
This is true for all | − x | < 1 or for |x | < 1.
Mark Woodard (Furman University) §8.7–Taylor and Maclaurin Series Fall 2007 18 / 23
Newton’s binomial theorem
Problem
Find the Maclaurin series for 1/√
9 + x. For which x does the seriesrepresent the function?
Solution
We have
1/√
9 + x = (9 + x)−1/2 =1
3
(1 + (x/9)
)−1/2
=∞∑
n=0
1
3
(−1/2
n
)(x/9)n
=∞∑
n=0
1
3 · 9n
(−1/2
n
)xn.
This series converges for |x/3| < 1 or |x | < 3.
Mark Woodard (Furman University) §8.7–Taylor and Maclaurin Series Fall 2007 19 / 23
Newton’s binomial theorem
Problem
Find the Maclaurin series for 1/√
9 + x. For which x does the seriesrepresent the function?
Solution
We have
1/√
9 + x = (9 + x)−1/2 =1
3
(1 + (x/9)
)−1/2
=∞∑
n=0
1
3
(−1/2
n
)(x/9)n
=∞∑
n=0
1
3 · 9n
(−1/2
n
)xn.
This series converges for |x/3| < 1 or |x | < 3.
Mark Woodard (Furman University) §8.7–Taylor and Maclaurin Series Fall 2007 19 / 23
Some uses of Taylor series
Problem
Express I =∫ 10 e−x2
dx as an infinite series. Approximate the integral towithin an error of .001.
Solution
We integrate term-by-term with the series for eu with u = −x2. Thus
I =∞∑
n=0
(−1)n
∫ 1
0
x2n
n!dx =
∞∑n=0
(−1)n 1
(2n + 1)n!= 1−1
3+
1
10− 1
42+· · ·
Since the series alternates, we can estimate the error using thealternating series test. The term 1/((2n + 1)n!) is first less than .001for n = 5. Thus
I ≈ 1− 1
3+
1
10− 1
42+
1
216= .747486772 . . .
Mark Woodard (Furman University) §8.7–Taylor and Maclaurin Series Fall 2007 20 / 23
Some uses of Taylor series
Problem
Express I =∫ 10 e−x2
dx as an infinite series. Approximate the integral towithin an error of .001.
Solution
We integrate term-by-term with the series for eu with u = −x2. Thus
I =∞∑
n=0
(−1)n
∫ 1
0
x2n
n!dx =
∞∑n=0
(−1)n 1
(2n + 1)n!= 1−1
3+
1
10− 1
42+· · ·
Since the series alternates, we can estimate the error using thealternating series test. The term 1/((2n + 1)n!) is first less than .001for n = 5. Thus
I ≈ 1− 1
3+
1
10− 1
42+
1
216= .747486772 . . .
Mark Woodard (Furman University) §8.7–Taylor and Maclaurin Series Fall 2007 20 / 23
Some uses of Taylor series
Problem
Express I =∫ 10 e−x2
dx as an infinite series. Approximate the integral towithin an error of .001.
Solution
We integrate term-by-term with the series for eu with u = −x2. Thus
I =∞∑
n=0
(−1)n
∫ 1
0
x2n
n!dx =
∞∑n=0
(−1)n 1
(2n + 1)n!= 1−1
3+
1
10− 1
42+· · ·
Since the series alternates, we can estimate the error using thealternating series test. The term 1/((2n + 1)n!) is first less than .001for n = 5. Thus
I ≈ 1− 1
3+
1
10− 1
42+
1
216= .747486772 . . .
Mark Woodard (Furman University) §8.7–Taylor and Maclaurin Series Fall 2007 20 / 23
Some uses of Taylor series
Problem
Express I =∫ 10 e−x2
dx as an infinite series. Approximate the integral towithin an error of .001.
Solution
We integrate term-by-term with the series for eu with u = −x2. Thus
I =∞∑
n=0
(−1)n
∫ 1
0
x2n
n!dx =
∞∑n=0
(−1)n 1
(2n + 1)n!= 1−1
3+
1
10− 1
42+· · ·
Since the series alternates, we can estimate the error using thealternating series test. The term 1/((2n + 1)n!) is first less than .001for n = 5. Thus
I ≈ 1− 1
3+
1
10− 1
42+
1
216= .747486772 . . .
Mark Woodard (Furman University) §8.7–Taylor and Maclaurin Series Fall 2007 20 / 23
Some uses of Taylor series
Problem
Evaluate the limit
limx→0
sin(x)− x
x3
using the Maclaurin series for sin(x).
Solution
We can replace sin(x) by its Maclaurin series:
limx→0
sin(x)− x
x3= lim
x→0
(x − x3/6 + x5/120− · · ·
)− x
x3
= limx→0
(−1
6+
x2
120− · · ·
)= −1
6
Mark Woodard (Furman University) §8.7–Taylor and Maclaurin Series Fall 2007 21 / 23
Some uses of Taylor series
Problem
Evaluate the limit
limx→0
sin(x)− x
x3
using the Maclaurin series for sin(x).
Solution
We can replace sin(x) by its Maclaurin series:
limx→0
sin(x)− x
x3= lim
x→0
(x − x3/6 + x5/120− · · ·
)− x
x3
= limx→0
(−1
6+
x2
120− · · ·
)= −1
6
Mark Woodard (Furman University) §8.7–Taylor and Maclaurin Series Fall 2007 21 / 23
Multiplication and division of series
Problem
Find the first three non-zero terms of the Maclaurin series for (1− x)−1ex .
Solution
We will multiply the power series and collect the terms up to x2:
1
1− xex = (1 + x + x2 + x3 + · · · )(1 + x + x2/2 + x3/6 + · · · )
= (1 + x + x2/2) + (x + x2) + (x2/2) + · · ·
= 1 + 2x +5
2x2 + · · ·
Mark Woodard (Furman University) §8.7–Taylor and Maclaurin Series Fall 2007 22 / 23
Multiplication and division of series
Problem
Find the first three non-zero terms of the Maclaurin series for (1− x)−1ex .
Solution
We will multiply the power series and collect the terms up to x2:
1
1− xex = (1 + x + x2 + x3 + · · · )(1 + x + x2/2 + x3/6 + · · · )
= (1 + x + x2/2) + (x + x2) + (x2/2) + · · ·
= 1 + 2x +5
2x2 + · · ·
Mark Woodard (Furman University) §8.7–Taylor and Maclaurin Series Fall 2007 22 / 23
Multiplication and division of series
Problem
Find the first three non-zero terms of the Maclaurin series for tan(x).
Solution
Using long division, we find that
tan(x) =sin(x)
cos(x)
=x − x3/3! + x5/5!− · · ·1− x2/2! + x4/4!− · · ·
= x +1
3x3 +
2
15x5 + · · ·
Mark Woodard (Furman University) §8.7–Taylor and Maclaurin Series Fall 2007 23 / 23
Multiplication and division of series
Problem
Find the first three non-zero terms of the Maclaurin series for tan(x).
Solution
Using long division, we find that
tan(x) =sin(x)
cos(x)
=x − x3/3! + x5/5!− · · ·1− x2/2! + x4/4!− · · ·
= x +1
3x3 +
2
15x5 + · · ·
Mark Woodard (Furman University) §8.7–Taylor and Maclaurin Series Fall 2007 23 / 23