- •Functions
- •The Concept of a Function
- •Trigonometric Functions
- •Inverse Trigonometric Functions
- •Logarithmic, Exponential and Hyperbolic Functions
- •Limits and Continuity
- •Introductory Examples
- •Continuity Examples
- •Linear Function Approximations
- •Limits and Sequences
- •Properties of Continuous Functions
- •The Derivative
- •The Chain Rule
- •Higher Order Derivatives
- •Mathematical Applications
- •Antidifferentiation
- •Linear Second Order Homogeneous Differential Equations
- •Linear Non-Homogeneous Second Order Differential Equations
- •Area Approximation
- •Integration by Substitution
- •Integration by Parts
- •Logarithmic, Exponential and Hyperbolic Functions
- •The Riemann Integral
- •Volumes of Revolution
- •Arc Length and Surface Area
- •Techniques of Integration
- •Integration by formulae
- •Integration by Substitution
- •Integration by Parts
- •Trigonometric Integrals
- •Trigonometric Substitutions
- •Integration by Partial Fractions
- •Fractional Power Substitutions
- •Numerical Integration
- •Integrals over Unbounded Intervals
- •Discontinuities at End Points
- •Improper Integrals
- •Sequences
- •Monotone Sequences
- •Infinite Series
- •Series with Positive Terms
- •Alternating Series
- •Power Series
- •Taylor Polynomials and Series
- •Applications
- •Parabola
- •Ellipse
- •Hyperbola
- •Polar Coordinates
- •Graphs in Polar Coordinates
- •Areas in Polar Coordinates
- •Parametric Equations
354 CHAPTER 8. INFINITE SERIES
Exercises 8.4 In problem 1–12, determine the Taylor series expansion for each function f about the given value of a.
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f(x) = e−2x, a = 0 |
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f(x) = cos(3x), a = 0 |
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f(x) = ln(x), a = 1 |
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f(x) = (1 + x)−2, a = 0 |
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5. |
f(x) = (1 + x)−3/2, a = 0 |
6. |
f(x) = ex, a = 2 |
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8. |
f(x) = cos x, a = |
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f(x) = sin x, a = |
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f(x) = x1/3, a = 8 |
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f(x) = cos x − 2 |
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f(x) = sin x − 2 |
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In problems 13-20, determine f(k)(a) |
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13. |
f(x) = ex2 , a = 0, n = 3 |
14. |
f(x) = x2e−x, a = 0, n = 3 |
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f(x) = |
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f(x) = arctan x, a = 0, n = 3 |
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1 − x2 |
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f(x) = e2x cos 3x, a = 0, n = 4 |
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f(x) = arcsin x, a = 0, n = 3 |
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f(x) = tan x, a = 0, n = 3 |
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f(x) = (1 + x)1/2, a = 0, n = 5 |
8.7Taylor Polynomials and Series
Theorem 8.7.1 (Taylor’s Theorem) Suppose that f, f0, · · · , f(n+1) are all continuous for all x such that |x − a| < R. Then there exists some c between a and x such that
f(x) = Pn(x) + Rn(x)
where
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(x − a)k |
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, Rn(x) = f(n+1)(c) |
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8.7. TAYLOR POLYNOMIALS AND SERIES |
355 |
The polynomial Pn(x) is called the nth degree Taylor polynomial approximation of f. The term Rn(x) is called the Lagrange form of the remainder.
Proof. We define a function g of a variable z such that
g(z) = [f(x) |
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f(z)] |
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(z)(x − z) |
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f00(z)(x − z)2 |
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Then
( n f(k)(a) ) X
g(a) = f(x) − (x − a)k + Rn(x) = 0, k!
k=0
and
g(x) = f(x) − f(x) = 0.
By the Mean Value Theorem for derivatives there exists some c between a and x such that g0(c) = 0. But
g0(z) = −f0(z) − [−f0(z) + f00(z)(x − z)] − −f00(z)(x − z) + |
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000( |
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z)2 |
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f(n+1)(z)(x − z)n |
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g0(c) = 0 = |
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Therefore, |
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R |
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f(n+1)(c) |
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(x − a)n+1 |
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as required. This completes the proof of this theorem.
Theorem 8.7.2 (Binomial Series) If m is a real number and |x| < 1, then
∞
(1 + x)m = 1 + X m(m − 1) · · · (m − k + 1) xk k!
m(m − 1) 2!
356 |
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CHAPTER 8. INFINITE SERIES |
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This series is called the binomial series. If we use the notation |
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m m |
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is called the binomial coe cient and |
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then k |
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If m is a natural number, then we get the binomial expansion |
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Proof. Let f(x) = (1 + x)m. Then for all natural numbers n,
f0(x) = m(1 + x)m−1, f00(x) = m(m − 1)(1 + x)m−2, · · · , f(n)(x) = m(m − 1) · · · (m − n + 1)(1 + x)m−n.
Thus, f(n)(0) = m(m − 1) · · · (m − n + 1), and
∞
X m(m − 1)(m − 2) · · · (m − n + 1) xn
n!
n=0
=X mn xn
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where 0 = 1 and |
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binomial coe cient. By the ratio test we get |
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lim |
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(m |
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n + 1)xn |
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8.7. TAYLOR POLYNOMIALS AND SERIES |
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and, hence, the series converges for |x| < 1. This completes the proof of the theorem.
Theorem 8.7.3 The following power series expansions of functions are valid.
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k! , e−x = 1 + |
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sin x = |
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cos x = |
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sinh x = |
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cosh x = |
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ln(1 + x) = |
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arctan x = |
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10. arcsin x = k=0 |
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358 CHAPTER 8. INFINITE SERIES
Proof.
Part 1. By the geometric series expansion, for all |x| < 1, we have
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xk and |
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Part 2. |
If f(x) = ex, then f(n)(x) = ex |
and f(n)(0) = 1 for each n = |
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0, 1, 2, · · · . Thus |
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∞ |
xn |
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X |
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ex = |
n! |
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n=0
By the ratio test the series converges for all x.
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n! |
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1 |
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(n + 1)! · |
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n + 1 |
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lim |
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Part 3. Let f(x) = sin |
x. Then f0(x) = cos x, |
f00(x) = |
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sin x, f |
(x) = |
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− cos x and f |
(4) |
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(x) = sin x. It follows that, for each n = 0, 1, 2, 3, · · · , we |
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have |
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f(4n)(0) = 0, f(4n+1)(0) = 1, |
f(4n+2)(0) = 0 |
and f(4n+3)(0) = −1. |
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Hence, |
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x3 |
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x5 |
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sin x = x − |
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3! |
5! |
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∞
X
=(−1)n
n=0
x2n+1
(2n + 1)!.
By the ratio test, the series converges for all |x| < ∞:
n→∞ |
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− |
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n+1 |
x2n+3 |
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(2n + 1)! |
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(2n + 3)! |
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x2n+1 |
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lim |
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1) |
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n→∞ |
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Part 4. By term-by-term di erentiation we get
∞ |
x2n |
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cos x = (sin x)0 = X(−1)n |
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, |x| < ∞. |
(2n)! |
n=0
8.7. TAYLOR POLYNOMIALS AND SERIES |
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359 |
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Part 5. For all |x| < ∞, we get |
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1 |
(ex − e−x) |
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sinh x = |
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= 2 |
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n=0 |
n! |
− n=0(−1)n n! |
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∞ |
xn |
∞ |
xn |
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X |
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X |
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∞ x2n+1
X
=(2n + 1)!.
n=0
Part 6. By di erentiating term-by-term, we get
∞ |
x2n |
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X |
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cosh x = (sinh x)0 = |
(2n)! |
, l |x| < ∞. |
n=0 |
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Part 7. For each |x| < 1, by performing term by integration, we get
ln(1 + x) = Z0 |
x |
1 + x dx |
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1 |
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= |
∞ |
∞ |
(−1)nxn! dx |
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Z |
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X |
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0n=0
∞ |
xn+1 |
X |
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=(−1)n n + 1.
n=0
Part 8. By Part 7, for all |x| < 1, we get
1 |
ln |
1 + x |
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[ln(1 + x) − ln(1 − x)] |
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2 |
1 |
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x |
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2 |
"n=0 |
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n + 1 |
− n=0 |
− |
n + 1 |
# |
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= 1 |
∞ |
( 1)n |
xn+1 |
∞ |
( 1)n |
(−x)n+1 |
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X |
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X |
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= 2 |
"n=0 |
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(n−+ 1 |
(1 − (−1)n+1)xn+1# |
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1 |
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∞ |
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1)n |
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X |
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∞ x2k+1
X
=2k + 1.
k=0
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1 + x |
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Recall that arctanh x = |
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ln |
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2 |
1 − x |