Explanation Given: The function f ( x ) = e − x . Result used: (1) Ratio test : Let ∑ k = 1 ∞ a k be an infinite series and let r = lim k → ∞ | a k + 1 a k | . If r < 1 , then the series ∑ k = 1 ∞ a k is converges. Calculation: Obtain the power series for the function f ( x ) = e − x . Substitute − x for x in e x = ∑ k = 0 ∞ x k k ! . e − x = ∑ k = 0 ∞ ( − x ) k k ! Hence, the power series representation of e − x is ∑ k = 0 ∞ ( − 1 ) k ( x ) k k ! . The interval of convergence of ∑ k = 0 ∞ ( − 1 ) k ( x ) k k ! is computed as follows, Let a k = ( − 1 ) k ( x ) k k ! . Then, a k + 1 = ( − 1 ) k + 1 ( x ) k + 1 ( k + 1 ) ! . Use the Ratio test to compute the value | a k + 1 a k |

Single Variable Calculus: Early Transcendentals & Student Solutions Manual, Single Variable for Calculus: Early Transcendentals & MyLab Math -- Valuepack Access Card Package

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ISBN: 9780133941760

Author: William L. Briggs, Lyle Cochran, Bernard Gillett

Publisher: PEARSON

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Chapter 9.2, Problem 69E

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To find: The power series for the given function and interval of convergence by using the exponential function centered at 0 is ex=∑k=0∞xkk!, for−∞<x<∞.

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Chapter 9.2, Problem 68E

Chapter 9.2, Problem 70E

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Question 4 Find an equation of (a) The plane through the point (2, 0, 1) and perpendicular to the line x = y=2t, z=3+4t. 3t, (b) The plane through the point (3, −2, 8) and parallel to the plane z = x+y. (c) The plane that contains the line x = parallel to the plane 5x + 2y + z = 1. 1+t, y2t, z = 43t and is (d) The plane that passes through the point (1,2,3) and contains the line x = 3t, y=1+t, and z = 2 – t. (e) The plane that contains the lines L₁ : x = 1 + t, y = 1 − t, z = = L2 x 2s, y = s, z = 2. 2t and

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Chapter 9 Solutions

Single Variable Calculus: Early Transcendentals & Student Solutions Manual, Single Variable for Calculus: Early Transcendentals & MyLab Math -- Valuepack Access Card Package

Ch. 9.1 - Prob. 1QC Ch. 9.1 - Prob. 2QC Ch. 9.1 - Prob. 3QC Ch. 9.1 - Prob. 4QC Ch. 9.1 - Prob. 5QC Ch. 9.1 - In Example 7, find an approximate upper bound for...Ch. 9.1 - Suppose you use a second-order Taylor polynomial...Ch. 9.1 - Does the accuracy of an approximation given by a...Ch. 9.1 - The first three Taylor polynomials for f(x)=1+x...Ch. 9.1 - Prob. 4E

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The power series for the given function and interval of convergence by using the exponential function centered at 0 is e x = ∑ k = 0 ∞ x k k ! , for − ∞ < x < ∞ .

Solution Summary: The author explains how to find the power series for the function f(x)=e-x.

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To find: The power series for the given function and interval of convergence by using the exponential function centered at 0 is ex=∑k=0∞xkk!, for−∞<x<∞.

Want to see the full answer?

Chapter 9 Solutions

The power series for the given function and interval of convergence by using the exponential function centered at 0 is e x = ∑ k = 0 ∞ x k k ! , for − ∞ &lt; x &lt; ∞ .

Solution Summary: The author explains how to find the power series for the function f(x)=e-x.

Videos

To find: The power series for the given function and interval of convergence by using the exponential function centered at 0 is ex=∑k=0∞xkk!, for−∞<x<∞.

Want to see the full answer?

Chapter 9 Solutions

The power series for the given function and interval of convergence by using the exponential function centered at 0 is e x = ∑ k = 0 ∞ x k k ! , for − ∞ < x < ∞ .