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Ch. 11 - Power Series
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
All textbooksBriggs 3rd EditionCh. 11 - Power SeriesProblem 11.2.73
Chapter 11, Problem 11.2.73

Exponential function In Section 11.3, we show that the power series for the exponential function centered at 0 is


eˣ = ∑ₖ₌₀∞ (xᵏ)/k!, for −∞ < x < ∞


Use the methods of this section to find the power series centered at 0 for the following functions. Give the interval of convergence for the resulting series.


f(x) = e⁻³ˣ

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Recall the power series expansion for the exponential function centered at 0: \[e^{x} = \sum_{k=0}^{\infty} \frac{x^{k}}{k!}\] which converges for all real numbers \[x\] (i.e., the interval of convergence is \[(-\infty, \infty)\]).
To find the power series for \[f(x) = e^{-3x}\], substitute \[-3x\] in place of \[x\] in the original series. This gives: \[e^{-3x} = \sum_{k=0}^{\infty} \frac{(-3x)^{k}}{k!}\].
Rewrite the terms inside the summation to separate powers of \[x\] and constants: \[e^{-3x} = \sum_{k=0}^{\infty} \frac{(-3)^{k} x^{k}}{k!}\].
Since the original exponential series converges for all real numbers, replacing \[x\] by \[-3x\] does not change the radius of convergence. Therefore, the interval of convergence remains \[(-\infty, \infty)\].
Summarize the power series for \[f(x)\] as \[f(x) = \sum_{k=0}^{\infty} \frac{(-3)^{k}}{k!} x^{k}\] with interval of convergence \[(-\infty, \infty)\].

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Power Series Representation of Functions

A power series expresses a function as an infinite sum of terms involving powers of the variable, typically centered at a point (here, 0). For example, the exponential function eˣ can be written as ∑ₖ₌₀∞ (xᵏ)/k!. Understanding this allows us to represent more complex functions by manipulating known series.
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Representing Functions as Power Series

Substitution in Power Series

To find the power series of a function like f(x) = e⁻³ˣ, we substitute the inner expression (−3x) into the known series for eˣ. This means replacing x by −3x in the series, resulting in ∑ₖ₌₀∞ ((−3x)ᵏ)/k!. This technique helps generate new series from existing ones.
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Interval of Convergence

The interval of convergence is the set of x-values for which the power series converges to the function. For the exponential series, the interval is all real numbers (−∞, ∞). After substitution, it is important to verify if the interval changes, ensuring the series accurately represents the function within that domain.
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Related Practice
Textbook Question

L'Hôpital's Rule by Taylor series Suppose f and g have Taylor series about the point a.

a. If f(a) = g(a) = 0 and g′(a) ≠ 0, evaluate lim ₓ→ₐ f(x)/g(x) by expanding f and g in their Taylor series. Show that the result is consistent withl’Hôpital’s Rule.

b. If f(a) = g(a) =f′(a) = g′(a) = 0 and g′′(a) ≠ 0, evaluate lim ₓ→ₐ f(x)/g(x) by expanding f and g in their Taylor series. Show that the result is consistent with two applications of 1'Hôpital's Rule.

Textbook Question

{Use of Tech} Maximum error Use the remainder term to find a bound on the error in the following approximations on the given interval. Error bounds are not unique.


tan x ≈ x on [−π/6, π/6]

Textbook Question

Radius and interval of convergence Determine the radius and interval of convergence of the following power series.


∑ₖ₌₁∞ (3x + 2)ᵏ/k

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Textbook Question

Remainders Find the remainder Rₙ for the nth−order Taylor polynomial centered at a for the given functions. Express the result for a general value of n.


f(x) = sin x, a = π/2

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Textbook Question

Inverse sine Given the power series

1/√(1 − x²) = 1 + (1/2)x² + (1 ⋅ 3)/(2 ⋅ 4) x⁴ + (1 ⋅ 3 ⋅ 5)/(2 ⋅ 4 ⋅ 6) x⁶ +⋯

for −1<x<1, find the power series for f(x) = sin ⁻¹ x centered at 0.

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Textbook Question

Limits Evaluate the following limits using Taylor series.

lim ₓ→₁ (x 1)/(ln x)