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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
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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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.

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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]

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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

Working with binomial series Use properties of power series, substitution, and factoring to find the first four nonzero terms of the Maclaurin series for the following functions. Use the Maclaurin series


(1 + x)⁻² = 1 − 2x + 3x² − 4x³ + ⋯, for −1 < x < 1.


(1 + 4x)⁻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)

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