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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 11.3.11c
Chapter 11, Problem 11.3.11c

Taylor series and interval of convergence


c. Determine the interval of convergence of the series.


f(x) = e⁻ˣ, a=0

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Recall that the Taylor series for the function \(f(x) = e^{-x}\) centered at \(a=0\) is given by the Maclaurin series expansion: \[f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} x^n\] Since \(f(x) = e^{-x}\), all derivatives are of the form \(f^{(n)}(x) = (-1)^n e^{-x}\), so at \(x=0\), \(f^{(n)}(0) = (-1)^n\).
Write the Taylor series explicitly using the derivatives at 0: \[e^{-x} = \sum_{n=0}^{\infty} \frac{(-1)^n}{n!} x^n = \sum_{n=0}^{\infty} \frac{(-x)^n}{n!}\]
To find the interval of convergence, apply the Ratio Test to the general term of the series: Let \(a_n = \frac{(-x)^n}{n!}\). Then consider \[L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \left| \frac{(-x)^{n+1} / (n+1)!}{(-x)^n / n!} \right| = \lim_{n \to \infty} \frac{|x|}{n+1}\]
Evaluate the limit \(L\): Since \(\lim_{n \to \infty} \frac{|x|}{n+1} = 0\) for all real \(x\), the Ratio Test tells us the series converges for all \(x \in \mathbb{R}\).
Conclude that the interval of convergence is the entire real line: \[(-\infty, \infty)\] This means the Taylor series for \(e^{-x}\) converges for every real number \(x\).

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Key Concepts

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

Taylor Series Expansion

A Taylor series represents a function as an infinite sum of terms calculated from the derivatives of the function at a single point. For f(x) = e^{-x} at a = 0, the series is formed using derivatives evaluated at 0, allowing approximation of the function near that point.
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Interval of Convergence

The interval of convergence is the set of x-values for which the Taylor series converges to the function. Determining this interval involves testing the series for convergence using methods like the ratio or root test, ensuring the series accurately represents the function within that range.
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Ratio Test for Convergence

The ratio test is a method to determine the convergence of an infinite series by examining the limit of the ratio of successive terms. If this limit is less than one, the series converges absolutely; if greater than one, it diverges. This test is commonly used to find the radius and interval of convergence for power series.
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Related Practice
Textbook Question

Sine integral function The function Si(x) = ∫₀ˣ f(t) dt, where f(t) = {(sin t)/t if t ≠ 0, 1 if t = 0, is called the sine integral function.

c. Approximate Si(0.5) and Si(1). Use enough terms of the series so the error in the approximation does not exceed 10⁻³.

Textbook Question

Taylor series and interval of convergence


c. Determine the interval of convergence of the series.


f(x)=2/(1−x)³, a=0

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Textbook Question

{Use of Tech} Small argument approximations Consider the following common approximations when x is near zero. 


b. Estimate f(0.2) and give a bound on the error in the approximation.


f(x) =√(1+x) ≈ 1 + x/2

Textbook Question

Taylor series


b. Write the power series using summation notation.


f(x)=sin x, a = π/2

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Textbook Question

Explain why or why not Determine whether the following statements are true and give an explanation or counterexample.


b. Let f(x)=x⁵−1 The Taylor polynomial for f of order 10 centered at 0 is f itself.

Textbook Question

Explain why or why not Determine whether the following statements are true and give an explanation or counterexample.


c. ∑ₖ₌₀∞ (ln 2)ᵏ/k! = 2