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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 11.3.65

Remainders Find the remainder in the Taylor series centered at the point a for the following functions. Then show that lim ₙ→∞ Rₙ(x)=0, for all x in the interval of convergence.


f(x) = e⁻ˣ, a = 0

Verified step by step guidance
1
Identify the function and the center of the Taylor series: here, the function is \(f(x) = e^{-x}\) and the series is centered at \(a = 0\).
Recall the Taylor series expansion formula for a function \(f(x)\) centered at \(a\): \[f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!} (x - a)^n + R_n(x),\] where \(R_n(x)\) is the remainder after \(n\) terms.
Find the \(n\)-th derivative of \(f(x) = e^{-x}\). Since the derivative of \(e^{-x}\) is \(-e^{-x}\), the \(n\)-th derivative is \[f^{(n)}(x) = (-1)^n e^{-x}.\] Evaluate this at \(a=0\) to get \[f^{(n)}(0) = (-1)^n e^{0} = (-1)^n.\]
Write the remainder term \(R_n(x)\) using the Lagrange form of the remainder: \[R_n(x) = \frac{f^{(n+1)}(c)}{(n+1)!} (x - 0)^{n+1} = \frac{(-1)^{n+1} e^{-c}}{(n+1)!} x^{n+1},\] where \(c\) is some number between \(0\) and \(x\).
To show that \(\lim_{n \to \infty} R_n(x) = 0\) for all \(x\) in the interval of convergence, analyze the absolute value: \[|R_n(x)| = \left| \frac{e^{-c}}{(n+1)!} x^{n+1} \right| \leq \frac{e^{-|c|}}{(n+1)!} |x|^{n+1}.\] Since \(e^{-c}\) is bounded and factorial growth in the denominator dominates any power of \(x\), the limit of \(R_n(x)\) as \(n \to \infty\) is zero for all real \(x\).

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Key Concepts

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

Taylor Series and Expansion

A Taylor series represents a function as an infinite sum of terms calculated from the derivatives of the function at a single point a. It approximates the function near a by polynomials, where each term involves higher-order derivatives evaluated at a, multiplied by powers of (x - a). Understanding this expansion is essential to express f(x) = e^{-x} around a = 0.
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Remainder Term in Taylor Series

The remainder term Rₙ(x) measures the error between the actual function and its nth-degree Taylor polynomial approximation. It quantifies how well the polynomial approximates the function and is often expressed using Lagrange's form involving the (n+1)th derivative. Analyzing Rₙ(x) helps determine the accuracy and convergence of the series.
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Limit of the Remainder and Interval of Convergence

Showing that limₙ→∞ Rₙ(x) = 0 within the interval of convergence proves that the Taylor series converges to the function f(x). The interval of convergence is the set of x-values where the series converges. For e^{-x}, the series converges for all real x, ensuring the remainder vanishes as n grows large.
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