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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 11.3.67d
Chapter 11, Problem 11.3.67d

Explain why or why not Determine whether the following statements are true and give an explanation or counterexample.
d. If p(x) is the Taylor series for f centered at 0, then p(x−1) is the Taylor series for f centered at 1.

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Recall that the Taylor series for a function \(f\) centered at a point \(a\) is given by the formula: \[f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!} (x - a)^n,\] where \(f^{(n)}(a)\) denotes the \(n\)-th derivative of \(f\) evaluated at \(a\).
Given that \(p(x)\) is the Taylor series for \(f\) centered at 0, it means \[p(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} x^n.\]
Now, consider \(p(x-1)\). Substituting \(x-1\) into the series, we get \[p(x-1) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} (x-1)^n.\]
To check if \(p(x-1)\) is the Taylor series for \(f\) centered at 1, compare it with the Taylor series centered at 1: \[\sum_{n=0}^{\infty} \frac{f^{(n)}(1)}{n!} (x-1)^n.\] Notice that the coefficients involve derivatives evaluated at 1, not at 0.
Therefore, \(p(x-1)\) uses derivatives at 0, not at 1, so it generally does not equal the Taylor series of \(f\) centered at 1. This shows that the statement is false, and \(p(x-1)\) is not the Taylor series for \(f\) centered at 1.

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Key Concepts

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

Taylor Series and Center of Expansion

A Taylor series represents a function as an infinite sum of terms calculated from the derivatives at a specific point called the center. The series approximates the function near this center, and changing the center changes the coefficients and form of the series.
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Effect of Function Translation on Taylor Series

Replacing x by (x - a) in a Taylor series centered at 0 shifts the input but does not automatically produce the Taylor series centered at a. The coefficients depend on derivatives at the new center, so simply substituting variables does not yield the correct series centered at a.
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Counterexamples in Series Expansion

To test if p(x−1) is the Taylor series of f centered at 1, one can use specific functions like e^x or sin x. These examples show that substituting x−1 into p(x) does not match the Taylor series centered at 1, illustrating the statement is false.
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Related Practice
Textbook Question

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

e. The Taylor series for an even function centered at 0 has only even powers of x.

Textbook Question

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

d. Suppose f'' is continuous on an interval that contains a, where f has an inflection point at a. Then the second−order Taylor polynomial for f at a is linear.

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Textbook Question

{Use of Tech} Fresnel integrals The theory of optics gives rise to the two Fresnel integrals

S(x) = ∫₀ˣ sin t² dt and C(x) = ∫₀ˣ cos t² dt

e. How many terms of the Maclaurin series are required to approximate C(−0.25) with an error no greater than 10⁻⁶?

Textbook Question

{Use of Tech} Fresnel integrals The theory of optics gives rise to the two Fresnel integrals

S(x) = ∫₀ˣ sin t² dt and C(x) = ∫₀ˣ cos t² dt

d. How many terms of the Maclaurin series are required to approximate S(0.05) with an error no greater than 10⁻⁴?

Textbook Question

Matching functions with polynomials Match functions a–f with Taylor polynomials A–F (all centered at 0). Give reasons for your choices.


d. 1/(1 + 2x)


A. p₂(x)= 1 + 2x + 2x²

B. p₂(x) = 1 − 6x + 24x²

C. p₂(x) = 1 + x − x²/2

D. p₂(x) = 1 − 2x + 4x²

E. p₂(x) = 1 − x + (3/2)x²

F. p₂(x) = 1 − 2x + 2x²

Textbook Question

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


If f(x)=∑ₖ₌₀∞ cₖ xᵏ=0, for all x on an interval (−a, a), then cₖ = 0, for all k.

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