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.
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Ch. 11 - Power Series
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243
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Table of contents
- Ch. 1 - Functions210
- Ch. 2 - Limits371
- Ch. 3 - Derivatives633
- Ch. 4 - Applications of the Derivative412
- Ch. 5 - Integration328
- Ch. 6 - Applications of Integration331
- Ch. 7 - Logarithmic, Exponential Functions, and Hyperbolic Functions177
- Ch. 8 - Integration Techniques394
- Ch. 9 - Differential Equations179
- Ch. 10 - Sequences and Infinite Series339
- Ch. 11 - Power Series218
- Ch.12 - Parametric and Polar Curves215
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
Chapter 11, Problem 11.1.67f
Matching functions with polynomials Match functions a–f with Taylor polynomials A–F (all centered at 0). Give reasons for your choices.
f. e⁻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
Verified step by step guidance1
Recall that the Taylor polynomial of a function \(f(x)\) centered at 0 (Maclaurin polynomial) up to degree 2 is given by:
\[p_2(x) = f(0) + f'(0)x + \frac{f''(0)}{2}x^2\]
Start by finding the derivatives of the function \(f(x) = e^{-2x}\):
- First derivative:
\[f'(x) = \frac{d}{dx} e^{-2x} = -2e^{-2x}\]
- Second derivative:
\[f''(x) = \frac{d}{dx} (-2e^{-2x}) = 4e^{-2x}\]
Evaluate the function and its derivatives at \(x=0\):
- \(f(0) = e^0 = 1\)
- \(f'(0) = -2e^0 = -2\)
- \(f''(0) = 4e^0 = 4\)
Substitute these values into the Taylor polynomial formula:
\[p_2(x) = 1 + (-2)x + \frac{4}{2}x^2 = 1 - 2x + 2x^2\]
Compare this polynomial with the given options and identify the matching polynomial for \(f(x) = e^{-2x}\).
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Taylor Polynomials
Taylor polynomials approximate functions near a specific point (here, 0) using derivatives. The nth-degree Taylor polynomial uses the function's value and its first n derivatives at the center to build a polynomial that closely matches the function locally.
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Taylor Polynomials
Derivatives and Their Role in Taylor Series
The coefficients of a Taylor polynomial come from the function's derivatives at the center point. Specifically, the coefficient of x^k is the kth derivative at 0 divided by k factorial, reflecting how the function's slope, curvature, and higher changes shape the polynomial.
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08:42Taylor Series
Matching Functions to Polynomials Using Derivative Values
To match a function with its Taylor polynomial, calculate the function's value and first two derivatives at 0, then compare these to the polynomial's coefficients. This ensures the polynomial accurately represents the function's behavior near the center.
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Related Practice
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?
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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
d. How many terms of the Maclaurin series are required to approximate S(0.05) with an error no greater than 10⁻⁴?
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