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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 11.3.33a

Taylor series


a. Use the definition of a Taylor series to find the first four nonzero terms of the Taylor series for the given function centered at a.


f(x) = 2ˣ, a = 1

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Recall the definition of the Taylor series of a function \(f(x)\) centered at \(a\): \[f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!} (x - a)^n,\] where \(f^{(n)}(a)\) is the \(n\)-th derivative of \(f\) evaluated at \(x = a\).
Identify the function and the center: here, \(f(x) = 2^x\) and \(a = 1\). We will need to find the derivatives of \(f(x)\) and evaluate them at \(x=1\).
Find the first four derivatives of \(f(x)\): - \(f(x) = 2^x\) - \(f'(x) = 2^x \ln(2)\) - \(f''(x) = 2^x (\ln(2))^2\) - \(f^{(3)}(x) = 2^x (\ln(2))^3\) - \(f^{(4)}(x) = 2^x (\ln(2))^4\)
Evaluate each derivative at \(x = 1\): - \(f(1) = 2^1\) - \(f'(1) = 2^1 \ln(2)\) - \(f''(1) = 2^1 (\ln(2))^2\) - \(f^{(3)}(1) = 2^1 (\ln(2))^3\) - \(f^{(4)}(1) = 2^1 (\ln(2))^4\)
Write the first four nonzero terms of the Taylor series using the formula: \[f(x) \approx \sum_{n=0}^3 \frac{f^{(n)}(1)}{n!} (x - 1)^n = \sum_{n=0}^3 \frac{2 \cdot (\ln(2))^n}{n!} (x - 1)^n,\] which explicitly is: \[2 + 2 \ln(2)(x-1) + \frac{2 (\ln(2))^2}{2!} (x-1)^2 + \frac{2 (\ln(2))^3}{3!} (x-1)^3.\]

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Key Concepts

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

Taylor Series Definition

A Taylor series represents a function as an infinite sum of terms calculated from the derivatives of the function at a single point. Each term involves the nth derivative evaluated at the center point, multiplied by (x - a)^n and divided by n!. This series approximates the function near the center a.
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Derivatives of Exponential Functions

For the function f(x) = 2^x, derivatives involve the natural logarithm of the base. Specifically, the nth derivative of 2^x is (ln 2)^n times 2^x. Understanding this pattern is essential to compute the terms of the Taylor series accurately.
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Derivatives of General Exponential Functions

Evaluating the Series at the Center Point

To find the Taylor series terms centered at a = 1, each derivative must be evaluated at x = 1. These values are then used in the formula for each term, ensuring the series accurately approximates the function near x = 1.
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Related Practice
Textbook Question

Probability: sudden−death playoff Teams A and B go into suddendeath overtime after playing to a tie. The teams alternate possession of the ball, and the first team to score wins. Assume each team has a 1/6 chance of scoring when it has the ball, and Team A has the ball first.


a. The probability that Team A ultimately wins is ∑ₖ₌₀∞ (1/6)(5/6)²ᵏ. Evaluate this series.

Textbook Question

Taylor series and interval of convergence


a. Use the definition of a Taylor/Maclaurin series to find the first four nonzero terms of the Taylor series for the given function centered at a.


f(x) = tan ⁻¹ (x/2), a = 0

Textbook Question

Taylor series


a. Use the definition of a Taylor series to find the first four nonzero terms of the Taylor series for the given function centered at a.


f(x) = ln x, a = 3

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Textbook Question

Radius and interval of convergence Use the Ratio Test or the Root Test to determine the radius of convergence of the following power series. Test the endpoints to determine the interval of convergence, when appropriate.



x +x³/3 +x⁵/5 +x⁷/7 + ...

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Textbook Question

Taylor series and interval of convergence


a. Use the definition of a Taylor/Maclaurin series to find the first four nonzero terms of the Taylor series for the given function centered at a.


f(x)=2/(1−x)³, a=0

Textbook Question

Taylor series and interval of convergence


a. Use the definition of a Taylor/Maclaurin series to find the first four nonzero terms of the Taylor series for the given function centered at a.


f(x) = e⁻ˣ, a=0