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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 11.3.21b

Taylor series and interval of convergence


b. Write the power series using summation notation.


f(x)=3ˣ, a=0

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Recall that the Taylor series of a function \( f(x) \) centered at \( a = 0 \) (also called the Maclaurin series) is given by the formula: \[ f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} x^n \] where \( f^{(n)}(0) \) is the \( n \)-th derivative of \( f(x) \) evaluated at \( x=0 \).
Identify the function given: \( f(x) = 3^x \). To find the Taylor series, we need to compute the derivatives \( f^{(n)}(x) \) and then evaluate them at \( x=0 \).
Calculate the first few derivatives of \( f(x) = 3^x \): - \( f(x) = 3^x \) - \( f'(x) = 3^x \ln(3) \) - \( f''(x) = 3^x (\ln(3))^2 \) - \( f^{(n)}(x) = 3^x (\ln(3))^n \) This pattern holds for all \( n \geq 0 \).
Evaluate the \( n \)-th derivative at \( x=0 \): \[ f^{(n)}(0) = 3^0 (\ln(3))^n = (\ln(3))^n \]
Substitute \( f^{(n)}(0) \) into the Taylor series formula to write the power series in summation notation: \[ f(x) = \sum_{n=0}^{\infty} \frac{(\ln(3))^n}{n!} x^n \]

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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, called the center (a). For f(x) = 3^x at a = 0, the series uses derivatives evaluated at 0 to approximate the function near that point.
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Power Series and Summation Notation

A power series is an infinite sum of terms in the form c_n(x - a)^n, where c_n are coefficients and a is the center. Summation notation compactly expresses this series as Σ c_n (x - a)^n, making it easier to write and analyze the series.
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Interval of Convergence

The interval of convergence is the set of x-values for which the power series converges to the function. Determining this interval ensures the series accurately represents the function within that range, often found using ratio or root tests.
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