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Chapter 11: Problem 23

Find the Taylor series for \(f(x)\) centered at the given value of \(a\). [ Assume that \(f\) has a power series expansion. Do not show that \(\left.R_{n}(x) \rightarrow 0 .\right]\) Also find the associated radius of convergence. $$ f(x)=e^{2 x}, \quad a=3 $$

Short Answer

Expert verified
The Taylor series is \( \sum_{n=0}^{\infty} \frac{2^n e^6}{n!} (x-3)^n \) with radius of convergence \( R = \infty \).

Step by step solution

01

Define the Function and Center

We are given that the function is \( f(x) = e^{2x} \). We need to find the Taylor series centered at \( a = 3 \).
02

Find the Derivatives of the Function

First, calculate the derivatives of \( f(x) = e^{2x} \):1. \( f'(x) = 2e^{2x} \)2. \( f''(x) = 4e^{2x} \)3. \( f^{(3)}(x) = 8e^{2x} \) and so on.The \( n \)-th derivative is \( f^{(n)}(x) = 2^n e^{2x} \).
03

Evaluate the Derivatives at Center \( a = 3 \)

Evaluate each derivative at \( x = 3 \):1. \( f(3) = e^6 \)2. \( f'(3) = 2e^6 \)3. \( f''(3) = 4e^6 \)In general, \( f^{(n)}(3) = 2^n e^6 \).
04

Write the Taylor Series Expression

The Taylor series for \( f(x) \) centered at \( a = 3 \) is given by:\[T(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(3)}{n!} (x-3)^n\]Substitute the derivatives:\[T(x) = \sum_{n=0}^{\infty} \frac{2^n e^6}{n!} (x-3)^n\]
05

Derive the Radius of Convergence

To find the radius of convergence \( R \), we use the formula for the convergence of a power series:\[\frac{1}{R} = \lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right|\]Here, \( a_n = \frac{2^n e^6}{n!} \). So,\[\frac{a_{n+1}}{a_n} = \frac{2^{n+1} e^6}{(n+1)!} \cdot \frac{n!}{2^n e^6} = \frac{2}{n+1}\]Taking the limit:\[\lim_{n \to \infty} \frac{2}{n+1} = 0\]Since this limit is zero, the series converges for all \( x \). Hence, \( R = \infty \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Radius of Convergence
The radius of convergence is a crucial concept when dealing with power series, particularly Taylor series. It tells us the interval where the series converges to the function. The radius of convergence, denoted as \( R \), can be determined using the ratio test. This test involves taking the limit as \( n \) approaches infinity of the absolute value of the ratio of successive coefficients.
  • The formula used is \( \frac{1}{R} = \lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right| \).
  • If this limit exists, \( R \) is positive, and the series converges absolutely whenever \( |x-a| < R \).
  • If the limit is zero (as in our case), the series converges for any \( x \), meaning the radius of convergence \( R = \infty \).
Understanding the radius of convergence helps us predict where the Taylor series representation is valid and matches the behavior of the original function.
Power Series Expansion
A power series is a series of the form \( \sum_{n=0}^\infty a_n (x-a)^n \). It's essentially an infinite polynomial, and it provides a way to represent functions around a certain center \( a \). In a Taylor series, the coefficients \( a_n \) are derived from the function's derivatives evaluated at \( a \).
  • For example, for the function \( f(x) = e^{2x} \), the Taylor series centered at \( a = 3 \) results in \( \sum_{n=0}^\infty \frac{2^n e^6}{n!} (x-3)^n \).
  • Each term in this series involves coefficients multiplied by powers of \( (x-a) \).
Using power series expansions allows us to approximate complex functions with polynomials, which are easier to work with analytically.
Derivatives
Derivatives are essential in forming a Taylor series, as they determine the coefficients of the series. A derivative measures how a function changes as its input changes. For a Taylor series, we calculate derivatives of all orders (e.g., first derivative, second derivative, etc.).
  • For \( f(x) = e^{2x} \), each derivative is \( f^{(n)}(x) = 2^n e^{2x} \).
  • These derivatives are then evaluated at \( x = a \) to get \( f^{(n)}(a) \), which become coefficients in the series \( \frac{f^{(n)}(a)}{n!} \).
Understanding derivatives helps in unraveling the behavior of functions locally and capturing that behavior in a polynomial form. This is what makes Taylor series so powerful.
Limit of a Sequence
The limit of a sequence is an important tool in determining the convergence of series. A sequence is a list of numbers in a specific order, and the limit describes the behavior of the sequence as it extends indefinitely. In the context of Taylor series, we often use limits to understand convergence.
  • To find the radius of convergence, we compute \( \lim_{n \to \infty} \frac{2}{n+1} \), which is zero.
  • This means the Taylor series converges for all \( x \), showing the limit of this sequence affects the entire series.
By grasping how limits work in sequences, students can better understand why certain series converge and how they relate to the functions they represent.

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Most popular questions from this chapter

Find the Taylor polynomial \(T_{3}(x)\) for the function \(f\) centered at the number \(a\). Graph \(f\) and \(T_{3}\) on the same screen. $$ f(x)=\cos x, \quad a=\pi / 2 $$

Find the Maclaurin series for \(f(x)\) using the definition of a Maclaurin series. [Assume that \(f\) has a power series expansion. Do not show that \(\left.R_{n}(x) \rightarrow 0 .\right]\) Also find the associated radius of convergence. $$ f(x)=2^{x} $$

Approximate \(f\) by a Taylor polynomial with degree \(n\) at the number \(a\). (b) Use Taylor's Inequality to estimate the accuracy of the approximation \(f(x)=T_{x}(x)\) when \(x\) lies in the given interval. (c) Check your result in part (b) by graphing \(\left|R_{n}(x)\right|\). $$ f(x)=\ln (1+2 x), \quad a=1, \quad n=3, \quad 0.5 \leqslant x \leqslant 1.5 $$

The resistivity \(\rho\) of a conducting wire is the reciprocal of the conductivity and is measured in units of ohm-meters \((\Omega-\mathrm{m}) .\) The resistivity of a given metal depends on the temperature according to the equation \(\rho(t)=\rho_{20} e^{\alpha(t-2 \omega)}\) where \(t\) is the temperature in \(^{\circ} \mathrm{C}\). There are tables that list the values of \(\alpha\) (called the temperature coefficient) and \(\rho_{20}\) (the resistivity at \(20^{\circ} \mathrm{C}\) ) for various metals. Except at very low temperatures, the resistivity varies almost linearly with temperature and so it is common to approximate the expression for \(\rho(t)\) by its first-or second-degree Taylor polynomial at \(t=20 .\) $$ \begin{array}{l}{\text { (a) Find expressions for these linear and quadratic }} \\ {\text { approximations. }} \\ {\text { (b) For copper, the tables give } \alpha=0.0039 /^{\circ} \mathrm{C} \text { and }} \\ {\rho_{20}=1.7 \times 10^{-8} \Omega \text { -m. Graph the resistivity of copper }} \\\ {\text { and the linear and quadratic approximations for }} \\ {-250^{\circ} \mathrm{C} \leqslant t \leqslant 1000^{\circ} \mathrm{C} \text { . }} \\\ {\text { (c) For what values of } t \text { does the linear approximation agree }} \\ {\text { with the exponential expression to within one percent? }}\end{array} $$

Use the definition of a Taylor series to find the first four nonzero terms of the series for \(f(x)\) centered at the given value of \(a .\) $$ f(x)=\frac{1}{1+x}, \quad a=2 $$
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