(1 point) Find the Taylor series for the function f(x) = 1/r centered at a = 3. The series has the form f(x) = E, Cn (x – 3)". Enter the first few coefficients below. Co = sqrt3 Ci = C2 C3 = C4 =
(1 point) Find the Taylor series for the function f(x) = 1/r centered at a = 3. The series has the form f(x) = E, Cn (x – 3)". Enter the first few coefficients below. Co = sqrt3 Ci = C2 C3 = C4 =
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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![### Taylor Series Exercise
**Problem Statement:**
(1 point) Find the Taylor series for the function \( f(x) = \frac{1}{x} \) centered at \( a = 3 \). The series has the form \( f(x) = \sum_{n=0}^{\infty} c_n (x - 3)^n \). Enter the first few coefficients below.
**Initial Conditions and Coefficients:**
```
c_0 = sqrt3
c_1 = _______
c_2 = _______
c_3 = _______
c_4 = _______
```
In this problem, students are asked to determine the first few coefficients of the Taylor series expansion of \( f(x) = \frac{1}{x} \) around \( x = 3 \). The Taylor series expansion is a way to represent the function as an infinite sum of terms calculated from the values of its derivatives at a single point.
### How to Approach:
1. **Identify the Function and Center:**
The function given is \( \frac{1}{x} \), and the Taylor series is centered at \( a = 3 \).
2. **Taylor Series Formula:**
The general Taylor series for \( f(x) \) centered at \( a \) is:
\[
f(x) = \sum_{n=0}^{\infty} c_n (x - a)^n
\]
where \( c_n = \frac{f^{(n)}(a)}{n!} \).
3. **Calculate Coefficients:**
For each \( n \) from 0 to 4, compute the coefficient \( c_n \) using the derivatives of \( f(x) \) evaluated at \( x = 3 \).
a. \( c_0 = f(3) = \frac{1}{3} = \sqrt{3} \)
b. \( c_1 = f'(3) = -\frac{1}{9} \cdot (x - 3)^1 \)
c. \( c_2 = f''(3) = 2 \cdot \frac{1}{27} \cdot (x - 3)^2 \)
d. \( c_3 = f'''(3) = -6 \cdot \frac{1}{81} \](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd7b1f94a-aeb1-4e21-a353-b14674dcbf02%2F5319dcc4-a305-4100-9fe8-1c36f294324e%2Fqfzf2f_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Taylor Series Exercise
**Problem Statement:**
(1 point) Find the Taylor series for the function \( f(x) = \frac{1}{x} \) centered at \( a = 3 \). The series has the form \( f(x) = \sum_{n=0}^{\infty} c_n (x - 3)^n \). Enter the first few coefficients below.
**Initial Conditions and Coefficients:**
```
c_0 = sqrt3
c_1 = _______
c_2 = _______
c_3 = _______
c_4 = _______
```
In this problem, students are asked to determine the first few coefficients of the Taylor series expansion of \( f(x) = \frac{1}{x} \) around \( x = 3 \). The Taylor series expansion is a way to represent the function as an infinite sum of terms calculated from the values of its derivatives at a single point.
### How to Approach:
1. **Identify the Function and Center:**
The function given is \( \frac{1}{x} \), and the Taylor series is centered at \( a = 3 \).
2. **Taylor Series Formula:**
The general Taylor series for \( f(x) \) centered at \( a \) is:
\[
f(x) = \sum_{n=0}^{\infty} c_n (x - a)^n
\]
where \( c_n = \frac{f^{(n)}(a)}{n!} \).
3. **Calculate Coefficients:**
For each \( n \) from 0 to 4, compute the coefficient \( c_n \) using the derivatives of \( f(x) \) evaluated at \( x = 3 \).
a. \( c_0 = f(3) = \frac{1}{3} = \sqrt{3} \)
b. \( c_1 = f'(3) = -\frac{1}{9} \cdot (x - 3)^1 \)
c. \( c_2 = f''(3) = 2 \cdot \frac{1}{27} \cdot (x - 3)^2 \)
d. \( c_3 = f'''(3) = -6 \cdot \frac{1}{81} \
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