### 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 = sqrt3c_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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Answered: (1 point) Find the Taylor series for…

(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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Question

### 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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