a. Find the Taylor polynomial of 6th order at a=- 2 b. Find error at x = π. X c. Graph the original function and T₂(x),T₁(x), and Tg(x). 2 6 f(x)=cos2x
a. Find the Taylor polynomial of 6th order at a=- 2 b. Find error at x = π. X c. Graph the original function and T₂(x),T₁(x), and Tg(x). 2 6 f(x)=cos2x
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 Polynomial Approximation and Error Analysis
#### Problem 1:
Consider the function \( f(x) = \cos(2x) \). Perform the following tasks:
**a.** Find the Taylor polynomial of 6th order at \( a = \frac{\pi}{2} \).
**b.** Find the error at \( x = \pi \).
**c.** Graph the original function \( f(x) = \cos(2x) \) along with the Taylor polynomials \( T_2(x) \), \( T_4(x) \), and \( T_6(x) \).
### Solution Approach:
#### a. Taylor Polynomial of 6th Order at \( a = \frac{\pi}{2} \)
The general formula for the Taylor polynomial of a function \( f(x) \) centered at \( a \) is given by:
\[ T_n(x) = \sum_{k=0}^n \frac{f^{(k)}(a)}{k!}(x - a)^k \]
Calculate the derivatives of \( f(x) = \cos(2x) \) and evaluate them at \( x = \frac{\pi}{2} \):
1. \( f(x) = \cos(2x) \)
- \( f\left(\frac{\pi}{2}\right) = \cos\left(\pi\right) = -1 \)
2. \( f'(x) = -2\sin(2x) \)
- \( f'\left(\frac{\pi}{2}\right) = -2\sin(\pi) = 0 \)
3. \( f''(x) = -4\cos(2x) \)
- \( f''\left(\frac{\pi}{2}\right) = -4\cos(\pi) = 4 \)
4. \( f^{(3)}(x) = 8\sin(2x) \)
- \( f^{(3)}\left(\frac{\pi}{2}\right) = 8\sin(\pi) = 0 \)
5. \( f^{(4)}(x) = 16\cos(2x) \)
- \( f^{(4)}\left(\frac{\pi}{2}\right) = 16\cos(\pi) = -16](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F4f3d9715-5083-4150-a4a6-02c5c7b14656%2Fb701f1c0-e2b1-43d3-a58e-3604a099d4cb%2Fy4yt2va_processed.png&w=3840&q=75)
Transcribed Image Text:### Taylor Polynomial Approximation and Error Analysis
#### Problem 1:
Consider the function \( f(x) = \cos(2x) \). Perform the following tasks:
**a.** Find the Taylor polynomial of 6th order at \( a = \frac{\pi}{2} \).
**b.** Find the error at \( x = \pi \).
**c.** Graph the original function \( f(x) = \cos(2x) \) along with the Taylor polynomials \( T_2(x) \), \( T_4(x) \), and \( T_6(x) \).
### Solution Approach:
#### a. Taylor Polynomial of 6th Order at \( a = \frac{\pi}{2} \)
The general formula for the Taylor polynomial of a function \( f(x) \) centered at \( a \) is given by:
\[ T_n(x) = \sum_{k=0}^n \frac{f^{(k)}(a)}{k!}(x - a)^k \]
Calculate the derivatives of \( f(x) = \cos(2x) \) and evaluate them at \( x = \frac{\pi}{2} \):
1. \( f(x) = \cos(2x) \)
- \( f\left(\frac{\pi}{2}\right) = \cos\left(\pi\right) = -1 \)
2. \( f'(x) = -2\sin(2x) \)
- \( f'\left(\frac{\pi}{2}\right) = -2\sin(\pi) = 0 \)
3. \( f''(x) = -4\cos(2x) \)
- \( f''\left(\frac{\pi}{2}\right) = -4\cos(\pi) = 4 \)
4. \( f^{(3)}(x) = 8\sin(2x) \)
- \( f^{(3)}\left(\frac{\pi}{2}\right) = 8\sin(\pi) = 0 \)
5. \( f^{(4)}(x) = 16\cos(2x) \)
- \( f^{(4)}\left(\frac{\pi}{2}\right) = 16\cos(\pi) = -16
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