Chapter 4: Problem 2
Find the 5 th Taylor polynomial, \(P_{5}(x)\), at \(\bar{x}=0\) for \(\cos x\). Determine an upper bound for the error \(\left|P_{5}(x)-\cos x\right|\) for \(x \in[-\pi / 2, \pi / 2]\).
Short Answer
Expert verified
The 5th Taylor polynomial is \(P_5(x) = 1 - \frac{x^2}{2} + \frac{x^4}{24}\), and the error bound is approximately 0.000028 for \(x \in [-\pi/2, \pi/2]\).
Step by step solution
01
Understand the Taylor Series Formula
The Taylor series for a function \(f(x)\) centered at \(\bar{x}\) is given by \(f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(\bar{x})}{n!} (x - \bar{x})^n\). For \(\cos x\), we will calculate the Taylor series at \(\bar{x} = 0\).
02
Find the Derivatives of \(\cos x\)
The derivatives of \(\cos x\) are cyclical. We have: - \(f(x) = \cos x\) - \(f'(x) = -\sin x\) - \(f''(x) = -\cos x\) - \(f'''(x) = \sin x\) - \(f^{(4)}(x) = \cos x\)These derivatives repeat every four terms.
03
Calculate the Initial Derivatives at \(x = 0\)
Evaluate the derivatives at \(x = 0\):- \(f(0) = \cos(0) = 1\)- \(f'(0) = -\sin(0) = 0\)- \(f''(0) = -\cos(0) = -1\)- \(f'''(0) = \sin(0) = 0\)- \(f^{(4)}(0) = \cos(0) = 1\) - \(f^{(5)}(0) = -\sin(0) = 0\)
04
Construct the 5th Taylor Polynomial
Using the derivatives found, construct the Taylor polynomial up to the 5th degree: \[ P_5(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} = 1 - \frac{x^2}{2} + \frac{x^4}{24}. \]
05
Determine the Error Bound Using the Remainder Term
The Lagrange remainder term for the nth Taylor polynomial is given by \[ R_n(x) = \frac{f^{(n+1)}(c)}{(n+1)!} (x - \bar{x})^{n+1}, \]where \(c\) is some point in the interval \([-\pi/2, \pi/2]\). For \(P_5(x)\), \(n=5\), so \[ R_5(x) = \frac{f^{(6)}(c)}{6!} x^6. \]Since \(f^{(6)}(x) = \cos(x)\), its maximum value in \([-\pi/2, \pi/2]\) is 1. Thus, \[ |R_5(x)| \leq \frac{1}{720} x^6. \] The maximum \(|x|\) within this interval is \(\pi/2\), so the error is at most \[ \left| \frac{1}{720} \left( \frac{\pi}{2} \right)^6 \right|. \]
06
Calculate the Numerical Error Upper Bound
Substitute \(x = \pi/2\) into the error formula:\[ \left| \frac{1}{720} \left( \frac{\pi}{2} \right)^6 \right| \approx \frac{1 \cdot (1.571)^6}{720} \approx 0.000028. \]Thus, the error is bounded by approximately 0.000028.
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Understanding the Cosine Function
The cosine function, denoted as \( \cos x \), is a fundamental trigonometric function that describes how a unit circle's horizontal coordinate changes as the angle \( x \) varies. It has a characteristic wave-like pattern that repeats every \( 2\pi \) radians, known as periodicity.
- At \( x = 0 \), \( \cos 0 = 1 \).
- Cosine has an amplitude range of -1 to 1.
- It is an even function, meaning \( \cos(-x) = \cos(x) \).
Exploring Derivatives
Derivatives help understand how a function changes as the input changes. They are critical for forming Taylor polynomials, which approximate functions around a certain point. For \( \cos x \), the derivatives cycle every four times, allowing for accurate approximations:
- The first derivative, \( \cos'(x) = -\sin x \), explains how cosine slopes downwards at \( x = 0 \).
- The second derivative, \( \cos''(x) = -\cos x \), indicates that cosine curves in the opposite direction compared to its original shape.
- The third derivative, \( \cos'''(x) = \sin x \), resembles the first derivative of sine and shifts the behavior to sin's characteristics.
- The fourth derivative brings us back to \( \cos x \), completing the cycle.
Understanding Error Bound
In Taylor polynomial approximations, the error bound provides a maximum deviation estimate from the true function value. Calculating this is crucial in deciding the accuracy of a Taylor polynomial. The error bound is rooted in the remainder term of the Taylor series, called the Lagrange remainder, which gives a theoretical grasp on how precise the approximation is.
- It reflects the influence of the next term in the series which is not included in the polynomial.
- This helps determine how closely \( P_5(x) \) approximates \( \cos x \) in the interval \([-\pi/2, \pi/2]\).
- Having a precise error bound ensures that calculations meet desired levels of accuracy, which is especially important in scientific computations.
Different Aspects of Lagrange Remainder
The Lagrange remainder is a crucial concept when dealing with Taylor series. It helps in understanding the potential error involved when approximating a function with a polynomial. Specifically, it gives a precise expression for the error of a Taylor polynomial, defined as:
\[ R_n(x) = \frac{f^{(n+1)}(c)}{(n+1)!} (x - \bar{x})^{n+1}, \]
where \( c \) lies somewhere between the center \( \bar{x} \) and \( x \). This remainder becomes particularly useful because:
\[ R_n(x) = \frac{f^{(n+1)}(c)}{(n+1)!} (x - \bar{x})^{n+1}, \]
where \( c \) lies somewhere between the center \( \bar{x} \) and \( x \). This remainder becomes particularly useful because:
- It provides a mathematical way to quantify how far off a Taylor polynomial is from the actual function.
- Gives a clear estimate, which, in turn, tells how many terms are needed for a desired precision.
- By calculating the maximum value of \( f^{(n+1)} \), one can find a worst-case error scenario over a particular interval.