Chapter 7: Problem 21
Compute the Taylor polynomial of degree \(n\) about a and compare the value of the approximation with the value of the function at the given point \(x\). $$ f(x)=\cos x, a=\frac{\pi}{2}, n=3 ; x=\frac{\pi}{3} $$
Short Answer
Expert verified
The Taylor polynomial at \(x=\frac{\pi}{3}\) approximates about 0.5236, close to \(\cos\left(\frac{\pi}{3}\right) = 0.5\).
Step by step solution
01
Understanding the Problem
We are tasked with finding the third-degree Taylor polynomial of the function \(f(x) = \cos x\) centered at \(a = \frac{\pi}{2}\). After finding the polynomial, we will evaluate it at \(x = \frac{\pi}{3}\) and compare it to the actual value of \(\cos\left(\frac{\pi}{3}\right)\).
02
Finding Derivatives of \(f(x)\)
To construct the Taylor polynomial, we need to find the derivatives of \(f(x) = \cos x\) up to the third order. First derivative: \(f'(x) = -\sin x\)Second derivative: \(f''(x) = -\cos x\)Third derivative: \(f'''(x) = \sin x\)
03
Evaluating Derivatives at \(a = \frac{\pi}{2}\)
Evaluate each derivative at \(a = \frac{\pi}{2}\):\[f\left(\frac{\pi}{2}\right) = \cos\left(\frac{\pi}{2}\right) = 0\]\[f'\left(\frac{\pi}{2}\right) = -\sin\left(\frac{\pi}{2}\right) = -1\]\[f''\left(\frac{\pi}{2}\right) = -\cos\left(\frac{\pi}{2}\right) = 0\]\[f'''\left(\frac{\pi}{2}\right) = \sin\left(\frac{\pi}{2}\right) = 1\]
04
Constructing the Taylor Polynomial
The Taylor polynomial of degree 3 for \(f(x)\) centered at \(a = \frac{\pi}{2}\) is given by:\[P_3(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3\]Substituting the values:\[P_3(x) = 0 - 1(x - \frac{\pi}{2}) + 0 + \frac{1}{6}(x - \frac{\pi}{2})^3\]Which simplifies to:\[P_3(x) = -(x - \frac{\pi}{2}) + \frac{1}{6}(x - \frac{\pi}{2})^3\]
05
Evaluating the Polynomial at \(x = \frac{\pi}{3}\)
Substitute \(x = \frac{\pi}{3}\) into the polynomial:\[P_3\left(\frac{\pi}{3}\right) = -\left(\frac{\pi}{3} - \frac{\pi}{2}\right) + \frac{1}{6}\left(\frac{\pi}{3} - \frac{\pi}{2}\right)^3\]Simplify the expression:\[= -\left(-\frac{\pi}{6}\right) + \frac{1}{6}\left(-\frac{\pi}{6}\right)^3\]\[= \frac{\pi}{6} - \frac{1}{6} \cdot \frac{\pi^3}{216} \approx \frac{\pi}{6}\]
06
Comparing with the Actual Function Value
The actual value of \(\cos\left(\frac{\pi}{3}\right)\) is \(\frac{1}{2}\).Compare this with the approximation \(P_3\left(\frac{\pi}{3}\right) \approx \frac{\pi}{6} \approx 0.5236\).This shows that the Taylor polynomial approximates \(\cos\left(\frac{\pi}{3}\right)\) fairly closely for a 3rd-degree polynomial, though there is some error.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Understanding Derivatives
When dealing with Taylor polynomials, derivatives are essential. A derivative estimates the rate of change of a function. In simple terms, it tells us how a function's output changes as its input changes slightly. This is crucial for creating accurate polynomial approximations.
- First derivative: Shows the slope or gradient of the function. For example, the first derivative of \(f(x) = \cos x\) is \(f'(x) = -\sin x\).
- Second derivative: Provides information about the curvature of the function. It's like seeing whether the slope is increasing or decreasing. In this exercise, \(f''(x) = -\cos x\).
- Third derivative: Continues to show changes in curvature, delving deeper into how the curve is shaped. Our third derivative is \(f'''(x) = \sin x\).
Exploring Trigonometric Functions
Trigonometric functions, such as \(\cos x\), are foundational in mathematics, connecting angles with ratios of a right triangle’s sides. They are periodic, meaning they repeat their values at regular intervals, which we can use to model many real-world phenomena such as waves.
In the specific exercise, we’re using:
In the specific exercise, we’re using:
- **Cosine Function**: \(f(x) = \cos x\), which ranges between -1 and 1, representing the cosine of the angle \(x\).
- **Sine Function**: Arises from the derivative of cosine, \(f'(x) = -\sin x\), indicative of how the cosine function's rate of change varies.
Approximation with Taylor Polynomials
Taylor polynomials offer a powerful tool for approximating functions, providing insight into function behavior around a particular point \(a\). This allows us to simplify complex trigonometric functions and their calculations.
- **Concept**: The Taylor polynomial approximates a function using derivatives evaluated at a specific point. Its accuracy improves with a higher degree (more derivatives).
- **Example in the Exercise**: The 3rd-degree Taylor polynomial for \(\cos x\) centered at \(\frac{\pi}{2}\) includes terms up to the third derivative, making it: \[P_3(x) = -\left(x - \frac{\pi}{2}\right) + \frac{1}{6}\left(x - \frac{\pi}{2}\right)^3. \]
- **Practical Use**: By evaluating this polynomial at \(x = \frac{\pi}{3}\), we see how closely the polynomial mimics the original function's behavior. As seen, \(P_3\left(\frac{\pi}{3}\right) \approx \frac{\pi}{6}\), providing a reasonably close approximation to the true value of \(\cos\left(\frac{\pi}{3}\right) = \frac{1}{2}\).