Problem 14
Question
Find the Taylor polynomial of degree \(n\) for \(x\) near the given point \(a\). $$\cos x, \quad a=\pi / 2, \quad n=4$$
Step-by-Step Solution
Verified Answer
The Taylor polynomial is \(P_4(x) = -(x - \pi/2) + \frac{(x - \pi/2)^3}{6}\).
1Step 1: Understanding the Taylor Polynomial
The Taylor polynomial of degree \(n\) for a function \(f(x)\) centered at \(a\) is given by: \[P_n(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \cdots + \frac{f^{(n)}(a)}{n!}(x-a)^n\]We need to find this polynomial for \(f(x) = \cos x\) at the point \(a = \pi/2\), and degree \(n = 4\).
2Step 2: Compute Function Values and Derivatives at \(a\)
First, determine the value of \(f(x) = \cos x\) and its derivatives at \(x = \pi / 2\):1. \(f(x) = \cos x \Rightarrow f(\pi/2) = 0\).2. \(f'(x) = -\sin x \Rightarrow f'(\pi/2) = -1\).3. \(f''(x) = -\cos x \Rightarrow f''(\pi/2) = 0\).4. \(f'''(x) = \sin x \Rightarrow f'''(\pi/2) = 1\).5. \(f^{(4)}(x) = \cos x \Rightarrow f^{(4)}(\pi/2) = 0\).
3Step 3: Plug Values into Taylor Polynomial Formula
Now, substitute the computed values of the function and its derivatives at \(x = \pi/2\) into the Taylor polynomial formula:\[P_4(x) = 0 - 1 \cdot (x-\pi/2) + 0 \cdot \frac{(x-\pi/2)^2}{2!} + 1 \cdot \frac{(x-\pi/2)^3}{3!} + 0 \cdot \frac{(x-\pi/2)^4}{4!}\]This simplifies to:\[P_4(x) = -(x - \pi/2) + \frac{(x - \pi/2)^3}{6}\]
4Step 4: Write Final Taylor Polynomial
The Taylor polynomial of degree 4 for \(\cos x\) centered at \(a = \pi / 2\) is:\[P_4(x) = -(x - \pi/2) + \frac{(x - \pi/2)^3}{6}\]
Key Concepts
DerivativesTrigonometric FunctionsFunction ApproximationDegree of Polynomial
Derivatives
Derivatives represent the rate at which a function changes. They are essential for finding the Taylor polynomial, as derivatives give us the slope of the tangent line at a specific point on the function.
Here's how it works:
Here's how it works:
- The first derivative, noted as \(f'(x)\), tells us how steep the line is. For \(f(x) = \cos x\), the first derivative is \(-\sin x\).
- The second derivative, \(f''(x)\), provides information on how the slope is changing. For \(\cos x\), it is \(-\cos x\).
- Continuing this way, we can find higher order derivatives, such as the third derivative \(f'''(x) = \sin x\), and the fourth derivative \(f^{(4)}(x) = \cos x\).
Trigonometric Functions
Trigonometric functions like \(\cos x\) and \(\sin x\) play an important role in many areas of mathematics. They are periodic functions, meaning they repeat their values in regular intervals. This periodic nature makes them highly predictable and useful in approximations.
When using a trigonometric function in a Taylor polynomial, it's crucial to know its behavior at specific points, especially important points like \(\pi/2\), \(\pi\), etc. For instance, \(\cos(\pi/2) = 0\), which was crucial in our example for determining the coefficient of the zeroth power term in the Taylor polynomial.
Understanding how these functions change as their input varies allows us to determine the derivatives that form the foundation of the Taylor polynomial.
When using a trigonometric function in a Taylor polynomial, it's crucial to know its behavior at specific points, especially important points like \(\pi/2\), \(\pi\), etc. For instance, \(\cos(\pi/2) = 0\), which was crucial in our example for determining the coefficient of the zeroth power term in the Taylor polynomial.
Understanding how these functions change as their input varies allows us to determine the derivatives that form the foundation of the Taylor polynomial.
Function Approximation
Function approximation seeks to represent complex functions using simpler forms, like polynomials. The Taylor polynomial is a fantastic tool for this purpose. It works by summing up the function's value and its derivatives at a point to create a polynomial that behaves similarly to the original function near that point.
The accuracy of this approximation depends on the degree of the polynomial used. Higher-degree polynomials can capture more features of the function. In our case, a fourth-degree Taylor polynomial was used to accurately approximate \(\cos x\) near \(\pi/2\).
This kind of approximation is incredibly beneficial in numerical analysis and various engineering applications, where working with simple polynomials instead of more complex functions can save time and computational resources.
The accuracy of this approximation depends on the degree of the polynomial used. Higher-degree polynomials can capture more features of the function. In our case, a fourth-degree Taylor polynomial was used to accurately approximate \(\cos x\) near \(\pi/2\).
This kind of approximation is incredibly beneficial in numerical analysis and various engineering applications, where working with simple polynomials instead of more complex functions can save time and computational resources.
Degree of Polynomial
The degree of a polynomial is the highest power of \(x\) in the equation. In the context of Taylor polynomials, the degree determines how many terms of the function's derivative are included in the approximation.
- A higher-degree polynomial generally provides a better approximation of the original function over a broader range of \(x\) values.
- For our example, a degree of 4 means that our polynomial includes terms up to the fourth derivative of \(\cos x\).
- Each term contributes to the accuracy of the polynomial, capturing more minor changes in the function's behavior.
Other exercises in this chapter
Problem 13
Find the first four terms of the Taylor series for the function about the point \(a\). $$1 / x, \quad a=1$$
View solution Problem 14
expand the quantity about 0 in terms of the variable given. Give four nonzero terms. $$\frac{1}{2+x} \text { in terms of } \frac{x}{2}$$
View solution Problem 14
Find the first four terms of the Taylor series for the function about the point \(a\). $$1 / x, \quad a=2$$
View solution Problem 15
You approximate \(f(t)=e^{t}\) by a Taylor polynomial of degree 0 about \(t=0\) on the interval [0,0.5]. (a) Reasoning informally, say whether the approximation
View solution