Problem 16 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the Taylor polynomial of order 4 based at 2 for \(f(x)=x^{4}\) and show that it represents \(f(x)\) exactly.

Step-by-Step Solution

Answer

The Taylor polynomial is exactly \( T_4(x) = x^4 \), matching \( f(x) \) exactly.

Step 1: Understand the Problem

We need to find the Taylor polynomial of order 4 for the function \( f(x) = x^4 \) centered at \( x = 2 \). This polynomial should represent \( f(x) \) exactly because the function is a polynomial of degree 4.

Step 2: Derivatives of the Function

Calculate the first four derivatives of \( f(x) = x^4 \): - \( f'(x) = 4x^3 \) - \( f''(x) = 12x^2 \) - \( f'''(x) = 24x \) - \( f^{(4)}(x) = 24 \) - All higher derivatives are zero since it's a polynomial of degree 4.

Step 3: Evaluate Derivatives at the Base Point

Evaluate each derivative at \( x = 2 \): - \( f(2) = 2^4 = 16 \) - \( f'(2) = 4 imes 2^3 = 32 \) - \( f''(2) = 12 imes 2^2 = 48 \) - \( f'''(2) = 24 imes 2 = 48 \) - \( f^{(4)}(2) = 24 \) - Higher order derivatives beyond the 4th evaluate to zero.

Step 4: Construct the Taylor Polynomial

Use the formula for the Taylor polynomial: \[ T_4(x) = f(a) + \frac{f'(a)}{1!}(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \frac{f^{(4)}(a)}{4!}(x-a)^4 \] Plug in the values obtained:\[ T_4(x) = 16 + \frac{32}{1}(x-2) + \frac{48}{2}(x-2)^2 + \frac{48}{6}(x-2)^3 + \frac{24}{24}(x-2)^4 \] Simplify:\[ T_4(x) = 16 + 32(x - 2) + 24(x-2)^2 + 8(x-2)^3 + 1(x-2)^4 \]

Step 5: Verify that it Represents the Original Function

Since \( f(x) = x^4 \) is itself a polynomial of degree 4, the Taylor polynomial will exactly equal \( f(x) \). After expanding \( T_4(x) \), we will indeed return to \( f(x) = x^4 \). This confirms that the Taylor polynomial represents \( f(x) \) perfectly, matching all terms.

Key Concepts

Derivatives

To find the Taylor polynomial for a function, you need to understand derivatives. Derivatives represent the rate at which a function changes. They're essential for constructing a Taylor polynomial, as each derivative contributes a term to the polynomial.
For the function \( f(x) = x^4 \), the first four derivatives are crucial. Let's break them down:

• First derivative (\( f'(x) \)): This measures the slope or rate of change of the function at any point. For \( f(x) = x^4 \), the first derivative is \( f'(x) = 4x^3 \).
• Second derivative (\( f''(x) \)): It tells us how the rate of change itself is changing. Here, it is \( f''(x) = 12x^2 \).
• Third derivative (\( f'''(x) \)): It can be thought of as the rate at which the second derivative changes. This result is \( f'''(x) = 24x \).
• Fourth derivative (\( f^{(4)}(x) \)): It shows the rate of change of the third derivative. The outcome is a constant \( f^{(4)}(x) = 24 \).

Interestingly, because \( x^4 \) is a polynomial of degree 4, all higher-order derivatives (5th, 6th, etc.) are zero. Understanding these derivatives helps build the Taylor polynomial.

Polynomial Degree

The degree of a polynomial is a fundamental concept in understanding Taylor polynomials. It's the highest power of the variable in a polynomial expression. For example, in the function \( f(x) = x^4 \), the highest power is 4. Thus, it is a degree 4 polynomial. Why is knowing the polynomial degree important?

• It tells us how many derivatives we need to check when crafting a Taylor polynomial. For \( x^4 \), it’s up to the 4th derivative.
• Polynomials of degree \( n \) can be perfectly represented by a Taylor polynomial of order \( n \).

Understanding the polynomial degree is vital for ensuring your Taylor polynomial accurately reflects the original function, especially since each term's importance diminishes as the order exceeds the original polynomial's degree.

Calculus Steps

Constructing a Taylor polynomial involves several calculus steps that ensure the polynomial closely approximates the original function. Let’s look at this process using our example function, \( f(x) = x^4 \), centered at \( x = 2 \):Firstly, evaluate the derivatives at the base point (\( x = 2 \)). This gives us specific values:

• \( f(2) = 16 \)
• \( f'(2) = 32 \)
• \( f''(2) = 48 \)
• \( f'''(2) = 48 \)
• \( f^{(4)}(2) = 24 \)

The calculus formula for a Taylor polynomial is:\[T_n(x) = f(a) + \frac{f'(a)}{1!}(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \cdots + \frac{f^{(n)}(a)}{n!}(x-a)^n\]Plugging in the relevant values for each term and simplifying, we achieve:\[T_4(x) = 16 + 32(x - 2) + 24(x-2)^2 + 8(x-2)^3 + 1(x-2)^4\]In this context, understanding each calculus step is essential for confirming that the Taylor polynomial mirrors the original function. Finally, note that once expanded, \( T_4(x) \) should perfectly match the original polynomial, \( f(x) = x^4 \), proving its exact representation.