Problem 9 - Calculus for Biology and Medicine

Question

Compute the Taylor polynomial of degree \(n\) about \(x=0\) for each function. $$ f(x)=x^{5}, n=6 $$

Step-by-Step Solution

Answer

The Taylor polynomial is \(P_6(x) = x^5\).

Step 1: Understanding the Problem

We need to compute the Taylor polynomial of degree 6 for the function \(f(x) = x^5\) about \(x = 0\). This involves calculating the derivatives of the function up to the 6th degree and evaluating them at \(x = 0\).

Step 2: Compute Derivatives

Let's compute the derivatives of \(f(x) = x^5\):- First derivative, \(f'(x) = 5x^4\).- Second derivative, \(f''(x) = 20x^3\).- Third derivative, \(f'''(x) = 60x^2\).- Fourth derivative, \(f^{(4)}(x) = 120x\).- Fifth derivative, \(f^{(5)}(x) = 120\).- Sixth derivative, \(f^{(6)}(x) = 0\). All derivatives beyond the fifth are zero.

Step 3: Evaluate Derivatives at x=0

Now, evaluate each derivative at \(x = 0\):- \(f(0) = 0^5 = 0\).- \(f'(0) = 5 imes 0^4 = 0\).- \(f''(0) = 20 imes 0^3 = 0\).- \(f'''(0) = 60 imes 0^2 = 0\).- \(f^{(4)}(0) = 120 imes 0 = 0\).- \(f^{(5)}(0) = 120\).- \(f^{(6)}(0) = 0\).

Step 4: Formulate the Taylor Polynomial

The Taylor polynomial of degree 6 is given by the formula:\[P_6(x) = \sum_{k=0}^{6} \frac{f^{(k)}(0)}{k!} x^k\]Substitute the evaluated derivatives:\[P_6(x) = 0 + 0\cdot x + 0\cdot x^2 + 0\cdot x^3 + 0\cdot x^4 + \frac{120}{5!} x^5 + 0\cdot x^6\]Simplifying, we obtain:\[P_6(x) = x^5\]

Step 5: Final Taylor Polynomial

The Taylor polynomial of degree 6 for \(f(x) = x^5\) about \(x = 0\) is simply \(P_6(x) = x^5\). All terms beyond \(x^5\) are zero.

Key Concepts

Degree of Polynomial

The degree of a polynomial is the highest exponent of the variable in the polynomial expression. It indicates the polynomial's order or power. In the context of Taylor polynomials, knowing the degree helps us to determine how many terms we need to consider.

In this exercise, our function is \( f(x) = x^5 \), a polynomial of degree 5. However, we're tasked with finding a Taylor polynomial of degree 6 about \( x = 0 \). This means we consider terms up to, and including, terms with the variable raised to the power of 6.

Here's why the degree is important:

• It tells us how many derivatives we need to take. If we're computing up to degree 6, we need derivatives up to the 6th order.
• It affects the accuracy and capability of the Taylor series to approximate functions around a chosen point.
• If the function is already a polynomial of lower degree, like our function, derivatives beyond its degree might be zero.

Derivatives

Derivatives are fundamental when working with Taylor polynomials as they help describe how a function changes. Calculating the derivative of a function involves breaking it down into its component rates of change.

For our function \( f(x) = x^5 \):

• The first derivative \( f'(x) = 5x^4 \), represents the rate of change of the function.
• The second derivative \( f''(x) = 20x^3 \), provides information about the curvature or concavity of the function.
• The process continues until the nth derivative, in this case up to the 6th derivative \( f^{(6)}(x) = 0 \).

Beyond the 5th derivative, the function derivatives become zero since \( x^5 \) cannot yield a non-zero term when continually differentiated past the power level of 5.

Derivatives help us determine how accurately the Taylor polynomial approximates the function, particularly when evaluating them at a specific point.

Evaluating Derivatives at a Point

Once we have the derivatives, evaluating them at a specific point fine-tunes our Taylor polynomial. This point is often referred to as "about" which in this exercise is \( x = 0 \) (also known as the Maclaurin series when "about" is zero).

Given our function \( f(x) = x^5 \), we evaluate derivatives at \( x=0 \):

• All derivatives except for the 5th evaluate to zero because any derivative of \( x^n \), when \( n > 0 \), at \( x=0 \) is zero.
• Specifically, \( f^{(5)}(0) = 120 \), which affects the term \( \frac{120}{5!} x^5 \) in the Taylor polynomial.

Evaluating derivatives at a point simplifies the polynomial, making it easier to see which terms are significant. It configures the polynomial to fit the original function as precisely as possible purely at that specific point.

Higher Order Derivatives

Higher order derivatives refer to derivatives taken multiple times. They give us further insight into the behavior and characteristics of a function.

In this exercise, we calculated derivatives beyond the first few orders:

• 4th and 5th derivatives show how sharply a function is changing, which is helpful in Taylor series as they account for more accuracy around the point of expansion.
• Once we hit the 6th derivative of \( x^5 \), and beyond, they become zero, meaning they don't contribute non-zero terms to the expansion.

Higher order derivatives allow us to capture more subtle elements of change in a function, making our Taylor polynomial a closer approximation. However, they also tell us when the polynomial is as accurate as it can get—when further derivatives don't yield new information, as they do here beyond the 5th derivative.