Problem 14 - Calculus for Biology and Medicine

Question

In Problems 11-16, compute the Taylor polynomial of degree \(n\) about \(a=0\) for the indicated functions and compare the value of the functions at the indicated point with the value of the corresponding Taylor polynomial. $$ f(x)=e^{-x}, n=4, x=0.3 $$

Step-by-Step Solution

Answer

The Taylor polynomial value is approximately 0.7408375, closely matching the actual value 0.740818.

Step 1: Identify Function and Expansion Parameters

Given the function is \( f(x) = e^{-x} \). We need to find the Taylor polynomial of degree \( n = 4 \) about \( a = 0 \).

Step 2: Find Derivatives of the Function

Calculate the first few derivatives of \( f(x) = e^{-x} \):- \( f(x) = e^{-x} \)- \( f'(x) = -e^{-x} \)- \( f''(x) = e^{-x} \)- \( f'''(x) = -e^{-x} \)- \( f''''(x) = e^{-x} \)

Step 3: Evaluate Derivatives at \( x = 0 \)

Evaluate each derivative at \( x = 0 \):- \( f(0) = e^{0} = 1 \)- \( f'(0) = -e^{0} = -1 \)- \( f''(0) = e^{0} = 1 \)- \( f'''(0) = -e^{0} = -1 \)- \( f''''(0) = e^{0} = 1 \)

Step 4: Construct Taylor Polynomial of Degree 4

The Taylor polynomial of degree 4 is given by:\[ P_4(x) = f(0) + \frac{f'(0)}{1!}x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \frac{f''''(0)}{4!}x^4 \]Substitute the values:\[ P_4(x) = 1 - x + \frac{x^2}{2} - \frac{x^3}{6} + \frac{x^4}{24} \]

Step 5: Evaluate Taylor Polynomial at \( x = 0.3 \)

Substitute \( x = 0.3 \) into the polynomial:\[ P_4(0.3) = 1 - 0.3 + \frac{(0.3)^2}{2} - \frac{(0.3)^3}{6} + \frac{(0.3)^4}{24} \]Calculate:\[ P_4(0.3) \approx 1 - 0.3 + 0.045 - 0.0045 + 0.0003375 \approx 0.7408375 \]

Step 6: Compute Actual Function Value at \( x = 0.3 \)

Calculate \( f(0.3) = e^{-0.3} \). Using a calculator, we find:\[ f(0.3) \approx 0.740818 \]

Step 7: Compare Results

Compare the value of the Taylor polynomial \( P_4(0.3) \approx 0.7408375 \) with the actual function value \( f(0.3) \approx 0.740818 \). The values are very close, showing the polynomial is a good approximation.

Key Concepts

Derivative Evaluation

To accurately create a Taylor polynomial, one needs to understand how to evaluate derivatives. Derivatives represent the rate at which a function changes at a given point. For the function \( f(x) = e^{-x} \), we compute the first few derivatives to feed into our polynomial.

• The first derivative \( f'(x) = -e^{-x} \).
• The second derivative \( f''(x) = e^{-x} \).
• The third derivative \( f'''(x) = -e^{-x} \).
• The fourth derivative \( f''''(x) = e^{-x} \).

Notice the alternating signs and repetitive exponential form. Understanding this pattern helps us recognize the function's behavior, which is crucial when evaluating derivatives. This cycle suggests a systematic method where derivatives alternate between positive and negative, simplifying computation when evaluated at specific points.

Function Approximation

Taylor polynomials are a powerful tool in calculus for approximating functions. Here, they are used to approximate \( f(x) = e^{-x} \) at a particular point, using a polynomial of a specified degree. Each degree of the polynomial improves the approximation around a point, typically chosen as \( a = 0 \).Constructing this polynomial involves:

• Calculating a series of derivatives evaluated at the point \( a \).
• Utilizing the Taylor series formula:\[P_n(x) = f(0) + \frac{f'(0)}{1!}x + \frac{f''(0)}{2!}x^2 + \cdots + \frac{f^{(n)}(0)}{n!}x^n\]
• For our specific case: \[P_4(x) = 1 - x + \frac{x^2}{2} - \frac{x^3}{6} + \frac{x^4}{24}\]

Function approximation with Taylor polynomials becomes increasingly precise as the polynomial degree \( n \) increases. This technique tells us how closely the polynomial, \( P_4(x) \), mimics \( f(x) = e^{-x} \) near \( x = 0 \).

Exponential Function

The exponential function, \( e^{-x} \), plays a foundational role in both calculus and real-world applications. Its defining characteristic is that it changes rapidly, especially near \( x = 0 \). When approximating this function, the Taylor polynomial provides a snapshot of how \( e^{-x} \) behaves over a small interval.Using the exponential function:

• In mathematical contexts often involves exploring behaviors near a point, like \( a = 0 \).
• The function's inherent smoothness and the repeating derivative pattern make it ideal for Taylor expansion.They reveal its dynamic nature as the degree of the Taylor polynomial increases.

In our example, computing \( f(0.3) = e^{-0.3} \) gives roughly \( 0.740818 \), which closely matches our fourth-degree Taylor polynomial approximation of \( 0.7408375 \). It's this close resemblance that highlights the power of Taylor expansions in capturing the essence of exponential functions within limited domains.