Problem 8 - Calculus for Biology and Medicine

Question

Compute the Taylor polynomial of degree \(n\) about \(x=0\) for each function. $$ f(x)=e^{-x}, n=3 $$

Step-by-Step Solution

Answer

The Taylor polynomial of degree 3 for \( f(x) = e^{-x} \) about \( x = 0 \) is \( P_3(x) = 1 - x + \frac{1}{2}x^2 - \frac{1}{6}x^3 \).

Step 1: Understand the function and derivatives

The function given is \( f(x) = e^{-x} \). We need to calculate its Taylor polynomial of degree 3 about \( x = 0 \). The first step is to compute the derivatives of \( f(x) \).

Step 2: Calculate the function derivatives at x=0

The function is \( f(x) = e^{-x} \). The derivatives are:1. \( f'(x) = -e^{-x} \Rightarrow f'(0) = -1 \)2. \( f''(x) = e^{-x} \Rightarrow f''(0) = 1 \)3. \( f'''(x) = -e^{-x} \Rightarrow f'''(0) = -1 \)

Step 3: Set up the Taylor polynomial formula

The nth-degree Taylor polynomial for a function \( f(x) \) about \( x = 0 \) is given by:\[ P_n(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots + \frac{f^{(n)}(0)}{n!}x^n \]

Step 4: Substitute the derivatives into the polynomial

Using the derivatives we calculated:\[ P_3(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 \]Substitute the values:\[ P_3(x) = 1 - x + \frac{1}{2}x^2 - \frac{1}{6}x^3 \]

Step 5: Write out the Taylor polynomial

The Taylor polynomial of degree 3 for \( f(x) = e^{-x} \) about \( x = 0 \) is:\[ P_3(x) = 1 - x + \frac{1}{2}x^2 - \frac{1}{6}x^3 \]

Key Concepts

Exponential Function

An exponential function is a mathematical expression where a constant base is raised to the power of a variable exponent. Commonly, this is expressed in the form of \(e^x\), where \(e\) is a mathematical constant approximately equal to 2.71828. The exponential function is unique because its rate of growth is proportional to its current value, leading to fast increases or decreases. It has crucial applications in various fields such as physics, economics, and biology.

When the base \(e\) is raised to the power of \(-x\), as in \(e^{-x}\), the function represents exponential decay. This means that as \(x\) increases, the value of the function decreases rapidly, which is the opposite of the exponential growth scenario.

• Exponential growth: \( e^x \)
• Exponential decay: \( e^{-x} \)

These properties are fundamental and help in scenarios like decay modeling, where substances decrease in quantity over time.

Derivatives

The concept of a derivative involves the rate at which a function changes. It is a core idea in calculus and describes how a function evolves as its inputs change. When calculating derivatives of a function, we understand how sensitive the function is to changes in input, influencing fields like physics and engineering.

For the function \(f(x) = e^{-x}\), we compute derivatives to find how the function behaves near specific points. The derivatives of exponential functions have interesting properties that make them easy to handle:- The derivative of \(e^{-x}\) is \(-e^{-x}\).- For higher-order derivatives of \(e^{-x}\), the signs alternate:

• First derivative: \(f'(x) = -e^{-x}\)
• Second derivative: \(f''(x) = e^{-x}\)
• Third derivative: \(f'''(x) = -e^{-x}\)

These derivatives reveal periodicity and symmetry in the function, which are essential when forming Taylor series.

Taylor Series

The Taylor series is a powerful mathematical tool used to approximate functions as infinite sums of terms calculated from the values of their derivatives at a single point. This concept is especially useful for approximating functions that are difficult to calculate directly.

The Taylor polynomial is a truncated version of the Taylor series, comprising a finite number of terms. For a function \(f(x)\), the Taylor series centered at \(x = a\) is given by:\[P(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \ldots\]For \(f(x) = e^{-x}\) about \(x = 0\), we calculate the Taylor polynomial of degree 3:- Use the derivatives at \(x = 0\):\(f(0) = 1\), \(f'(0) = -1\), \(f''(0) = 1\), \(f'''(0) = -1\).- The Taylor polynomial is assembled as:\[P_3(x) = 1 - x + \frac{1}{2}x^2 - \frac{1}{6}x^3\]The Taylor series is invaluable in representing functions in an easily analyzable form, allowing for approximations over intervals. It simplifies many computational processes, highlighting the significance of calculus in solving real-world problems.