Problem 6 - Calculus for Biology and Medicine

Question

In Problems \(6-10\), compute the Taylor polynomial of degree \(n\) about \(a=0\) for the indicated functions. $$ f(x)=\frac{1}{1+x}, n=4 $$

Step-by-Step Solution

Answer

The Taylor polynomial of degree 4 for \( f(x) = \frac{1}{1+x} \) about \( a=0 \) is \( P_4(x) = 1 - x + x^2 - x^3 + x^4 \).

Step 1: Identify the Function and Expansion Point

We are given the function \( f(x) = \frac{1}{1+x} \) and we need to compute the Taylor polynomial about \( a = 0 \) of degree \( n = 4 \). The expansion point \( a \) is the value at which the function is expanded.

Step 2: Recall the Taylor Series Formula

The Taylor series expansion of a function \( f(x) \) around a point \( a \) is given by \( P_n(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \cdots + \frac{f^n(a)}{n!}(x-a)^n \).

Step 3: Find the Derivatives of the Function

Calculate the derivatives of \( f(x) = \frac{1}{1+x} \): 1. \( f'(x) = -\frac{1}{(1+x)^2} \)2. \( f''(x) = \frac{2}{(1+x)^3} \)3. \( f'''(x) = -\frac{6}{(1+x)^4} \)4. \( f^{(4)}(x) = \frac{24}{(1+x)^5} \).

Step 4: Evaluate the Derivatives at \( a = 0 \)

Calculate each derivative at \( x = 0 \):1. \( f(0) = 1 \)2. \( f'(0) = -1 \)3. \( f''(0) = 2 \)4. \( f'''(0) = -6 \)5. \( f^{(4)}(0) = 24 \).

Step 5: Form the Degree 4 Taylor Polynomial

Use the evaluated derivatives to construct the Taylor polynomial:\[ P_4(x) = 1 - 1x + \frac{2}{2!}x^2 - \frac{6}{3!}x^3 + \frac{24}{4!}x^4 \].

Step 6: Simplify the Polynomial

Simplify the expressions in the polynomial:1. \( 1 \)2. \( -x \)3. \( + \frac{2}{2}x^2 = x^2 \)4. \( - \frac{6}{6}x^3 = -x^3 \)5. \( + \frac{24}{24}x^4 = x^4 \).Thus, the polynomial is: \( P_4(x) = 1 - x + x^2 - x^3 + x^4 \).

Key Concepts

Taylor Series

In mathematics, a **Taylor series** is a way to express a function as an infinite sum of terms calculated from the values of its derivatives at a single point. It's like breaking down a complex curve into a simpler polynomial that hugs the curve closely. The point about which the function is expanded is called the expansion point, often denoted as \(a\).

The general formula for a Taylor series of a function \(f(x)\) around an expansion point \(a\) is given by:

• \(P_n(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \cdots + \frac{f^n(a)}{n!}(x-a)^n \)

This formula shows that each term of the series is derived from the function’s derivatives evaluated at \(a\), weighted by a factorial term, and powers of \((x-a)\).

When you're asked to find a Taylor polynomial of degree \(n\), it means you're looking for the polynomial that includes all terms up to \(n\). This simplifies the concept of a Taylor series by ending the series at a certain number of terms. It's a crucial tool in approximation and is widely used in calculus.

Derivatives

Understanding **derivatives** is essential when building a Taylor polynomial. A derivative represents the rate at which a function is changing at any point along its curve. To find the Taylor polynomial, you need several derivatives of the function, as these form the coefficients in the polynomial.

Let's break down the process using the function \( f(x) = \frac{1}{1+x} \):

• The first derivative \(f'(x) = -\frac{1}{(1+x)^2}\) demonstrates how quickly \(f(x)\) is decreasing since it is negative.
• The second derivative \(f''(x) = \frac{2}{(1+x)^3}\) indicates how the slope of \(f(x)\) is changing, revealing the function’s concavity.
• Further derivatives, like \(f'''(x) = -\frac{6}{(1+x)^4}\) and \(f^{(4)}(x) = \frac{24}{(1+x)^5}\), continue to refine the curve's description, highlighting more intricate behaviors such as the rate of change of the slope's rate of change.

Each derivative provides a piece of the puzzle, telling us about the function's shape and changes at the chosen point, helping us construct a more accurate polynomial approximation.

Polynomial Simplification

The final stage in forming a Taylor polynomial is **simplifying the polynomial** expression. This involves substituting the evaluated derivatives back into the formula, adjusting for factorials, and simplifying any algebraic expressions.

To illustrate, after computing the derivatives of \( f(x) = \frac{1}{1+x} \) and evaluating them at \(a = 0\), we formed the degree 4 polynomial:

• \( P_4(x) = 1 - 1x + \frac{2}{2!}x^2 - \frac{6}{3!}x^3 + \frac{24}{4!}x^4 \)

Simplifying each term gives us:

• \( -1x \rightarrow -x \)
• \( + \frac{2}{2}x^2 \rightarrow x^2 \)
• \( - \frac{6}{6}x^3 \rightarrow -x^3 \)
• \( + \frac{24}{24}x^4 \rightarrow x^4 \)

Thus, the simplification leads us to the polynomial \( P_4(x) = 1 - x + x^2 - x^3 + x^4 \).Breaking down the polynomial ensures precision and helps to better approximate the original function within a certain interval.