Problem 30 - Applied Calculus: For Business, Economics, and the Social and Life Sciences

Question

Find the Taylor series for the given function at the specified value of \(x=a\). $$ f(x)=\frac{1-x}{1+x} ; a=0 $$

Step-by-Step Solution

Answer

The Taylor series for \(f(x) = \frac{1 - x}{1 + x}\) at \(x = 0\) is \[ -x + x^2 - x^3 + x^4 - x^5 + \text{...}\].

Step 1: Function Evaluation at a = 0

First, evaluate the function at the given point, which is \(a = 0\). For \(f(x) = \frac{1 - x}{1 + x}\), substitute \(x = 0\): \[ f(0) = \frac{1 - 0}{1 + 0} = 1 \]

Step 2: First Derivative

Find the first derivative of the function. We have \[ f(x) = \frac{1 - x}{1 + x} \] Using the quotient rule \( (u/v)' = \frac{u'v - uv'}{v^2} \), where \(u = 1 - x\) and \(v = 1 + x\): \[ f'(x) = \frac{-(1+x) - (1-x)}{(1+x)^2} = \frac{-2}{(1+x)^2} \] Evaluate it at \(x = 0\): \[ f'(0) = \frac{-2}{(1+0)^2} = -2 \]

Step 3: Second Derivative

Find the second derivative of the function. Start from the first derivative: \[ f'(x) = \frac{-2}{(1+x)^2} \] Use the chain rule for the second derivative: \[ f''(x) = -2 \times -2 (1+x)^{-3} \times (1+x)' \] Simplify: \[ f''(x) = 4 (1+x)^{-3} \] Evaluate it at \(x = 0\): \[ f''(0) = 4 (1+0)^{-3} = 4 \]

Step 4: Higher Order Derivatives

Continue finding higher-order derivatives and evaluating them at \(x = 0\) for as many terms as needed. For better understanding, calculate the third derivative. \[ f''(x) = 4 (1+x)^{-3} \] Using the chain rule again: \[ f'''(x) = 4 \times -3 (1+x)^{-4} \times (1+x)' = -12 (1+x)^{-4} \] Evaluate it at \(x = 0\): \[ f'''(0) = -12 (1+0)^{-4} = -12 \]

Step 5: General Derivatives and Pattern Recognition

Observe the pattern in the derivatives. For \(n \geq 1\), it appears \[ f^{(n)}(x) = (-1)^{n+1} n! (1+x)^{-(n+1)} \] Evaluate at \(x = 0\): \[ f^{(n)}(0) = (-1)^{n+1} n! \]

Step 6: Taylor Series Expansion

Construct the Taylor series using the formula: \[ f(x) = \frac{1 - x}{1 + x} = \text{sum from n=0 to }\text{infinity of} \frac{f^{(n)}(0)}{n!} x^n \] For our function: \[ f(x) = \text{sum from n=0 to }\text{infinity of} \frac{(-1)^{n+1} n!}{n!} x^n = \text{sum from n=0 to }\text{infinity of} (-1)^{n+1} x^n \]

Key Concepts

Taylor series

A Taylor series allows us to represent a function as an infinite sum of terms calculated from the values of its derivatives at a single point. Using Taylor series, complex functions can be approximated by polynomials, which are often much simpler to work with. The general formula for the Taylor series of a function \(f(x)\) about the point \(x = a\) is: \[ f(x) = \text{sum from n=0 to infinity of} \frac{f^{(n)}(a)}{n!} (x-a)^n \].
This series expands the function into an infinite series of terms, each governed by the function's \(n\)-th derivative at the point \(a\). The Taylor series is highly useful in various fields like physics, engineering, and computer science, as it helps in solving differential equations and in numerical analysis.

Function derivatives

Derivatives measure the rate of change of a function. For constructing a Taylor series, the first few derivatives of the function at a given point are crucial. Let's revisit our example:
Step 1: Start by evaluating the function at \(a = 0\). We have \(f(0) = \frac{1-0}{1+0} = 1\).
Step 2: Find the first derivative using the quotient rule. For \(f(x) = \frac{1-x}{1+x}\), \[ f'(x) = \frac{-2}{(1+x)^2} \]. Evaluate it at \(x = 0\): \(f'(0) = -2\).
Step 3: For the second derivative, apply the chain rule to \(f'(x)\). \[ f''(x) = 4 (1+x)^{-3} \] and evaluate at \(x=0\): \(f''(0) = 4\).
Higher-order derivatives follow the same pattern. Recognizing patterns in the derivatives is key to constructing the series efficiently.

Series expansion

Series expansion is a way of representing functions as sums of simpler components, typically powers of \(x\). In our problem, we use the Taylor series to expand \(f(x) = \frac{1-x}{1+x}\) around \(x = 0\). By substituting derivatives into the Taylor series formula, we obtain a power series:\[ f(x) = \frac{1-x}{1+x} = \text{sum from n=0 to infinity of} (-1)^{n+1} x^n \].
Each term in the series \((-1)^{n+1} x^n\) gives us an approximation of the function. The more terms we include, the closer the approximation to the actual function. This technique is powerful in approximations that require high precision.

Pattern recognition in derivatives

Recognizing patterns in derivatives is invaluable for deriving Taylor series efficiently. By observing the derivatives, we can establish a formula for each derivative:\[ f^{(n)}(x) = (-1)^{n+1} n! (1+x)^{-(n+1)} \]. Evaluating this at \(x = 0\): \[ f^{(n)}(0) = (-1)^{n+1} n! \].
This pattern allows us to write the Taylor series compactly without computing every individual derivative from scratch. Recognizing these patterns simplifies and accelerates the process of series expansion.
Pattern recognition extends beyond Taylor series and is useful in numerous calculus problems where functions exhibit regular behavior in their differential properties.