Problem 43 - Calculus

Question

Explain what is wrong with the statement. Let \(f(x)=\frac{1}{1-x} .\) The coefficient of the \(x\) term of the Taylor polynomial of degree 3 approximating \(f(x)\) near \(x=0\) is -1

Step-by-Step Solution

Answer

The coefficient of the \(x\) term is +1, not -1.

Step 1: Find the Taylor Series Formula

The Taylor series of a function \( f(x) \) centered at \( x = a \) is given by \( f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \ldots \). In this exercise, we need to find the Taylor series for \( f(x) = \frac{1}{1-x} \) near \( x=0 \). This means \( a=0 \).

Step 2: Find the Derivatives at \( x = 0 \)

The function \( f(x) = \frac{1}{1-x} \) is a geometric series \( 1 + x + x^2 + x^3 + \ldots \). Therefore, its derivatives are: \( f'(x) = \frac{1}{(1-x)^2} \), \( f''(x) = \frac{2}{(1-x)^3} \), and \( f'''(x) = \frac{6}{(1-x)^4} \). Notice at \( x = 0 \) these are \( f(0) = 1 \), \( f'(0) = 1 \), \( f''(0) = 2 \), and \( f'''(0) = 6 \).

Step 3: Build the Taylor Polynomial of Degree 3

Substitute the values into the Taylor series equation: \[ f(x) \approx 1 + (1)x + \frac{2}{2!}x^2 + \frac{6}{3!}x^3 \]. Simplifying, we get \( f(x) \approx 1 + x + x^2 + x^3 \).

Step 4: Identify the Coefficient of the \(x\) Term

In the approximated Taylor polynomial, \( f(x) \approx 1 + x + x^2 + x^3 \), the coefficient of the \(x\) term is \(+1\).

Step 5: Compare with the Statement

The statement claims that the coefficient of \( x \) in the Taylor polynomial is \(-1\), but as calculated, the correct coefficient for the \( x \) term in the Taylor polynomial of \( f(x) \) near \( x=0 \) is \(+1\).

Key Concepts

Taylor Series

The Taylor Series is a way to represent a function as an infinite sum of terms. Each term is calculated from the derivatives of the function at a single point. When approximating functions near a particular point, Taylor series can be very useful because they allow for polynomial approximations that are easier to work with. The general formula for a Taylor series centered at a point \( a \) is:\[f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \ldots\]For a Taylor polynomial of degree \( n \), simply stop after the \( n \)-th term. This allows us to approximate a function with a polynomial up to degree \( n \), offering a more straightforward expression to calculate.In this exercise, the focus is on the function \( f(x) = \frac{1}{1-x} \). This function is being evaluated around \( x=0 \), meaning it involves a Maclaurin series, a specific Taylor series where \( a=0 \). Understanding Taylor series is vital as it helps simplify complex functions for easier analysis and calculation.

Geometric Series

A geometric series is a type of series where each term is a constant multiple of the previous term. It has a straightforward formula if summed to infinity, provided that the common ratio is between -1 and 1:\[s = \frac{a}{1-r}\]Where \( a \) is the first term and \( r \) is the common ratio. In the context of the problem, \( f(x) = \frac{1}{1-x} \) is the sum of a geometric series where each term is derived from the previous one, multiplied by \( x \). This results in:\[f(x) = 1 + x + x^2 + x^3 + \cdots\]When analyzing functions like this, understanding the geometric series allows us to quickly grasp the structure and behavior of the function without deriving every term from scratch. Simply by recognizing it as a geometric series, we have a powerful tool to approximate and explore the behavior of functions like \( \frac{1}{1-x} \). By using geometric series, we can recognize that the Taylor series for this particular function matches the sum of this infinite geometric series.

Derivative Calculation

Derivatives are fundamental in finding Taylor series since they help determine the coefficients of the terms in the series. For a function \( f(x) \), its derivatives at a specific point give us the rates at which the function value is changing. Here, determining derivatives of \( f(x) = \frac{1}{1-x} \) is key to building its Taylor series.The first few derivatives are:

• First derivative: \( f'(x) = \frac{1}{(1-x)^2} \)
• Second derivative: \( f''(x) = \frac{2}{(1-x)^3} \)
• Third derivative: \( f'''(x) = \frac{6}{(1-x)^4} \)

Evaluating these at \( x = 0 \) gives us coefficients used in the Taylor polynomial:

• \( f(0) = 1 \)
• \( f'(0) = 1 \)
• \( f''(0) = 2 \)
• \( f'''(0) = 6 \)

Understanding how to compute these derivatives and their values at specific points is crucial for constructing the Taylor polynomial. Derivative calculations ensure the accuracy of the polynomial in approximating the function near a particular point. This way, we can confidently sum the series to approximate \( f(x) \) around \( x=0 \) accurately.