Problem 29 - Algebra Form and Function

Question

Refer to the functions \(f(x)\) and \(g(x),\) where the function $$ g(x)=1+\frac{1}{2} x+\frac{3}{8} x^{2}+\frac{5}{16} x^{3} $$ is used to approximate the values of $$ f(x)=\frac{1}{\sqrt{1-x}} $$ Evaluate \(f(x)\) and \(g(x)\) at \(x=0.1,0.2,0.3\) and record your answers to three decimal places in a table. Does your table support the claim that \(g(x)\) is a good approximation to \(f(x)\) for these values of \(x\) ?

Step-by-Step Solution

Answer

Answer: Yes, g(x) is a good approximation for f(x) for x=0.1, 0.2, and 0.3, as the values of g(x) are close to the values of f(x) and the differences between them are small.

Step 1: Evaluate f(x) at x = 0.1, 0.2, and 0.3

To evaluate f(x) at the given values of x, simply substitute the values into f(x) and compute: $$ f(0.1) = \frac{1}{\sqrt{1-0.1}} = \frac{1}{\sqrt{0.9}} \approx 1.053 $$ $$ f(0.2) = \frac{1}{\sqrt{1-0.2}} = \frac{1}{\sqrt{0.8}} \approx 1.118 $$ $$ f(0.3) = \frac{1}{\sqrt{1-0.3}} = \frac{1}{\sqrt{0.7}} \approx 1.195 $$

Step 2: Evaluate g(x) at x = 0.1, 0.2, and 0.3

To evaluate g(x) at the given values of x, substitute the values into g(x) and compute: $$ g(0.1) = 1 + \frac{1}{2}(0.1) + \frac{3}{8}(0.1^2) + \frac{5}{16}(0.1^3) \approx 1.051 $$ $$ g(0.2) = 1 + \frac{1}{2}(0.2) + \frac{3}{8}(0.2^2) + \frac{5}{16}(0.2^3) \approx 1.104 $$ $$ g(0.3) = 1 + \frac{1}{2}(0.3) + \frac{3}{8}(0.3^2) + \frac{5}{16}(0.3^3) \approx 1.147 $$

Step 3: Create the table

Now, we can record the values of f(x) and g(x) for the given values of x in a table: | x | f(x) | g(x) | |-------|--------|--------| | 0.1 | 1.053 | 1.051 | | 0.2 | 1.118 | 1.104 | | 0.3 | 1.195 | 1.147 |

Step 4: Analyze the results

By comparing the values of f(x) and g(x) in the table, we can see that g(x) is indeed a good approximation for f(x) for \(x=0.1,0.2,0.3\). The values of g(x) are close to the values of f(x), and the differences between them are small. Therefore, the claim that \(g(x)\) is a good approximation for \(f(x)\) for these values of x is supported by the table.

Key Concepts

Function Evaluation

When dealing with Taylor series and function approximations, evaluating a function is a crucial step. Function evaluation involves substituting specific values into a given function to find its output. In this exercise, we evaluate the functions \( f(x) \) and \( g(x) \). The function \( f(x) = \frac{1}{\sqrt{1-x}} \) is the original mathematical function, while \( g(x) = 1 + \frac{1}{2}x + \frac{3}{8}x^2 + \frac{5}{16}x^3 \) is an approximation of \( f(x) \).
To evaluate \( f(x) \) at different points, substitute the values of \( x = 0.1, 0.2, \) and \( 0.3 \) into \( f(x) \). This involves basic arithmetic operations and square root calculations:

• \( f(0.1) = \frac{1}{\sqrt{1-0.1}} \approx 1.053 \)
• \( f(0.2) = \frac{1}{\sqrt{1-0.2}} \approx 1.118 \)
• \( f(0.3) = \frac{1}{\sqrt{1-0.3}} \approx 1.195 \)

Evaluating \( g(x) \) involves substituting the same values into the polynomial approximation:

• \( g(0.1) \approx 1.051 \)
• \( g(0.2) \approx 1.104 \)
• \( g(0.3) \approx 1.147 \)

This sets the stage for understanding how close the approximation \( g(x) \) is to the original \( f(x) \).

Approximation Accuracy

Accuracy in approximation is all about how closely the approximation function, \( g(x) \), matches the original function, \( f(x) \). Especially in mathematical modeling, obtaining an accurate approximation is vital. It helps in simplifying complex functions for easier calculations while still retaining acceptable precision.

In this example, we approximate \( f(x) \) using a Taylor series expansion \( g(x) \). The accuracy can be checked by comparing the values of both functions at certain points:

• At \( x = 0.1 \), \( f(x) \approx 1.053 \) and \( g(x) \approx 1.051 \)
• At \( x = 0.2 \), \( f(x) \approx 1.118 \) and \( g(x) \approx 1.104 \)
• At \( x = 0.3 \), \( f(x) \approx 1.195 \) and \( g(x) \approx 1.147 \)

The small differences between \( f(x) \) and \( g(x) \) demonstrate that \( g(x) \) has high accuracy for these values of \( x \). The better the accuracy, the more reliable \( g(x) \) is as an approximation, validating the mathematical effort behind constructing \( g(x) \).

Mathematical Functions

Mathematical functions are fundamental in mathematics, providing a way to express relationships between variables. In this context, \( f(x) = \frac{1}{\sqrt{1-x}} \) represents a common mathematical function found in calculus and analysis, often related to binomial expansions.

Approximating complex functions like \( f(x) \) often involves creating polynomial functions, as polynomials are easier to compute and analyze.
Taylor series, like \( g(x) \) in this problem, offer a method for approximating functions by breaking them down into polynomial expressions:

• They capture the behavior of a complex function around a specific point.
• The approximation becomes more accurate with more terms.
• They simplify calculations by reducing the need for complicated operations.

Understanding these functions and their approximations can make solving problems in mathematics more approachable and teaches important concepts about the behavior of functions and the idea of convergence in sequences of functions.