Problem 17 - Essential Calculus

Question

In Exercises \(7-18\), find the Maclaurin polynomial of degree \(n\) for the function. $$ f(x)=\sec x, \quad n=2 $$

Step-by-Step Solution

Answer

Maclaurin polynomial of degree 2 for \(f(x) = \sec x\) is \(1 + \frac{x^2}{2}\).

Step 1: Calculation First and Second Derivatives at 0

First, the first and second derivatives of the function \(f(x) = \sec x\) at \(x=0\) need to be calculated. Using the chain rule and the definition of the derivative, we start by finding \(f'(x) = \sec x \tan x\) and \(f''(x) = \sec x \tan^2 x + \sec^3 x\). Now, substituting \(x=0\) into these, we have: \(f'(0) = \sec(0) \tan(0) = 1*0 = 0\) and \(f''(0) = \sec(0) \tan^2(0) + \sec^3(0) = 1*0 + 1 = 1\).

Step 2: Maclaurin Polynomial Formulation

Next we use these derivatives to form our Maclaurin polynomial. A Maclaurin series is a Taylor series expansion of a function about 0, in the form of: \( f(x) = f(0) + \frac{f'(0)}{1!} x + \frac{f''(0)}{2!} x^2 + \cdots\). Using our calculated derivatives and \(f(0)\), our polynomial with degree \(n = 2\) becomes: \(1 + 0 \cdot x + \frac{1}{2} \cdot x^2\).

Key Concepts

Taylor Series

The Taylor Series is a powerful mathematical tool used to approximate functions. It is essentially an infinite series of calculations that can represent functions as polynomials. The series is centered around a particular point, which is often chosen to make calculations straightforward. When the center point is zero, the expansion is called a Maclaurin Series. This is exactly what we're dealing with in the current exercise. The Maclaurin Series is just a special case of the Taylor Series.

Here's the general form of a Taylor series expansion for a function \(f(x)\) about a point \(a\):

• \[f(x) = f(a) + \frac{f'(a)}{1!}(x - a) + \frac{f''(a)}{2!}(x - a)^2 + \cdots\]

In the case of a Maclaurin series, \(a = 0\), simplifying the expression to:

• \[f(x) = f(0) + \frac{f'(0)}{1!}x + \frac{f''(0)}{2!}x^2 + \cdots\]

Using Taylor Series to approximate functions is common when attempting to simplify complex calculations, making them easier to handle.

Polynomial Derivatives

Polynomial derivatives are straightforward yet crucial when dealing with Taylor and Maclaurin series. Derivatives describe the rate at which a function's output changes concerning its input. Calculating derivatives is necessary for finding coefficients in the Taylor or Maclaurin polynomial expansion. For instance, while finding the Maclaurin polynomial for \(\sec x\) in the exercise, the first two derivatives were computed:

• First derivative: \(f'(x) = \sec x \tan x\)
• Second derivative: \(f''(x) = \sec x \tan^2 x + \sec^3 x\)

At \(x = 0\):

• \(f'(0) = 0\)
• \(f''(0) = 1\)

Knowing how to perform these calculations is fundamental in calculus and lays the groundwork for more advanced concepts. Each derivative contributes to the corresponding term in the polynomial, enabling us to build an approximation of the function.

Calculus Education

Calculus education often focuses on how to break down complex problems into manageable parts. Understanding series like Taylor and Maclaurin is integral in grasping the broader scope of calculus. These series are an entry point into more advanced topics like differential equations and numerical analysis.

Students learn to:

• Enhance their problem-solving skills by practicing derivative calculations.
• Understand how infinite processes can represent finite solutions.
• Apply theoretical knowledge practically, approximating real-world problems.

The step-by-step approach demonstrated in finding the Maclaurin polynomial not only enhances comprehension but also shows the power of calculus in providing solutions to problems that initially seem daunting. This aspect of calculus education is crucial as it equips students with a systematic method of tackling mathematical challenges.