Question

Find the first few terms of the Maclaurin series for each of the following functions and check your results by computer. $$\sec x=\frac{1}{\cos x}$$

Short answer

The first few terms of the Maclaurin series for \(\sec x\) are \( 1 + \frac{x^2}{2} \).

Step-by-step solution

- Know the Maclaurin Series Definition

The Maclaurin series for a function can be written as: \[ f(x) = f(0) + f'(0)x + \frac{f''(0)x^2}{2!} + \frac{f'''(0)x^3}{3!} + \cdots \] The coefficients are derived from the function's derivatives evaluated at 0.

- Find the Function's Derivatives

Identify the first few derivatives of \(\sec x\). Start by writing \( \sec x = \frac{1}{\cos x} \): \( f(x) = \sec x \) Calculate the first derivative: \( f'(x) = \sec x \tan x \) Second derivative: \( f''(x) = \sec x (\tan^2 x + \sec^2 x) \) Third derivative: \( f'''(x) = \sec x (\tan^3 x + 5 \sec^2 x \tan x) \)

- Evaluate the Derivatives at x=0

Now evaluate each derivative at \( x = 0 \): \( f(0) = \sec 0 = 1 \) \( f'(0) = \sec 0 \tan 0 = 0 \) \( f''(0) = \sec 0 (\tan^2 0 + \sec^2 0) = 1 \) \( f'''(0) = \sec 0 (\tan^3 0 + 5 \sec^2 0 \tan 0) = 0 \)

- Construct the Maclaurin Series

Combine the results into the Maclaurin series form: \[ \sec x = 1 + \frac{1x^2}{2!} + \cdots \] Therefore, the first few terms are: \[ \sec x \approx 1 + \frac{x^2}{2} \]

- Verify Using Computer Software

Use computer algebra software like WolframAlpha or Python's SymPy to confirm the series expansion of \( \sec x \). For WolframAlpha, type 'Taylor series of sec(x) at x=0', and for Python, use the `series` function from SymPy library.

Key concepts

Derivatives

Derivatives are fundamental in calculus. They measure how a function changes as its input changes. Think of it as the slope of a function at a point. For the function \(\frac{d}{dx} f(x)\), it refers to the rate at which \( f(x) \) changes with respect to \( x \).
For the given problem, we need to find the derivatives of \( \sec x \). First, we recognize that \( \sec x \) is \( \frac{1}{\cos x} \). Taking derivatives of this is slightly more complex than for simple functions:
1. First derivative: \( f'(x) = \sec x \tan x \).
2. Second derivative: \( f''(x) = \sec x (\tan^2 x + \sec^2 x) \).
3. Third derivative: \( f'''(x) = \sec x (\tan^3 x + 5 \sec^2 x \tan x) \).
With these derivatives, we can generate the Maclaurin series by evaluating them at \( x = 0 \).

Series Expansion

A series expansion lets us express functions as an infinite sum of terms. The Maclaurin series is a type of series expansion centered at zero. It follows the form:
\[ f(x) = f(0) + f'(0)x + \frac{f''(0)x^2}{2!} + \frac{f'''(0)x^3}{3!} + \cdots \]
Each term in this series involves a derivative of the function evaluated at zero (\f(0), f'(0), etc.).
For \( \sec x \), we calculate a few terms of the series as follows:
- \( f(0) = \sec 0 = 1 \)
- \( f'(0) = \sec 0 \tan 0 = 0 \)
- \( f''(0) = \sec 0 (\tan^2 0 + \sec^2 0) = 1 \)
- \( f'''(0) = \sec 0 (\tan^3 0 + 5 \sec^2 0 \tan 0) = 0 \)
Adding these up, we get:
\[ \sec x \approx 1 + \frac{x^2}{2} \]

Secant Function

The secant function, denoted as \( \sec x \), is the reciprocal of the cosine function: \( \sec x = \frac{1}{\cos x} \).
In the unit circle context, when \( x \) represents an angle, the secant function gives the ratio of the hypotenuse to the adjacent side. Therefore, it can also be understood as a measure of the stretch factor required in the circle.
Secant is especially useful in trigonometry and calculus because it frequently appears in integrals and series expansions. Secant’s behavior near zero is interesting because \( \cos(0) = 1 \), so \( \sec(0) = 1 \).
This makes the computation simpler initially, but as angles grow, the secant function increases quicker than both sine and cosine.

Calculus

Calculus is the mathematical study of continuous change. It is divided mostly into differential calculus and integral calculus.
- **Differential Calculus** deals with the concept of the derivative. It provides tools to determine the rate of change of functions.
- **Integral Calculus** deals with the concept of the integral. It helps us in finding areas, volumes, and accumulations.
The problem involving the Maclaurin series is a typical case utilizing differential calculus. By differentiating a function multiple times and then evaluating at zero, we unravel the function’s behavior around that point.
Finally, using these derivatives to build a series helps approximate complex functions such as \( \sec x \) with simple polynomials. This method is a powerful tool in both theoretical and applied mathematics.