Problem 32 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

Use any method to find the Maclaurin series for \(f(x) .\) (Strive for efficiency.) Determine the radius of convergence. $$ f(x)=\cos ^{2} x $$

Step-by-Step Solution

Answer

The Maclaurin series for \(f(x) = cos^2(x)\) is \(f(x) = \frac{1+\sum_{n=0}^\infty (-1)^n \frac{(2x)^{2n}}{(2n)!}}{2}\). The radius of convergence is \(\infty\).

Step 1: Express the function in a simpler form

First express the function \(f(x)\) in a form that's easy to find its Maclaurin series. The function \(f(x) = \cos^2(x)\) can be rewritten as \(f(x) = \frac{1+\cos(2x)}{2}\).

Step 2: Find the Maclaurin series

The Maclaurin series for \(\cos(x)\) is already known, \(cos(x) = \sum_{n=0}^\infty (-1)^n \frac{x^{2n}}{(2n)!}\) . Thus, the Maclaurin series for \(\cos(2x)\) is \(cos(2x) = \sum_{n=0}^\infty (-1)^n \frac{(2x)^{2n}}{(2n)!}\). Substituting this into the simplified formula of \(f(x)\), the Maclaurin series for \(f(x)\) becomes \(f(x) = \frac{1+\sum_{n=0}^\infty (-1)^n \frac{(2x)^{2n}}{(2n)!}}{2}\).

Step 3: Determine the radius of convergence

For the Maclaurin series \(f(x) = \frac{1+\sum_{n=0}^\infty (-1)^n \frac{(2x)^{2n}}{(2n)!}}{2}\), we use the ratio test to find the radius of convergence. The absolute value of the ratio between the (n+1)th term and the nth term should be less than 1. In this case, because the series is derived from the well-known cosine function whose radius of convergence is \(\infty\), thus the radius of convergence of the series for \(f(x)\) is also \(\infty\).

Key Concepts

Radius of Convergence

The radius of convergence is a key concept in understanding series. It tells us where the series converges, or adds up to a meaningful number. Think of it as the distance within which the series is valid around the center point, known as the origin when dealing with Maclaurin series.
Why does it matter? Well, outside this radius, the series might not represent a function accurately. To determine the radius of convergence, we often use the ratio test. For a series derived from the cosine function, like in this problem, the radius is infinite. This means the series works for all real numbers, capturing the essence of the cosine function everywhere.

Cosine Function

The cosine function, \(\cos(x)\), is fundamental in trigonometry and appears in various mathematical analyses. It oscillates between -1 and 1, portraying the wave-like behavior which is crucial in fields like physics and engineering.
In this exercise, we break it down further by exploring \(\cos^2(x)\). By expressing \(\cos^2(x)\) as \(\frac{1 + \cos(2x)}{2}\), we utilize a trigonometric identity. This simplification helps us find the Maclaurin series more efficiently, as the double angle formula reduces the complexity involved in deriving the series directly from \(\cos^2(x)\).
Understanding these properties lets us seamlessly transition between different forms of the function, opening the door to a deeper understanding of wave patterns and oscillatory behaviors.

Series Expansion

Series expansion is the process of expressing a function as a sum of its terms, typically involving powers of variables. The Maclaurin series is a specific type that expands a function around zero, making it particularly useful for evaluating functions at small values.

Here's how it works in this context:

• We rewrite \(\cos^2(x)\) using a double angle identity to simplify the function.
• We then use the known Maclaurin series for \(\cos(x)\), which is \( \sum_{n=0}^\infty (-1)^n \frac{x^{2n}}{(2n)!} \).
• Substituting \(2x\) gives us the series for \(\cos(2x)\).
• Combine these results to construct the series for \(f(x)\).

The series expansion lets us approximate functions succinctly and accurately within their convergence radius, making complex functions much easier to work with for calculations and predictions. It simplifies both theoretical analysis and practical application in various scientific fields.