Using the Maclaurin Series of sin(x) to Demonstrate Its Bounds Between -1 and 1

In this article, we will explore how the Maclaurin series of the sine function, sin(x), can be used to demonstrate that this function lies between -1 and 1. This analysis highlights the power and utility of the Maclaurin series in understanding the behavior of trigonometric functions.

Introduction to the Maclaurin Series of sin(x)

The Maclaurin series for the sine function is given by:

sin(x) x - frac{x^3}{3!} frac{x^5}{5!} - frac{x^7}{7!} cdots

This series can be written in summation notation as:

sin(x) sum_{n0}^{infty} frac{(-1)^n x^{2n 1}}{(2n 1)!}

The series expands sin(x) as an infinite sum of terms, each of which is a polynomial of increasing degree. Importantly, the series converges for all x due to the nature of the factorial in the denominator, which grows extremely quickly.

Step 1: Analyzing the Series

To show that sin(x) lies between -1 and 1, we begin by considering the behavior of the function for values of x in the interval [-1, 1]. This is a common approach in calculus to understand the range of a function.

Step 2: Considering the Derivative

In addition to analyzing the series itself, we can also look at its derivative, which is the cosine function, cos(x). This will help us further understand the behavior and properties of sin(x).

Step 3: Evaluating sin(x) at Specific Points

Let's evaluate the function at several key points:

sin(0) 0 sin(frac{pi}{2}) 1 sin(-frac{pi}{2}) -1

These evaluations help us understand the range and extreme values of the function.

Step 4: Using Properties of Sine

The sine function is periodic with a period of 2pi and is continuous. Importantly, the maximum value of sin(x) is 1 and the minimum is -1. These properties can be demonstrated using the first and second derivatives:

critical points occur where cos(x) 0, which are at x frac{pi}{2} npi, leading to sin(x) pm 1. the second derivative test confirms these are maxima or minima.

Step 5: Concluding Boundedness

Combining these insights, we conclude that:

-1 leq sin(x) leq 1 quad text{for all } x in mathbb{R}

This result shows that the Maclaurin series of sin(x) converges to values that demonstrate the function's bounded nature.

Revisiting the Function Through Taylor Series

Given the Taylor series of an odd function, such as our sin(x), we can further show that this function is bounded between -1 and 1 for all x in mathbb{R}. This approach leverages the properties of complex exponentials and the periodicity of the sine function.

First Method: Using Taylor Series and Euler's Formula

We start by expressing sin(x) as:

sin(x) frac{e^{ix} - e^{-ix}}{2i}

Evaluating at x frac{pi}{2} gives sin(frac{pi}{2}) 1 and sin(-frac{pi}{2}) -1. Considering the function's periodicity and the fact that sin(x) is bounded by its second derivative, it follows that:

sin(x) leq 1

Second Method: Numerical Evidence

We can also verify this through numerical evidence. For example, consider the series:

sin(x) x - frac{x^3}{3!} frac{x^5}{5!} - frac{x^7}{7!} cdots

Evaluating this series for values of x in the interval [0, frac{pi}{2}]), we observe that the maximum value approached is:

sin_3(x) x - frac{x^3}{3!} frac{x^5}{5!} - frac{x^7}{7!} leq 1 sin_5(x) sin_3(x) - frac{x^9}{9!} frac{x^{11}}{11!} leq 1 sin_7(x) sin_5(x) - frac{x^{13}}{13!} frac{x^{15}}{15!} leq 1

The numerical evidence supports the theoretical understanding that sin(x) is indeed bounded between -1 and 1.

Conclusion

This exploration of the Maclaurin series and Taylor series of the sine function demonstrates its bounded nature. By understanding and analyzing the series, we can confidently state that sin(x) lies between -1 and 1 for all real values of x. This knowledge is fundamental in many areas of mathematics and its applications.