Question

In each exercise, determine the largest positive integer \(r\) such that \(q(h)=O\left(h^{r}\right) .\) [Hint: Determine the first nonvanishing term in the Maclaurin expansion of \(q .\) ] \(q(h)=\sin 2 h\)

Short answer

Answer: The largest positive integer r is 1.

Step-by-step solution

Recall the Maclaurin series formula

The Maclaurin series for a function f(h) is given by the formula: \(f(h) = f(0) + f'(0)h + \frac{f''(0)}{2!}h^2 + \frac{f'''(0)}{3!}h^3 + \cdots\) We will now find the Maclaurin series for \(q(h) = \sin(2h)\).

Calculate derivatives of q(h)

Let's find the first few derivatives of q(h), evaluated at h = 0: \(q(h) = \sin(2h)\) \(q'(h) = \cos(2h) \cdot 2\) \(q''(h) = -\sin(2h) \cdot 4\) \(q'''(h) = -\cos(2h) \cdot 8\) \(q^{(4)}(h) = \sin(2h) \cdot 16\) Now, we will evaluate these derivatives at h = 0: \(q(0) = \sin(2 \cdot 0) = 0\) \(q'(0) = \cos(2 \cdot 0) \cdot 2 = 2\) \(q''(0) = -\sin(2 \cdot 0) \cdot 4 = 0\) \(q'''(0) = -\cos(2 \cdot 0) \cdot 8 = -8\) \(q^{(4)}(0) = \sin(2 \cdot 0) \cdot 16 = 0\)

Write down the Maclaurin series for q(h)

Using the evaluated derivatives, we can now write down the Maclaurin series for q(h): \(q(h) = 0 + 2h + \frac{0}{2!}h^2 + \frac{-8}{3!}h^3 + \cdots\) Simplifying the Maclaurin series, we have: \(q(h) = 2h - \frac{4}{3}h^3 + \cdots\)

Determine the order of q(h)

The first non-vanishing term in the Maclaurin expansion of q(h) is \(2h\). Since the coefficient of the first non-vanishing term is 2, we have \(q(h) = O(h)\). Therefore, the largest positive integer \(r\) such that \(q(h) = O(h^r)\) is \(r = 1\).

Key concepts

Differential Equations

Differential equations play a pivotal role in mathematics, physics, engineering, and many other disciplines. They are used to describe situations where there is a continuously changing dynamic, such as growth rates, motion, or electrical currents, and they serve as the foundation for understanding phenomena in these fields.

A differential equation is a type of equation that involves the derivatives of a function. These derivatives represent rates of change, and the differential equation defines a relationship between the original function and its derivatives. Solving a differential equation typically means finding a function (or set of functions) that satisfy the given relationship. For instance, the equation presented in the exercise, \( q(h) = \sin(2h) \), can be thought of as a solution to a differential equation, where the function describes the position of a particle over time undergoing simple harmonic motion.

To connect the concept back to the exercise, by finding the Maclaurin series of \( q(h) \) and evaluating its derivatives at zero, we essentially investigate the behavior of the particle at the initial moment, giving us insights into the nature of the underlying differential equation.

Power Series Expansion

The concept of a power series expansion is a cornerstone of calculus and applies to many areas of mathematics, including solving differential equations. A power series is an infinite series in the form \( \sum_{n=0}^\infty a_n x^n \), where \( a_n \) are coefficients, and \( x \) is the variable of the function.

The Maclaurin series is a specific type of power series expansion where we expand a function around the value 0. This series is especially useful because it can represent a wide range of functions through polynomials, making it easier to handle them algebraically or numerically. As seen in the exercise, the Maclaurin series for the sine function, when multiplied by a constant (in this case, 2), provides a polynomial approximation that we can truncate at any degree of accuracy we need.

The goal when constructing the Maclaurin series is to find the coefficients \( a_n \) by deriving the function multiple times, setting the variable to zero, and then using these coefficients to build the series. The power of such expansions lies in their ability to approximate complex functions with simple polynomials over certain intervals, making them a powerful tool in solving differential equations and analyzing functions near a point.

Order of Approximation

In the study of functions and their approximations, the term 'order of approximation' refers to the degree of accuracy of a polynomial when approximating a function. It indicates the highest power of the variable for which the approximation and the original function's expansion match. Higher orders generally mean better approximations.In the context of our exercise, the order of approximation is determined by the first non-vanishing term in the Maclaurin series of \( q(h) \). As you refine the expansion of \( q(h) \), which is \( q(h) = 2h - \frac{4}{3}h^3 + \cdots \), you can see that the approximation begins with the term \( 2h \). This term indicates that the first significant effect on the approximation occurs linearly with \( h \). Thus, the order of approximation is one, signified by the notation \( q(h) = O(h) \).

Understanding the order of approximation has practical implications. For example, in numerical methods, selecting an appropriate order of approximation affects the accuracy and performance of the calculations. In the dynamic systems described by differential equations, the order of approximation can provide insights into system behaviors close to the initial conditions, guiding predictions, and analyses.