Problem 9 - Advanced Engineering Mathematics

Question

In Problems \(9-12\), the indicated number is a zero of the given function. Use a Maclaurin or Taylor series to determine the order of the zero. $$ f(z)=z\left(1-\cos z^{2}\right) ; z=0 $$

Step-by-Step Solution

Answer

The order of the zero at \(z=0\) is 5.

Step 1: Review Function and Zero Definition

The function given is \( f(z) = z(1 - \cos(z^2)) \) with \( z = 0 \) as the zero. We need to investigate the function at this point to determine the order of the zero.

Step 2: Determine the Taylor Series for \(\cos(z^2)\)

The Taylor series expansion of \(\cos(x)\) at \(x = 0\) is \(1 - \frac{x^2}{2} + \frac{x^4}{24} - \cdots\). Substituting \(z^2\) for \(x\), the expansion for \(\cos(z^2)\) becomes \(1 - \frac{(z^2)^2}{2} + \frac{(z^2)^4}{24} - \cdots\). Simplifying this gives: \[\cos(z^2) = 1 - \frac{z^4}{2} + \frac{z^8}{24} - \cdots\].

Step 3: Substitute the Series into the Function

Substitute the series expansion into \(f(z)\). This gives: \[f(z) = z\left(1 - \left(1 - \frac{z^4}{2} + \frac{z^8}{24} - \cdots \right) \right) = z\left(\frac{z^4}{2} - \frac{z^8}{24} + \cdots\right).\]

Step 4: Simplify the Series

Simplify \(f(z)\) by distributing \(z\): \[f(z) = \frac{z^5}{2} - \frac{z^9}{24} + \cdots\].

Step 5: Identify the Order of the Zero

The function \(f(z)\) starts with \(\frac{z^5}{2}\), implying that it has its first non-zero term with \(z^5\). Therefore, the smallest power of \(z\) is 5, which means the order of the zero at \(z = 0\) is 5.

Key Concepts

Maclaurin series

Understanding the concept of a Maclaurin Series is crucial when analyzing functions about the zero point. The Maclaurin Series is essentially a special case of Taylor Series and is used for function expansion at zero. It can be expressed as follows:

• A Taylor Series expanded around zero.
• Follows the formula: \(f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \cdots \)
• Takes advantage of known derivatives at zero to approximate functions.

In many cases, especially when dealing with polynomials or trigonometric functions like \( \cos(x) \), the Maclaurin Series provides a clear path to understand behavior near zero. For instance, using a Maclaurin series for \( \cos(x) \) at zero we get \( 1 - \frac{x^2}{2} + \frac{x^4}{24} - \cdots \). These expressions help us factor and simplify complex functions around the origin.

order of a zero

The 'order of a zero' in functions refers to the first non-zero derivative term in a power series expansion. This concept is critical to understand

• The number of times a function's output becomes zero consecutively as you evaluate closer to the zero point.
• How steep or flat a curve is at the zero point.
• The lowest power of the variable with a non-zero coefficient in the series.

For example, in determining the order of a zero for the function \( f(z) = z(1 - \cos(z^2)) \), we find its expansion: \[ f(z) = \frac{z^5}{2} - \frac{z^9}{24} + \cdots \]Here, the first term \( \frac{z^5}{2} \) tells us that the order of the zero at \( z = 0 \) is 5. This indicates that the function's value and its first four derivatives at \( z=0 \) are all zero, and the fifth derivative is the first non-zero term.

function expansion

Function expansion using series like the Maclaurin or Taylor Series allows mathematicians to break down complex functions into manageable polynomials. This serves several purposes:

• Simplifying calculations near a specific point, often zero.
• Enabling precise approximations of functions with infinite terms.
• Providing insights into the function's behavior and properties.

To expand a function, substituting a series (such as the expansion of \( \cos(x) \) in powers of \( z \)) transforms the function into a form that is easier to analyze and differentiate. This method reduces intricate equations into simpler polynomial forms by: - Replacing part of a function with its series equivalent.- Distributing and simplifying to find non-zero terms. For instance, in expanding \( f(z) = z(1 - \cos(z^2)) \), we use \( \cos(z^2) = 1 - \frac{z^4}{2} + \cdots \), simplifying the function further and identifying significant terms such as \( \frac{z^5}{2} \). This expansion helps in assessing the function's order of zero and is invaluable in calculus and real-world applications like physics where such approximations are frequently needed.