Problem 9 - Advanced Engineering Mathematics

Question

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

Step 1: Identify the Function and Its Derivatives

The given function is \( f(z) = z (1 - \cos(z^2)) \). To find the order of the zero, we need to expand \( f(z) \) into a power series and determine the smallest degree of \( z \) with a non-zero coefficient.

Step 2: Expand 1 - cos(z^2) as a Power Series

The cosine function \( \cos(z^2) \) can be expanded as a power series: \[ \cos(z^2) = 1 - \frac{z^4}{2!} + \frac{z^8}{4!} - \cdots \] Therefore, \( 1 - \cos(z^2) \) becomes: \[ 1 - \cos(z^2) = \frac{z^4}{2!} - \frac{z^8}{4!} + \cdots \]

Step 3: Multiply by z to Get the Series of f(z)

Substitute the series from Step 2 into \( f(z) = z(1 - \cos(z^2)) \): \[ f(z) = z \left( \frac{z^4}{2!} - \frac{z^8}{4!} + \cdots \right) = \frac{z^5}{2!} - \frac{z^9}{4!} + \cdots \]

Step 4: Determine the Order of the Zero

The series for \( f(z) \) starts with the term \( \frac{z^5}{2!} \). Thus, the smallest power of \( z \) with a non-zero coefficient is 5, indicating that the zero at \( z=0 \) is of order 5.

Key Concepts

Maclaurin series

A Maclaurin series is a specific type of power series expansion used to express a function near zero. It's a special case of the Taylor series, where the expansion is centered at zero.It breaks down complex functions into an infinite sum of terms calculated from the values of their derivatives at zero.

• The general form of a Maclaurin series is: \[f(x) = f(0) + f'(0) \cdot x + \frac{f''(0)}{2!} \cdot x^2 + \frac{f'''(0)}{3!} \cdot x^3 + \cdots \]
• Each term involves a derivative of the function and a factorial, which helps in approximating the function's behavior near zero.
• For example, common functions like \( e^x \), \( \sin x \), and \( \cos x \) have well-known Maclaurin series that are used frequently in calculus.

In the context of the problem, we used the Maclaurin series to determine the order of the zero of \( f(z) \) at \( z = 0 \). The key is finding the first non-zero coefficient in the series expansion, which tells us the order of the zero.

Taylor series

A Taylor series is similar to a Maclaurin series but more general, as it can represent a function around any point \( a \), not just zero.This enables more flexibility in analyzing a function near different values.

• Its general form is: \[f(x) = f(a) + f'(a) (x - a) + \frac{f''(a)}{2!} (x - a)^2 + \frac{f'''(a)}{3!} (x - a)^3 + \cdots \]
• By adjusting the point \( a \), a Taylor series can approximate the function in different regions of its domain.
• The concept is widely used in numerical analysis and engineering for approximating difficult functions.

While we focused on the Maclaurin series since \( z = 0 \) in the current problem, understanding both forms is valuable as Taylor series expansions provide insight into the behavior and approximation of functions at various points.

Power series expansion

Power series expansions give us a tool to express functions as an infinite sum of terms involving powers of a variable.These series are invaluable in mathematical analysis because they simplify the representation and calculation of complex functions.

• A power series generally takes the form: \[S(x) = c_0 + c_1 x + c_2 x^2 + c_3 x^3 + \cdots \]
• The coefficients \( c_i \) accompany powers of \( x \), and finding the right coefficients is essential to accurately approximate a function.
• Power series are used to represent not only elementary functions but also transcendental functions like exponential, sin, and cosine.

In practical terms, power series are used to model real-world problems where approximate solutions are acceptable. In the given exercise, we used a power series expansion of \( f(z) \) to determine the zero's order, showcasing its utility in solving mathematical problems.