Problem 10 - Advanced Engineering Mathematics

Question

From \(\quad f(z)=z-\sin z=\frac{z^{3}}{3 !}-\frac{z^{5}}{5 !}+\cdots=z^{3}\left(\frac{1}{3 !}-\frac{z^{2}}{5 !}+\cdots\right)\) we see that \(z=0\) is a zero of order three.

Step-by-Step Solution

Answer

The zero at \(z=0\) is of order three because \(f(z) = z^3g(z)\) with \(g(0) \neq 0\).

Step 1: Understand the problem statement

We are given a function \(f(z) = z - \sin z\) and need to confirm that \(z = 0\) is a zero of order three. This means we must show that \(f(z)\) is of the form \(z^3 g(z)\) where \(g(z)\) is analytic and \(g(0) eq 0\).

Step 2: Expand the Taylor Series for \(\sin z\)

The Taylor series expansion for \(\sin z\) around 0 is:\[ \sin z = z - \frac{z^3}{3!} + \frac{z^5}{5!} - \cdots \]

Step 3: Substitute the Series into \(f(z)\)

Substitute the series from Step 2 into the function \(f(z)\):\[ f(z) = z - \left( z - \frac{z^3}{3!} + \frac{z^5}{5!} - \cdots \right) \] Simplifying, this becomes:\[ f(z) = \frac{z^3}{3!} - \frac{z^5}{5!} + \cdots \]

Step 4: Factor \(z^3\) from the Series

Factor \(z^3\) from all terms in \(f(z)\):\[ f(z) = z^3 \left( \frac{1}{3!} - \frac{z^2}{5!} + \cdots \right) \] This indicates that \(z^3\) is a common factor, confirming \(z=0\) as a zero of order three.

Step 5: Evaluate \(g(0)\)

The function \(g(z) = \frac{1}{3!} - \frac{z^2}{5!} + \cdots\) is analytic, and evaluating at \(z = 0\), we find \(g(0) = \frac{1}{3!} eq 0\). Thus, it confirms the order of the zero.

Key Concepts

Taylor Series

The Taylor Series is a way to represent a function as an infinite sum of terms calculated from values of its derivatives at a single point. It's like building a function piecemeal, using easier pieces each time. The Taylor Series for a function \( f(z) \) centered about 0 (often termed as Maclaurin series) takes the form:

\[ f(z) = f(0) + f'(0)z + \frac{f''(0)}{2!}z^2 + \frac{f'''(0)}{3!}z^3 + \cdots \]

This structure is immensely useful in approximations and in determining aspects like zeros of functions. For example, expanding the well-known function sine, \( \sin z \), gives us:

• \( \sin z = z - \frac{z^3}{3!} + \frac{z^5}{5!} - \cdots \)

Understanding and manipulating Taylor Series is essential for exploring and solving many problems in mathematics, including the exercise we're examining.

Zero of a function

A zero of a function is simply a point where the function equals zero, meaning the graph of the function touches or crosses the x-axis at that point. If \( f(z) = 0 \) for \( z = a \), then \( a \) is a zero. In our context, the function \( f(z) = z - \sin z \) has a zero at \( z = 0 \). This means substituting \( z = 0 \) into the function also results in zero:

• \( f(0) = 0 - \sin(0) = 0 \)

Zeros are often key points, especially when determining the behavior of a function. They hold special significance when a zero has repeated factors, leading into our next concept.

Analytic Function

An analytic function is one that's differentiable at every point in its domain and can be represented by a Taylor Series. This property allows it to be expressed with extreme precision within a radius from the point known as the radius of convergence. When a function is analytic, it exhibits smoothness and can be manipulated using series expansion without losing underlying characteristics.

In the case of our function \( g(z) \), derived from factoring \( f(z) \), we see:

• \( g(z) = \frac{1}{3!} - \frac{z^2}{5!} + \cdots \)

Because \( g(z) \) is analytic and \( g(0) eq 0 \), it validates the form of \( f(z) = z^3 g(z) \). The analyticity of \( g(z) \) ensures it's a valid function to pair with \( z^3 \) to confirm the zero's order.

Order of a zero

The order of a zero refers to how many times a particular zero is repeated at a point. In other words, it tells us how often a function's value smoothly approaches zero at that point. For a function \( f(z) \) expressed as \( z^n g(z) \), \( z = 0 \) is a zero of order \( n \) if \( g(z) \) is analytic and \( g(0) eq 0 \).

In our function \( f(z) = z^3 g(z) \), \( z=0 \) is described as having a zero of order three due to the presence of \( z^3 \) as its leading term. The reason it's significant is that as you approach \( z = 0 \), the function becomes \( 0 \) more precisely than weaker zeros like order one or two, profoundly influencing the function’s behavior near that point.

• It's a topological property closely related to the concept of multiplicity in algebraic curves.
• It helps in understanding how functions behave near zeros.