Problem 11 - 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. $$ f(z)=1-e^{z-1} ; z=1 $$

Step-by-Step Solution

Answer

The order of the zero at \( z=1 \) is 1.

Step 1: Identify the Function and Point

We have the function \( f(z) = 1 - e^{z-1} \) and we are given that \( z = 1 \) is a zero of the function. This means when \( z = 1 \), \( f(1) = 0 \). Our goal is to determine the order of this zero by expanding the function around this point.

Step 2: Expand Using Taylor Series

The Taylor series for \( e^{x} \) around \( x = 0 \) is given by \( e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots \). We will use this to expand \( e^{z-1} \) around \( z=1 \). Substituting \( x = z-1 \), the expansion becomes:\[e^{z-1} = e^0 + (z-1) + \frac{(z-1)^2}{2!} + \frac{(z-1)^3}{3!} + \cdots = 1 + (z-1) + \frac{(z-1)^2}{2} + \frac{(z-1)^3}{6} + \cdots \]

Step 3: Substitute into the Function

Substitute the Taylor series expansion of \( e^{z-1} \) into \( f(z) \):\[f(z) = 1 - \left( 1 + (z-1) + \frac{(z-1)^2}{2} + \frac{(z-1)^3}{6} + \cdots \right) = 1 - 1 - (z-1) - \frac{(z-1)^2}{2} - \frac{(z-1)^3}{6} - \cdots \]Simplifying, we get:\[f(z) = -(z-1) - \frac{(z-1)^2}{2} - \frac{(z-1)^3}{6} - \cdots \]

Step 4: Determine the Order of the Zero

From the expression \( f(z) = -(z-1) - \frac{(z-1)^2}{2} - \frac{(z-1)^3}{6} - \cdots \), we observe that the lowest power term in the expansion is \( (z-1) \). Thus, the order of the zero at \( z=1 \) is 1, since the term \( (z-1)^1 \) is the first non-zero term in the expansion.

Key Concepts

Order of Zero

Understanding the order of a zero is crucial when dealing with functions and their expansions using series. The 'order of zero' refers to the smallest positive integer, say \( n \), such that \( (z-z_0)^n \) is the first term in the series that does not vanish. In simpler terms, it indicates how many times the graph of the function "touches" or "crosses" the x-axis at a zero point.

• The order of a zero is 1 if the function linearly passes through the zero.
• If the zero is of order 2, the graph "bounces" at that point and does not cross the axis.

In our exercise, the function \( f(z) = 1 - e^{z-1} \) at \( z = 1 \) is such that the linear term \(- (z-1)\) is the first non-zero term when expanded, hence the order of the zero at \( z = 1 \) is 1.

Maclaurin Series

The Maclaurin series is a special case of the Taylor series and is centered at zero. It expresses a function as an infinite sum of its derivatives at this center point. Mathematically, it is given by: \[ f(x) = f(0) + f'(0)x + \frac{f''(0)x^2}{2!} + \frac{f'''(0)x^3}{3!} + \cdots\]

• It is used for approximating functions.
• Maclaurin series help simplify complex functions at points close to zero.
• The series applies to functions with derivatives of all orders.

In the given exercise, though centered at \( z = 1 \), it serves as a foundation for understanding series expansions around different points, resembling a Maclaurin series when \( z_0 = 0 \). By substituting \( (z-1) \), this transforms our series into a more conventional Taylor expansion with the same principle of using derivatives and their sums.

Exponential Function Expansion

One of the most common expansions involves the exponential function \( e^x \). The Taylor series expansion for \( e^x \), centered around zero, shows:\[e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots\]

• This series is simple to remember and crucial for various calculations.
• The exponential function is unique due to its infinite radius of convergence.
• This expansion helps transform complex exponential terms into simple polynomial-like expressions.

In the exercise, expanding \( e^{z-1} \) using this concept allows us to simplify \( f(z) = 1 - e^{z-1} \) by breaking it down into terms that vividly illustrate the function's behavior around \( z = 1 \). Understanding how to expand exponential functions is key to analyzing and solving many equations in calculus and differential equations.