Problem 3 - Complex Analysis for Mathematics and Engineering

Question

Find the Taylor series centered at \(a=1\) and state where it converges for (a) \(f(z)=\frac{1-x}{x-2}\). (b) \(f(z)=\frac{1-1}{x-3}\). Hint: \(\frac{1-s}{x-1}=\left(\frac{1}{2}\right) \frac{x-1}{1-\frac{1-1}{2}}=\left(\frac{1}{2}\right)(z-1) \frac{1}{1-\frac{s-1}{2}}\).

Step-by-Step Solution

Answer

(a) Taylor series: \(-(x-1) - (x-1)^2 - (x-1)^3 - \ldots\), converges for \(|x-1|<1\). (b) Zero function, converges everywhere.

Step 1: Express the function for part (a)

For the function \( f(z) = \frac{1-x}{x-2} \), rewrite it in a form that can be expanded as a Taylor series. Write it as: \[ f(z) = \frac{1-x}{x-2} = -\frac{x-1}{x-2} = -(x-1)\left(\frac{1}{x-2}\right) \].

Step 2: Re-express the fraction for part (a)

Express \( \frac{1}{x-2} \) in terms of \( z-1 \), where \( z = x \). Since we need a Taylor series centered at \( a=1 \), let's express our function as: \[ -\frac{x-1}{x-2} = -(x-1) \frac{1}{(1+((x-1)-1))} \].

Step 3: Expand using geometric series part (a)

The term \( \frac{1}{1-(x-1)} \) can be expanded using the formula for geometric series: \( \frac{1}{1-y} = 1 + y + y^2 + y^3 + \ldots \) for \(|y| < 1\). We have, \( 1-(x-1) \), so the series becomes: \[ \frac{1}{1-(x-1)} = 1 + (x-1) + (x-1)^2 + \ldots \].

Step 4: Multiply and simplify for part (a)

Substitute the series back into the expression: \[ -\frac{x-1}{x-2} = -(x-1)(1 + (x-1) + (x-1)^2 + \ldots) \]. Thus, the expansion becomes: \[ -(x-1) - (x-1)^2 - (x-1)^3 - \ldots \].

Step 5: Identify convergence for part (a)

Given \( \frac{1}{1-y} \) converges for \(|y| < 1\), our series converges for \(|x-1| < 1\).

Step 6: Express the function for part (b)

For \( f(z) = \frac{1-1}{x-3} \), this simplifies further since the numerator is effectively zero. Thus \( f(z) = 0 \).

Step 7: State Taylor Series Expansion for part (b)

Since \( f(z) = 0 \), the Taylor series expansion centered at \( a=1 \) is just 0 for all terms.

Step 8: Convergence region for part (b)

For a constant function (zero in this case), the Taylor series converges everywhere or for all \( x \).

Key Concepts

Geometric Series

A geometric series is a sum of terms that have a constant ratio between successive terms. The classic form of a geometric series is expressed as:

• \( a + ar + ar^2 + ar^3 + \ldots \)

where \( a \) is the first term and \( r \) is the common ratio. The series can be conveniently summed to:\[ \frac{a}{1-r} \]provided \(|r| < 1\). This condition ensures the series converges, meaning that as you add more and more terms, the sum approaches a finite value.
In the solution outlined above, the geometric series idea is applied to transform functions into Taylor series around a point. Here, we substitute variables and manipulate expressions until we can fit them into the geometric series format. This allows us to re-write functions as infinite sums of powers of \( x-1 \). In the case of:

• \( \frac{1}{1-(x-1)} = 1 + (x-1) + (x-1)^2 + \ldots \)

this follows the geometric series formula with \( y = x-1 \) serving as the part that transforms the series into

• \( \sum_{n=0}^{\infty} (x-1)^n \).

Convergence of Series

Convergence is a key concept when dealing with infinite series. A series converges when the sum of its terms approaches a specific value as more terms are added. For geometric series, convergence is determined by the common ratio \( r \). The series converges if \(|r| < 1\). In our example:

• \( \frac{1}{1-(x-1)} = 1 + (x-1) + (x-1)^2 + \ldots \)

the convergence condition becomes \(|x-1| < 1\).
This implies that the series will provide meaningful, finite sum results when \( x \) is within a unit distance from 1 on the complex plane.
For part (a), this means that values of \( x \) like 0.5 and 1.5 will work, but not 2.1, since it is outside the interval for convergence.

• In part (b), the Taylor series for a constant zero function trivially converges everywhere since adding zero terms is consistent for all \( x \).

Complex Functions

Complex functions are those that involve complex numbers, of the form \( z = x + yi \) where \( i \) represents the imaginary unit with the property \( i^2 = -1 \). These numbers extend the one-dimensional number line to a two-dimensional plane making analysis versatile.
In mathematical practice, complex functions allow integration, differentiation, and expansion, such as how Taylor series is applied to complex functions of \( z \).
For example, when studying functions like \( f(z) = \frac{1-x}{x-2} \), we can consider them in the complex plane to uncover more profound insights and solutions.
It enables examination of not just real values but those that involve the imaginary component \( yi \). This offers a complete view of the function's behavior over a broader domain. Understanding which regions in the complex plane will converge, helps solve real-world problems across engineering and physics where such functions are frequently applied.