Problem 13 - Advanced Engineering Mathematics

Question

Expand the given function in a Taylor series centered at the indicated point. Give the radius of convergence of each series. \(f(z)=1 / z, z_{0}=1\)

Step-by-Step Solution

Answer

The series is \( \sum_{n=0}^{\infty} (-1)^n (z - 1)^n \) with a radius of convergence of 1.

Step 1: Function Differentiation

Start with the function \( f(z) = \frac{1}{z} \). We need to find the derivatives \( f^{(n)}(z) \) of the function to expand it into a Taylor series centered at \( z_0 = 1 \). Differentiate the function multiple times to recognize a pattern in the derivatives.\[ f'(z) = -\frac{1}{z^2}, \quad f''(z) = \frac{2}{z^3}, \quad f'''(z) = -\frac{6}{z^4}, \quad \text{and so on.} \]

Step 2: Evaluate Derivatives at the Center

Evaluate the derivatives at the center \( z_0 = 1 \). This will help to write the terms of the Taylor series.\[ f^{(n)}(1) = (-1)^n n! \]

Step 3: Write the Taylor Series

Use the Taylor series formula centered at \( z_0 = 1 \): \[ f(z) = \sum_{n=0}^{\infty} \frac{f^{(n)}(1)}{n!} (z - 1)^n \] Substitute the derivatives evaluated in Step 2: \[ f(z) = \sum_{n=0}^{\infty} (-1)^n (z - 1)^n \]

Step 4: Determine the Radius of Convergence

The radius of convergence \( R \) for a Taylor series is given by \( \frac{1}{\limsup_{n \to \infty} |a_n|^{1/n}} \). In this case, \( a_n = (-1)^n \), so \[ |a_n| = 1 \text{ for all } n \] and \( \limsup_{n \to \infty} |a_n|^{1/n} = 1 \). Hence, the radius of convergence is \( R = \frac{1}{1} = 1 \).

Step 5: Conclusion

The Taylor series expansion of \( f(z) = \frac{1}{z} \) centered at \( z_0 = 1 \) is \( \sum_{n=0}^{\infty} (-1)^n (z - 1)^n \) with a radius of convergence of 1.

Key Concepts

Function Differentiation

The process of function differentiation involves finding the derivatives of a function, which means calculating how the function's output changes as its input changes. This is useful for Taylor series expansions, where we need these derivatives at a specific point.

For our function, which is given by \( f(z) = \frac{1}{z} \), we need to determine its derivatives at the center point, \( z_0 = 1 \), to construct its Taylor series. By differentiating the function repeatedly, we identify a pattern:

• First derivative: \( f'(z) = -\frac{1}{z^2} \)
• Second derivative: \( f''(z) = \frac{2}{z^3} \)
• Third derivative: \( f'''(z) = -\frac{6}{z^4} \)

We can generalize these derivatives using a pattern: \[ f^{(n)}(z) = (-1)^n \cdot \frac{n!}{z^{n+1}}. \]Evaluating these at \( z_0 = 1 \), they simplify to \( (-1)^n n! \), which are the coefficients used in the Taylor series expansion formula.

Radius of Convergence

In the context of Taylor series, the radius of convergence is the range within which the series representation faithfully approximates the function.

To find this radius, we rely on the concept of \( \limsup \) (limit superior) and the coefficients \( a_n \) of the Taylor series.

For our function, those coefficients are \( (-1)^n \), which means the absolute value is always 1.With this information, we utilize the radius of convergence formula: \[ R = \frac{1}{\limsup_{n \to \infty} |a_n|^{1/n}}. \]Because \( \limsup_{n \to \infty} |1|^{1/n} = 1 \), we find that

\[ R = \frac{1}{1} = 1. \] This tells us that the Taylor series expansion will converge within a unit circle around \( z_0 = 1 \), so it's valid for all \( z \) such that \( |z - 1| < 1 \). Thus, understanding the radius of convergence is crucial for knowing where the series will effectively approximate the function.

Series Expansion

Series expansion, specifically Taylor series, allows us to express a function as an infinite sum of terms calculated from the derivatives at a single point. Here, we center our series at \( z_0 = 1 \).

In a Taylor series, each term is derived from the function's derivatives, involving powers of \( (z - z_0) \).For the function \( f(z) = \frac{1}{z} \), the series is expressed as: \[ f(z) = \sum_{n=0}^{\infty} \frac{f^{(n)}(1)}{n!} (z - 1)^n. \]Using the pattern recognized in the derivatives:

• \( f^{(n)}(1) = (-1)^n n! \)

We substitute these into the formula to derive the expansion: \[ f(z) = \sum_{n=0}^{\infty} (-1)^n (z - 1)^n. \]This gives us a practical way to approximate \( f(z) \) within the radius where the expansion converges. Taylor series expansions are a cornerstone of analysis, providing powerful tools for approximating functions around specific points while preserving their differentiable nature.