Problem 16 - Advanced Engineering Mathematics

Question

In Problems 13-22, expand the given function in a Taylor series centered at the indicated point. Give the radius of convergence of each series. $$ f(z)=\frac{1}{1+z}, z_{0}=-i $$

Step-by-Step Solution

Answer

The Taylor series is \( \sum_{n=0}^{\infty} \frac{(-i)^n}{(1-i)^{n+1}} (z+i)^n \) and the radius of convergence is 1.

Step 1: Recall the Taylor Series Formula

The Taylor series of a function \( f(z) \) centered around a point \( z_0 \) is given by \( f(z) = \sum_{n=0}^{\infty} \frac{f^{(n)}(z_0)}{n!} (z-z_0)^n \). Here \( f^{(n)}(z_0) \) represents the \( n\)-th derivative of \( f(z) \) evaluated at \( z=z_0 \).

Step 2: Derive the Necessary Derivatives

For \( f(z) = \frac{1}{1+z} \), calculate the first few derivatives: \( f'(z) = \frac{-1}{(1+z)^2} \), \( f''(z) = \frac{2}{(1+z)^3} \), and so on, using the power rule for differentiation.

Step 3: Evaluate the Derivatives at \( z_0 = -i \)

Substitute \( z_0 = -i \) in the derivatives. For instance, \( f(-i) = \frac{1}{1-i} \), \( f'(-i) = \frac{-1}{(1-i)^2} \), etc. Simplify these expressions using complex arithmetic.

Step 4: Formulate the Taylor Series

Using the derivatives at \( z_0 = -i \), construct the Taylor series: \( f(z) = \sum_{n=0}^{\infty} \frac{f^{(n)}(-i)}{n!} (z + i)^n \). Substitute the values from previously calculated derivatives to write out the first few terms.

Step 5: Determine the Radius of Convergence

The function \( f(z) = \frac{1}{1+z} \) is defined and analytic on the open disk where \( |1+z| eq 0 \). Hence the radius of convergence is given by the distance to the nearest singularity, which is \( |z - (-(-1))| = 1 \).

Key Concepts

Radius of Convergence

The radius of convergence is an essential concept when dealing with Taylor series. It helps us determine where our series representation of a function is valid. For a series centered at a point \( z_0 \), the radius of convergence \( R \) is the distance from \( z_0 \) to the nearest singularity of the function.
In simpler terms, it tells us how far we can move from this center point and still accurately approximate the function with our series.

• For the given function \( f(z) = \frac{1}{1+z} \) centered at \( z_0 = -i \), we identify the singularity by setting the denominator to zero, which is at \( z = -1 \).
• This singular point is crucial as it determines the limits of our series's accuracy.
• The radius of convergence \( R \) can be calculated as: \[ R = |z_0 - (-1)| = |-i - (-1)| = 1.\]

With the radius of \( R = 1 \), the Taylor series accurately represents \( f(z) \) within the circle of radius 1 centered at \( -i \). Understanding the radius of convergence ensures that we apply the Taylor series correctly, avoiding regions where the series could misrepresent the function.

Complex Derivatives

Calculating derivatives for complex functions is similar to real functions, although we use complex numbers. The concept of derivatives extends smoothly into the complex domain. That means if you know how to differentiate real functions, complex derivatives are just one step ahead.
For a function \( f(z) = \frac{1}{1+z} \), we calculate successive derivatives to generate terms for the Taylor series:

• The first derivative is \( f'(z) = \frac{-1}{(1+z)^2} \).
• This is derived using the power rule, similar to real differentiation.
• The second derivative becomes \( f''(z) = \frac{2}{(1+z)^3} \), achieved through continued differentiation.

Calculating derivatives at the center point \( z_0 = -i \) involves substituting \( z_0 \) into these derivative expressions and performing complex arithmetic. At complex points, the calculations involve not only the common rules of derivatives but also assumptions of the function's behavior in the complex plane. This understanding forms the foundation of generating each term of the Taylor series.

Analytic Functions

A function is called analytic if it possesses a derivative at all points in its domain, not just at individual points. Analytic functions, like \( f(z) = \frac{1}{1+z} \), are extremely important for functions of a complex variable.
Such functions can be represented by power series similar to Taylor series, which shows their expansion around some point \( z_0 \):

• Being analytic implies that these series will converge to the actual function within some radius.
• For \( f(z) \) given in the exercise, since it is analytic everywhere except where the series denominator equals zero, we identify the series converges inside the computed radius of convergence.
• This ensures that the function behaves predictably and can be expanded into a Taylor series within some region of its domain.

Having a function that is analytic is a powerful property, allowing complex functions to be studied with the same tools used for polynomials or basic power series, thus simplifying the complexity.