Problem 22 - A First Course in Complex Analysis with Applications

Question

Expand the given function in a Taylor series centered at the indicated point \(z_{0}\). Give the radius of convergence \(R\) of each series. $$ f(z)=\sin z, \quad z_{0}=\pi / 2 $$

Step-by-Step Solution

Answer

The Taylor series is: \( 1 - \frac{1}{2}(z-\pi/2)^2 + \frac{1}{4!}(z-\pi/2)^4 - \cdots \), \( R = \infty \).

Step 1: Recall the Taylor Series Formula

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

Step 2: Determine the Derivatives of \( \sin z \)

Calculate several derivatives of \( f(z) = \sin z \) evaluated at \( z_0 = \pi/2 \):1. \( f(z) = \sin z \) implies \( f'(z) = \cos z \)2. \( f''(z) = -\sin z \)3. \( f'''(z) = -\cos z \)Evaluate these at \( z = \pi/2 \):- \( \sin(\pi/2) = 1 \)- \( \cos(\pi/2) = 0 \)- \( -\sin(\pi/2) = -1 \)- \( -\cos(\pi/2) = 0 \)

Step 3: Substitute into Taylor Series Formula

Substitute these derivatives into the Taylor series formula:- \( n = 0 \): \( \frac{1}{0!} (z - \pi/2)^0 = 1 \)- \( n = 1 \): \( \frac{0}{1!} (z - \pi/2)^1 = 0 \)- \( n = 2 \): \( \frac{-1}{2!} (z - \pi/2)^2 = -\frac{1}{2} (z - \pi/2)^2 \)- \( n = 3 \): \( \frac{0}{3!} (z - \pi/2)^3 = 0 \)The pattern repeats every four terms.

Step 4: Write the Taylor Series

The Taylor series of \( \sin z \) centered at \( \pi/2 \) becomes:\[ \sin z = 1 - \frac{1}{2!}(z - \pi/2)^2 + \frac{1}{4!}(z - \pi/2)^4 - \cdots \]This uses the fact that the function and its derivatives have alternating non-zero terms which follow the sine function's parity.

Step 5: Determine Radius of Convergence

The radius of convergence \( R \) of the Taylor series for \( \sin z \) around any point is infinite, since sine is an entire function. Thus, \( R = \infty \).

Key Concepts

Understanding the Radius of Convergence

The radius of convergence is a central concept when dealing with Taylor series. It defines the interval in which the series converges to the function it represents. To understand this, imagine the Taylor series as a smooth approximation of your function within a circular region around the center point, called the expansion point. This region's radius is what we call the radius of convergence.

For any power series, this is determined using the formula: \[ R = \frac{1}{L} \] where \( L \) is the limit of the absolute value of the ratio of successive coefficients as \( n \) approaches infinity. Alternatively, we determine convergence using the root test, specifically for the function \( \sin z \). But fascinatingly, trigonometric functions like \( \sin z \) have an essential property: their convergence extends to the entire complex plane.

• This means the series converges for any complex number \( z \).
• The radius of convergence is thus infinite, indicating an all-encompassing domain for sine's Taylor series centered at any point.

Derivatives of Trigonometric Functions

To construct the Taylor series for \( \sin z \), we must first understand the derivatives of trigonometric functions. Knowing how a function behaves as we take its derivatives is crucial for accurate series expansion.

For \( \sin z \), the function follows a cyclic pattern of derivatives:

• Start with the base function \( \sin z \). Its first derivative is \( \cos z \).
• The second derivative returns us to \(-\sin z \).
• The third derivative yields \(-\cos z \).
• The fourth derivative naturally circles back to \( \sin z \) again.

This cycle repeats every four derivatives. Evaluating these derivatives at specific points, such as \( z = \pi/2 \), helps us spot the non-zero terms:

• \( \sin(\pi/2) = 1 \)
• \( \cos(\pi/2) = 0 \)
• \( -\sin(\pi/2) = -1 \)

This consistent pattern is why trigonometric functions are excellent candidates for elegant Taylor series expressions.

The Phenomenon of Infinite Radius of Convergence

The infinite radius of convergence is an exciting property that characterizes entire functions. Entire functions are holomorphic (analytic) everywhere on the complex plane.

Such a property means that their Taylor series can approximate the function at any point in the complex plane with perfect precision, with no worries about convergence issues.

• For \( \sin z \), this indicates that no matter where you center the Taylor series, whether at \( z_0 = \frac{\pi}{2} \) or any other point, the series will always converge.
• This universality marks \( \sin z \) as a function with an infinite radius of convergence, offering stability and robustness in mathematical modeling and calculations.

Understanding that some series enjoy this property helps when solving complex analytical problems, as it grants flexibility in mathematical approaches.