Problem 20 - 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. \(f(z)=\sin z, z_{0}=\pi / 2\)

Step-by-Step Solution

Answer

The Taylor series for \(\sin z\) around \(z_0=\pi/2\) is \(1 - \frac{(z-\pi/2)^2}{2} + \frac{(z-\pi/2)^4}{24} - \cdots\) with infinite radius of convergence.

Step 1: Identify the Function and Point

We need to expand the function \(f(z) = \sin z\) in a Taylor series around the point \(z_0 = \pi/2\). Taylor series around a point \(z=a\) is given by: \[ f(z) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!} (z-a)^n \] where \(f^{(n)}(a)\) is the \(n\)-th derivative evaluated at \(z = a\).

Step 2: Find the Derivatives of the Function

Compute the first few derivatives of \(f(z) = \sin z\):- 1st derivative: \(f'(z) = \cos z\)- 2nd derivative: \(f''(z) = -\sin z\)- 3rd derivative: \(f'''(z) = -\cos z\)- 4th derivative: \(f^{(4)}(z) = \sin z\)Continue this pattern to identify that derivatives repeat every four terms.

Step 3: Evaluate Derivatives at \(z_0 = \pi/2\)

Substitute \(z_0 = \pi/2\) into the derivatives:- \(f(\pi/2) = \sin(\pi/2) = 1\)- \(f'(\pi/2) = \cos(\pi/2) = 0\)- \(f''(\pi/2) = -\sin(\pi/2) = -1\)- \(f'''(\pi/2) = -\cos(\pi/2) = 0\)- \(f^{(4)}(\pi/2) = \sin(\pi/2) = 1\)These values are cyclical with a period of four.

Step 4: Construct the Taylor Series

Substitute the derivative values into the Taylor series formula:\[ \sin z = 1 - \frac{(z-\pi/2)^2}{2!} + \frac{(z-\pi/2)^4}{4!} - \cdots \]This results in:\[ -\frac{(z-\pi/2)^2}{2} + \frac{(z-\pi/2)^4}{24} - \cdots \]

Step 5: Determine the Radius of Convergence

The radius of convergence for the Taylor series of \(\sin z\) is infinite. This is because \(\sin z\) is an entire function (differentiable everywhere in the complex plane), and hence its Taylor series converges for all \(z\).

Key Concepts

Radius of Convergence

The radius of convergence is a key concept in understanding the behavior of Taylor series. It tells us within what range around the center point the series will accurately represent the function.

• The radius of convergence, often denoted by \( R \), is the distance from the center of the series to the edge of convergence.
• If \( R \) is infinite, the series converges everywhere in the complex plane.
• If the series has a finite \( R \), then it converges only within a disk of radius \( R \).

For the sine function \( \sin z \), its Taylor series has an infinite radius of convergence because it is an entire function. This means that it is differentiable everywhere in the complex plane. Consequently, its Taylor series converges for all complex numbers \( z \). Understanding this can significantly simplify complex analysis.

Function Derivatives

Derivatives help us find the terms of a Taylor series. For the sine function \( \sin z \), derivatives are crucial because they define each term of the series expansion.

• First derivative: \( f'(z) = \cos z \)
• Second derivative: \( f''(z) = -\sin z \)
• Third derivative: \( f'''(z) = -\cos z \)
• Fourth derivative: \( f^{(4)}(z) = \sin z \)

Notice that the sine function's derivatives repeat every four terms. This periodicity is a key feature that helps streamline the calculation of Taylor series. Evaluating these derivatives at the given point, such as \( z = \pi/2 \), gives the coefficients for each term.

Sin Function

The sine function, \( \sin z \), is one of the most recognized periodic functions in mathematics. Its Taylor series showcases its trigonometric identity through infinite sums.

• The general Taylor series expansion of \( \sin z \) around a point \( z_0 \) is constructed using its derivatives at \( z_0 \).
• For example, when expanded at \( \pi/2 \), the sine function's Taylor series becomes \( 1 - \frac{(z-\pi/2)^2}{2} + \frac{(z-\pi/2)^4}{24} - \cdots \).
• This expansion uses only even powers of \( (z - \pi/2) \) because the derivatives evaluated at odd powers are zero.

Understanding these properties allows for better comprehension of how functions behave at specific points. This series effectively approximates \( \sin z \) around \( \pi/2 \).

Complex Analysis

Complex analysis extends real analysis into the complex plane, where numbers have both real and imaginary parts. It's vital for understanding functions like \( \sin z \) in their full context.

• Functions like \( \sin z \) remain analytical across the entire complex plane, classifying them as 'entire functions'.
• These functions are infinitely differentiable, meaning you can compute derivatives at any point in the complex plane.
• Complex analysis involves using these derivatives to expand functions into Taylor series, encompassing the entire plane.

A major benefit of exploring complex analysis is learning how entire functions, such as the sine function, have properties like infinite radius of convergence. This highlights their powerful utility in mathematical modeling and problem-solving.