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

Step-by-Step Solution

Answer

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

Step 1: Identify the function and center point

The given function is \( f(z) = \sin z \) and the center point for the Taylor series expansion is \( z_0 = \frac{\pi}{2} \). This expansion will be centered around \( z = \frac{\pi}{2} \).

Step 2: Calculate derivatives of the function

Find the derivatives of \( f \) up to the desired order. The derivatives of \( \sin z \) are:- \( f'(z) = \cos z \)- \( f''(z) = -\sin z \)- \( f'''(z) = -\cos z \)- \( f^{(4)}(z) = \sin z \)This pattern repeats every fourth derivative.

Step 3: Evaluate derivatives at the center point

Calculate the value of each derivative at \( z = \frac{\pi}{2} \):- \( f\left(\frac{\pi}{2}\right) = 1 \)- \( f'\left(\frac{\pi}{2}\right) = 0 \)- \( f''\left(\frac{\pi}{2}\right) = -1 \)- \( f''\left(\frac{\pi}{2}\right) = 0 \)The pattern of values repeats: 1, 0, -1, 0.

Step 4: Construct the Taylor series

The Taylor series for \( f(z) \) centered at \( z = \frac{\pi}{2} \) is given by:\[ f(z) = \sum_{n=0}^{\infty} \frac{f^{(n)}(z_0)}{n!} (z - z_0)^n\]Substituting the derivatives values:\[ f(z) = 1 - \frac{1}{2!}(z - \frac{\pi}{2})^2 + \frac{1}{4!}(z - \frac{\pi}{2})^4 - \cdots\]

Step 5: Determine the radius of convergence

The function \( \sin z \) is entire, which means the Taylor series expansion converges for all complex numbers. Therefore, the radius of convergence is \( R = \infty \).

Key Concepts

Radius of Convergence

The radius of convergence of a Taylor series indicates the interval over which the series converges to the original function. For the Taylor series, the radius of convergence is calculated based on the function's derivatives and their behavior.

In this exercise, the function given is the sine function, \( \sin z \), which is known as an entire function. Entire functions have some special characteristics:

• They are infinitely differentiable.
• They are analytic everywhere on the complex plane.
• Their Taylor series expansions converge everywhere (for all complex numbers).

Because \( \sin z \) is entire, its Taylor series has an infinite radius of convergence. This means it perfectly represents the sine function \( \sin z \) for any value of \( z \). To put it simply, the approximation will work everywhere, no limits!

Derivatives of Trigonometric Functions

Computing the derivatives of trigonometric functions like \( \sin z \) is critical in forming Taylor series. The derivatives help us understand how the function behaves around the point of expansion. Let's break down the derivatives:

- The first derivative of \( \sin z \) is \( \cos z \).- Following, the second derivative, we get\( -\sin z \).

- The third derivative results in\( -\cos z \).

- Interestingly, with the fourth derivative, we return to \( \sin z \) again.

This sequence of derivatives repeats every four steps: \( \sin z \), \( \cos z \), \( -\sin z \), \( -\cos z \). Recognizing this pattern simplifies computations, allowing us to easily predict higher-order derivatives, a helpful hint when working with trigonometric functions in calculus.

Center of Expansion

In a Taylor series, the center of expansion (or expansion point) is the value of \( z \) around which we expand the function. In our exercise, the function \( \sin z \) expands about the center point \( z_0 = \frac{\pi}{2} \).

Why is this center important? It affects the accuracy of the Taylor series approximation around this point. Specifically, the approximation is:- Most accurate near the center.- Gets less accurate as you move away, but with a larger radius of convergence, this effect is minimized.

For \( \sin z \) centered at \( z_0 = \frac{\pi}{2} \), the series is constructed by evaluating the function and its derivatives at \( \frac{\pi}{2} \):

• \( \sin(\frac{\pi}{2}) = 1 \)
• \( \cos(\frac{\pi}{2}) = 0 \)
• \( -\sin(\frac{\pi}{2}) = -1\)

This chain of values (1, 0, -1, 0) gives us the series terms, ensuring the Taylor series smoothly aligns with the sine wave's behavior at and around \( \frac{\pi}{2} \).