Problem 19 - 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. $$ f(z)=\cos z, z_{0}=\pi / 4 $$

Step-by-Step Solution

Answer

The Taylor series is built using the evaluated derivatives at \( \frac{\pi}{4} \), and the radius of convergence is infinite.

Step 1: Recall the Taylor Series Formula

The Taylor series expansion for a function \( f(z) \) centered at a point \( z_0 \) is given by \( \sum_{n=0}^{\infty} \frac{f^{(n)}(z_0)}{n!}(z-z_0)^n \).

Step 2: Derive the Derivatives of \( f(z) = \cos z \)

Find the derivatives of \( \cos z \): 1. \( f(z) = \cos z \) 2. \( f'(z) = -\sin z \) 3. \( f''(z) = -\cos z \) 4. \( f'''(z) = \sin z \) 5. \( f^{(4)}(z) = \cos z \) This cycle repeats every four derivatives.

Step 3: Evaluate Derivatives at \( z_0 = \frac{\pi}{4} \)

Calculate the values of the first few derivatives at \( z_0 = \frac{\pi}{4} \): 1. \( f(z_0) = \cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} \) 2. \( f'(z_0) = -\sin\left(\frac{\pi}{4}\right) = -\frac{\sqrt{2}}{2} \) 3. \( f''(z_0) = -\cos\left(\frac{\pi}{4}\right) = -\frac{\sqrt{2}}{2} \) 4. \( f'''(z_0) = \sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} \) 5. \( f^{(4)}(z_0) = \cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} \)

Step 4: Construct the Taylor Series

Each term in the Taylor series is constructed using \( \frac{f^{(n)}(z_0)}{n!}(z-z_0)^n \): \(\begin{align*}f(z) &= \cos z \approx \frac{\sqrt{2}}{2} + \frac{1}{1!} \left(-\frac{\sqrt{2}}{2}\right) (z - \frac{\pi}{4}) \&+ \frac{1}{2!}\left(-\frac{\sqrt{2}}{2}\right) (z - \frac{\pi}{4})^2 + \frac{1}{3!}\frac{\sqrt{2}}{2} (z - \frac{\pi}{4})^3 \&+ \frac{1}{4!}\frac{\sqrt{2}}{2} (z - \frac{\pi}{4})^4 + \cdots \end{align*}\) This pattern continues, with expressions from step 3 filling in the coefficients.

Step 5: Determine the Radius of Convergence

For the function \(\cos z\), since it is entire (analytic everywhere in the complex plane), the radius of convergence \( R \) is infinite (\( R = \infty \)).

Key Concepts

Radius of Convergence

The radius of convergence is an important concept when working with Taylor series. It describes how far away from the center of the series (in this case, the point \( z_0 = \frac{\pi}{4} \) ) the series will accurately represent the function. For many functions, determining this radius involves looking at the behavior of the function and where it becomes non-analytic, such as singularities or discontinuities.

However, for the trigonometric function \( \cos z \), it's quite different. Why? Because \( \cos z \) is **entire**. An entire function is analytic everywhere on the complex plane.

This is why the radius of convergence is infinite, meaning you can use the Taylor series expansion of \( \cos z \) to represent the function accurately everywhere, without any restrictions on the value of \( z \).

• Analytic everywhere = Entire function
• Infinite radius of convergence

Understanding the radius of convergence is key to knowing where a Taylor series can be applied confidently.

Function Expansion

Function expansion, like using the Taylor series, is a powerful tool for turning complex functions into polynomial expressions. This makes them easier to work with, especially for calculations and approximations.

A Taylor series expansion takes a function and expresses it as an infinite sum of terms, which depend on the derivatives of the function at a specific point \( z_0 \). The general formula is: \[\sum_{n=0}^{\infty} \frac{f^{(n)}(z_0)}{n!}(z-z_0)^n\]where \( f^{(n)}(z_0) \) is the \( n \)-th derivative of \( f \) evaluated at \( z_0 \). Each term in the series becomes more intricate as \( n \) increases, but thanks to the factorial denominator, their contribution to the sum decreases.

This series captures the behavior of a function around \( z_0 \) accurately, as seen with \( \cos z \) in the provided example, where every four terms repeat in a cycle. Taylor series essentially allows you to approximate or represent functions fully in a polynomial format, which is often more manageable.

Derivatives Calculation

Calculating derivatives is a crucial step in forming a Taylor series. For a function like \( f(z) = \cos z \), this involves differentiating repeatedly until a pattern emerges.

Here’s how it typically unfolds:

• The first derivative, \( f'(z) = -\sin z \)
• The second derivative, \( f''(z) = -\cos z \)
• The third derivative, \( f'''(z) = \sin z \)
• The fourth derivative, \( f^{(4)}(z) = \cos z \)

After four derivatives, it repeats back to the original function due to the cyclical nature of trigonometric derivatives.

Once these derivatives are calculated for \( z_0 = \frac{\pi}{4} \), they are used as coefficients in the Taylor series expansion. Calculating derivatives accurately ensures the Taylor series correctly approximates the function around a specific point, which is the heart of Taylor series methods.

Trigonometric Functions

Trigonometric functions like \( \cos z \) are prevalent in mathematics and engineering. They describe oscillatory patterns such as waves and vibrations. When these functions are part of exercises involving Taylor series, the process gives you a polynomial approach to these periodic functions.

Interestingly, trigonometric functions have repeating derivatives. With \( \cos z \), every four derivatives return back to the start of the cycle. This periodicity simplifies their analysis and makes them an ideal candidate for function expansion through Taylor series.

Additionally, understanding the behavior and properties of these functions allows us to better exploit their uses in mathematical models. It shows the recurring nature of trigonometric derivatives, helping in deriving series expansions that maintain the same patterns and symmetries. They are not only theoretically interesting, but play a significant role in modeling real-world phenomena.

Using Taylor series to expand trigonometric functions converts complex wave-like behaviors into easier manipulated algebraic terms. This conversion is useful in calculations, predictions, and simulations across various fields.