Problem 10 - Advanced Engineering Mathematics

Question

In Problems 1-12, expand the given function in a Maclaurin series. Give the radius of convergence of each series. $$ f(z)=\sin 3 z $$

Step-by-Step Solution

Answer

The Maclaurin series for \(\sin 3z\) is \(3z - \frac{9}{2}z^3 + \frac{81}{40}z^5 - \ldots \) with an infinite radius of convergence.

Step 1: Identify the Maclaurin series formula

The Maclaurin series for a function \(f(z)\) is given by \(f(z) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} z^n\), where \(f^{(n)}(0)\) is the \(n\)-th derivative of \(f(z)\) evaluated at \(z=0\).

Step 2: Find derivatives of \(\sin 3z\)

First, identify the derivatives of \(f(z) = \sin 3z\): 1. \(f(z) = \sin 3z\)2. \(f'(z) = 3\cos 3z\)3. \(f''(z) = -9\sin 3z\)4. \(f'''(z) = -27\cos 3z\)5. \(f^{(4)}(z) = 81\sin 3z\)As we can see, the pattern repeats every four derivatives. Thus, \(f^{(4k)}(z) = (3^4)^k \sin 3z\).

Step 3: Evaluate derivatives at \(z=0\)

Now calculate the derivatives at \(z=0\):1. \(f(0) = \sin 0 = 0\)2. \(f'(0) = 3\cos 0 = 3\)3. \(f''(0) = -9\sin 0 = 0\)4. \(f'''(0) = -27\cos 0 = -27\)5. Continue repeating every four terms.

Step 4: Write the Maclaurin series

Using the derivatives evaluated at \(z=0\), substitute into the Maclaurin formula:\[ \sin 3z = 0 + \frac{3}{1!} z - \frac{27}{3!} z^3 + \frac{0}{4!} z^4 + \frac{3(3^4)}{5!} z^5 - \ldots \]Simplify:\[ \sin 3z = 3z - \frac{27}{6} z^3 + \frac{243}{120} z^5 - \ldots \]Or:\[ \sin 3z = 3z - \frac{9}{2} z^3 + \frac{81}{40} z^5 - \ldots \]

Step 5: Determine the radius of convergence

To find the radius of convergence for \( \sin 3z \), note that the series is derived from a standard power series expansion:The radius of convergence for the series \( \sin z \) is infinite, thus, replacing \(z\) with \(3z\), the radius of convergence remains infinite. The coefficients still satisfy the convergence condition.

Key Concepts

Radius of Convergence

The radius of convergence is a crucial concept when dealing with power series like the Maclaurin series. To understand this, consider a power series of the form \( \sum_{n=0}^{\infty} c_n (z - a)^n \). The radius of convergence, denoted by \( R \), is the distance within which the series converges to a value. It tells us where the series behaves nicely and sums to a finite value.

For the sine function, such as \( \sin 3z \), replacing \( z \) with \( 3z \) does not alter the convergence radius because it is scaled uniformly. Here's the key takeaway:

• For a function like \( \sin z \), the radius of convergence is infinite since the sine function is entire (analytic at all points in the complex plane).
• Therefore, for \( \sin 3z \), the radius of convergence remains infinite, implying it converges everywhere in the complex plane.

Understanding this helps determine how far out in the complex plane the series gives valid outputs.

Derivatives

Derivatives are central to forming the Maclaurin series. The nth derivative of a function tells us the rate at which the function's value changes. For sine functions, derivatives follow a predictable cyclical pattern due to their trigonometric nature.

Consider \( f(z) = \sin 3z \):

• 1st derivative: \( f'(z) = 3\cos 3z \)
• 2nd derivative: \( f''(z) = -9\sin 3z \)
• 3rd derivative: \( f'''(z) = -27\cos 3z \)
• 4th derivative: \( f^{(4)}(z) = 81\sin 3z \)

These derivatives show a repeating cycle every four terms, alternating between sine and cosine with increasing powers of 3. Knowing this pattern is vital for constructing the series quickly.

When evaluated at \( z=0 \), these derivatives help populate the terms of the Maclaurin series. For example, \( f'(0) = 3 \) and \( f'''(0) = -27 \). Such patterns ease the computation process.

Sine function

The sine function is one of the basic trigonometric functions and is periodic with a period of \( 2\pi \). Its Maclaurin series expansion can express it as an infinite series, enabling calculations that go beyond what's easily computable directly.

Here's why \( \sin x \) is particularly important:

• An entire function: The sine function is analytic everywhere, meaning its series converges for all complex numbers.
• Produces familiar repeating patterns: \( \sin x \) derivatives result in familiar cosine and sine forms, making it easier to compute its series.
• Its symmetry and periodic nature simplify series manipulations, aiding in calculations that require smoothing periodic data.

The series for \( \sin 3z \), gives a deeper insight into predictable mathematical behaviors like oscillation, allowing engineers and scientists to model wave-like phenomena effectively.

Power Series Expansion

Power series expansion, including the Maclaurin series, is a fundamental tool in mathematics. It represents a function as an infinite sum of terms calculated from the function's derivatives at a single point, typically zero.

For the sine function, the Maclaurin series expansion for \( \sin 3z \) is:

• Use the formula \( f(z) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} z^n \).
• Calculate and insert each derivative value at \( z=0 \), forming terms in the series.
• For \( \sin 3z \), results in terms like \( 3z \), \( -\frac{9}{2} z^3 \), etc.

This expansion breaks a complex function into a manageable series of simpler polynomial parts.

Why expand functions like \( \sin 3z \) this way? It turns complicated functions into infinite polynomial series, easier to compute and analyze in practice applications such as physics or engineering. Recognizing the power of series expansion is key to leveraging this technique effectively in problem-solving.