Problem 11 - Advanced Engineering Mathematics

Question

Expand the given function in a Maclaurin series. Give the radius of convergence of each series. \(f(z)=\sin z^{2}\)

Step-by-Step Solution

Answer

The Maclaurin series for \( \sin(z^2) \) is \( z^2 - \frac{z^6}{3!} + \frac{z^{10}}{5!} - \cdots \), with an infinite radius of convergence.

Step 1: Recall 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 \( f \) evaluated at \( z = 0 \). Let's apply this for \( \sin(z^2) \).

Step 2: Identify Function and Substitute

Given \( f(z) = \sin(z^2) \), substitute \( w = z^2 \) into the Taylor series for \( \sin(w) \): \[ \sin(w) = \sum_{n=0}^{\infty} (-1)^n \frac{w^{2n+1}}{(2n+1)!} \] Thus, \[ \sin(z^2) = \sum_{n=0}^{\infty} (-1)^n \frac{(z^2)^{2n+1}}{(2n+1)!} = \sum_{n=0}^{\infty} (-1)^n \frac{z^{4n+2}}{(2n+1)!} \].

Step 3: Determine Maclaurin Series Terms

To write the expansion, note that each term \( (-1)^n \frac{z^{4n+2}}{(2n+1)!} \) is derived directly from substituting \( w = z^2 \) into the series for \( \sin(w) \). Thus, the Maclaurin series for \( \sin(z^2) \) becomes: \[ \sin(z^2) = z^2 - \frac{z^6}{3!} + \frac{z^{10}}{5!} - \frac{z^{14}}{7!} + \cdots \]

Step 4: Calculate Radius of Convergence

The series of \( \sin(w) \) has an infinite radius of convergence, \(|w| < \infty\). Since \( w = z^2 \), we apply this directly to the modified series, concluding that \( \sin(z^2) \) has an infinite radius of convergence, \(|z^2| < \infty\) implies \(|z| < \infty\).

Key Concepts

Radius of Convergence

The radius of convergence of a power series is a crucial concept that determines where the series converges to a finite sum. Let's explore this in more depth. For a series \( \sum_{n=0}^{\infty} a_n z^n \), the radius of convergence \( R \) tells us the range of values for \( z \) where the series converges. It is calculated using the formula \( R = \frac{1}{\limsup_{n \to \infty} \sqrt[n]{|a_n|}} \).
- If \( R = \infty \), the series converges for all \( z \).
- If \( R = 0 \), the series only converges at \( z = 0 \).
- If \( 0 < R < \infty \), the series converges when \( |z| < R \).
Understanding the radius of convergence helps us predict where a series will behave well (converges) and where it might diverge. In the exercise provided, the Maclaurin series for \( \sin(z^2) \) has an infinite radius of convergence, meaning it converges for any \( z \). This infinite radius is typical for trigonometric functions like sine, breaking down their behavior into well-defined intervals no matter how large \( z \) becomes.

Taylor Series Expansion

In calculus, the Taylor series expansion is a powerful tool to express functions as infinite sums of their derivatives at a certain point. For the Maclaurin series, this point is zero, simplifying calculation by centering the expansion around zero.
When a function \( f(z) \) has derivatives that are well-defined at \( z = 0 \), its Maclaurin series is written as \[ f(z) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} z^n \].
Using this, we can approximate functions with polynomials, simplifying complex calculations. In the case of \( \sin(z^2) \), the Maclaurin series involves substituting \( w = z^2 \) into the known series for the sine function. This leads to the series:
\( \sin(z^2) = z^2 - \frac{z^6}{3!} + \frac{z^{10}}{5!} - \frac{z^{14}}{7!} + \cdots \)
This series represents \( \sin(z^2) \) using powers of \( z \). Complete with infinite terms, it can perfectly replicate the original function's values in its convergence region, thanks to the power of mathematical limits and series.

Trigonometric Functions

Trigonometric functions, such as \( \sin \), \( \cos \), and \( \tan \), play a vital role in mathematics, describing relationships in right triangles and periodic phenomena like waves. In the context of calculus and series, they are often expanded using power series to handle complex trigonometric equations more easily.
The function \( \sin(z^2) \) was specifically chosen in our exercise to show how substituting variables into well-known series can simplify computations. For example:

• The general Maclaurin series for \( \sin(w) \) is \( \sum_{n=0}^{\infty} (-1)^n \frac{w^{2n+1}}{(2n+1)!} \).
• By replacing \( w \) with \( z^2 \), we tailor the series to represent \( \sin(z^2) \).

This approach not only reduces calculus problems into polynomials but is a critical technique in solving differential equations and modeling physical systems. Trigonometric functions' behavior is universally understood, aiding in these complex transformations and series expansions.