Problem 54 - Calculus: One and Several Variables

Question

Derive a series expansion in \(x\) for the function and specify the numbers \(x\) for which the expansion is valid. Take \(a>0\). $$f(x) \equiv \sin a x$$

Step-by-Step Solution

Answer

The series expansion for the function \(f(x) = \sin(ax)\) is given by the Maclaurin series: \[f(x) \approx \sum_{n=0}^{\infty} (-1)^n \frac{a^{2n+1}}{(2n+1)!} x^{2n+1}\] This expansion is valid for all values of \(x \in \mathbb{R}\).

Step 1: Find the first few derivatives of the function

We will need to find the first few derivatives of \(f(x) = \sin(ax)\) with respect to \(x\). We'll compute the 1st, 2nd, 3rd, and 4th derivatives. \[f'(x) = \frac{d}{dx}(\sin(ax)) = a\cos(ax)\] \[f''(x) = \frac{d^2}{dx^2}(\sin(ax)) = -a^2 \sin(ax)\] \[f'''(x) = \frac{d^3}{dx^3}(\sin(ax)) = -a^3 \cos(ax)\] \[f^{(4)}(x) = \frac{d^4}{dx^4}(\sin(ax)) = a^4 \sin(ax)\]

Step 2: Apply the Taylor series expansion formula for sin(ax) at x=0

Now let's plug the derivatives into the Taylor series expansion formula. Since we are deriving a series expansion in \(x\), we'll take \(a = 0\): \[f(x) \approx \sin(0) + (a\cos(0))(x-0) - \frac{(a^2\sin(0))}{2!}(x-0)^2 - \frac{(a^3\cos(0))}{3!}(x-0)^3 + \cdots\]

Step 3: Simplify the expression and write out the power series

Replace \(\sin(0)\) and \(\cos(0)\) with their values in the series and simplify: \[f(x) \approx (0) + (ax) - \frac{(0)}{2!}(x)^2 - \frac{(a^3)}{3!}(x)^3 + \cdots\] This simplifies to: \[f(x) \approx ax - \frac{a^3}{3!}x^3 + \frac{a^5}{5!}x^5 - \frac{a^7}{7!}x^7 + \cdots\] Now we have obtained a series expansion for our function \(f(x) = \sin(ax)\). This is a Maclaurin series, which is a special case of Taylor series centered at \(x=0\): \[f(x) \approx \sum_{n=0}^{\infty} (-1)^n \frac{a^{2n+1}}{(2n+1)!} x^{2n+1}\]

Step 4: Determine the range of validity for x values

The given Maclaurin series converges for all values of x for a sine function. The reason is that sine function is an entire function, meaning it is analytic everywhere on the complex plane. Thus, the range of validity for \(x\) is: \[x \in \mathbb{R}\] In conclusion, the series expansion for the function \(f(x) = \sin(ax)\) is: \[f(x) \approx \sum_{n=0}^{\infty} (-1)^n \frac{a^{2n+1}}{(2n+1)!} x^{2n+1}\] And the expansion is valid for all values of \(x\) in the real domain.

Key Concepts

Maclaurin series

The Maclaurin series is a specific type of Taylor series that's centered at zero. The goal of this series is to express complex functions as infinite sums of simpler terms. This involves taking derivatives of the function at zero and using these to create a polynomial.

For the Maclaurin series, the formula is given by: \[f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} x^n\] where \( f^{(n)}(0) \) represents the \( n \)-th derivative of the function evaluated at zero.

Maclaurin series is particularly useful because it simplifies the function into an infinitely long polynomial, making it easier to approximate and analyze at points close to the center \( x=0 \). In this way, we can understand the behavior of complex functions using simple polynomial terms.

derivatives

Derivatives are crucial in constructing the Maclaurin series. They provide the various coefficients of the terms in the polynomial expansion of a function.

When constructing a series expansion for the function \( f(x) = \sin(ax) \), finding derivatives of this function shows how the sine wave changes at different points. Here's how each derivative helps in forming the series:

• \(f'(x) = a\cos(ax)\) gives the slope or rate of change at any point.
• \(f''(x) = -a^2\sin(ax)\) gives the curvature, showing how steeply it turns.
• Higher derivatives like \(f'''(x)=-a^3\cos(ax)\) and \(f^{(4)}(x)=a^4\sin(ax)\) further refine the shape.

Each derivative is like peeling back a layer of the function, allowing us to see deeper changes as the function evolves. These insights are then captured as numerical coefficients in the Maclaurin series.

sine function

The sine function \( \sin(x) \) is a periodic trigonometric function that oscillates between -1 and 1. It's highly useful in modeling phenomena with wave-like behavior because of its periodic nature.

For example, in our exercise, we considered \(f(x) = \sin(ax)\), a scaled version of the sine function. The constant \(a\) adjusts the frequency of oscillation. Larger \(a\) values make the function oscillate more rapidly:

• The basic shape remains the same: rising to a peak, falling to a trough, and repeating.
• The series expansion for \(\sin(ax)\) using the Maclaurin series allows us to approximate this complex function using simple polynomial terms.

This practical method provides reliable results suitable for both theoretical and real-world applications. The beauty of the sine function lies in its predictability and universal presence in nature, physics, and engineering.