Problem 15 - Calculus: One and Several Variables

Question

Expand \(g(x)\) as indicated and specify the values of \(x\) for which the expansion is valid. \(g(x)=\sin x \quad\) in powers of \(x-\pi\).

Step-by-Step Solution

Answer

The series expansion of \(g(x) = \sin x\) about \(x - \pi\) is given by: \(\sin(x-\pi) = \sum_{n=0}^{\infty} (-1)^n \frac{(x-\pi)^{2n+1}}{(2n+1)!}\). It is valid for all real values of \(x\).

Step 1: Memorizing the Taylor Series Expansion of \(\sin x\) about 0

The Taylor series expansion of \(\sin x\) about 0 is given by: \(\sin(x) = \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{(2n+1)!}\). This is an infinite series where n changes values from 0 to infinity.

Step 2: Replace \(x\) with \(x-\pi\)

Now we replace every \(x\) in the series with \(x-\pi\). The series now becomes: \(\sin(x-\pi) = \sum_{n=0}^{\infty} (-1)^n \frac{(x-\pi)^{2n+1}}{(2n+1)!}\). This is the series expansion of \(\sin x\) about \( x - \pi\).

Step 3: Identify the domain of validity for the series expansion

The series expansion of a function is valid in a neighborhood around the point about which the series is expanded. Since \(-1 \leq \sin(x-\pi) \leq 1\), the series expansion of \(\sin x\) about \(x - \pi\) will converge for all real values of \(x\).

Key Concepts

sin x expansion

When dealing with the function \( \sin x \), one of the most powerful tools we have is its Taylor series expansion. This expansion represents the function as an infinite sum of terms calculated from the values of its derivatives at a single point. For \( \sin x \), the standard expansion about 0 (also called a Maclaurin series) is given by:

• \( \sin(x) = \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{(2n+1)!} \)

This means that \( \sin x \) can be expressed as a series of odd powers of \( x \), where each power has a coefficient determined by its position in the sum. The beauty of Taylor series is that they provide a powerful approximation tool, allowing us to express functions in a simplified form that is often easier to work with in calculus.
Replacing \( x \) with \( x - a \) in the series moves the center of the expansion to \( a \). In this exercise, transforming \( \sin x \) by substituting \( x - \pi \) allows us to create an expansion centered at \( \pi \) rather than at 0.

series convergence

Convergence of a series is a crucial topic in understanding power series like the Taylor series. In simple terms, a series is said to converge if the sum of its terms approaches a specific value as more and more terms are added. For our series involving \( \sin(x - \pi) \), the goal is to determine over what range of \( x \) the series correctly represents the function.
The series expansion of \( \sin(x - \pi) \) converges for all real values of \( x \) because the cosine and sine functions are bounded between -1 and 1 regardless of their argument. This particular characteristic of trigonometric functions implies that their Taylor series expansion around any center will converge for all real number inputs.
This uniform convergence across the entire set of real numbers is a hallmark feature of periodic functions like \( \sin x \) and \( \cos x \). Thus, the range of validity for the convergence of \( \sin(x - \pi) \) expanding into its series is the set of all real numbers.

power series about a point

Power series are mathematical expressions that resemble a sum of powers of a variable, each accompanied by a coefficient. When we expand a function like \( \sin x \) about a point \( a \), we express \( \sin x \) in terms of a new variable \( u \) where \( u = x-a \). Specifically, this results in a series known as a Taylor series, centered around \( a \). In this exercise, our power series is shifted to work around \( \pi \).
To write \( \sin x \) as a power series about \( \pi \), we replace \( x \) with \( x-\pi \), thus moving the center of expansion from 0 to \( \pi \). This results in the series:

• \( \sin(x-\pi) = \sum_{n=0}^{\infty} (-1)^n \frac{(x-\pi)^{2n+1}}{(2n+1)!} \)

By expanding about \( \pi \), you effectively modify the inputs around which the infinite series approximation is most accurate. This approach is particularly useful in calculus when dealing with patterns or approximations near particular values, other than zero, in function behavior.
Using power series, particularly the Taylor series, helps in simplifying complex analytic operations, solving differential equations, and approximating functions in physics and engineering problems.