Problem 94 - Calculus Single Variable

Question

Suppose that \(f(x)\) is an odd function: \(f(-x)=-f(x)\). Let $$ b_{n}=\frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \sin (n x) d x \quad(n \geq 1) $$ The function $$ \sum_{n=1}^{N} b_{n} \sin (n x) $$ is said to be the order \(N\) Fourier sine-polynomial approximation of \(f\). Using the viewing window \([-2,2] \times\) \([-2,2],\) plot the order \(N\) Fourier sine-polynomial approximation of \(f(x)=x\) for \(N=3\) and 5 .

Step-by-Step Solution

Answer

For \(N=3\) and \(N=5\), plot the Fourier sine series: \( f(x) \approx \sum_{n=1}^{N} \frac{2(-1)^{n+1}}{n} \sin(nx) \) on \([-2,2] \times [-2,2]\).

Step 1: Understanding Odd Functions

An odd function has the property that for all values of its domain, it satisfies \( f(-x) = -f(x) \). As a consequence, the Fourier series for an odd function only contains sine terms.

Step 2: Fourier Sine Coefficients

For an odd function, the Fourier sine coefficients are given by \( b_{n} = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \sin(nx) \, dx \). This expression is used to determine the coefficients for the sine terms in the Fourier series.

Step 3: Calculate \(b_n\) for \(f(x) = x\)

For \( f(x) = x \), substitute into the formula: \( b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} x \sin(nx) \, dx \). This integral can be solved using integration by parts.

Step 4: Integration by Parts Formula

Recall the integration by parts formula: \( \int u \, dv = uv - \int v \, du \). Let \( u = x \) and \( dv = \sin(nx) \, dx \). Then, \( du = dx \) and \( v = -\frac{1}{n}\cos(nx) \). Apply this to compute the integral.

Step 5: Solve for \(b_n\)

Substituting \( u \), \( dv \), \( du \), and \( v \) into the integration by parts formula, compute \[ b_n = \frac{1}{\pi} \left[ -\frac{x}{n}\cos(nx) \Big|_{-\pi}^{\pi} + \frac{1}{n}\int_{-\pi}^{\pi} \cos(nx) \, dx \right] \]. Simplify the result to \( b_n = \frac{2(-1)^{n+1}}{n} \).

Step 6: Construct the Fourier Sine Polynomial

The Fourier sine-polynomial approximation of order \( N \) is given by \( \sum_{n=1}^{N} b_{n} \sin(nx) \). Substitute the known \( b_n \) values into this sum for \( N=3 \) and \( N=5 \).

Step 7: Plot for \(N = 3\) and \(N = 5\)

Using software like Desmos or Python with Matplotlib, plot \( \sum_{n=1}^{3} \frac{2(-1)^{n+1}}{n} \sin(nx) \) and \( \sum_{n=1}^{5} \frac{2(-1)^{n+1}}{n} \sin(nx) \) over the interval \([-2,2]\). Use plot range \([-2,2]\) for both x and y axes.

Key Concepts

odd functions

One fundamental concept in Fourier analysis is that of odd functions. An odd function is characterized by a specific type of symmetry across the origin. For any function \(f(x)\), if we find that \(f(-x) = -f(x)\) for all values in its domain, then the function is considered odd. This means that when you flip the function about the y-axis and then the x-axis, you get the original function. This property impacts the structure of its Fourier series representation.
In the context of Fourier series, only sine terms appear in the expansion of odd functions. This happens because sine itself is an odd function \( \sin(-x) = -\sin(x)\). Thus, the cosine terms (which are even functions) vanish, leaving only the sine components. Understanding this property is critical when working with the Fourier series, as it allows us to focus solely on finding the sine coefficients.

Fourier sine coefficients

The Fourier sine coefficients are crucial when constructing a Fourier series for an odd function. These coefficients determine the amplitude of each sine wave component contributing to the Fourier series representation of the function. For an odd function \(f(x)\), the Fourier sine coefficients are derived using the formula:

• \(b_{n} = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \sin(nx) \, dx\)

This integral calculates how much of the sine function \( \sin(nx) \) is needed to approximate the given function \( f(x) \) over the interval \([-\pi, \pi]\).
By computing these coefficients, you can build a sine-series representation that captures the essence of the original function, especially when the function is odd. It's a step-by-step process where each coefficient \(b_n\) enhances the accuracy of the approximation.

integration by parts

Integration by parts is a valuable technique in calculus, particularly useful when integrating the product of two functions. This method is instrumental in calculating Fourier sine coefficients for complex functions like \(f(x) = x\). The integration by parts formula is expressed as:

• \( \int u \, dv = uv - \int v \, du \)

When applying this to find the Fourier sine coefficients for \(f(x) = x\), we choose:

• \( u = x \)
• \( dv = \sin(nx) \, dx \)
• This gives \( du = dx \) and \( v = -\frac{1}{n}\cos(nx) \)

The integration becomes:

• \[ \int_{-\pi}^{\pi} x \sin(nx) \, dx = -\frac{x}{n}\cos(nx) \Big|_{-\pi}^{\pi} + \frac{1}{n}\int_{-\pi}^{\pi} \cos(nx) \, dx \]

This process simplifies the integral by reducing a complex problem into simpler parts, yielding a precise value for each coefficient \(b_n\). Learning to utilize integration by parts effectively is key in solving these integrals and calculating the coefficients accurately.

Fourier sine-polynomial approximation

The Fourier sine-polynomial approximation is a technique for approximating a function by summing its Fourier sine series terms. It's particularly effective for odd functions, as it leverages their inherent properties to simplify calculations. The approximation of an order \(N\) looks like:

• \( \sum_{n=1}^{N} b_{n} \sin(nx) \)

This form sums the first \(N\) sine terms, with each term proportionally weighted by its respective coefficient \(b_n\).
The more terms included in the summation, the more accurately the approximation can replicate the original function. For example, using \(N=3\) or \(N=5\) provides a truncated series that captures the primary behavior of the function \(f(x) = x\) across the interval \([-2, 2]\). Ultimately, this approximation helps visualize complex functions using simpler periodic components, aiding both analysis and visualization in many mathematical and engineering applications.