Question

Show that the functions \(e^{i n \pi x / l}, n=0,\pm 1,\pm 2, \cdots,\) are a set of orthogonal functions on \((-l, l)\).

Short answer

The functions \(e^{i n \pi x / l}\) are orthogonal for \(n \eq m\) on \(-l, l\).

Step-by-step solution

- Understand Orthogonality

Two functions \(f(x)\) and \(g(x)\) are considered orthogonal on the interval \([-l, l]\) if their inner product is zero. Mathematically, this is expressed as: \( \int_{-l}^{l} f(x) g(x) \, dx = 0 \). We need to show this for the given functions.

- Identify the Functions

The functions given are \(e^{i n \pi x / l}\) for different values of \(n\), where \(n = 0, \pm 1, \pm 2, \cdots\). Let’s denote these functions as \(f_n(x) = e^{i n \pi x / l}\).

- Form the Inner Product

To show that \(f_n(x)\) and \(f_m(x)\) are orthogonal, we need to compute their inner product: \(\int_{-l}^{l} f_n(x) \overline{f_m(x)} \, dx\). Here, \(\overline{f_m(x)}\) is the complex conjugate of \(f_m(x)\).

- Compute the Complex Conjugate

The complex conjugate of \(f_m(x)\) is \(\overline{e^{i m \pi x / l}} = e^{-i m \pi x / l}\). So, the inner product to evaluate is \(\int_{-l}^{l} e^{i n \pi x / l} e^{-i m \pi x / l} \, dx \).

- Simplify the Expression

Simplify the exponent by combining the terms: \(e^{i n \pi x / l} e^{-i m \pi x / l} = e^{i (n - m) \pi x / l}\). Now, we need to evaluate \(\int_{-l}^{l} e^{i (n - m) \pi x / l} \, dx\).

- Solve the Integral

Evaluate the integral: \(\int_{-l}^{l} e^{i (n - m) \pi x / l} \, dx\). Case 1: If \(n = m\), then the integral simplifies to \(\int_{-l}^{l} 1 \, dx = 2l\). Case 2: If \(n \eq m\), use the result that \(\int_{-l}^{l} e^{i k x} \, dx = 0\) for any non-zero \(k\). Here \(k = (n - m) \pi / l\). Thus, the result is 0.

- Conclude Orthogonality

If \(n \eq m\), \(\int_{-l}^{l} e^{i (n - m) \pi x / l} \, dx = 0\), indicating that the functions are orthogonal. If \(n = m\), the result is non-zero, indicating they are not orthogonal when identical.

Key concepts

Complex Exponential Function

The complex exponential function is a powerful concept in mathematics, particularly in the field of complex analysis. It is denoted as \( e^{ix} \), where \( i \) is the imaginary unit (athrm{i} = athsf{\frac{\text 1}{ athrm{\text sqrt -1} }} ). This function can be described using Euler's formula, which states:

• \( e^{ix} = athsf{\text cos x} + athrm i \text sin x \).

This form is useful because it combines both cosine and sine functions into a single complex exponential. In the exercise, the functions are given as \(e^{i athpi x / l}\). Each value of \( n \) creates a different function, transforming it into \( e^{i n athpi x / l} \). The next sections will delve further into how these functions work together and why they are orthogonal.

Inner Product

Understanding inner products is crucial when dealing with orthogonal functions. The inner product is a measure of the 'overlap' between two functions over a given interval. For functions \( f(x) \) and \( g(x) \) defined on an interval \([-l, l]\), the inner product is calculated as:
athint_{-l}^{l} f(x) g(x) athrm dx.

However, for functions that can be complex, like the complex exponential functions in the exercise, we need to use the complex conjugate of one of the functions. Thus, the inner product becomes:
athint_{-l}^{l} f_n(x) overline\text_f_m\text(x) athrm dx.

• Where \(\overline\text_f_m(x) = athtext\text complex conjugate\text\text of f_m(x)\).

This means we compute the inner product **between** \( f_n \text(x) \) **and** \( athrm\text the complex conjugate of f_m(x)\).
This concept is essential for showing the orthogonality of the given functions.

Orthogonality

Orthogonality is a key concept in linear algebra and functional analysis, particularly in the context of complex functions. Two functions \( f(x) \) and \( g(x) \) are said to be orthogonal on the interval \([-l, l]\) if their inner product is zero:
athint_{-l}^{l} f(x) athoverline{g(x)} athrm dx = 0

In our problem, we need to show that for the given set of functions \( e^{i n athpi x / l} \), the inner product is zero whenever \( n eq m \). If \( n = m \), the inner product will not be zero, demonstrating they would not be orthogonal in this special case.

Complex Conjugate

The complex conjugate of a complex number or function is found by changing the sign of the imaginary part. For a complex number \(z = a + bi\), its complex conjugate is \(\overline{z} = a - bi\). For the function in our problem, \( f_m(x) = e^{i m athpi x / l} \), the complex conjugate is:

• \( f_m(x) \)= \( \overline{e^{i m athpi x / l}} = e^{-i m athpi x / l} \).

Applying the complex conjugate is an important step in verifying orthogonality. It allows us to correctly compute the inner product of the complex functions.

Integral Evaluation

Evaluating integrals is a crucial step in proving the orthogonality of the given functions. To determine if \( e^{i n athpi x / l} \) and \( e^{i m athpi x / l} \) are orthogonal, we need to compute the integral:
athint_{-l}^{l} e^{i (n - m) athpi x / l} athrm dx.

Let's consider the two cases:

• If \( n = m \), the exponent becomes zero, and the integral simplifies to athint_{-l}^{l} 1 athrm dx, which evaluates to \(2l\).


• If \( n eq m \), we can utilize the result athint_{-l}^{l} e^{i k x} athrm dx = 0 for any non-zero \( k \). In our case, \( k = \frac{(n - m) athpi}{l} \), making the integral equal to zero.

Therefore, the functions are orthogonal when \( n eq m \) since the integral evaluates to zero.