# Show that set is an orthonormal set in PC(0,l)

• A
• Shackleford
In summary, the conversation discusses the process of proving that the set {(2/l)^{1/2} sin(n-1/2)(\pi x/l)}^\infty_1 is an orthonormal set in the interval (0,l). The conversation also addresses a potential error in the derivation and suggests using the original form of the integral when m=n.

#### Shackleford

Show that set $$\{ (2/l)^{1/2} sin(n-1/2)(\pi x/l)\}^\infty_1$$ is an orthonormal set in PC(0,l).
Of course, I need to show that $$<\phi_m, \phi_n> = \delta_{mn}$$
$$<\phi_m, \phi_n> = (2/l) \int_0^l sin(m-1/2)(\pi x/l) sin(n-1/2)(\pi x/l)\, dx \\ = 1/l \int_0^l cos(m-n)(\pi x/l) cos(m+n+1)(\pi x/l)\, dx \\ = \left( \frac 1 l \right) \frac {l} {\pi(m-n)} sin(m-n)(\pi x/l) \Big|_0^l - \left( \frac 1 l \right) \frac {l} {\pi(m-n)} sin(m+n+1)(\pi x/l)\ \Big|_0^l \\ = \frac {1} {\pi(m-n)} sin(m-n)(\pi x/l) \Big|_0^l - \frac {1} {\pi(m-n)} sin(m+n+1)(\pi x/l)\ \Big|_0^l$$

I used the trig identity:
$$2 sin \theta sin \psi = cos(\theta - \psi) - cos(\theta + \psi)$$

What is your question? The expression you have (you should check it out) = 0 for ##m\ne n## and =1 for ##m=n##.

mathman said:
What is your question? The expression you have (you should check it out) = 0 for ##m\ne n## and =1 for ##m=n##.

Thanks for the reply. When I looked at it, I could swear it's zero in both cases. Let me check again.

$$\\ = \frac {1} {\pi(m-n)} sin(m-n)(\pi x/l) \Big|_0^l - \frac {1} {\pi(m-n)} sin(m+n+1)(\pi x/l)\ \Big|_0^l$$

When you evaluate at x = l and x = 0, the argument for sine is either zero or a multiple of pi. When m = n, there's an indeterminate form in the common term.

When m=n, I suggest you don't use the cos identity, but use the original form, where the integral is ##2/l\int_0^lsin^2((m-1/2)(\pi x/l))dx##.

I think you have some errors in the derivation. You have the product of cos terms, rather than the difference, although this seems to have been corrected for the next line. When you integrated the cos (m+n+!)...) term, you divided by m-n.

## 1. What does it mean for a set to be orthonormal in PC(0,l)?

In PC(0,l), a set is orthonormal if all of its elements are orthogonal (perpendicular) to each other and have a magnitude of 1. This means that the inner product of any two elements in the set is 0, and the norm (or length) of each element is 1.

## 2. How do you prove that a set is orthonormal in PC(0,l)?

To prove that a set is orthonormal in PC(0,l), you must show that the inner product of any two elements in the set is 0 and that the norm of each element is 1. This can be done using the properties of inner product and the Pythagorean theorem, respectively.

## 3. Can a set be both orthonormal and linearly dependent in PC(0,l)?

No, in PC(0,l), a set cannot be both orthonormal and linearly dependent. If a set is orthonormal, it means that all of its elements are linearly independent. This is because if they were linearly dependent, then at least one of the elements would have a norm of 0, which would violate the definition of an orthonormal set.

## 4. What is the significance of orthonormal sets in PC(0,l)?

In PC(0,l), orthonormal sets have many applications, including in signal processing, quantum mechanics, and linear algebra. They allow for easier and more efficient calculations and can simplify complex problems.

## 5. Can a set be orthonormal in PC(0,l) if it contains infinite elements?

Yes, a set can be orthonormal in PC(0,l) even if it contains an infinite number of elements. This is because the concept of orthonormality is defined in terms of inner product and norm, which can be calculated for infinite sets using integrals. However, in practical applications, a finite orthonormal set is often used for computational simplicity.