Gram-Schmidt Orthonormal Functions

• Robsta
In summary, the function f(x) = xe-3x2 is expressed as a linear combination of the basis functions un(x), which are orthogonal and normalised from minus infinity to infinity. It is expressed by xe-3x2 = ∑anun(x) the un(x)'s are even functions of x for n = 0,2,4 and are odd functions of x for n = 1,3,5 Calculate a0, a1 and a2 given that: u1(x) = (4sqrt(2)/sqrt(pi))1/2xe-x2 and f(x) is odd.

Homework Statement

The function f(x) = xe-3x2 is expressed as a linear combination of the basis functions un(x), which are orthogonal and normalised from minus infinity to infinity.

It is expressed by xe-3x2 = ∑anun(x)

the un(x)'s are even functions of x for n = 0,2,4 and are odd functions of x for n = 1,3,5

Calculate a0, a1 and a2 given that:

u1(x) = (4sqrt(2)/sqrt(pi))1/2xe-x2

I'm also told that the integral from -infinity to infinity of x2e-a2x2dx = sqrt(pi)/2a3

Homework Equations

I'm aware of Gram Schmidt orthogonalisation but am at a bit of a loss as to how to do it from the second term and how to guess the next terms. I get the process but don't you need a set of linearly independent functions to do it to?

The Attempt at a Solution

It's asking for the various coefficients and not the u(x) terms, but do I have to work those out with G-S orthogonalisation? If so I need to find the set of LI functions to orthogonalise.

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Robsta said:
the un(x)'s are even functions of x for n = 0,2,4 and are odd functions of x for n = 1,3,5
Can't you use that by itself?

I'm not sure what you mean. I know that u0 and u2 will be even functions?

But there are so many even functions! Which ones do I choose?

Can you write an equation for a coefficient ##a_n## in terms of ##u_n## and ##f(x)##?

Either I'm misreading it or u1 is a constant multiple of f. Doesn't that make it somewhat trivial?

Sorry haruspex I typed the question wrong. I've now rewritten u1 as it should be.

Dr Claude, I don't think I can. I could write a particular phase of G-S orthogonalisation but not a general one in terms of n. If I integrated the product of f(x) and un(x) would I get an?

Robsta said:
Dr Claude, I don't think I can. I could write a particular phase of G-S orthogonalisation but not a general one in terms of n. If I integrated the product of f(x) and un(x) would I get an?

Yep, okay. So if you integrate the product of a function and one of it's basis functions, it gives you the amount that it projects along that basis function (using the vector analogy). So to find a0 I have to integrate f(x)u0(x) from minus infinity to infinity. So how do I guess what u(x) is so that I can put it into the integral.

Robsta said:
Sorry haruspex I typed the question wrong. I've now rewritten u1 as it should be.
OK. Is f odd, even or neither? What does that suggest about the contributions from the basis functions?

Robsta said:
Yep, okay. So if you integrate the product of a function and one of it's basis functions, it gives you the amount that it projects along that basis function (using the vector analogy). So to find a0 I have to integrate f(x)u0(x) from minus infinity to infinity. So how do I guess what u(x) is so that I can put it into the integral.
You don't need to guess. Go back to post #2.

So I know that I have to integrate f(x)u0(x)dx and that u0(x) is even. Is there some kind of dimensional grounds that I can work out what sort of even function it is?

f(x) is odd, since it's a product of an odd function and an even function.

So what happens when you integrate an odd function for ##-\infty## to ##\infty##?

Does that imply that it can be made with just odd functions so a0 and a2 are zero?

Robsta said:
Does that imply that it can be made with just odd functions so a0 and a2 are zero?
Exact!

Fantastic! Thanks for your help, both of you. I'm just going to compute the integral to find a1 then. Will reply to let you know of success.

Robsta said:
Does that imply that it can be made with just odd functions so a0 and a2 are zero?
It's a bit stronger than that - it can only be made from the odd functions. If you imagine summing the odd and even basis functions separately into F, G respectively, you have f = F+G. But to make f odd, G must be zero. Since the even functions are orthonormal, their coefficients must all be zero.

Great, I got it done. Pretty straightforward and not nearly as daunting as it first seemed! Thanks for your help :)

1. What is the Gram-Schmidt process?

The Gram-Schmidt process is a method used for transforming a set of linearly independent vectors into an orthonormal set of vectors. This process is commonly used in linear algebra and functional analysis.

2. What are orthonormal functions?

Orthonormal functions are a set of functions that are both orthogonal and normalized. This means that they are perpendicular to each other and have a magnitude of 1. These functions are commonly used in mathematical and scientific fields, such as signal processing and quantum mechanics.

3. How are Gram-Schmidt orthonormal functions calculated?

The Gram-Schmidt process involves three steps: 1) choosing a set of linearly independent vectors, 2) orthogonalizing the vectors using the Gram-Schmidt process, and 3) normalizing the resulting vectors to obtain an orthonormal set. The specific calculations for each step can vary depending on the context and application.

4. What is the significance of Gram-Schmidt orthonormal functions?

Gram-Schmidt orthonormal functions have many applications in mathematics, physics, and engineering. They are commonly used in solving systems of linear equations, orthogonal projection, and finding the best fit for a set of data. They also play a crucial role in quantum mechanics and signal processing.

5. Are there any limitations to using Gram-Schmidt orthonormal functions?

While Gram-Schmidt orthonormal functions have many applications, they do have some limitations. The process can be computationally expensive, especially for large sets of vectors. Additionally, the resulting orthonormal set may not be unique, and the calculated functions may not accurately represent the original set of vectors in some cases.