B GramSchmidt process for Taylor basis

[Somali_Physicist]

Why are the limits as so for the integral?

 

[fresh_42]

In the end it depends which topological space you consider and how the inner product is defined. Those two information set the conditions. As it is an example, we can simply consider $L_2([-1,1])$. Taylor series are often developed around $x=0$, so the choice of this interval makes some sense.

 

[Somali_Physicist]

In the end it depends which topological space you consider and how the inner product is defined. Those two information set the conditions. As it is an example, we can simply consider $L_2([-1,1])$. Taylor series are often developed around $x=0$, so the choice of this interval makes some sense.

This arrives at the legendary basis so It is abit fishy that it is a “coincidence “.So I could pic any numbers around 0?

 

[fresh_42]

This arrives at the legendary basis so It is abit fishy that it is a “coincidence “.So I could pic any numbers around 0?

I haven't checked the algorithm. You might get additional scaling factors for your basis vectors by a different inner product, i.e. other spaces. Depends on whether you want to have a orthonormal or orthogonal basis.

 

[Somali_Physicist]

I haven't checked the algorithm. You might get additional scaling factors for your basis vectors by a different inner product, i.e. other spaces. Depends on whether you want to have a orthonormal or orthogonal basis.

I don’t believe that matters here, I just want to clarify why the limits were used and can we change them ad hocly.

 

[fresh_42]

I don’t believe that matters here, I just want to clarify why the limits were used and can we change them ad hocly.

As I said, the limits are determined by the space you consider, which again is determined by the elements and the inner product. You must now reason backwards. First is the space, then algorithm and at the end the basis. Another space results in another basis.

 

[Somali_Physicist]

As I said, the limits are determined by the space you consider, which again is determined by the elements and the inner product. You must now reason backwards. First is the space, then algorithm and at the end the basis. Another space results in another basis.

Ahh okay, well we want orthanormal bases, is it because of that we chose -1 to 1?

 

[fresh_42]

Ahh okay, well we want orthonormal bases, is it because of that we chose -1 to 1?

No. One possibility for an inner product for say, continuous functions on a real interval $[a,b]$ is e.g. $\langle f,g \rangle = \int_a^b f(x)g(x)\,dx\,.$ This determines how the algorithm is done and what basis finally drops out. You start with the Hilbert space which you want to consider, not the other way around. At the end of this article here https://www.physicsforums.com/insights/hilbert-spaces-relatives/ you can find some other examples. E.g. 5.10 is worth a look. There the integration is over the entire manifold. Of course in the above example we would get infinite integrals if we regarded $\mathbb{R}$ as our domain, so we need some restrictions on the allowed functions. But in general there is a variety of examples. As soon as you define and thus tell us, which Hilbert space you use, as soon can we talk about the algorithm and the borders of integration: different borders means different Hilbert spaces. So before the question even occurs, it has to be said, what we are talking about. I admit that this information isn't part of what you copied and in my opinion sloppy.