# I Completeness of a basis function

1. Oct 24, 2018

### joshmccraney

Hi PF!

I'm somewhat new to the concept of completeness, but from what I understand, a particular basis function is complete in a given space if it can create any other function in that space. Is this correct?

I read that the set of polynomials is not complete (unsure of the space, since Taylor series can represent all continuous functions) but that Legendre polynomials are complete. Can anyone correct, finesse, or provide a working example of this?

Thanks!

Last edited: Oct 24, 2018
2. Oct 24, 2018

### Staff: Mentor

No.
At least your question isn't specifically enough to be answered clearly. Completeness usually relates to a certain set, and there are more than one way to complete such a set. E.g. $\mathbb{R}$ is the topological completion of $\mathbb{Q}$, whereas $\mathbb{C}$ is the algebraic completion (better: closure) of $\mathbb{R}$.
Taylor series are series, not polynomials.
What you wrote sounds to be the completion of a few linear independent functions to a basis in a given space; or the change to an orthonormal basis; or the description of the solution space to some equations. This usage of completion is misleading, at least if not explicitly stated what has to be completed to achieve what.

Last edited: Oct 24, 2018
3. Oct 24, 2018

### joshmccraney

Thanks for the response.

Hmmmm so how would you interpret the statement "complete basis functions"? Application here is with a functional, and evidently the input functions must be complete (nothing further given, I guess in physics/engineering it's assumed the reader knows the space and norm)?

4. Oct 24, 2018

### Staff: Mentor

This depends on the context. In (this) topological forum, I would assume a topological completeness, which is the standard usage of the term. It means to expand a given metric space with all possible limits of its Cauchy sequences. Functions of various types form topological, metric spaces, like smooth functions, or square integrable functions, or just continuous functions. So my first question would be: Which space? Which metric? As you see by my examples, there is no unique answer to this, even for engineers. Smooth and continuous are both important and very different. In quantum physics, I would assume the square integrable Lebesgue space.

However, you mentioned a single function to be completed, which doesn't fit in here. You also mentioned Legendre polynomials, which point to some differential equation, in which case the solution space might be meant and completion would indicate a basis of said space. "The input functions must be complete" doesn't make sense to me, except for "the [set of all] input functions must be complete [in order to ...]" and the terms in brackets are clear by context. In any case, I would ask for the area the formulation is taken from. Completion has an inherent meaning of adding something new to something given, so two questions have to be answered beforehand: what is given, and how is new defined. In the usual sense, the given is some metric space and the new are limits of Cauchy sequences. Since this appears not to be the case here, more information (context) is definitely needed.

5. Nov 9, 2018

### joshmccraney

Thanks for the response, and sorry it's taken me so long to reply. I'll do my best to give you precise insight on what I'm doing.

I am trying to solve an ODE that looks like this $$L(\phi_n(s)) = \lambda \phi(s):s\in[0,1]$$ where $L(\phi_n)\equiv \phi_n''(s)+\phi_n(s)$ and the subscript $n$ denotes a normal derivative to a surface (rather than go into details here, let's just think of that subscript as being one more derivative with some extra complications). I can't solve the ODE exactly, but I can solve the weak formulation, which looks like this $$(L(\phi_n),\phi_n) = \lambda(\phi,\phi_n) : (f,g) = \int_0^1 fg\, ds.$$ I will solve the weak form through an eigenfunction expansion, so I'll let $\phi = f_i$ for some predetermined $f_i$, so we could think of $f_i = \sin(i \pi x)$ or perhaps $x^i(1-x)$. Then we see the weak formulation is now an algebraic eigenvalue problem with matrices.

I didn't mention BC's and this is where my question of completeness enters. With some BC's it's obvious how to formulate the trial functions, such as $\phi(0)=\phi(1) = 0$; in this case we can let $f_i = x(1-x)^i$. However, in general it's not so simple how to build the BCs into the function space, so I have a technique, where basically I superimpose combinations of my trial functions to automatically solve the BCs.

For example, if I'm trying to solve the BCs $\phi(0)=\phi(1)=0$ and I'm using 3 trial functions $f_i = x^i$, then I need to take linear combinations of $\{x,x^2,x^3\}$ to satisfy the BCs, perhaps $\{x-x^2, x-x^3\}$. In this scenario, these two new functions will be my trial functions, specifically $f_1 = x-x^2$ and $f_2 = x-x^3$.

All of this for my final question: looking at the weak formulation, it seems what I am looking for is assurance that any $L^2(0,1)$ function can be expanded as linearly independent, linear combinations of a chosen function space.

I should specify that I am using a computer algebra package to determine the ways I should superimpose my selected trial functions. On problems where I do not have to recombine the trial functions, I get good results. However, sometimes when I recombine basis functions everything goes wrong. Do you think this is at all related to the completeness of the trial functions, once recombined?

6. Nov 10, 2018

### Svein

7. Nov 12, 2018

### WWGD

I believe some of these function spaces are not metrizable.