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Question regarding basis of function space

  1. Jan 26, 2012 #1
    I only possess a rudimentary understanding of Linear Algebra so I'm not going to be rigorous in my explanation, but is the concept of an infinite basis well defined? More specifically, I was thinking about how the polynomials could form a basis for function space, given that every function has a Taylor expansion which is a linear combination of the polynomials.
     
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  3. Jan 26, 2012 #2

    LCKurtz

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    If you want to say the set of polynomials are to represent a basis, you need to talk about the notion of convergence or what topology you are using on the function space. For example, consider C[0,1], the continuous real functions on [0,1] as your function space where convergence ##f_n\rightarrow f## means the sequence ##f_n## converges uniformly to ##f##. Now, it is true that the set of polynomials form a basis in the sense that given any ##f## and ##\epsilon > 0##, you can find a polynomial p with ##\|f-p\|<\epsilon##. We say that the set of polynomials is dense in C[0,1] and they form a topological basis. But that is not the same thing as saying every function in C[0,1], even if it has derivatives of all orders, can be well approximated by a Taylor polynomial. There are infinitely smooth functions whose Taylor expansion just gives 0. Taylor polynomials are too specialized for that particular job.
     
  4. Jan 27, 2012 #3

    HallsofIvy

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    In infinite dimensional spaces there may be two different types of "bases".

    A "Hamel basis" is an infinite set such that any vector in the vector space can be written as a linear combination of a finite number of vectors in the basis. The functions 1, x, [itex]x^2[/itex], ...,[itex]x^n[/itex], ... form a Hamel basis for the space of all polynomials but not for the set of (real) analytic functions (a real valued function on the real numbers is "analytic" if and only if it is equal to its Taylor series). It can be shown (assuming axiom of choice) that every vector space has a Hamel basis.

    If you have a topology on your vector space, and so a notion of "convergence", a more general concept of "basis" is, as LKurtz said, a set of vectors such that any vector can be written as a possibly infinite linear combination. The functions 1, x, [itex]x^2[/itex], ...,[itex]x^n[/itex], ... form a basis, in this sense, for the space of all (real) analytic functions.
     
  5. Jan 28, 2012 #4

    Bacle2

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    I don't know what you mean by well-defined, but a (Hamel) basis is a linearly-independent set that spans the space , or as said in both replies, a L.I set such
    that for all f in the space, there is a linear combination of the base elements that converges to f. You can check that for any n, the set {1,x,x2,...,xn} is linearly-independent in C[0,1], and spans in the Schauder sense, and, in this sense, it is infinite.
     
  6. Jan 28, 2012 #5

    morphism

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    No - there should be no mention of convergence here.

    {1, x, x^2, ...} isn't a Hamel basis for C[0,1], but as you somewhat indicate, it's a Schauder basis.
     
  7. Jan 28, 2012 #6

    Bacle2

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    Yes, I did sort-of mix up both cases; my bad--I should stop posting at 3 a.m.

    Still, for any n, the mentioned set is linearly-independent.
     
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