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Vector Space Question

  1. Mar 11, 2004 #1
    The 3 dimensional space that we inhabit must have a basis of 3 vectors which is fair enough.

    But in my partial differential equations class in which fourier series was introduced, it was said that piecewise smooth function space has a basis of an infinite number of vectors. If there is a simple enough answer to this, I am curious to why this is. Does it have to do with the discontinuities that can arise in PWS space?
     
  2. jcsd
  3. Mar 11, 2004 #2

    matt grime

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    No, it's because there are an infinite number of linearly independent smooth functions (forget even piecewise)

    Take Functions from R to R

    set f_i(x) = x^i, i in N

    if the space were finite dimensional, of say dimension n, then given any n+1 of those functions, say just the first n+1, for ease, we could find real numbers a_i with

    sum 0 to n a_ix^i equal to the zero function on R ie zero for all x in R. But poly of degree n has only n roots, doesn't it? so it can't be identically zero.
     
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