Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Even Continuous Functions

  1. Feb 12, 2009 #1
    The problem statement, all variables and given/known data
    Let F be the set of all continuous functions with domain [-1,1] and codomain R. Let A be the algebra of all polynomials that contain only terms of even degree (A is a subset of F). Show that the closure of A in F is the set of even functions in F.

    The attempt at a solution
    I have to show that (i) if f in F is even, then f is in the closure of A and (ii) if f is in the closure of A, then f is even. I don't have problems proving (ii), rather I'm stuck proving (i). Here's what I have so far:

    Let f in F be even. By the Weierstrass Approx. Theorem, there is a sequence of polynomials {p_n} that converge uniformly to f. Now let q_n be the polynomial derived from p_n by squaring each term so that all the degrees are even. For x in [0,1], q_n(sqrt(x)) = p_n(x), so {q_n(sqrt(x))} converges to f(x). Since f is even, {q_n(sqrt(x))} converges to f(-x). Now it would be nice to show that {q_n} converges to f, but this is not the case. If anything, {q_n} converges to f(x^2). How do I proceed from here?
     
  2. jcsd
  3. Feb 12, 2009 #2

    Office_Shredder

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    If qn is converging to f(x2) maybe you want to try approximating f(sqrt(|x|)) instead?
     
  4. Feb 12, 2009 #3
    But what comes after that? I'm interested in approximating, f(x), not f(sqrt(|x|)). This is the part I can't figure out.
     
  5. Feb 12, 2009 #4

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    If f(x) is even and continuous, then it's equal to F(x^2) where F is continuous on [0,1] (sure F(x)=f(sqrt(|x|))). Approximate F(x) with a polynomial p(x) on [0,1]. Then p(x^2) approximates f(x), right?
     
  6. Feb 12, 2009 #5
    Ah, OK. My error was that I started approximating f(x) first. Silly me.
     
  7. Feb 13, 2009 #6
    Now suppose we change A so that it is the set of all polynomials in F whose terms all have odd degree. Show that the closure of A is the set of odd function in F.

    Again, (ii) is easy to show, but (i) is not. Let f in F be an odd function. We need to find two continuous functions g and h such that g(h(x)) = f(x). By the Weierstrass Approx. Theorem, there is a sequence of polynomials {p_n} that converge uniformly to g. We will need for p_n(h(x)) to converge to g(h(x)) = f(x), such that p_n(h(x)) is a polynomial with odd degree terms. It is not clear to me how we can make p_n(h(x)) into a polynomial with odd degree terms. For example, how do we even get rid of the constant term in p_n(h(x))?
     
  8. Feb 13, 2009 #7

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    If p(x) approximates f(x) on [-1,1], then p(-x) also approximates f(-x). Can you think of a way to combine the two approximations to get another approximation with only odd powers?
     
  9. Feb 13, 2009 #8
    Smart. p(x) - p(-x) approximates f(x) - f(-x) = f(x) + f(x) = 2f(x). p(x) - p(-x) has odd degree terms. And so 1/2[p(x) - p(-x)] is a polynomial in A that approximates f(x).

    What made you think about this?
     
  10. Feb 13, 2009 #9

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Hard to say. But if g(x) is any function (g(x)-g(-x))/2 is odd. If g is already odd it just gives you g back again. So if p is 'almost' g. Then (p(x)-p(-x))/2 must be 'almost' g. You could have done the even problem the same way.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook