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Deduce that (P[0,1], norm(inf)) is not complete.

  1. Feb 22, 2012 #1
    1. The problem statement, all variables and given/known data
    Deduce that (P[0,1], norm(inf)) is not complete.

    2. Relevant equations

    3. The attempt at a solution
    I am not really sure how to start this.

    My strongest instinct is to find a polynomial that converges to something that isn't a polynomial (discontinuous, trigonometric, or whatever). I have no tools to find such a polynomial without the 'needle in a haystack' approach.

    Every Cauchy sequence is bounded, so trying to find an unbounded element of P[0,1] is out.

    Any direction to steer this ship in would be appreciated.
     
  2. jcsd
  3. Feb 22, 2012 #2

    jbunniii

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    Your instinct is good. However, you are not trying to find ONE polynomial that converges to something that is not a polynomial. (Not sure what that even means.) You are trying to find a SEQUENCE of polynomials that converges to something other than a polynomial.

    And since you're using the infinity norm, you need the convergence to be uniform, right?

    What kinds of theorems do you know about polynomials? E.g. do you know any spaces that they are dense in? What do you know about Taylor series? etc.
     
  4. Feb 22, 2012 #3
    Agree. My sloppy use of words here.

    Yes.

    I would guess that polynomials are dense in space of continuous functions? (This wasn't covered in book or notes however.)

    I hadn't thought about Taylor series being a polynomial. So Taylor series that converges to exp^t would be an example to use.

    Taylor Expansion of exp^t is 1+t+...+t^n/n!+...

    I could use p(t)=1+t+...+t^n/n!. And guess that it converges to p=Taylor expansion.

    Need to show that inf[p(t)-p]-->0. Then since p is not an element of P[0,1], the space is not complete.
     
  5. Feb 22, 2012 #4

    jbunniii

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    Yes, they are. This is the Stone-Weierstrass theorem. But if you haven't covered that yet, then don't worry about it - you can handle it using Taylor series.


    Each partial sum of a Taylor series has only finitely many terms, and is therefore a polynomial! And the sum of the series is the limit of the sequence of partial sums.

    Right, although I would call it [itex]p_n(t)[/itex] since you're building a sequence.

    [tex]\lim_{n \rightarrow \infty} p_n(t) = p(t) = e^t[/tex]

    By the way, you might want to call the limit something other than [itex]p(t)[/itex] since the notation suggests that it's a polynomial, and the whole point is that you don't want it to be a polynomial. But that's just a nitpick.

    Don't you mean sup, not inf? And writing it more carefully, it would be

    [tex]\sup_{t \in [0,1]} [p_n(t) - p(t)] \rightarrow 0[/tex]

    as [itex]n \rightarrow \infty[/itex]

    i.e. you need to establish that the convergence is uniform. Do you know any theorems about the uniformity of convergence of Taylor series (or, more generally, power series)?

    Correct.

    Good question. They haven't appeared explicitly yet, but having established that [itex]p_n \rightarrow p[/itex] uniformly in the larger space of continuous functions, you immediately know that the sequence [itex]p_n[/itex] is uniformly Cauchy. (If you haven't covered this fact already, you should prove it, but it's an easy proof.)
     
  6. Feb 22, 2012 #5
    A power series converges uniformly to its limit in the interval of convergence?
    ^^^ This comes by Google and not text or notes (again). I am not familiar with an 'interval of convergence' though. Is it referring to [0,1] in this case?
     
  7. Feb 22, 2012 #6

    jbunniii

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    Yes, this is correct. Every power series has an interval of convergence, of the form (-R,R), such that the series converges for values of t inside the interval, and diverges for values of t outside the interval. Convergence is uniform on any closed interval of the form [-A,A] where A < R. In the case of the Taylor series for e^t, the interval of convergence is actually (-infinity,infinity), and the series converges uniformly on any bounded closed interval.

    If you don't have this theorem to work with, you can prove it directly for your specific series. Do you know any theorems about uniform convergence? How about the Weierstrass M test?
     
  8. Feb 22, 2012 #7

    jbunniii

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    P.S. If you don't know any theorems about uniform convergence, what do you know about the remainder term of a Taylor series?
     
  9. Feb 23, 2012 #8
    The remainder term of a Taylor series would go to zero as the number of terms approaches infinity?
     
  10. Feb 23, 2012 #9

    jbunniii

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    It does, but this only proves that the series converges pointwise. Do you know a formula or bound for the remainder term which you could use to show that the convergence is uniform on your interval?
     
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