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Polynomial bounded w.r.t supremum norm

  1. Jan 25, 2010 #1
    1. The problem statement, all variables and given/known data

    E1 = {pn(t) = nt(1-t)n:n in N};

    E2 = {pn(t) = t + (1/2)t2 +...+(1/n)tn: n in N};

    where N is set of natural numbers

    is the polynomial bounded w.r.t the supremum norm on P[0,1]?

    2. Relevant equations

    supremum norm = ||*|| = sup{|pn(t)|: t in [0,1]}

    3. The attempt at a solution

    I know that the set is bounded on P[0,1] if I can show that the supremum norm is less than some constant for all n. Can someone give me some advice on how to show either polynomial is bounded or not?

    I think E1 is unbounded, no matter your choice of t, since the polynomial's value will forever increase as n in increases. I feel the same can be said of E2 but I know this is wrong.
     
  2. jcsd
  3. Jan 25, 2010 #2
    You can answer this fairly directly, by computing the maximum and minimum values of each [tex]p_n[/tex]; because polynomials are smooth functions, the extreme values of [tex]p_n[/tex] on [tex][0,1][/tex] must occur at a critical point of [tex]p_n[/tex] or at one of the endpoints.

    The members of [tex]E_2[/tex] share a particular property which will save you even this computation.
     
  4. Jan 26, 2010 #3
    I'm not seeing the special property of E2. Its min value must be at t = 0 for all n, and max when t = 1, but I can't pin down a value for sup. It looks like an expanded function but it doesn't look familiar.

    factoring out a t makes p2(t) = t(1 + t + (1/2)t + (1/3)t2 + ... + (1/n)tn-1), but I'm still not seeing the trick.
     
    Last edited: Jan 26, 2010
  5. Jan 26, 2010 #4
    Also, at E1, the critical points are t = 0 and t = 1. At both of these points, no matter the value of N, the polynomial is zero. Does this mean that is my sup? That doesn't make sense to me b/c the shape of this function spikes in the middle between t = 0 and t = 1.
     
  6. Jan 26, 2010 #5
    for E2, p'(t) gives the geometric series, which equals 1/(1-t) on the interval |t|<1. The second derivative is 1/(1-t)2 < 0

    What can I do with this information? The series doesn't converge at either t = 0 or t = 1.
     
  7. Jan 26, 2010 #6
    If the function is zero at 0 and 1, and positive in the middle, then there must be another critical point between, no?

    For [tex]E_2[/tex], put [tex]q_n(t) = t^n/n[/tex], so that [tex]p_n(t) = \sum_{1\leq j\leq n} q_n(t)[/tex]. Where does each [tex]q_n[/tex] reach its maximum absolute value? What does this tell you about the [tex]p_n[/tex]?
     
  8. Jan 26, 2010 #7
    yes, and it'll be a function of n, but as n -> inf, isn't the max value of the function going to diverge to inf as well?


    qn reaches its max value at t = 1 => [tex]p_n(1) = \sum_{1\leq j\leq n}q_n(1) = \sum_{1\leq j\leq n}\frac{1}{n} [/tex]

    so the value of this polynomial is the sum of this series and is finite and therefore this polynomial is bounded.
     
  9. Jan 26, 2010 #8
    That's precisely what the question asks you to determine.

    Bounded? Uniformly in [tex]n[/tex]? Look again.
     
  10. Jan 26, 2010 #9
    I understand that, but I am not confident about my answer.

    you're right, I got mixed up with our indexes on our summation notation. Anyway, the sum of (1/n) to infinity is harmonic sequence which does not converge, and therefore THIS polynomial is not bounded. Am I getting close?
     
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