1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Norms and orthogonal Polynomials

  1. Sep 15, 2011 #1
    1. The problem statement, all variables and given/known data
    Thanks very much for reading.
    I actually have two problems, I hope it's ok to state both of the in the same thread.

    1. Let Vn be the space of all functions having the n'th derivitve in the point x0.
    I've been given the semi-norm (holds all the norm axioms other than ||v|| = 0 => v = 0) defined by:
    ||f|| = [itex]\sum^{n}_{k=0}\frac{1}{k!}|f^{(k)}(x_{0})|[/itex]

    I need to show that [itex]||fg|| \leq ||f|| ||g||[/itex]

    2. I need to prove that if the set of polynomials {Pi}, i=0,...,n is orthonormal on [a,b] in respect to the inner product defined by: [itex]<f,g> = \int ^{b}_{a}f(x)g(x)dx[/itex]
    then they hold the recurrance relation:
    Pk+1(x) = (akx + bk)Pk(x) + ckPk-1(x)

    3. The attempt at a solution

    1. This one's been a nightmare for me. I've tried proving it straightforward, and then using induction, but I just get lost in a sea of indexes. I'm thinking there might be a more elegant way, but can't see it.
    I'm of course trying to use the triangle inequality, which works great for the cases n=1 and n=2, but I just cannot generlize it.
    I could write the whole development I've done here, it's like one whole page, but it'll take me ages and therefore think it's rather unnescesarry. I'll greatly appreciate a hint. If there's no other way but an ugly induction I think I'll skip it :-)
    (I've used the fact that [itex](fg)^{(k)}(x_{0}) = \sum^{k}_{i=0}(\frac{k!}{i!(k-i)!})f^{(k-i)}g^{(i)}(x_{0})[/itex]

    2. Well, I've been thinking polynomial division. If I assume the set of given Pi's is of ascending degrees (0,1,2...), I can always divide Pk+1(x) in Pk(x) and write:
    [itex]P_{k+1}(x) = (a_{k}x + b_{k})P_{k}(x) + L(x) [/itex]
    Where deg(L(x)) < deg(Pk(x))
    I don't know how to take it from here. How do I use the orthogonality? I would greatly appreciate hints :-)
     
  2. jcsd
  3. Sep 15, 2011 #2

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    OK, for (a). You've come to the conclusion already that

    [tex]
    \begin{eqnarray*}
    \|fg\|
    & = & \sum_{k=0}^n{\frac{1}{k!}\left|\sum_{i=0}^k\frac{k!}{i!(k-i)!}f^{(k-i)}(x_0)g^{(i)}(x_0)\right|}\\
    & \leq & \sum_{k=0}^n{\sum_{i=0}^k\frac{1}{i!(k-i)!}|f^{(k-i)}(x_0)g^{(i)}(x_0)|}
    \end{eqnarray*}
    [/tex]

    The thing to do now is to switch those two sums appropriately.

    For (b). The convention for orthogonal polynomials is that [itex]P_k[/itex] always has degree k. So I shall follow that convention here. (I'm pretty sure the result is false if the convention is not followed).

    Anyway. If you pick a suitable [itex]a_k[/itex], then [itex]P_{k+1}-a_kxP_k[/itex] has degree [itex]\leq k[/itex]. So we can write

    [tex]P_{k+1}-a_k xP_k=b_kP_k+c_kP_{k-1}+\sum_{j=1}^{k-2}{d_jP_{k-2}}[/tex]

    Our job is to prove [itex]d_j=0[/itex]. Do this by taking appropriate inner products.
     
  4. Sep 16, 2011 #3
    Thanks, I'll try working on that.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Norms and orthogonal Polynomials
Loading...