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What do these new symbols mean ?I can't start this without knowing

  1. Sep 17, 2011 #1
    1. The problem statement, all variables and given/known data


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    3. The attempt at a solution

    What is the [tex]P^n _{k}[/tex] part thing?

    Someone should probably go over the i) for me too...
     
  2. jcsd
  3. Sep 17, 2011 #2

    micromass

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    The [itex]P_n^k[/itex] is just a symbol. It is a name for a polynomial defined by

    [tex]P_n^k(x)=\binom{n}{k}x^k(1-x)^{n-k}[/tex]

    So it's just like we define [itex]P(x)=x^2[/itex]. But now our polynomial depends of n and k.
     
  4. Sep 17, 2011 #3
    For the i)

    [tex]\sum^{n}_{k=0} \left( \alpha f\left(\frac{k}{n}\right) \binom{n}{k} x^k(1-x)^{n-k} + \beta g\left(\frac{k}{n}\right) \binom{n}{k} x^k(1-x)^{n-k}\right) = \alpha \sum^{n}_{k=0} f\left(\frac{k}{n}\right) \binom{n}{k} x^k(1-x)^{n-k} + \beta \sum^{n}_{k=0} g\left(\frac{k}{n}\right) \binom{n}{k} x^k(1-x)^{n-k}[/tex]

    Summation property??
     
  5. Sep 17, 2011 #4
    Something is wrong...do I need induction? This was from proof class
     
  6. Sep 17, 2011 #5

    micromass

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    For (i), you need to start from

    [tex]B_n(\alpha f+\beta g)=\sum_{k=0}^n{(\alpha f+\beta g)(\frac{k}{n})x^k(1-x)^{n-k}}[/tex]
     
  7. Sep 17, 2011 #6
    What's wrong with what I did...? GO on grill me!!
     
  8. Sep 17, 2011 #7

    micromass

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    Nothing is wrong, it's all correct, but it's not finished yet.

    You still need to show that the left-hand-side of your post equals

    [tex]B_n(\alpha f+\beta g)[/tex]

    and that the right-hand-side equals

    [tex]\alpha B_n(f)+\beta B_n(g)[/tex]
     
  9. Sep 17, 2011 #8
    [tex]B_n(\alpha f+\beta g)=\sum_{k=0}^n{(\alpha f+\beta g)(\frac{k}{n})x^k(1-x)^{n-k}} [/tex]

    [tex]= \sum_{k=0}^n{(\alpha f (\frac{k}{n})x^k(1-x)^{n-k}} + \beta g(\frac{k}{n})x^k(1-x)^{n-k} ) = \sum_{k=0}^n{\alpha f (\frac{k}{n})x^k(1-x)^{n-k}} + \sum_{k=0}^n \beta g(\frac{k}{n})x^k(1-x)^{n-k} ) = \alpha B_n(f)+\beta B_n(g)[/tex]

    okay done, got lazy with the long tex
     
  10. Sep 17, 2011 #9

    micromass

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    Yeah, that looks good!!
     
  11. Sep 17, 2011 #10
    How do I start ii) then?
     
  12. Sep 17, 2011 #11

    micromass

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    You need to prove

    [tex]B_n(f)\leq B_n(g)[/tex]

    Start by writing these things out according to the definition of the [itex]B_n[/itex].
     
  13. Sep 17, 2011 #12
    [tex] \sum^{n}_{k=0} f\left(\frac{k}{n}\right) \binom{n}{k} x^k(1-x)^{n-k} \leq \sum^{n}_{k=0} g\left(\frac{k}{n}\right) \binom{n}{k} x^k(1-x)^{n-k}/tex]

    SOme kinda of cancellations...?
     
  14. Sep 17, 2011 #13

    micromass

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    Well, you know that

    [tex]f(k/n)\leq g(k/n)[/tex]

    Now try to introduce the terms needed to conclude that [itex]B_nf\leq B_ng[/itex].
     
  15. Sep 17, 2011 #14
    Can you multiply P to both sides...? I don't know if P is always positive
     
  16. Sep 17, 2011 #15

    micromass

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    That's the idea.

    Try to prove it then. Prove that [itex]P_k^n(x)[/itex] is positive if [itex]x\in [0,1][/itex]...
     
  17. Sep 17, 2011 #16
    So I have to prove two things...

    OKay since [tex]x \in [0, 1] \leq 0[/tex], then it doesn't matter what i put in right? Now how do do that in proper English...?
     
  18. Sep 17, 2011 #17

    micromass

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    This makes no sense to me...
     
  19. Sep 17, 2011 #18
    I meant to say [tex]x \in [0, 1] \geq 0[/tex]

    sorry
     
  20. Sep 17, 2011 #19

    micromass

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    Yeah, that also makes no sense. How can [itex][0,1]\geq 0[/itex]?? [0,1] is a set.
     
  21. Sep 17, 2011 #20
    OKay I wanted to say that numbers in [0,1] are alll positive, so we never had to worry about odd powers messing up with negative numbers
     
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