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Looking for P(x)

  1. Aug 13, 2014 #1
    1. The problem statement, all variables and given/known data
    Find all polynomials P(x), such that P(x² - 2x) = [P(x-2)]²

    2. Relevant equations
    Polynomial equations

    3. The attempt at a solution
    Let x = y+2, then
    P(y²+2y) = P²(y)

    I tried to analyse by derivating
    P'(y²+2y)(2y+2) = 2P(y)P'(y)
    P'(y²+2y)(y+1) = P'(y)P(y)
    I am not sure if I derivated correctly, though

    Yet, that's it. Any ideas?
     
  2. jcsd
  3. Aug 14, 2014 #2
    I would consider what ##\deg\big( P(x)\big)## says about ##\deg\big(P'(y²+2y)(y+1)\big)## and ##\deg\big( P'(y)P(y)\big)##. I'm not 100% certain, but I'm guessing you will get a good constraint on ##\deg\big( P(x)\big)## which will make it easier to just brute force the coefficients.

    Also, and this is just a pet peeve of mine, the appropriate verb and its tenses are differentiate, differentiating, and differentiated.
     
  4. Aug 14, 2014 #3
    Good call, let deg(P(x)) = n, then
    2(n-1) + 1 = n(n-1)
    n² - 3n +1 = 0??
    Did I differentiated correctly?
    By the way, not peeve at all, my english is quite rusty and I miss sometimes some english technical terms, thanks for the correction xD
     
    Last edited: Aug 14, 2014
  5. Aug 15, 2014 #4

    haruspex

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    Think again about the right hand side of that equation. (I think the corrected form will be tautology.)

    Try some other lines:
    - find a value of x for which P can be evaluated (or, at least, for which you can limit the possible values to a small set).
    - suppose P has a root, α. What other roots are implied?
    - what can you say about a local minimum of P?
     
    Last edited: Aug 15, 2014
  6. Aug 15, 2014 #5

    haruspex

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    ... or consider P[x] = Q[x+c]. See if you can find a value of c which yields a simpler equation.
     
  7. Aug 16, 2014 #6
    Why P(x) = Q(x+c)? Did you mean, if the function were periodic of period c, like P(x) = P(x + c)?

    By the way, Could you say if I differentiated correctly at post #3? Shouldn't I get an integer n?
    Thanks
     
  8. Aug 16, 2014 #7
    The answer in my cousin's book seems to be

    [itex](x+1)^{n}[/itex]

    I really thought the answer would be a monster Oo
     
  9. Aug 16, 2014 #8

    haruspex

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    No, not periodic:
    P(y²+2y) = P²(y)
    Q(y²+2y+c) = Q²(y+c)

    Can you spot a value of c that makes that interesting?
    You differentiated ok, but wrongly evaluated the highest power that would arise on the right hand side. If you do it correctly, you'll find that the equation doesn't tell you anything.
     
  10. Aug 16, 2014 #9
    Oh, I see now, thanks :redface:
     
  11. Aug 17, 2014 #10

    haruspex

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    Did you find the value of c in my Q()? I think I can help you find how to solve the problem.
     
  12. Aug 17, 2014 #11
    Let me retry the problem
     
  13. Aug 17, 2014 #12
    c = 1, so that Q(y²+2y+1) = Q[(y+1)²]
    and, therefore,
    Q[(y+1)²]=Q²[(y+1)]
    Great, haruspex!
    Now, Let y+1 = z
    Q(z²)=Q²(z)
    Now, I got stucked, how can I really prove Q(z) = (z+1)^n from it? I can guess the answer, but I couldn't really demonstrate it
     
  14. Aug 18, 2014 #13

    haruspex

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    Good progress. Now you're in a better position to try one of my first suggestions. (It might have worked with P(), but it's easier with Q().) Suppose α is a root (not necessarily real). What other roots are implied?
     
  15. Aug 18, 2014 #14

    haruspex

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    Edit: There's an easier way: just consider the lowest two degree terms in Q, ##a_r x^r + a_s x^s##, say. Equate the lowest two terms of Q(x2) with the lowest two of Q2(x).
     
  16. Aug 22, 2014 #15
    Great! Thanks, haruspex!
     
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