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A polynome question

  1. Aug 20, 2004 #1
    [SOLVED] A polynome question

    (x+1)*P(x)=x*P(x+1) v (x+1)*P(x)=x*P(x-1) ==> P(x)=?
     
  2. jcsd
  3. Aug 20, 2004 #2
    I think your polynome must be 0.

    1: [tex]P(0) = 0[/tex]
    2: for x>-1 : [tex]P(x+1) = \frac{x}{x+1}P(x)[/tex]

    recurence : P is identically 0 on the natural integers.
    I guess your polynome has finite order and infinitely many zeros
    q.e.d. ?
     
  4. Aug 20, 2004 #3

    HallsofIvy

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    Assuming that the "v" mean "and", since the left side of each equation is the same, you have P(x+1)= P(x-1). The only polynomial satisfying that is P(x)= a constant. In that case, the requirement that (x+1)P(x)= xP(x+1) becomes
    (x+1)c= xc so c=0. The only polynomial satisfying (x+1)P(x)= xP(x+1) and
    (x+1)P(x)= xP(x-1) is P(x)= 0.

    "v" more commonly means "or". We could interpret that as meaning one equation is true for some values of x and the other for other values of x or as meaning that there are two questions for two different polynomials. I don't think there is enough information to solve either way.
     
    Last edited: Aug 20, 2004
  5. Aug 20, 2004 #4
    I was concerned about the fact that the "v" should mean "or". But I could not make sens of the assumption with the "or". On the other hand, if it means "and", then there is too much information to solve the problem :confused:
     
  6. Aug 20, 2004 #5
    Actually, i meant ''and''. I found P(0)=0 but when i put the ''0'' instead of ''x'', it seems impossible, because, lets write one of these polynomes, (x+1)*P(x)=x*P(x+1) we can write also like [(x+1)*P(x)]/x=P(x+1) then we put ''0'' instead of ''x'' but the result is found roughly infinite (that is through little). But, i might be wrong, what do you think?
     
  7. Aug 20, 2004 #6

    Zurtex

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    When you divide through by x it is only valid for [itex]x \neq 0[/itex]...
     
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