What is the Solution to the Polynomial Equation (x+1)P(x) = xP(x+1)?

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Discussion Overview

The discussion revolves around the polynomial equation (x+1)P(x) = xP(x+1) and its implications. Participants explore potential solutions for P(x), examining the conditions under which the equation holds true, and whether the interpretation of the symbol "v" affects the outcome.

Discussion Character

  • Exploratory
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant suggests that P(x) must be identically zero, citing that P(0) = 0 and a recurrence relation for x > -1 leads to infinitely many zeros.
  • Another participant argues that if "v" means "and," then P(x+1) = P(x-1) must hold, implying P(x) is a constant, which leads to the conclusion that P(x) = 0.
  • Concerns are raised about the interpretation of "v" as "or," suggesting it complicates the problem and may not provide enough information to reach a solution.
  • A participant expresses confusion about substituting x = 0 into the equations, leading to an undefined situation, and questions the validity of dividing by x when x = 0.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the interpretation of "v" or the implications for P(x). Multiple competing views remain regarding the nature of the polynomial and the validity of certain substitutions.

Contextual Notes

There are limitations regarding the assumptions made about the symbol "v" and the conditions under which the polynomial is evaluated, particularly concerning the substitution of x = 0 and the implications of dividing by x.

canopus
[SOLVED] A polynome question

(x+1)*P(x)=x*P(x+1) v (x+1)*P(x)=x*P(x-1) ==> P(x)=?
 
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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. ?
 
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.
 
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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:
 
Actually, i meant ''and''. I found P(0)=0 but when i put the ''0'' instead of ''x'', it seems impossible, because, let's 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?
 
canopus said:
Actually, i meant ''and''. I found P(0)=0 but when i put the ''0'' instead of ''x'', it seems impossible, because, let's 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?
When you divide through by x it is only valid for [itex]x \neq 0[/itex]...
 

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