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

[canopus]

[SOLVED] A polynome question

(x+1)*P(x)=x*P(x+1) v (x+1)*P(x)=x*P(x-1) ==> P(x)=?

 

[humanino]

I think your polynome must be 0.

1: $$P(0) = 0$$
2: for x>-1 : $$P(x+1) = \frac{x}{x+1}P(x)$$

recurence : P is identically 0 on the natural integers.
I guess your polynome has finite order and infinitely many zeros
q.e.d. ?

 

[HallsofIvy]

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.

 

[humanino]

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

 

[canopus]

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?

 

[Zurtex]

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 $x \neq 0$...