# A polynome question

1. Aug 20, 2004

### canopus

[SOLVED] A polynome question

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

2. Aug 20, 2004

### 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. ?

3. Aug 20, 2004

### 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.

Last edited by a moderator: Aug 20, 2004
4. Aug 20, 2004

### 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

5. Aug 20, 2004

### canopus

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?

6. Aug 20, 2004

### Zurtex

When you divide through by x it is only valid for $x \neq 0$...