What is the general form of a polynomial p(x) such that that p(x^2)=p(x)p(x+2)
no constant.....
Not constant.
i mean it doesnt have a constant
x^2+x is a polynomial without a constant, but does not satisfy this problem.
Even simpler, x is a polynomial without a constant, but does not satisfy this problem.
It could have a constant. I need the general form of the solution.
@nbouscal find one that satisfies the condition.
I am attempting to, but this is a rather tricky problem.
Or at least, it seems to be.
The zero polynomial is obviously a trivial solution. I have not yet found a non-trivial solution.
it should have either all even powers or all odd powers
@eliassaab, are you sure there are non-trivial solutions?
x^2-x?
naah i missed it....shd be near that is wrong all even and odd thing may not be true sry
let x=0 then p(0)=p(0)p(2) then p(0)=0 or p(2)=1 if P(0)=0 then let x=-2 it gives p(4)=p(-2)p(0)=0 P(16)=p(4)Pp(6)=0 .... so p(0)=/=0 and p(2)=1
Okay, I'm dumb. I was doing the problem wrong. p(x)=x-1 is a non-trivial solution. p(x^2) there is x^2-1 and p(x)p(x+2) is (x-1)(x+1)=x^2-1
lol yea h i got it to :P
the trick is aw we go on for higher powers things will not be cancelled coz of lame constants and powers should be consecutive so
It's aggravating to me that p(x)=x-1 is the first polynomial that I tried, and because I computed it wrong I threw it out.
well that was a good qn :)
Still is a good question. I still do not know what the general form of the solution is.
Yup. \(p(x)=x-1\) is the only solution, I am pretty sure.
p(x)=x-1 if the polynomial is of the form x+a with a constant. That much I am certain of. It being the only solution I have not yet been convinced of.
I do wish that wolf would show steps on that. It's hard to trust wolf because it is so often wrong =/
Avinash's reasoning convinced me. It's vague.
edit: It's vague, but his point makes sense.
Even if p(x)=x-1 is the only solution, I would like to be able to prove that formally.
I can formally prove that it is the only solution of the form x+a, but past that, I do not know.
I think a proof would involve polynomial factorization.
only answer is \[\huge P(x)=(x-1)^n\]
Confirmed that \((x-1)^2\) is a solution. Nice work, mukushla.
\[ \begin{align} p(x^2)&=(x^2-1)^n\\ p(x)p(x+2)&=(x-1)^n(x+1)^n\\ &=((x-1)(x+1))^n\\ &=(x^2-1)^n \end{align} \]
so, mukushla's answer is correct for all \(n\).
I don't see a proof that this is the only solution, however.
i proved that p(0)=/=0 now Suppose p(a) = 0 for some complex number a. Then p(a^2) = p(a^4) =· · · = 0. then |a| = 1. Similarly, if a is any such root, then P((a −2)^2) = P(a − 2) P(a) = 0. So (a − 2)^2 is also a root. It follows that |a − 2| = 1. there is only 1 complex numbers such that |a| = 1 and |a −2 | = 1 and thats a=1 so the only root of P(x) is x=1 then we have \[P(x)=a(x-1)^n\] put this in the equation that gives a^2=a ---> a=1 because leading coefficient of 2 sides must be equal so u have \[\large P(x)=(x-1)^n\] for any positive integer n
*** Suppose p(a) = 0 for some complex number a. Then p(a^2) = p(a)p(a+2) =0 p(a^4) = p(a^2)p(a^2+2) =0 ............. . However, a polynomial can only have a finite number of roots. So a must be a root of unity. Therefore, |a| = 1
Similarly, we can prove that |a + 2| = 1
am i right guys?
That all looks right to me, mukushla.
Quite right.
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