Determine all polynominals (real coeffitients of course) such that for arbitrary number there is:
First implies .
Now say is non-constant so . Where are not necessrily distinct. (Note must be a monic polynomial. This is easily to see. Just expand LHS and RHS to the leading term).
Now here is the result that we need: The equation has complex zeros if and only if .
Look at the LHS at the factors . These has complex zeros if . But if this is the case then the RHS cannot have the same zeros because it has only real solutions. A contradiction! Thus, for all . And hence the only possibility is .
And this works because if we have .
I have some questions to this solution:
1. I cannot see why is monic polynominal - I mean I see that it must have first coefficient equal to 1, but I cannot see why it must have integer coefficients!
2. I do not understand that have complex zeros when is not equal to .
3. I do not understand why you mean that RHS has got only real zeros! - It has when is monic??(why??) For example is monic polynominal - but is does not have real solution. That is why you cannot say that are ALL real zeros of
4. Why do you say that it is only a partial solution - maybe I did not said the is not real? So I am sorry, is REAL.
Mayby someone has another solution??
Note: I made a mistake, I should have been more careful. If the first coefficient is then this equality leads to . Thus, . We omit the solution because we are working with a polynomial of non-zero coefficient. Thus, and hence where is the 3rd-root of unity.
Hence the first coeffcient can be . But once you factor that coefficient you have a monic polynomial (the coefficients are not necessarily integers).
Do you something about complex numbers? The solution to the equation is where is the 3rd-root of unity. Now if these solutions are non-zero and hence complex (provided that ).2. I do not understand that have complex zeros when is not equal to .
We are considering all the polynomials that have only real zeros. The polynomials that have complex zeros are not being considered here.4. Why do you say that it is only a partial solution?????
By what is answered to #4 the polynomial here has only real zeros. So (the RHS) has only real zeros.3. I do not understand why you mean that RHS has got only real zeros! - It has when is monic - but I cannot see it, as I have written above.
So with my minor mistake fixed the solutions should be this time (for polynomials having zero reals) .
EDIT: Ignore what I just said on the bottom. I forgot that you said "polynomials with real coefficients". So my original solution is correct.
#1 - ok.
#2 , my stupid mistake, ignore it, i did not see sth, sorry
#3What if we have - this polynominal has only complex solutions, and: it's coefficients are real, and x is real, and it is monic.
I just do not understand why you have written:
maybe some are complex??
expanding these as far as the two highest order terms on each side:
So by equating exponents of the second heighest order terms we conclude
that a contradiction. So if , and then we must
have , so if is real .
This all assumed that , so we need also to check
if and are solutions, which they are.