Determine all polynominals (real coeffitients of course) such that for arbitrary number there is:

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- Sep 1st 2007, 08:54 AMUnunuquantiumPolynominal
Determine all polynominals (real coeffitients of course) such that for arbitrary number there is:

- Sep 2nd 2007, 09:42 AMThePerfectHacker
Here is an partial solution. We shall determine all real polynomial having real zeros that satisfy this.

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).

Thus,

.

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 .

Hence, . - Oct 19th 2007, 10:20 AMUnunuquantium
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.

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Mayby someone has another solution?? - Oct 19th 2007, 10:49 AMThePerfectHacker
The word "monic" means the first coefficient is 1 it does not say anything about the other coefficients.

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).

Quote:

2. I do not understand that have complex zeros when is not equal to .

Quote:

4. Why do you say that it is only a partial solution?????

Quote:

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. - Oct 19th 2007, 11:04 AMUnunuquantium
#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.

??

#4 ok

I just do not understand why you have written:

maybe some are complex?? - Oct 19th 2007, 11:49 AMThePerfectHacker
- Oct 19th 2007, 12:36 PMUnunuquantium
Ok, sorry, I forgot what was the question exactly.

Thank you very much - Oct 19th 2007, 01:26 PMCaptainBlack
Let be the highest order

terms, then:

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.

RonL - Oct 19th 2007, 01:29 PMCaptainBlack