Say you choose just pi. If our polynomial is p(x) = x we get pi. If our polynomial is p(x) = pi, then we also get pi, thus not identifying the polynomial completely. Using e (or any other number, it doesn't have to be e) we get e for p(x) = x, and pi for p(x) = pi. That helps us determine between these two polynomials.

@Thomas Klemm, I didn't even notice that the original question asked you to choose ANY two values, so any variable would work too. Good eye. The original question posed on the NCTM websites says we can ask for any two REAL number values to plug in.

Better would be to limit the guesses to integral values.

- Pauli

It makes for a more interesting answer, yes, but look at all the work you save using a transcendental! LOL

(12-27-2013 05:18 AM)Han Wrote: [ -> ]A computer generates a random polynomial whose coefficients are non-negative integers.

(12-29-2013 01:46 AM)Les_Koller Wrote: [ -> ]Say you choose just pi. If our polynomial is p(x) = x we get pi. If our polynomial is p(x) = pi, then we also get pi, thus not identifying the polynomial completely.

That can not be: \(p(x)=\pi\) doesn't have coefficients that are non-negative

integers. \(\pi\) isn't an integer.

OTOH: if \(p(\pi)=q(\pi)\) for some distinct polynomials \(p(x)\) and \(q(x)\) then \(p(\pi)-q(\pi)=0\).

But \(\pi\) can't be the root of a polynomial with integer coefficients since it's transcendent.

Thus you can distinguish these kind of polynomials based on the evaluation at \(\pi\). Or any other transcendent number.

Cheers

Thomas

(12-29-2013 01:54 AM)Paul Dale Wrote: [ -> ]Better would be to limit the guesses to integral values.

You could use \(p(10^{-k})\) as your 2nd guess. That makes reading the first coefficient \(a_n\) easier. Well, it depends on the order you have to tell them.

@Thomas again, yes, of course you are right. For p(x) = pi, the coefficient of the x^0 term is not integral. Foiled again! Thanks for being persistent...