Math brain teaser
12-29-2013, 01:46 AM
Post: #21
 Les_Koller Junior Member Posts: 22 Joined: Dec 2013
RE: Math brain teaser
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.

I'm a math teacher. Of course I have problems.
12-29-2013, 01:50 AM
Post: #22
 Les_Koller Junior Member Posts: 22 Joined: Dec 2013
RE: Math brain teaser
@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.

I'm a math teacher. Of course I have problems.
12-29-2013, 01:54 AM
Post: #23
 Paul Dale Senior Member Posts: 1,839 Joined: Dec 2013
RE: Math brain teaser
Better would be to limit the guesses to integral values.

- Pauli
12-29-2013, 01:58 AM
Post: #24
 Les_Koller Junior Member Posts: 22 Joined: Dec 2013
RE: Math brain teaser
It makes for a more interesting answer, yes, but look at all the work you save using a transcendental! LOL

I'm a math teacher. Of course I have problems.
12-29-2013, 08:33 AM
Post: #25
 Thomas Klemm Senior Member Posts: 2,142 Joined: Dec 2013
RE: Math brain teaser
(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, 11:25 AM
Post: #26
 Thomas Klemm Senior Member Posts: 2,142 Joined: Dec 2013
RE: Math brain teaser
(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.
12-29-2013, 05:52 PM
Post: #27
 Les_Koller Junior Member Posts: 22 Joined: Dec 2013
RE: Math brain teaser
@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...

I'm a math teacher. Of course I have problems.
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