Math brain teaser

12292013, 01:46 AM
Post: #21




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. 

12292013, 01:50 AM
Post: #22




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. 

12292013, 01:54 AM
Post: #23




RE: Math brain teaser
Better would be to limit the guesses to integral values.
 Pauli 

12292013, 01:58 AM
Post: #24




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. 

12292013, 08:33 AM
Post: #25




RE: Math brain teaser
(12272013 05:18 AM)Han Wrote: A computer generates a random polynomial whose coefficients are nonnegative integers. (12292013 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 nonnegative 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 

12292013, 11:25 AM
Post: #26




RE: Math brain teaser  
12292013, 05:52 PM
Post: #27




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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