Find the number of polynomials, f(x) with integer coefficients such that f(7) =11; f(11)=13
No. I don't know how to do this.
@hartnn
@UnkleRhaukus @.Sam.
i wouldn't know where to begin
What kind of interpolation is this? :-O
Wow. http://math.stackexchange.com/questions/375161/finding-the-number-of-polynomial-when-fa-fb-is-given
I was just browsing through the active questions and I got this.
@Hoa actually, the asker asked the same question there as well. :-)
Hi friend, i don't know about your stuff, so I asked my prof about that . I've just received his message, I copy it for you in case you still need. this is my prof's answer A polynomial with integer coefficients of degree d would be a sum a(d)x^d + a(d-1)x^(d-1) + ... + a(0) where a(0),a(1),...,a(d) are integers. To have values 11 at x = 7 and 13 at x = 11 the coefficients would have to solve the two linear equations with coefficients the powers of 7 (for one equation) and 11 (for the other) with right-hand sides 11 and 13 (7^d)*a(d) + (7^(d-1))* a(d-1) + ... + 7a(1) + a(0) = 11 (11^d) * a(d) + (11^(d-1))*a(d-1) + ... + 11a(1) + a(0) = 13 the solutions are required to be integers, so you could consider this a question about linear diophantine equations. Note there are no integer solutions for d = 1 (because the unique solution is not integral). SO if the question were "how many linear polynomials with integer coefficients... " the answer would be 0. hope this makes sense to you. I read it as if japanese language :)
One more thing, If you want to ask any question and want to ask directly my prof, I'll give you his email. he is a Ph.D and very generous.
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