actually, your proof is fine. a polynomial of degree n, cannot have more than n roots.
so you have proved that V(x) is NOT of degree 0,1, or 2. what's left?
This problem is from my Gelfand's Algebra book.
Problem 164. Prove that a polynomial of degree not exceeding 2 is defined uniquely by three of its values.
This means that if and are polynomials of degree not exceeding 2 and for three different numbers then the polynomials and are equal.
I'm not very good at proofs, so I have questions. If I show that if is true and and are polynomials of degree not exceeding 2, then and are equal, will it prove that a polynomial of degree not exceeding 2 is defined uniquely by three of its values?
Is this the way to go?
Let , After that , because , so are roots of which can't have more than two roots. Here I'm confused, I shouldn't have assumed that was true, right?
Yes, but does that prove that polynomial of degree not exceeding 2 is defined uniquely by three of its values?
Yes, I think that's right, I'm assuming too much and I don't really understand how I can prove that polynomial of degree not exceeding 2 is defined uniquely by three of its values.
well this is only true in a field, of course, but presumably we are talking about polynomials in R[x].
and one can argue by degree:
proof(by induction): if p(x) in R[x] is of degree n > 0 then p has at most n real roots.
(the reason for requiring n > 0 will be explained later).
base case: n = 1.
in this case p(x) = ax + b, which has the sole root x = -b/a (we may assume a ≠ 0, or else p is not of degree 1). since exactly one root is a stronger condition that at most one root, the theorem holds in this case.
(strong) induction hypothesis: suppose whenever 0 < k < n, deg(p) = k implies p has at most k roots.
let deg(p) = n.
it could be that p has no roots at all. since 0 < n, the theorem holds in this case.
otherwise, let a be a root of p.
writing p(x) = q(x)(x - a) + r(x), where either r(x) is identically 0, or deg(r) < deg(x-a) = 1, we see that r is a constant polynomial, or the 0-polynomial.
so p(x) = q(x)(x - a) + r, for some real number r. thus:
0 = p(a) = q(a)(a - a) + r = q(a)0 + r = r.
hence p(x) = q(x)(x - a).
since deg(p) = deg(q) + deg(x-a), we have:
n = deg(q) + 1, that is:
deg(q) = n-1 < n, so q has at most n-1 roots.
lemma: if p(x) = f(x)g(x) for real polynomials p,f, and g, then if a is a root of p, either a is a root of f, or a is a root of g.
proof: 0 = p(a) = f(a)g(a), so either f(a) = 0, or g(a) = 0.
main proof continued:
let b be any root of p(x) = q(x)(x - a).
then either b is a root of q(x) (of which we have at most n - 1), or b is a root of x - a (which has the single root a).
thus we have at most n - 1 + 1 = n roots of p, QED.
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to answer PaulAdreinMauriceDirac's question:
you have proved that if P = Q at 3 points (and deg(P), deg(Q) < 3), P = Q at every point.
the only thing remaining to show is that given any 3 points, there is a polynomial of degree < 3 that goes through those 3 points. can you do this?
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OOPS! i almost forgot my explanation for why we take n > 0:
a polynomial of degree 0 is a constant polynomial:
p(x) = c.
unless c = 0, p has no roots at all (which is ok according to our theorem).
but if p(x) = 0, for all x, we have infinitely many roots (very bad).
on the other hand:
p(x) = 0 = 0 + 0x = 0 + 0x + 0x^{2}, etc.
is sort of a SPECIAL polynomial, it really doesn't "have" a degree. it is this special case we seek to avoid by stating deg(p) > 0. depending on which book you read you get:
deg(0) = undefined, or sometimes:
deg(0) = -∞ (this is to make the formula deg(fg) = deg(f) + deg(g) always work out).
Because you are specifically asked about "polynomials of degree not greater than 2", I don't think you need to be very "sophisticated"! You can use the fact that any such polynomial can be written in the form . So suppose , , and and that takes on those same values. That gives three equations you can solve to show that , , and .