Given n+1 (x,y) data pairs, with all x values being unique, then a polynomial of order n or less passes thru the (n+1) data points. How can we prove that this polynomial is unique?
I am going to show you the proof for a particular case and you can extend it to polynomials of any order n.
Lets suppose you are given three data points (x1,y1), (x2,y2), (x3,y3) where x1 \ne x2 \ne x3.
Then if a polynomial P(x) of order 2 or less passes thru the three data points, we want to show that P(x) is unique.
We will prove this by contradiction.
Let there be another polynomial Q(x) of order 2 or less that goes thru the three data points. Then R(x)=P(x)-Q(x) is another polynomial of order 2 or less. But the value of P(x) and Q(x) is same at the three x-values of the data points x1, x2, x3. Hence R(x) has three zeros, at x=x1, x2 and x3.
But a second order polynomial only has two zeros; the only case where a second order polynomial can have three zeros is if R(x) is identically equal to zero, and hence have infinite zeros. Since R(x)=P(x)-Q(x), and R(x) \equiv 0, then P(x) \equiv Q(x). End of proof.
But how do you know that a second order polynomial with three zeros is identically zero.
R(x) is of the form a0+a1*x+a2*x^2 and has three zeros, x1, x2, x3. Then it needs to satisfy the following three equations
a0+a1*x1+a2*x1^2=0
a0+a1*x2+a2*x2^2=0
a0+a1*x3+a2*x3^2=0
The above equations have the trivial solution a0=a1=a2=0 as the only solution if
det(1 x1 x1^2; 1 x2 x2^2; 1 x3 x3^2) \ne 0.
That is in fact the case as
det(1 x1 x1^2; 1 x2 x2^2; 1 x3 x3^2) = (x1-x2)*(x2-x3)*(x3-x1),
and since x1 \ne x2 \ne x3, the
det(1 x1 x1^2; 1 x2 x2^2; 1 x3 x3^2) \ne 0
So the only solution is a0=a1=a2=0 making R(x) \equiv 0
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