**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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