Proof: Let p(x) be a polynomial of degree at most d, passing through d+1 distinct points, and q(x) is a different polynomial of degree d, passing through these points.

Let

r(x)=p(x)-q(x). [1.1]

r(x) is a non-zero polynomial, because p(x) and q(x) are different. Also, r(x) is a polynomial of degree at most d, because it is the difference between two polynomials of degree at most d.

According to the Fundamental Theorem of Algebra, a polynomial of degree d has exactly d roots, and, at most, d distinct roots.

At each of these d+1 distinct points where the two polynomials, p(x) and q(x), agree:

r(x)=p(x)-q(x)=0 [1.2]

Because r(x) is zero at d+1 distinct points, it has d+1 zeros (or roots). But a polynomial of degree at most d, can have at most d distinct zeros. Therefore, the two polynomials, p(x) and q(x) are not different. These polynomials are, therefore, the same polynomial. This is true because if they are not different, they can only be the same (obvious logical deduction). Associated polynomials, which differ by the multiple of a constant, agree at n points, but not at n+1.

If the polynomials are of degree less than d, then the argument is all the more powerful, because r(x) has even more "impossible" roots!

Whatever, of course, p(x) and q(x) will be of the same degree, because they are identical.

We can also conclude that if two polynomials, p(x) and q(x) agree at d+1 distinct points, they agree everywhere (of course, to state the obvious), because they are the same polynomial!

If these polynomials agree at one place, then their difference is zero, so their coefficients have cancelled out, and one of the a's is equal to one of the b's. For instance:

(for some value) [2.2]

The constant factor could, of course be one (which it will be when the polynomials are identical, and naturally have identical coefficients), or the x value at that place is a root of both equations.

If they agree at more than n different points (n+1 points), then they are exactly the same and are identical polynomials, because the equality at the (n+1)th value cannot be a root and must mean the coefficients are equal. The two polynomials are therefore identical.

Therefore, (obvious conclusion) they have the same values for any x.

If two polynomials of degree at most d agree at d+1 distinct points, then they agree everywhere, because they are the same.

One use of this is to show that if both sides of an equation in
polynomials are equal for d+1 values, then the equation is valid for
all values, generally, real and complex.There are variations...

A polynomial of degree of at most d, has at most d identical values. For instance, y=x

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