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First published on Friday, Sep 4, 2026 and last modified on Saturday, Sep 5, 2026
In that test, we will prove the following result about the polynomials.
Theorem 1
There is at most one polynomial function of degree \( n-1\) or less whose graph passes through \( n\) points in the plane with distinct \( x\) -coordinates.
We will start with the example of a quadratic polynomial with 3 zeros, proving that it is the null polynomial.
Namely, we will prove that if a quadratic polynomial function \( P(x)=a_0 + a_1x + a_2x^2\) is zero for \( x = - 1\) , \( x = 0\) , and \( x = 1\) , then \( a_0 = a_1= a_2 = 0\) .
That is, calculate \( P(-1)\) , \( P(0)\) and \( P(1)\) as functions of \( a_0\) , \( a_1\) and \( a_2\) .
And obtain the system to solve in the variables \( a_0\) , \( a_1\) and \( a_2\) .
Find \( a_0\) and add the 1st equation to the 3rd equation.
We may generalize the previous result the following way.
Theorem 2
If a polynomial function \( P(x) = a_0 + a_1x + … + a_{n-1}x^{n-1}\) is zero for \( n\) \( x\) -values or more, then \( a_0 = a_1 = …= a_{n-1} = 0\) .
Consider \( n\) points of the plane \( (x_1,y_1)\) , \( (x_2,y_2)\) ,…,\( (x_n,y_n)\) , with distinct abscissa \( x_i\) .
Consider the polynomials of degree less or equal to \( n-1\) , \( P(x)=a_0+a_1x+…+a_{n-1}x^{n-1}\) and \( Q(x)=b_0+b_1x+…+b_{n-1}x^{n-1}\) .
Assume that the graphs of the polynomial functions \( P(x)\) and \( Q(x)\) both pass through the points \( (x_1,y_1)\) , \( (x_2,y_2)\) ,…,\( (x_n,y_n)\) .
Consider the polynomial of degree less or equal than \( n-1\) , \( D(x)=c_0+c_1x+…+c_{n-1}x^{n-1}\) , where \( c_i=a_i-b_i\) for \( i=0,1,…,n-1\) .
Prove that for any real number \( x\in\mathbb{R}\) , \( D(x)=P(x)-Q(x)\) .
Use the commutativity and associativity of the addition in \( \mathbb{R}\) , as well as the distributivity of the multiplication on the subtraction and of the ’-’ sign on the addition in \( \mathbb{R}\) .
Calculate \( P(x_i)\) , \( Q(x_i\) ) and \( D(x_i)\) for \( i=1,2,…,n-1\) , using the question 2.1 for \( D(x_i)\) .
Prove that \( a_i=b_i\) for \( i=1;2,…,n-1\) , using the facts that \( D(x)\) is of degree \( n-1\) or less and that the \( x_i\) are all distinct.
We have thus proved the theorem 1.
We have discovered the problem of polynomial interpolation of Lagrange. Indeed, if there is at most one polynomial function of degree \( n-1\) or less whose graph passes through \( n\) points \( (x_1,y_1)\) , \( (x_2,y2)\) ,…,\( (x_n,y_n)\) in the plane with distinct \( x\) -coordinates, that polynomial exists and it is called the Lagrange polynomial for that list of points.
The Lagrange polynomial \( L(x)\) is equal to \( L(x)=\Sigma_{1\le j\le n}y_jl_j(x)\) , where
\( l_j(x)=\frac{(x-x_1)…(x-x_{j-1})(x-x{j+1})…(x-x_n)} {(x_j-x_1)…(x_j-x_{j-1})(x_j-x{j+1})…(x_j-x_n)} =\Pi_{1\le i\le n,i\ne j}\frac{x-x_i}{x_j-x_i}\) .
As each \( l_j(x)\) has \( n-1\) factors of degree \( 1\) at the numerator and a constant at the denominator, it is of degree \( n-1\) . So that \( L(x)\) is of degree less or equal to \( n-1\) .
Moreover, for any \( j\) between \( 1\) and \( n\) , \( l_j(x_j)=1\) , then for any \( j\) , \( L(x_j)=y_j\) and thus \( L(x)\) solves the problem of polynomial interpolation of the list of points \( (x_1,y_1)\) , \( (x_2,y2)\) ,…,\( (x_n,y_n)\) .