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It is easy to prove that it is impossible.
Given any set of n input and output values, of x and f(x) values, it is easy to prove that there is at least one polynomial of degree n-1 which will fit them. There are, therefore, infinitely many polynomials that will fit these n pairs and any additional pair of the infinitely many choices for the "next" x and f(x).
It is easy to prove that it is impossible.
Given any set of n input and output values, of x and f(x) values, it is easy to prove that there is at least one polynomial of degree n-1 which will fit them. There are, therefore, infinitely many polynomials that will fit these n pairs and any additional pair of the infinitely many choices for the "next" x and f(x).
It is easy to prove that it is impossible.
Given any set of n input and output values, of x and f(x) values, it is easy to prove that there is at least one polynomial of degree n-1 which will fit them. There are, therefore, infinitely many polynomials that will fit these n pairs and any additional pair of the infinitely many choices for the "next" x and f(x).
It is easy to prove that it is impossible.
Given any set of n input and output values, of x and f(x) values, it is easy to prove that there is at least one polynomial of degree n-1 which will fit them. There are, therefore, infinitely many polynomials that will fit these n pairs and any additional pair of the infinitely many choices for the "next" x and f(x).
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TaigaIt is easy to prove that it is impossible.
Given any set of n input and output values, of x and f(x) values, it is easy to prove that there is at least one polynomial of degree n-1 which will fit them. There are, therefore, infinitely many polynomials that will fit these n pairs and any additional pair of the infinitely many choices for the "next" x and f(x).
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