The “Fundamental Theorem of Algebra” states that a polynomial of degree n has n roots. Its first assertion in a different form is attributed to Peter Rothe in 1606 and later Albert Girard in 1629. Euler gave a clear statement of the theorem in a letter to Gauss in 1742 and at different times Gauss gave four different proofs (see  , p. 292-306).
A nearly as important property of a polynomial is the constancy of the nth‑differences of its subsequent values. To clarify this point let us begin with some demonstrations. While it is customary to use polynomials with real coefficients, here a second-order polynomial with complex coefficients is considered first,
where is the imaginary unit. Taking a real starting point and a real step value the following Table 1 of differences can be established for the subsequent values of the polynomial.
The first differences are computed by taking the differences of the subsequent values of the polynomial as in .
Table 1. Second-order differences for a sample second-order polynomial.
The second differences are obtained similarly using the first difference values: .
Expressing the first differences in terms of polynomial values and , the first value of the second differences may be written as
which is a particular form, , of the general theorem presented in Section 2. The constant value of the second-differences can be calculated from the general expression where n is the degree of the polynomial and the coefficient of the nth-order term. For this particular example and hence the constant becomes as found above.
Another example is now given for a third-order polynomial with real coefficients
In this example a complex starting point and a complex step value are used so that Table 2 of differences is constructed, where the constant value of the third-differences can be calculated from the general formula as .
A direct connection with the finite-difference approximation of derivatives of a polynomial is of course possible. Finite-difference approximation of the third-derivative of a third-order polynomial is given by
where s is the incremental step. If the third-order polynomial is defined as its third-derivative is . Now using this in (4) results in
which exactly corresponds to the tabulated constant of third-differences. A remarkable point is that while finite-difference approximations are typically formulated for real and relatively small incremental step sizes, for the general expression no such restrictions apply: the incremental step s may be complex and arbitrarily large while the result is always exact.
Table 2. Third-order differences for a sample third-order polynomial.
Finally, a possible application of (5) or its general form for an nth-order polynomial, is its use as a recurrence formula for evaluating a given polynomial at equal intervals once the polynomial is evaluated at n distinct points. For instance for a third-order polynomial it is sufficient to know , , and to obtain from (5). Then, by setting to in (5), can be obtained from the same recurrence relationship and continuing in this manner gives , , etc. with considerably less arithmetic operations compared to straightforward evaluation of polynomial.
2. Main Theorem and Proof
The main theorem which expresses the constancy of nth-order differences for an nth-order polynomial is stated first and then proven by the method of induction.
For an nth-order polynomial with the following relationship holds
where and 's, , or .
Proof of Theorem 1. The base case: Setting in (6) results in
Substituting and gives
which is correct.
The inductive step: Assuming that the statement (6) holds true for any integer n it is now proven that it also holds true for :
can be expressed in terms of as
Letting in (10) and using it in (9) result in
Since for any n (odd or even) the summation proportional to the constant vanishes, reducing (11) to
Making use of (  , p. 882) in the first summation above results in
By re-defining the running index in the second summation (13) becomes
Since x may be assigned to any value, substituting x + s in place of x in the base case (6) reveals that is too equal to the same quantity: . Noting in the second summation in (14) that renders the terms in square brackets zero. Thus, to complete the proof the remaining equality in the second line of (14) must be proven:
For m = 0 the first term of the summation in (15) is zero hence bringing no contribution. Therefore, we can start the summation from m = 1 without any error. Then, the summation may be expressed as
where an obvious change of running index m = p + 1 has been implemented in the final stage. Employing the last expression obtained above after changing p to m for the summation of (15) results in
As indicated above, the main theorem may also be stated as . Using this in (16) yields
which proves that the proposition holds true for (n + 1) as well. ☐