If P(x) denotes a polynomial of degree n such that

for k = 0, 1, 2, ... n, determine P(n + 1).

Solution

Every polynomial can be written in Newton's form

The coefficients a_{0}, a_{1}, ... are the so called *Newton's divided differences* that have a well developed theory and find numerous application in both theory and practice of interpolation. However, for this particular problem, the theory is not needed as the values of the coefficients are easily guessed following a few examples. The formula shows that

for some coefficient a_{n}, Obviously, P_{0}(x)\equiv{0} . Further, P_{1}(x) = \frac{x}{2}.

To find P_{2}(x) = \frac{x}{2} + a_{2}x(x-1) we only need to determine the coefficient b. We know that P_{2}(2) = \frac{2}{3} which gives

which gives a_{2} = -\frac{1}{6}.

Since P_{3}(x) = P_{2}{x} + a_{3}x(x-1)(x-2), we again need only to find one coefficient, a_{3}. This is found from P_{3}(3) = \frac{3}{4}:

which gives a_{3} = \frac{1}{24} = \frac{1}{4!}. We may now guess that the generic coefficient a_{n} = \frac{(-1)^{n+1}}{(n+1)!} so that

The convenience of Newton's formula is apparent from the following property: For all k \lt n, P_{n}(k) = P_{n-1}(k).

More explicitly these polynomials are written as

Next we verify that P_{n}(n) = \frac{n}{n+1}. Multiplying and dividing by n+1 does a magic

If we recollect the binomial formula and the fact that (1-1)^{n+1} = 0 , we obtain

Just as stipulated in the problem. On the other hand, multiplying and dividing by n+2,

So that