Let p(x) be the real polynomial 1 + α_{1}x^{m1} + α_{2}x^{m2} + ... + α_{n}x^{mn}, where 0 < m_{1} < m_{2} < ... < m_{n}. For some real polynomial q(x) we have p(x) = (1 - x)^{n}q(x). Find q(1) in terms of m_{1}, ... , m_{n}.

**Solution**

Answer: m_{1}m_{2}...m_{n}/n!

*Moderately hard, unless you are used to manipulating determinants.*

Notice that n occurs in two places: p(x) has n terms apart from 1, and it also has an n-fold root x = 1. The first key (but elementary) point is the fact that p(x) has a factor (1 - x)^{n} iff p(1) = p'(1) = ... = p^{(n-1)}(1) = 0. This gives us n linear equations for the n variables α_{i}, so the α_{i} are completely determined in terms of the m_{i}. Moreover, differentiating p(x) one more time (n in all) gives p^{(n)}(1) = (-1)^{n} n! q(1). So having solved for the α_{i} we just have to substitute them into p^{(n)}(1) to get the answer. That is all straightforward, the only difficulty of the question is actually carrying out that programme. If you are not familiar with determinants, it can seem sufficiently intimidating to make you give up and try another approach.

It is convenient to have some notation for m_{1}(m_{1} - 1) ... (m_{1} - r + 1). The traditional notation is Pochhammer's symbol (m_{1})_{r}, but Graham et al are pushing m_{1}^{r} (read as m_{1} to the r falling) [see Ronald Graham, Donald Knuth, Oren Patashnik, Concrete Mathematics, 2nd Ed 1994, Addison-Wesley, ISBN 0201558025, especially section 2.6 Finite Calculus], which I also prefer. So, we have:

α_{1} + α_{2} + ... + α_{n} = -1;

m_{1}^{1} α_{1} + m_{2}^{1} α_{2} + ... + m_{n}^{1} α_{n} = 0;

m_{1}^{2} α_{1} + m_{2}^{2} α_{2} + ... + m_{n}^{2} α_{n} = 0;

**. . . **

m_{1}^{n-1} α_{1} + m_{2}^{n-1} α_{2} + ... + m_{n}^{n-1} α_{n} = 0;

The above n equations determine the α_{i}, and we then substitute them into:

m_{1}^{n} α_{1} + m_{2}^{n} α_{2} + ... + m_{n}^{n} α_{n} = (-1)^{n} n! q(1); (*)

Let D be the n x n determinant with first row 1, 1, ... ,1; and ith row m_{1}^{i-1}, m_{2}^{i-1}, ... , m_{n}^{i-1}; etc) for i > 1. Let D_{i} be the determinant derived from D by replacing the ith column by the values on the other side of the n simultaneous equations (-1 in the top row and 0s below). [Sorry, that is hard to typeset in the type of HTML that browsers currently accept]. Then we know that α_{i} = D_{i}/D. (**)

It is also convenient to define E as the determinant derived from D by replacing the first row by m_{1}^{n}, m_{2}^{n}, ... , m_{n}^{n} and F as the determinant derived from D by deleting its first row and adding m_{1}^{n}, m_{2}^{n}, ... , m_{n}^{n} as a new bottom row. Note that E = (-1)^{n-1}F since it is derived from it by n-1 row transpositions.

Having got the solution (**) we just substitute the values into (*). But notice that because the ith column of D_{i} is all zeros except for -1 in the first row, its value is just -D_{1i}, where D_{1i} is the cofactor of the (1, i) element of D [D_{1i} is (-1)^{1+i} times the value of the (n-1) x (n-1) determinant obtained from D by deleting the 1st row and ith column]. So the left hand side of (*) is just: -(m_{1}^{n} D_{11} + m_{2}^{n} D_{12} + ... + m_{n}^{n} D_{1n}) / D. But the expression in parentheses is just the expansion of the determinant E, so (remembering that E = (-1)^{n-1} F) we get that q(1) n! = F/D.

m^{k} is a polynomial in m with highest term m^{k} and no constant term m^{0} (unless i > m, in which case it is zero). So by adding multiples of rows 2, 3, ... , (k - 1) to row k successively for k = 2, 3, ... , n we can convert D into the Vandermonde determinant V which has m_{i}^{k} replaced by m_{i}^{k}. Since such row operations do not change the value of a determinant, we have D = V. Similarly, F = W, the determinant derived from F by replacing m_{i}^{k} by m_{i}^{k}. Now observe that every element in column i of W has a factor m_{i}. We can take out these factors to get that W = m_{1}m_{2} ... m_{n} V. Hence result.

© John Scholes

jscholes@kalva.demon.co.uk

3 Oct 1999