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POLYA001: Degree n polynomial P_n(x) for which   x^(n+1) P_n(x) - 1 is divisible by (x-1)^(n+1) (proposed by Paul Yiu, 11/1/01).

Find the polynomial P_n(x) of degree n such that x^(n+1) P_n(x) - 1  is divisible by (x-1)^(n+1).

Discussion. [PY, 11/7/01]:  I saw the question for n=2 in Honsberger's book ``In Polya's Footstep'', and naturally asked this more general question. I observed some sort of pattern but did not try hard to find one. But I expect there is a good way of handling the problem using calculus rather than undetermined coefficients.

[JPE, 11/7/01]: Putting x = 1+t, there exists a polynomial q such that (1+t)^(n+1) p(1+t)-1 = t^(n+1)q(t). Hence, (1+t)^(-n-1) = p(1+t) + O(t^(n+1)), and

 p(1+t) = sum((-1)^k C(n+k,k) t^k, k = 0..n).

[PY, 11/8/01]: Is it possible to rewrite this as a polynomial in t, with coefficients given in simple closed forms?

[SCL, 11/9/01]: Here is an approximate answer  for simplification of  the coefficients of the polynomial (except possibly for the smallest case(s)).

P(n,x)= sum (  (-1)^j * C(2*n+2,n-j) * C(n+j+1,j) * (n+j+2) / (n+j+1) /2 * x^j , j=0..n)

At this stage I am not providing a proof. This is from empirical results, using the pretty formula that was already displayed by JPE. My technique is simple-minded -- look at the factors of the coefficients. Since the coefficients factor nicely, they might be products of factorials and powers.That seems to be the case here, and then it is a simple matter of aligning the factorials and powers to match the numbers.

Of course, even such a simple-minded technique is used by some famous people: Tutte would demonstrate this as a first step in getting hold of the coefficients of generating functions when it seemed too hard to use analytic methods.

[JPE, 11/9/01]: Here is a possible way to find a closed form for your polynomial. (x-1)^n divides d/dx[x^(n+1)p(x)-1]; hence, using Gauss theorem, there exists a constant K such that (n+1)p(x) + xp'(x) = K(x-1)^n. Looking at the coefficient of x^n, we get K = (-1)^n (2n+1) C(2n, n). Now, if p = sum(a[k] x^k, k=0..n), an identification gives a[k] = (-1)^k (2n+1)/(n+k+1) C(2n,n)C(n,k).

[JPE, 11/9/01]: It seems that your polynomial is closely related to less or more classical ones, called Bezout polynomials, which are the polynomials
of the Bezout identity for (x-a)^p and (x-b)^q, but I have no reference about these polynomials. In any case, using exactly the same method, it is not difficult to find the expression of these "Bezout polynomials".


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