2009 Putnam Exam, Question B4?

Question

The 2009 Putnam Exam may be found at http://amc.maa.org/a-activities/a7-problems/putnam/-pdf/2009.pdf A presumably reliable answer sheet may be found at http://amc.maa.org/a-activities/a7-problems/putnam/-pdf/2009s.pdf But the answer I come up with differs. 

The question is:

Say that a polynomial with real coefﬁcients in two variables, x, y, is balanced if the average value of the polynomial on each circle centered at the origin is 0. The balanced polynomials of degree at most 2009 form a vector space V over R. Find the dimension of V.

My answer is as follows:

Divide the monomials x^j * y^k of degree <= 2009 into 4 disjoint sets:

S1: j, k both are even.
S2: j is even, k is odd
S3: j is odd, k is even
S4: j, k are both odd

The union of S1, S2, S3, and S4 is obviously a basis for the entire vector space of polynomials in x,y with degree <= 2009.

Now, the members of S1 are all positive at any point that is not the origin, while the members of S2, S3, and S4 all average to 0 on any circle centered at the origin. So the union of S2, S3, and S4 is a basis for V, and dimension of V is just the sum of their cardinalities -- i.e., 3 * 1005^2.

But the answer sheet says the answer is 2 * 1005^2. Where did I go wrong?

♣ K-Dub ♣ · Accepted Answer

I don't think that I have it all here, and this could be completely wrong......

The sets S1, S2, and S3 have cardinality C(1006,2) (the binomial coefficient) by a "stars and bars" argument. For S1 we have

2j + 2k ≤ 2009
2(j + k) ≤ 2008 (as number on LHS is even)
j + k ≤ 1004.

Out of a lineup of 1006 stars, we replace any two with bars, forming three groups. The first group represents half the power on x, and the second half the power on y. 

Adding an x (for S3) or a y (for S2) will still result in a degree at most 2009, so the cardinalities of S2 and S3 are the same as that of S1. However, S4 has a different cardinality:

(2j + 1) + (2k + 1) ≤ 2009
2(j + k) ≤ 2007
2(j + k) ≤ 2006
j + k ≤ 1003.

S4 has a cardinality of C(1005,2).

But 2*C(1006,2) + C(1005,2) = 1,515,540, short of the correct answer.

I believe that there are, in fact, some missing. 

Consider x^{2j} y^{2k}, and the averages over the circles. I imagine that this average would be the same as that of the monomial x^{2k} y^{2j}, by the reflection across the line y=x. Then

x^{2j} y^{2k} - x^{2k} y^{2j} 

would average to zero on each circle (I think), and these cannot be covered by those in S2-S4.

Anyway, hope this helps.

♣ ♣

barff · Answer

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2009 Putnam Exam, Question B4?

2 Answers