Timeline for Signed factors of harmonic polynomials
Current License: CC BY-SA 3.0
9 events
when toggle format | what | by | license | comment | |
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Nov 22, 2023 at 21:00 | answer | added | Erik Lundberg | timeline score: 2 | |
Jul 8, 2011 at 6:48 | comment | added | Denis Serre | @Noam. This is not a problem. Just use $\Delta(fg)=f\Delta g+g\Delta f+2\nabla f\cdot\nabla g$. Then $\Delta |X|^\alpha$ and $\nabla|X|^\alpha$ are easy by using spherical coordinates. At last $X\cdot\nabla Q=mQ$ is the Euler's identity for homogeneous functions. | |
Jul 8, 2011 at 2:44 | comment | added | Noam D. Elkies | Yes, that's nice, thanks; but still not all that easy — one must somehow come up with the formula for $\Delta |X|^{2k} Q$ (and under the constraints of a Putnam exam!). | |
Jun 29, 2011 at 7:55 | comment | added | Denis Serre | @Noam. Here is a short and easy proof. If $P\neq0$, write $P=|X|^{2k}Q(X)$, with $Q$ not divisible by $|X|^2$. The degree of $Q$ is denoted $m$. Then $$\Delta P=2k(2k+n-2+2m)|X|^{2(k-1)}Q+|X|^{2k}\Delta Q.$$ Since $\Delta P=0$, this shows that $|X|^2$ divides $Q$, a contradiction. | |
Jun 28, 2011 at 17:46 | comment | added | Noam D. Elkies | Veterans of the 2005 Putnam exam may dispute your assessment of the difficulty of the result that 0 is the only harmonic polynomial divisible by $X_1^2 + \cdots + X_n^2$... This was that year's Problem B-5, and was the hardest on the exam, solved by only five of the top 200 scorers. See the Monthly article on that Putnam exam by Klosinski, Alexanderson, and Larson (Oct.2006 = Vol.**113** #8, pages 733-743). | |
Jun 28, 2011 at 9:30 | answer | added | Denis Serre | timeline score: 7 | |
Jun 12, 2011 at 10:28 | answer | added | Vladimir Dotsenko | timeline score: 5 | |
Jun 10, 2011 at 9:17 | history | edited | Denis Serre | CC BY-SA 3.0 |
added 18 characters in body
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Jun 10, 2011 at 8:01 | history | asked | Denis Serre | CC BY-SA 3.0 |