- #1
SeReNiTy
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Hey guys, I was doing some Putnam questions for fun and came across something strage to me. In the 2005 Putnam competition, question B3, link provided below:
http://www.unl.edu/amc/a-activities/a7-problems/putnam/-pdf/
I solved it by guessing a polynomial solution and then verifying the co-efficients, but in their solutions they have a more analytical approach where they let x = a/x. Now I'm a bit uncomfortable with this substitution so i let y = a/x, and preceeded to compute their solution. The point i get stuck is after they calculate the second derivative, they somehow eliminate all the a's. Could something explain that step? Its probably trivial and I'm just not putnam material!
http://www.unl.edu/amc/a-activities/a7-problems/putnam/-pdf/
I solved it by guessing a polynomial solution and then verifying the co-efficients, but in their solutions they have a more analytical approach where they let x = a/x. Now I'm a bit uncomfortable with this substitution so i let y = a/x, and preceeded to compute their solution. The point i get stuck is after they calculate the second derivative, they somehow eliminate all the a's. Could something explain that step? Its probably trivial and I'm just not putnam material!
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