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Mar 11, 2015 at 18:15 vote accept Vladimir
Mar 11, 2015 at 14:55 comment added Fedor Petrov @TheMaskedAvenger my final result is an estimate for logarithm, of course
Mar 11, 2015 at 14:53 comment added The Masked Avenger This strikes me as too small. In particular, one has a factor of 2^k to account for signs when m is the sum of k nonzero squares. Also, one can pad by zeroes to generate O(k^d) different sets of vectors where each vector has one nonzero component. I expect growth more like exp(m).
Mar 11, 2015 at 14:41 comment added Fedor Petrov Yes. They are actually close enough by some standard reasoning (function is not too much oscillating.)
Mar 11, 2015 at 14:29 comment added Vladimir Thanks! So just to be clear - to turn this into a proof we would need to estimate how far the integral is from the sum?
Mar 11, 2015 at 14:29 vote accept Vladimir
Mar 11, 2015 at 14:55
Mar 11, 2015 at 13:57 history answered Fedor Petrov CC BY-SA 3.0