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 Binary partition at oo ? tommy1729 Ultimate Fellow Posts: 1,354 Threads: 328 Joined: Feb 2009 10/03/2014, 09:11 PM (This post was last modified: 10/03/2014, 09:14 PM by tommy1729.) Once again I wonder about the binary partition function and Jay's approximation. For both it seems 0 is a fixpoint ( of the recurrence ! ). So it seems intuitive to conjecture that for both functions we have f ( a + oo i ) = 0. This is similar to what happens to the fixpoints of the recursions for tetration , gamma and others. Im not sure about solutions to f(z) = 0. Maybe some initial conditions/parameters matter here. Im aware this is all very informal , but that is the issue here : making things formal. Or maybe Im wrong ? regards tommy1729 jaydfox Long Time Fellow Posts: 440 Threads: 31 Joined: Aug 2007 10/06/2014, 07:17 PM (10/03/2014, 09:11 PM)tommy1729 Wrote: So it seems intuitive to conjecture that for both functions we have f ( a + oo i ) = 0. Actually, now that I've had more time to analyze things, I realize that the function grows at about the same rate in all directions. For example, in the negative direction, there are infinitely many zeroes. However, in between the zeroes, the function is oscillating, with each local minimum or local maximum being about 0.0016185 times the value of the function at the equivalent positive value. For example, there are zeroes at approximately: -1.8822219377154e30 -3.8040468193666e30 In between, there is a local minimum at about -2.6756571755e30. The value at the minimum, compared to the respective positive value: f(-2.6756571755e30) ~= -1.672619088e1396 f(2.6756571755e30) ~= 9.295773339e1398 The ratio is about -0.0017993329 As we go further and further in the negative direction, this ratio comes down a little, bit seems to bottom out at +/- 0.0016185. This latter value can be calculated as follows: $ \frac{\sum_{k=-\infty}^{\infty} \left((-1)^k\, 2^{-k^2/2}\right)}{\sum_{k=-\infty}^{\infty} \left(2^{-k^2/2}\right)}$ This evaluates to approximately (0.004872868560797)/(3.01076739115959), which is approximately 0.001618480582427. I'll show the way I derived those two summations in a later post, but it's basically a consequence of the formulas I showed in this link: http://math.eretrandre.org/tetrationforu...53#pid7453 By the way, if we go off in the imaginary direction, we see a similar pattern, namely that the function grows at a near constant rate, relative to the respective values in the positive direction. Using a similar summation, we can even calculate that the constant in the imaginary direction is approximately 0.168663. ~ Jay Daniel Fox tommy1729 Ultimate Fellow Posts: 1,354 Threads: 328 Joined: Feb 2009 10/07/2014, 07:22 PM (This post was last modified: 10/07/2014, 07:23 PM by tommy1729.) (10/06/2014, 07:17 PM)jaydfox Wrote: $ \frac{\sum_{k=-\infty}^{\infty} \left((-1)^k\, 2^{-k^2/2}\right)}{\sum_{k=-\infty}^{\infty} \left(2^{-k^2/2}\right)}$ This evaluates to approximately (0.004872868560797)/(3.01076739115959), which is approximately 0.001618480582427. In general $\sum_{k=-\infty}^{\infty} \left((-1)^k\, f(-k^2)\right)$ equals $2 \sum_{k=1}^{\infty} \left((-1)^k\, f(-k^2)\right)-f(0)$. You probably already know that , but maybe it simplifies matters ? Are you claiming that there are no zero's off the real line ? Very intresting stuff. I think we are onto a general thing ; No zero's in the upper complex plane Good entire approximations ( fake function , J(x) for the binary p ) ... with all derivatives positive. It all seems connected. Id like to see how you arrived at these things. Thanks. regards tommy1729 « Next Oldest | Next Newest »

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