[math-fun] Sec@n/n →0??
Abs exceeds 9 for four of the first thousand convergents of π/2: [62]:= Select[Sec[#]/# & /@ Numerator@Convergents[π/2, 999], Abs[#] > 9 &] //tim During evaluation of In[62]:= 0.213593,4 Out[62]= {Sec[11]/11, Sec[214112296674652]/214112296674652, Sec[1428599129020608582548671]/1428599129020608582548671, Sec[63008132762960627316194351129]/63008132762960627316194351129} And no others among the next hundred thousand! In[63]:= Select[Sec[#]/# & /@ Numerator@Convergents[π/2, 99999], Abs[#] > 9 &] //tim During evaluation of In[63]:= 25.228223,4 Out[63]= {Sec[11]/11, Sec[214112296674652]/214112296674652, Sec[1428599129020608582548671]/1428599129020608582548671, Sec[63008132762960627316194351129]/63008132762960627316194351129} !?! In[64]:= N[N[%, 33], 9] Out[64]= {20.5411872, 18.0078002, 11.5073819, 16.8247791} And none negative. Wouldn't Gauss's Log[2,1+tail] probability predict a fairly steady, non-tapering stream of qualifiers? --rwg On Sat, Jun 2, 2018 at 12:00 AM, Bill Gosper <billgosper@gmail.com> wrote:
Out[42]= Sec[18190586279576483]/18190586279576483
In[43]:= {N@%, N[%, 22]}
Out[43]= {-6.53302256185329*10^-17, 2.154106494388454341277} --rwg Does |Sec@n/n| (integer n) have a supremum, or even a simple max? Hopelessly difficult: The minimal p for which Sec@n/n^p has a limit.
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Bill Gosper