Fw: why the mantissas used below are all roots of powers of 10

*To*: mathgroup at smc.vnet.net*Subject*: [mg131887] Fw: why the mantissas used below are all roots of powers of 10*From*: Marvin Burns <marvin at marvinrayburns.com>*Date*: Thu, 24 Oct 2013 23:48:10 -0400 (EDT)*Delivered-to*: l-mathgroup@mail-archive0.wolfram.com*Delivered-to*: l-mathgroup@wolfram.com*Delivered-to*: mathgroup-outx@smc.vnet.net*Delivered-to*: mathgroup-newsendx@smc.vnet.net*References*: <1382136837.63494.YahooMailNeo@web2801.biz.mail.ne1.yahoo.com>*Reply-to*: Marvin Burns <marvin at marvinrayburns.com>

We get nearly identical answers without the sine function Try Table[{x, (10 MantissaExponent[N[10^(-100 - 1/x)], 10][[1]])^x}, {x,1,10}] // TableForm replacing the 1/x with 2/x,3/x and 4/x, etc. I think that can be explained! Anyone care to do it? ----- Forwarded Message ----- From: Marvin Burns <marvin at marvinrayburns.com> To: "mathgroup at smc.vnet.net" <mathgroup at smc.vnet.net> Sent: Friday, October 18, 2013 6:53 PM Subject: [mg131887] why the mantissas used below are all roots of powers of 10 I sent this in yesterday but only part of the message was posted. From Marvin Ray Burns to mathgroup We will use the Mathematica code Table[{x, (10 MantissaExponent[N[Sin[10^(-100 - 1/x)], 10]][[1]])^ x}, {x, 1, 10}] // TableForm I was playing around with small sines and posted this on another blog and didn't get much response. However, I would like to know why the mantissas used below are all roots of powers of 10. . There seems to be patterns for sin(10^-k) for rational k; Here we have the "floats." n sin(10^(-n-1/2)) 1 0.03161750640 2 0.003162272390 3 0.0003162277607 4 0.00003162277660 5 0.000003162277660 6 0.0000003162277660 7 0.00000003162277660 Basically the mantissa of 316227766 is being "floated out," because for small x sin(x)~=x. Here we use the mantissa: Noticing that 3.162277660^2~=10 we have below a more subtle and beautiful pattern for sin(10^-k), using sufficiently large integral value for k. Here we use 100 but 9 is usually sufficient. x (10 Mantissa[sin(10^(-100 - 1/x))])^x 1 1.* 10^1 2 1.* 10^1 3 1.* 10^2 4 1.* 10^3 5 1.* 10^4 6 1.* 10^5 etc. x (10 Mantissa[sin(10^(-100 - 2/x))])^x 1 1.* 10^0 2 1.* 10^2 3 1.* 10^1 4 1.* 10^2 5 1.* 10^3 6 1.* 10^4 etc. x (10 Mantissa[sin(10^(-100 - 3/x))])^x 1 1.* 10^1 2 1.* 10^1 3 1.* 10^3 4 1.* 10^1 5 1.* 10^2 6 1.* 10^3 etc. x (10 Mantissa[sin(10^(-100 - 4/x))])^x 1 1.* 10^1 2 1.* 10^0 3 1.* 10^2 4 1.* 10^4 5 1.* 10^1 6 1.* 10^2 7 1.* 10^3 etc. x (10 Mantissa[sin(10^(-100 - 5/x))])^x 1 1.* 10^1 2 1.* 10^1 3 1.* 10^1 4 1.* 10^3 5 1.* 10^4 6 1.* 10^1 7 1.* 10^2 8 1.* 10^3 etc. The Mathematica code for this is Table[{x, (10 MantissaExponent[N[Sin[10^(-100 - 1/x)], 10]][[1]])^ x}, {x, 1, 10}] // TableForm Change 1/x to 2/x,3/x, etc . Can anyone figure out the pattern here? Replacing 1/x with (3/2)/x and ^x to ^(2x) we find x (10 Mantissa[sin(10^(-100 - (3/2)/x))])^(2x) 1 1.* 10^1 2 1.* 10^1 3 1.* 10^3 4 1.* 10^5 5 1.* 10^7 6 1.* 10^9 etc. Replacing 1/x with (5/2)/x and ^x to ^(2x) we find x (10 Mantissa[sin(10^(-100 - (5/2)/x))])^(2x) 1 1.* 10^1 2 1.* 10^3 3 1.* 10^1 4 1.* 10^3 5 1.* 10^5 6 1.* 10^7 etc. Replacing 1/x with (5/3)/x and ^x to ^(3x) we find x (10 Mantissa[sin(10^(-100 - (5/3)/x))])^(3x) 1 1.* 10^1 2 1.* 10^1 3 1.* 10^4 4 1.* 10^7 5 1.* 10^10 6 1.* 10^12 etc.