(10 Mantissa[sin(10^(-100 - r1/x))])^(r2x)
The reason for the argument form .10^[-n-(1/x)] is the beautiful pattern found in sin(10^-n) for positive integer n.
$$ \begin{array}{| c | r |} \hline n& sin(10^{-n}) \\ \hline \\ \hline 1& 9.98334166\cdot10^{-2}\\ \hline 2& 9.99983333416666468\cdot10^{-3}\\ \hline 3& 9.9999983333334166666646825\cdot10^{-4}\\ \hline 4& 9.9999999833333333416666666646825396\cdot10^{-5}\\ \hline 5& 9.99999999983333333333416666666666468253968254\cdot10^{-6}\\ \hline 6& 9.999999999998333333333333416666666666664682539682539100\cdot10^{-7}\\ \hline 7& 9.9999999999999833333333333333416666666666666646825396825396828152\cdot10^{-8}\\ \hline 8& 9.99999999999999983333333333333333416666666666666666468253968253968254243827\cdot10^{-9} \\ \hline \end{array} $$
etc.
I have some difficulty summarizing the following into a clear theory but here are the two parts I have observed and written about thus far.
Even if you can't help me, you might find this interesting. But hopefully someone can help!
First, we have the lesser part, which I can explain. This part should be well known.
Look at the following tables.
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
0.3162277660... is 1/sqrt(10)
n sin(10^(-n-1/3))
1 0.0463992
2 0.00464157
3 0.000464159
4 0.0000464159
5 0.0000464159 etc
Basically the same mantissa is being "floated out" to 0. because, for small x, sin(x)~=x and Limit(sin(x),x=0)=0.
Here is the second part which might not be so well known. We use the mantissa. Noticing that 3.162277660^2~=10 and 4.64159^3 ~=100 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 to see the same result.) I have used the Mathematica code Table[{x, (10 MantissaExponent[N[Sin[10^(-100 - 1/x)], 10]][[1]])^ x}, {x, 1, 10}] // TableForm (Changing 1/x to 2/x,3/x, etc .)
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.
Again the Mathematica code for this is Table[{x, (10 MantissaExponent[N[Sin[10^(-100 - 1/x)], 10]][[1]])^ x}, {x, 1, 10}] // TableForm 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.
You can even use decimal expansions in place of the "1" in "1/x.". Table[{x, (10 MantissaExponent[N[Sin[10^(-100 - 3.14000000000000000/x)], 10]][[1]])^(50 x)}, {x, 1, 10}] // TableForm
gives
1 1.*10^43
2 1.*10^42
3 1.*10^143
4 1.*10^43
5 1.*10^93
6 1.*10^143
7 1.*10^193
8 1.*10^243
9 1.*10^293
10 1.*10^343
It seems if you replace "1/x" with "any rational number/x" there exists a "^rational number *x" to replace the "^x" that gives only powers of 10. I would like to summarize what I have done in one mathematical statement. Any help here?
Marvin Ray Burns
The answer to my question was staring me in the face. I guess a few of you were just waiting for me to figure that out!
As you know the sin(x)~=x for small x, so instead of (10 Mantissa[sin(10^(-100 - 1/x))])^x we are really confronted with (10 Mantissa[10^(-100 - 1/x)])^x.= (10 Mantissa[10^(- 1/x)])^x = 10^(x-1). Thus we have the roots of the powers of 10 that I was concerned about. If I missed anything that you notice let me know.