Now if Druff doesn't mind, too much, I will finish off my post above, about how pi relates to the physical fine-structure constants. All of the known mathematical constants do.

Note as well that things above might have ended up with 1691__1271_1721__1961 instead, which is even more in line with the results at
https://vegascasinotalk.com/forum/sh...l=1#post171482

And, that (-2*9.01222434314022^2 + 2*9.01222434314022 + 113) = -31.41592653589791 ---> 3.141592653589791 was chosen because it's the spot in the thus equations that naturally begins with the first digits of pi, namely, -31 ---> 3.1, given that (-2*9^2 + 2*9 + 113) = -31, on the 113-side, which has to do with the fine-structure constants 137, and 142.

Now to finish things off by bringing the numeral, 113, from the 9's, and 4's, of [9].0122243[4]3422210[9], which is the usual way of numerologically expanding the digits in those positions. Interestingly, there's 9 = [3^2 + (1 - 1)], 0122243 = 11*11113, and, [³√(11 - 3)]^2, which together is 311__11_11113_3113 ---> 31111_11113__3113 ---> 311_113__311_113, by overlapping the sets of four 1's, on the left-hand side, and, splitting off the sets of two 1's, on the right right. Note that exponents under other exponents, here 3 under 2, become part of the result.

What is the basic relation between pi, and the numeral, 113? "335/113 is the best rational approximation of pi with a denominator of four digits or fewer, being accurate to six decimal places. It is within 0.000009% of the value of pi, or in terms of common fractions overestimates π by less than 1/3748629. The next rational number (ordered by size of denominator) that is a better rational approximation of pi is 52163/16604, though it is still only correct to six decimal places. To be accurate to seven decimal places, one needs to go as far as 86953/27678. For eight, 102928/32763 is needed."

And, so, another very curious set of equations with their own corresponding forms, to do with the digits of 1961, and 1721, again, matter/antimatter pairings of the results.


#1. [(19 + 0^0=0)^2 - 6 + 1 - 0^0] = 355, or, 355 + 1,

#2. {[12 - (antilog(-0^0)=1)^-1] + 7^3 + 1 - 0^0} = 355, or, 355 - 1,

#3. {[(0100 + ²√(900) + antilog(0^0)=10] + 6^3 - 1 + 0^0} = 355, or, 355 + 1,

#4. {[0010^(2^0) + 2^(²√(009)) + 0^0=01]^2 - 7 + 1 + 0^0} = 355, or, 355 + 1. And,


#5. [(-1 - 9 - 0^0=01)^2 - 6 - 1 - 0^0] = 113, or, 113 + 1,

#6. {[-(1 - 2) + (antilog(-0^0)=10)^-1]^2 - 7 - 1 + 0^0} = 113, or, 113 + 1,

#7. [(-0010^(2^1) - ²√9 - 0^0=0) + 6^3 + 1 - 0^0] = 113, or, 113 - 1,

#8. {[-0100 - 2^(9 - ²) - antilog(0^0)=1] + 7^3 - 1 + 0^0} = 113, or, 113 + 1.


Note that, in equation #3, the 0^0 = 01 reverses to 10 = anti-log(0^0). And, that, in equations #3, #4, and #6, #8, respectively, that the 009 reverses to 900, and, the ²√9 reverses to (9 - ²). The latter reversal involves being able to reverse a numeral, but, not a square root (in the exponent on 2), versus, not being able to reverse a single-digit numeral, but, being able to reverse the square root, and the thus numeral, by writing both as numerals (in the exponent on 2).

Again, I had to resort to the anti-matters numerals of 113, and 355, in terms of symbolically creating them from the left-over amounts, by 0^0 for 0, or 1. To write the above equations with 113_311, and 355_553, on their respective right-hand sides. Then (10 - 1 - 3^2) = [3 - 1 - log(100)] = 0 ---> 113, or, (3 - 1 - 1) = 1 ---> 113, and, [3 - log(500 + 500)] = 0 ---> 355, or, (-3^2 + 5 + 5) = 1 ---> 355.

As well, the 012243 part of [9].0122243[4] = [(-3 + 5 + 5) + (5 - 3)^2 * (30550 + 3^2)] ---> 355_553__355_553, which corresponds to 311_113__311_113, by stripping a second 5 out of the (5-3)^2 term, because it's squared, and, by splitting off, or shedding of, the two 5's overlapped in the (30550 + 3^2) term. As well, 0122243 ---> 342221 = 11*311110 .

And, oh, I had to go with ²√ for the square root of, and, ³√ for the cube root of, because the formatting here doesn't allow for as compact a thus sign for square root, as it does for cube root as ∛.