A formula of infinitesimal powers

[Ploppernz]

(N^(P = (a, b, c, d, e...))) = 1.999.. or 2.000...
If N^P = 1.999... or 2.000...
then we can say the opposite or positive of P (N^1P) = 2N or 2^(n+1)

(a, b, c, d, e...) = (1/2, 1/3, 1/4, 1/5...)

For example:

N^21 (2^21) = 2097152
P = 1/21 (0.047619047619)

N^P = 1.999... or 2.000...? Yup 1.999...

Then (N^1P = 2N or 2(n + 1)

2(n + 1) = 2^22
2N = 4194304
N^1P = 4194303.999... (1.047619047619)

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The largest number I've done is 2^2016 and have found its power to reach every number, but typing that would take way to long.

The next largest number I've done is 2^1000 (a millinillion number) which turns out to be 1.07 millinillion to be exact.
The power to get this massive number back to 2.000 or 1.999 was 0.001. With intervals of 1 at the 100th decimal place you can change the number to be any number (1, 2, 3, 4, 5, 6)

On a side note, not all decimals derived from fractions give an exact calculation. For this reason the sums with either be close to the destination or dead on. A lot of numbers (past the millions, billions, trillions) have undeterminable sums, what I call continuum (1.333... 3.14159265359...) sums.

A lot of the powers have patterns for the intervals to reach the next number, and others do not (or it is to large/small to calculate)


The real question now is what does this lead to, and what can this be applied to...

I am in no way a mathematician, but somehow feel that this could potentially lead to something much bigger.

Think my equation is wrong? Go ahead and leave a number below (can even be a decimal base number if you want) and I will calculate the power for it to reach the next step and to get back to 2 (Do not have a clear calculation for how to get decimals to the next step yet - will have to figure that out next)

[Renee]

Ploppernz, did you use a calculator, or a computer, or longhand to calculate these out?

If you used a computer or a calculator, I am afraid your finding is not accurate. To see this, enter "2" then take the square root, the square the result, and you're already not at "2".

These devices can be deceptive. This is why teachers warn students to not to have to rely on a calculator, if higher accuracy than exactness at 254 significant digits is required.

[Ploppernz]

I have an update, the formula remains the same, I have just found the pattern between the powers.

What you need to understand and remember is this formula changes and can be off slightly. I can get sums accurate up to a decimal but if I increase the decimals shown the possibility for error goes up and the powers require more accuracy. I do to the 1000th decimal place and a base reference point.

2^(P*PoP) = N^P

N = any number
P = any number (rule only applies to base 2 numbers)

PoP = Place of P (the sum of 2^(P*PoP))

I have tested this formula on small to large numbers ranging from 0 - 2^1000000 including numbers that are not base 2 for P.
I will have another update with the addition of how to account for non-base 2 numbers another time.

Some more interesting numbers I have been able to crack to a decimal point is Pi, Tau, and e.

The first of the 3 (most interesting)

281474976710656 (2^48)^Pi = 16,777,712 (2^24)^Tau

Why is this most interesting? This means that I could theoretically derive the next power of 2 to the power of Pi^3 ^4 ^5 ^6 ^7 ...

The second one is e

Its hard to base your calculation when most calculators cannot handle more than 16 decimal places (on average). In a Go lang program that was constructed by a friend of mine and constantly modified, I am able to do very high decimal calculations, from which I can derive accurate powers.

I calculated e to 2.71828182846

All numbers below are base 2 numbers that = e with the powers derived from 2^P

2^1.44269503885
4^0.72134751942
8^0.48089834628
16^0.36067375971
32^0.28853900777
.64
.128
.256
...

Pi^0.8735685256 = e

This brings me to the third, Pi

I figured if this formula can apply to e and every other number then it should work for Pi

All numbers below are base 2 numbers that = Pi with the powers derived from 2^P

My power for 2 is the most accurate at 3.14159265359 (I could do 58 like its supposed to be but I would probably then change the decimal place to display 10,000)
My least accurate but highest tested is 16,777,712 at just 3.14 (with adjustments can get more accurate)

2^1.65141961294769
4^ 0.82574806473
8^0.55049870982
16^0.41287403236
32^0.33029922589
.64
.128
.256
...
and the big boy

16,777,712^0.06880915053

[Ploppernz]

No I am using a custom Go lang program designed only to take a base number (N), a power (P) and a decimal place display (dp). I understand most calculators round, this is not a stock microsoft calculator limited to e+8192 before overflow occurs. My calculations are done with large float data type numbers, and can do up to 100000 decimal place before my computer takes to long. If you want to test the accuracy of the numbers to the decimal places be my guest, the numbers are accurate to a point on scientific calculators.

[Renee]

No I am using a custom Go lang program designed only to take a base number (N), a power (P) and a decimal place display (dp). I understand most calculators round, this is not a stock microsoft calculator limited to e+8192 before overflow occurs. My calculations are done with large float data type numbers, and can do up to 100000 decimal place before my computer takes to long. If you want to test the accuracy of the numbers to the decimal places be my guest, the numbers are accurate to a point on scientific calculators.

Thanks for the explanation.

I take your points.

My new point is that you are trying to show a theoretical mathematical proposition with examples. That is fine, but it won't convince a mathematician. And the other point, which is a bit of a reiteration of what I said earlier, honed up: if you have a calculator that has an accuracy of even ten billion significant digits, it is still prone to error when infinitely repeating decimal fractions are parts of the calculations. That can't be avoided by increasing the decimal places or the number of significant digits.

So I see three problems: one is that yours is a demonstration, and while it may generate some interest, without a proof it is not a significant finding. The other problem is you're using numerical methods instead of pure math to illustrate parts of the demonstration, and that will throw your calculations off, no matter what. The third challenge is that you offer no pattern that you've discovered; you are reporting a type of curious mathematical calculation, which -- without trying to dismiss your interest or disrespect it -- is neither here nor there.