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1/9998 = 0.0001 0002 0004 0008 0016 0032 0064 0128 0256..

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Re: 1/9998 = 0.0001 0002 0004 0008 0016 0032 0064 0128 0256..

#92
post #21

The pattern will break down once you get past 8192, which is 2^13. That means that the pattern continues for an impressive 52 significant figures (well, it actually breaks down on the 52nd digit, which will be a 3 instead of a 2). The reason it works is that 9998 = 10^4 - 2. You can expand as 1 / (10^n - 2) = 1/10^n * 1/(1 - 2/10^n) = 1/10^n * (1 + 2/10^n + 2^2 /10^2n + 2^3 /10^3n + ...) which gives the observed patt…

> The pattern will break down It doesn't actually: 4096 8193 6387 = 4096+8192 + 1 6384 + …

You can seemingly increase the digit space by doing 1/99998 or so.

Re: 1/9998 = 0.0001 0002 0004 0008 0016 0032 0064 0128 0256..

#93

In high school, I was pretty fond of plugging 11^n to get rows of Pascal's triangle. It breaks down at row 5, but inserting 0's in the middle extends it (e.g. 101^n, 1001^n, 10001^n). 11^0 1 11^1 1 1 11^2 1 2 1 11^3 1 3 3 1 11^4 1 4 6 4 1

While bored in middle school algebra, I figured out on my TI-30 which number, raised to itself as a power, would equal 9.9999999E99 (not sure on the precise number of nines after the decimal point, but basically it flooded the screen with all nines). 56.96124843225 ^ 56.96124843225 Wolfram confirms that it's pretty close to a full googol. Of course, you can keep adding digits to the end of the number to make it even…

I was wondering if there was an inverse operation for tetration, and it turns out there is: https://en.wikipedia.org/wiki/Tetration#Square_super-root (what you're basically finding is 56.96124843225⇈1, which is apparently ssrt(1googol))

Re: 1/9998 = 0.0001 0002 0004 0008 0016 0032 0064 0128 0256..

#94
post #53

Earlier quoted context omitted.

On a vaguely similar, yet less practical note: 111,111,111 * 111,111,111 = 12,345,678,987,654,321 :)

On most pocket calculators, 11111111×= will yield 12345678. People are often surprised that that sequence is accepted. To me, it exposes something about the calculator's internal architecture. It's also a useful self-test if you think the battery might be going.

The sequence is a shortcut accepted by the vast majority of regular calculators for most of operations. It simply takes the second operand to be the same as the first and repeated pressing of the = key repeats the operation ad infinitum. Ie. 1+== is 3.

I have written an iOS calculator app and had very interesting times trying to find and mimic these shortcuts. I have thought for a long time they had to follow from some simple implementation detail, as all the calculators got them precisely the same, but I never found this one consistent rule, I had to implement the features in a series of hacks.

Re: 1/9998 = 0.0001 0002 0004 0008 0016 0032 0064 0128 0256..

#95
post #53

Earlier quoted context omitted.

On a vaguely similar, yet less practical note: 111,111,111 * 111,111,111 = 12,345,678,987,654,321 :)

On most pocket calculators, 11111111×= will yield 12345678. People are often surprised that that sequence is accepted. To me, it exposes something about the calculator's internal architecture. It's also a useful self-test if you think the battery might be going.

i'm interested now - what does that expose about the internal architecture for you?

Re: 1/9998 = 0.0001 0002 0004 0008 0016 0032 0064 0128 0256..

#96

The pattern will break down once you get past 8192, which is 2^13. That means that the pattern continues for an impressive 52 significant figures (well, it actually breaks down on the 52nd digit, which will be a 3 instead of a 2). The reason it works is that 9998 = 10^4 - 2. You can expand as 1 / (10^n - 2) = 1/10^n * 1/(1 - 2/10^n) = 1/10^n * (1 + 2/10^n + 2^2 /10^2n + 2^3 /10^3n + ...) which gives the observed patt…

If you'd like to continue the pattern beyond 52 digits, just keep adding 9s to the original fraction... 1/9999999999998 = 1.0000000000002 0000000000004 0000000000008 0000000000016 0000000000032 0000000000064 0000000000128 0000000000256 0000000000512 0000000001024 0000000002048 0000000004096 0000000008192 0000000016384 0000000032768 0000000065536 0000000131072 00000002621440... × 10^-13

interestingly, this works the other way around too: 1/98 = 0. 01 02 04 08 16 32 65...

and even further? 1/8 = 0.125

Re: 1/9998 = 0.0001 0002 0004 0008 0016 0032 0064 0128 0256..

#97

Earlier quoted context omitted.

If you'd like to continue the pattern beyond 52 digits, just keep adding 9s to the original fraction... 1/9999999999998 = 1.0000000000002 0000000000004 0000000000008 0000000000016 0000000000032 0000000000064 0000000000128 0000000000256 0000000000512 0000000001024 0000000002048 0000000004096 0000000008192 0000000016384 0000000032768 0000000065536 0000000131072 00000002621440... × 10^-13

For the fibonacci, add a 9 on both sides of the denomator 1/998999 1/99989999 1/9999899999 To get more 0 spacing and avoid overflow

This should be denominator ;-)

Re: 1/9998 = 0.0001 0002 0004 0008 0016 0032 0064 0128 0256..

#98
post #76

You are all genius. The first person found it is definitely outstanding. There are so much mathematic magic out there that always made me feel fascinating. I'm amazed by the nature of this world which can be described by math. It's unbelievable. So I made my logo utilizing one of it. http://bit.ly/1gre9Bh

You can just do 1/0.0001000200040008001600320064... to find it.

Re: 1/9998 = 0.0001 0002 0004 0008 0016 0032 0064 0128 0256..

#100

maxima code follows fpprintprec:100; fpprec:100; s : string(bfloat(1)/bfloat(9998)); makelist(substring(s,3+4 i,7+4 i),i,0,15); [0002, 0004, 0008, 0016, 0032, 0064, 0128, 0256, 0512, 1024, 2048,4096, 8193, broken pattern ,6387, 2774, 5549]

with_bigfloat_precision(500) do BigInt(1)/BigInt(9998) end
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