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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..

#21

Post by lelf »

crntaylor wrote: post #10

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
    +           …

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

#22

Post by [deleted] »

rsiqueira wrote: post #20
crntaylor wrote: post #10

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 is not really breaking. What happens is that 16384 doesn't fit in a 4 digit space so it's first digit "1" jumps to 8192 and it becomes 8193. Then the next number (32768) add it's first digit "3" to 16384 and it becomes 16387 and so on, so the sequence appears strange after 4096: ...409681936387...

[deleted]

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

#23

Post by VeryVito »

crntaylor wrote: post #10

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

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

#25

Post by qdpb »

rsiqueira wrote: post #20
crntaylor wrote: post #10

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 is not really breaking. What happens is that 16384 doesn't fit in a 4 digit space so it's first digit "1" jumps to 8192 and it becomes 8193. Then the next number (32768) add it's first digit "3" to 16384 and it becomes 16387 and so on, so the sequence appears strange after 4096: ...409681936387...

I think that's what you would call a break in a pattern though. Of course, the infinite series will stay infinite.

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

#27

Post by dsugarman »

everyone seems to be concerned with the point at which this breaks. You determine this, the size of the denominator gives you the precision you can calculate:

1 / 99998 will return:

0.00001 00002 00004 00008 00016 ....

[0]http://www.wolframalpha.com/input/?i=1%2F99998&dataset=&equa...

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

#30

Post by [deleted] »

lelf wrote: post #21
crntaylor wrote: post #10

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 + …

[deleted]
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