room topic changed to decimal notations for fractions: Discussion of infinite series of decimal numbers equivalent to a given fraction [decimal-expansion] [fractions] [notation]

I created this room for the discussion of math.stackexchange.com/q/1371400/139123

Thanks. Plain text of the markdown document I tried to share is in: gist.github.com/markomanninen/2eec41d340c0c0970c86

I don't know if there is an online reader/editor that can show mathjax equations properly, I tried with stackedit.io...

Maybe you can copy and paste the raw text to your own stackedit.io document or markdown editor and see it properly. It clumsy to read just plain text. Anyway hope to hear your comments, if complicated log part is justified after all of this.

OK, so actually 1/7 is one of the fractions you were interested in, apparently. And you are looking not just at the fact that there is a repetition group, but you are looking for patterns within patterns.

At first 1/7, but then 1/49 and 1/89 as well. And eventually with 1/x to find out how patterns evolve.

So for 1/7 the pattern that I see is 1/7 = 0.14 + 0.0028 + 0.000056 + 0.00000112 + 0.0000000224 + ... which shows the "doubling" phenomenon. But it doesn't really show to me the "repetition" phenomenon, that is, the first group of digits 142857 is evident from the first four terms of the sequence, but it is somewhat less evident that the next six digits will be the same; and what about the 1000th group of 6 digits? Also, why does it start with 0.14?

(more to say but message gets too long for one upload)

Its still intriquing how product of the continuous calculation carried backward on progression, which seems to mess simple string of progression. I have an advocate for baseless summation for x/y, I try to link it to wolfram so its easier to see

yeah, it doesn't tell why it starts with 14, but .3+.09+.0027+.00081 tells why it transforms to 14... of course not being evident visually but after some manual calculation

plus for greater numbers even manual calculation goes difficult. which is same as this is not a way to expand decimals in a school exam. just one way to show how they progress from simple base x^0 and up to infinity.

I find if you actually write out the long division in grammar-school fashion, the pattern emerges. Adapting the notation for linear format, we get 1.0 = 7*0.1 + 0.3 = 7*0.1 + 7*0.04 + 0.02. So in two steps we got to 1.0 = 7*0.14 + 0.02, and the remainder is equal to the original number shifted two places to the right and multiplied by 2. Now the long division has us dividing 2 by 7, so the next terms should be twice as "large" but shifted two places: 1 = 7*0.14 + 7*0.0028 + 0.0004.

The remainder has "doubled" again, so 1 = 7*0.14 + 7*0.0028 + 7*0.000056 + 0.000008. And now we notice happily that the remainder (times suitable power of 10) is congruent to 1 modulo 7; we can write 8=7+1, so 1 = 7*0.14 + 7*0.0028 + 7*0.000056 + 7*0.000001 + 0.000001 and now we begin the cycle over again.

Interesting yes... thats shows the sequence in very early stage.

Of course another way to look at it is that you don't take the 7 out of the 8 in the remainder 0.000008, but just let it keep doubling, and the carries we get each time we do three more steps (getting a number six digits longer) will make up for the fact that the previous steps did not end in the repeating group 142857 but in something less.

So the "just let it keep doubling" is somehow analogous to the summation you wrote, except that for 1/7 you have a tripling with division by 10 each time instead of doubling with division by 100. And yet it must add up to the same thing, which is a bit surprising (and therefore interesting).

I hope this link works for you: bit.ly/1VEPU4o it goes to wolfram alpha...

these should be there in the link. are these conversations permanent so I can come here and see what we talked about, or should I copy and paste them to my computer? I haven't really used SE for chat, just once

Sorry about the flood, but this is also interesting, related to 1/7 and 1/49:

n/49	length	sum	sequence
1	42	189	020408163265306122448979591836734693877551
2	42	189	040816326530612244897959183673469387755102
3	42	189	061224489795918367346938775510204081632653
4	42	189	081632653061224489795918367346938775510204
5	42	189	102040816326530612244897959183673469387755
6	42	189	122448979591836734693877551020408163265306
7	6	27	142857
8	42	189	163265306122448979591836734693877551020408
9	42	189	183673469387755102040816326530612244897959

42 digits long sequence of n/49 suddenly truncates to 6 digit length. 7/49 = 1/7. these are the two repeating sequences of n/49 digits counting 6+42 = 48 = 49-1

Thinking about it more, the the series you get for 1/7 with ratio 3/10 (so a "tripling" phenomenon) also shows why the final sum repeats every 6 digits: 3^6 is congruent to 1 mod 7, so after 6 divisions, if you add the integer part of (3^6)/7 to the 6 digits you have so far, you can start over (and generate the same 6 digits again, etc.). And while 0.1 + 0.03 + 0.009 = 0.139 doesn't look much like 0.142, the integer part of 27/7 is 3, which is where the missing 0.003 comes from.

Noticing that 27 = 7*4 - 1, we can write 0.139 + 0.004 = 0.143, and 1/7 = 0.143 - 0.000143 + 0.000000143 -+... .

Hmh... this brings one possibility to my mind, what i was thinking earlier. Maybe there are infinite ways of doubling, tripling, adding each fraction decimals? Depending where do you start?

At least for 1/7 I have several different summation functions ready. It would be nice to know, what actually is the limit?

The summations are all equal, of course. You can use summation of geometric series (the same thing I relied upon to simplify the complicated summation in the question) to prove it. At least, you can do this for each series, one at a time, to show their limits are the same number. If there's a general summation formula that fits all the "interesting" series, which also is a geometric series, we can also prove all the series equal in one proof.

I come back to read and ask more of this later. I have to pack for moving. Thanks for your input so far. If you want to add something on your SE answer after this conversation, please feel free. I will approve it as an answer tomorrow or this weekend.