Nobody else knows this about dividing fractions!

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29 Sep 2008, 3:06 am

Whenever you divide a fraction,
you might see interesting stuff and wonder why it happens that way.
There is a pattern and a reason that I will show you now.

When you divide a fraction you calculate all the powers of a number.
It's not the number in the fraction though.
The pattern of it is also often hidden by carries which is why it's not obvious.

1/7=.142857142857...
It looks like it's trying to double 7 and then gets caught in a loop.
What it's actually doing is trying to calculate the powers of 3.
That is more obvious with 1/97 = .01 03 09 27 8xxxxxxxxx

1/89=.011235xxxxx
It looks like it tries to make fibonacci numbers.
What it's really doing is calculating the powers of 11.
1/99989=.0000100011001210133114641xxxxxxx

The powers of 11 look like pascal's triangle until they are mixed up with carries.

If you want a better pascal's triangle then you put zeroes in between the two ones
so the powers of that number don't carry as soon.
The powers of 101 and 1001 make better pascal's triangles than 11 does.

Here is the formula for the pattern of getting all powers from only one division:
1
-----------
(10^p)-n

p is the precision you want. That means how many digits you need for the powers.
n is the number you want to calculate ALL of the powers of with just one divide

Here's a simple example for the powers of 2:
1/(1000-2)=.001 002 004 008 016 032 064 128 256 513024xxxxxxxxxx
See how 1024 carried to 512 and messed it up.

To do this you need a calculator for big numbers. Not one with only 8 digits.

This is weird because you can do one divide instead of lots of multiplies to get powers.
This is also weird because normal functions don't work like this.
I do weird math and make weird machines.

TallyMan
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29 Sep 2008, 3:44 am

Interesting. Thanks for sharing.

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n4mwd
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29 Sep 2008, 4:05 am

ValMikeSmith wrote:

If you want a better pascal's triangle then you put zeroes in between the two ones
so the powers of that number don't carry as soon.
The powers of 101 and 1001 make better pascal's triangles than 11 does.

Here is the formula for the pattern of getting all powers from only one division:
1
-----------
(10^p)-n

p is the precision you want. That means how many digits you need for the powers.
n is the number you want to calculate ALL of the powers of with just one divide

Here's a simple example for the powers of 2:
1/(1000-2)=.001 002 004 008 016 032 064 128 256 513024xxxxxxxxxx
See how 1024 carried to 512 and messed it up.

To do this you need a calculator for big numbers. Not one with only 8 digits.

This is weird because you can do one divide instead of lots of multiplies to get powers.
This is also weird because normal functions don't work like this.
I do weird math and make weird machines.

You may have found a slick way to invert a number 1/x too. Are you sure you don't mean "multiples" 7-14-28-56-etc. When I think of powers I think 7-49-343-etc.

An interesting comment about calculators too. With this $700B bailout package looming, I wonder if most Americans realize that most calculators are only 8 digits and can't add up to 700B.

I'm using the windows calculator and it seems to work pretty good with big numbers.

ValMikeSmith
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29 Sep 2008, 11:49 am

Quote:

You may have found a slick way to invert a number 1/x too. Are you sure you don't mean "multiples" 7-14-28-56-etc. When I think of powers I think 7-49-343-etc.

Powers!

1/(10000-7)=.0001 0007 0049 0343 2401xxxxxxxxxxxxx

There are some weird number theory stuffs going on when it looks like something else like
in the examples of 1/7 looking like 7 14 28 56 and 1/89 looking like fibonacci numbers.
I showed a way to exploit that in my example of using this to make pascal triangles.

lau
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29 Sep 2008, 1:18 pm

1/(1-x) = 1 + x + x^2 + x^3 + x^4 + ....

Taylor series.

When you set x = k/10^p, you get your results, but with the decimal point just after the first non-zero digit (the one, which is k^0).

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ValMikeSmith
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29 Sep 2008, 3:23 pm

I'll have to think about that one because it doesn't look easier but maybe it really is.

NUMBERS 1:23 "Quinquaginta Novem..."
60499999499/490050000000

lau
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29 Sep 2008, 6:08 pm

ValMikeSmith wrote:

...
1/89=.011235xxxxx
It looks like it tries to make fibonacci numbers.
...

and... I thought, what a coincidence... but it bugged me a little, and ...

... that is exactly what it is doing.

See the closed form expression for Fibonacci numbers.

With some fairly horrible messing about with the series summation, you can fairly easily show that this sequence:

0.0
0.01
0.001
0.0002
0.00003
0.000005
0.0000008
0.00000013
0.000000021
0.0000000034
....

does indeed sum to exactly 1/89.

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lau
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29 Sep 2008, 6:13 pm

PS. If you want a calculator that handles quite a few digits, you could try the ancient one off my website.

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Eggman
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30 Sep 2008, 12:32 am

ValMikeSmith wrote:

Whenever you divide a fraction,
you might see interesting stuff and wonder why it happens that way.
There is a pattern and a reason that I will show you now.

When you divide a fraction you calculate all the powers of a number.
It's not the number in the fraction though.
The pattern of it is also often hidden by carries which is why it's not obvious.

1/7=.142857142857...
It looks like it's trying to double 7 and then gets caught in a loop.
What it's actually doing is trying to calculate the powers of 3.
That is more obvious with 1/97 = .01 03 09 27 8xxxxxxxxx

1/89=.011235xxxxx

It looks like it tries to make fibonacci numbers.
What it's really doing is calculating the powers of 11.
1/99989=.0000100011001210133114641xxxxxxx

The powers of 11 look like pascal's triangle until they are mixed up with carries.

If you want a better pascal's triangle then you put zeroes in between the two ones
so the powers of that number don't carry as soon.
The powers of 101 and 1001 make better pascal's triangles than 11 does.

Here is the formula for the pattern of getting all powers from only one division:
1
-----------
(10^p)-n

p is the precision you want. That means how many digits you need for the powers.
n is the number you want to calculate ALL of the powers of with just one divide

Here's a simple example for the powers of 2:
1/(1000-2)=.001 002 004 008 016 032 064 128 256 513024xxxxxxxxxx
See how 1024 carried to 512 and messed it up.

To do this you need a calculator for big numbers. Not one with only 8 digits.

This is weird because you can do one divide instead of lots of multiplies to get powers.
This is also weird because normal functions don't work like this.
I do weird math and make weird machines.

no one else in the world knows this?

ValMikeSmith
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30 Sep 2008, 10:22 am

Quote:

no one else in the world knows this?

Well I've talked about it with many people who should know but were surprised.
This thread is only the second time someone said it reminded them of the Taylor series.
---
Apparently the fractions that make fibonacci numbers are among ones that fail to
make powers of 11 (or (10^q)+1) or pascal's triangles, because the precision is wrong
in such a way that the powers can't appear.
1/998999 makes many more fibos without carries than 1/89.
(Those are 1/(10^6 -1001) and 1/(10^2 -11))

Although I'm still holding to the division=powers pattern, I've just noticed a few
new things that are somewhat unverified and less defined in my mind as a pattern.

I don't think it is impossible, but I haven't been able to yet find another fraction
like 1/7 which appears to double as in the case of the number 7. I found that
1/49 and 1/499 do double 2 but trying to tweak the numbers for 7 hasn't
worked.

n/9801 appears to make multiplication tables for n. I haven't tested it to it's limitations,
which it most likely has. It certainly holds well the pattern for the multiplication table.

n/49005 may work better except for the fact that the table generated is not for n.
n=1 gives the multiplications of 2
n=5 gives the multiplications of 1 (although a large fraction given previously
also does, which I just assume does it better at the moment.)
n=35 gives the multiplications of 7
I'm yet unaware of the relationship between n and the table generated here.

The inverse of the following numbers are interesting but could be arbitrary and
coincidental: 3621 and also 19602 (make multiples of 3 and 5 ?)

The information in this post is is all new discovery to me within the past day.

lau
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30 Sep 2008, 11:27 am

There's one thing that you are not taking note of, particularly, at the moment, ValMikeSmith, and that is that you are only using base ten arithmetic.

For your "doubling" effect, you want the series expansion whose terms were (2/(10^k))^i. I.e the sum would be 1/(1-2/(10^k)). Hence...

5/4 = 1.2
50/49 = 1.0204081632653061224489795....
500/499 = 1.002004008016032064128256513026052104....
5000/4999 = 1.0002000400080016003200640128025605121024204840968193638...

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lau
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30 Sep 2008, 12:41 pm

Whoops. Browser crash....

Anyway, the "arithmetic tables" come from 1/(1-1/(10^k))^2.

(10/9)^2 = 100/81 = 1.234567901234567901234567...
(100/99)^2 = 10000/9801 = 1.0203040506070809101112131415161718192021222...
(1000/999)^2 = 1000000/998001 = 1.00200300400500600700800901001101201301401...

Your "3 times table" was probably a mistype for 9801/3 = 3267.
1/3267 = 0.000306091215182124273033363...

All rationals, p/q, eventually recur. Think - for each digit, you can only have at most (q-1) remainders, to carry on to the next digit. You can't have zero as a carry, except if the expansion terminates. For decimal expansions, that would only happen if the denominator is of the form 2^a*5^b.

I forget the detail at this point.... but IIRC the period is a factor of the product of one less than each prime factor of the denominator?

The other way to get at it is that, whatever the period is, the denominator must now be a factor of the number containing that many 1s (or rather, 9s).
factor 1 -> 1:
factor 11 -> 11: 11
factor 111 -> 111: 3 37
factor 1111 -> 1111: 11 101
factor 11111 -> 11111: 41 271
factor 111111 -> 111111: 3 7 11 13 37
factor 1111111 -> 1111111: 239 4649
factor 11111111 -> 11111111: 11 73 101 137
factor 111111111 -> 111111111: 3 3 37 333667
factor 1111111111 -> 1111111111: 11 41 271 9091

E.g. 1/239 = 0.0041841004184100....

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Last edited by lau on 30 Sep 2008, 4:29 pm, edited 1 time in total.

ValMikeSmith
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30 Sep 2008, 2:45 pm

Quote:

There's one thing that you are not taking note of, particularly, at the moment, ValMikeSmith, and that is that you are only using base ten arithmetic.

I realize that and I'm also speaking English.
Just for laughs we could speak latin and use roman numerals but does not seem
appropriate especially on WP. Seriously I'm familiar with the other bases (binary and up).
I thought permutation was just another base but I didn't get very far discussing it.

Base e, I'm not as familiar with. Logs and Trigs are calculable, and i is only calculable
in 2-dimensional numbers such as a+bi, but not when e or i are used to mix the
dimensions. Euler's identity is interesting but I don't know how to use it to translate
complex functions to something calculable. You did translate above, so thanks.
I said logs and trigs are calculable (to me) because I know their arithmetic series.

On my permutation thread, someone thought a summation from 1 to infinity would be
good for calculating permutations, but I understand, and recall from a previous problem,
that that means the same as counting the set from 1 to infinity (which I'm already doing),
and not deriving the member of the set from it's ordinal number as desired.

lau, you may wonder of what interest it is to me that division calculates powers.
Do you suppose that certain numbers I have, which have over 300,000 digits,
could be expressed as a short sum of power-towers?
It does seem to me that a lot of the very large primes are arithmetically compressed,
although obviously the case of the mersennes are simply powers of 2, minus 1.
If we know all the digits of a number then it is calculable.

Omnibus ex nihilo ducendis sufficit unum.

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30 Sep 2008, 3:07 pm

ValMikeSmith wrote:

Quote:

no one else in the world knows this?

Then I must bow before you, not often something new under the sun comes around.

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