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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..

#52

Post by dwaltrip »

kachnuv_ocasek wrote: post #35

I gotta be that guy: Why is this the first post on the front page of Hacker News? Is basic arithmetic really so fascinating to the computer people?

There is a little more going on here than basic arithmetic. See the top comment for further evidence: https://news.ycombinator.com/item?id=7144804. But yes, non-math-geeks will probably find this uninteresting. Apparently HN has a lot of math geeks ;)

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

#53

Post by diydsp »

antimatter15 wrote: post #37

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

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

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

#54

Post by tryeng »

geoffhotchkiss wrote: post #46
antimatter15 wrote: post #37

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

It actually doesn't break, you just have to do the carries as you would during normal addition. You have to read from right to left (1's place, 10's place, 100's place, and so on). So you're really just converting to base 10. 5th row: 1 5 10 10 5 1 Writing this a bit backwards, 1 * 1 + 5 * 10 + 10 * 100 + 10 * 1000 + 5 * 10000 + 1 * 100000 = 161051 = 11^5.

You can of course do this trick in any base. If we choose e.g. base 2^n for the n-th row of pascals triangle, we can use the following code for getting the n-th row of pascals triangle:

    def pascal(n):
      base = max(2, 2**n)
      row = (base+1)**n
      return [row/base**i % i for i in range(n+1)]
Nice, but hopelessly inefficient. :) You can also calculate a binomial coefficient the same way without any looping construct (the exponential operator does the looping for you).

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

#57

Post by Xcelerate »

Here's a generalization for any arithmetic sequence. With first term a0, difference d, and digit "padding" of n, the fraction that will result is:

(a0 + (d - a0)(1/10^n)) / (1 - 1/10^n)^2

For instance the sequence 1, 4, 7, 10, 13...

(1 + (3 - 1)(1/10^2)) / (1 - 1/10^2) = 1.02 / 0.9801 = 3400/3267 = 1.004 007 010 013 016...

For any kind of recursive sequence, you can find its generating function G(x) and then substitute some integer power of 0.1 for x to generate cool decimal expansions like this.

The generating function for the Fibonacci sequence is:

G(x) = x / (1 - x - x^2)

Substituting in 0.001 gives 0.001 / 0.998999 = 0.001 001 002 003 005 008...

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

#58

Post by sushirain »

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…

Perhaps this explanation is helpful:

    1 / (10000 - 2) = 1/10000 * 1/(1 - 2/10000)
Notice that the sum of a geometric series is:

    1/(1 - x) = sum_k( x^k )
    1/(1 - 2/10000) = sum_k( (2/10000)^k )
So:

    1/10000 * 1/(1 - 2/10000) = 1/10000 * (1 + 2/10000 + 2^2/10000^2 + 2^3/10000^3 + ...)

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

#60

Post by pitchups »

Xcelerate wrote: post #57

Here's a generalization for any arithmetic sequence. With first term a0, difference d, and digit "padding" of n, the fraction that will result is: (a0 + (d - a0)(1/10^n)) / (1 - 1/10^n)^2 For instance the sequence 1, 4, 7, 10, 13... (1 + (3 - 1)(1/10^2)) / (1 - 1/10^2) = 1.02 / 0.9801 = 3400/3267 = 1.004 007 010 013 016... For any kind of recursive sequence, you can find its generating function G(x) and then substitu…

Neat! Not familiar though with generating functions - can you pls explain how the generating function for the Fibonnaci sequence is x/(1 - x - x^2 ) ?
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