#decimal expansion

[somethingusefulfromflorida]

By random chance, the decimal expansion of pi briefly begins repeating itself after 6954 digits before resuming its randomness

3.1415926535...987571415957811...

That's the longest repeated string in the first million digits, so a really good rational approximation would have the first 6954 digits repeat infinitely. You can create any rational number with n repeating digits by taking a string of length n and dividing it by an equally long string of nines.

0.123123123... = 123/999

0.141591415914159... = 14159/99999

Of course, these may not be in simplest form

0.123123123... = 123/999 = 41/333

0.363636... = 36/99 = 12/33 = 4/11

0.102911029110291... = 10291/99999 = 251/2439

So the first 6954 digits of pi over 6954 nines could very likely be simplified. At that magnitude, the odds that they are co-prime (share no factors) is very slim

Scratch that. I plugged the first 6954 digits into a prime factorization calculator, and the first three factors it was able to find were 99551 × 18298401827 × 104229615391 (100k, 18.3 billion, 104.1 billion). It took 30 minutes to find those three, so it could take HOURS or DAYS to calculate the remaining factors because it still has over 6,900 digits to get through, but this makes it seem more likely that the string is co-prime with 9999999...9999999 just because there are multiple orders of magnitude between factors, so the odds of any two random 12+ digit numbers being the same are slim. I don't know, I'm not a statistician.

3 + (1.4159...98757 x 10^6953) / (10^6954 - 1) ≈ pi

[stackexplorer]

[wiskwa-en]

0.9999... (once more)

Is 0.99999etc (so, with infinitely many nines after the decimal point) equal to 1?

This was the first question that was written about in this blog. The answer was and is: yes, 0.99999…=1.

You can look at this equality in a slightly different way and justify it by starting from 1 and find a decimal expansion of that number; I write `a' because some numbers have more than one such expansion. The one that you find is determined by the way you make it.

Take a number, x, in the interval (0,1). We make sequences of digits, a1, a2, a3, … and numbers, x1, x2, x3, … as follows.

There is i<10 such that i/10≤x<(i+1)/10. We write a1=i and x1=a1/10. The difference x-x1 is less than 1/10, so we can find j<10 with j/100≤x-x1<(j+1)/100. We write a2=j and x2=x1+a2/100. The difference x-x2 is less than 1/100, so we can find k<10 with k/1000≤x-x2<(k+1)/1000. We write a3=k en x3=x2+a3/1000.

We keep following this recipe: at stage n we have xn with 0≤x-xn<1/10n; so there is an an+1<10 such that an+1/10n+1 ≤ x-xn < (an+1+1)/10n+1. Using this we make xn+1=xn+an+1/10n+1.

According to the definition given in this post the sequence of xn's converges to x and we express this, rather suggestively, as x=0,a1a2a3…. Note well that for a mathematician this equality represents the whole story given above.

Sometimes you are lucky and you find only finitely many i; that happens when there is an n such that x=xn; in that case ai=0 for i>n. If you do not like that then you can adapt the recipe and make sure you will keep getting non-zero ai.

Given our number, x, in the interval (0,1) we make sequences of digits, b1, b2, b3, … and numbers, y1, y2, y3, … as follows.

There is i<10 such that i/10<x≤(i+1)/10. We write b1=i and y1=b1/10. We have 0<x-y1≤1/10, so there is j<10 with j/100<x-y1≤(j+1)/100. We write b2=j and y2=y1+b2/100. We have 0<x-y2≤1/100, so there is k<10 with k/1000<x-y2≤(k+1)/1000. We write b3=k and y3=y2+b3/1000.

We keep following this recipe: at stage n we have yn such that 0<x-yn≤1/10n; so there is a bn+1<10 such that bn+1/10n+1 < x-yn ≤ (bn+1+1)/10n+1. Using this we make n+1=yn+bn+1/10n+1.

The sequence of yn's converges to x too and we express this as above, rather suggestively, as x=0,b1b2b3…. There is a whole story behind this equality too.

The difference between the two recipes is that possibly xn=x for some n but that yn<x for all n. If xn=x never occurs then we find xn=yn and an=bn for all n and we find just one decimal expansion for x. However, if xn=x then the sequences go their separate ways: you can check that bn=an-1 in that case and also bi=9 for i>n. For example when x=1/2: then a1=5 and b1=4. So 1/2=0.5 and 1/2=0.4999….

The Dutch version of the question mentions the fractions 1/9, 2/9, … 8/9. For these both recipes yield the same expansion: i/9=0,iiiii…; the nice thing is that the second recipe also works for 1=9/9: you will find bn=9 for all n. The equality 1=0,999… is thus a direct result of the second recipe.

[aaronwarwick]

Visit Original Link

I don’t think I’m even going to upload this to soundcloud. 

There are three synths in this track. Each plays a four bar loop of eighth notes determined by the following:

Pick an irrational number

Write out its decimal expansion to 128 places (including the digit in the units place)

Group the digits into pairs (e.g. 3.1415926535897... becomes 31 41 59 26 53 58 97...)

Each two digit number is matched with its corresponding key on the piano, where 01 -> A1, 02 -> A#1, ... 88 -> C9. Numbers greater than 88 are treated as rests. 00 maps to A1 (I considered having it be a rest as well, but I just ended up not doing that).

The numbers used are √(2), √(3), √(5), π, e, and φ (golden ratio).

(this was also a chance to mess around with the plug-in gross beat, since it recently became free with a certain license of fl studio)