Favorite numeral system?

[angus]

other:  Maya

Also, 8 has the letter a in German.  40 has the letter a in Italian, Spanish, French, and Portuguese.  In English, one thousand has the letter a, as does one thousand one, one thousand two, one thousand three, as do an infinite number of other integers. 

[angus]

Base 3 doesn't come up much, but there is an argument that it is the most efficient numeral system.

Suppose that the cost of a number is related to the product of the number of digits in a number times the number of choices for a digit (the base). For example the number 134₁₀ has a cost of 3*10=30, 10000110₂ has a cost of 8*2=16, so it's more efficient to store that number in binary than decimal. This model of cost is actually not unreasonable given our knowledge of information storage.

The number of digits in an arbitrary number k is equal to the logarithm of a number in the base n it is represented by (rounded down).  Eliminating the rounding factor I can write the cost in terms of the natural logarithm (ln) as

C(k,n) = (ln k/ln n) * n.

The cost as a function of the base is minimized when the derivative of the cost function is zero.

dC(k,n)/dn = 0

Finding the derivative reduces this equation to

ln n - 1 = 0

The solution is n = e, the base of the natural logarithm. e is an irrational number approximately equal to 2.718 and can't be used as a base for discrete counting numbers. However, the closest integer to e is 3, so that is the most efficient integer base. To use my earlier example 134₁₀ = 11222₃ (81+27+2*9+2*3+2*1). The cost of 11222₃ is 5*3 = 15, so it comes up slightly better than binary.

I follow, except that in the third paragraph you might need the phrase "the ratio of the natural log of the number to the natural log of the base" after the word equal.  What you have in prose is not quite correct, I think.

The math looks okay.  I started with dC/dn=0 and applied both the product and chain rule and got ln(n)=1.  (fun exercise, by the way!)

This yields n=e.  

Also, I'm wondering about the next bit.  (I actually started to post "e" when I initially posted, to be a smart-aleck, but then I remembered that we need an integer for the base.  Then I googled Klingon thinking that they might have a cool numbering system; alas they use base 10 as well.  Then I remembered the Maya have a nifty system based on 1, 5, and multiples of 20 and 40.)  Anyway, you then say that since e won't work, we must choose 3, which is the nearest integer.  But does it follow that the if the problem is minimized for n=e, and since we need an integer, then n=3 is necessarily the integer?

[angus]

...my supposition seems to hold.

seems so, thus far.  I guess what we want to demonstrate is something like:

 lim  C_n=3  <  lim  C_n=2
n→∞               n→∞

Or, more broadly,

 lim  C_n=3  <  lim  C_{n∈{N ∀ n≠3}}
n→∞               n→∞

Proof by the Second Principle of Mathematical Induction, perhaps, assuming that you can demonstrate conclusively for 2 and 4? 

Well, it's an interesting subject.  I hadn't encountered the idea of the cost of expressing numbers before.  The next time I'm at a cocktail party during which the subject of favorite bases comes up, I'll be able to suggest 3 in a way that makes me look really smart.

[angus]

yes, I meant as k approaches infinity, not n.  

It's clear that n/ln(n) increases without bound for n≥3.  This is independent of k.  I made smoothed curve Microsoft Excel plot of ln(n)/n for the first 15 integers, where I used 1.2 for the first independent value to avoid overflow.  



I guess I was thinking of an elegant syllogistic proof.  It has been a long time since I have thought about those things.  The proofs and demonstrations we do in thermodynamics are algebraic derivations involving simple substitutions with differential and integral calculus manipulations.  Nothing too complicated.   

It turns out that Wikipedia has a fairly accessible article on Radix Economy.  It shows, in tabular form, that e minimizes C, and that 3 gives the lowest C among the integers under 60.  It also offers a direct proof that e is most efficient, as you did.  Then it simply states that "it follows that 3 is the integer base with the lowest average radix economy."  I'm not sure that isn't a non-sequitur, but I'd have to give it more thought.  

[angus]

The actual cost goes as floor(log_nk)

That's what the Wikipedia article I mentioned does, except that it defines Economy as k*floor(log_nk + 1).  In this case, the limit as k approaches infinity is infinity times zero, which is indeterminate.  Can we say whether a quantity does not converge if in the large limit it is indeterminate?