Crafting grafting, part 4

(I’ve gone back and revised my notation. a becomes j and b becomes 10^m.)

Check out this grafting number, the second biggest I’ve got at the moment:


There’s some structure to understand here. This is generated using j = 9261, m = 8, and n = 7. You can read those values off the numbers above, though. There are 22 digits in the integer on the left, and that’s 2n+m. On the right, those 22 digits recur starting 7 digits left of the decimal point; that’s n. Before them are the digits 9261, which is j.

Likewise, examine this one:

\sqrt{4110105646} = 64110.105646457...

You can see 2n+m = 10 and n = 4, so m = 2, and j = 6.

Though here’s an evil one:

\sqrt{576276646} = 24005.76276646922...

Here 2n+m = 9 and n = 1, so m = 7, and j = 2400. Or… wait, is it? If you write the integer on the left as 0576276646 then 2n+m = 10 and n = 2 so m = 6 and j = 240. Or should you write it as 00576276646 and conclude 2n+m = 11 and n = 3 so m = 5 and j = 24? Turns out all three interpretations are valid; they lead to three different equations yielding the same grafting number.

What about this miserable attempt at a grafting number?

\sqrt{100} = 10.000...

Well, 2n+m = 3 and n = 2, so m = , um, -1? And j = 0? Those give x = 0.1000... (with the + sign in the quadratic formula, unlike all normal grafting numbers, because the - sign gives x=0) and the equation 10^2(0+0.1) = \sqrt{10^4(10^{-1}(0.1))} or 10 = \sqrt{100}. Yes, that sort of fits the profile, but in an ugly way.

Likewise, \sqrt{98} = 9.8995... and \sqrt{99} = 9.9498... have similarly pathological analyses. Those I think really do need to be counted as grafting numbers, but not normal ones.

Two things I have not figured out yet is why it’s always the ceiling, not the floor, that gives the grafting number, and why j = 2 and m = 1 gives grafting numbers for, apparently, all values of n while other values of j and m don’t. I suspect these are related. Observe

\sqrt{\lceil y \rceil}  = \sqrt{y-\{y\}+1} = \sqrt{y}\sqrt{1-\frac{\{y\}-1}{y}}

(with braces denoting fractional part) which is, to first order

\sqrt{y}\left (1-\frac{\{y\}-1}{2y} \right ) = \sqrt{y} - \frac{\{y\}-1}{2\sqrt{y}}

and \{y\} is whatever it is, but it’s in 0<\{y\}<1 and for an approximate result we can take \{y\}\approx1/2 so

\sqrt{\lceil y \rceil}  \approx \sqrt{y} + \frac{1}{4\sqrt{y}}.

By taking the square root of the ceiling of a number y instead of y itself, we add approximately 1/(4\sqrt{y}) to the square root. By the same token, using floor subtracts about 1/(4\sqrt{y}).

That’s sort of an average, though, and in the case of grafting numbers, it’s often an overestimate. For instance, for n = 3, j = 8079, we get x=49991903-\sqrt{2499190300000000}=6557202816420999992418.... Then with m = 8 we end up with grafting number \lceil 65572028164209.99992418... \rceil. Clearly the ceiling in this case changes the number by several orders of magnitude less than \approx 1/2. But why don’t numbers looking hypothetically like \lfloor 65572028164209.00002418... \rfloor work out?


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