23 Jun
2004
23 Jun
'04
11:05 a.m.
> How many can do a *cube* root without a calculator
Well...I can, though I'd ask to be allowed a four-function calculator.
Doing it by hand pushes the error rate higher than I would be
comfortable with.
I'd wager
"long division" sets many scratching their heads...
Now, what about
non-restoring long division? How many even know the square root of 2 and 3?
2.236+ ;-) The square root of 2 AND 3
...which is another of the possible meanings of that text...
For what it may be worth, I know the square root of 2 to about four
significant digits (1.4142+), of 3 to two (1.7+). There are other
important constants I know better (I don't use sqrt(3) much); pi to 26,
log(2) to five, e to ten....
Amusingly, I know the square root of 15? to 17 significant digits
(integer part plus 16 decimal places), because of some family
silliness. (My father went so far as to work it out longhand, using a
four-function calculator to help, basically doing a longhand square
root - yes, he knew how! - in base 10000. He carried it to some
faintly insane number of places, something like 50 or 100 decimal
digits. It gives me a rather "sic transit gloria mundi" feeling to now
type a couple dozen keystrokes and have the result to over a thousand
places. Far more precise and accurate but not nearly as much fun.)
Interestingly, it begins 3.93700393700, which looks suspiciously
rational; fortunately for my faith in arithmetic :), it continues
59055, dispelling the apparent rationality. Looking at
3.93700393700393700393700..., I find its continued fraction form is
[3; 1 14 1 6 1 14 1 6 1 14 1 3], which leads me to suspect that
sqrt(15.5) is [3; 1 14 1 6 1 14 1 6 1 14 1 6...], a conjecture borne
out numerically; a little scribbling verifies its theoretical truth.
(I keep meaning to build software to work with numbers represented as
continued fractions....)
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