23 Jun
2004
23 Jun
'04
2:33 p.m.
On Wed, 23 Jun 2004 19:58:38 -0400 (EDT)
der Mouse <mouse(a)rodents.montreal.qc.ca> wrote:
Knowing how to do it, though, involves learning enough
of
the theory
behind it to stand you in good stead in other cases,
though. (Assuming
you actually understand it. Learning how to do it by
rote, without
understanding why, _is_ pretty pointless.)
Exactly. Ditto for all the other roots (though it
gets obscene *real* quick)
I expect that there may well be a similar method
for
the cube root.
There is, but it is significantly more elaborate to
perform.
Longhand square root extraction is based on (X+d)^2 = X^2
+ 2Xd + d^2. For example, to work out sqrt(15.5) (in decimal), we
start with the
first digit. (If you can't get the first digit right
without
assistance, you have no business trying something this
elaborate! :-)
___3.___________
) 15.50 00 00 00
9 6
--
6 50
Now, we have written 15.5 = (3^2 + 2?3?d + d^2), with d
to be
determined. The 6.50 is 2?3?d + d^2. The 6 sitting off
at the right
is what I saw described as the memorandum column; it's
twice the
I record these on the *left* -- as if I was preparing a
long division and the "6x" was the divisor. Instead of
"twice the root-so-far", I use "20 * root-so-far" -- same
difference since this makes the "decimal shift left"
more explicit.
An example:
3 5 1 3
===========
12 34 56 78
9
-----
65) 3 34
3 25
--------
701) 9 56
7 01
--------
7020) 2 55 78
2 10 69
-----
45 09
root-so-far. Now, we look for a digit X such that 6X
* X
<= 650 (this
is 2?3?d+d^2). We want the largest digit that fits; 9
works, so we
fill it in, and write in and subtract off 69*9:
[snip]
Usually, you can get the next digit right by looking
at
just the high
digit or two of the memorandum column; only when it's
close to the edge
may you have to try multiplying out different choices in
full. In the
next position, for example, I'd look and say "55/7 = 7+,
let's try 7",
and indeed, 7 works (barely - the remainder is so low the
next two
digits of the root are zero).
You can do the same thing for cube roots, writing
(N+d)^3=N^3+3N^2d+3Nd^2+d^3, but the arithmetic with the
memorandum
column becomes a good deal messier.
Yes. And higher roots are reserved for monks with nothing
else to pass their days... :-/
One of the subtle advantages of learning algorithms like
these is the perspective it gives you on the design of
other algorithms. E.g. iterative graphic drawing
algorithms, etc.
If I needed a high-precision root, of course the last
steps would be
N-R. But under no-tools circumstances, it's unlikely the
root is
needed to more than about three significant digits
anyway.
Yup. And, in many cases, what is more important is to
get a "feel" for the answer -- not the answer itself.
A gallon (US) of water weighs *about* 8 pounds. There
are *about* 8 gallons in a cubic foot (to within 10%).
So, a cubic foot of water weighs *about* 60 pounds.
And, that 8*20*30 foot swimming pool holds *about*
350,000 pounds of water. (am I still within 10%??
<shrug>)
One exercise I like doing with "kids" having problems
with math is to pick *something* (a bushel of apples,
a ceiling tile, a slab of cement in the sidewalk out
front, etc.) and say, "If this is 10, show me what
100 would be" (trivial example). The point is just
to get them to get an idea as to magnitudes and relative
scales, etc.
So, when the kid at the local grocery store rings up
your *two* $9.95 items and tells you the total is *$40*,
you can just smile and say "What's wrong with this
picture?"
--don