The Lost Art of Manually Calculating Square Roots
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http://www.ee.ryerson.ca/~elf/abacus/feynman.html
However, the correct Portuguese for "cube roots" is "raízes cúbicas", not "raios cubicos"(—edited with thanks to JetSpiegel for the correction!). I've thought that Feynman may have been better at speaking Portuguese than at spelling it, or that most of Surely You're Joking may have been transcribed from tapes by Ralph Leighton, who probably didn't speak Portuguese.)
Edit: Also, I was partly inspired to learn Portuguese by Richard Feynman. I didn't imagine that that would lead to complaining about his spelling. :-)
1. Pick a number HI whose square is obviously higher than N.
2. Pick a number LO whose square is obviously lower than N.
3. Choose a number MED which is roughly (HI+LO)/2.
4. If MED * MED > N, replace HI with MED, else replace LO with MED.
5. Go to 3.
6. MED converges to your answer.
What didn't you understand about manually calculating square roots? Because your method requires tons of troublesome multiplication of numbers that get longer with each step.
Why would you be both insulting and wrong in the same post?
Or did you think that multiplication somehow isn't manual?
Next time work on phrasing your comment a bit better. It makes the world a better place.
The math part had this weird American division symbol I had never seen before, and the closest thing I had seen was square root. Which I found a bit ambitious, but I went ahead and did the square roots.
They accepted me despite the fact that I couldn't do division, but put me in the "remedial math" class. Got out of it when I did the in-class "calculator exercises" (weird concept, that) in my head, faster than anyone with the calculator.
[1] [ https://en.wikipedia.org/wiki/Methods_of_computing_square_ro... ]
Start with a ball park estimate of the root of x, then take an average of (estimate + x / the estimate). That's your new estimate. Rinse and repeat until convergence to an acceptable degree.
The biggest drawback if doing this by hand is the division problem can be a pain.
I see the attraction and "purity" of the proposed lost art.
n-th roots are not quite as nice, but still work fairly simply: ((n-1) a + x/a^(n-1) )/n
You can see that the fixed point (a = nth root of x) works out as it should.
Calculators of any kind were forbidden at school, which sucked, if you had one of those cool Casio watch calculators. Shorts were also forbidden, because suffering is good for you, apparently.
I've never had to do it since, of course. Though I do now wear shorts on occasion.
FWIW, I did too in the 90s/00s, though I confess I don't remember the method now. I think it was Newton-Rhaphson.
Estimate sqrt(17):
Clearly it has to be between 4 and 5.
Try 4.5. Result: 20.25 Try 4.2. Result: 17.64 Try 4.1. Result: 16.81 Try 4.15. Result: 17.01. Probably good enough for most practical purposes.
In this case the value is sanity checking a calculator / program.
Were any memory tricks involved (e.g. pegs, palaces) or did he just go through a table of square roots every morning?
His justification for starting was trying to keep track of the order of magnitude of everything he calculated to catch fat finger mistakes. And a sanity check. However, you need a few digits on the front of a number before you can really safely keep track of exponents that way.
There are 12 numbers from 4.0 through 5.0 inclusive - in steps of 0.1, and there are 10 integer numbers from 16 through 25 inclusive - clearly there's an uneven, but still somewhat smooth mapping from one through the other via the square function (or root function, for that matter). It's easy to see (note that python rounds down up to .5 inclusive):
for i in range(40, 51):
n=i/10.0
print(n, round(n**2))
4.0 16
4.1 17
4.2 18
4.3 18
4.4 19
4.5 20
4.6 21
4.7 22
4.8 23
4.9 24
5.0 25
This is indeed a bit different when approximating a more complicated function, eg one that doesn't monotonically increase/decrease.
Since the term q for solving the remainder r for any step as in the binomial r = x2 - 4p2 - q(4p + q) is either 1 or 0 in a base-2 system, evaluating the "fit" becomes merely a matter of testing a boolean value. (Therefor we can go for a straight-forward algorithm. No need for interpolations, here.)
So 4^2 = 16 is under by 1, add 0.1. 4.1^2 = 16.81 is off by around .2, add 0.02. 4.12^2 = 16.9744 so add .003. You're basically adding a digit at a time without needing division.
17 is a bit over 5·5·2/3, which is a bit under 2^(1.4 + 1.4 + 1 – 1.7) = 2^4.1 [base 12], now to take the square root we get 2^2.06 ~ 4·35/34 [base 10], but we want something a bit higher, so for a round number to cancel the 4 we can guess 4·33/32 = 33/8 [base 10].
3 ~ 2^(19/12) -> lg 3 ~ 1;7 or to be more precise 1;703
5 ~ 2^(7/3) = 2^(28/12) -> lg 5 ~ 2;4 or to be more precise 2;3a4
This makes 2^0;7 just about exactly a “perfect fifth” 3/2 (off by about .1%), and 2^0;4 almost a “major third” 5/4 (off by .8%).
If you step up a scale by 2^(1/12) or in log scale 0;1 increments, you can find close fractional approximations to most of the steps. Then I happen to also know that 0;06 is ~lg 35/34, but that one obviously isn’t nearly as important or easy to work with.
That's because in Dutch we just call it 'root' (wortel).
I really wish teachers used more graphic forms in math because it helps to understand what kind of problems it can solve.
I am asking because in France we have the official term racine carrée (square root), abbreviated to racine (root) when the "square" is implied.
This is part of an excursus trying to explain, how basic math was implemented on the DEC PDP-1, namely in Spacewar!, the first digital video game (1961/62). The excursus starts at http://www.masswerk.at/spacewar/inside/insidespacewar-pt6-gr... (there's a tab on the bottom right for an instruction list of the PDP-1, mind that the PDP-1 uses 1's-complement for negative numbers).
This describes how to find the side length of the square, who's area is smaller than the rectangle.
I don't think its important to the calculation though
: something usually requiring some skill that not many people do any more
(SNL had a whole bit on this back in the 90's: "a chickpea is neither a chick nor a pea, discuss")
https://en.wikipedia.org/wiki/Binary_search_algorithm
rather than representing numbers or performing arithmetic in base 2.
sqrt(7720.17)
first guess 50:
2500
Go up
second guess 100:
10000
too high go down
third guess 75
5625
too low go up
fourth guess 86 (for ease of calculation)
6400 480 + 480 36
7396
too low go up: 93
8529
too high go down: 90
8100
too high go down: 88
7744
Etc.
It's not a great method, but it's what first came to mind. I do it sometimes in my head for a quick estimate.
7720 in binary is 1111000101000. This is 13 bits. Square root is at most half as long, so first guess is 10000000 in binary, or 128.
Square that, realize that it's too high, set the highest bit to 0 and next bit to 1.
Square 1000000 binary or 64, realize that it's too low. Keep the top bit, raise the next bit.
Square 1100000 binary or 96, too high, drop that bit back to zero, raise the next bit.
Square 1010000 binary or 80, too low, raise the next bit.
Square 1011000 binary or 88, too high, drop this bit and raise the next one.
Square 1010100 binary or 84, too low, raise the next bit.
Square 1010110 binary or 86, too low, raise the next bit.
Got 1010111 binary or 87.
I've never heard of anyone below the age of 30 learning in school how to do them manually.
(Of course, I learned how to calculate square roots on paper before I went to college, but only because I asked my father to show me how.)
nyah!