Conference rusure::math

    Obviously, Newton's method is not well adapted to human use.
    I'll have to go get my 7th grade math book.
    Thanks for the problem.
    

    Well,
    
        here is my solution to the square root of 79 by hand.
    
    
         79    |  8.9700
      -  64    |--------------------------------------------
      -------- |  8 |  8.9  |   8.97 |   8.970 |   8.9700
         15    |  8 |   .9  |     .7 |       0 |        0
       -  8.01 |----|-------|--------|---------|----------
      -------- | 64 |  8.01 |  6.279 |       0 |        0
       6.990    
    -  6.279
      --------
       0.711
    

    This means that my answer is:  8.9700
    and                           (8.9700) * (8.9700) = 80.4609
                                  which is greater than 79.
    
    Enjoy,
    
    Kostas G.
    

    Re .1, .2:
    
    > Obviously, Newton's method is not well adapted to human use.
    
    I can assure you Newton did not design the method for computers.  :-)
    
    There is an algorithm for taking square roots which is akin to the
    algorithm for performing long division.  I was hoping somebody still
    remembered such things, but if nobody enters it in a few days, I will. 
    
    
    				-- edp

    I think Gilbert got lucky on that one.  I don't think it can be
    made into a general method.  Here is the algorithm I learned:
    
    Write the number whose square root is to be taken and determine
    the first digit:
    
           8.
         ____________________________
    	V 79
    
    Square the first digit, subtract it from the number, and add two
    more zeroes:
             
           8.
         ____________________________
    	V 79
          64
         ---
          15 00
    
    Double the current root, to get 16, and find the largest digit d such
    that 16d times d is less than 1500.  169 times 9 is 1521, but 168
    times 8 is 1344.  Subtract this and add two more zeroes.
             
           8. 8
         ____________________________
    	V 79
         ...
          15 00
          13 44
          -----
           1 56 00
    
    Again, double the current root, to get 176, and find the next digit.
    1769 times 9 is 15921 but 1768 times 8 is 14144.  Subtract again.
             
           8. 8  8
         ____________________________
    	V 79
         ...
           1 56 00
           1 41 44
           -------
             15 56 00
    
    Repeat this one more time, to find that 17765 times 5 is 88825,
    so the next digit is at least five, and we should round up the second
    digit, so the answer to two digits is 8.89.
    
    
    				-- edp

   The root of 79 is about 9, but that's a bit high, so the root is
   really about (9-x);  estimate x by squaring, (9-x)**2 = 81-18*x+x**2 = 79,
   or x = 2/(18-x), giving a rapidly converging sequence of approximations,

   x1 = 2/18 = 1/9,
		sqrt(79) ~= 9-1/9 = 8.88888888...,

   x2 = 2/(18-x1) = 2/(18-1/9) = 18/(162-1) = 18/161,
		sqrt(79) ~= 9-18/161 = 9*(1-2/161),

   x3 = 2/(18-x2) = 2/(18-18/161) = 1/(9*(1-1/161)) = (161/160)/9,
		etc

   the second x is already close enough.  You can see how a continued
   fraction results.

   I used a binary equivalent of the pencil and paper method mentioned
   above in the RT-11 Fortran square root routine in the floating point
   package (looooong ago).  It was actually faster than using
   Newton's method using fixed point multiply divide hardware (but loses
   if you have floating point of course).

   When I was a kid I remember using a trick to get a couple of additional
   digits out of the pencil and paper root method - just divide the remainder
   with twice the current estimate of the root on my slide rule and stick
   the digits on the end.

   There are a class of 'pseudo multiplication' algorithms for calculating
   logs and doing rotations (sometimes referred to as 'CORDIC' methods)
   that have a little flavor of the pencil and paper method.  So much so that
   I think I read that Briggs or Napier used such techniques for calculating
   their tables.

    - Jim

No, it wasn't luck, except that the numbers turned out to be easy to handle.

I took my chances posting 485.4 before checking with la computer.  Apparently,
there are no silly errors:

	sqrt(79) = 8.88819441732	(by hand)
	sqrt(79) = 8.888194417315588850	(VAX Basic, using /Hfloat)

Oddly, *generating* the answer by hand seemed like less work than would
be required to square the answer to verify the result!

    Using the first five terms of the continued fraction for sqr(79),
    namely [8 1 7 1 16 . . .], we get
    
    	sqr(79) = 1351/152 as a best approximation.
    
    Squaring this is 78.9994.
    
    This is a "best" approximation because it's the most accurate fraction
    possible with such a small denominator.
    
    /Eric

                                                   
    When I saw the basenote, I decided to try to do the problem WITHOUT
    paper.  I succeeded, though I admit that I read the replies before
    I entered this note.  Here's the method I used:
    
    I first noted that 9^2 = 81, and did a first-order approximation
    	off of 9 to get the first digit.
    Since 79 = (9 - d)^2 ~= 81 - 18d = 81 - 79 = 2,
    	I expect by first-order approximation that d = 2/18 ~= 0.1 .
    Thus the square root is approximately 9 - 0.1 = 8.9 .
    8.9^2 = 79.21 = 79 + 0.21 (I did that in my head).
    8.9^2 - 8.8^2 = 0.89 + 0.88 = 1.77 (one of my favorite shortcuts).
    By linear interpolation:  8.90^2 = 79 + .21
    			      8.8x^2 = 79
    			      8.80^2 = 79 - (1.77 - .21)
    x is about 1/10 of the way down, so it's a 9, and the answer is
    8.89 .
    
    Pretty slimy, huh?
    John