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  Square Roots Without a Calculator  

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How do you find the square root of a number by hand?
What about cube roots?

The square root of a number is just the number which when multiplied by itself gives the first number. So 2 is the square root of 4 because 2 * 2 = 4.

Start with the number you want to find the square root of. Let's use 12. There are three steps:

  1. Guess
  2. Divide
  3. Average.

    ... and then just keep repeating steps 2 and 3.

First, start by guessing a square root value. It helps if your guess is a good one but it will work even if it is a terrible guess. We will guess that 2 is the square root of 12.

In step two, we divide 12 by our guess of 2 and we get 6.

In step three, we average 6 and 2: (6+2)/2 = 4

Now we repeat step two with the new guess of 4. So 12/4 = 3

Now average 4 and 3: (4+3)/2 = 3.5

Repeat step two: 12/3.5 = 3.43

Average: (3.5 + 3.43)/2 = 3.465

We could keep going forever, getting a better and better approximation but let's stop here to see how we are doing.

    3.465 * 3.465 = 12.006225

That is quite close to 12, so we are doing pretty well.

Square Roots Using Infinite Series

Another way of computing square roots is to use the Binomial Theorem. The most general form of this gives the following infinite series:
(1+x)t = 1 + tx/1! 
           + t(t-1)x2/2!
           + t(t-1)(t-2)x3/3! + . . .
           + t(t-1)...(t-k+1)xk/k! + . . .
where t is real.

Notice that to get the xk+1 term of the series from the xk term, just multiply by (t-k)x/(k+1).

This only converges if |x| < 1. You want to apply this to taking the square root of any positive real number a. That means that t = 1/2. One way to proceed is as follows.

Find positive rational numbers b and c such that

max[(b-1)/c,b/(c+1)]2 < a < (b/c)2,
max[(b-1)/b,c/(c+1)]2 < a(c/b)2 < 1.
This means that b/c is a reasonable approximation to a1/2, and a little larger than it. Let d = a(c/b)2. Then d1/2b/c = a1/2, and the problem is reduced to finding d1/2 where
1/4 <= max[(b-1)/b,c/(c+1)]2 < d < 1.  
Now let x = d - 1. Then at worst -3/4 < x < 0. Now apply the infinite series:
d1/2 = (1+x)1/2
= 1 + (1/2)x/1! + (1/2)(-1/2)x2/2! + (1/2)(-1/2)(-3/2)x3/3! + (1/2)(-1/2)(-3/2)(-5/2)x4/4! + (1/2)(-1/2)(-3/2)(-5/2)(-7/2)x5/5! + . . .
= 1 + (1/2)x - (1/8)x2 + (1/16)x3 - (5/128)x4 + (7/256)x5 - (21/1024)x6 + (33/2048)x7 - ...
This converges faster than a geometric series with common ratio -x, so the number of terms to give m-place accuracy is at most m/log10(-x). If convergence is too slow for your taste, go back and pick a different b and c with larger values. This will reduce the size of -x. Once you have d1/2, you can find
a1/2 = d1/2b/c.

Example: Find the square root of 821 to four decimal places.

We find that one choice of b and c is b = 28, c = 1, because

282 = 784 < 821 = a < 841 = 292.
Then we will find the square root of d = 821(1/29)2 = 821/841. x = d - 1 = -20/841 = -0.0237812128. Using the series, we find that
(821/841)1/2 = 1 - 0.0118906064 
                 - 0.0000706933 
                 - 0.0000008406
                 - 0.0000000125 
                 - 0.000000002

            = 0.9880378470

     8211/2 = (0.9880378470)29 

            = 28.65309756

To four decimal places, 8211/2 = 28.6531.

Choosing perfect squares on each side of x works very well for large a, like the 821 used in the example, but very poorly for 0 < a < 1.

One just needs to start with some pretty good rational approximation b/c, larger than the square root, so that both (b-1)/c and b/(c+1) are smaller than the square root. The better the approximation, the better the convergence rate. That usually means the larger b and c are chosen, the better the rate. One way to choose b and c is to let c be a power of 10, c = 10-e, with e an integer such that a/100 < 100e < a.

If a > 1, e = [[(n-1)/2]] >= 0, where n is the number of digits in a to the left of the decimal point. ([[x]] here means the greatest integer no larger than x.)

If a < 1, then e = [[(-m-1)/2]] < 0, where m is the number of zeroes in a between the decimal point and the first significant digit. Then ac2 lies between 1 and 100.

Pick b such that b2 is the next square larger than ac2. Then 2 <= b <= 10. This requires knowing the squares of numbers in this range. This choice of b and c works for every a.

Using this method for a = 821, e = 1 and c = 1/10. Then we get ac2 = 8.21, so b = 3 works, and

 
20 = (b-1)/c < a1/2 < b/c = 30.
Then d = ac2/b2 = 821/900 and d - 1 = x = -79/900 = -0.0877777777, which is plenty small enough to give rapid convergence. Picking b = 29 and c = 1 gave better convergence, since then x = -0.02378.

Example:

Here is an example with a small value of a. To find the square root of a = 0.000000379, you can choose c = 104, so a*c^2 = 37.9. Then b = 7 will do, and d = 37.9/49 = 0.7734694, so x = -0.2265306 Then

d1/2 = 1 - 0.1132653
         - 0.0064145
         - 0.0007265
         - 0.0000163
         - 0.0000028
         - 0.0000005
         - 0.0000001

      = 0.879471
and therefore
          a1/2 = 0.879471*7/10000,

0.0000003791/2 = 0.0006156297,

approximately.



There are more answers to this question in the Dr. Math archives. See:

For square roots in general, see



Cube Roots

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