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:
 Guess
 Divide
 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(t1)x^{2}/2!
+ t(t1)(t2)x^{3}/3! + . . .
+ t(t1)...(tk+1)x^{k}/k! + . . .
where t is real.
Notice that to get the x^{k+1} term of the series from the
x^{k} term, just multiply by (tk)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[(b1)/c,b/(c+1)]^{2} < a < (b/c)^{2},
max[(b1)/b,c/(c+1)]^{2} < a(c/b)^{2} < 1.
This means that b/c is a reasonable approximation to a^{1/2}, and
a little larger than it. Let d = a(c/b)^{2}. Then
d^{1/2}b/c = a^{1/2}, and the problem is reduced to
finding d^{1/2}
where
1/4 <= max[(b1)/b,c/(c+1)]^{2} < d < 1.
Now let x = d  1.
Then at worst 3/4 < x < 0. Now apply the infinite series:
d^{1/2} = (1+x)^{1/2}
= 1 + (1/2)x/1!
+ (1/2)(1/2)x^{2}/2!
+ (1/2)(1/2)(3/2)x^{3}/3!
+ (1/2)(1/2)(3/2)(5/2)x^{4}/4!
+ (1/2)(1/2)(3/2)(5/2)(7/2)x^{5}/5! + . . .
= 1 + (1/2)x
 (1/8)x^{2}
+ (1/16)x^{3}
 (5/128)x^{4}
+ (7/256)x^{5}
 (21/1024)x^{6}
+ (33/2048)x^{7}  ...
This converges faster than a geometric series with common ratio x, so the
number of terms to give mplace accuracy is at most
m/log_{10}(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 d^{1/2}, you can find
a^{1/2} = d^{1/2}b/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
28^{2} = 784 < 821 = a < 841 = 29^{2}.
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
821^{1/2} = (0.9880378470)29
= 28.65309756
To four decimal places, 821^{1/2} = 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 (b1)/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 < 100^{e} < a.
If a > 1,
e = [[(n1)/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 = [[(m1)/2]] < 0, where m is
the number of zeroes in a between the decimal point and the first
significant digit. Then ac^{2} lies between 1 and 100.
Pick b such that b^{2} is
the next square larger than ac^{2}. 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
ac^{2} = 8.21, so b = 3 works, and
20 = (b1)/c < a^{1/2} < b/c = 30.
Then d = ac^{2}/b^{2} = 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 = 10^{4}, so a*c^2 =
37.9. Then b = 7 will do, and d = 37.9/49 = 0.7734694, so x =
0.2265306 Then
d^{1/2} = 1  0.1132653
 0.0064145
 0.0007265
 0.0000163
 0.0000028
 0.0000005
 0.0000001
= 0.879471
and therefore
a^{1/2} = 0.879471*7/10000,
0.000000379^{1/2} = 0.0006156297,
approximately.
There are more answers to this question in the Dr. Math archives. See:
For square roots in general, see
Cube Roots
From the Dr. Math archives:
