Just for fun - my various calculators are all in working order - But how many of you remember how to calculate square roots without a calculator and without an antique math book? I, myself, have long forgotten.
I 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. (] 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.
Just something I pulled from the top of my head.
I hope I’m right.
When I went to school it was called ‘arithmetic’ and a ‘square route’ was the sidewalk around a city block.
I know what you mean. When I went to school, they hadn’t invented the number “7”, yet. We just counted to six…hummed a little bit…then started again, at “8”. It was really tough, especially that long period between my 6th and 8th birthday…when I wasn’t anything.
I’ll never forget that feeling.
James, you have completely lost me! Are all those numbers for real? If I ever understood any of this, I was a lot smarter 45 years ago than I am now - in math, at least. My husband often reminds me that I sat behind Jesus in 2nd grade.
They are for real. I googled them, myself.
Find all the primary number factors of the number. Group all the pairs of numbers together, multiple one copy of each pair and pull it outside the square root, multiple everything left over and put it under.
Square root of 99
99 * 3 = 33
33 * 3 = 11
11 * 11 = 1
So 3 * 3 * 11 = 99. We have a pair of 3’s so pull one out and leave the 11 under. We end up with 3 root 11.
Dom, you don’t count. You have only been out of high school a couple years. Thread is …’’ For Us Old Folks" Like Bushart…
Ahh, for the old ‘folks’. Sorry, I taught HS math for several years as well and couldn’t resist