Estimating square roots

I recently came upon the following video, which shows a method to find the square root of a number from 100 to 10,000 as long as that number is a perfect square.


Direct: https://www.youtube.com/watch?v=d9BuWzlFoz8

However, it made me recall this Far Side comic strip. My apologies to Gary Larson if he finds this usage objectionable. I am just trying to promote math, which I think he would approve of.



I will just say for now that the square root of 5,248 is not a whole number.

What is a method to estimate it to the nearest whole number?

Here is what I'll call the Wizard method.

1. Start out by ignoring the last two digits, so we're left with 52.
2. As in the video above, find the largest number n where it's square is less than or equal to the number from step 1. In this case, n=7, because 7^2=49, which is the largest square under or equal to 52. Note that 8^2=64, which is over 52. That number will be the first digit(s). In this case 7.
3. Find the range of numbers that would map to this number. In this case, 4900 to 6399 would map to a first digit of 7.
4. Find the size of that range. Let's keep it simple and go from 4900 to 6400, which is a range of 1500.
5. Find the size of the range between the original number and the lower bound of the range in step 5. In this case, 5248-4900 = 348. However, since we're just estimating, let's round that up to 350.
6. Divide the result of step 5 by that of step 4. In this case 350/1500. We can reduce that to 35/150 and further to 7/30. It's easy to see this is a little less than 25%. Let's go with 24%.
7. Multiple the result of step 6 by 10 and round to the nearest integer. This will be the final digit in our estimate. In this case we get round(10*0.24) = round(2.4) = 2.
8. Multiply the result of step 2 by 10 and add the result of step 7. In this case we get 72.

The actual square root of 5,248 is 72.4431.

Let x by the highest number you have the square memorized. In my case that would be 20. This method will work for 100 to 100*n^2. In my case, 100 to 40,000.

Any comments? Suggestions for improvement?

Here's an approximation method:
1. Ignore the last two digits.
2. Find the largest number whose square <= the remaining digits
3. Multiply that number by 20 and add 5
4. Subtract (the square of the number from 2) x 100 from the original number, then divide by the number from 3.
5. The solution is (the number from 2) x 10 + (the number from 4)

For example, for 5248:
1. Ignore the 48; this leaves 52
2. The largest number whose square <= 52 is 7
3. 7 x 20 + 5 = 145
4. (5248 - 4900) / 145 = 2.4
5. The solution is 72.4

This is a simplification of the "one digit at a time" method I learned for solving square roots:
1. Break the number into "groups" of two digits on each side of the decimal point. For example, 5248 would be {52}{48}.{00}{00}{00}...
2. Start with the leftmost group, and find the largest number whose square <= the number in that group; in this case, 7
3. Subtract the square of that number from the group; 52 - 49 = 3
For each remaining group:
1. Multiply the "remainder" from the previous step by 100, and add the next group; 3 x 100 + 48 = 348
2. Multiply the partial solution, ignoring any decimal point, by 20; 7 x 2 = 140
3. Find the largest integer n such that (n + the number in 2) x n <= the number in 1
141 x 1 = 141; 142 x 2 = 284; 143 x 3 = 429 > 348, so n = 2
4. Add n to the partial solution: 72
5. Subtract the number in 3 from the number in 1: 348 - 284 = 64
Repeat until either the number in 5 is zero, in which case the square root is exact, or you have enough digits in your solution.
The next "group of steps" would be this:
1. 64 x 100 + 0 = 6400
2. 72 x 2 = 144
3. 1444 x 4 = 5776 <= 6400; 1445 x 5 = 7225 > 6400, so n = 4
4. The partial solution is now 72.4
5. 6400 - 5776 = 624

I like the Babylonian method

1) Make a reasonable estimate of 70 = a
2) 5248 / a = 74.9714 = b
3) (a + b)/2 = 72.4857 = c
4) 5248 / c = 72.400 = d
5) (c + d) / 2 = 72.443, already accurate to five digits

This method converges quickly, even with a poor initial estimate

Incidentally, I believe the first logarithm tables were made using this kind of method. If you need to know what x^ 0.3 is, x^(19/64) will get you close, x^(77/256) even closer. As they say, you don’t calculate logarithms, you find them

Quote: Ace2

Incidentally, I believe the first logarithm tables were made using this kind of method. If you need to know what x^ 0.3 is, x^(19/64) will get you close, x^(77/256) even closer. As they say, you don’t calculate logarithms, you find them

Interesting

Personally I find there is not too much to learn from "trickery" unless it leads to such an insight

On the other hand, what if somebody did hold a gun to your head and demand the square root of some number? Better know some tricks!

Quote: odiousgambit

Interesting

Personally I find there is not too much to learn from "trickery" unless it leads to such an insight

On the other hand, what if somebody did hold a gun to your head and demand the square root of some number? Better know some tricks!

Before calculators were invented this wasn’t trickery...it was how people did it. You could say they used slide rules or log tables, but the numbers printed on them must be calculated first

Maybe calculators still use this kind of method

Quote: Ace2

I like the Babylonian method

1) Make a reasonable estimate of 70 = a
2) 5248 / a = 74.9714 = b
3) (a + b)/2 = 72.4857 = c
4) 5248 / c = 72.400 = d
5) (c + d) / 2 = 72.443, already accurate to five digits

This method converges quickly, even with a poor initial estimate

This is an application of the Newton-Raphson method (also known as Householder's Method of Order 1), which actually applies to finding the roots of any function. The "quick version" is:
(a) Draw the graph y = f(x) where you want to find a root of the function; in this case, to find the square root of 5248, the function is f(x) = x^2 - 5248
(b) Choose an estimate for x
(c) Draw a line tangent to the graph at (x, f(x))
(d) Where the line crosses the x-axis is the new estimate for x
Repeat (c) and (d) until the value for x converges; that is the solution.

Note that step (c) is calculating x - f(x) / (df(x)/dx)
For f(x) = x^2 - 5248, df(x)/dx = 2x, and x - f(x) / (df(x)/dx) = x - (x^2 - 5248) / 2x = x - x / 2 + 5248 / 2x = (x^2 + 5248) / 2x = (x + 5248/x) / 2, which is step 3 (and 5) in Ace2's description.

Thank you for all the replies. I am still processing them.

Quote: Wizard

I recently came upon the following video, which shows a method to find the square root of a number from 100 to 10,000 as long as that number is a perfect square.

Direct: https://www.youtube.com/watch?v=d9BuWzlFoz8

However, it made me recall this Far Side comic strip. My apologies to Gary Larson if he finds this usage objectionable. I am just trying to promote math, which I think he would approve of.



I will just say for now that the square root of 5,248 is not a whole number.

What is a method to estimate it to the nearest whole number?

Here is what I'll call the Wizard method.

1. Start out by ignoring the last two digits, so we're left with 52.
2. As in the video above, find the largest number n where it's square is less than or equal to the number from step 1. In this case, n=7, because 7^2=49, which is the largest square under or equal to 52. Note that 8^2=64, which is over 52. That number will be the first digit(s). In this case 7.
3. Find the range of numbers that would map to this number. In this case, 4900 to 6399 would map to a first digit of 7.
4. Find the size of that range. Let's keep it simple and go from 4900 to 6400, which is a range of 1500.
5. Find the size of the range between the original number and the lower bound of the range in step 5. In this case, 5248-4900 = 348. However, since we're just estimating, let's round that up to 350.
6. Divide the result of step 5 by that of step 4. In this case 350/1500. We can reduce that to 35/150 and further to 7/30. It's easy to see this is a little less than 25%. Let's go with 24%.
7. Multiple the result of step 6 by 10 and round to the nearest integer. This will be the final digit in our estimate. In this case we get round(10*0.24) = round(2.4) = 2.
8. Multiply the result of step 2 by 10 and add the result of step 7. In this case we get 72.

The actual square root of 5,248 is 72.4431.

Let x by the highest number you have the square memorized. In my case that would be 20. This method will work for 100 to 100*n^2. In my case, 100 to 40,000.

Any comments? Suggestions for improvement?

I have to try it later when I can concentrate. But I will say we were never taught in school and always wondered how to do it so if it works thanks in advance, if it doesn't my teacher lied when she said I will not always have a calculator thanks to the smart phone.

If it works I will buy you a beer so we can discuss if math is natural or synthetic.