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Topic: Root
Replies: 21   Last Post: Jul 14, 1999 3:46 PM

 Messages: [ Previous | Next ]
 philnovi Posts: 11 Registered: 12/12/04
Re: Root (EXACT Square Root Algorithm)
Posted: Jul 12, 1999 6:24 PM

This method for square roots is EXACT for the number of digits that you have
the patience to use it to.

To view this file properly, convert it/view it with a 'non-proportional'
font (eg Courier New in Windows).

This method was formerly taught in primary school, but the algorithm seems
to be 'getting lost' today!!!
----------------------------------------------------------------------------
--------------------------------------------------------------------------

To find the square root of any number to any number of digits:
****************************************************************************

(This algorithm looks like long division, but a bit more tedious)

Write down the number:

1234567890.

Starting at the decimal point and working to the left, separate the number
into groups of two:

12 34 56 78 90.

leave lots of room to the left:
12 34 56 78 90

________________
| 12 34 56 78 90

1. What is the largest number squared that will subtract from the first
group?
(It is obviously 3 here)
Square this number and perform the subtraction:
3
________________
| 12 34 56 78 90
-9
___
3
(This sequence of digits above the bar will be referred to as the 'top
row'.)

2. Drop down the next group of two digits:

3
________________
| 12 34 56 78 90
| -9
| ___
| 3 34

3. Double the top row and bring this doubled value down to the left,
(we'll call this the left row) leaving a placeholder for a less significant
digit:

3
________________
| 12 34 56 78 90
| -9
| ___
6_| 3 34

4. Now, we need to find the one digit such that the product of this digit
and
the current left row with this digit appended is just less than the result
of
the last subtraction. Put this digit in the placehoder and append it to the
top row.
( i.e. in this particular case, the digit is 5, since 5*65<334, but
6*66>334)
Perform this multiplication and subtraction:

So we now have:
35
________________
| 12 34 56 78 90
| -9
| ___
65| 3 34
3 25
____
9
5. Loop starting at step 2 until you have achieved the desired precision.

(continuing the process so you can see the result develop:)

35
________________
| 12 34 56 78 90
| -9
| ___
65| 3 34
| 3 25
| ____
Step 2: 9 56

35
________________
| 12 34 56 78 90
| -9
| ___
65| 3 34
| 3 25
| ____
Step 3: 70_| 9 56

Step 4: Next digit is 1 since 1*701<956, but 2*702>956)

351
________________
| 12 34 56 78 90
| -9
| ___
65| 3 34
| 3 25
| ____
701| 9 56
| 7 01
| ____
| 2 55

continuing on:

351
________________
| 12 34 56 78 90
| -9
| ___
65| 3 34
| 3 25
| ____
701| 9 56
| 7 01
| ____
702_| 25578

3513
________________
| 12 34 56 78 90
| -9
| ___
65| 3 34
| 3 25
| ____
701| 9 56
| 7 01
| ____
7023| 25578
21069
_____
4509

3513
________________
| 12 34 56 78 90
| -9
| ___
65| 3 34
| 3 25
| ____
701| 9 56
| 7 01
| ____
7023| 25578
| 21069
| _____
7026_| 450990

35136
________________
| 12 34 56 78 90
| -9
| ___
65| 3 34
| 3 25
| ____
701| 9 56
| 7 01
| ____
7023| 25578
| 21069
| _____
70266| 450990
421596
______
29394.00

so, at this point, to an approximation, sqrt(1234567890) is 35136.

You are only limited by your patience as to how many digits you require.

Thomas Chrapkiewicz 06ap98

----------------------------------------------------------------------------
--------------------------------------------------------------------------

User wrote in message ...
>
>Does anyone know how to go about doing a square root or cube root
>calculation?
>
>If I gave you a number (no calculators), and a blank piece of paper,
>can you describe the steps one by one how you go about calculating
>the cube or square root? (or any root for that matter).
>
>I also assume I can say... I want it to a precision of 80 decimal points.
>You would go about using your method and you would come up with
>80 decimal points (or less if it rounds off perfectly).
>
>Where can I find procedures on doing this? Is there a very simply
>way to do this? using basic operators (no log sin cos, etc).
>
>Thank you.
>
>Thank you.
>
>
>

Date Subject Author
7/10/99 User
7/10/99 Prasanth A. Kumar
7/11/99 William L. Bahn
7/11/99 Kaimbridge
7/11/99 carel
7/11/99 chri0562@sable.ox.ac.uk
7/11/99 spamless@nil.nil
7/12/99 test@test.com
7/14/99 Chris J. Bennardo
7/12/99 John Popelish
7/12/99 philnovi
7/12/99 Ken Cox
7/12/99 randy
7/13/99 John Savard
7/13/99 Paul J. Gans
7/13/99 Jack Silver
7/13/99 Paul Schlyter
7/13/99 philnovi
7/12/99 Gareth McCaughan
7/12/99 randy