A Method Of Deriving Cube Roots: Part 2 3Root002.rtf p1 of 5
For this method to work it's not necessary to determine the nearest perfect
cube to the problem cube. It's enough to use cubes based on the powers of
ten. One way to know when we're done is to follow the amount of adjustment
after each pass. If the amount is trivial enough that the current value has the
precision we want then we're done.
Problem: Extract 3√n
Method: 1.) Find the nearest perfect cube (limit) to the problem value (n) which
is a power of ten.
2.) The root of this perfect cube =n1
3.) Find the difference (diff) between n and limit. This will be n -limit or limit -n
depending on whether n is above or below limit; diff will be a positive
integer.
4.) Calculate (n1) (diff) to get the adjustment. (a)
3 (limit)
5.) If n > limit then root =n1+ a
If n < limit then root =n1 - a
The example will derive the cube root of a value which is near the opposite end of the
range of perfect cubes. This will show the error creep move from near maximum in
the beginning to near zero after four passes. A much more sane approach would be
to choose the power of ten below 234567 (1000) because it's much closer to n.
Example : Extract 3√234567 n =234567
Pass 1:
1.) Nearest power-of-ten perfect cube above 234567 is 1,000,000
limit =1 000 000
2.) n1 = 3√1 000 000 =100
3.) diff =(1 000 000) -(234 567) =765433
4.) (n1) (diff) = (100) (765 433) =76 543 300 =25.51443333 is the adjustment.
3 (limit) (3) (1 000 000) 3 000 000
adjustment (a) =25.51443333
5.) n is less than the limit so root =n1- a =(1 000 000) -(25.51443333) =74.48556667
Without resorting to a calculator or a longhand derivation of the root we don't know
at this point how close we are to an acceptable value. So we'll use the outputs of
pass 1 as inputs into pass 2. We'll assume that the estimated root is still too high
after just one pass so we'll create the new high limit by cubing this estimated root.
Our estimated root of 74.48556667 becomes our new n1
n13 which is 413 256.3457 becomes our new limit
(413 256.3457) -(234567) =178689.3457 becomes our new diff
Now, make another pass.
3Root002.rtf p2 of 5
Pass 2:
1.) limit =413 256.3457
2.) n1 =74.48556667
3.) diff =178689.3457
4.) (n1) (diff) = (74.48556667) (178689.3457) =10.7356909 is the adjustment
3 (limit) (3) (413 256.3457)
5.) n is less than the limit so root =n1- a =(74.48556667) -(10.7356909) =63.74987577
Given the large difference in the estimated root between pass one and pass two it
seems we aren't anywhere near a stable value yet. So make another pass.
Pass 3:
Our estimated root of 63.74987577 becomes our new n1
n1 3 which is 259082.4697 becomes our new limit
(259082.4697) -(234567) =24 515.4697 becomes our new diff
1.) limit =259 082.4697
2.) n1 = 63.74987577
3.) diff =24 515.4697
4.) (n1) (diff) = (1 562 858.151) =2.010 760 194 is the adjustment
3 (limit) 777 247.4091
5.) . . . root =n1- a =(63.74987577) -(2.010 760 194) =61.73911558
This still is a significant drop from the last estimated root so make another pass.
Pass4:
Our estimated root of 61.73911558 becomes our new n1
n1 3 which is 235 332.1224 becomes our new limit
(235 332.1224) -(234567) =765.122401 becomes our new diff
1.) limit =235 332.1224
2.) n1 = 61.73911558
3.) diff =765.122401
4.) (n1) (diff) = 47 237.97757 =0.06690966 is the adjustment
3 (limit) 705.9963672
5.) . . . root =n1- a =(61.73911558) -(0.06690966) =61.6722059
We had no way of knowing how close we were after pass 3 but because of the
miniscule change after this pass we can conclude that we are very close. The cube
root of 234567 on a calculator shows 61.67213327.
Our last pass shows 61.6722059.
afk May 16, 2024