A Method Of Deriving Cube Roots: Part 1 3Root001.rtf p1 of 5
This method is illustrated at https://www.youtube.com/watch?v=Mv0FUD7AiI0
Unfortunately, the math tutor is speaking in a foreign language (Hindi?) so my under-
standing is based only on copying down the visual presentation.
Note that unlike most of the math clickbait concerning root extraction, this method can
be made to work for any cube value.
But there are two apparent limitations -
First: This method as demonstrated in the video requires the location of the nearest
perfect cube. This is because the method uses the entire cube value at once rather
than dividing it into three-digit periods. So without an extensive list of perfect cubes
the method would appear to be inconvenient at best.
Second: The method has error creep which increases with the distance the problem
cube is from the perfect cube.
But there is a solution which solves the second issue nicely and that is to repeat the
method as many times as needed using the output of the current pass as input to the
next until the required precision and accuracy is reached. This also allows us to use any
known perfect cube above or below the problem cube.
This file covers the method of finding the nearest perfet cube while the file 3Root002.rtf shows how
perfect cubes based on powers of 10 can be used instead of trying to find or manufacture
the closest perfect cube to the problem value.
Problem: Extract 3√n
Method: 1.) Find the nearest perfect cube (limit) above or below the problem value (n).
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
Example 1: Extract 3√63 n =63
1.) Nearest perfect cube (above) 43 =64; limit =64
2.) n1 =4
3.) diff =64 -63 =1
4.) (n1) (diff) = (4) (1) = 4 =0.0208333
3 (limit) (3) (64) 192
adjustment (a) =0.0208333
5.) n is less than the limit so root =n1- a =(4.0 - .0208333) =3.9791666
3√63 on a TI-34 =3.979057208
3Root001.rtf p2 of 5
Example 2: Extract 3√63 n =63
In this example the lower perfect cube will be used. The calculations are the same but
the values will be different. This example is only to show you the error creep as diff in-
creases. If perfect cubes are used to solve the problem then you would always go to
the nearest perfect cube as we did in example 1.
n =63
limit =33 =27 (use 33 instead of 43)
n1 =3
diff =(63-27) =36
The end calculations are :
(n1) (diff) = (3) (36) = 108 =1.33333333 (a)
3 (limit) (3) (27) 81
adjustment =1.333333
n is greater than the limit so root =n1 + a =(3.0 +1.333333) =4.333333
3√63 on a TI-34 =3.979057208
0.354275792 is quite a bit off from 3.979057208. ( 8 percent off actually. )
Discrepancies such as this (and even the tiny one in example 1) can be shaved finer
and finer by modifying the inputs to the method.
One way to do this is by "manufacturing" a cube value (the limit) which isn't a "perfect"
cube but is closer to the problem value n.This is shown in example 2a.
A similar way is to run the sequence again using the cube of the resulting inaccurate root
estimate (n13) as a new limit. In example 2b we'll take the root result above (4.33333) and
cube it to get 81.37018 and use this cube as a new limit. It doesn't need to be a whole
integer like 27 or 64 to be useful as long as we have a reasonably accurate cube root of
the limit.
3Root001.rtf p3 of 5
Example 2a: Extract 3√63 (Start by setting a new lower limit closer to n)
As we saw in example 2, the error creep increases the farther away n is from the
next perfect cube. This is expressed as the diff. In example 1 we had a minimal diff
because we used 43 =64 as the perfect cube with a diff of 1 (64 -63). This was ideal
and the error discrepancy was very acceptable.
In this contrived example a value for limit will be selected that results in a lower value
for diff than example 2.
The cube of 3.5 is 42.875. This gives us a limit which is closer to 63 than was 27 but
not as close as was 64.
Again:
n=63
limit =3.53 =42.875
n1 =3.5
diff =(63 -42.875) =20.125
The end calculations are:
(n1) (diff) = (3.5) (20.125) =70.4375 =.547619 (a)
3 (limit) (3) (42.875) 81
adjustment =.547619
n is greater than the limit so root =n1 + a =(3.5 +.547619) =4.047619
3√63 on a TI-34 =3.979057208
Using 33 in example 2 resulted in an over-error of 0.35427 or about 8%
Using 3.53 reduced the error to 0.068561 or about 1.7%
So as limit cube values are made closer to n, error creep is reduced.
3Root001.rtf p4 of 5
Example 2b. 3√63 (Repeat the procedure using output values as input for next pass.)
Example 2 produced a root of 4.333333 which was unacceptably inaccurate. This was
addressed in example 2a by selecting a cube closer to n. In this example we'll take the
resulting (inaccurate) cube root of 4.333333 and cube it to create a new limit of
81.370182. Then we repeat the method using these new values.
n=63
limit =4.333333 =81.370182
n1 =4.33
diff =(81.370182 - 63) =18.370182
The end calculations are:
(n1) (diff) = (4.33) (18.370182) =79.542888 =.0.325847 (a)
3 (limit) (3) (81.370182) 244.110546
adjustment =0.325847
n is less than the limit so root =n1 - a =(4.3333 -0.325847.) =4.0007153
3√63 on a TI-34 =3.979057208
4.33333 was off by ~8% from 3.979057208
4.0007153 is off by ~0.5% from 3.979057208
Better but still not as good as example 1
Third Pass:
n=63
limit =4.00071533 =64.034340
n1 =4.00
diff =(64.034340 - 63) =1.034340
The end calculations are:
(n1) (diff) = (4.00) (1.034340) =4.1373621 =0.0215304
3 (limit) (3) (64.034340) 192.16362
adjustment =0.0215304
n is less than the limit so root =n1 -a =(4.0007153 -0.0215304) =3.9792119
3√63 on a TI-34 =3.979057208
4.33333 was off by ~8% from 3.979057208
4.0007153 is off by ~0.5% from 3.979057208
3.9792119 is off by ~0.003% from 3.979057208
3Root001.rtf p5 of 5
Fourth Pass:
n=63
limit =3.97921193 =63.00734796
n1 =3.9793119
diff =(63.00734796 - 63) =0.007347957
The end calculations are:
(n1) (diff) = (3.9792119) (0.007347957) =0.029239077 = 0.000154686
3 (limit) (3) (63.00734796) 189.0220439
adjustment =0.000154686
n < limit so root =n1 -a =(3.9792119 -0.000154686) =3.979057214
3√63 on a TI-34 =3.979057208
4.33333 was off by ~8% from 3.979057208
4.0007153 is off by ~0.5% from 3.979057208
3.9792119 is off by ~0.003% from 3.979057208
3.979057214 is off by ~0.0000006% from 3.979057208
All this work for something that's virtually useless. But it was fun.
afk October 18, 2018
afk May 14, 2024