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What is the point of this section ? This is an encyclopedia, not a bunch of random stuff from high school math problems. I think this part should be deleted.
User:Shen has replaced the old picture with a newer one (that is probably better). Ah well... I don't like images (especially ones I made) to disappear into oblivion. :-) — Ambush Commander(Talk) 17:03, 29 January 2006 (UTC)
Both the new and the old image suffer from the same problem, the numbers on the axes are too small. Would be nice if the picture is redrawn with biggerfont sizes. Oleg Alexandrov (talk) 03:29, 30 January 2006 (UTC)
dxvs\fg ]\]u90-[u90] uj[u90- —Preceding unsigned comment added by 86.154.161.191 (talk) 19:38, 11 September 2008 (UTC)
I found a cube root method in a book on Vedic Mathematics. There is an algebraic digit schedule. The method is simple enough to be done by mental math. If one has already memorized the cubes in the first twelve digits then you can obtain the first and last digit on sight of the perfect cube. If you have the digit schedule then you can calculate the middle digits. Example: four Cube Root digits: HJKL ==> Start with the units digit in the root, L, as the units digit in the cube root of L3 in the number.
A Hindu woman traveled to San Jose, California about 20 years ago and beat a computer at the 23rd root of a 50?-digit number in a demonstration of mental math and Vedic algorithms. Apparently she had the general formula. More later. Larry R. Holmgren 09:40, 1 March 2007 (UTC)
It is for sure that the issue on the extremely simple and general high-order arithmetical root-solving methods shown in the aforementioned webpage: http://mipagina.cantv.net/arithmetic/rmdef.htm are of so much relevance to the Cube-Root's article and its history, indeed.
The history of root-solving is plenty of Tricks and Patches (as for example: those Vedic tricks you have brought to light), however the true is that no natural arithmetical methods habe been developed --since Babylonian times up to now-- by far similar to these new general high-order methods shown in the aforementioned webpage, and frankly this is so worrying. Author: D. Gomez.
Cube Roots of exact cubes, mainly by inspection and argumentation, page 316-317. Place-value analysis of a three-digit cube root, cba, page 318. Analytic sorting of the expansion into place-values, the three-digit cube root (cba) schedule of subtractions for cube roots of a given number, page 319, 321. First divide out any factors of 8, or 64, or 512, from the number to remove ambiguity in digit selection, page 323. Knowing the cubes of the first 10 counting numbers enables one to match the units digit of the given number (the perfect cube) to the cube root. A similar analysis is done to infer the next digit, b, by seeing the right-most digit in the cube. If the remainders grow too large, go back and increase the quotient digit. The following step's deductions will be larger.
What book is this coming from?Nickalh50 (talk) 12:23, 11 February 2011 (UTC)
The units place is determined by a3. (Subtract it to eliminate the units digit.) The tens place is determined by 3a2b. The hundreds place is formed by 3ab2 + 3a2b. The thousands place is formed by b3 + 3b2c. The ten-thousands place is given by 3ac2 + 3b2c. The hundred-thousands place is contributed by 3bc2. The millions place is constituted by c3.
What is the cube root of the perfect cube —The preceding unsigned comment was added by Larry R. Holmgren (talk • contribs) 00:56, 6 March 2007 (UTC).
Given the cube of a number, we work from right to left. To determine the units place, L, examine the units digit in the cube. Deduct L3. The tens digit is K. Evaluate and analyse 3L2K. Deduct 3L2K. For J, the hundreds digit, evaluate and analyse 3L2J + 3LK2. Deduct 3L2J + 3LK2. For H, the thousands digit, evaluate and analyse 3L2H + 6LKJ + K3. And 3L2H + 6LKJ + K3 is the portion to deduct.
Four-digit cube root, HJKL Method
1. Subtract (L3) to eliminate the units digit. We have L by the final digit pattern in the cube.
2. Subtract (3L2K) to eliminate the penultimate digit to find the cube root digit K.
3. Subtract (3L2J+3LK2) to eliminate the pre-penultimate digit to find the cube root digit J.
4. Subtract (3L2H+6LKJ+K3) to eliminate the prior digit to find the cube root digit H.
Example: Find an incompletely given perfect cube. Find the cube root of a 12-digit perfect cube whose last four digits are 6741. Thus, on sight, we known that the number of cube root digits = 12/3 = 4; Root digits H = ? and L = 1. Framework: ___,___,___6,741. = (HJKL)3 For L : L=1, Therefore L3 = 1. Subtract it. ………6741 For K: Deduct 3L2K = 3K (ends in 4). ……-……1 The 3rd multiple of 8 ends in 4. K=8. Deduction:3K=24. ……-…24 For J: Deduct 3L2J+3LK2 = 3J+192 (ends in 5). ………65x So, 3J ends in 3, J=1. Deduction = 3+192=195. ……-…195 For H: Deduct 3L2H+6LKJ+K3 = 3H+48+512 = 3H+560. ………7x (ends in 7). So, 3H ends in 7. …-……587 The 9th multiple of 3 ends in 7. So H=9. ………x (No need to deduct 27+560=587. Therefore, (HJKL)3 = (9181)3 = ___,___,__6,741.
General cube roots can be calculated using a long division framework and rightward, column-wise division. The sequence of digits for a three digit schedule, for cube roots, abc, given the number (100a+10b+1c)3 + R, page 327:
The millions place is determined by a3. The hundred-thousands place is determined by 3a2b. The ten-thousands place is formed by 3ab2 + 3a2b. The one-thousands place is formed by b3 + 3b2c. The hundreds place is given by 3ac2 + 3b2c. The tens place is contributed by 3bc2. The units place is constituted by c3.
The Dividends, Quotients, and Remainders: The divisor in cube roots is 3c3. The units digit, c, is obtained by inspection of the first group of digits in the cube after the digits of the given number are grouped in threes. Since we know (or have a reference list of) the first ten cubes we may see the nearest cube below the first group at sight.
The first D, Q, and R are available on sight. Use the following adjustments to the gross dividend at each step. From the second dividend no deduction is made. From the third dividend subtract 3ab2. Note the change. From the fourth dividend subtract 6abc + b3. Note the change. From the fifth dividend subtract 3ac2 + 3b2c. From the sixth dividend subtract 3bc2. From the seventh dividend subtract c3.
The divisor should not be small, try two groups (4, 5, or 6 digits) as the first group, page 330. The four-digit schedule for cube roots, abcd, given the cube (1000a+100b+10c+1d)3, or an incomplete cube with a remainder. pages 337-338.
First digit, billions = a3 Second digit, hundred-millions = 3a2b. Third digit, ten-millions place = 3ab2 + 3a2c Fourth digit, one-millions place = 6abc + b3 + 3a2d Fifth digit, hundred-thousands place = 6abd + 3ac2 + 3b2c Sixth digit, ten-thousands place = 6acd + 3bc2 + 3b2d Seventh digit, one-thousands place = 6bcd + 3ad2 + c3 Eighth digit, hundreds place = 3bd2 + 3c2d Ninth digit, tens place = 3cd2 Tenth digit, ones place = d3
The divisor in cube roots is 3a3. The deduction schedule for the long division, rightward, column-wise method for cube roots.
For the first dividend, Q1 and R1 are obtained by inspection, a3. From the second dividend Q2 and R2 by simple division, no other subtraction! From the third gross dividend subtract 3a2b. From the fourth gross dividend subtract 6abc + b3 From the fifth gross dividend subtract 6abd + 3ac2 + 3b2c From the sixth gross dividend subtract 6acd + 3bc2 + 3b2d From the seventh gross dividend subtract 6bcd + 3ad2 + c3 From the eighth gross dividend subtract 3bd2 + 3c2d From the ninth gross dividend subtract 3cd2 From the tenth gross dividend subtract d3
An alternative method is to cut the four-digit schedule to three-digits by equating d=0, page 338, and substituting (c+d) for c, page 340.
No Vedic sutra mentioned for cube roots by the author of Vedic Mathematics: Sixteen Simple Mathematical Formulae from the Vedas, by Swami Sankaracarya (1880-1960), Motilal Banarsidass Indological Publishers and Booksellers, Varnasi, India, 1965; reprinted in Delhi, India, 1975, 1978. 367 pages.
Examples and several cube root digit schedules are given. 5th root, 7th root, and nth root general formula only hinted at by the binomial expansion of that power. Page 340, Vedic Mathematics, 1965, 1978. Larry R. Holmgren 20:03, 3 March 2007 (UTC)
The introductory paragraph begins: "In mathematics, the cube root of a number, denoted or x1/3, is the number a such that a3 = x. All real numbers have exactly one real cube root and 2 complex roots, and all nonzero complex numbers have 3 distinct complex cube roots."
As there are generally three cube roots, shouldn't we replace the with a? The new text would read: "In mathematics, a cube root of a number, denoted or x1/3, is a number a such that a3 = x. All real numbers have exactly one real cube root and 2 complex roots, and all nonzero complex numbers have 3 distinct complex cube roots." DRE 18:01, 20 February 2007 (UTC)
[Query by Joe user] 18:30, 22 October 2007 (UTC)
I have another question about the introductory paragraph. How many cube-roots does zero have?
Also, some explanation of the 3D plot would be appreciated. The vertical axis is clearly the imaginary part of the complex cube-root. Which of the two horizontal axes is the real number whose cube-root is being evaluated? Obviously the remaining axis is the one that is real (part of the complex) cube-root.
[End query by Joe user] 18:30, 22 October 2007 (UTC)
I had to revert the algorithm to the original one I posted on January 2006. Unfortunately in this edit someone changed the order of the operation, resuting in a loss of precision. Notice that trying to fix an non-existent bug he introduced one. Best regards, Alpertron (talk) 01:05, 8 November 2008 (UTC)
The text:
makes no sense vis a vis the calculator available on a Windows XP platform. The algorithm works when you execute it as indicated in the text preceding this paragraph, at which point the display shows the correct cube root. Taking one last square root destroys this result.
It is unclear what numbers are actually being multiplied when the * key is pressed on the Windows calculator, and from my own personal experience, different commodity calculators do different things in this context of key sequences. This section of the article therefore needs some clarification, and the assumed behaviour of the calculator keys should be explicated in the "Why the algoithm works" section. —Preceding unsigned comment added by 199.199.129.35 (talk) 17:26, 2 July 2009 (UTC)
The section Cube root#Continued fraction is out of place here. It should be moved to a different article. The generalized continued fraction article already has a section for continued fractions of roots, and so this material is much more germane there. We should be observant here both of WP:WEIGHT and summary style. Having long and ugly sections with many continued fraction expansions does not improve this article. I don't see so much of a problem mentioning that such and so expansions exist. Le Docteur (talk) 18:09, 10 November 2009 (UTC)
I wonder what makes a person so "priviliged" that his/her external-link is allowed to appear at wikipedia. I mean, if Wikipedia does not like EXTERNAL LINKS then WIKIPEDIA SHOULD NOT mantain the external-links section. I think it is an obvious public-outrage for many people to realize that only two or three guys are "priviliged" by wikipedia, i do not see any justice on that. Wikipedia should either mantain the EXTERNAL LINKS in another related page or definitely eliminate all of them. What makes some few guys so "privileged"? What makes some guy to be the judge for selecting only two or three external links? Mirificium (talk) 22:39, 20 March 2010 (UTC)
Hi, about this identity:
I think it's necessary to mention how to obtain it, or at least link to which kind of identity this is. --UAwiki (talk) 12:40, 21 October 2011 (UTC)
The section "Complex numbers" contains the following images:
Not already knowing the material presented by the graphs, I have no idea what they are saying. What is measured on the vertical and horizontal axes? What do the shades of colors represent? What are "leaves"? The first picture "shows the main branch" of what? None of this is explained in the article. And "the main branch which is described in the text" must refer to the text of some other article that this image appears in.
Can we just delete these images? Duoduoduo (talk) 15:25, 26 November 2011 (UTC)
The section which begins "For complex numbers, the principal cube root is usually defined as the cube root that has the greatest real part" is completely unsourced. I have never come across this concept of complex principal cube root anywhere else, so I wonder is it really "usually" defined that way, or only occasionally? Can anybody provide any sources, or justify the statement? The editor who uses the pseudonym "JamesBWatson" (talk) 11:01, 10 July 2018 (UTC)
References
I am wondering if we can put in a section on the methods of calculating all cube roots of real numbers (including the two complex conjugates) and imaginary numbers (including the two complex conjugates). I was a math minor in college, and I drafted something that can be added as a section to this post:
requested addition
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The key to computing all three cube roots of a real number or an imaginary number is to set up a formula of the cube of a complex number, using the coefficients “a” for the “real” component and “b” for the “imaginary”. Then, in the case of a real number, get the formula for each coefficient with the “real” component set to the cube and the “imaginary” component set to 0. In the case of an imaginary number, get the formula for each coefficient with the “real” component set to 0 and the “imaginary” component set to the cube. • The formula for the cube of a complex number: (a + bi) 3 = a3 + 3a2bi + 3ab2i2 + b3i3 • The third term contains “i2” equal to -1, and the fourth term contains “i3” equal to -i: = a3 + 3a2bi - 3ab2 - b3i • Combining “like” terms based on “i”: = a3 - 3ab2 + 3a2bi - b3i • Factor out “i” and group “real” and “imaginary” components: = (a3 - 3ab2) + (3a2b - b3) i • Factor out “a” in the “real” component and “b” in the “imaginary” component: = a(a2 - 3b2) + b(3a2 – b2) i For a real number k where “imaginary” component = 0: b(3a2 – b2) = 0 (“imaginary” component) Get the values of the coefficients “a” and “b” based on each term above set to 0. A. b = 0 a(a2 – 3b2) = k (“real” component) a(a2 – 0) = k a3 = k a = 3√k B. 3a2 – b2 = 0 v) b2 = 3a2 a(a2 – 3b2) = k (“real” component) a(a2 – 3[3a2]) = k a(a2 – 9a2) = k a(–8a2) = k –8a3 = k a = – 3√k / 2 b2 = 3 (– 3√k / 2 )2 (equation v) above) b2 = 3( (3√k)2 / 4 ) b = ±√3 ( 3√k / 2 ) ⇒ ±3√k / 2 ⋅ √3 The roots are: 3√k, – (3√k / 2) + (3√k / 2 ⋅ √3)i, – (3√k / 2) – (3√k /2 ⋅ √3) i For examples, the following are cube roots for noted values: 3√1 ⇒ 1, – 1/2 + 1/2√3i, – 1/2 – 1/2√3i 3√–1 ⇒ –1, 1/2 + 1/2√3i, 1/2 – 1/2√3i 3√8 ⇒ 2, – 1 + √3i [ – 2/2 + (2/2 ⋅ √3)i ], – 1 – √3i [ – 2/2 – (2/2 ⋅ √3)i ] 3√–8 ⇒ –2, 1 + √3i [– (–2/2) + (2/2 ⋅ √3)i ], 1 – √3i [– (–2/2) – (2/2 ⋅ √3)i ] 3√27 ⇒ 3, – 3/2 + 3/2√3i, – 3/2 – 3/2√3i 3√–27 ⇒ –3, 3/2 + 3/2√3i, 3/2 – 3/2√3i For an imaginary number ki where “real” component = 0: a(a2 - 3b2) = 0 (“real” component) Get the values of the coefficients “a” and “b” based on each term above set to 0. A. a = 0 b(3a2 – b2) = k (“imaginary” component) b(0 – b2) = k – b3 = k b = 3√(–k) = – 3√k B. a2 - 3b2 = 0 w) a2 = 3b2 b(3a2 – b2) = k (“imaginary” component) b(3[3b2] – b2) = k b(9b2 – b2) = k b(8b2) = k 8b3 = k b = 3√k / 2 a2 = 3( 3√k / 2 )2 (equation w) above) a2 = 3( (3√k)2 / 4 ) a = ±√3( 3√k / 2 ) ⇒ ±3√k / 2 ⋅ √3 The roots are: – 3√k i, + (3√k / 2 ⋅ √3) + (3√k / 2)i, – (3√k / 2 ⋅ √3) + (3√k / 2)i For examples, the following are cube roots for noted values: 3√i ⇒ – i [ – (+1)i ], 1/2√3 + 1/2i, – 1/2√3 + 1/2 i 3√–i ⇒ i [ – (–1)i ], 1/2√3 – 1/2i, – 1/2√3 – 1/2 i 3√8i ⇒ –2i [ – (+2)i ], √3 + i [ 2/2√3 + 2/2i ], –√3 + i [ – 2/2√3 + 2/2i ] 3√–8i ⇒ 2i [ – (–2)i ], √3 – i [ 2/2√3 – 2/2i ], –√3 – i [ – 2/2√3 – 2/2i ] 3√27i ⇒ – 3i [ – (+3)i ], 3/2√3 + 3/2i, – 3/2√3 + 3/2i 3√–27i ⇒ 3i [ – (–3)i ], 3/2√3 – 3/2i, – 3/2√3 – 3/2i |
Dkf12 (talk) 21:20, 29 March 2020 (UTC)