Talk:Cube root

message: 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)
replies: text: Aha, yeah... Sorry about the confusion. I'll get round to telling people if I do it again (really) Shen 21:29, 29 January 2006 (UTC)
replies: text: Uh... no need to be sorry. :-D If it's a better picture, it's a better picture. — Ambush Commander(Talk) 21:34, 29 January 2006 (UTC)

text: Well, part of the problem is the resizing of the plots. — Ambush Commander(Talk) 03:36, 30 January 2006 (UTC)
replies: text:  Good point. The first picture does not scale the axes, and is more natural. Unless the new picture fixes that and the font sizes, I would suggest using the old one. Oleg Alexandrov (talk) 21:07, 30 January 2006 (UTC)
replies: text:  I gave it some bigger fonts (increased the size of everything by 2). It should look a lot nicer! Shen 21:11, 30 January 2006 (UTC)
replies: text:  Thanks, looks good! Oleg Alexandrov (talk) 02:23, 31 January 2006 (UTC)
thread_title: Related image

message: 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)
replies: text: This method is only useful on TV shows. Most numbers have no integer cube root so the algorithm cannot be applied. Thus it is only a curiosity. It cannot be extended to work for any N-th root (N odd) as claimed above: for N=11 there are 8 possible endings for eleventh roots of ...7056: namely 0006, 1256, 2506, 3756, 5006, 6256, 7506, 8756. --Alpertron 12:59, 30 March 2007 (UTC)

text: Example one: 
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)

text: How much space is appropriate for examples? Larry R. Holmgren 19:02, 4 March 2007 (UTC)
thread_title: The Cube Root and Root-Solving History

message: 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)
thread_title: '''CUBE ROOTS'''

message: 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.
:Example one: 
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)
thread_title: '''Digit schedule of subtractions for a three-digit cube root, cba'''

message: 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)
replies: text: How much space is appropriate for examples? Larry R. Holmgren 19:02, 4 March 2007 (UTC)
thread_title: '''The analytic digit schedule:'''

message: The introductory paragraph begins: "In mathematics, the cube root of a number, denoted \sqrt[3]{x}  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 \sqrt[3]{x}  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)
replies: text: Sounds good. I've went ahead and made the change. — Edward Z. Yang(Talk) 00:18, 4 March 2007 (UTC)
replies: text: Technically, the complex numbers are the set of numbers in the form a + bi where a and b are real numbers.  So, all real numbers are also complex numbers, if b = 0.  Thus, perhaps it would be to say "exactly one real cube root and 2 nonreal roots..." StatisticsMan 20:03, 21 April 2007 (UTC)
replies: text: Except when a is also zero. In this case all three roots are real. --Alpertron 13:18, 30 April 2007 (UTC)
replies: text:  Well, distinct cube roots, anyway. — Edward Z. Yang(Talk) 19:13, 30 April 2007 (UTC)
thread_title: "''The'' cube root" vs. "''A'' cube root"

message: 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)
replies: text: Very well. It has been so moved. I'll not press the point any further. ---Glenn L (talk) 23:32, 10 November 2009 (UTC)

text: Update: since making the move it has been determined that, as far as generalized continued fractions are concerned, the cube root is simply a case of the nth root with n = 3 so was never needed here in the first place. Only the square root is really a special case. Thanks for helping me "see outside the box". Glenn L (talk) 02:04, 12 November 2010 (UTC)
thread_title:  Proposed removal of "Continued fraction" material to main article [[generalized continued fraction]] 

message: 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)
thread_title: EXTERNAL LINKS

message: 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)
thread_title: Unexplained images

message: 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:
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)
replies: text: Wikipedia does not accept our wp:original research. - DVdm (talk) 18:17, 29 March 2020 (UTC)
replies: text: Without providing anything like above, one will need to leave Wikipedia and search for the solution that will be difficult for most out there.  I did that via BING.com and stumbled on an posting on the cube root of "I" that started with solving cubes with the coefficients, and I took it from there. Dkf12 (talk) 18:56, 29 March 2020 (UTC)
replies: text: I reviewed the section on wp:original research, but what I put is not anything like political opinions or the like. What I stated are mathematical facts. Please note I did not update the post and instead put it all in a talk page, to make sure I clear any hurdles or whatever. Please advise on how I can get a hold of wp:reliable sources, because I believe if this posts delves more on the solutions to get ALL cubes of both real an imaginary numbers, that will be good for people like me who can be called a math nerd! Dkf12 (talk) 19:39, 29 March 2020 (UTC)

text: Could we use "Template:Original research"? Dkf12 (talk) 19:45, 29 March 2020 (UTC)
replies: text:  See wp:CALC to learn about routine calculations as exceptions to wp:NOR. The above clearly does not qualify. - DVdm (talk) 20:16, 29 March 2020 (UTC)

text: I read "wp:CALC" and again "wp:NOR", but I disagree.  My formulas use nothing but routine calculations.  I appreciate how you want to keep a strict set of standards for updates to posts and Wikipedia does have a much better reputation then from even ten years ago or so.  But, how can we make the article better?  Listen, when someone like me, whom my wife calls a bit of a "math nerd", reads the post, that someone should get a clear explanation of how one calculates all three cube roots of real numbers.  The section on "Complex Numbers" uses formulas like "3√r * exp(I* Θ/3 + 2i*e/3)" that was vague to me since I never took complex analysis, and I wonder exaxtly how that leads to "-1 + √3" for one of the cube roots of 8.  Dkf12 (talk) 20:48, 29 March 2020 (UTC)
replies: text:  You are now disrupting this talk page—see wp:Talk page guidelines. Second level warning on your talk page. - DVdm (talk) 21:38, 29 March 2020 (UTC)

text:  @DVdm, I apologize.  I am trying to get the hang of editing the talk page.  Dkf12 (talk) 22:00, 29 March 2020 (UTC)

text: Please advise if we can make the post more clear, or put a link to a new post with the detail. I am a math major but never took complex analysis, yet I do like to see details. I solved it on my own with the formulas in this talk page and I wanted to share it with others out there. I understand we have an issue with "wp:original research" and I cannot update the post although my formulas use nothing but routine calculations. I just wish we could make the post more clear regardless. Dkf12 (talk) 19:40, 31 March 2020 (UTC)
replies: text:  The only advise we can give you, is to stop trying to share your own work.  Wikipedia does not want our own work. - DVdm (talk) 07:20, 1 April 2020 (UTC)
replies: text: My question now if we can make the article more clear.  I had to go to BING to get some insights into how one calculates all cube roots of real numbers without knowing complex analysis.  I found a clue and took it from there.  Most people out there won't have the mathematical skills I have and won't be able solve the problem.  I understand I can't share my work and I never attempted it outside of this talk page.  I was just hoping someone can make the article more clear for others out there.  Dkf12 (talk) 14:57, 1 April 2020 (UTC)
thread_title:  All cube roots of real numbers and imginary numbers