Talk:Nth root algorithm - Revision history

Legobot: Bot: Fixing lint errors, replacing obsolete HTML tags: `(6x)`

2024-09-30T20:05:25Z Bot: Fixing lint errors, replacing obsolete HTML tags: <tt> (6x) ← Previous revision Revision as of 20:05, 30 September 2024 Line 11: Line 11: return x return x For example, <tt>nthroot(12345678901234567890123456789012345**100,100)</tt> takes several minutes to finish. For example, <code>nthroot(12345678901234567890123456789012345**100,100)</code> takes several minutes to finish. Using binary search, the solution can be found in seconds. Using binary search, the solution can be found in seconds. Line 21: Line 21: ::No, no special library is required. I have tried various different initial guesses, but they offer no improvement. [[User:Fredrik|Fredrik]] | [[User talk:Fredrik|talk]] 20:55, 6 Jun 2005 (UTC) ::No, no special library is required. I have tried various different initial guesses, but they offer no improvement. [[User:Fredrik|Fredrik]] | [[User talk:Fredrik|talk]] 20:55, 6 Jun 2005 (UTC) :::I thought a bit more about it now. Basically, you are right: Newton's method is inefficient for large n. There are two reasons for it. Firstly, the computation of x**(n-1) in exact arithmetics is expensive if x is a large integer, so every iteration takes a long time. Secondly, the method needs a lot of iterations. For instance, <tt>nthroot(1234567890123456789012345**10, 10)</tt> requires 4725 iterations (note that I replaced 100 in your example with 10), while a binary search requires about 100 iterations. However, the method improves if you give a good initial guess. After replacing the first line of the <tt>nthroot</tt> function with <tt>x, xp = int(y**(1.0/n)), -1</tt>, the above example gets the correct answer in only 3 iterations. :::I thought a bit more about it now. Basically, you are right: Newton's method is inefficient for large n. There are two reasons for it. Firstly, the computation of x**(n-1) in exact arithmetics is expensive if x is a large integer, so every iteration takes a long time. Secondly, the method needs a lot of iterations. For instance, <code>nthroot(1234567890123456789012345**10, 10)</code> requires 4725 iterations (note that I replaced 100 in your example with 10), while a binary search requires about 100 iterations. However, the method improves if you give a good initial guess. After replacing the first line of the <code>nthroot</code> function with <code>x, xp = int(y**(1.0/n)), -1</code>, the above example gets the correct answer in only 3 iterations. :::The current article is rather bad: it only presents the algorithm, which is trivial if you know Newton's method. There is no analysis, no discussion on how to get the initial guess and no references. I am not convinced that the algorithm is used that often; I thought the common approach is to use exponential and logarithms. Bignum arithmetics is a separate issue. I hope you can find the time to improve the article. -- [[User:Jitse Niesen|Jitse Niesen]] 23:37, 7 Jun 2005 (UTC) :::The current article is rather bad: it only presents the algorithm, which is trivial if you know Newton's method. There is no analysis, no discussion on how to get the initial guess and no references. I am not convinced that the algorithm is used that often; I thought the common approach is to use exponential and logarithms. Bignum arithmetics is a separate issue. I hope you can find the time to improve the article. -- [[User:Jitse Niesen|Jitse Niesen]] 23:37, 7 Jun 2005 (UTC) ::::The catch with that solution is that <tt>y**(1.0/n)</tt> overflows for large y. Fortunately this can be avoided by using <tt>B**int(log(y, B)/n)</tt> with some integer B instead. However, this still leaves Newton (as implemented above) significantly slower than bsearch for very large n (a test with n=300 is nearly instantaneous with bsearch, but takes several seconds with Newton). I'll do some more detailed benchmarking when I get time, hopefully soon. For reference, I'll provide my bsearch implementation here: - [[User:Fredrik|Fredrik]] | [[User talk:Fredrik|talk]] 00:24, 8 Jun 2005 (UTC) ::::The catch with that solution is that <code>y**(1.0/n)</code> overflows for large y. Fortunately this can be avoided by using <code>B**int(log(y, B)/n)</code> with some integer B instead. However, this still leaves Newton (as implemented above) significantly slower than bsearch for very large n (a test with n=300 is nearly instantaneous with bsearch, but takes several seconds with Newton). I'll do some more detailed benchmarking when I get time, hopefully soon. For reference, I'll provide my bsearch implementation here: - [[User:Fredrik|Fredrik]] | [[User talk:Fredrik|talk]] 00:24, 8 Jun 2005 (UTC) # Returns a tuple of the (floor, ceil) values of the exact solution # Returns a tuple of the (floor, ceil) values of the exact solution

87.26.160.199: /* Complicate modulus method */ 2020-06-03T18:42:54Z Complicate modulus method ← Previous revision Revision as of 18:42, 3 June 2020 Line 127: Line 127: The n-th root of any Integer can be found by hand, knowing that [P^n-(P-1)^n] is the complicate modulus of any power of integer. This follows from the [[Telescoping series|Telescoping Sum property]]: any power of an Integer C can be written as a Sum of Gnomons: The n-th root of any Integer can be found by hand, knowing that [P^n-(P-1)^n] is the complicate modulus of any power of integer. This follows from the [[Telescoping series|Telescoping Sum property]]: any power of an Integer C can be written as a Sum of Gnomons: Will be used "x" as Index since it helps to stick the result on a Cartesian plane and understand that this is a method to square the area bellow the derivate (not just the first) of any function of the type: $Y=mX^n$. Will be used "x" as Index since it helps to stick the result on a Cartesian plane and understand that this is a method to square the area bellow the derivate (not just the first) of any function of the type: <math>Y=mX^n</math>. Since: Since: Line 200: Line 200: Where the new index '''x''' starts from '''1/10''' and rise '''Step 1/10''' (so 1/10,2/10,3/10...) till '''C''' (that can be an Integer or a 1 digit Rational). Where the new index '''x''' starts from '''1/10''' and rise '''Step 1/10''' (so 1/10,2/10,3/10...) till '''C''' (that can be an Integer or a 1 digit Rational). So the ne Sum, I've called Step Sum works with Rational index in this way: :<math>C^2 =\sum_{x=1/K}^{C}M_{2,K}= \sum_{x=1/K}^{C} \frac{2x}{K}-\frac{1}{K^2}</math>. So f.example if K=10 and C=2.1 we have: :<math>C^2 =\sum_{x=1/10}^{2.1} \frac{2x}{10}-\frac{1}{10^2}</math>. that is (nothing more than a scaled version of the 10 times bigger integer sum) : x 2x/k-1/k^2 Sum 0,1 0,01 0,01 0,2 0,03 0,04 0,3 0,05 0,09 0,4 0,07 0,16 0,5 0,09 0,25 0,6 0,11 0,36 0,7 0,13 0,49 0,8 0,15 0,64 0,9 0,17 0,81 1 0,19 1 1,1 0,21 1,21 1,2 0,23 1,44 1,3 0,25 1,69 1,4 0,27 1,96 1,5 0,29 2,25 1,6 0,31 2,56 1,7 0,33 2,89 1,8 0,35 3,24 1,9 0,37 3,61 2 0,39 4 2,1 0,41 4,41 etc... [[User:StefanoMaruelli|StefanoMaruelli]] ([[User talk:StefanoMaruelli|talk]]) 08:44, 16 June 2017 (UTC)StefanoMaruelli [[User:StefanoMaruelli|StefanoMaruelli]] ([[User talk:StefanoMaruelli|talk]]) 08:44, 16 June 2017 (UTC)StefanoMaruelli 87.26.160.199: /* Complicate modulus method */ 2020-06-03T18:30:11Z Complicate modulus method ← Previous revision Revision as of 18:30, 3 June 2020 Line 127: Line 127: The n-th root of any Integer can be found by hand, knowing that [P^n-(P-1)^n] is the complicate modulus of any power of integer. This follows from the [[Telescoping series|Telescoping Sum property]]: any power of an Integer C can be written as a Sum of Gnomons: The n-th root of any Integer can be found by hand, knowing that [P^n-(P-1)^n] is the complicate modulus of any power of integer. This follows from the [[Telescoping series|Telescoping Sum property]]: any power of an Integer C can be written as a Sum of Gnomons: Will be used "x" ad Index since it helps to stick the result on a Cartesian plane and understand that this is a method to square the area bellow the derivate (not just the first) of any function of the type: $Y=mX^n$. Will be used "x" as Index since it helps to stick the result on a Cartesian plane and understand that this is a method to square the area bellow the derivate (not just the first) of any function of the type: $Y=mX^n$. Since: Since: StefanoMaruelli: /* Complicate modulus method */ 2017-06-16T08:44:35Z Complicate modulus method ← Previous revision Revision as of 08:44, 16 June 2017 Line 181: Line 181: In the following pictures 2 example of how the Two-Hand-Clock works for numbers from 0 to 11, in the case of Square and Cubic Root or Power. In the following pictures 2 example of how the Two-Hand-Clock works for numbers from 0 to 11, in the case of Square and Cubic Root or Power. [[File:COMPLICATE-MODULUS-CLOCK-N2-0-11.jpg|thumb]] — Preceding [[Wikipedia:Signatures|unsigned]] comment added by [[User:StefanoMaruelli|StefanoMaruelli]] ([[User talk:StefanoMaruelli#top|talk]] • [[Special:Contributions/StefanoMaruelli|contribs]]) 05:45, 16 June 2017 (UTC)  [[File:COMPLICATE-MODULUS-CLOCK-N2-0-11.jpg|thumb]] [[File:COMPLICATE-MODULUS-CLOCK-N3-0-11.jpg|thumb]] — Preceding [[Wikipedia:Signatures|unsigned]] comment added by [[User:StefanoMaruelli|StefanoMaruelli]] ([[User talk:StefanoMaruelli#top|talk]] • [[Special:Contributions/StefanoMaruelli|contribs]]) 05:45, 16 June 2017 (UTC)  I call this "Complicate Modulus" clock calculator (now is the Two-Hand-Clock). I call this "Complicate Modulus" clock calculator (now is the Two-Hand-Clock). Line 196: Line 200: Where the new index '''x''' starts from '''1/10''' and rise '''Step 1/10''' (so 1/10,2/10,3/10...) till '''C''' (that can be an Integer or a 1 digit Rational). Where the new index '''x''' starts from '''1/10''' and rise '''Step 1/10''' (so 1/10,2/10,3/10...) till '''C''' (that can be an Integer or a 1 digit Rational). [[User:StefanoMaruelli|StefanoMaruelli]] ([[User talk:StefanoMaruelli|talk]]) 08:44, 16 June 2017 (UTC)StefanoMaruelli ==Maclaurin Series== ==Maclaurin Series== SineBot: Signing comment by StefanoMaruelli - "/* Complicate modulus method */" 2017-06-16T05:48:01Z Signing comment by StefanoMaruelli - "Complicate modulus method: " ← Previous revision Revision as of 05:48, 16 June 2017 Line 181: Line 181: In the following pictures 2 example of how the Two-Hand-Clock works for numbers from 0 to 11, in the case of Square and Cubic Root or Power. In the following pictures 2 example of how the Two-Hand-Clock works for numbers from 0 to 11, in the case of Square and Cubic Root or Power. [[File:COMPLICATE-MODULUS-CLOCK-N2-0-11.jpg|thumb]] — Preceding [[Wikipedia:Signatures|unsigned]] comment added by [[User:StefanoMaruelli|StefanoMaruelli]] ([[User talk:StefanoMaruelli#top|talk]] • [[Special:Contributions/StefanoMaruelli|contribs]]) 05:45, 16 June 2017 (UTC)  [[File:COMPLICATE-MODULUS-CLOCK-N2-0-11.jpg|thumb]] I call this "Complicate Modulus" clock calculator (now is the Two-Hand-Clock). I call this "Complicate Modulus" clock calculator (now is the Two-Hand-Clock). StefanoMaruelli: /* Complicate modulus method */ 2017-06-16T05:45:29Z Complicate modulus method ← Previous revision Revision as of 05:45, 16 June 2017 Line 178: Line 178: For the "n" root just do the same keeping "n" Tartaglia's line For the "n" root just do the same keeping "n" Tartaglia's line In the following pictures 2 example of how the Two-Hand-Clock works for numbers from 0 to 11, in the case of Square and Cubic Root or Power. [[File:COMPLICATE-MODULUS-CLOCK-N2-0-11.jpg|thumb]] I call this "Complicate Modulus" clock calculator (now is the Two-Hand-Clock). I call this "Complicate Modulus" clock calculator (now is the Two-Hand-Clock). StefanoMaruelli: /* Complicate modulus method */ 2017-06-15T04:59:31Z Complicate modulus method ← Previous revision Revision as of 04:59, 15 June 2017 Line 125: Line 125: == Complicate modulus method == == Complicate modulus method == The "n" root can be found by hand, knowing that [P^n-(P-1)^n] is the comlpicate modulus of any power of integer. Will be used "x" ad Index since it helps to stick the result on a Cartesian plane and understand that this a method to square the area bellow the derivate of any function of the type: $Y=mX^n$, thanks to the [[Telescoping series|Telescoping Sum property]]. The n-th root of any Integer can be found by hand, knowing that [P^n-(P-1)^n] is the complicate modulus of any power of integer. This follows from the [[Telescoping series|Telescoping Sum property]]: any power of an Integer C can be written as a Sum of Gnomons: Will be used "x" ad Index since it helps to stick the result on a Cartesian plane and understand that this is a method to square the area bellow the derivate (not just the first) of any function of the type: $Y=mX^n$. Since: Since: Line 131: Line 133: :<math>C^n =\sum_{x=1}^{C}[x^n-(x-1)^n]</math>. :<math>C^n =\sum_{x=1}^{C}[x^n-(x-1)^n]</math>. So to have the sqaure root of 9 just follow this: So to have the sqaure root of 9 just follow this Recursive Difference: 1) wrote Tartaglia's line for (x-1)^2 (line n=2): 1) wrote Tartaglia's line for (x-1)^2 (line n=2): StefanoMaruelli: /* Complicate modulus method */ 2017-06-13T09:31:38Z Complicate modulus method ← Previous revision Revision as of 09:31, 13 June 2017 Line 190: Line 190: Where the new index '''x''' starts from '''1/10''' and rise '''Step 1/10''' (so 1/10,2/10,3/10...) till '''C''' (that can be an Integer or a 1 digit Rational). Where the new index '''x''' starts from '''1/10''' and rise '''Step 1/10''' (so 1/10,2/10,3/10...) till '''C''' (that can be an Integer or a 1 digit Rational). Stefano Maruelli — Preceding [[Wikipedia:Signatures|unsigned]] comment added by [[Special:Contributions/94.81.217.96|94.81.217.96]] ([[User talk:94.81.217.96|talk]]) 16:08, 19 October 2011 (UTC)  ==Maclaurin Series== ==Maclaurin Series== StefanoMaruelli: /* Complicate modulus method */ 2017-06-13T09:28:14Z Complicate modulus method ← Previous revision Revision as of 09:28, 13 June 2017 Line 189: Line 189: :<math>M_{2,10}=2x/K-1/K^2 </math>. :<math>M_{2,10}=2x/K-1/K^2 </math>. Where the new index x start from 1/10 and rise step 1/10 till C (that can be an Integer or a 1 digit Rational). Where the new index '''x''' starts from '''1/10''' and rise '''Step 1/10''' (so 1/10,2/10,3/10...) till '''C''' (that can be an Integer or a 1 digit Rational). Stefano Maruelli — Preceding [[Wikipedia:Signatures|unsigned]] comment added by [[Special:Contributions/94.81.217.96|94.81.217.96]] ([[User talk:94.81.217.96|talk]]) 16:08, 19 October 2011 (UTC)  Stefano Maruelli — Preceding [[Wikipedia:Signatures|unsigned]] comment added by [[Special:Contributions/94.81.217.96|94.81.217.96]] ([[User talk:94.81.217.96|talk]]) 16:08, 19 October 2011 (UTC)  StefanoMaruelli: /* Complicate modulus method */ 2017-06-13T09:26:58Z Complicate modulus method ← Previous revision Revision as of 09:26, 13 June 2017 Line 185: Line 185: Where to have '''m''' digit choose :<math>K=10^m </math>. Where to have '''m''' digit choose :<math>K=10^m </math>. Example of Square Root of 10, with 1 digit precision: Example of Rational Complicate Modulus, for 1 digit precision: :<math>M_{2,10}=2x/K-1/K^2 </math>. :<math>M_{2,10}=2x/K-1/K^2 </math>.

← Previous revision		Revision as of 20:05, 30 September 2024
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	return x		return x

	For example, <tt>nthroot(12345678901234567890123456789012345**100,100)</tt> takes several minutes to finish.		For example, <code>nthroot(12345678901234567890123456789012345**100,100)</code> takes several minutes to finish.

	Using binary search, the solution can be found in seconds.		Using binary search, the solution can be found in seconds.
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	::No, no special library is required. I have tried various different initial guesses, but they offer no improvement. [[User:Fredrik\|Fredrik]] \| [[User talk:Fredrik\|talk]] 20:55, 6 Jun 2005 (UTC)		::No, no special library is required. I have tried various different initial guesses, but they offer no improvement. [[User:Fredrik\|Fredrik]] \| [[User talk:Fredrik\|talk]] 20:55, 6 Jun 2005 (UTC)

	:::I thought a bit more about it now. Basically, you are right: Newton's method is inefficient for large n. There are two reasons for it. Firstly, the computation of x(n-1) in exact arithmetics is expensive if x is a large integer, so every iteration takes a long time. Secondly, the method needs a lot of iterations. For instance, <tt>nthroot(123456789012345678901234510, 10)</tt> requires 4725 iterations (note that I replaced 100 in your example with 10), while a binary search requires about 100 iterations. However, the method improves if you give a good initial guess. After replacing the first line of the <tt>nthroot</tt> function with <tt>x, xp = int(y**(1.0/n)), -1</tt>, the above example gets the correct answer in only 3 iterations.		:::I thought a bit more about it now. Basically, you are right: Newton's method is inefficient for large n. There are two reasons for it. Firstly, the computation of x(n-1) in exact arithmetics is expensive if x is a large integer, so every iteration takes a long time. Secondly, the method needs a lot of iterations. For instance, <code>nthroot(123456789012345678901234510, 10)</code> requires 4725 iterations (note that I replaced 100 in your example with 10), while a binary search requires about 100 iterations. However, the method improves if you give a good initial guess. After replacing the first line of the <code>nthroot</code> function with <code>x, xp = int(y**(1.0/n)), -1</code>, the above example gets the correct answer in only 3 iterations.
	:::The current article is rather bad: it only presents the algorithm, which is trivial if you know Newton's method. There is no analysis, no discussion on how to get the initial guess and no references. I am not convinced that the algorithm is used that often; I thought the common approach is to use exponential and logarithms. Bignum arithmetics is a separate issue. I hope you can find the time to improve the article. -- [[User:Jitse Niesen\|Jitse Niesen]] 23:37, 7 Jun 2005 (UTC)		:::The current article is rather bad: it only presents the algorithm, which is trivial if you know Newton's method. There is no analysis, no discussion on how to get the initial guess and no references. I am not convinced that the algorithm is used that often; I thought the common approach is to use exponential and logarithms. Bignum arithmetics is a separate issue. I hope you can find the time to improve the article. -- [[User:Jitse Niesen\|Jitse Niesen]] 23:37, 7 Jun 2005 (UTC)

	::::The catch with that solution is that <tt>y(1.0/n)</tt> overflows for large y. Fortunately this can be avoided by using <tt>Bint(log(y, B)/n)</tt> with some integer B instead. However, this still leaves Newton (as implemented above) significantly slower than bsearch for very large n (a test with n=300 is nearly instantaneous with bsearch, but takes several seconds with Newton). I'll do some more detailed benchmarking when I get time, hopefully soon. For reference, I'll provide my bsearch implementation here: - [[User:Fredrik\|Fredrik]] \| [[User talk:Fredrik\|talk]] 00:24, 8 Jun 2005 (UTC)		::::The catch with that solution is that <code>y(1.0/n)</code> overflows for large y. Fortunately this can be avoided by using <code>Bint(log(y, B)/n)</code> with some integer B instead. However, this still leaves Newton (as implemented above) significantly slower than bsearch for very large n (a test with n=300 is nearly instantaneous with bsearch, but takes several seconds with Newton). I'll do some more detailed benchmarking when I get time, hopefully soon. For reference, I'll provide my bsearch implementation here: - [[User:Fredrik\|Fredrik]] \| [[User talk:Fredrik\|talk]] 00:24, 8 Jun 2005 (UTC)

	# Returns a tuple of the (floor, ceil) values of the exact solution		# Returns a tuple of the (floor, ceil) values of the exact solution

@@ Line 127: / Line 127: @@
 The n-th root of any Integer can be found by hand, knowing that [P^n-(P-1)^n] is the complicate modulus of any power of integer. This follows from the [[Telescoping series|Telescoping Sum property]]: any power of an Integer C can be written as a Sum of Gnomons:
- Will be used "x" as Index since it helps to stick the result on a Cartesian plane and understand that this is a method to square the area bellow the derivate (not just the first) of any function of the type: $Y=mX^n$.
+ Will be used "x" as Index since it helps to stick the result on a Cartesian plane and understand that this is a method to square the area bellow the derivate (not just the first) of any function of the type: <math>Y=mX^n</math>.
 Since:
@@ Line 200: / Line 200: @@
 Where the new index '''x''' starts from '''1/10''' and rise '''Step 1/10''' (so 1/10,2/10,3/10...) till '''C''' (that can be an Integer or a 1 digit Rational).
 [[User:StefanoMaruelli|StefanoMaruelli]] ([[User talk:StefanoMaruelli|talk]]) 08:44, 16 June 2017 (UTC)StefanoMaruelli

← Previous revision		Revision as of 18:30, 3 June 2020
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	The n-th root of any Integer can be found by hand, knowing that [P^n-(P-1)^n] is the complicate modulus of any power of integer. This follows from the [[Telescoping series\|Telescoping Sum property]]: any power of an Integer C can be written as a Sum of Gnomons:		The n-th root of any Integer can be found by hand, knowing that [P^n-(P-1)^n] is the complicate modulus of any power of integer. This follows from the [[Telescoping series\|Telescoping Sum property]]: any power of an Integer C can be written as a Sum of Gnomons:

	Will be used "x" ad Index since it helps to stick the result on a Cartesian plane and understand that this is a method to square the area bellow the derivate (not just the first) of any function of the type: $Y=mX^n$.		Will be used "x" as Index since it helps to stick the result on a Cartesian plane and understand that this is a method to square the area bellow the derivate (not just the first) of any function of the type: $Y=mX^n$.

	Since:		Since:

← Previous revision		Revision as of 08:44, 16 June 2017
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	In the following pictures 2 example of how the Two-Hand-Clock works for numbers from 0 to 11, in the case of Square and Cubic Root or Power.		In the following pictures 2 example of how the Two-Hand-Clock works for numbers from 0 to 11, in the case of Square and Cubic Root or Power.

	[[File:COMPLICATE-MODULUS-CLOCK-N2-0-11.jpg\|thumb]] <!-- Template:Unsigned --><small class="autosigned">— Preceding [[Wikipedia:Signatures\|unsigned]] comment added by [[User:StefanoMaruelli\|StefanoMaruelli]] ([[User talk:StefanoMaruelli#top\|talk]] • [[Special:Contributions/StefanoMaruelli\|contribs]]) 05:45, 16 June 2017 (UTC)</small> <!--Autosigned by SineBot-->		[[File:COMPLICATE-MODULUS-CLOCK-N2-0-11.jpg\|thumb]]

			[[File:COMPLICATE-MODULUS-CLOCK-N3-0-11.jpg\|thumb]]

			<!-- Template:Unsigned --><small class="autosigned">— Preceding [[Wikipedia:Signatures\|unsigned]] comment added by [[User:StefanoMaruelli\|StefanoMaruelli]] ([[User talk:StefanoMaruelli#top\|talk]] • [[Special:Contributions/StefanoMaruelli\|contribs]]) 05:45, 16 June 2017 (UTC)</small> <!--Autosigned by SineBot-->

	I call this "Complicate Modulus" clock calculator (now is the Two-Hand-Clock).		I call this "Complicate Modulus" clock calculator (now is the Two-Hand-Clock).
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	Where the new index '''x''' starts from '''1/10''' and rise '''Step 1/10''' (so 1/10,2/10,3/10...) till '''C''' (that can be an Integer or a 1 digit Rational).		Where the new index '''x''' starts from '''1/10''' and rise '''Step 1/10''' (so 1/10,2/10,3/10...) till '''C''' (that can be an Integer or a 1 digit Rational).

			[[User:StefanoMaruelli\|StefanoMaruelli]] ([[User talk:StefanoMaruelli\|talk]]) 08:44, 16 June 2017 (UTC)StefanoMaruelli

	==Maclaurin Series==		==Maclaurin Series==

← Previous revision		Revision as of 05:48, 16 June 2017
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	In the following pictures 2 example of how the Two-Hand-Clock works for numbers from 0 to 11, in the case of Square and Cubic Root or Power.		In the following pictures 2 example of how the Two-Hand-Clock works for numbers from 0 to 11, in the case of Square and Cubic Root or Power.

			[[File:COMPLICATE-MODULUS-CLOCK-N2-0-11.jpg\|thumb]] <!-- Template:Unsigned --><small class="autosigned">— Preceding [[Wikipedia:Signatures\|unsigned]] comment added by [[User:StefanoMaruelli\|StefanoMaruelli]] ([[User talk:StefanoMaruelli#top\|talk]] • [[Special:Contributions/StefanoMaruelli\|contribs]]) 05:45, 16 June 2017 (UTC)</small> <!--Autosigned by SineBot-->
	[[File:COMPLICATE-MODULUS-CLOCK-N2-0-11.jpg\|thumb]]

	I call this "Complicate Modulus" clock calculator (now is the Two-Hand-Clock).		I call this "Complicate Modulus" clock calculator (now is the Two-Hand-Clock).

← Previous revision		Revision as of 05:45, 16 June 2017
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	For the "n" root just do the same keeping "n" Tartaglia's line		For the "n" root just do the same keeping "n" Tartaglia's line

			In the following pictures 2 example of how the Two-Hand-Clock works for numbers from 0 to 11, in the case of Square and Cubic Root or Power.

			[[File:COMPLICATE-MODULUS-CLOCK-N2-0-11.jpg\|thumb]]

	I call this "Complicate Modulus" clock calculator (now is the Two-Hand-Clock).		I call this "Complicate Modulus" clock calculator (now is the Two-Hand-Clock).

← Previous revision		Revision as of 04:59, 15 June 2017
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	== Complicate modulus method ==		== Complicate modulus method ==

	The "n" root can be found by hand, knowing that [P^n-(P-1)^n] is the ~~comlpicate~~ modulus of any power of integer. ~~Will~~ be ~~used~~ ~~"x"~~ ad ~~Index~~ ~~since~~ it ~~helps~~ to ~~stick~~ ~~the~~ ~~result~~ on a ~~Cartesian~~ ~~plane~~ ~~and understand that this~~ a ~~method to square the area bellow the derivate~~ of ~~any function of the type~~: ~~$Y=mX^n$, thanks to the [[Telescoping series\|Telescoping Sum property]].~~		The n-th root of any Integer can be found by hand, knowing that [P^n-(P-1)^n] is the complicate modulus of any power of integer. This follows from the [[Telescoping series\|Telescoping Sum property]]: any power of an Integer C can be written as a Sum of Gnomons:

			Will be used "x" ad Index since it helps to stick the result on a Cartesian plane and understand that this is a method to square the area bellow the derivate (not just the first) of any function of the type: $Y=mX^n$.

	Since:		Since:
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	:<math>C^n =\sum_{x=1}^{C}[x^n-(x-1)^n]</math>.		:<math>C^n =\sum_{x=1}^{C}[x^n-(x-1)^n]</math>.

	So to have the sqaure root of 9 just follow this:		So to have the sqaure root of 9 just follow this Recursive Difference:

	1) wrote Tartaglia's line for (x-1)^2 (line n=2):		1) wrote Tartaglia's line for (x-1)^2 (line n=2):

← Previous revision		Revision as of 09:31, 13 June 2017
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	Where the new index '''x''' starts from '''1/10''' and rise '''Step 1/10''' (so 1/10,2/10,3/10...) till '''C''' (that can be an Integer or a 1 digit Rational).		Where the new index '''x''' starts from '''1/10''' and rise '''Step 1/10''' (so 1/10,2/10,3/10...) till '''C''' (that can be an Integer or a 1 digit Rational).

	Stefano Maruelli <span style="font-size: smaller;" class="autosigned">— Preceding [[Wikipedia:Signatures\|unsigned]] comment added by [[Special:Contributions/94.81.217.96\|94.81.217.96]] ([[User talk:94.81.217.96\|talk]]) 16:08, 19 October 2011 (UTC)</span><!-- Template:Unsigned IP --> <!--Autosigned by SineBot-->

	==Maclaurin Series==		==Maclaurin Series==

← Previous revision		Revision as of 09:28, 13 June 2017
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	:<math>M_{2,10}=2x/K-1/K^2 </math>.		:<math>M_{2,10}=2x/K-1/K^2 </math>.

	Where the new index x ~~start~~ from 1/10 and rise ~~step~~ 1/10 till C (that can be an Integer or a 1 digit Rational).		Where the new index '''x''' starts from '''1/10''' and rise '''Step 1/10''' (so 1/10,2/10,3/10...) till '''C''' (that can be an Integer or a 1 digit Rational).

	Stefano Maruelli <span style="font-size: smaller;" class="autosigned">— Preceding [[Wikipedia:Signatures\|unsigned]] comment added by [[Special:Contributions/94.81.217.96\|94.81.217.96]] ([[User talk:94.81.217.96\|talk]]) 16:08, 19 October 2011 (UTC)</span><!-- Template:Unsigned IP --> <!--Autosigned by SineBot-->		Stefano Maruelli <span style="font-size: smaller;" class="autosigned">— Preceding [[Wikipedia:Signatures\|unsigned]] comment added by [[Special:Contributions/94.81.217.96\|94.81.217.96]] ([[User talk:94.81.217.96\|talk]]) 16:08, 19 October 2011 (UTC)</span><!-- Template:Unsigned IP --> <!--Autosigned by SineBot-->

← Previous revision		Revision as of 09:26, 13 June 2017
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	Where to have '''m''' digit choose :<math>K=10^m </math>.		Where to have '''m''' digit choose :<math>K=10^m </math>.

	Example of ~~Square~~ ~~Root~~ ~~of 10~~, ~~with~~ 1 digit precision:		Example of Rational Complicate Modulus, for 1 digit precision:

	:<math>M_{2,10}=2x/K-1/K^2 </math>.		:<math>M_{2,10}=2x/K-1/K^2 </math>.

Talk:Nth root algorithm - Revision history

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87.26.160.199: /* Complicate modulus method */

87.26.160.199: /* Complicate modulus method */

StefanoMaruelli: /* Complicate modulus method */

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StefanoMaruelli: /* Complicate modulus method */

StefanoMaruelli: /* Complicate modulus method */

StefanoMaruelli: /* Complicate modulus method */

StefanoMaruelli: /* Complicate modulus method */

StefanoMaruelli: /* Complicate modulus method */

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