Nth root algorithm: Difference between revisions

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== N-th perfect root comes from Tartaglia's triangle and "Complicate Modulus" Method ==
 
 
Given C and n as integer, is possible to write that:
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How to have the Complicate modulus M(n):
M(n) comes from Tartaglia's developeenvelope of <math>(x-1)^n</math> in this way:
 
Remembering Tartaglia's triange:
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n=4 1 4 6 4 1 M(4) = (4x^3-6x^2+4x-1) etc...
 
So our complicate modulsmodulus M(n) is the n row, less first element, changeingchanging all the singssigns.
 
Example if n=2 :
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Because this is a special clock characterized by:
 
- 2 hands: the shortest one show the Numbers of Turns, the logerlonger one the Rest
 
- at each turn the numbers of division of the clock turn change by the complicate modulus that is the known function.
 
- The complicate modulus must be declared onto the clock to understand witchwhich function we are using.
 
=== How to make the n-th root: ===
 
If we wannawant to make the 3-th root of 27 we:
 
1) From Tartaglia's debvelopeenvelope of (x-1)^n we keep n=3 row,
 
n=3 1 -3 +3 -1
 
- we elimiateellimiate the first value
 
-3 3 -1
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We call X the numbers of complete turns of the short hand and
 
M(3)x [or M(3) calculated in x], is the number of division we will sesee onto the clock
 
So at the 1st turn we will sesee:
turn: x=1
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M(3)x = 3x^2 - 3x + 1 = 19 division etc...
 
=== To make the e-th3rd root of 27 so we have to tabulate: ===
 
x <math> M(3)= 3x^2 - 3x + 1 </math> 27
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<math> 27 = 3M(3)+ 0 </math> or <math> 27= 3^3</math>
 
In case we have the 3th3rd root of 28 without make other calculation we can say that will be:
 
<math> 31 = 3m(3) + 1 </math> or <math> 31 = 3^3 +1 </math> etc….
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<math> 28 = 3m(3) + 1 </math> or <math> 28 = 3^3 +1 </math> etc….
 
Was clear that this method is intriguing since in case we have to make a very long series of root calculation we never loose of precision durringduring the calculation...
 
(we are noticed, but is not jet not proved, that, probably, Daniel Bernoulli do, or try to do, somethink of similar)
 
=== Note ===
 
We also note that M(n-1)= n* derivatederivative of (M(n)
 
Example:
 
M(2) = 3 * derivatederivative of M(3) = 3* derivatederivative (3x^2 - 3x + 1) = 3* (2x-1)
 
Infact M(2) = (2x-1)
 
InfactIn fact M(2) = (2x-1)
Stefano Maruelli
 
==References==