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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
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
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
- 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
=== How to make the n-th root: ===
If we
1) From Tartaglia's
n=3 1 -3 +3 -1
- we
-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
So at the 1st turn we will
turn: x=1
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M(3)x = 3x^2 - 3x + 1 = 19 division etc...
=== To make the
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
<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
=== Note ===
We also note that M(n-1)= n*
Example:
M(2) = 3 *
Infact M(2) = (2x-1) ▼
==References==
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