Nth root algorithm: Difference between revisions

Revision as of 19:41, 18 April 2011

Revision as of 09:40, 6 November 2011

Line 43:

leading to the general ''n''^th root algorithm.

== N-th perfect root comes from Tartaglia's triangle and "Complicate Modulus" Method ==

Given C and n as integer, is possible to write that:

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

Since this sum can be used for any integer C and n, we call M(n) Complicate modulus:

:<math> M(n)=[x^n -(x-1)^n] </math>.

How to have the Complicate modulus M(n):

M(n) comes from Tartaglia's develope of <math>(x-1)^n</math> in this way:

Remembering Tartaglia's triange:

n=0 1

n=1 1 1

n=2 1 2 1 M(2) = (2x-1)

n=3 1 3 3 1 M(3) = (3x^2-3x+1) etc...

n=4 1 4 6 4 1 M(4) = (4x^3-6x^2+4x-1) etc...

So our complicate moduls M(n) is the n row, less first element, changeing all the sings.

Example if n=2 :

:<math>C^2 =\sum_{x=1}^{C}[x^2-(x-1)^2]</math>.

That is the very well known odds sum:

:<math> C^2 = 1+3+5+7+9+ ... +[C^2-(C-1)^2)</math>.

Example if n=3 :

:<math>C^3 =\sum_{x=1}^{C}[x^3-(x-1)^3]</math>.

Or special odds sum:

:<math> C^3 = 1+7+19... +[x^3-(x-1)^3)</math>.

etc...

=== Why "Complicate Modulus" ? ===

Because this is a special clock characterized by:

- 2 hands: the shortest one show the Numbers of Turns, the loger 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 witch function we are using.

=== How to make the n-th root: ===

If we wanna make the 3-th root of 27 we:

1) From Tartaglia's debvelope of (x-1)^n we keep n=3 row,

n=3 1 -3 +3 -1

- we elimiate the first value

-3 3 -1

- we change all the signs:

3 -3 1

So we have the complicate modulus M(3):

:<math> M(3)= 3x^2 - 3x + 1 </math>.

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 se onto the clock

So at the 1st turn we will se:

turn: x=1

M(3)x = 3x^2 - 3x + 1 = 1 division

at the 2th we will see:

turn: x=2

M(3)x = 3x^2 - 3x + 1 = 7 division

at the 3th we will see:

turn: x=3

M(3)x = 3x^2 - 3x + 1 = 19 division etc...

=== To make the e-th root of 27 so we have to tabulate: ===

x <math> M(3)= 3x^2 - 3x + 1 </math> 27

1 1 27-1 = 26

2 7 26-7 = 19

3 19 19-19 = 0 <- REST

So the 3th root of 27 is 3 and since the rest is zero this is a perfect cube.

So we can wrote

<math> 27 = 3M(3)+ 0 </math> or <math> 27= 3^3</math>

In case we have the 3th 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….

To confirm we can just tabulate again:

x <math> M(3)= 3x^2 - 3x + 1 </math> 28

1 1 28-1 = 27

2 7 27-7 = 20

3 19 20-19 = 1 <- REST

So:

<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 durring 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* derivate of (M(n)

Example:

M(2) = 3 * derivate of M(3) = 3* derivate (3x^2 - 3x + 1) = 3* (2x-1)

Infact M(2) = (2x-1)

Stefano Maruelli

==References==