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- Posted December 4, 2012 by
- PCSTEFANIDES

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## SPIRALOGARITHM – POWERS,SERIES,SIMPLY,BY COMPASS AND RULER By Panagiotis Stefanides

SPIRALOGARITHM – POWERS,SERIES,SIMPLY,BY COMPASS AND RULER

By Panagiotis Stefanides

http://www.stefanides.gr/why_logarithm.htm

SPIRALOGORITHM [ name coinned by Panagiotis Stefanides].

Configuration giving simple relationships between spirals and logarithms.

Introduced here is the STEPHANOID CURVE[ authors proposal of nomination] , the mirror images of two Archimedes curves of a special form,

passing Zero x-y point and on which lie vector length logarithms, of vector line lengths on the spiral[ one on top of the other].

Let X=e^[ Θ/90], be a point on the curve of this Spiral of Logarithmic base e ,on the plane of x-y co-ordinate system of Cartesian axes.

Phasor OX is at angle Θ from the x-axis taking anticlockwise direction as positive(Θ on x-axis is taken zero )and clockwise negative.

This Spiral crosses the [-y ] axis at point [ 1/e ],then the x axis at point 1 [moving anticlockwise] ,following crosses the y axis at point with value e ,then the [-x] axis at e^2... etc...

This gives ln(X)=[ Θ/90], OR in RADIANS ln(X)= Θ/[Pi/2].

The 0 of the x y cross section is the assymptotic point of the Spiral.

This theory is illustrated under:

http://www.stefanides.gr/why_logarithm.htm

TRANSFORMING [e^x] EXPANSION INVOLVING OTHER THAN [e ] BASES :

e^x = 1+x+(x^2/2!)+(x^3/3!)+(x^4/4!)+(x^5/5!)

may be transformed to other than e bases.

For instance,

Lets for reasons that serve my web work(references below) , instead of x use Z

where Z is a ral positive or negative number.

So e^Z= 1+ Z+ [Z^2]/2!+…+.....

Let X=e^Z , AND Log_e[X]=Z

So X= 1 + Log_e[X] + {[Log_e[X] ]^2}/2!+....

Raise e^Z to the power of N (POSITIVE REAL NUMBER),we get

e^[NZ]=1+NZ+{[NZ]^2}/2!+...

Let e^N = T SO THAT Log_e[T]=N

WHERE T is a POSITIVE REAL NUMBER

[BASE OF ANOTHER THAN e LOGARITHM ] , then ,

e^[NZ]= T^Z=1+Z[Log_e[T]] + [{Z[Log_e[T]]}^2]/2!+.....

Let this T^Z = Y, or Log_T[Y]=Z

AND ( from above ) Log_e[X]=Z = Log_T[Y] ,so

X=e^{Log_T[Y]} =1+{Log_T[Y]} + =…. AND

Y=T^{ Log_e[X] }=1+ [{ Log_e[X]}*{Log_e[T]} ]+…+….or

Y= 1+[{ Log_T[Y]}*{Log_e[T]} ] +…+…

According to this involved theory :

Log_e[X] = Log_T[Y] ==Z= Θ( deg)/90=Θ(rad)/[Pi/2]

For BASE T=1 then Y= T^Z=1 [ MODULUS (1) ,ARGUMENT (Θ) ] or

1^Z=1 [ MODULUS{sqrt[cos^2[theta] +sin^2[theta] }=1 ] ,

ARGUMENT (Θ)

or 1^{ Θ/[Pi/2] }=1[ MODULUS ( 1) ,ARGUMENT (Θ) ]

i.e. the ROOTS and POWERS of UNITY ,

And Log_1[1] = Z = Θ/90=Θ/[Pi/2].

This is the case of a CIRCLE, a DEGENERATED SPIRAL OF BASE 1

This CIRCLE passes points on x and y axes :

x=1 , y=1

REF:

http://www.stefanides.gr/why_logarithm.htm

Powers POWERS,SERIES,SIMPLY,BY COMPASS AND RULER.

Powers of a Length A[ab] Graphically by Compass and Ruler.

[The theory simply lies on the "similar orthogonal triangles"]

Draw a line horizontally of the Unit length of the Ruler.

Let it be [ej] ,and A avertical line [jh] ,at j corresponding to the Length A ,picked up by the compass.

Then [eh] will be the hypotenuse of the orthogonal triangle [ejh].

Let Θ be the angle whose tangent is: [hj]/[ej]=A/1=A.

We extend the line [ej] to [k],and [eh] to [q].

We use the compass and place the length A on [ek] ,

so we get a point on it [m] , that [em]=A.

Then , we draw a vertical line on [em] at [m],which meets

the line [eq] at [n].

Then [mn]=[A^2] , and tan(Θ) = A.

So we have graphically obtained the square of the Length A , GRAPHICALLY BY COMPASS and RULER .

We pick up the length [A^2] lay it along line [ek] so

we get a point on it [r] ,so that [er]=A^2 , the vertical

on it ,meets [eq] at [s] ,so [rs]=A^3, and tan(Θ)=A.

By doing so we get the powers we need of a length graphically by compass and ruler.

So any series that involves powers of a length x

[ such as e^x=1+x+[1/2!]x^2+... , log x , trigs etc ], may be drawn graphically.

Of course a similar method is involved for the product

of two lengths , inverses ,etc.

[The inverse 1/A is obtained

from the original triangle taking as vertical the Unit Length of the ruler.

Then the horizontal length is 1/A and tan(Θ)=A]

Ref:

http://mathforum.org/kb/thread.jspa?threadID=1191778&messageID=3882495#3882495

http://www.stefanides.gr/Html/logarithm.htm

http://www.stefanides.gr/Html/Nautilus.htm

© Copyright Panagiotis Stefanides

Panagiotis Stefanides

http://www.stefganides.gr

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