CLASSICAL PROBLEMS et alii with web Links

 

By:Eur Ing Panagiotis Stefanides CEng MIET

http://www.stefanides.gr    panamars@otenet.gr

 

1.            Tracing ARCHIMEDES Method Of Polygons for “CIRCLE MEASUREMENT”

2.            The Delian DOUBLING of the CUBE

3.            A Non Linear Trigonometric IDENTIRY

4.            Non Linear Trigonometric EXPRESSION for P i

5.            A Non Linear Expression of THREE NUMBERS in a ROW

6.            TRISECTION of an Angle ITERATIVELY

7.            Links

 

 

 

 

 

1       Tracing ARCHIMEDES Method Of Polygons for “CIRCLE MEASUREMENT”

By:Eur Ing Panagiotis Stefanides CEng MIET

© Copyright Panagiotis C. Stefanides.

 

y(n+1)=0.5SQRT[A^2+B^2] ,   A=y(n) ,B=[A^2]/C

C=SQRT[R^2-A^2] +R ,  R=RADIUS  OF CIRCLE=2 units

THETA[R]rad =2[THETA(0)]Pi/180[2^n] ,THETA(0)=30

2y(0)=2 units ,corresponds to one of the six sides of a regular

HEXAGON inscribed in the CIRCLE of radius 2 units,

The arc of which subtends 60 deg ,at the center of theCIRCLE.

y(0)=1

y(1)=0.517638090205041524697797675248097

y(2)=0.261052384440103183096812455790978

y(3)=0.130806258460286133630631117550351

y(4)=0.0654381656435522841273198526345763

y(5)=0.0327234632529735632859438469683461

y(6)=0.016362279207874258570398246589218

y(7)=0.00818120805246957918924210834302384

y(8)=0.00409061258232819022882611784626426

y(9)=0.00204530736067660908238592229206388

y(10)=0.00102265381402739500247163869946902

y(11)=5.11326923724832989668067056305212e-4

y(12)=0.0002556634639513086612958129429003

y(13)=1.27831732236765852255655919873982e-4

y(14)=6.39158661510218663537974854526572e-5

y(15)=3.19579330795908007059262098055451e-5

y(16)=1.59789665403053837941159531748569e-5

y(17)= 7.98948327017683312504301824151782e-6

y(18)= 3.99474163509638505378952308041447e-6

y(19)= 1.99737081754918858830326403024717e-6

y(20)= 9.98685408774718801827694849662372e-7

2THETA30/[180(2^n)]=

For     n        0 = 0.333333333333333333333333333333333

“        1=0.166666666666666666666666666666667

“        2=0.0833333333333333333333333333333333

“        3=  0.0416666666666666666666666666666667

“        4=  0.0208333333333333333333333333333333 

“        5= 0.0104166666666666666666666666666667

“        6= 0.00520833333333333333333333333333333

“        7= 0.00260416666666666666666666666666667

“        8= 0.00130208333333333333333333333333333

“        9= 6.51041666666666666666666666666667e-4

“        10=3.25520833333333333333333333333333e-4

“        11=1.62760416666666666666666666666667e-4

“        12=8.13802083333333333333333333333333e-5

“        13=4.06901041666666666666666666666667e-5

“        14=2.03450520833333333333333333333333e-5

“        15=1.01725260416666666666666666666667e-5 

“        16=5.08626302083333333333333333333333e-6

“        17=2.5431315104166666666666666666665e-6

“        18=1.27156575520833333333333333333325e-6 

“        19=6.357828776041666666666666666665e-7

“        20=3.1789143880208333333333333333325e-7

 

y(16)/( 2th30/[180(2^16)] )   = 3.14159265355636089699354932180887

y(17)/( 2th30/[180.(2^17)] )  = 3.14159265356585361409691546085749

y(18)/( 2th30/[180.(2^18)] )  = 3.14159265357212029062180221517697

y(19)/( 2th30/[180.(2^19)] )  = 3.14159265357368695975302507567124

y(20)/( 2th30/[180.(2^20] )   = 3.14159265357407862703583086403904

 

[ BACK CHECKING CIRCLE  RADIUS   FOR ACCURACY OF COMPUTER CALCULATIONS:

 

h(19)=9.9737254569952718968759340441557e-13

h(19)^2=9.9475199491515545244016386648656e-25

y(19)=9.9475199491515545244016386648656e-25

y(19)^2=3.9894901827971140079143423452449e-12

y(19)^2+h(19)^2=3.9894901827971140079143423452449e-12

y(19)^2+h(19)^2/2h(19)=2.00000000000000000058096940421553  =R=Radius of Circle

Actual R=2  ]

 

© Copyright Panagiotis C. Stefanides

 

Links

 

http://mathforum.org/kb/thread.jspa?forumID=13&threadID=87711&messageID=429947#429947

e j pi

http://mathforum.org/kb/thread.jspa?forumID=13&threadID=43562&messageID=548825#548825

http://mathforum.org/kb/message.jspa?messageID=314852&tstart=0

 

 

 

 

 

 

2       The Delian DOUBLING of the CUBE

 

By:Eur Ing Panagiotis Stefanides CEng MIET

© Copyright Panagiotis C. Stefanides

 

Let

1]  SIGMA = SUM

2]  X=SIGMA{([2M] ^n )/n!}for  n=0   to infinity  the SIGMA limits

3]  M=SIGMA{(1/[2N-1])*[K]^2N } ,  for  K=1/3  ,and  for  N =1 to infinity  the SIGMA limits

 

a) Find M [summing the  first nine terms ] .

          M=   [1/1]*[K^2]        +

+     [1/3]*[K^4 }      +

+     [1/5]*[K^6]        +

+     [1/7]*[K^8 ]       +

+     [1/9]*[K^10 ]     +

+     [1/11]*[K^12]    +

+     [1/13]*[K^14]    +

+     [1/15]*[K^16]    +

+     [1/17]*[K^18 ]   +

M=0.11552453.. ,         2M=0.23104906..

 

Find  X  [summing the  first nine terms ]

          X=    [1/0!]* [2M]^0   +

          +     [1/1!]*[2M]^1    +

          +     [1/2!]*[2M]^2    +

          +     [1/3!]*[2M]^3    +

          +     [1/4!]*[2M]^4    +

          +     [1/5!]*[2M]^5    +

          +     [1/6!]*[2M]^6    +

          +     [1/7!]*[2M]^7    +

          +     [1/8!]*[2M]^8    +

X = 1.25992105..

[1.25992105..]^3 = 2

X = CUBIC ROOT OF  2

Using  X as the side of  a CUBE  ,its VOLUME  X^3 =2  is DOUBLE   in volume of  the CUBE whose side has length of  UNITY and volume 1*1*1=1  ,  i.e. we have a solution  to the DELIAN PROBLEM  of  DOUBLING OF THE CUBE.

 

© Copyright Panagiotis C. Stefanides

 

Links

 

http://mathforum.org/kb/thread.jspa?forumID=130&threadID=357981&messageID=1094568#1094568

 

 

 

 

 

 

3       A Non Linear Trigonometric IDENTITY

By :Eur Ing Panagiotis Stefanides CEng MIET

© Copyright Panagiotis C. Stefanides

 

SQRT[X] =Tan{ArcCos[SQRT(1/ [ (1/B)+1] ) ] }

 

Where,

 

B=[1/D^2]-1

 

D = Sin{ArcTan[SQRT(X)] }             [SQRT=SQUARE ROOT ]

 

[For X positive real number ]

EXAMPLE:

X=365.6  , D = 0.998635184.., B = 0.00273523…

SQRT365.6=19.1206693.. = Tan{ArcCos[SQRT(1/ [ (1/B)+1] ) ] }

Tan{ArcCos[SQRT(1/ [ (1/0.00273523…)+1] ) ] }=19.1206693..

 

 © Copyright Panagiotis C. Stefanides

 

Links

 

http://mathforum.org/kb/thread.jspa?forumID=13&threadID=111770&messageID=576366#576366

 

 

 

 

 

4          A Non Linear Trigonometric EXPRESSION for P i .

By:Eur Ing Panagiotis Stefanides CEng MIET

© Copyright Panagiotis C. Stefanides

 

For any positive real number X:

 

Pi= 4*ArcTanX + 2*ArcTan[(1-X^2)/2X] , Working in Radians , or ,

 

180= 4*ArcTanX + 2*ArcTan[(1-X^2)/2X] ,Working in Degrees ].

 

EXAMPLE:

X=365.6  , [Working in Degrees]

4*ArcTan[365.6] + 2*ArcTan[(1-[365.6] ^2)/2[365.6]] = 180

X=365.6  , [Working in Radians]

4*ArcTan[365.6] + 2*ArcTan[(1-[365.6] ^2)/2[365.6]]=3.141592654..

=Pi = 3.141592654..

 

© Copyright Panagiotis C. Stefanides

 

Links

 

http://mathforum.org/kb/thread.jspa?forumID=13&threadID=1085185&messageID=3400097#3400097

The Square Root of the Golden Section By an Iterative Method.

http://mathforum.org/kb/thread.jspa?forumID=13&threadID=101817&messageID=516846#516846

 

 

 

 

 

5          A Non Linear  Expression of THREE NUMBERS in  a ROW

By:Eur Ing Panagiotis Stefanides CEng MIET

© Copyright Panagiotis C. Stefanides

 

(n+1)tan[arctan(n)-arctan(1)]=(n-1)

This NON LINEAR expression  is  of inerest , as it relates 3 numbers (n+1),n,(n-1)]  in row,differing by unity.

The geometric proof of :

(n+1)tan[arctan(n)-arctan(1)]=(n-1) ,

requires proving,that:

angle[arctan(n)-arctan(1)]=angle{arctan[(n-1)/(n+1)]}

The values for n, may be positive,negative integers, or

fractions,e.g.

3tan[arctan(2)-45deg]=1

4tan[arctan(3)-45deg]=2 ....

Geometric proof that: arctan(n)-arctan(1)=arctan[(n-1)/(n+1)]

1.Draw X,Y, axes,with origon 0.

2.With center o,draw 3 concentric circles:

a).The radius of the first circle is unity(1),crossing the X-axis,

say,at E,and the Y-axis,say,at F.

b).The radius of the second is n(representing a number on the Xand Y-axis),crossing the X-axis,say at B,and the Y-axis,say,

at A(make it greater than the first circle ,for drawing clarity)

c).The radius of the third circle exceeds n, by unity(n+1),crossing

X=axis, at D and Y-axis at C.

3.Draw another circle,with center A , and radius unity(1),crossing the Y-axis at G and point C.

4.Join DC,DG,AE and EF.

5.Draw a perpendicular from G to DC, crossing DC at H.

Angle CDO=angle DCO=45deg=angle FEO= angle EFO ,

(DC is parallel with EF).

Call angle GDO=P, and angle AEF=R,

Angle HDG=(45-P), angle AEO=(45+R),and

angle EAO=90- (45+R)=(45-R)   

OA=n , OC=n+1 , OF=1 , OG=n-1

OB=n , OD=n+1 , OE=1

*Triangle CHG:Angle HCG=45deg=angle HGC. Since GA=AC=1, GC=2,

thus : HC=HG=SQRT(2)     [i.e. point H on circle center A ]

*Triangle HDG: Since angle HDG=(45-P), then,

angle HGD=90-(45-P)= =(45+P).

DC=SQRT[OD^2+OC^2]= SQRT[2*(n+1)^2]=(n+1)SQRT(2).

DH=DC-HC=(n+1)SQRT(2)-SQRT(2)=n*SQRT(2) .

Ratio : HG/DH=SQRT(2)/n*SQRT(2)=1/n=tan(45-P) .

*Triangle AEO : Angle EAO=90-(45+R)=(45-R) .

Ratio : EO/OA=1/n=tan(45-R), thus :

1/n=tan(45-P)=1/n=tan(45-R), i.e. angle R= angle P.

Since ratio:OG/OD=(n-1)/(n+1)=tan(P) , or arctan[(n-1)/(n+1)]=P,

and ,ratio:OA/OE=n/1=tan(45+R)=n, or arctan(n)=(45+R),

R=[arctan(n)-45], then since R=P :

[arctan(n)-45]=arctan[(n-1)/(n+1)] or,

(n+1)tan[arctan(n)-arctan(1)]=(n-1).

EXAMPLE:  n = 365.6  , 

366.6tan[ 89.84328327 – 45 ] =364.6

 

© Copyright Panagiotis C. Stefanides

 

Links

 

http://mathforum.org/kb/thread.jspa?forumID=13&threadID=70265&messageID=307651#307651

http://mathforum.org/kb/message.jspa?messageID=307652&tstart=0

http://mathforum.org/kb/message.jspa?messageID=307653&tstart=0

http://mathforum.org/kb/thread.jspa?forumID=130&threadID=357422&messageID=1093401#1093401

http://mathforum.org/kb/thread.jspa?forumID=130&threadID=357422&messageID=1093400#1093400

 

 

 

 

 

6          TRISECTION of an Angle ITERATIVELY ****

By:Eur Ing Panagiotis Stefanides CEng MIET

© Copyright Panagiotis C. Stefanides

 

 

 VALUE OF

SIN[THETA/3] :

SIN[THETA/ 3]=( SIN[THETA] )/ [3-4K^2]

WHERE

K =( SIN[ THETA ] )/ L

WHERE

L=3-4M^2

WHERE

M=( SIN[ THETA ] )/ N

WHERE

N=3-4P^2

WHERE

P=SIN[THETA]

 

ETC. ETC.

 

 

The algorithm is an iterative formula

and works for values :

0 DEG < THETA < 45DEG

The compass and ruler is also an approximate solution

making use the formula:

[SIN[THETA]]^2 =[1-COS(2THETA)]/2

 

For THETA=45 DEG ,SUBTRACT A 30 DEG ANGLE ,

OR BISECT 30 DEG ANGLE ,

as constructable for a 30 DEG - 60 DEG Ortogonal

Triangle.

 

For greater angles one manipulates them accordingly ,

making use of the 45 DEG OR OTHER SMALLER

THAN 45 DEG TRISECTIONS.

 

 

 

 

 

EXAMPLE FOR THETA = 30 DEG :

SIN[THETA]=0.5

SIN[30]=0.5

 

{SIN[30]}^2=0.25

 

1        3-4{SIN[30]}^2=2

2        (SIN[30])/2=0.25

3        (0.25)^2=0.0625

4        3-4(0.0625) = 2.75

5        (SIN[30])/2.75=0.181818182 ..

6        (0.181818182.. )^2=0.033057851..

7        3-4(0.033057851..)=2.867768595..)

8        (SIN[30])/ 2.867768595..=0.174351585..

9        (0.174351585..)^2=0.030398475..

10      3-4(0.030398475..)=2.878406099..

11      (SIN[30])/ 2.878406099..=0.173707247..

12      (0.173707247..)^2 =0.030174208..

13      3-4(0.030174208..)= 2.879303169..

14      (SIN[30])/ 2.879303169..=0.173653127..

15      (0.173653127.. )^2 =0.030155409..

16      3-4(0.030155409..) = 2.879378365..

17      (SIN[30])/ 2.879378365.. = 0.173648592..

18      (0.173648592..)^2=0.030153834..

19      3-4(0.030153834.. ) = 2.879384666..

20      (SIN[30])/ 2.879384666.. = 0.173648212..

21      [0.173648212..]^2 = 0.030153702..

22      3-4[0.030153702..] =2.879385193..

23      (SIN[30])/ 2.879385193.. = 0.173648181..

24      [0.173648181..]^2 = 0.030153691..

25      3-4[0.030153691..] = 2.879385238..

26      (SIN[30])/ 2.879385238..= 0.173648178..

27      [0.173648178..]^2 = 0.03015369..

28      3-4[0.03015369..] = 2.879385241..

29      (SIN[30])/ 2.879385241.. =0.173648178..

30      ArcSin [0.173648178..] = 10 DEG

 

SIN[THETA/3]= [0.173648178..]

 

 [THETA/3]= 10 DEG

 

For the ruler and compass Geometrical Solution ,

and for 0 <angle theta <45

 

1        DRAW ANGLE ,UNIT HYPOTENUSE,

2        FIND SINE MAGNITUDE ---------------------[A]

3        DOUBLE ANGLE ,FIND COSINE MAGNITUDE [FOR HYPOTENUSE ALSO UNITY]. SUBTRACT IT FROM UNITY ,AND TAKE HALF.TAKE FOUR TIMES THIS HALF AND  SUBTRACT THE RESULT FROM 3 UNITS.---[B]

4        WITHIN ORIGINAL ANGLE DRAW A NEW ORTHOGONAL TRIANGLE WITH [B] AS COSINE AND [A] AS SINE .THE NEW ANGLE WILL BE THE TANGENT OF THIS TRIANGLE AND THIS WILL BE ITS FIRST APPROXIMATION.

 

ALL ABOVE WILL BE REPEATED FOR THE

NEW , GRAPHICALLY FOUND, ANGLE ,

FIND NEW VALUE FOR [B] CALL IT [C]

AND FIND NEW ANGLE AS ARCTANGENT{[A]/[C])

AND CONTINUE UNTILL SATURATION

OF THE PROCESS.

 

 

**** [ ADDENDUM

An other interesting solution which

I, circulated in handouts ,

to colleagues of mine ,in 1993, is:

 

Assume angle akl.

 Draw isosceles triangle akl ,with ak=kl.

With ak as radius draw arc al ,and then

bisect it

into two equal angles ae , and el .

Then continue BISECTING BISECTIONS .

to total four , and eight equal angles.

Let arc al be ,as such ,

divided into eight eight equal arcs:

ab , bc , cd , de , ef , fg , gh , and hl.

 

The first approximation of trisecting akl , is

obtained by adding akb to bkc ,dke to ekf ,and

gkh to hkl.

Angle ckd is divided into four equal angles

two of which are addet to angle akc , one to dkf

leaving one more for further divisions by four.

Similarly angle fkg is divided into four

equal angles,

two of which are added to angle gkl ,one to fkd,

leaving one more for further divisions by four.

And so on untill saturation of the process.

It is realized that the continuous process is :

[1/4]^1+[1/4]^2 + [1/4]+^3+..+..[1/4]^n = [1/3]

which gives the trisection approximation . ]

 

 

[ SECOND ADDENDUM

 

TRISECTION OF THETA AS LOGARITHMIC [TRIG/ e ] FUNCTION

 

THETA/3 =Ln[B^60]                 [Ln is the Natural Logarithm base e]

B=Tan{ArcSin[D]}

D=SQRT[F]

F=1/[G]

G=[e^h]+1

h= -THETA/90

Where THETA is in degrees.

 

examplum gratias:

for THETA=65 deg

h=- 0.722222222..

G=0.485671785 +1=1.485671785..

F=1/1.485671785 = 0.673096178..

D=0.820424389..

B=1.434922888..

B^60 = 2568702300..

Ln[2568702300]=21.66666667.. 

THETA/3 = 21.66666667..

 

© Copyright Panagiotis C. Stefanides

 

Links

 

http://mathforum.org/kb/search!execute.jspa?q=eur ing panagiotis stefanides trisection of an angle&rankBy=10001&threadID=349956

http://mathforum.org/kb/message.jspa?messageID=1072470&tstart=0

 

 

 

 

 

7       Links

A.   LINKS to “ The Math Forum @ Drexel”

Click  on any of the following  URLs  to open and then  introduce to  the

“Search  All Discussions”window :   PANAGIOTIS STEFANIDES .

Then click on  GO  to find “Math Forum” discussions of PANAGIOTIS STEFANIDES  .

http://mathforum.org/kb/forum.jspa?forumID=13&start=405

http://mathforum.org/kb/forum.jspa?forumID=130

http://mathforum.org/kb/forum.jspa?forumID=149

http://mathforum.org/kb/forumcategory.jspa?categoryID=2

 

 

 

 B.LINKS to the   NATIONAL CURVE BANK Mathematics

     Department Team of the    CALIFORNIA STATE UNIVERSITY ,   

     Los Angeles CA:   http://curvebank.calstatela.edu/log/log.htm

(Mathematical Curves-relating Spirals to Logarithms )

http://curvebank.calstatela.edu/home/home.htm  (Recently Submitted Curves )            

http://curvebank.calstatela.edu/greatlinks/greatlinks.htm  ( Great Liks to Other Sites )

 

 © Copyright Panagiotis C. Stefanides

 

Links

 

http://www.2dcurves.com/links.html

 

 

C.LINKS to the  UNIVERSITY OF PORTO:

-        http://www.direito.up.pt/IJI/node/62   (VOL II/III)

-        http://www.antigona.web.pt

 

 

 

http://www.stefanides.gr          panamars@otenet.gr