Analysis of simple branching trees with TI-92

Transcription

1 Analysis of simple banching tees with TI-9 Dušan Pagon, Univesity of Maibo, Slovenia Abstact. In the complex plane we stat at the cente of the coodinate system with a vetical segment of the length one unit. Then we tun to the left and to the ight fo a positive angle ϑ, not exceeding 90 degees, and add two new segments of length <. The last two steps ae epeated infinitely many times, to obtain a self simila factal object, which we call simple banching tee (SBT). Typical questions about SBT ae like: at what elation between and ϑ the banches of ou tee will meet (ovelap), and what will be the bounday cuve, when thee is no ovelapping. The integation of algebaic and geometic tools that gaphic calculato TI-9 possesses, allows us to find the answes to most of these questions. Fo instance, by summation of geometic pogessions, an explicit connection between and ϑ is found fo the bounday case and the obtained polynomial equations ae solved numeically. Plane and spatial factals ae becoming quite popula in diffeent aeas of science and technology. Ou pupose, in this pape, is to discuss a special class of plana factal sets, which ae used as a model in medicine, with TI-9 gaphic calculato. Factals ae usually defined as objects with nonintege dimension. We can not hee go into details of geneal Hausdoff-Bezicovich dimension, but will shotly intoduce a special case, the so called self-simila dimension. A compact set in a metic space is called self-simila (in a stong sense) with a similaity coefficient p, when it can be divided into N conguent sets, each of which is exactly p-times smalle copy of the oiginal set. As an example, we can take a squae in IR, o a cube in IR 3, divided into an abitay numbe of equal pieces. N=6, p= N=7, p=3 As we can notice, in all such examples the atio ln n ln p is equal to the dimension of the consideed objects. Theefoe, the idea appeas, to define the dimension of any self-simila compact set of points with this atio. Let s see two examples, the fist of which is known in mathematics as the Siepinski capet. d = ln 3 ln =,5850. d = ln ln 3 =,68.

2 As we have aleady mentioned, in what follows, we will study a special plana factal, which is a bounday set fo all banches of a simple banching tee. In the complex plane we stat at the cente of the coodinate system an place a vetical segment (in the diection of the imaginay axe) of unit length. Then we tun to the left (in positive sense) and to the ight fo an angle ϑ, and each time we add a segment, which has the length of the pevious segment, multiplied with the coefficient <. The last step is epeated instantly to constuct the simple banching tee T (, ϑ). Thus, the total length of each complete banch of this tee will be =. Below we show two examples: T (0.6, 5 ) and T (0.65, ).

3 Clealy, the tees themselves ae not self simila, but the esulting bounday sets ae, and thei dimension, accoding to ou definition is ln ln 0.5 = ln ln. This is only tue, when the banches of ou tee ae not ovelapping. So, the next step is to find the values of and ϑ, fo which the two pats of ou tee touch n+ The fist way, to each the symmety axe, is to move up (in the diection of the imaginay axe) fom cente of the coodinate system fo one unit and then, afte the left (positive) tun fo an angle ϑ, go on fo <. Next, we tun fo angle ϑ in the negative sense (eaching the oiginal diection of the imaginay axe), and continue fo units. Fo an angle n ϑ (n ),n we epeat the last move n times, always tuning fo the same angle ϑ and always shotening ou step with the facto. In this way, we will obtain the closest point to the imaginay axe (afte the mentioned numbe of steps), and, at the same time, we each the most hoizontal diection. Depending on whethe n ϑ n o n ϑ (n ), one of the diffeences (n )ϑ o nϑ will be smalle, indicating the next move towads the imaginay axe. Afte this, the altenation of positive and negative tuns fo the angle ϑ will take us the most fa in the desied diection (towads the imaginay axe). Thus, we will each the symmety axe if and only if the following inequality holds: sin ϑ + 3 sin ϑ + sin ϑ n sin(n )ϑ + n+ sin(n )ϑ+ + n+ sin nϑ + n+3 sin(n )ϑ + n+ sin nϑ + n+5 sin(n )ϑ The above expession contains imaginay pats of the sums of geometic sequences: n 3 I. sin ϑ + 3 sin ϑ n sin(n )ϑ =Im( e iϑ k e ikϑ )= = Im( e iϑ n e i(n )ϑ e iϑ ) = Im( e iϑ ( n e i(n )ϑ )(e iϑ ) (e iϑ )(e iϑ ) )= = Im( e iϑ ( n e i(n 3)ϑ n e i(n )ϑ e iϑ +)) ( cos ϑ ) +( sin ϑ) = n+ sin(n )ϑ n sin(n )ϑ+ sin ϑ cos ϑ+. Finally, subtacting the fist them in ou inequality (sin ϑ), we obtain cos ϑ+ (n+ sin(n )ϑ n sin(n )ϑ +sin ϑ cos ϑ sin ϑ). By CAS of TI-9 we fist find the sum of a finite geometic pogession with abitay coefficient x, and then substitute x = e iϑ. The obtained expession is fist simplified by specifying the ange of vaiables and ϑ, and then we can apply the tcollect command. II. The emaining two infinite geometic sequences contibute with: 3

4 ( n sin(n )ϑ + n+ sin nϑ)( )= n sin(n )ϑ+ n+ sin nϑ. So, multiplying the initial inequality with the positive common denominato ( cos ϑ + )( )/, we finally obtain: n+ sin(n )ϑ n sin(n )ϑ n+3 sin(n )ϑ + n+ sin(n )ϑ +( ) sin ϑ( cos ϑ )+ + n ( sin(n )ϑ + 3 sin nϑ sin(n )ϑ cos ϑ sin nϑ cos ϑ + sin(n )ϑ + sin nϑ) 0, o: n+3 (sin nϑ sin(n )ϑ)+ n+ (sin(n )ϑ sin nϑ cos ϑ)+( ) sin ϑ( cos ϑ ) 0 n+3 cos(n )ϑ n+ cos nϑ +( )( cos ϑ ) 0. () To eliminate the facto sin ϑ, we have to use a combination of commands texpand and tcollect. Simila calculations show, that: n n k cos(kϑ) =Re( k e ikϑ )= n+ cos(n )ϑ n cos nϑ cos ϑ +, yielding the equality: cos ϑ + ( cos ϑ +) n k cos(kϑ) = so an equivalent inequality to () is: = n+3 cos(n )ϑ n+ cos nϑ +( )( cos ϑ )+ cos ϑ +, n Example. Fo k =, 3 we, espectively, obtain: 3 cos ϑ + 0 and cos ϑ + 3 cos ϑ + 0. k+ cos(kϑ) 0. () Polynomials P n in left pat of inequalities () pass though point (0, ), while the initial conditions (n )ϑ nϑ yield that P n() = cos(n )ϑ cos nϑ > 0. The fist deivative of the left side in () is clealy positive fo 0(as0<ϑ n ). So, each of the polynomials P n has a unique zeo on the inteval (0, ). Fo the maximal angle ϑ = banches of ou tee touch if =, while fo vey small ϑ>0 the citical value of uns to.

5 The expession at the left side of () is a function of two vaiables, theefoe, to plot it we can use the 3D gaphing capability of TI-9. But, to obseve the behavio of the citical eal oot, belonging to the inteval (, ), it is moe useful to plot the gaph of the polynomial of vaiable fo the value of vaiable ϑ fom the elevant inteval ( n+, n ). Fo small values of n (coesponding to elatively big angles ϑ) we can solve by TI-9 the inequality () symbolically, to expess explicitely one of the vaiables, ϑ with the othe. As the ode of the discussed equation exceeds, the numeical solving gives us the equied citical value. In the following table we list the polynomials in vaiable and the bounday values fo angles ϑ = n, n =, 3,...0. angle ϑ polynomial with citical oot ( 5 ) + 3 ( 5+) sin + sin cos ( ) (+ ) sin cos cos ( 5 ) ( 5+) As we see, the maximal possible value of, fo which the obtained tee-set is not self-ovelapping, is in the case of the staight angle ϑ. But then the dimension of the obtained object is ln ln =! So this is not any moe a factal! In fact, as we will see a little late, in this case we obtain a whole ectangle with sides and. A staightfowad explanation of this phenomena is based on the fact, that the coodinates of each point in the unit squae can be uniquely witten as a decimal sequence of 0 and in a binay base. Fo instance, fo each, standing at an abitay 5

6 m-th position afte the decimal comma in such a notion fo x =0,d d..., which is peceeded by k 0 zeos, we move to the ight at (m k)-th step, and then etun to the left duing the next k even steps. In this way, we will obtain the equied component:... m k m k+ = m. m Now we can pass to the second case of possible self ovelapping. Again, we stat at the cente of the coodinate system, and go up (in the diection of the imaginay axe) fo one unit. Then we tun to the ight (in negative sense) fo the angle ϑ and 3 move fowad fo the length. If fo the angle ϑ we know, that 3 m ϑ 3 m,m, we will epeat the last move m times (always tuning fo the same angle ϑ and always shotening ou step with the facto ). In this way, afte the mentioned numbe of steps, we m will obtain the neaest possible point to the imaginay axe at the basement of ou tee. Afte this, the altenation of positive and negative tuns fo the angle ϑ will take us the most fa in the desied diection - towads the imaginay axe and thus, towads the symmetical banch of ou tee, with which the ovelapping 0 can take place. So this time the condition, that in ou tee ovelapping occus, is sin ϑ + sin ϑ m sin(m )ϑ + m sin mϑ + m+ sin(m )ϑ + m+ sin mϑ We can ewite this inequality in the fom Re(e m 3 iϑ k e ikϑ )+( m sin(m )ϑ + m sin mϑ) k 0 m sin(m )ϑ m sin(m )ϑ+ sin ϑ cos ϑ+ + m sin(m )ϑ+ m sin mϑ 0 and simply it futhe to: m+ cos(m )ϑ m cos mϑ + 0. (3) It follows, that fo ϑ = banches of ou tee will touch in this way when =, while fo 6

7 vey small positive values of ϑ the citical value of uns towads. Keeping in mind that < (m )ϑ 3 mϑ, we notice that the left side of the last inequality will have a negative deivative (m +) m cos(m )ϑ m m cos mϑ fo all > 0. On the othe hand, the polynomials in (3) take a positive value at = 0 and negative value cos(m )ϑ cos mϑ when =. Thus, fo each of these polynomials on the inteval (0, ) exists a unique zeo. We have aleady shown, that ( cos ϑ +) m k cos(kϑ) = m+ cos(m )ϑ m cos mϑ cos ϑ +, so the conditions (3) can also be witten in an equivalent fom: m Fo k =, 5 we, espectively, obtain the following inequalities: 3 cos 3ϑ + cos ϑ + cos ϑ + 0, cos ϑ + 3 cos 3ϑ + cos ϑ + cos ϑ + 0. k= k cos(kϑ)+ 0. () Again, we can plot the suface, given by the left side of the inequality (), as well as the membes of the family of polynomials, obtained fom the coesponding equality, when is explicitly expessed fo diffeent values of angle ϑ = 3 n The positive zeos of the fist ten such polynomials ae given in the following table, indicating, that the self touching (and ovelapping) at the top of the tee will always peceed such contact at the basement of the same tee angle uppe intesection lowe intesection occus, if is geate then occus, if is geate then = = = Now we pass to the discussion of some geometical figues, which contain ou tee factal set. Fist - 7

8 we will show, that in all but one case (when we have obtained a ectangle) the measue of the esulting factal set is equal to 0. Then, we will out scibe the tee factal pieces of logaithmic spials. G C a 0 B E D a x θ x 0 θ F h 0 h A a The cosine theoem, applied to the tiangle AEF, shows that x 0 = ( + +cos ϑ). On the othe side, x = x 0 + x 0 + x = x 0. Calculating x by Pythagoean theoem fom ectangula tiangles ABC, ABD and compaing the obtained expessions, we aive to the following system of equations: { ( a 0 ) + h 0 = x ( a ) +(h + ) = x o Theefoe, h 0 += ( )x { ( a 0 ) + h 0 = x ( a 0 ) +(h 0 +) =( x ). = + + cos ϑ, yielding h 0 = + cos ϑ and a 0 = x h 0 = sin ϑ. The aea of the initial tiangle ABG equals S = a 0h 0, while the aea of the obtuse tiangle ABF, with which we fill the gap, is T = a 0. So, fo the total aea S 0 of all tiangles, which contain ou factal, we obtain: S 0 = S +(T + S + T + S +...)= S+T S = a 0( +h 0 ( +)) ( ) = sin ϑ( cos ϑ++cos ϑ) ( ). 3 To minimize the aea of all tiangles, in which the factal is embedded, we can apply the above pocedue sepaately to each of its k components fo any positive intege k. In this case, the total aea of polygons, containing ou factal, becomes S k = k S 0 k = S 0 ( ) k, and, thus, uns towads 0, as we incease k, fo evey <. In the exceptional case, when equals (and ϑ must be to avoid ovelapping), we fo all k IN obtain S k = S 0 =, what coesponds to the aea of the ectangle with sides ( +( ) 3 +( ) )= = and + +( ) +...= =, which is entiely filled with the consideed factal. 8

9 Befoe we descibe the second aea, that includes the tee factal set, we must efesh ou geometical knowledge about a cetain class of cuves. The logaithmic spials ae given in pola coodinates with the equation R = ae bϕ,a,b>0. Each of these cuves has its cente at the beginning of the coodinate system (ϑ R 0). As R = ae b(ϕ+ϕ 0) = ae bϕ e bϕ 0 = k 0 ae bϕ, the adius R of the point on logaithmic spial always inceases k 0 = e bϕ 0 -times, as we add ϕ 0 to the agument ϕ of any point on the spial. So, if we notice the same popety duing the study of simple banching tees, it will be possible to daw a logaithmic spial though all the vetices of ou tee, that satisfy this condition. The slope of the tangent line to the logaithmic spial at point (R cos ϕ, R sin ϕ) is (R sin ϕ) (R cos ϕ) = R sin ϕ+r cos ϕ R cos ϕ R sin ϕ b sin ϕ+cos ϕ tan α = = b cos ϕ sin ϕ, yielding, that the angle, between the adius vecto of the point on the spial and tangent line, will be: tan α tan ϕ tan α ϕ = +tan α tan ϕ = sin ϕ b sin ϕ cos ϕ+cos ϕ+b sin ϕ cos ϕ b sin ϕ+sin ϕ cos ϕ+b cos ϕ sin ϕ cos ϕ = b. Theefoe, we found the second basic popety: this angle is constant and equal to actan b. Let us find the coodinates of the point S, to which we will aive, fom the beginning of the coodinate system, with only ight tuns fo angle ϑ<, and stetching the initial vetical unit segment on each step with the facto <. Theefoe, we conside the sum: e i + e i( ϑ) + e i( ϑ) +...= i( + e iϑ +(e iϑ ) +...)= i i ie = = iϑ sin ϑ+i( cos ϑ) e iϑ ( cos ϑ) +( sin ϑ) = + cos ϑ. Next, we move the cente of the coodinates to the point S, obtaining that the bounday points sin ϑ of the fist segment in ou tee (the unit segment) ae A 0 = ( + cos ϑ, cos ϑ + cos ϑ ) and 9

10 sin ϑ + cos ϑ, cos ϑ A =( + cos ϑ ). By induction, we obtain that A i S = i +. Fom this infomation we can conclude, that cos ϑ all the tiangles A j A j S, j IN ae simila and thei angle at vetex S always equals ϑ. This popety guaantees, that all the ends A j of the segments, obtained in the tee T,ϑ by only ight tuns will indeed belong to the logaithmic spial with cente at the point S and the equation R = ae ln ϑ ϕ. Finally, we find the coefficient a, by applying the last equation to the point A 0 : e ln cos ϑ (+actan + ϑ sin ϑ ) = cos ϑ (+actan ϑ sin ϑ ) cos ϑ + cos ϑ a = So, the equation of the inne spial, which contains all points A j, is: R = ( ) ϑ (ϕ actan. cos ϑ sin ϑ ) + cos ϑ In a simila way, the equation of the logaithmic spial, containing all the vetices B i, i 0of the bounday polygon fo ou factal set, can be found. We shall shotly efe to this cuve as to the elated oute spial. Again, we stat with the coodinates of this vetices in the coodinate system with cente at point S, and the axe of coodinates paallel to the initial unit segment: B 0 = G = ( )(+ cos ϑ) ( sin ϑ( cos ϑ), sin ϑ), B = B = ( )(+ cos ϑ) ( ( cos ϑ), sin ϑ). We obtain: B S = sin ϑ ( ) + cos ϑ and B 0S = sin ϑ ( ) = + cos ϑ B S = B S =... So, the two spials, elated to the same factal set, will only diffe by the multiplicative constant, which we can easily calculate by substituting the point B, and obtain the final equation: R = sin ϑ ( ) ( + cos ϑ ) ϑ (ϕ actan sin ϑ cos ϑ ) Using the fomulae fo calculating the aea in pola coodinates, we see that the two pats of both ou logaithmic spials suound the aea p = ( )dϕ = (a a )ϑ ( ln ) ϑ. Finally, we can discuss the possibilities of ceating simple banching tees with TI-9. This constuction is independent of all pevious esults, so it could, as well, be a pat of the intoduction, used as motivation fo all othe discussions and calculations. Stating with a segment, and using the Rotation and Dilation commands, we can poduce a maco, which will constuct the next 0

11 level of ou tee. Fist of all, we otate the oiginal segment (of a unit length) aound one of its vetices fo the angle α = ϑ. Then we stetch the obtained copy towads the common point with the facto. And finally, we otate the last segment fo the angle β =ϑ =( α). In this way, the initial objects of the basic maco ae the two vetices of the oiginal segment, and the thee numeical values: α,, β. The final objects of the same maco ae the two stetched segments of the length. As each segment is defined by its vetices, we can not avoid the tiplication of the common point of ou thee segments. This does not affect ou figue visually, but takes some additional memoy of ou TI-9. Afte the basic maco is pepaed, we can join thee o fou such steps togethe in a highe maco, a consecutive use of which will give us a good appoximation of the final infinite tee. The easonable numbe of steps fo a paticula tee depends on the facto, as the TI-9 sceen makes it impossible to see vey shot segments. All the advantages of the dynamic geomety can be demonstated, by vaying the values of and α and poducing vey quickly many diffeent tees. We can even include the Animation tool, to expeimentally find the appoximate connection between the stetching facto and the angle ϑ. REFERENCES. Factals in Biology and Medicine, Bikhäuse Velag, Band I, II, Mandelbot B. Die Faktale Geometie in de Natu, Bikhäuse Velag, Mandelbot B. The Canopy and the Shotest Path in Selfcontacting Factal, Math. Intelligence (999); 8-7. Peitgen H., Jügens H., Saupe D. Factals fo the Classoom, pat I, II, New Yok Schoede M. Faktale, Chaos und Selbstähnlichkeit, Spektum Velag Zeitle H., Pagon D. Faktale Geometie, Vieweg Velag Zeitle H. Bonhien und Baumstuktuen (pepint)