Fractals on non-euclidean metric

Transcription

1 Frctls on non-eucliden metric Yery Cchón Sntn April, 8 As fr s I know, there is no study on frctls on non eucliden metrics.this pper proposes rst pproch method bout generting frctls on non-eucliden metric. The ide is to extend the clculus of frctls on non-eucliden metrics. Using the Riemnn metric, there will be dened non-eucliden modulo of complex number in order to check the divergence of the series generted by the Mndelbrot set. It lso shown tht the frctls re not invrint versus rottions. The study will be extended to the quternions, where is shown tht the study of frctls might not be extended to quternions with generl metric becuse of the high divergence of the series condition in order to generte frctl is selecting bounded opertors). Finlly, Jv progrm will be found s exmple to show those kind of frctls, where ny metric cn be dened, so it will be helpful to study those properties. Contents Riemnn distnce Non-eucliden metric 3. Invrints Quternions 5 4 Conclusions 9 5 Jv progrm 9

2 Riemnn distnce As n introduction, let's using the length of the rc on Riemnn geometry: L = b dx g i dx j ij Once chosen bse, the metric tensor will be represented by mtrix G. g = n g ij dx i dx j i,j= g g g n. G = g g m g mn For exmple, using the eucliden metric g ij = L = b dx dx + dy dy = b dx + dy When considering the Mndelbrot set: ), the length of the rc will be the eucliden metric: z =,where c is complex constnt z n+ = zn + c This set is constructed following those steps: ) Get point of the complex plne ) Iterte z n+ = zn + c 3) If the succession diverges, this element doesn't belong to Mndelbrot set. If the succession converges, the element belongs. Lemm: The distnce on eucliden metric is the the modulo of the complex number. z n = x n + iy n x n + y n z = L = Proof: Tking the prmetric eqution: x n = t y n = t, the distnce will be: L = x n dx ) + dy ) = x + = x + = x + x) = x + y ` A compct topologicl spce X is self-similr if there exists nite set S indexing set of non-surjective homeomorphisms f s : s Sfor which X = f s X). If X Y. we cll X self-similr if it's only non-empty s S subset of Y such the previous equtions for f s : s S. Let's cll L = X, S, f s : s S) self-similr structure. The Mndelbrot set is self-similr round Misiurewicz points. A prmeter c is Misiurewicz point if M k,n if it stises the equtions: f k) c f k) c z cr ) nd c z cr ) z cr ) = f c k+n) z cr ) f k+n) so M k,n = c : f c k) z cr ) = f c k+n) z cr ), where z cr is criticl point k nd n re positive integers f k c denotes the k-th iterte of f c It doesn't depend on the metric chosen so, in non-eucliden metric, the properties of the frctl my be conserved. This llows to extend the study of the frctls to non-eucliden geometry.

3 Non-eucliden metric In this exmple, let's use non-eucliden metric, s it ws shown previously: ij n g = g ij dx i dx j i,h= ) cxy =, Firstly, let's clculte the term: dx g i dx j ij = cxy ) dx + dy L = x n ) cxy ) dx ) + dy Using the prmetric eqution: x = V t y = V t, considering the complex number z n = x n + iy n, the distnce will be: L = t n cxy dx ) + dy ) = V tn With the substitution u = tv c L = V u n V c + u du = un c + u du u + du = u u + + ln u + + u ) So, L = u n u n + c + u ln n + ) + u n Now, tking in considertion: = yn V t = yn x n nd u = tv c = x n L = ym x n yn cx n cxn y n : cv t + = V t n x n yn c x 4 n + cx n y n + ln x cxn n y n + ) + cx3 n y n + cv t = V t n + tv c ) So, the modulo on complex number x+iy considering non-eucliden metric is done by: L = ym yn ) x n x n cx n yn c x 4 n + cx n y n + ln x cxn n y n + + cx3 n y n Once the metric is dened, the Mndelbrot set: z = z n+ = zn, + c The divergence of the succession will be ccording to the new metric dened. So, ccording to the new modulo dened, the succession will converge or the will diverge. On prctice wy, tht it's done is set threshold vlue so, once the modulo is greter thn the threshold, the number of itertions needed to rech the threshold will dene the color of the pixel. This llows to crete computer progrms in order to grph those sets.. Invrints In this prt, let's check if the modulo nd the frctls re invrint versus rottions. Tking n eucliden metric: z n = x n + iy n z = L = x n + y n Considering rottion of ngle ϕ, w = ze iϕ, following the sme method: z = z n+ = e iϕ zn + c ),where c is complex constnt This set is constructed following those steps: ) Get point of the complex plne ) Iterte z n+ = e iϕ zn + c ) = e iϕ zn + c ϕ 3) If the succession diverges, this element doesn't belong to Mndelbrot set. If the succession converges, the element belongs. So, the rottion won't chnge the nture of divergence or convergence. Now, the question is if the number of itertions will be the sme fter rottion. z n+ = zn + c n zn + c n < L w n+ = e iϕ zn + c n e iϕ zn + cn = z n + cn < L The number of itertions is not invrint versus rottions. In order to see this, let's check this tble, which summrizes the process: 3

4 z n z o z = z o + c z = z + c = z o + c) + c = z 4 o + z c + c ) + c w n w = z o e iϕ w = w + c = ze iϕ + c = z e iϕ + c ce iϕ w = w + c = z e iϕ + c ce iϕ) + c = z + c ) e iϕ + c ce iϕ) + c = z + c ) e 4iϕ + c ce iϕ) + z + c ) e iϕ c ce iϕ) + c = z 4 o + z c + c )e 4iϕ + c + c e 4iϕ c e iϕ + z + c) ce iϕ c ) + c I = z e 4iϕ ce 4iϕ + I w = z e 4iϕ ce 4iϕ + I z e 4iϕ ce 4iϕ + I z e 4iϕ + ce 4iϕ + I = z + c + I So, fter itertions, the modulo won't mtch. As the threshold L is dened for the complex z, so the threshold is not respected, the thresholds dier by c + I. So the frctls re not invrint versus rottion. 4

5 3 Quternions This section will show study bout the possibility of using quternions to generte frctls. Firstly, let's dene the quternion: q = + bî + cĵ + dˆk The conjugte: q = bî cĵ dˆk, so the norm n) of quternion is dened by nq) = q q = + b + c + d. Dening the metric tensor G s G = cn be represented s: q = q α q α = q α G β αq β = b c d ) b c d, the modulo of quternion = + b + c + d, s q β = G βα q α = q In prticulr cse, considering the complex number z = + bi, this cn be represented on tensor wy s z α, so the squre modulo will be dened s: G = ), z = z z = z α G β αz β = b ) In this cse, frctl set on quternion cse will be like this: q = ) b ) = + b, s z β = G βα z α = z q n+ = qn + q c,where q c is quternion constnt This set is constructed following those steps: ) Get point of the quternion spce ) Iterte q n+ = qn + q c 3) If the succession diverges, this element doesn't belong to this set. If the succession converges, the element belongs. Bering in mind the quternion q n = n + b n î + c n ĵ + d nˆk, let's clculte q n. In order to do this, tke in considertion the tble of quternions multipliction: x i j k i j k i i - k -j j j -k - i k k j -i - so ij=k, ji=-k,jk=i,kj=-i,ki=j,ik=-j ) ) So, qn = n + b n î + c n ĵ + d nˆk n + b n î + c n ĵ + d nˆk = n b n c n dn) +n b n î+ n c n ĵ+ n d nˆk, Let's check the itertion with severl vlues: 5

6 itertion n b n c n d n,94,69959,549, ,9478,448,9497, , ,4554 -, , , , , , , ,4635 -,9647 -, , ,899,8568, , , ,36844,784 7, ,99489,77363, , ,8583,438675, , , , ,537858, ,583,467446, , ,488584,387965, , , , , , ,7769,5979, , , , , ,93 -, , , , , , , ,38897, , , ,76576,45975, ,7688 9,7735 -, , , , , ,3583 -,55347,633664,334974,6636,535583,8756,66837,538577, ,69595,3665,8596, ,959986, ,4968, ,89366, , , ,867685,936495,74488, , , ,959 -, ,444757, ,5489, , ,637696, , ,3685, , , As shown on the tble, the succession diverges very quickly. q n = n b n c n d n) + 4 nb n + 4 nc n + 4 nd n At the sme wy tht on the Mndelbrot set, it need to be checked if the succession converges or diverges. As prticulr cse, where c n = d n =, qn = n + b n = z n Following up the sme method s the section, let's try to dene non-eucliden metric in 4 dimensions. u Prior to this, let's consider the quternion q = x y nd let's consider the mtrix G giving by: z G = In prticulr cse, considering the complex number z = + bi, this cn be represented on tensor wy s z α, so the squre modulo ) will be dened s: G =, z = z z = z α G β αz β = b ) ) ) = b + b, s z β = G βα z α = z n g = g ij dx i dx j i,h= g = g g g 33 6

7 As this opertor is bounded, the metric will be bounded. Let's check this following the sme exmple s previously: q = q α q α = q α G β αq β = b c d ) g g g g b g g c = g + gb + g 33 g 33 d g g gc + g33d, s q β = q = G βα q α = g b g c = g b g c g 33 d g 33 d Let's dene this metric: g = n g ij dx i dx j i,h= = bxy cxz, dyz Becuse of the mtrix G is symmetric, it cn be found bse B n where the mtrix is digonl. Firstly, let's clculte the term q n : q n = q α G β αq β = q α G β α G βα q α ) = dx g i dx j ij = ) du + b x y ) dx ) + c x z dy + d y z ) dz ij L = t n du ) + b x y dx ) ) + c x z dy + d y z ) dz Using the prmetric equtions: u = V t x = V p t y = V p t z = V p 3 t considering the quternion q n = n + b n î + c n ĵ + d nˆk, ) + b x y dx ) + c x z dy du d V 4 p p 3t 4 V p 3 = V + b p 4 p V 6 + c p p p 3V 6 + d p p 4 3V 6 t 4 the distnce will be: L = t n V + b p 4 p V 6 + c p p p 3 V 6 + d p p4 3 V 6 t 4 L = V t n + b p 4 p +c p p p 3 +d p p4 3 V 4 t 4 Let's cll M 4 = b p 4 p +c p p p 3 +d p p4 3 V 4. + M 4 t 4 ) + d y z dz ) = V + b p p V 4 t 4 V p + c V p V p 3t 4 V p + L = V t n This integrl doesn't hve n ntiderivte, so let's try to do seril development: + M 4 t 4 + M 4 t 4 M 8 t M t , where M t 4 So, L = V t n + M 4 t 4 V t n + M 4 t 4 M 8 t M t = V This formul needs to be set in function of u n, x n, y n, z n ), so t = V u n p = xn u n p = yn u n p 3 = zn u n So, mking those substitutions, the distnce of 4d point will be: L L u n + u n + b p 4 p +c p p p 3 +d p p4 3 u 5 n 7 b x 4 n y n +c x n y n z n +d y n z4 n u n 7 b x 4 n y n +c x n y n z n +d x n y4 n b p 4 p +c p p p 3 +d p p4 3 ) u 9 n + b x 4 n y n +c x n y n z n +d y n z4 n ) 3 u 5 n +... t n + M 4 t 5 n M 8 t 9 n 7 + M t ) u 3 n ) 3 u 3 n +... b p 4 p +c p p p 3 +d p p4 3 As the succession diverges very quickly, could it be possible to clculte the modulo of the succession in order to crete frctl? Is there prcticl wy to crete those frctls? As the metrics depends on the coordintes, the sme criteri cnnot be pplied on generl cse. The frctl set on quternion cse will be like this: q =,where q n+ = qn q c is quternion constnt + q c 7

8 ) qn) k = q n q n...) k times In the lst exmple, it shown tht, in generl cse, the succession diverges very quickly, so there is no wy to clculte the modulo of the succession. Let's see wht re the conditions where the norm cn converge or diverge ccording to the point. Considering the mtrix g s n opertor, the succession will be bounded if the opertor is bounded. Let's see on the previous exmples, if the opertors re bounded or not. n Ex : g = g ij dx i dx j i,h= = b c d An opertor is bounded if, nd only if, Lv ψ M v χ. A trce clss opertor is compct opertor for which trce my be dened, such tht the trce is nite nd independent of the choice of bsis. g = T r g = b c d So, s the trce is bsolutely convergent, the opertor g is bounded nd, hence, the succession of the severl point cn be clculted to crete frctl. n Ex : g = g ij dx i dx j i,h= = bxy cxz dyz The trce g = T r g = bxy cxz dyz, is not bounded, so cnnot be clculte frctl becuse the succession will diverge in every point. Hving seen this, condition in order to generte frctl is selecting bounded opertors. n Ex 3:g = g ij dx i dx j i,h= = bcosx) ccosy) dcosz) This metric is bounded, g = T r g = bcosx) ccosy) dcosz), b c d g + b + c + d. Also, let's dene positive metric, so > b+c+d. Following the sme procedure: q n = q α G β αq β = dx g i dx j ij = ) du + b cos x) ) dx ) + c cos y) dy + d cos z) ) dz ij L = t n ) du + b cos x) ) dx ) + c cos y) dy + d cos z) ) dz Using the prmetric equtions: u = V t x = V p t y = V p t z = V p 3 t considering the quternion q n = n + b n î + c n ĵ + d nˆk, L = t n V + b cos V p t)v p + c cos V p t)v p + d cos V p 3 t)v p 3 In fct, we're interested to know if the length hs nite vlue. In this cse, it's known tht mb ) b fx) Mb ) where m = min fx) nd M = mx fx). L = V t n + b cos V p t)p + c cos V p t)p + d cos V p 3 t)p 3 Let's cll fx) = + b cos V p t)p + c cos V p t)p + d cos V p 3 t)p 3, so, in order to check if this function is bounded. f E = f = + b cos V p t)p + c cos V p t)p + d cos V p 3 t)p 3 f E = V b p 3 sinv p t) V c p 3 sinv p t) V d p 3 3sinV p 3 t) f E = V b p 4 cosv p t) V c p 4 cosv p t) V d p 4 3cosV p 3 t) t= -> f' E = nd f E <, so t t= is the mximum of the function, f mx = + b p + c p + d p 3, so the function hs mximum it's bounded). 8

9 4 Conclusions This pper proposes method bout crete frctls uses non-eucliden metric, which llows to extend the possibilities of study of frctls. It ws shown how to build distnce on Riemnn geometry in order dene the modulo of complex number. Then, it ws extended the study to the quternions, where it ws shown some conditions where the frctls re strictly divergent, so it cnnot be included. I hope tht this pper form the bsis for extending the study for those objects. At the end of this pper you cn nd Jv progrm, which llows to test severl formuls in order to see the imges generted. Feel free to send me ny comments to ycchon@gmil.com bout this pper. 5 Jv progrm This section shows n exmple of Jv which llows to show frctls using whtever modulo metric you wnt to set: pckge frctld; import jv.mth.*; import jv.wt.grphics; import jv.wt.imge.bueredimge; import jvx.swing.jfrme; public clss Frctld extends JFrme privte nl int MAX_ITER = 8; privte nl double ZOOM = 3; privte BueredImge I; privte double zx, zy, cx, cy, new_zx; privte int cr,cg,cb; // Color RGB privte double r,thet,vl; nl privte double centerx = -; nl privte double centery = 5; double modulo double zx,double zy) // Here you set the modulo you wnt to dene return zx * zx + zy * zy; public frctl_set) setbounds,, 8, 6); setresizbleflse); setdefultcloseopertionexit_on_close); I = new BueredImgegetWih), getheight), BueredImge.TYPE_INT_RGB); vl = ; for int y = ; y < getheight); y++) for int x = ; x < getwih); x++) zx = zy = ; cx = x - centerx) / ZOOM; cy = y - centery) / ZOOM; int iter = MAX_ITER; while modulozx,zy) < && iter > ) new_zx = zx * zx - zy * zy + cx; zy =. * zx * zy + cy; zx = new_zx; iter; cr=iter; cg=iter*; cb=iter*4; I.setRGBx, y, cr<<6) cg<<8) cb); public void pintgrphics g) 9

10 g.drwimgei,,, this); public sttic void minstring[] rgs) new Frctld).setVisibletrue); Here there is n exmple s center 48,6), zoom 4 iter using the modulo L = ym yn x n x n cx n yn c x 4 n + cx n y n + ln x cxn n y n + + cx3 n y n ). On the previous progrm, it should be use this function: double modulo double zx,double zy,double c) return zy/zx*mth.sqrtzy/zx/c)*.5*zx/zy*mth.sqrtc*c*mth.powzx, 4)+c*zx*zy) +.5*Mth.logzx*Mth.sqrtc*zx/zy)+Mth.sqrt+c*Mth.powzx, 3)/zy))); Using the eucliden metric: Using the non-eucliden metric:

11 References [ [ [ [ [] []