Prof. Dr. Ludwig Paditz,

University of Applied Sciences

Germany

Workshop I (Ciechocinek, Poland, 27th June 2004)

Several aspects of 2D-graphics with the ClassPad300

      Let’s beginning with some nice graphics (cartesian, polar or parametric): 

      1. a snowflake  r1 = r(q) = 3 - cos(6q) ,  0 < q < 2p , -8 < x < 8 , -4 < y < 4.

      2. a butterfly swimmer  r2 = r(q) = q , r3 = r(q) = 1,  0 < q < p , -20 < x < 20 , -10 < y < 10 ,

                                              y4 = y(x) = - (x-5)^6 + 5,  y5 = y(x) = - (x+5)^6 + 10, y6 = y(x) < 0.

       3. a spider  r7 = r(q) = q * (sin(q))^2 ,   0 < q < 6p , -10 < x < 10 , -6 < y < 6 .

      4.  the number 8  r8 = r(q) = cos(q/2)* sin(q) ,   0 < q < 4p , -1.6 < x < 1.6 , -0.8 < y < 0.8 .

5.  a hibiscus flower  r9 = r(q) = 2,  r10 = r(q) = sin(22q)+3+lg(9) ,

      r11 = r(q) = sin(13q)+5+lg(5) , 0 < q < 2p , -8 < x < 8 , -7 < y < 7 .

6. a bear  r12 = r(q) = 4*sin(q/5),  0 < q < 5p , -6.8 < x < 6.8 , -3.5 < y < 4.5,

             xt13 = x(t) = cos(t/5+p) ,                 yt13 = y(t) = sin(t/5+ p) – 1,

             xt14 = x(t) = cos(t)/3 + 1.6 ,             yt14 = y(t) = sin(t)/3 + 1.6,

             xt15 = x(t) = cos(t)/3 - 1.6,               yt15 = y(t) = sin(t)/3 + 1.6,

             xt16 = x(t) = cos(t/5-p/4) + 2.8 ,      yt16 = y(t) = sin(t/5-p/4) + 2.8,

             xt17 = x(t) = cos(t/5+p/4) - 2.8,       yt17 = y(t) = sin(t/5+p/4) + 2.8.

7. a CD-rom r18 = r(q) = seq(R, R, 1.1, 4, 0.145), 0 < q < 2p , -9 < x < 9, -4.5 < y < 4.5.

8. a disk-capacitor  List1 = {-5, -4, -3, -2 ,-1, 0, .5, 1, 1.25, 1.5, 1.75, 2} ,

      List2 = {-.75, -.5, -.25, 0, .25, .5, .75}*p , (input in main-menu)

y21= y(x) = p*signum((-x)^.5),  y22 = y(x) = -y21(x), (the disk-capacitor)

xt23 = x(t) = e^List1 * cos( (2t+1)p / 11) + List1 + 1 ,  (the potential-lines)

yt23 = y(t) = e^List1 * sin((2t+1)p / 11) + (2t+1)p / 11 ,

xt24 = x(t) = e^t * cos( List2 ) + t + 1 ,  (the field-lines)

yt24 = y(t) = e^t * sin( List2 ) + List2,     -6 < q,t < 5 , -5 < x < 10, -5 < y < 5 .

9. a four-leaf clover r25 = r(q) = 3*cos(2q), 0 < q < 2p , -8 < x < 8, -4 < y < 4 .

10. an exponential spiral r26 = r(q) = 3*e^(ln(0.8)* q /(2p)), 0 < q < 8p , -6 < x < 6, -3 < y < 3 .

Now consider a random function, using the random generator of ClassPad300:

     y27 = y(x) = rand(0,1) , y28 = y27(x) , -10 < x < 10, -1 < y < 2 .

Now consider a random hatching in a parallelogram:

     y27 = y(x) = rand(0,1) , y29 = y(x) = y27(x)*(2-|x-2|) + 2 + (|x-2|-x)/2,  0 < x < 4, 0 < y < 4 .

Now consider a periodic random signal with time out (pause):

y27 = y(x) = rand(0,1),   y30 = y(x) = ln(cos(x))/ln(cos(x)),   y31 = y(x) = y27(x)*y30(x) ,

-10 < x < 10, -1 < y < 2 .

Nonlinear regression in a scatter plot (log-log-scaling of the view-window):

     xlist = {61, 62, 64, 66, 68, 71, 75, 80, 90, 99, 110, 130, 145, 175, 220, 300, 390} ,

     ylist = {1000, 500, 200, 100, 50, 20, 10, 5, 2, 1, 0.5, 0.2, 0.1, 0.05, 0.02, 0.01, 0.005} .

     view-window :   10 < x < 400 ,   -100 < y < 1100 .

     view-window (log-log-scaling):   50 < x < 500 ,    0.001 < y < 1500 .

The power-regression in the log-log-scaling shows the bad approximation:

Problem:         How to improve the nonlinear regression?

Solution:         Transformation of the pairs (xi, yi) into (log(xi), log(yi)+3) ,

                        where “+3” to get positive values again.

Power-regression gives  y19 = y(x) = a*x^b, i.e. log(y(x)) + 3 = a*log(x)^b , thus

                        y20 = y(x) = 10^(y19(log(x) – 3) = 10^y19(log(x)) / 10^3 .

Summary: the scatter-plot with nonlinear regression:

            left:       the simple power-regression   y(x) = a * x^b

right:     the power regression with log-log-data: log(y(x)) + 3 = A * (log(x))^B

            and back-transformation:  y(x) = 10^( A * (log(x))^B – 3)