@Carl Love It's faster but quality of resolution does not satisfy me.

@Carl Love 

thank u my French friend but I know what is it I am just looking for revolution surface of polar coords to get a wrinkled circle I used this Nash torus bq I wanted to recreate it (not perfect but only to get some similar plot) 

Edit:

so if you said that there could not be such option to rotate this curve it's sad but ok maybe you know how to wrinkle given torus? or use polar coords in 3D to get this bq I found something about bumpu spheres:

I didnt know about Borrelli so this is new information for me I learnt that it is called Nash torus so u must be worry about ur French heritage so thank u for this artcile I will spread this in my sorroundings thanks once again

@Carl Love 

I thought about use parametrization of torus:

display(plot3d([(R + r*cos(t)*cos(s), (R + r*cos(t)*sin(s), r*sin(t)], t = 0 .. 2*Pi, s = 0 .. 2*Pi), view = [-2 .. 2, -2 .. 2, -2 .. 2])

and than x=rcos(t) y=rsin(t) so if I could get x, y equation of this wrinkled circle I would sketch this bq I dont need excactly to get that wave fractal shape

Edit:

I started to think that it is naive approach bq this cartesian equation could be implicit function and the problem is not gonna be solved like that

@Preben Alsholm 

I have found an idea for this equation from the link below. But I cannot say that it is not a gradient if he wrote that Delta is a local average as laplace operator.

https://www.youtube.com/watch?v=bOn7m03ei0o&ab_channel=NilsBerglund

@rcorless 

Do you know maybe how to change this to a number?

"0,002441406"

I used parse to elimante string but I don't know how to change this "," to a dot.

Edit: I've already found this parse(SubstituteAll("0,002441406", ",", "."))

only problem is with minus sign because parse delete this

@acer 

Thanks, that's the idea.

@gkokovidis 

this plot if from analytic solution

dsolve({diff(x(t), t, t) = 0, diff(y(t), t, t) = -g, x(0) = 0, y(0) = 0, D(x)(0) = 10, D(y)(0) = 2})

plot([10*t, -981/200*t^2 + 2*t, t = 0 .. 10])

and this from numeric solution looks like only x(t) plot

odeplot(dsolve({diff(x(t), t, t) = 0, diff(y(t), t, t) = -g, x(0) = 0, y(0) = 0, D(x)(0) = 10, D(y)(0) = 2}, numeric), numpoints = 1000)

@Rouben Rostamian  

I used taylor expantion, because I couldn't calculate this for sin(x) (it worked only for polynomials)

@Carl Love 

I wrote the previous message without seeing this additional comment thanks. It's excatcly what I wanted.

@Carl Love 

Thanks, nice plot It's something I wanted but I used TallyInto correctly?

x[0] := 0.2;
for i from 0 to 10^4 do
    x[i + 1] := 3.7*x[i]*(1 - x[i]);
end do;
h:=<seq(x[i], i = 0 .. 10^4)>
with(Statistics);
TallyInto(h, [0 .. 1]);
Histogram(h, 'binbounds' = [seq(n/1000, n = 1 .. 1000)])

@mmcdara 

you are right I shoulded write (I wanted separate this by unit circle (in this case this unit circle is an interval)):

if abs(x(t)) > 1 then 
   diff(x(t), t $ 2) = 2
elif abs(x(t)) < 1 then 
   diff(x(t), t $ 2) = 0
end if

@Rouben Rostamian  

I have done your method:

restart;
with(LinearAlgebra);
with(plots);
assume(t, real);

K := plot3d(1, theta = 0 .. 2*Pi, phi = 0 .. Pi, coords = spherical):

f := (theta, phi) -> (cos(theta)*sin(phi), sin(theta)*sin(phi), cos(phi)):

P := pointplot3d([[f(Pi/4, Pi/8)], [f(0, Pi/8)], [f(Pi/4, (3*Pi)/8)]], color = [black, white, white], symbol = solidsphere):

A := <f(0, Pi/8)>

B := <f(Pi/4, (3*Pi)/8)>

N := simplify(CrossProduct(A, B))

N1 := simplify(N/Norm(N))

M := simplify(CrossProduct(N1, A))

A*cos(t) + M*sin(t)

Vector[column](3, [cos(t)*sin(1/8*Pi) + sin(t)*(-1 + 2*sqrt(2))*cos(1/8*Pi)/(1 + sqrt(2)), 2*sin(t)*cos(1/8*Pi)*sqrt(2)/(1 + sqrt(2)), cos(t)*cos(1/8*Pi) + sin(t)*(1 - 2*sqrt(2))*sin(1/8*Pi)/(1 + sqrt(2))])

S2 := display(spacecurve([cos(t)*sin(t + Pi/8), sin(t + Pi/8)*sin(t), cos(t + Pi/8)], t = 0 .. 2*Pi, thickness = 3, color = green), spacecurve([cos(t)*sin(Pi/8) + sin(t)*(-1 + 2*sqrt(2))*cos(Pi/8)/(1 + sqrt(2)), 2*sin(t)*cos(Pi/8)*sqrt(2)/(1 + sqrt(2)), cos(t)*cos(Pi/8) + sin(t)*(1 - 2*sqrt(2))*sin(Pi/8)/(1 + sqrt(2))], t = 0 .. 2*Pi, thickness = 3, color = green))

what should be scale to get nice great circle?

Edit: I guessed this equation for green line but now I see it's not the great circle so probably you are right because you got shorter path

@Rouben Rostamian  

how did you calculate this? I used parametrization alpha(t)= (cos(t)*sin(t + Pi/8), sin(t + Pi/8)*sin(t), cos(t + Pi/8)) where 0<=t<=Pi/4 for green line and calculate integral int(sqrt(alpha'(t)*alpha'(t)),t=0..Pi/4) where "*" is symbol for dot product and alpha'(t) is derivative of our parametrization.

@acer I'd like to calculate this interesection point for parametrization [f(t),g(t)]. This point isn't equal [2,2] or [exp(1)+exp(-1),exp(-1)+exp(1)]

@one man 

I meant this red line intersection point