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

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

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)

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

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

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)])

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

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))