janhardo

765 Reputation

12 Badges

11 years, 164 days

MaplePrimes Activity


These are answers submitted by janhardo

The pattern recognition of simplify falls short for this expression and must first be converted into a recognizable form for simplify via collect.
That seems like a good explanation to me.

This is working in a worksheet 

This is the secant method as a procedure for not singular elliptic curves.
I did some examples, but th eprocedure code is not yet fully functional...
Example 1 is single point and example 2 is doubling example 1 

fxy := y^2 - 2*y + 14 = 2*x^3 + 11*x^2 - 29*x - 17:
P := [3, 7];

# Test met jouw kromme
fxy := y^2 - 2*y + 14 = 2*x^3 + 11*x^2 - 29*x - 17:
P := [3, 7];

printf("=== TEST 1: POINT DOUBLING 2P ===\n");
P2 := EllipticCurveSecantMethod(fxy, x, y, P, P, -10..10, -10..15, 
                               steps = true, showplot = true);
printf("2P = (%a, %a)\n", P2[1], P2[2])


Use : m > 1 and not 3m (m= 1... n )  


Found k  with 2 ai's , but the solution to this is the interesting part

From the Maple Programming Guide, section "5.3 Constructors":
?object,ModuleCopy:

it was rather easy with ai to come up with this animation , but now the details ...:-)

restart;
with(PDEtools);
with(LinearAlgebra);
with(SolveTools);
_local(gamma);
K := 2*k[i]*exp((-(alpha*l[i]^3 + (-k[i]^2 - b - c)*l[i] - a)*k[i]*Int(1/g(_z1), _z1 = 0 .. t) + 2*(k[i]*(y + z)*l[i] + x*k[i] + eta[i])*beta)/(2*beta))/(1 + exp((-(alpha*l[i]^3 + (-k[i]^2 - b - c)*l[i] - a)*k[i]*Int(1/g(_z1), _z1 = 0 .. t) + 2*(k[i]*(y + z)*l[i] + x*k[i] + eta[i])*beta)/(2*beta)));
K1 := eval(K, i = 1);
Fig1params := {a = 1, alpha = 1, b = 1, beta = 1, c = 1, y = 1, z = 1, eta[1] = 1, k[1] = 1, l[1] = -2, r[1] = 1};
test := value(eval(K1, {g = (t -> cos(t)), i = 1}));
unum := eval(test, Fig1params);
ux := eval(unum, t = 0);
printf("=== CONTROLE ===\n");
printf("test = ");
print(test);
printf("unum = ");
print(unum);
printf("ux = ");
print(ux);
plot(ux, x = -10 .. 10, numpoints = 300, title = "ux plot");
plot3d(unum, t = -10 .. 10, x = -10 .. 10, shading = ZHUE, grid = [50, 50], lightmodel = light4, style = surfacecontour, title = "unum 3D plot");
=================================================================
restart;
with(plots);
K := 2*k[i]*exp((-(alpha*l[i]^3 + (-k[i]^2 - b - c)*l[i] - a)*k[i]*Int(1/g(_z1), _z1 = 0 .. t) + 2*(k[i]*(y + z)*l[i] + x*k[i] + eta[i])*beta)/(2*beta))/(1 + exp((-(alpha*l[i]^3 + (-k[i]^2 - b - c)*l[i] - a)*k[i]*Int(1/g(_z1), _z1 = 0 .. t) + 2*(k[i]*(y + z)*l[i] + x*k[i] + eta[i])*beta)/(2*beta)));
K1 := eval(K, i = 1);
Fig1params := {a = 1, alpha = 1, b = 1, beta = 1, c = 1, y = 1, z = 1, eta[1] = 1, k[1] = 1, l[1] = -2, r[1] = 1};
K1_num := (x_val, t_val, g_func) -> eval(eval(K1, {g = g_func, t = t_val, x = x_val, Int(1/g(_z1), _z1 = 0 .. t) = evalf(Int(1/g_func(s), s = 0 .. t_val))}), Fig1params);
ux_num := x -> K1_num(x, 0, cos);
printf("=== TEST EENVOUDIGE PUNTEN ===\n");
for i from -2 to 2 do
    result := ux_num(i);
    printf("ux_num(%d) = %a\n", i, result);
end do;
plot(ux_num(x), x = -10 .. 10, numpoints = 300, title = "2D Plot: ux_num(x)");
plot3d(K1_num(x, t, cos), x = -10 .. 10, t = -5 .. 5, shading = ZHUE, grid = [30, 30], title = "3D Plot: K1_num(x,t,cos)");

generalized hyperbolic distribution
Generalized Hyperbolic Distributions - Jim Killingsworth
GENERALIZED_HYPERBOLIC_DISTRIBUTIONmprimes7-10-2025.mw


We can use the max­i­mum like­li­hood method to fit a gen­er­al­ized hy­per­bol­ic dis­tri­b­u­tion to a giv­en set of data
Let me try   for stock exchance in Amsterdam (AEX) ?, no i  take first the example data his­tor­i­cal stock prices of Mi­crosoft Cor­po­ra­tion

GH_Distribution_Plot := proc(mu, delta, alpha, lambda, {beta := 0, xmin := -10, xmax := 10})
restart;
f := 419*x^2 + 116*x*y - 426*x*z + 78*y^2 - 142*y*z + 133*z^2 - 1604*x - 682*y + 1086*z + 2306;
df_dx := diff(f, x);
df_dy := diff(f, y);
df_dz := diff(f, z);
centrum := solve({df_dx = 0, df_dy = 0, df_dz = 0}, {x, y, z});
                      "maple.ini in user"

             df_dx := 838 x + 116 y - 426 z - 1604

              df_dy := 116 x + 156 y - 142 z - 682

             df_dz := -426 x - 142 y + 266 z + 1086

               centrum := {x = 7, y = 11, z = 13}

It is a quadratic surface, with this point : centrum ? : point symmetry around centrum, needs more info.

eqt1 := u(x, y, 0, t) = 4*alpha*(-lambda[1]*t*(lambda[1]^2 - 3*lambda[2]^2)*alpha + (b*lambda[1] + ((r[1] + r[2])*c)/2 + a)*t + beta*(y*lambda[1] + x))*lambda[2]^2*beta/(lambda[2]^2*t^2*(lambda[1]^2 + lambda[2]^2)^3*alpha^3 + 3*t*lambda[2]^2*alpha^2*(2*(b*t + beta*y)*lambda[2]^4/3 + t*c*(r[2] - r[1])*lambda[2]^3*I/3 + 2*lambda[1]*((((r[1] + r[2])*c)/2 + a)*t + x*beta)*lambda[2]^2 - lambda[1]^2*t*c*(r[2] - r[1])*lambda[2]*I - (2*lambda[1]^3*((b*lambda[1] + ((r[1] + r[2])*c)/2 + a)*t + beta*(y*lambda[1] + x)))/3) + lambda[2]^2*((b*t + beta*y)^2*lambda[2]^2 + t*c*(b*t + beta*y)*(r[2] - r[1])*lambda[2]*I + ((b*lambda[1] + c*r[2] + a)*t + beta*(y*lambda[1] + x))*((b*lambda[1] + c*r[1] + a)*t + beta*(y*lambda[1] + x)))*alpha + beta^2)

Can this function be used now ?
1 2 3 4 5 6 7 Page 1 of 7