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POTTA SATEESH
am 3 Mai 2021
Kommentiert: POTTA SATEESH
am 4 Mai 2021
u1 = 0.32;
v1 = 0.032;
u2 = 1400000;
v2 = 70000;
u3 = 100;
v3 = 40;
a1 = -0.5;
a2 = -0.5;
a3 = 0.707;
A1 =0;
A2 = 0;
A3 = 0;
U3 = 0;
V3 = 0;
S = 1;
K = 0;
B1 = 5;
delta = 0.001;
LM = (u3^2)/(sqrt((u3^2)+(v3^2)));
LS = sqrt(log(1+((v3^2)/(u3^2))));
X1 = @(z1,u1,v1) (u1+(v1*z1));
X2 = @(z2,u2,v2) (u2+(v2*z2));
X3 = @(z3,U,V) (U+(V*z3));
f = @(X1,X2,X3,z1,u1,v1,z2,u2,v2,z3,U,V) ((X1(z1,u1,v1)*X2(z2,u2,v2))-(2000*X3(z3,U,V)));
while (abs(A1-a1)>0.01 && abs(A2-a2)>0.01 && abs(A3-a3)>0.01 && abs(U3-u3)>0.01 && abs(V3-v3)>0.01)
if S~=1
a1 = A1;
a2 = A2;
a3 = A3;
u3 = U3;
v3 = V3;
end
U = u3;
V = v3;
% B1 = (-((44800*a1)+(22400*a2)-(2000*V*a3))+sqrt(((44800*a1)+(22400*a2)-(2000*V*a3))^2-(4*(2240*a1*a2)*(-2000*U))))/(2*2240*a1*a2);
% B2 = (-((44800*a1)+(22400*a2)-(2000*v*a3))-sqrt(((44800*a1)+(22400*a2)-(2000*v*a3))^2-(4*(2240*a1*a2)*(-2000*u))))/(2*2240*a1*a2);
z1 = a1*B1;
z2 = a2*B1;
z3 = a3*B1;
A1 = (f(X1,X2,X3,z1+delta,u1,v1,z2,u2,v2,z3,U,V)-f(X1,X2,X3,z1,u1,v1,z2,u2,v2,z3,U,V))/(delta);
A2 = (f(X1,X2,X3,z1,u1,v1,z2+delta,u2,v2,z3,U,V)-f(X1,X2,X3,z1,u1,v1,z2,u2,v2,z3,U,V))/(delta);
A3 = (f(X1,X2,X3,z1,u1,v1,z2,u2,v2,z3+delta,U,V)-f(X1,X2,X3,z1,u1,v1,z2,u2,v2,z3,U,V))/(delta);
K = (sqrt((A1)^2+(A2)^2+(A3)^2));
A1 = -(A1)/K;
A2 = -(A2)/K;
A3 = -(A3)/K;
X3 = U+(A3*B1*V);
V3 = (X3*LS);
U3 = X3*(1-log(X3)+log(LM));
S = S+1;
end
error:
Index in position 1 is invalid. Array indices must be positive integers or logical values.
Error in newtrail_1>@(X1,X2,X3,z1,u1,v1,z2,u2,v2,z3,U,V)((X1(z1,u1,v1)*X2(z2,u2,v2))-(2000*X3(z3,U,V))) (line 24)
f = @(X1,X2,X3,z1,u1,v1,z2,u2,v2,z3,U,V) ((X1(z1,u1,v1)*X2(z2,u2,v2))-(2000*X3(z3,U,V)));
Error in newtrail_1 (line 40)
A1 = (f(X1,X2,X3,z1+delta,u1,v1,z2,u2,v2,z3,U,V)-f(X1,X2,X3,z1,u1,v1,z2,u2,v2,z3,U,V))/(delta);
0 Kommentare
Akzeptierte Antwort
Walter Roberson
am 3 Mai 2021
X3 = @(z3,U,V) (U+(V*z3));
So X3 is a function handle.
X3 = U+(A3*B1*V);
So X3 is a numeric variable now.
But you are inside a loop, and when you go back, X3 is still a numeric value but will be expected to be a function handle.
3 Kommentare
Walter Roberson
am 4 Mai 2021
u1 = 0.32;
v1 = 0.032;
u2 = 1400000;
v2 = 70000;
u3 = 100;
v3 = 40;
a1 = -0.5;
a2 = -0.5;
a3 = 0.707;
A1 =0;
A2 = 0;
A3 = 0;
U3 = 0;
V3 = 0;
S = 1;
K = 0;
B1 = 5;
delta = 0.001;
LM = (u3^2)/(sqrt((u3^2)+(v3^2)));
LS = sqrt(log(1+((v3^2)/(u3^2))));
X1 = @(z1,u1,v1) (u1+(v1*z1));
X2 = @(z2,u2,v2) (u2+(v2*z2));
X3 = @(z3,U,V) (U+(V*z3));
f = @(X1,X2,X3,z1,u1,v1,z2,u2,v2,z3,U,V) ((X1(z1,u1,v1)*X2(z2,u2,v2))-(2000*X3(z3,U,V)));
while (abs(A1-a1)>0.01 && abs(A2-a2)>0.01 && abs(A3-a3)>0.01 && abs(U3-u3)>0.01 && abs(V3-v3)>0.01)
if S~=1
a1 = A1;
a2 = A2;
a3 = A3;
u3 = U3;
v3 = V3;
end
U = u3;
V = v3;
% B1 = (-((44800*a1)+(22400*a2)-(2000*V*a3))+sqrt(((44800*a1)+(22400*a2)-(2000*V*a3))^2-(4*(2240*a1*a2)*(-2000*U))))/(2*2240*a1*a2);
% B2 = (-((44800*a1)+(22400*a2)-(2000*v*a3))-sqrt(((44800*a1)+(22400*a2)-(2000*v*a3))^2-(4*(2240*a1*a2)*(-2000*u))))/(2*2240*a1*a2);
z1 = a1*B1;
z2 = a2*B1;
z3 = a3*B1;
A1 = (f(X1,X2,X3,z1+delta,u1,v1,z2,u2,v2,z3,U,V)-f(X1,X2,X3,z1,u1,v1,z2,u2,v2,z3,U,V))/(delta);
A2 = (f(X1,X2,X3,z1,u1,v1,z2+delta,u2,v2,z3,U,V)-f(X1,X2,X3,z1,u1,v1,z2,u2,v2,z3,U,V))/(delta);
A3 = (f(X1,X2,X3,z1,u1,v1,z2,u2,v2,z3+delta,U,V)-f(X1,X2,X3,z1,u1,v1,z2,u2,v2,z3,U,V))/(delta);
K = (sqrt((A1)^2+(A2)^2+(A3)^2));
A1 = -(A1)/K;
A2 = -(A2)/K;
A3 = -(A3)/K;
X3_var = U+(A3*B1*V);
V3 = (X3_var*LS);
U3 = X3_var*(1-log(X3_var)+log(LM));
S = S+1;
end
However, if you need X3 to be a function that is changing in time, then you would need different code.
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