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Topic:
Handle function implicitly accounting for independent variables
Replies:
2
Last Post:
Jan 7, 2013 2:08 PM




Handle function implicitly accounting for independent variables
Posted:
Jan 7, 2013 4:06 AM


I have that
clc, clear all, close all tic k1 = 1E02:0.1:1E+02; k2 = 1E02:0.1:1E+02; k3 = 1E02:0.1:1E+02; k = sqrt(k1.^2+k2.^2+k3.^2); c = 1.476; gamma = 3.9; colors = {'b'}; Ek = (1.453*k.^4)./((1 + k.^2).^(17/6)); E = @(k) (1.453*k.^4)./((1 + k.^2).^(17/6)); E_int = zeros(1,numel(k)); E_int(1) = 1.5;
for i = 2:numel(k) if k(i) < 400 E_int(i) = E_int(i1)  integral(E,k(i1),k(i)); elseif k(i) > 400 E_int(i) = 2.180/(k(i)^(2/3)); end %end if end %end i
beta = (c*gamma)./(k.*sqrt(E_int)); figure plot(k,beta,colors{1})
count = 0; %F_11 = zeros(1,numel(k1)); F_33 = zeros(1,numel(k1));
Afterwards, I compute F_33 as
for i = 1:numel(k1) count = count + 1; phi_33 = @(k2,k3) (1.453./(4.*pi)).*(((k1(i)^2+k2.^2+(k3 + beta(i).*k1(i)).^2).^2)./((k1(i)^2+k2.^2+k3.^2).^2)).*((k1(i)^2+k2.^2)./((1+k1(i)^2+k2.^2+(k3+beta(i).*k1(i)).^2).^(17/6))); F_33(count) = 4*integral2(phi_33,0,1000,0,1000); end
Now let's come to my question. I know from a paper that:
k = sqrt(k1.^2+k2.^2+k3.^2); k30 = k3 + beta.*k1; k0 = sqrt(k1.^2+k2.^2+k30.^2); E_k0 = 1.453.*(k0.^4./((1+k0.^2).^(17/6)));
Therefore the expression for phi_33 would result in
phi_33 = (E_k0./(4*pi.*(k.^4))).*(k1.^2+k2.^2);
The question is: how can I make use of this final expression insted of the long one I'm using at the moment (within the for loop)?
The last expression for phi_33 is easier to handle (especially because of reckless mistakes in writing the former) and it would "pass by reference" (k2,k3), which are the independent variables.
Any hint is more than welcome.
Best regards, fpe



