```Date: Jan 7, 2013 4:06 AM
Author: Francesco Perrone
Subject: Handle function implicitly accounting for independent variables

I have thatclc, clear all, close alltick1 =  1E-02:0.1:1E+02;k2 =  1E-02:0.1:1E+02;k3 =  1E-02:0.1:1E+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(i-1) - integral(E,k(i-1),k(i));    elseif k(i) > 400        E_int(i) = 2.180/(k(i)^(2/3));    end %end ifend %end ibeta = (c*gamma)./(k.*sqrt(E_int));figureplot(k,beta,colors{1})count = 0;%F_11 = zeros(1,numel(k1));F_33 = zeros(1,numel(k1));Afterwards, I compute F_33 asfor 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);    endNow 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 inphi_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
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