I need to put the roots of the equation shown (f) into a vector of one column. I have managed to go some way with this however i keep getting the following error:
_Attempted to access x(6,1); index must be a positive integer or logical.
Error in sec_C1 (line 24)
x((a+0.01)*100,1)= (fzero(f,[0,0.1])_
How can I get this to work, and is there a better way to obtain the vector?
w =0.001;x = ones(1,61)for a = 0:0.01:0.6g = @(v) (2048*((1+v).^4) + 128*((1+v).^2).*a.^6 + a.^12 + 16*(1+v).*sqrt(16384*((1+v).^6) + 2048*((1+v).^4).*a.^6 + 80*((1+v).^2).*a.^12 + a.^18)).^(1/3);r =@(v) (1/(16*(1+v))).*( a.^4 + (a.^8)./g(v) + g(v) );alpha= @(v) (a./r(v));A= @(v) 2*pi*(r(v).^2)*(1+sqrt(1-alpha(v).^2));epsilon = @(v) 1/2*(A(v)/(4*pi-pi*a.^2)-1);f= @(v) (1-sqrt(1-alpha(v).^2)).*epsilon(v) + (epsilon(v).^2) - w;format long % shows more d.p
%find roots
x((a+0.01)*100,1)= (fzero(f,[0,0.1]))end
Best Answer