using a 'for' loop to create a vector

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 = 0:0.01:0.6

g = @(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

Answer 1

[Kye Taylor]

You are parametrizing your for-loop using a vector of nonintegers, then trying to index into the variable x. I would instead parametrize the for-loop with a vector of integers, then use those integers to index into both x and the vector a. For example,

Instead of doing this

for a = 0:0.01:0.6
  x((a+0.01)*100,1) = a
end

I would try

a = 0:0.01:0.6;
for j = 1:length(a)
  x(j,1) = a(j);
end