store values from a for loop in a column vector?

I have this function and have to store the values of b(x) over the interval

x=0.01:0.01:5
b0=10;
K=2;
m=3;
for x=0.01:0.01:5
    b= b0.*(x.^(m)/(K.^(m)+ x.^(m)));
end

I don't know how to store every value in a column vector, only the value of b(5). any ideas on how to correct this code?

Answer 1

[James Tursa]

No need for the for-loop, just a one-liner:

b = b0.*(x.^(m)./(K.^(m)+ x.^(m)));  % <-- changed the / to a ./

[Margaret Winding]

It worked, thank you! But after that I am required to use the function fminbnd to find the max of the function b on the interval [0,5], but it won't let me as it says b isn't being saved as function. Do you know how to resolve that? Thank you for all your help!

[James Tursa]

Please show the code you are using for fminbnd.

[Margaret Winding]

[xmax, ymax]= fminbnd(-b, 0, 5)

[James Tursa]

The first argument of fminbnd should be a function handle so that fminbnd can evaluate the function internally to find the solution. So you could use something like this:

b = @(x)b0.*(x.^(m)./(K.^(m)+ x.^(m)));  % <-- create the function handle
[xmax, ymax]= fminbnd(b, 0, 5);  % <-- call fminbnd

Or if you really wanted -b make that part of the function handle:

b = @(x)-b0.*(x.^(m)./(K.^(m)+ x.^(m)));  % <-- create the function

[Margaret Winding]

thank you so much for your help!

Answer 2

[Jan]

James' answer solves the problem efficiently. But if you or anybody else requires a loop, the pre-allocation is important:

x  = 0.01:0.01:5
b0 = 10;
K  = 2;
m  = 3;
b  = zeros(size(x));  % Pre-allocate!
for ix = 1:numel(x)   % Use index according to vector x
  b(ix) = b0 * (x(ix) ^ m / (K ^ m + x(ix) ^ m));
end

Letting an array grow iteratively wastes a lot of resources: In the first iteration a scalar double is reserved and assigned. In the second iteration, a vector of two doubles is reserved, the former contents is copied and the last value is assigned. If you have 1000 elements, Matlab has to reserve sum(1:1000) elements by this way, which are 500'500 and copy almost the same number of doubles. For 1 million elements, we are talking about 2TB of data already, although the final result occupies 8MB RAM only (8 byte per double).

Answer 3

[John BG]

may be you want to keep the for loop because there may be more omitted lines in the loop

x=0.01:0.01:5
b0=10;
K=2;
m=3;
for x=0.01:0.01:5
    b= [b;b0.*(x.^(m)/(K.^(m)+ x.^(m)))];
end