Find argmin of function with unique vector

Hi,

I want to solve the next problem:

For example, for and

I tried to do some things like:

x=[theta,1-theta];
x0 =[0.0001,0.0001];
fun = @(x)sum((x-y)./x);
x = fminsearch(fun,x0);

Any idea how to minimize vector with respect to this scalar?

I know I can just write it like that:

x0 =[0.0001,0.0001];
fun = @(x) (x-y(1))/x + ((1-x)-y(2))/x 
x = fminsearch(fun,x0);

But there are more dimensions (then 2) in the original problem so it is impractical to do so.

Thanks!

[infinity]

Hello,

What is "n" in the formula?

Since in your formula, , where is a vector.

How can we understand ?

[DM]

Hi,

N is the dimenstion of the problem and n is the n'th value of the vector.

In the example above N=2 and and

[Kaustav Bhattacharya]

If you know n can you not use

x0 = ones(1,n)*0.0001

and then

x = fminsearch(fun,x0) ?

Answer 1

[Matt J]

You need to write code that generates in vectorized form. Then the rest is easy. The correct way to code your original example is,

f= @(theta) [theta,1-theta];

objective = @(theta)sum((f(theta)-y)./f(theta));

x = fminsearch(objective,0.0001);

[DM]

That's what I needed, thank you!!

Answer 2

[infinity]

Hello,

Here is an example that you can refer

I find mininum of function , the initial guess is given by . The code could be like that

clear 

N = 5;
fun = @(t) norm(t,2);
x0 = 1+zeros(1,N);

x = fminsearch(fun,x0) 

where I choose N = 5, but it can be changed.

[DM]

You can write

x0=ones(1,N);

But it doesn't help me since I need to find the minimum of theta from the minimum probelm above.

[infinity]

Yes, it is an idea how to solve problem with mutil-dimensiona. You may adapt the code for your case.