Find argmin of function with unique vector

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DM
DM 2019-6-23
评论： DM 2019-6-26
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!
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DM
DM 2019-6-25
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
Kaustav Bhattacharya 2019-6-25
If you know n can you not use
x0 = ones(1,n)*0.0001
and then
x = fminsearch(fun,x0) ?

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采纳的回答

Matt J
Matt J 2019-6-25
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);

更多回答（1 个）

infinity
infinity 2019-6-25
编辑：infinity 2019-6-25
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.
  2 个评论
DM
DM 2019-6-25
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
infinity 2019-6-25
Yes, it is an idea how to solve problem with mutil-dimensiona. You may adapt the code for your case.

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