Geek asks and answers

Using the fminunc function

I am trying to use the fminunc function for convex optimization. However, in my case, I take a gradient based on logx. Let my objective function be F. Then the gradient will be

dF/dx = (dF/dlogx) * (1/x)
= > dF/dlogx = (dF/dx) * x

So,

logx_new = logx_old + learning_rate * x * (dF/logx)
x_new = exp(logx_new)

How can I implement this in fminunc

+3
source share
1 answer

This is possible and is described in the documentation :

If the fun gradient can also be calculated, and the GradObj 'on' option, as set by options = optimset ('GradObj', 'on'), then the fun function should return the gradient value g, vector, in x in the second output argument.

fminunc with custom gradient

So for example: if f = @(x) x.^2;then df/dx = 2*x, and you can use

function [f df] = f_and_df(x)
    f = x.^2;
    if nargout>1
        df = 2*x;
    end
end

fminunc:

options = optimset('GradObj','on');
x0 = 5;
[x,fval] = fminunc(@f_and_df,x0,options);

fminunc logx

logx :

function [f df] = f_and_df(x)
    f = ...;
    if nargout>1
        df =  x * (dF/logx);
    end
end

fminunc .

fminunc

, :

f_and_df2 = @(x) deal(x(1).^2+x(2).^2,[2*x(1)  2*x(2)]);
[x,fval] = fminunc(f_and_df2,[5, 4],optimset('GradObj','on'))

fminunc logx

f = (log(x))^2

function [f df_dlogx] = f_and_df(x)
    f = log(x).^2;

    df_dx = 2*log(x)./x;
    df_dlogx = df_dx.* x;
end

:

>>x0=3;
>>[x,fval] = fminunc(@f_and_df,x0,optimset('GradObj','on'))
x =
   0.999999990550151

fval =
   8.92996430424197e-17

fminunc

, . f (x, y), , :

function [f df_dx] = f_and_df(x)
    f = x(1).2 + x(2).^2;

    df_dx(1) = 2*x(1);
    df_dx(2) = 2*x(2);
end

. , , : x0 = [- 5 3]

+7

More articles:


All Articles