Numerical answer to Partial derivative of Hopfield-style Energy in symbolic expression
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Hello,
I'm implementing a gradient descent optimizer that uses Hopfield-style energy as function,
the hopfield-style energy is a function of p: ![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/1080810/image.png)
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/1080810/image.png)
The hopfield energy equation can also be expressed as:
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/1080815/image.png)
The gradient vector is defined as: ![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/1080820/image.png)
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/1080820/image.png)
I'm attempting to calculate the gradient vector using the Symbolic Toolbox as follows:
r_mat = sym('r',[4,4]);
r_mat = r_mat(:);
p_mat = sym('p',[4*4,1]);
c_mat = sym('c',[4*4,1]);
Energy = -0.5*p_mat'*r_mat*p_mat' - c_mat'*p_mat;
%the first partial derivative:
q1 = -1*diff(testi1,p_mat(1,1));
which yields as the first term:
conj(c1) + r1_1*conj(p1) + (r1_2*conj(p5))/2 + (r1_3*conj(p9))/2 + (r2_1*conj(p2))/2 + (r1_4*conj(p13))/2 + (r2_2*conj(p6))/2 + (r2_3*conj(p10))/2 + (r3_1*conj(p3))/2 + (r2_4*conj(p14))/2 + (r3_2*conj(p7))/2 + (r3_3*conj(p11))/2 + (r4_1*conj(p4))/2 + (r3_4*conj(p15))/2 + (r4_2*conj(p8))/2 + (r4_3*conj(p12))/2 + (r4_4*conj(p16))/2
How can I replace the symbols with numbers to yield the q vector? I need the q-vector for the gradient descent
the algorithm I'm implementing: https://doi.org/10.1117/1.JEI.23.1.013007 (relevant part starts at Equation 5)
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采纳的回答
Torsten
2022-7-28
编辑:Torsten
2022-7-28
p is (4x1), not (16x1).
And keep r_mat as matrix, then
E = -0.5*p.'*r_mat*p - c.'*p
( r_mat can be assumed symmetric ).
By the way:
dE/dp = -r_mat*p - c
assuming r_mat is symmetric.
To replace symbols by numerical values, use "subs".
5 个评论
Torsten
2022-8-2
I used symbolic to calculate the partial derivative and then used that to think how to write the equation in vector form
That's the best you can do, I guess.
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