Derive function handle with a vector input argument

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Majeed
Majeed 2024-1-11
评论： Majeed 2024-1-12
when I run this piece of code
q=[3 2 2 3];
f=@(p,x,y) p(1)*x.^2+p(2).*y.^2+p(3)*x.*y+p(4);
syms 'x'
df=diff(f,x);
dfdx=matlabFunction(df);
A=feval(f,q,2,2);
I get an error says
Index exceeds the number of array elements (1).
What I am trying to do is to cluster my equation coefficients into one vector to make my code compact. In my real application I have 15 coefficients. Any advice if this process is possible?

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采纳的回答

Walter Roberson
Walter Roberson 2024-1-11
移动：Walter Roberson 2024-1-11
Ran in:
q=[3 2 2 3];
f=@(p,x,y) p(1)*x.^2+p(2).*y.^2+p(3)*x.*y+p(4);
syms p [1 4]
syms x y
df=diff(f(p,x,y),x)
df = 
dfdx = matlabFunction(df, 'vars', {p, x, y})
dfdx = function_handle with value:
@(in1,x,y)in1(:,1).*x.*2.0+in1(:,3).*y
A = feval(f,q,2,2)
A = 31

更多回答（2 个）

Paul
Paul 2024-1-11
编辑：Paul 2024-1-11
Ran in:
Hi Majeed,
Maybe you're looking for something like this?
q = [3 2 2 3];
p = sym('p',[1 4]);
syms x y
f = p(1)*x.^2+p(2).*y.^2+p(3)*x.*y+p(4)
f = 
df=diff(f,x)
df = 
A = subs(f,[p x y],[q,2,2])
A = 
31
Matt J
Matt J 2024-1-11
编辑：Matt J 2024-1-11
Ran in:
If you have the Deep Learning Toolbox, you can do automatic differentiation of vector-valued dlarrays,
p=1:15;
x0=dlarray(rand(size(p)));
[fx,gradfx] = dlfeval(@(x) somefunc(x,p),x0)
fx =
1×1 dlarray 68.8215
gradfx =
1×15 dlarray 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
function [y,dydx]=somefunc(x,p)
y=p*x';
dydx=dlgradient(y,x);
end

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