function handle array Problem

Question

Martin 2023-3-23

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评论： Martin 2023-3-24

Hi

The following code is a short expample of my real code. I want to compute iterativly U and V. And at the end i have a intergral where i want to divide the V by the U of the last iteration and then integrate that over the variable a.

How can i do this?

Thank you?

m = [1,2];
U(3) = @(a) a + 2;
V(3) = @(a) a - 2;
for j = 2:-1:1
    a_i(j) = @(a) sqrt(a.^2 + 1i * m(j));
    U{j} = @(a) + a_i(j) .* V{j+1};
    V{j} = @(a) - a_i(j) .* U{j+1}; 
end
result = integral(V{1}(a) ./ U{1}(a), 0, 1000)

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Martin 2023-3-23

This is the error message:

"Conversion to cell from function_handle is not possible."

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Answer 1

Torsten 2023-3-23

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在 MATLAB Online 中打开

result = integral(@fun,0,1000,'ArrayValued',1)
result = 9.7513e+02 + 8.3131e-17i
function value = fun(a)
  m = [1,2];
  U(3) = a + 2;
  V(3) = a - 2;
  for j = 2:-1:1
    a_i =  sqrt(a^2 + 1i * m(j));
    U(j) = a_i * V(j+1);
    V(j) = a_i * U(j+1); 
  end
  value = V(1)/U(1);
end

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Torsten 2023-3-23

编辑：Torsten 2023-3-23

在 MATLAB Online 中打开

And if i use other variables (constants defined at the begin of the program) in the for loop, how can i pass them to the function?

As usual:

constant1 = 1;
constant2 = pi;
constant3 = exp(1);
result = integral(@(a)fun(a,constant1,constant2,constant3),0,1000,'ArrayValued',1)
function value = fun(a,constant1,constant2,constant3)
   ...
end

Or use a structure:

params.constant1 = 1;
params.constant2 = pi;
params.constant3 = exp(1);
result = integral(@(a)fun(a,params),0,1000,'ArrayValued',1)
function value = fun(a,params)
   constant1 = params.constant1;
   constant2 = params.constant2;
   constant3 = params.constant3;
   ...
end

Martin 2023-3-24

Thank you! That works perfectly!

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Answer 2

Steven Lord 2023-3-23

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https://ww2.mathworks.cn/matlabcentral/answers/1934085-function-handle-array-problem#answer_1199435

在 MATLAB Online 中打开

MATLAB no longer allows non-scalar arrays of function handles; I think the last release in which that was supported was release R13SP1 (MATLAB 6.5.1) or R13SP2 (MATLAB 6.5.2) back in 2003 though I could be wrong. I'm fairly certain it started at least issuing warnings if not throwing errors when we introduced anonymous functions in release R14 in 2004.

You can make a cell array of function handles like you did on this line (commented out so I can run code later in the answer):

% U{j} = @(a) + a_i(j) .* V{j+1};

but this doesn't do what you think it does. The + operator in this function handle does not add a_i(j).*V{j+1} to the function handle stored in either U{j} or U{j-1}. It is the unary plus operator. In addition, this will error when evaluated. You can't multiply a number by a function handle. You could multiply the result of evaluating a function handle by a number.

f = @(x) sin(x);
gWorks = @(x) 2.*f(x); % 2 times the result of evaluating f works
gDoesNotWork = @(x) 2.*f; % 2 times f does not work
[gWorks(1:5); 2*sin(1:5)]
ans = 2×5
    1.6829    1.8186    0.2822   -1.5136   -1.9178
    1.6829    1.8186    0.2822   -1.5136   -1.9178
gDoesNotWork(1:5)
Operator '.*' is not supported for operands of type 'function_handle'.

Error in solution>@(x)2.*f (line 4)
gDoesNotWork = @(x) 2.*f;

What's the mathematical equation you're trying to integrate? It may be more straightforward to write a function in a file and integrate that function rather than trying to iteratively assemble a tower of anonymous functions.