function handle array Problem

2 visualizzazioni (ultimi 30 giorni)
Martin
Martin il 23 Mar 2023
Commentato: Martin il 24 Mar 2023
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)
  1 Commento
Martin
Martin il 23 Mar 2023
This is the error message:
"Conversion to cell from function_handle is not possible."

Accedi per commentare.

Risposta accettata

Torsten
Torsten il 23 Mar 2023
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
  3 Commenti
Torsten
Torsten il 23 Mar 2023
Modificato: Torsten il 23 Mar 2023
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
Martin il 24 Mar 2023
Thank you! That works perfectly!

Accedi per commentare.

Più risposte (1)

Steven Lord
Steven Lord il 23 Mar 2023
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.

Categorie

Scopri di più su Loops and Conditional Statements in Help Center e File Exchange

Prodotti


Release

R2019b

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by