how can i use integral function with vector limits by using another for loop ?

Question

work wolf 2023-7-3

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1991188-how-can-i-use-integral-function-with-vector-limits-by-using-another-for-loop

评论： work wolf 2023-7-5

% Define the function for integration
f = @(t, y,m) y + (t-m);
% Create sample data
m=[0 0.2 0.3 0.4 ];
datalower = [0.1 0.5 0.6];
dataupper = [0.9 2 0.5];
% Define the integral function handle
IntV = @(y,k,m, lower, upper) integral(@(t) f(t, y(k),m(k)), lower(k), upper(k));
% To use it within a a recursive formula without using for loop as follows:
k = 1:numel(m)-1; 
y(1)=0;
y(k+1) = y(k) +  IntV(y,k,m, lower, upper)./m(k-2) ;

by looking for integral function, it's not accept vector limitis, Thus can i convert integral to matrix or summations ( with respect to same result) to aviod using for loop or symbolic 'int' ?

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Answer 1

Star Strider 2023-7-3

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https://ww2.mathworks.cn/matlabcentral/answers/1991188-how-can-i-use-integral-function-with-vector-limits-by-using-another-for-loop#answer_1266378

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All the vectors have to be the same size ()so I shortened ‘m’ here), then arrayfun works (and so would a loop, indexing each vector) —

% Define the function for integration
f = @(t, y,m) y + (t-m);
% Create sample data
m=[0 0.2 0.3];
datalower = [0.1 0.5 0.6];
dataupper = [0.9 2 0.5];
% Define the integral function handle
IntV = @(y,m, lower, upper) integral(@(t) f(t, y,m), lower, upper);
% To use it within a a recursive formula without using for loop as follows:
% k = 1:numel(m)-1; 
% y(1)=0;
% y(k+1) = y(k) +  IntV(y,k,m, lower, upper)./m(k-2) ;
y = randn(size(m));
y = arrayfun(IntV,y,m,datalower,dataupper)
y = 1×3
   -0.2075    1.7554   -0.0391

.

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work wolf 2023-7-4

编辑：work wolf 2023-7-4

在 MATLAB Online 中打开

@Star Strider thank you so much . The problem when i apply it inside a recursive formula, becasue y is not previously known, but by a recursive formula with initial value and fill gradually according to previous value.

% Define the function for integration
f = @(t, y,m) y + (t-m);
% Create sample data
m=[0 0.2 0.3];
datalower = [0.1 0.5 0.6];
dataupper = [0.9 2 0.5];
% Define the integral function handle
IntV = @(y,m, lower, upper) integral(@(t) f(t, y,m), lower, upper);
%  Define the function handle dep. on y to use in a recursive formula by using arrayfun
intV_Fun = @(y) arrayfun(IntV,y,m,datalower,dataupper); %y = datalower; %randn(size(m)); test
%-------------------------------------------------------------------------------------------------
% IntV_Fun = @(y) splitapply(@(k) IntV(y(k),m(k),datalower(k),dataupper(k)), k, k);
% IntV_Fun = @(y) cell2mat(arrayfun(@(k) IntV(y(k),m(k),datalower(k),dataupper(k)), k,'UniformOutput', false) );
% IntV_Fun = @(y) accumarray(k.', k, [], @(k) IntV(y(k),m(k),datalower(k),dataupper(k) ) );
% IntV_Fun = @(y) groupsummary(k, k, @(k) IntV(y(k),m(k),datalower(k),dataupper(k) ) );
% IntV_Fun = @(y) groupsummary(k', k', @(k) IntV(y(k),m(k),datalower(k),dataupper(k) ) );
% IntV_Fun = @(y) groupsummary({y,m,datalower,dataupper}, k,...
%     @(y,m,datalower,dataupper) IntV(y,m,datalower,dataupper) ) ;
% IntV_Fun = @(y) groupsummary({y' , m', datalower', dataupper'}, k',...
%     @(y,m,datalower,dataupper) IntV(y,m,datalower,dataupper ) ); 
%
%  all gives 1x3 vector when y previously known for example y =y = datalower; (!).
%  ** The problem when i apply integral inside a recursive formula, 
%  becasue y is not previously known, but by a recursive formula with 
%  initial value and fill gradually according to previous value.
%  y(k+1) = y(k) +  intV_Fun(y(k))./m(k-2) ;
%---------------------------------------------------------------------------------------------------------------------------
% To use it within a a recursive formula without using for loop as follows:
k = 1:numel(m)-1; % k = 1:numel(m); same Error 
 y(1)=0;
 y(k+1) = y(k) +  intV_Fun(y(k))./m(k-2) ;
 
 % gives same Error in all cases, i.e, splitapply, accumarray & groupsummary
 
 Error : Index exceeds the number of array elements. Index must not exceed 1.
 

please , can you help me ! ^_^

Star Strider 2023-7-4

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My pleasure!

I cannot help because what you want to do does not make sense.

As in my code, ‘y’ must be the same size as the other elements.

% Define the function for integration
f = @(t, y,m) y + (t-m);
% Create sample data
m=[0 0.2 0.3];
datalower = [0.1 0.5 0.6];
dataupper = [0.9 2 0.5];
% Define the integral function handle
IntV = @(y,m, lower, upper) integral(@(t) f(t, y,m), lower, upper);
%  Define the function handle dep. on y to use in a recursive formula by using arrayfun
intV_Fun = @(y) arrayfun(IntV,y,m,datalower,dataupper); %y = datalower; %randn(size(m)); test
%-------------------------------------------------------------------------------------------------
% IntV_Fun = @(y) splitapply(@(k) IntV(y(k),m(k),datalower(k),dataupper(k)), k, k);
% IntV_Fun = @(y) cell2mat(arrayfun(@(k) IntV(y(k),m(k),datalower(k),dataupper(k)), k,'UniformOutput', false) );
% IntV_Fun = @(y) accumarray(k.', k, [], @(k) IntV(y(k),m(k),datalower(k),dataupper(k) ) );
% IntV_Fun = @(y) groupsummary(k, k, @(k) IntV(y(k),m(k),datalower(k),dataupper(k) ) );
% IntV_Fun = @(y) groupsummary(k', k', @(k) IntV(y(k),m(k),datalower(k),dataupper(k) ) );
% IntV_Fun = @(y) groupsummary({y,m,datalower,dataupper}, k,...
%     @(y,m,datalower,dataupper) IntV(y,m,datalower,dataupper) ) ;
% IntV_Fun = @(y) groupsummary({y' , m', datalower', dataupper'}, k',...
%     @(y,m,datalower,dataupper) IntV(y,m,datalower,dataupper ) ); 
%
% all gives 1x3 vector when y know for example y =y = datalower; (!)
%---------------------------------------------------------------------------------------------------------------------------
% To use it within a a recursive formula without using for loop as follows:
% I can make this work (sort of) however nothing further — 
k = 3;
y(3) = 0;
intV_Fun(y(k)*ones(size(m)))./m(k-2) 
ans = 1×3
   Inf   Inf  -Inf
% NOTE — It expands the scalar value for 'y' to a vector to match the sizes of the rest
% of the arguments.  
% This is not recursive because  there is no recursion of any kind — 
k = 1:numel(m)-1; % k = 1:numel(m); same Error 
 y(1)=0;
 y(k+1) = y(k) +  intV_Fun(y(k)*ones(size(m)))./m(k-2) ;
Error using *
Incorrect dimensions for matrix multiplication. Check that the number of columns in the first matrix matches the number of rows in the second matrix. To operate on each element of the matrix individually, use TIMES (.*) for elementwise multiplication.
 
 % gives same Error in all cases, i.e, splitapply, accumarray & groupsummary
 
 % Error : Index exceeds the number of array elements. Index must not exceed 1.
 

Describe what you want to do and it may be possible to do it (no promises). I cannot figure it out from the code.

.

work wolf 2023-7-4

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@Star Strider My regards, thank you so much for your time. I modified the code within your answers and m(k) insted of m(k-2) to avoid Inf, but problem it's not gives same results as follows:

clear
clc
% Define the function for integration
f = @(t,y,m) y + (t-m);
% Create sample data
m=[0.2 0.3 0.4 ];
datalower = [0.1 0.5 0.6];
dataupper = [0.9 2 0.5];
% Define the integral function handle
IntV = @(y,m,lower, upper) integral(@(t) f(t,y,m), lower, upper);
%---------------------------------------------------------
y1(1) = 0;
for k = 1:numel(m)-1
  y1(k+1) = y1(k) + IntV(y1(k),m(k),datalower(k),dataupper(k))/m(k); % /m(k-1+(k==1));
end
y1 % with m(k)
%----------------------------------------------------------
intV_Fun = @(y) arrayfun(IntV,y,m,datalower,dataupper); 
y2=zeros(size(m));
y2(1) = 0;
k = 1:numel(m);
y2 (k+1)= y2(k) + intV_Fun(y2(k))/m(k);
y2=y2(1:end-1)
%-------------------------------------------------------
y3(1) = 0;
for k = 1:numel(m)-1
  y3(k+1) = y3(k) + IntV(y3(k),m(k),datalower(k),dataupper(k)); % /m(k-1+(k==1));
end
y3
%----------------------------------------------------------
intV_Fun = @(y) arrayfun(IntV,y,m,datalower,dataupper); 
y4=zeros(size(m));
y4(1) = 0;
k = 1:numel(m);
y4 (k+1)= y4(k) + intV_Fun(y4(k));
y4=y4(1:end-1)
%-------------
%   result
%------------
y1 =
         0    1.2000   11.9500
y2 =
         0    1.6190    1.6190
%------------------------------------
y3 =
         0    0.2400    2.0250
y4 =
         0    0.2400    1.4250

work wolf 2023-7-4

编辑：work wolf 2023-7-4

在 MATLAB Online 中打开

@Star Strider

First and foremost, I would like to express my sincere gratitude to you for your attention and insightful words, especially considering that each problem is unique and requires a different approach.

The crux of the matter lies in how to convert the "for loop" system into the array system, which significantly simplifies the process for me, particularly in terms of time. Hence, I kindly request your collaboration in assisting me with this conversion, as your extensive expertise in MATLAB makes it a straightforward task, rather than a major challenge.

Despite my exhaustive efforts and attempts, I still encounter a predicament: how to make the "integral" function accept integration limits in the form of an array. As you may have noticed, I have experimented with various functions such as "arrayfun", "splitapply", "accumarray" and "groupsummary" However, they pose challenges regarding the size of arrays when applied within a recursive formula.

Consequently, I have resorted to posing my question here and seeking advice from experts like yourself. Please note that I have summarized my attempts here through these observations:

Summary of Issues:

1. The "integral" function does not accept integration limits as an array.

2. The size of arrays poses challenges when using the following functions:"arrayfun", "splitapply", "accumarray" and "groupsummary" within a recursive formula.

The issue lies in the application of intervals. Each partial interval requires a distinct "for loop" which is illogical to apply each step using a separate "for loop." This approach consumes time during implementation and impacts performance. Although it resolves the problem, it resembles a manual solution.

For instance, in the case of partial intervals, when obtaining the value for each iteration before integration:

1. 1/2 integral(f, m(1), m(2)) when k == 1

2. 1/3 integral(f, m(k-1), m(k+1)) when k = 2:numel(m)-1

3. 1/2 integral(f, m(end-1), m(end)) when k = numel(m)

Using arrays or logical inference alone requires only one definition, whereas employing a "for loop" necessitates three definitions due to the varying pre-integration values and integration limits for each interval.

I apologize for the lengthy message; however, I wanted to convey the information accurately in written form.

I hope I don't confuse the ideas because I feel a bit lost in explaining the problem, especially through writing.

I sincerely hope that you will assist me and take an interest in this issue, even if it may not seem important to you. In any case, I appreciate any help you can provide, regardless of the approach. Really, i need it so much.

In short, , it seems impossible to convert the previous system into the array system without resorting to some form of "for loop".

%---------------------------------------------------------
y1(1) = 0;
for k = 1:numel(m)-1
    y1(k+1) = y1(k) + IntV(y1(k), m(k), datalower(k), dataupper(k))/m(k); % /m(k-1+(k==1));
end
y1