Firstly, I'm quite new to Matlab.
I am currently trying to do a definite integral with respect to y of a particular function. The function that I want to integrate is
(note that the big parenthesis is multiplying with the first factor - I can't get the latex to not make it look like power)
I have tried plugging the above integral into Desmos and it worked as intended. My plan was to vary the value of x and y and will be using for loop via matlab.
However, after trying to use the int function to calculate the definite integral with the code as follow:
h = 5;
a = 2;
syms y
x = 3.8;
p = 2.*x.^2+2.*a.*y;
q = x.^2+y.^2;
r = x.^2+a.^2;
f = (-1./sqrt(1-(p.^2./(4.*q.*r)))).*(2.*sqrt(q).*sqrt(r).*2.*a-p.*2.*y.*sqrt(r)./sqrt(q))./(4.*q.*r);
theta = int(f,y,a+0.01,h) %the integral is undefined at y=2, hence the +0.01
the result is not quite as expected
theta =
int(-((8*461^(1/2)*(y^2 + 361/25)^(1/2))/5 - (461^(1/2)*y*(8*y + 1444/25))/(5*(y^2 + 361/25)^(1/2)))/((1 - (4*y + 722/25)^2/((1844*y^2)/25 + 665684/625))^(1/2)*((1844*y^2)/25 + 665684/625)), y, 21/10, 5)
After browsing through various posts, the common mistake is the undefined interval but the +0.01 should have fixed it. Any guidance on what went wrong is much appreciated.
The Definite Integrals example in the docs shows exactly this type of output when a closed form cannot be computed. You can approximate it numerically using vpa, i.e.
F = int(f,y,a,h);
theta = vpa(F);
Or you can do a numerical computation directly
theta = vpaintegral(f,y,a,h);
From the docs:
The vpaintegral function is faster and provides control over integration tolerances.
I'm trying to implement this integral representation of Bessel function of the first kind of order n.
here is what I tried:
t = -pi:0.1:pi;
n = 1;
x = 0:5:20;
A(t) = exp(sqrt(-1)*(n*t-x*sin(t)));
B(t) = integral(A(t),-pi,pi);
plot(A(t),x)
the plot i'm trying to get is as shown in the wikipedia page.
it said:
Error using * Inner matrix dimensions must agree.
Error in besselfn (line 8) A(t) = exp(sqrt(-1)*(n*t-x*sin(t)));
so i tried putting x-5;
and the output was:
Subscript indices must either be real positive integers or logicals.
Error in besselfn (line 8) A(t) = exp(sqrt(-1)*(n*t-x*sin(t)));
How to get this correct? what am I missing?
To present an anonymous function in MATLAB you can use (NOT A(t)=...)
A = #(t) exp(sqrt(-1)*(n*t-x.*sin(t)));
with element-by-element operations (here I used .*).
Additional comments:
You can use 1i instead of sqrt(-1).
B(t) cannot be the function of the t argument, because t is the internal variable for integration.
There are two independent variables in plot(A(t),x). Thus you can display plot just if t and x have the same size. May be you meant something like this plot(x,A(x)) to display the function A(x) or plot(A(x),x) to display the inverse function of A(x).
Finally you code can be like this:
n = 1;
x = 0:.1:20;
A = #(x,t) exp(sqrt(-1)*(n*t-x.*sin(t)));
B = #(x) integral(#(t) A(x,t),-pi,pi);
for n_x=1:length(x)
B_x(n_x) = B(x(n_x));
end
plot(x,real(B_x))
So I'm just trying to plot 4 different subplots with variations of the increments. So first would be dx=5, then dx=1, dx=0.1 and dx=0.01 from 0<=x<=20.
I tried to this:
%for dx = 5
x = 0:5:20;
fx = 2*pi*x *sin(x^2)
plot(x,fx)
however I get the error inner matrix elements must agree. Then I tried to do this,
x = 0:5:20
fx = (2*pi).*x.*sin(x.^2)
plot(x,fx)
I get a figure, but I'm not entirely sure if this would be the same as what I am trying to do initially. Is this correct?
The initial error arose since two vectors with the same shape cannot be squared (x^2) nor multiplied (x * sin(x^2)). The addition of the . before the * and ^ operators is correct here since that will perform the operation on the individual elements of the vectors. So yes, this is correct.
Also, bit of a more advanced feature, you can use an anonymous function to aid in the expressions:
fx = #(x) 2*pi.*x.*sin(x.^2); % function of x
x = 0:5:20;
plot(x,fx(x));
hold('on');
x = 0:1:20;
plot(x,fx(x));
hold('off');
The problem in the link:
can be integrated analytically and the answer is 4, however I'm interested in integrating it numerically using Matlab, because it's similar in form to a problem that I can't integrate analytically. The difficulty in the numerical integration arises because the function in the two inner integrals is a function of x,y and z and z can't be factored out.
by no means, this is elegant. hope someone can make better use of matlab functions than me. i have tried the brute force way just to practice numerical integration. i have tried to avoid the pole in the inner integral at z=0 by exploiting the fact that it is also being multiplied by z. i get 3.9993. someone must get better solution by using something better than trapezoidal rule
function []=sofn
clear all
global x y z xx yy zz dx dy
dx=0.05;
x=0:dx:1;
dy=0.002;
dz=0.002;
y=0:dy:1;
z=0:dz:2;
xx=length(x);
yy=length(y);
zz=length(z);
s1=0;
for i=1:zz-1
s1=s1+0.5*dz*(z(i+1)*exp(inte1(z(i+1)))+z(i)*exp(inte1(z(i))));
end
s1
end
function s2=inte1(localz)
global y yy dy
if localz==0
s2=0;
else
s2=0;
for j=1:yy-1
s2=s2+0.5*dy*(inte2(y(j),localz)+inte2(y(j+1),localz));
end
end
end
function s3=inte2(localy,localz)
global x xx dx
s3=0;
for k=1:xx-1
s3=s3+0.5*dx*(2/(localy+localz));
end
end
Well, this is strange, because on the poster's similar previous question I claimed this can't be done, and now after having looked at Guddu's answer I realize its not that complicated. What I wrote before, that a numerical integration results in a number but not a function, is true – but beside the point: One can just define a function that evaluates the integral for every given parameter, and this way effectively one does have a function as a result of a numerical integration.
Anyways, here it goes:
function q = outer
f = #(z) (z .* exp(inner(z)));
q = quad(f, eps, 2);
end
function qs = inner(zs)
% compute \int_0^1 1 / (y + z) dy for given z
qs = nan(size(zs));
for i = 1 : numel(zs)
z = zs(i);
f = #(y) (1 ./ (y + z));
qs(i) = quad(f, 0 , 1);
end
end
I applied the simplification suggested by myself in a comment, eliminating x. The function inner calculates the value of the inner integral over y as a function of z. Then the function outer computes the outer integral over z. I avoid the pole at z = 0 by letting the integration run from eps instead of 0. The result is
4.00000013663955
inner has to be implemented using a for loop because a function given to quad needs to be able to return its value simultaneously for several argument values.
I created a function in matlab that returns a vector like
function w = W_1D(x,pos,h)
w=zeros(1,length(x));
if (h~=0)
xmpos = x-pos;
inds1 = (-h <= xmpos) & (xmpos < 0);
w(inds1) = xmpos(inds1)./h + 1;
inds2 = (0 <= xmpos) & (xmpos <= h);
w(inds2) = -xmpos(inds2)./h + 1;
else
error('h shouldn't be 0')
end
end
Thus, in the end, there is a vector w of size length(x).
Now i created a second function like
function f = W_2D(x,y,pos_1,pos_2,h)
w_x = W_1D(x,pos_1,h);
w_y = W_1D(y,pos_2,h);
f = w_x'*w_y;
end
where length(x)=length(y). Thus, the function W_2D obviously returns a matrix.
But when I now try to evaluate the integral over a rectangular domain like e.g.
V = integral2(#(x,y) W_2D(x,y,2,3,h),0,10,0,10);
matlab returns some errors:
Error using integral2Calc>integral2t/tensor (line 242)
Integrand output size does not match the input size.
Error in integral2Calc>integral2t (line 56)
[Qsub,esub] = tensor(thetaL,thetaR,phiB,phiT);
Error in integral2Calc (line 10)
[q,errbnd] = integral2t(fun,xmin,xmax,ymin,ymax,optionstruct);
Error in integral2 (line 107)
Q = integral2Calc(fun,xmin,xmax,yminfun,ymaxfun,opstruct);
I also tried to vary something in the W_2D-function: instead of f = w_x'*w_y;
I tried f = w_x.'*w_y;
or w_y = transpose(w_y); f = kron(w_x,w_y);, but there is always this error with the Integrand output size-stuff.
Can anyone explain, where my fault is?
EDIT: After Werner's hint with the keyboard debugging method, I can tell you the following.
The first step returns w_x of type <1x154 double>, w_y is <1x192 double>, x and y are both <14x14 double>. In the next step, f appears with a value of <154x192 double>. Then everything disappears, except x and y and the matlab-function integral2Calc.m appears in the editor and it jumps to the Function Call Stack integral2t/tensor and after some more steps, the error occurs here
Z = FUN(X,Y); NFE = NFE + 1;
if FIRSTFUNEVAL
if ~isfloat(Z)
error(message('MATLAB:integral2:UnsupportedClass',class(Z)));
end
% Check that FUN is properly vectorized. This is important here
% because we (otherwise) always pass in square matrices, which
% reduces the probability of the user generating an error by
% using matrix functions instead of elementwise functions.
Z1 = FUN(X(VTSTIDX),Y(VTSTIDX)); NFE = NFE + 1;
if ~isequal(size(Z),size(X)) || ~isequal(size(Z1),size(VTSTIDX))
% Example:
% integral2(#(x,y)1,0,1,0,1)
error(message('MATLAB:integral2:funSizeMismatch'));
end
Hope that information is detailed enough...I have no idea what happenes, because my example is exact as it is given on the mathworks site about integral2, isn't it?
Maybe I should precise a bit more, what I wanna do: since W_2D gives me a surface w(x,y) of a compactly supported 2-dimensional hat-function, stored in a matrix w, I want to calculate the volume between the (x,y)-plane and the surface z=w(x,y)...
EDIT2: I still do not understand how to handle the problem, that integral2 creates matrices as inputs for my W_1D-functions, which are called in W_2D and intended to have a <1xn double>-valued input and return a <1xn double> output, but at least I can simply use the following to solve the integration over the tensor product by using two one-dimensional integral-calls, that is
V = integral(#(x)integral(#(y)W_1D(y,3,h),0,10).*W_1D(x,2,h),0,10);
This first function is quite wrong. You are not indexing the array positions while you are doing w = x inside for.
Besides, if that would work, you are returning a line vector, that is, size 1xlength(x) and when you do w_x'*w_y you are doing length(x)x1 times 1xlength(y), which would give you a matrix length(x)*length(y).
Consider correcting your function:
function w = W_1D(x,pos)
w = zeros(length(x),1); % Allocate w as column vector, so that the product gives a scalar (as I suppose that it is what you want.
for ii=1:length(x) % Here, so that is indexes w and x elements as you need
w(ii)=x(ii) - pos; % I changed your code to something that makes sense, but I don't know if that is what you want to do, you have to adapt it to work correctly.
end
end
You may also want to debug your functions, consider adding keyboard before your operations and check what they are returning using dbstep. I.e:
function f = W_2D(x,y,pos_1,pos_2)
w_x = W_1D(x,pos_1);
w_y = W_1D(y,pos_2);
keyboard
f = w_x'*w_y;
end
Execution will stop at keyboard, then you can check w_x size, w_y size, and do dbstep to go after f = w_x'*w_y and see what it returned. After you finish debug, you can do dbcont so that it will continue execution.
This answer is a draft as it is quite difficult to help you with the information you have provided. But I think you can start working the things out with this. If you have more doubts feel free to ask.