When trying to use the integral2 function, I get the following warning:
"Integrand function outputs did not match to the required tolerance when the same input values were supplied in two separate calls with different size input matrices. Check that the function is vectorized properly."
My code is:
z1=0; z2=5;
t1=0; t2=10;
a=1:500;
b=linspace(0.15,0.02,length(a));
c=[a;b];
d=4.9e-07;
func = #(z,t) my_func((z + t),c) .* exp(-d.*t);
int_x = integral2(func,z1,z2,t1,t2,'RelTol',1e-3,'AbsTol',1e-3) ./ (z2 - z1);
my_func is:
function out = my_func(z_in,c)
z=reshape(z_in,1,numel(z_in));
for i=1:length(z)
if (z(i)<0.0)
z(i)=0.0;
end
end
expfactor=exp(-z/150);
sB=5.5*0.9*4*expfactor;
mB=0.9*interpolate(c(1,:),c(2,:),z);
out=sB+mB;
out=reshape(out,size(z_in,1),size(z_in,2));
end
z1=0, z2=9.6, t1=0, t2=3000 (i.e. all have the size (1,1)), and I have used element-wise operations, so not sure what the problem is.
When debugging, my_func is called three times, with z and t having different sizes each time (although same as each other; (14,14), (2,3) and (14,14)). When sizes are (14,14), the values of z are unique for each column, but the same for each row (e.g. [3,2,1;3,2,1]), and vice versa for t (e.g. [4,4,4;5,5,5]). But when sizes are (2,3), all values in the z and t matrices are unique. The warning comes after the second and third call.
What could be the reasons for the warning?
Edit: to accommodate the dimension requirements in my_func, I've reshaped z to a vector at the start, then reshaped the output back to matrix at the end, copying a similar function I've seen. But this returns the warning:
"Non-finite result. The integration was unsuccessful. Singularity likely."
Thanks
Related
i am solving a mathematical problem and i can't continue do to the error.
I tried all constant with sin^2(x) yet its the same.
clear
clc
t = 1:0.5:10;
theta = linspace(0,pi,19);
x = 2*sin(theta)
y = sin^2*(theta)*(t/4)
Error using sin
Not enough input arguments.
Error in lab2t114 (line 9)
y = sin^2*(theta)*(t/4)
sin is a function so it should be called as sin(value) which in this case is sin(theta) It may help to consider writing everything in intermediate steps:
temp = sin(theta);
y = temp.^2 ...
Once this is done you can always insert lines from previous calculations into the next line, inserting parentheses to ensure order of operations doesn't mess things up. Note in this case you don't really need the parentheses.
y = (sin(theta)).^2;
Finally, Matlab has matrix wise and element wise operations. The element wise operations start with a period '.' In Matlab you can look at, for example, help .* (element wise multiplication) and help * matrix wise calculation. For a scalar, like 2 in your example, this distinction doesn't matter. However for calculating y you need element-wise operations since theta and t are vectors (and in this case you are not looking to do matrix multiplication - I think ...)
t = 1:0.5:10;
theta = linspace(0,pi,19);
x = 2*sin(theta) %times scalar so no .* needed
sin_theta = sin(theta);
sin_theta_squared = sin_theta.^2; %element wise squaring needed since sin_theta is a vector
t_4 = t/4; %divide by scalar, this doesn't need a period
y = sin_theta_squared.*t_4; %element wise multiplication since both variables are arrays
OR
y = sin(theta).^2.*(t/4);
Also note these intermediate variables are largely for learning purposes. It is best not to write actual code like this since, in this case, the last line is a lot cleaner.
EDIT: Brief note, if you fix the sin(theta) error but not the .^ or .* errors, you would get some error like "Error using * Inner matrix dimensions must agree." - this is generally an indication that you forgot to use the element-wise operators
Why does the error message of this code return: "Subscript indices must either be real positive integers or logicals.", when I am using ceil for every subscript?
A=1:1:100;
B=1:1:100;
C=1;
D=1:1:100;
E=2;
F=1:1:100;
G=1:1:100;
H=0.1:0.1:10;
fun_1=#(t)integral(#(ti)G(ceil(ti)).*H(ceil(t-ti)),0.1,t-1);
fun_2=#(t)integral(#(ti)G(ceil(ti)).*B(ceil(ti)).*(C.*D(t).^E)./F(t).*...
exp(-integral(#(x)(C.*D(ceil(x)).^E)./F(ceil(x)),ti,5)-K.*(t-ti)),0.1,t-
1,'ArrayValued',true);
I=500;
J=1000;
K=2;
fun_3=#(t)I*integral(#(ti)min(fun_2(ceil(ti)),J).*exp(-(K+I).*(t-ti)),0.1,t-
1);
t=1:1:5;
figure(1)
fplot(fun_1,t);
figure(2)
fplot(fun_2,t);
figure(3)
fplot(fun_3,t);
fplot see documentation Called as fplot(f,xinterval) evaluates your function handle f over the interval xinterval. IT will evaluate f at automatically determined steps along that given interval.
From the docs:
xinterval — Interval for x [–5 5] (default) | two-element vector of
form [xmin xmax]
You seem to be trying to specify exactly where you want your functions evaluated
t=1:1:5;
...
fplot(fun_1,t);
But it doesn't work that way. What is happening is that fplot is evaluating the function from 1 to 2 (the first 2 elements of t). So for example it might feed values of t = 1, 1.05, 1.1,... ,2 into your fun_# functions.
You can tell this because you first function which does work actually plots over the x-range of 1 to 2.
The reason you are getting a subscript indices error is because in fun_2 you have this ...(C.*D(t).^E)./F(t).*... Since fplot is feeding in values for t which are spaced between 1 and 2 (ex. 1.1) that is not a valid index.
If you really just want the values of your functions at t = 1:1:5 The you probably do not want to use fplot and just want evaluate the functions at those times and plot it.
y = feval(fun_1,t);
plot(t,y)
EDIT: The above code doesn't work
You will need to do something like the code below. This is because the 2nd & 3rd trems to the intergral function need to be scalar (1x1). If you feed them an array for t then they crash. So evaluate at each t not all at once.
figure(1)
y_1 = arrayfun(fun_1,t);
plot(t,y_1);
figure(2)
y_2 = arrayfun(fun_2,t);
plot(t,y_2);
figure(3)
y_3 = arrayfun(fun_3,t);
plot(t,y_3);
Note: the Third function still errors ... and I'm not 100% sure why. I didn't really look at it.
Doing homework and there's a question that's giving me trouble. The questions is:
a)create a function M-file called nmoles that requires two vector inputs—
the mass and molecular weight—and returns the corresponding number
of moles. Because you are providing vector input, it will be necessary
to use the meshgrid function in your calculations.
b) Test your function for the compounds shown in the following table, for
masses from 1 to 10 g:
In my function file, i've got:
function [ n ] = nmoles(m, MW)
%% Finds number of moles
m = (1:10); %% mass range
MW = [78.115 46.07 102.3]; %% Values from the table
n=m/MW; %%formula provided by the textbook
My main file only has:
nmoles(m,MW)
I'm getting an error: "Error using / Matrix dimensions must agree"
Also: Error in nmoles (line 10) n=m/MW;
I'm inexperienced with MATLAB and still learning syntax but I assume that my formula is incorrect and i'm using the wrong symbol to divide, though i'm not sure how to correct this. Also, how would i incorporate the meshgrid function into my anwser?
Here is how your function should probably look like:
function [ n ] = nmoles(m, MW)
% Finds number of moles
[mv,MWv] = meshgrid(m, MW); % extends m and MW to match all combinations.
n = mv./MWv; %formula provided by the textbook
end
And then your main script should be:
m = 1:10; % mass range
MW = [78.115 46.07 102.3]; % Values from the table
nmoles(m, MW)
You had 3 mistakes:
If you get m and MW as input to the function, your not suppose to define them within it.
Because you want the number of moles for each combination of mass and molar weight, you use meshgrid.
When you want to perform an elementwise division or multiplication on 2 arrays, put a . before the / or *. In MATLAB everything is a matrix by default, and so this operations are interpreted as matrix multiplication and division by default. If you want to do a regular division or multiplication you use .* and ./. This is true also for power (^ and .^).
Matlab fit is no doubt useful but it is not clear how to use it as a function
apart from trivial integration and differentiation given on the official website:
http://uk.mathworks.com/help/curvefit/example-differentiating-and-integrating-a-fit.html
For example given a fit stored in the object 'curve' one can evaluate
curve(x) to get a number. But how one would, e.g. integrate |curve(x)|^2 (apart from clumsily creating a new fit)? Trying naively
curve = fit(x_vals,y_vals,'smoothingspline');
integral(curve(x)*curve(x), 0, 1)
gives an error:
Output of the function must be the same size as the input. If FUN is an array-valued integrand, set the 'ArrayValued' option to true.
I have also tried a work around by definining a normal function and an implicit function for the integrand (below) but both give the same error.
func=#(x)(curve(x))...; % trial solution 1
function func_val=func(curve, x)...; % trial solution 2
Defining function for the integrand followed by integration with option 'ArrayValued' set to 'true' works:
func=#(x)(curve(x)*curve(x));
integral(func,0,1,'ArrayValued',true)
You need to have the function vectorized, i.e use element-wise operations like curve(x).*curve(x) or curve(x).^2.
Also make sure that the shape of the output matches the input, i.e a row input gives a row output, similarly a column comes out as a column. It seems that evaluating the fit object always returns a column vector (e.g. f(1:10) returns a 10x1 vector not 1x10).
With that said, here is an example:
x = linspace(0,4*pi,100)';
y = sin(x);
y = y + 0.5*y.*randn(size(y));
f = fit(x, y, 'smoothingspline');
now you can integrate as:
integral(#(x) reshape(f(x).^2,size(x)), 0, 1)
in this case, it can be simplified as a simple transpose:
integral(#(x) (f(x).^2)', 0, 1)
here is my code:
C=#(k) [k,k,2.*k;3,2.*k,5;1,k,k];
AV=#(k,t) [3*t, 6, 9]*C(k)*[3*t ;6 ;9];
avaint=#(k,a,b) quadgk(#(k) AV(k,t),a,b);
AVAR=#(t) avaint(t,0,87600);
Is shows:
Error using vertcat
Dimensions of matrices being concatenated are not consistent.
when I want to print AVAR(3)
Source of Error
The quadgk function passes a vector of integration points to the function handle given to it.
From the documentation:
The function y = fun(x) should accept a vector argument x and return a vector result y, where y is the integrand evaluated at each element of x.
This creates the dimension mismatch causing the error.
Solutions
To get around this implementation, you can perform the numerical integration using the integral function with the ('ArrayValued',true) option pair:
avaint = #(t,a,b) integral(#(k) AV(k,t),a,b,'ArrayValued',true);
Or, you can use arrayfun within AV to abide by the requirement of quadgk:
AV = #(k,t) arrayfun(#(k_el) [3*t, 6, 9]*(C(k_el)*[3*t ;6 ;9]),k);