I'm trying to implement the forward Euler method using matlab, but don't understand the error I'm getting. This is what I have written:
function y = ForwardEulerMethod(f,y0,T,N)
h=T/N;
t=zeros(N+1,1);
for i=0:N
t(i)=i.*h; %line 5
end
y=zeros(N+1,1);
y(0)=y0;
for i=1:N
y(i)=y(i-1)+h.*f(t(i-1),y(i-1));
end
end
My error is with line 5 and says, "Subscript indices must either be real positive integers or logicals." I am familiar with this rule, but don't see how I'm breaking it. I'm just trying to replace a zero at each location in t with a numerical value. What am I missing?
Agree with #vijoc above. You are indexing with 0 at multiple places. You could either change how you are indexing the values or get rid of the for loop altogether, like below:
function y = ForwardEulerMethod(f,y0,T,N)
h=T/N;
t=0:N .* h; % this takes the place of the first for-loop
y=zeros(N+1,1);
y(1)=y0;
for i=2:N+1
y(i)=y(i-1)+h.*f(t(i-1),y(i-1));
end
end
You could even replace the second loop if the function f takes vector inputs like so:
y(1) = y0;
y(2:end) = y(1:end-1) + h .* f(t(1:end-1), y(1:end-1));
You're iterating over i = 0:N and using it as t(i)=i.*h, so you're trying to access t(0) during the first iteration. Matlab indexing starts from 1, hence the error.
You also have other lines which will cause the same error, once the execution gets that far.
Related
I am trying to code something in Matlab and it involves a lot of accessing elements in vectors. Below is a snippet of code that I am working on:
x(1)=1;
for i=2:18
x(i)=0;
end
for i=1:18
y(i)=1;
end
for i = 0:262124
x(i+18+1) = x(i+7+1) + mod(x(i+1),2);
y(i+18+1) = y(i+10+1) + y(i+7+1) + y(i+5+1) + mod(y(i+1), 2);
end
% n can be = 0, 1, 2,..., 262142
n = 2;
for i = 0: 262142
z(i+1) = x(mod(i+n+1, 262143)); %error: Subscript indices must either be real positive integers or logicals.
end
In the last "for" loop where I am initialising vector z(), I get an error saying: "Subscript indices must either be real positive integers or logicals." However, when I do not suppres z(i+1) by ommiting the semi colon, the program is able to run, and I can see the values of z in the workspace. Why is this?
The code I am writing in Matlab is based upon the series of instructions shown in the image below. However, I can't seem to track down my error which leads to me not being able to access the elements of x() (without not suppressing the output of z()).
I appreciate any ideas :-) Thank you!
The code breaks at that loop last iteration because , for i=262140 you get
(mod(i+n+1, 262143)) = 0
so you cant access x(0) in matlab. the first elements of any variable is x(1).
In addition, and not related to your question, this code doesn't use the advantages matlab has, instead of
for i=2:18
x(i)=0;
end
you can just write:
x(2:18)=0;
etc
To solve one dimensional advection equation denoted by
u_t+u_x = 0, u=u(x,t), and i.c. u(x,0)= 1+H(x+1)+H(x-1)
using Lax Wanderoff method,
I need to write a Heaviside step function H(x) and it needs to be zero when x <= 0, 1 when x>0 . The problem is I also need to use that function writing H(x-t+1), H(x-t-1) as I will compare what I find by the exact solution:
u(x,t) =1 + H(x-t+1) -H(x-t-1)
Here, the "x" and "t" are vectors such that;
x=-5:0.05:5
t=0:0.05:1
I wrote the Heaviside step function as the following; however, I need it without the for loop.
L=length(x)
function H_X= heavisidefunc(x,L)
H_X=zeros(1,L);
for i= 1:L
if x(i)<= 0
H_X(i)=0;
else
H_X(i)=1;
end
end
end
I get "Dimensions must agree." error if I write
H_X3 = heavisidefunc(x-t+1,L);
H_X4 = heavisidefunc(x-t-1,L);
The Heavyside function is really easy to program in Matlab
Heavyside=#(x) x>= 0;
The easiest way to get rid of the dimensions must agree error is to transpose one of the vectors. This will cause Matlab to construct a matrix of length(x1) by length(x2)
Heavyside(x-t'+1);
I came up with a solution. My new Heaviside function is;
function H_X= heavisidefunc(x)
if x<= 0
H_X=0;
else
H_X=1;
end
end
The problem I had was because I was storing the output as a vector and it just complicated things.
Now,writing H(x-t+1), H(x-t-1) is easier. Just put "heavisidefunc(x(i)-t(j)-1)" and loop from 1 to the length of x and l in two loops. Thanks to everyone!
I want to compute using MATLAB all the integers x such that x^2-2x<2000, but I am having problems in the displaying part because when I run my code MATLAB seems not to finish. My code is as follows:
x=100;
while x^2+2x-2000>=10^-6
x=x-20;
if x^2+2x-2000<10^-6
disp(x)
end
end
I think the wrong part is when I type disp(x) in the while loop, but I don't know how to fix it.
NOTE: I am using the 10^-6 to avoid rounding errors
Can someone help me to fix this code please?
Computationally, finding all integers that satisfy this condition will need some help from a quick insight into this problem. Otherwise, you'll have to test all the integers, which is impossible since there is infinite number of integers. Analytically, finding all the integers that satisfy the condition x^2-2x < 2000 means finding the integers that lies within the intersection of the curve x^2 - 2x and y = 2000.
Let's first take a look at the problem by plotting it:
x = -500:.1:500;
y = x.^2 - 2*x;
plot(x, y, 'color', 'g')
hold on
line([-200 200], [2000 2000], 'color', 'r')
You can easily see that you can limit your search to at least between -100 and 100. You can store the value in an array
results = [] % Declare empty array first and append value later.
for x = -100:100
if x^2 - 2*x < 2000
disp(x)
results = [results,x]; % Using a bracket to append value to the array.
end
end
And a faster way to get results using logical indexing.
x = -100:100
results = x(x.^2 - 2*x<2000)
In the above code, x.^2 - 2*x < 2000 generates a logical array that has the same size as x. Each element stores the logical value that comes from the evaluating each element in x with the expression. You then use this as a mask to pick out only elements in x that satisfy this condition.
If you add in some parentheses and use the correct syntax for 2*x, it works.
x=100;
while (x^2+2*x-2000)>= (10^-6)
x=(x-20);
if (x^2+2*x-2000) <10^-6
disp(x)
end
end
Building upon #User3667217's solution, you can also vectorise this:
x = -100:100;
y = x.^2-2*x;
tmp = y<2000;
results = y(tmp);
this will give you a speed up over the for-loop solution.
I am having issues with a code of mine dealing with decay. The error "Subscript indices must either be real positive integers or logicals" continues to occur no matter how many times I attempt to fix the line of code: M=M(t)+h.*F
Here is the complete code so that it may be easier to solve the issue:
M=10000;
M=#(t) M*exp(-4.5*t);
F=-4.5*M(t);
h=.1;
t(1)=0;
tmax=20;
n=(tmax-t(1))/h;
i=1;
while h<=.5
while i<=n
t=t+h;
M=M(t)+h.*F;
data_out=[t,M];
dlmwrite('single_decay_euler_h.txt',data_out,'delimiter','\t','-append');
i=i+1;
end
h=h+.1;
end
Thanks for any help.
In the start, you're setting M = 5000;. In the following line, you're creating an anonymous function also called M:
M=#(t) M*exp(-4.5*t);
Now, your initial M = 5000 variable has been overwritten, and is substituted by the function:
M(t) = 5000 * exp(-4.5*t); %// Note that the first M is used to get 5000
Thereafter you do F = -4.5*M(t). I don't know what the value t is here, but you're giving F the value -4.5 * 5000 * exp(-4.5*t), for some value of t. You are not creating a function F.
In the first iteration of the loop, M=M(t)+h.*F; is interpreted as:
M = 5000 * exp(-4.5*0) + 0.1*F %// Where F has some value determined by previous
%// the function above and the previous value of t
%// -4.5*0 is because t = 0
M is now no longer a function, but a single scalar value. The next iteration t = 0.1. When you do: M=M(t)+h.*F; now, it interprets both the first and second M as a variable, not a function. t is therefore used as an index, instead of being an input parameter to the function M (since you have overwritten it).
When you are writing M(t), you are trying to access the 0.1'th element of the 1x1 matrix (scalar) M, which obviously isn't possible.
Additional notes:
The outer while loop has no purpose as it stands now, since i isn't reset after the inner loop. When you're finished with the first iteration of the outer loop, i is already >n, so it will never enter the inner loop again.
You shouldn't mix variable and function names (as you do with M. Use different names, always. Unless you have a very good reason not to.
data_out=[t,M]; is a growing vector inside a loop. This is considered very bad practice, ans is very slow. It's better to pre-allocate memory for the vector, for instance using data_out = zeros(k,1), and insert new values using indexes, data_out(ii) = M.
It's recommended not to use i and j as variable names in MATLAB as these also represent the imaginary unit sqrt(-1). This might cause some strange bugs if you're not paying attention to it.
You can almost certainly do what you're trying to do without loops. However, the function you have written is not functioning, and it's not explained all too well what you're trying to do, so it's hard to give advice as to how you can get what you want (but I'll give it a try). I'm skipping the dlmwrite-part, because I don't really understand what you want to output.
M = 5000;
t0 = 0;
tmax = 20;
h = 0.1; %// I prefer leading zeros in decimal numbers
t = t0: h: tmax;
data_out = M .* exp(-4.5 * t);
The problem is caused by M(t) in your code, because t is not an integer or logical (t=1,1.1,1.2,...)
You need to change your code to pass an integer as a subscript. Either multiply t by 10, or don't use the matrix M if you don't need it.
I used nlfilter for a test function of mine as follows:
function funct
clear all;
clc;
I = rand(11,11);
ld = input('Enter the lag = ') % prompt for lag distance
A = nlfilter(I, [7 7], #dirvar);
% Subfunction
function [h] = dirvar(I)
c = (size(I)+1)/2
EW = I(c(1),c(2):end)
h = length(EW) - ld
end
end
The function works fine but it is expected that nlfilter progresses element by element, but in first two iterations the values of EW will be same 0.2089 0.4162 0.9398 0.1058. But then onwards for all iterations the next element is selected, for 3rd it is 0.4162 0.9398 0.1058 0.1920, for 4th it is 0.9398 0.1058 0.1920 0.5201 and so on. Why is it so?
This is nothing to worry about. It happens because nlfilter needs to evaluate your function to know what kind of output to create. So it uses feval once before starting to move across the image. The output from this feval call is what you see the first time.
From the nlfilter code:
% Find out what output type to make.
rows = 0:(nhood(1)-1);
cols = 0:(nhood(2)-1);
b = mkconstarray(class(feval(fun,aa(1+rows,1+cols),params{:})), 0, size(a));
% Apply fun to each neighborhood of a
f = waitbar(0,'Applying neighborhood operation...');
for i=1:ma,
for j=1:na,
x = aa(i+rows,j+cols);
b(i,j) = feval(fun,x,params{:});
end
waitbar(i/ma)
end
The 4th line call to eval is what you observe as the first output from EW, but it is not used to anything other than making the b matrix the right class. All the proper iterations happen in the for loop below. This means that the "duplicate" values you observe does not affect your final output matrix, and you need not worry.
I hope you know what the length function does? It does not give you the Euclidean length of a vector, but rather the largest dimension of a vector (so in your case, that should be 4). If you want the Euclidean length (or 2-norm), use the function norm instead. If your code does the right thing, you might want to use something like:
sz = size(I,2);
h = sz - (sz+1)/2 - ld;
In your example, this means that depending on the lag you provide, the output should be constant. Also note that you might want to put semicolons after each line in your subfunction and that using clear all as the first line of a function is useless since a function will always be executed in its own workspace (that will however clear persistent or global variables, but you don't use them in your code).