I'm trying to plot the Yukawa Potential in Matlab and I want to have my program go through user inputs for the values alpha (called alph in my program) and l. The values I need to use are 0.1, 0.2 and 0.3 for alpha with values of 0, 1 and 2 of l for each value of alpha. I know I could set up a loop for this but it doesn't have to be pretty and I want to test the values one at a time. Anyway I keep getting an error after I input the values for alpha, the error I keep getting is in my function, saying that I don't have enough input arguments. The output should be the T matrix, the Hamiltonian matrix and a plot of the first 10 eigenfunctions.
I've tried going in and simply defining alpha as the numbers I want to look at and the program works fine with displaying the output I'm looking for. I just want to be able to change the values for alpha without having to change the program itself. I haven't had any problems with the l inputs.
r = linspace(0.05,19.95,1999)
n = 1999
dr = 0.05
a = full(gallery("tridiag",n,1,-2,1))
T = -0.5*a/(dr^2)
l = input('Input a value for l.')
alph = input('Input a value for alpha.')
v = arrayfun(#(r) yuk_pot(r,l),r);
V = diag(v)
H = T + V
[O,D] = eig(H);
plot(r,O(:,1),r,O(:,2),r,O(:,3),r,O(:,4),r,O(:,5),r,O(:,6),r,O(:,7),r,O(:,8),r,O(:,9),r,O(:,10))
function v = yuk_pot(r,alph,l)
v = (-exp(-alph*r)/r) + 0.5*(l*(l+1)/(r^2));
end
your function function v = yuk_pot(r,alph,l) has 3 input arguments.
you call it with 2 arguments (r and l)
v = arrayfun(#(r) yuk_pot(r,l),r);
what about the second alph argument?
Related
I want to find the maximum value using the second derivative of the the expression when x is between 0 and 1. In other words I am taking the derivative of cox(x^2) twice to get the second derivative resulting in - 2*sin(x^2) - 4*x^2*cos(x^2), then I want to evaluate this second derivative at x = 0 to x = 1, and display the maximum value of the populated values.
I have:
syms x
f = cos(x^2);
secondD = diff(diff(f));
for i = 0:1
y = max(secondD(i))
end
Can someone help?
You can do it easily by subs and double:
syms x
f = cos(x^2);
secondD = diff(diff(f));
% instead of the for loop
epsilon = 0.01;
specified_range = 0:epsilon:1;
[max_val, max_ind] = max(double(subs(secondD, specified_range)));
Please note that it is a numerical approach to find the maximum and the returned answer is not completely correct all the time. However, by increasing the epsilon, you can expect a better result in general (again in some cases it is not completely correct).
I am trying to write a function that thresholds a grey-level image F and a threshold value t (0 ≤ t ≤ 255) such that r = 0 for r < t and r = 255 otherwise.
I have tried to implement this, but imshow(r) does not produce an output.
function f = imgThreshold(img, t)
f = img;
if (f < t)
f = 0;
else
f = 1;
end
img = imread('https://i.stack.imgur.com/kP0u2.png');
t = 20;
r = imgThreshold(img, t);
imshow(r);
This should threshold this image. However, it does not do so. What am I doing wrong?
Best would be to use logical indexing:
f(f<t)=0; % set all elements of f<t to 0
f(~(f==0))=1; % Set all elements where f is not 0 (i.e. the rest) to 1
f<t nicely produces a logical matrix adhering to the condition, but subsequently you do either f=1 or f=0, meaning that you set the entirety of f to be a scalar (one or zero), which of course just plots a black or white square. Instead, use the logical matrix as indexing operation into the matrix itself, then assigning the desired value to each true entry, like above.
Also a function definition either goes in its own file, or on the bottom of the script. Thus either you save the function as imgThreshold.m and leave the rest for the script, or first call the script and place function f = imgThreshold(img, t) etc after the call to imshow
I want to pass a function that has arguments to another function that takes the function handler (pointer) as an argument, but some of the arguments to the function handler need to be fixed. How does one do this in MATLAB?
Concretely I want to minimize a function h(eta) using fminsearch(fun,x0). One obvious issue is that if I pass all the arguments in x0 it will minimize it with respect to all the arguments but that isn't what I want to do. What I really want to do is instead:
I have tried the following and it seems to work as far as I tested it:
function [ h ] = function_returner( x )
%
t = -10000;
h = #(eta)-10*x + t + eta^2;
end
and the script to test it:
h = function_returner(0);
h(3) %-9991
[x,fval,exitflag] = fminsearch(h,[10]) %returns [0, -10000, 1]
returns the correct minimum and the correct value of x at which the minimum is obtained. The thing that I wanted to make sure is that the variables that are fixed of h indeed remain fixed after I returned the function handler. i.e. if h is passed around, how does one make sure that the variables don't change value just because they go into an environment that has the same variables names as in the function handler?
I also wrote:
x = inf;
t = inf;
eta = inf;
h = function_returner(0);
h(3) %-9991
[x,fval,exitflag] = fminsearch(h,[10]) %returns [0, -10000, 1]
and it seems to go unaffected (i.e. it works as I want it to work). But just to be on the safe side, I was hoping there was not a some weird edge condition that this doesn't work.
You are worrying unnecessarily. :-) The thing is, when you create an anonymous function, all the outer variables that are used for its definition are frozen, so you don't need to worry that the function will change behavior behind your back via modified parameters. To test this, try this script:
a = 1;
b = 2;
f = #(u,v) u*a + v*b;
x = f(1,1); %// x is 3 = 1*1 + 1*2
a = 2;
b = 3;
g = #(w) f(w,a)
y = g(1); %// y is 5 = f(1,2) = 1*1 + 2*2
So I am trying to go through a for loop that will increment .1 every time and will do this until the another variable h is less than or equal to zero. Then I am suppose to graph this h variable along another variable x. The code that I wrote looks like this:
O = 20;
v = 200;
g = 32.2;
for t = 0:.1:12
% Calculate the height
h(t) = (v)*(t)*(sin(O))-(1/2)*(g)*(t^2);
% Calculate the horizontal location
x(t) = (v)*(t)*cos(O);
if t > 0 && h <= 0
break
end
end
The Error that I keep getting when running this code says "Attempted to access h(0); index must be a positive integer or logical." I don't understand what exactly is going on in order for this to happen. So my question is why is this happening and is there a way I can solve it, Thank you in advance.
You're using t as your loop variable as well as your indexing variable. This doesn't work, because you'll try to access h(0), h(0.1), h(0.2), etc, which doesn't make sense. As the error says, you can only access variables using integers. You could replace your code with the following:
t = 0:0.1:12;
for i = 1:length(t)
% use t(i) instead of t now
end
I will also point out that you don't need to use a for loop to do this. MATLAB is optimised for acting on matrices (and vectors), and will in general run faster on vectorised functions rather than for loops. For instance, your equation for h could be replaced with the following:
O = 20;
v = 200;
g = 32.2;
t = 0:0.1:12;
h = v * t * sin(O) - 0.5 * g * t.^2;
The only difference is that you have to use the element-wise square (.^2) rather than the normal square (^2). This means that MATLAB will square each element of the vector t, rather than multiplying the vector t by itself.
In short:
As the error says, t needs to be an integer or logical.
But your t is t=0:0.1:12, therefore a decimal value.
O = 20;
v = 200;
g = 32.2;
for t = 0:.1:12
% Calculate the height
idx_t = 1:numel(t);
h(idx_t) = (v)*(t)*(sin(O))-(1/2)*(g)*(t^2);
% Calculate the horizontal location
x(idx_t) = (v)*(t)*cos(O);
if t > 0 && h <= 0
break
end
end
Look this question's answer for more options: Subscript indices must either be real positive integers or logical error
I'm trying to calculate how the errors depend on the step, h, for the trapezoidal rule. The errors should get smaller with a smaller value of h, but for me this doesn't happen. This is my code:
Iref is a reference value calculated and verified with Simpson's method and the MATLAB function quad, respectively
for h = 0.01:0.1:1
x = a:h:b;
v = y(x);
Itrap = (sum(v)-v(1)/2-v(end)/2)*h;
Error = abs(Itrap-Iref)
end
I think there's something wrong with the way I'm using h, because the trapezoidal rule works for known integrals. I would be really happy if someone could help me with this, because I can't understand why the errors are "jumping around" the way the do.
I wonder if maybe part of the problem is that not all intervals - for each step size h - have the same a and b just because of the way that x is constructed. Try the following with the additional fprintf statement:
for h = 0.01:0.1:1
x = a:h:b;
fprintf('a=%f b=%f\n',x(1),x(end));
v = y(x);
Itrap = (sum(v)-v(1)/2-v(end)/2)*h;
Error = abs(Itrap-Iref);
end
Depending upon your a and b (I chose a=0 and b=5) all the a values were identical (as expected) but the b varied from 4.55 to 5.0.
I think that you always want to keep the interval [a,b] the same for each step size that you choose in order to get a better comparison between each iteration. So rather than iterating over the step size, you could instead iterate over the n, the number of equally spaced sub-intervals within [a,b].
Rather than
for h = 0.01:0.1:1
x = a:h:b;
you could do something more like
% iterate over each value of n, chosen so that the step size
% is similar to what you had before
for n = [501 46 24 17 13 10 9 8 7 6]
% create an equally spaced vector of n numbers between a and b
x = linspace(a,b,n);
% get the step delta
h = x(2)-x(1);
v = y(x);
Itrap = (sum(v)-v(1)/2-v(end)/2)*h;
Error = abs(Itrap-Iref);
fprintf('a=%f b=%f len=%d h=%f Error=%f\n',x(1),x(end),length(x),h,Error);
end
When you evaluate the above code, you will notice that a and b are consistent for each iteration, h is roughly what you chose before, and the Error does increase as the step size increases.
Try the above and see what happens!