folks,
I am wondering if it is possible to write the following function of r as an inline function in matlab. I tried to include the condition as a separate factor such as *(r>a) and I got NaN due to the division of 1/r^3 when r is 0.
I fould a simple way out. It's basically what Shai and Jigg suggested, i.e. using an extra multiplicative factor of (r>a).
To get rid of NaN, we just need to add eps to the denominator of 1/r3, i.e.
1/(r+eps)^3 *(r>a)
First, you haven't stated what should actually happen if r = 0. Mathematically the term gets infinity. I assumed you rather want to set it to zero. And what should happen for r = a? Just another ill-defined case, are you sure your formula is correct?
If you have the Statistics Toolbox you can use nansum. If not, I'd say there is no way around to write your own function similar to nansum, which can't be done inline.
r = -5:1:5;
a = 1;
R = 42; %// rest of your function
%// not working, or removing of nan afterwards required
X = #( r ) (r>=a).*(a./r).^3*R;
%// inline solution with statistics toolbox
Y = #( r ) arrayfun(#(x) nansum( (x>=a)*(a/x)^3*R ), r);
output = [X(r)' Y(r)']
nansum is not vectorized, if you still want to use it for vectors wrap it into arrayfun.
The code of nansum does exactly what was suggested in the comments (output(isnan(output))=0), I'm probably not allowed to copy&paste it here. It filters out all NaN and then sums the input. Use open nansum to have insight.
As pointed out by Jigg, similar functions like nanmean would do the trick as well.
You can try
chi = 1; %// arbitrary value
a = 1; %// arbitrary value
theta = pi/3; %// arbitrary value
nu = #( r ) (r>a).*( (chi/3).*((a.^3)./(r.^3)).*(3*cos(theta).^2 -1);
Related
I am trying to automatize a process in MATLAB. I have a symbolic function that has to be converted into an anonymous function.
Basically, I have something like this:
syms x y z t
var = [x y z t];
f = x.^2+y;
gf = gradient(f, var);
gf_fun = matlabFunction(gf, 'vars', {var});
giving as output
gf_fun =
function_handle with value:
#(in1)[in1(:,1).*2.0;1.0;0.0;0.0]
Now, I'd like to evaluate this gf_fun in several points at a time, but, of course, I got strange results, due to how gf_fun is written. For example, if I want to evaluate gf_fun in 6 (different) points simultaneously, what I get is
rng('deafult')
vv = ones(6,4);
gf_fun(vv)
ans =
1.3575
1.5155
1.4863
0.7845
1.3110
0.3424
1.0000
0
0
instead of a matrix with dimensions 4x6, with each colomn being the evaluation of a single point.
I know that a workaround will be the use of a for loop, that is
results = zeros(4,6);
for i = 1:6
results(:,i) = gf_fun(vv(i,:));
end
but I must avoid it due to code performances reasons.
Is there a way to automatize all the process, having a matrix as output of gf_fun while evaluating different point at a time? So, basically, there is a simple way to replace the 0.0 and 1.0 in gf_fun with a more general zeros(size(in1)) and ones(size(in1)) automatically?
Thank you for you help!
I have this task to create a script that acts similarly to normcdf on matlab.
x=linspace(-5,5,1000); %values for x
p= 1/sqrt(2*pi) * exp((-x.^2)/2); % THE PDF for the standard normal
t=cumtrapz(x,p); % the CDF for the standard normal distribution
plot(x,t); %shows the graph of the CDF
The problem is when the t values are assigned to 1:1000 instead of -5:5 in increments. I want to know how to assign the correct x values, that is -5:5,1000 to the t values output? such as when I do t(n) I get the same result as normcdf(n).
Just to clarify: the problem is I cannot simply say t(-5) and get result =1 as I would in normcdf(1) because the cumtrapz calculated values are assigned to x=1:1000 instead of -5 to 5.
Updated answer
Ok, having read your comment; here is how to do what you want:
x = linspace(-5,5,1000);
p = 1/sqrt(2*pi) * exp((-x.^2)/2);
cdf = cumtrapz(x,p);
q = 3; % Query point
disp(normcdf(q)) % For reference
[~,I] = min(abs(x-q)); % Find closest index
disp(cdf(I)) % Show the value
Sadly, there is no matlab syntax which will do this nicely in one line, but if you abstract finding the closest index into a different function, you can do this:
cdf(findClosest(x,q))
function I = findClosest(x,q)
if q>max(x) || q<min(x)
warning('q outside the range of x');
end
[~,I] = min(abs(x-q));
end
Also; if you are certain that the exact value of the query point q exists in x, you can just do
cdf(x==q);
But beware of floating point errors though. You may think that a certain range outght to contain a certain value, but little did you know it was different by a tiny roundoff erorr. You can see that in action for example here:
x1 = linspace(0,1,1000); % Range
x2 = asin(sin(x1)); % Ought to be the same thing
plot((x1-x2)/eps); grid on; % But they differ by rougly 1 unit of machine precision
Old answer
As far as I can tell, running your code does reproduce the result of normcdf(x) well... If you want to do exactly what normcdf does them use erfc.
close all; clear; clc;
x = linspace(-5,5,1000);
cdf = normcdf(x); % Result of normcdf for comparison
%% 1 Trapezoidal integration of normal pd
p = 1/sqrt(2*pi) * exp((-x.^2)/2);
cdf1 = cumtrapz(x,p);
%% 2 But error function IS the integral of the normal pd
cdf2 = (1+erf(x/sqrt(2)))/2;
%% 3 Or, even better, use the error function complement (works better for large negative x)
cdf3 = erfc(-x/sqrt(2))/2;
fprintf('1: Mean error = %.2d\n',mean(abs(cdf1-cdf)));
fprintf('2: Mean error = %.2d\n',mean(abs(cdf2-cdf)));
fprintf('3: Mean error = %.2d\n',mean(abs(cdf3-cdf)));
plot(x,cdf1,x,cdf2,x,cdf3,x,cdf,'k--');
This gives me
1: Mean error = 7.83e-07
2: Mean error = 1.41e-17
3: Mean error = 00 <- Because that is literally what normcdf is doing
If your goal is not not to use predefined matlab funcitons, but instead to calculate the result numerically (i.e. calculate the error function) then it's an interesting challange which you can read about for example here or in this stats stackexchange post. Just as an example, the following piece of code calculates the error function by implementing eq. 2 form the first link:
nerf = #(x,n) (-1)^n*2/sqrt(pi)*x.^(2*n+1)./factorial(n)/(2*n+1);
figure(1); hold on;
temp = zeros(size(x)); p =[];
for n = 0:20
temp = temp + nerf(x/sqrt(2),n);
if~mod(n,3)
p(end+1) = plot(x,(1+temp)/2);
end
end
ylim([-1,2]);
title('\Sigma_{n=0}^{inf} ( 2/sqrt(pi) ) \times ( (-1)^n x^{2*n+1} ) \div ( n! (2*n+1) )');
p(end+1) = plot(x,cdf,'k--');
legend(p,'n = 0','\Sigma_{n} 0->3','\Sigma_{n} 0->6','\Sigma_{n} 0->9',...
'\Sigma_{n} 0->12','\Sigma_{n} 0->15','\Sigma_{n} 0->18','normcdf(x)',...
'location','southeast');
grid on; box on;
xlabel('x'); ylabel('norm. cdf approximations');
Marcin's answer suggests a way to find the nearest sample point. It is easier, IMO, to interpolate. Given x and t as defined in the question,
interp1(x,t,n)
returns the estimated value of the CDF at x==n, for whatever value of n. But note that, for values outside the computed range, it will extrapolate and produce unreliable values.
You can define an anonymous function that works like normcdf:
my_normcdf = #(n)interp1(x,t,n);
my_normcdf(-5)
Try replacing x with 0.01 when you call cumtrapz. You can either use a vector or a scalar spacing for cumtrapz (https://www.mathworks.com/help/matlab/ref/cumtrapz.html), and this might solve your problem. Also, have you checked the original x-values? Is the problem with linspace (i.e. you are not getting the correct x vector), or with cumtrapz?
I am working on an assignment that requires me to use the trapz function in MATLAB in order to evaluate an integral. I believe I have written the code correctly, but the program returns answers that are wildly incorrect. I am attempting to find the integral of e^(-x^2) from 0 to 1.
x = linspace(0,1,2000);
y = zeros(1,2000);
for iCnt = 1:2000
y(iCnt) = e.^(-(x(iCnt)^2));
end
a = trapz(y);
disp(a);
This code currently returns
1.4929e+03
What am I doing incorrectly?
You need to just specify also the x values:
x = linspace(0,1,2000);
y = exp(-x.^2);
a = trapz(x,y)
a =
0.7468
More details:
First of all, in MATLAB you can use vectors to avoid for-loops for performing operation on arrays (vectors). So the whole four lines of code
y = zeros(1,2000);
for iCnt = 1:2000
y(iCnt) = exp(-(x(iCnt)^2));
end
will be translated to one line:
y = exp(-x.^2)
You defined x = linspace(0,1,2000) it means that you need to calculate the integral of the given function in range [0 1]. So there is a mistake in the way you calculate y which returns it to be in range [1 2000] and that is why you got the big number as the result.
In addition, in MATLAB you should use exp there is not function as e in MATLAB.
Also, if you plot the function in the range, you will see that the result makes sense because the whole page has an area of 1x1.
I got two sparse matrices A and B, which have identical sparsity pattern (all nonzeros are at exactly the same locations):
i = randi(1000,[50,1]);
j = randi(1000,[50,1]);
a = rand(50,1);
b = rand(50,1);
A = sparse(i,j,a,1000,1000);
B = sparse(i,j,b,1000,1000);
I would like to calculate efficiently exp(A-B) only for the nonzeros, and save it back into A.
I tried to use spfun for that task:
f = #(x,y) exp(x-y);
A = spfun(f,A,B);
but I got an error in spfun saying: "Too many input arguments."
Can anyone suggest an efficient way to calculate it?
It should be calculated many times.
Thanks!
Edit: mikkola suggested A = spfun(#f,A-B) which solves the problem, but the question remains how to do it with a function of two variables that can't be solved using the same trick. For example:
g = #(x,y) x.*cos(y);
A = spfun(#g,A,B);
You can't use
A = spfun(#exp, A-B);
because for entries where A an B are equal you will get 0 instead of 1.
To solve that, you can compute the vector of exponentials at the nonzero components, and then build a sparse matrix from that:
A = sparse(i,j,exp(nonzeros(A)-nonzeros(B))); %// if you have i, j stored
or
A(find(A)) = exp(nonzeros(A)-nonzeros(B));
Edit
According to the documentation, spfun can take only two inputs: a function and one sparse matrix.
So you cannot do directly what you want to do. The best solution is probably what has been suggested in the comments, i.e.:
res = spfun(#exp, A-B);
Best,
I have one file with the following code:
function fx=ff(x)
fx=x;
I have another file with the following code:
function g = LaplaceTransform(s,N)
g = ff(x)*exp(-s*x);
a=0;
b=1;
If=0;
h=(b-a)/N;
If=If+g(a)*h/2+g(b)*h/2;
for i=1:(N-1)
If=If+g(a+h*i)*h;
end;
If
Whenever I run the second file, I get the following error:
Undefined function or variable 'x'.
What I am trying to do is integrate the function g between 0 and 1 using trapezoidal approximations. However, I am unsure how to deal with x and that is clearly causing problems as can be seen with the error.
Any help would be great. Thanks.
Looks like what you're trying to do is create a function in the variable g. That is, you want the first line to mean,
"Let g(x) be a function that is calculated like this: ff(x)*exp(-s*x)",
rather than
"calculate the value of ff(x)*exp(-s*x) and put the result in g".
Solution
You can create a subfunction for this
function result = g(x)
result = ff(x) * exp(-s * x);
end
Or you can create an anonymous function
g = #(x) ff(x) * exp(-s * x);
Then you can use g(a), g(b), etc to calculate what you want.
You can also use the TRAPZ function to perform trapezoidal numerical integration. Here is an example:
%# parameters
a = 0; b = 1;
N = 100; s = 1;
f = #(x) x;
%# integration
X = linspace(a,b,N);
Y = f(X).*exp(-s*X);
If = trapz(X,Y) %# value returned: 0.26423
%# plot
area(X,Y, 'FaceColor',[.5 .8 .9], 'EdgeColor','b', 'LineWidth',2)
grid on, set(gca, 'Layer','top', 'XLim',[a-0.5 b+0.5])
title('$\int_0^1 f(x) e^{-sx} \,dx$', 'Interpreter','latex', 'FontSize',14)
The error message here is about as self-explanatory as it gets. You aren't defining a variable called x, so when you reference it on the first line of your function, MATLAB doesn't know what to use. You need to either define it in the function before referencing it, pass it into the function, or define it somewhere further up the stack so that it will be accessible when you call LaplaceTransform.
Since you're trying to numerically integrate with respect to x, I'm guessing you want x to take on values evenly spaced on your domain [0,1]. You could accomplish this using e.g.
x = linspace(a,b,N);
EDIT: There are a couple of other problems here: first, when you define g, you need to use .* instead of * to multiply the elements in the arrays (by default MATLAB interprets multiplication as matrix multiplication). Second, your calls g(a) and g(b) are treating g as a function instead of as an array of function values. This is something that takes some getting used to in MATLAB; instead of g(a), you really want the first element of the vector g, which is given by g(1). Similarly, instead of g(b), you want the last element of g, which is given by g(length(g)) or g(end). If this doesn't make sense, I'd suggest looking at a basic MATLAB tutorial to get a handle on how vectors and functions are used.