How do I plot the value of Approximation - Answer as s varies in the code below? If you look at my code below, you can see the method I used (I put it in a separate file).
However, it does not show me a graph from 1 to 1000. Instead the graph is from 999 to 1001 and does not have any points on it.
for s = 1:1000
error = LaplaceTransform(s,5) - (antiderivative(1,s)-antiderivative(0,s));
end
plot(s,error);
title('Accuracy of Approximation');
xlabel('s');
ylabel('Approximation - Exact Answer');
The functions used:
function g = LaplaceTransform(s,N);
% define function parameters
a=0;
b=1;
h=(b-a)/N;
x = 0:h:1;
% define function
g = ff(x).*exp(-s*x);
% compute the exact answer of the integral
exact_answer=antiderivative(b,s)-antiderivative(a,s)
% compute the composite trapezoid sum
If=0;
for i=1:(N-1)
If=If+g(i).*h;
end;
If=If+g(1).*h/2+g(N).*h/2;
If
with
function fx=ff(x)
fx=x;
and
function fx=antiderivative(x,s);
fx= (-exp(-s*x)*(s*x+1))/(s^2);
Any help would be appreciated. Thanks.
The following
for s = 1:1000
error = LaplaceTransform(s,5) - (antiderivative(1,s)-antiderivative(0,s));
end
plot(s,error);
already has several issues. The two main ones are that error is getting overwritten at each iteration, as #Amro has pointed out, and that s, your loop variable, is a scalar.
Thus, you need to write
difference = zeros(1000,1); %# preassignment is good for you
for s = 1:1000
difference(s) = LaplaceTransform(s,5) - (antiderivative(1,s)-antiderivative(0,s));
end
plot(1:1000,difference);
There is another error in the LaplaceTransform function
function g = LaplaceTransform(s,N);
[...]
g = ff(x).*exp(-s*x); %# g is an array
[...]
If %# If is calculated, but not returned.
I assume you want to write
function If = LaplaceTransform(s,N);
instead, because otherwise, you try to assign the array g to the scalar difference(s).
Related
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 want to write a function that approximates integrals with the trapezoidal rule.
I first defined a function in one file:
function[y] = integrand(x)
y = x*exp(-x^2); %This will be integrand I want to approximate
end
Then I wrote my function that approximates definite integrals with lower bound a and upper bound b (also in another file):
function [result] = trapez(integrand,a,b,k)
sum = 0;
h = (b-a)/k; %split up the interval in equidistant spaces
for j = 1:k
x_j = a + j*h; %this are the points in the interval
sum = sum + ((x_j - x_(j-1))/2) * (integrand(x_(j-1)) + integrand(x_j));
end
result = sum
end
But when I want to call this function from the command window, using result = trapez(integrand,0,1,10) for example, I always get an error 'not enough input arguments'. I don't know what I'm doing wrong?
There are numerous issues with your code:
x_(j-1) is not defined, and is not really a valid Matlab syntax (assuming you want that to be a variable).
By calling trapez(integrand,0,1,10) you're actually calling integrand function with no input arguments. If you want to pass a handle, use #integrand instead. But in this case there's no need to pass it at all.
You should avoid variable names that coincide with Matlab functions, such as sum. This can easily lead to issues which are difficult to debug, if you also try to use sum as a function.
Here's a working version (note also a better code style):
function res = trapez(a, b, k)
res = 0;
h = (b-a)/k; % split up the interval in equidistant spaces
for j = 1:k
x_j1 = a + (j-1)*h;
x_j = a + j*h; % this are the points in the interval
res = res+ ((x_j - x_j1)/2) * (integrand(x_j1) + integrand(x_j));
end
end
function y = integrand(x)
y = x*exp(-x^2); % This will be integrand I want to approximate
end
And the way to call it is: result = trapez(0, 1, 10);
Your integrandfunction requires an input argument x, which you are not supplying in your command line function call
I wrote some code that works just fine to evaluate theta on its own with some test input. However, I would like to take this code and turn it into a function that I can call within another matlab file. I keep getting the error message, "Function definitions are not permitted in this context."
I want to be able to define four vectors in another matlab file and call SP1 to evaluate theta for those inputs. I'm not sure where I'm going wrong, though. Please help!
Thanks so much.
clc
clear all
function theta = SP1(p,q1,w1,r)
% INPUT:
%function theta = SP1(p,q1,w1,r)
% p = [5; -7; 12];
% q1 = [17.3037; -3.1128; 2.48175];
% w1 = [1/sqrt(8); sqrt(3/8); 1/sqrt(2)];
% r = [1; 2; -3];
% Define vectors u and v as well as u' and v'.
u = p - r;
v = q1 - r;
w1_t = transpose(w1);
u_prime = u - w1 * w1_t * u;
v_prime = v - w1 * w1_t * v;
% Calculate theta if conditions are met for a solution to exist.
if (abs(norm(u_prime)-norm(v_prime))<0.01) & (abs((w1_t * u)-(w1_t * v))<0.01)
X = w1_t*cross(u_prime,v_prime);
Y = dot(u_prime,v_prime);
theta = atan2(X,Y)
else if (norm(u_prime) == 0 | norm(v_prime) == 0)
disp('Infinite Number of Solutions')
else
disp('Conditions not satisfied to find a solution')
end
end
I think you can just remove the top two lines,
clc
clear all
and save the rest of the code starting with function as SP1.m file.
Then you should be able to call this function as SP1 from other m files.
I think you're confused about how functions work. The first line of a function definition defines how many inputs and outputs MATLAB expects:
function theta = SP1(p,q1,w1,r)
This means that calling a function SP1 will require you to give four inputs, and will return one output. It doesn't mean that:
Your inputs need to be named p, q1 and so on
Your output will be called theta automatically
The function will automatically take in the input variables p, q1, etc if they exist in the workspace.
It also doesn't do any checking on the inputs; so if you require that inputs be of a certain type, size, etc. you need to write your own error checking at the start of the file. You might intend that those inputs be 3x1 vectors, but there's nothing in the function to tell MATLAB that. So, SP1(1,2,3,4) will work, to some extent - it will take those inputs and try to run them through the function, and if they don't cause an error it will give you an output. The output might be wrong, but the computer doesn't know that.
Once you have a function you can call it multiple ways from the command line or from within other functions or scripts. As previously mentioned you don't have to stick to the naming of variables within the function, as long as input variables exist when the function is called MATLAB will accept them:
theta = SP1(p8,q27,w35,not_r);
myoutput = SP1(any,variable,I,like);
I don't necessarily have to give an output (but then the first output will be routed to ans)
SP1(this,will,also,work);
If I have some variables stored in a *.mat file (the case you seem to be asking about), I can do it like this:
load('mydata.mat'); %this file contains stored variables p, q1, w1 and r
theta = SP1(p,q1,w1,r);
My main script contains following code:
%# Grid and model parameters
nModel=50;
nModel_want=1;
nI_grid1=5;
Nth=1;
nRow.Scale1=5;
nCol.Scale1=5;
nRow.Scale2=5^2;
nCol.Scale2=5^2;
theta = 90; % degrees
a_minor = 2; % range along minor direction
a_major = 5; % range along major direction
sill = var(reshape(Deff_matrix_NthModel,nCell.Scale1,1)); % variance of the coarse data matrix of size nRow.Scale1 X nCol.Scale1
%# Covariance computation
% Scale 1
for ihRow = 1:nRow.Scale1
for ihCol = 1:nCol.Scale1
[cov.Scale1(ihRow,ihCol),heff.Scale1(ihRow,ihCol)] = general_CovModel(theta, ihCol, ihRow, a_minor, a_major, sill, 'Exp');
end
end
% Scale 2
for ihRow = 1:nRow.Scale2
for ihCol = 1:nCol.Scale2
[cov.Scale2(ihRow,ihCol),heff.Scale2(ihRow,ihCol)] = general_CovModel(theta, ihCol/(nCol.Scale2/nCol.Scale1), ihRow/(nRow.Scale2/nRow.Scale1), a_minor, a_major, sill/(nRow.Scale2*nCol.Scale2), 'Exp');
end
end
%# Scale-up of fine scale values by averaging
[covAvg.Scale2,var_covAvg.Scale2,varNorm_covAvg.Scale2] = general_AverageProperty(nRow.Scale2/nRow.Scale1,nCol.Scale2/nCol.Scale1,1,nRow.Scale1,nCol.Scale1,1,cov.Scale2,1);
I am using two functions, general_CovModel() and general_AverageProperty(), in my main script which are given as following:
function [cov,h_eff] = general_CovModel(theta, hx, hy, a_minor, a_major, sill, mod_type)
% mod_type should be in strings
angle_rad = theta*(pi/180); % theta in degrees, angle_rad in radians
R_theta = [sin(angle_rad) cos(angle_rad); -cos(angle_rad) sin(angle_rad)];
h = [hx; hy];
lambda = a_minor/a_major;
D_lambda = [lambda 0; 0 1];
h_2prime = D_lambda*R_theta*h;
h_eff = sqrt((h_2prime(1)^2)+(h_2prime(2)^2));
if strcmp(mod_type,'Sph')==1 || strcmp(mod_type,'sph') ==1
if h_eff<=a
cov = sill - sill.*(1.5*(h_eff/a_minor)-0.5*((h_eff/a_minor)^3));
else
cov = sill;
end
elseif strcmp(mod_type,'Exp')==1 || strcmp(mod_type,'exp') ==1
cov = sill-(sill.*(1-exp(-(3*h_eff)/a_minor)));
elseif strcmp(mod_type,'Gauss')==1 || strcmp(mod_type,'gauss') ==1
cov = sill-(sill.*(1-exp(-((3*h_eff)^2/(a_minor^2)))));
end
and
function [PropertyAvg,variance_PropertyAvg,NormVariance_PropertyAvg]=...
general_AverageProperty(blocksize_row,blocksize_col,blocksize_t,...
nUpscaledRow,nUpscaledCol,nUpscaledT,PropertyArray,omega)
% This function computes average of a property and variance of that averaged
% property using power averaging
PropertyAvg=zeros(nUpscaledRow,nUpscaledCol,nUpscaledT);
%# Average of property
for k=1:nUpscaledT,
for j=1:nUpscaledCol,
for i=1:nUpscaledRow,
sum=0;
for a=1:blocksize_row,
for b=1:blocksize_col,
for c=1:blocksize_t,
sum=sum+(PropertyArray((i-1)*blocksize_row+a,(j-1)*blocksize_col+b,(k-1)*blocksize_t+c).^omega); % add all the property values in 'blocksize_x','blocksize_y','blocksize_t' to one variable
end
end
end
PropertyAvg(i,j,k)=(sum/(blocksize_row*blocksize_col*blocksize_t)).^(1/omega); % take average of the summed property
end
end
end
%# Variance of averageed property
variance_PropertyAvg=var(reshape(PropertyAvg,...
nUpscaledRow*nUpscaledCol*nUpscaledT,1),1,1);
%# Normalized variance of averageed property
NormVariance_PropertyAvg=variance_PropertyAvg./(var(reshape(...
PropertyArray,numel(PropertyArray),1),1,1));
Question: Using Matlab, I would like to optimize covAvg.Scale2 such that it matches closely with cov.Scale1 by perturbing/varying any (or all) of the following variables
1) a_minor
2) a_major
3) theta
I am aware I can use fminsearch, however, how I am not able to perturb the variables I want to while using this fminsearch.
I won't pretend to understand everything that you are doing. But it sounds like a typical minimization problem. What you want to do is to come up with a single function that takes a_minor, a_major and theta as arguments, and returns the square of the difference between covAvg.Scale2 and cov.Scale1. Something like this:
function diff = minimize_me(a_minor, a_major, theta)
... your script goes here
diff = (covAvg.Scale2 - cov.Scale1)^2;
end
Then you need matlab to minimize this function. There's more than one option here. Since you only have three variables to minimize over, fminsearch is a good place to start. You would call it something like this:
opts = optimset('display', 'iter');
x = fminsearch( #(x) minimize_me(x(1), x(2), x(3)), [a_minor_start a_major_start theta_start], opts)
The first argument to fminsearch is the function you want to optimize. It must take a single argument: a vector of the variables that will be perturbed in order to find the minimum value. Here I use an anonymous function to extract the values from this vector and pass them into minimize_me. The second argument to fminsearch is a vector containing the values to start searching at. The third argument are options that affect the search; it's a good idea to set display to iter when you first start optimizing, so that you can get an idea of well the optimizer is converging.
If your parameters have restricted domains (e.g. they must all be positive) take a look at fminsearchbnd on the file exchange.
If I have misunderstood your problem, and this doesn't help at all, try posting code that we can run to reproduce the problem ourselves.
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.