I'm looking to integrate this expression:
However I seem to be having problems setting up the function. As outlined in this MATLAB explanation, I've defined a separate function named 'NDfx.m' which looks like this:
function [ y ] = NDfx(x)
y = (1/sqrt(2*pi))*exp(-.5*x^2); % Error occurs here
end
However when I call it within my main function I get an error at the commented line above. My main function looks like this:
function[P] = NormalDistro(u,o2,x)
delta = x-u;
dev = abs((delta)/o2); % Normalizes the parameters entered into function
P_inner = quad(#NDfx,-dev,dev); % Integrates function NDfx from -dev to dev (error here)
P_outer = 1 - P_inner; % Calculation of outer bounds of the integral
if delta > 0
P = P_inner + (P_outer/2);
elseif delta < 0
P = P_outer/2;
elseif dev == 0
P = .5;
end
end
The specific error that I get is:
Error in ==> mpower
Inputs must be a scalar and a square matrix
You've setup everything correctly for integration. The error is in the definition of the function itself. When using variables for function that will be integrated, a "." (period) must precede operators like ^ and * when they are applied to the variable:
function [y] = NDfx(x)
y = (1/sqrt(2*pi))*exp(-.5*(x.^2));
end
Krono and user57368 are correct. They have already correctly answered your actual question. My answer is merely to answer the question you did not ask. That is, why are you using quad here at all? The point is that many people want to integrate a function of that form, and it has been done already! Use existing tools to solve your problems, as those tools will often have been written by someone who knows how to solve the problem accurately and efficiently.
In this case, the existing tool is comprised of the functions erf and erfc. They provide an accurate, efficient, vectorized solution to your problem. The only thing you will need to do is figure out how to transform those integrals to your current problem, done by a simple scaling of the input to erf and the output.
Related
I'm trying to use the MATLAB function fzero properly but my program keeps returning an error message. This is my code (made up of two m-files):
friction_zero.m
function fric_zero = friction_zero(reynolds)
fric_zero = 0.25*power(log10(5.74/(power(reynolds,0.9))),-2);
flow.m
function f = flow(fric)
f = 1/(sqrt(fric))-1.873*log10(reynolds*sqrt(fric))-233/((reynolds*sqrt(fric))^0.9)-0.2361;
f_initial = friction_zero(power(10,4));
z = fzero(#flow,f_initial)
The goal is to return z as the root for the equation specified by f when flow.m is run.
I believe I have the correct syntax as I have spent a couple of hours online looking at examples. What happens is that it returns the following error message:
"Undefined function or variable 'fric'."
(Of course it's undefined, it's the variable I'm trying to solve!)
Can someone point out to me what I've done wrong? Thanks
EDIT
Thanks to all who helped! You have assisted me to eventually figure out my problem.
I had to add another file. Here is a full summary of the completed code with output.
friction_zero.m
function fric_zero = friction_zero(re)
fric_zero = 0.25*power(log10(5.74/(power(re,0.9))),-2); %starting value for fric
flow.m
function z = flow(fric)
re = power(10,4);
z = 1/(sqrt(fric))-1.873*log10(re*sqrt(fric))-233/((re*sqrt(fric))^0.9)-0.2361;
flow2.m
f_initial = friction_zero(re); %arbitrary starting value (Reynolds)
x = #flow;
fric_root = fzero(x,f_initial)
This returns an output of:
fric_root = 0.0235
Which seems to be the correct answer (phew!)
I realised that (1) I didn't define reynolds (which is now just re) in the right place, and (2) I was trying to do too much and thus skipped out on the line x = #flow;, for some reason when I added the extra line in, MATLAB stopped complaining. Not sure why it wouldn't have just taken #flow straight into fzero().
Once again, thanks :)
You need to make sure that f is a function in your code. This is simply an expression with reynolds being a constant when it isn't defined. As such, wrap this as an anonymous function with fric as the input variable. Also, you need to make sure the output variable from your function is z, not f. Since you're solving for fric, you don't need to specify this as the input variable into flow. Also, you need to specify f as the input into fzero, not flow. flow is the name of your main function. In addition, reynolds in flow is not defined, so I'm going to assume that it's the same as what you specified to friction_zero. With these edits, try doing this:
function z = flow()
reynolds = power(10,4);
f = #(fric) 1/(sqrt(fric))-1.873*log10(reynolds*sqrt(fric))-233/((reynolds*sqrt(fric))^0.9)-0.2361;
f_initial = friction_zero(reynolds);
z = fzero(#f, f_initial); %// You're solving for `f`, not flow. flow is your function name
The reason that you have a problem is because flow is called without argument I think. You should read a little more about matlab functions. By the way, reynolds is not defined either.
I am afraid I cannot help you completely since I have not been doing fluid mechanics. However, I can tell you about functions.
A matlab function definition looks something like this:
function x0 = f(xGuess)
a = 2;
fcn =#(t) a*t.^3+t; % t must not be an input to f.
disp(fcn);
a = 3;
disp(fcn);
x0 = fsolve(fcn1,xGuess); % x0 is calculated here
The function can then ne called as myX0 = f(myGuess). When you define a matlab function with arguments and return values, you must tell matlab what to do with them. Matlab cannot guess that. In this function you tell matlab to use xGuess as an initial guess to fsolve, when solving the anonymous function fcn. Notice also that matlab does not assume that an undefined variable is an independent variable. You need to tell matlab that now I want to create an anonymous function fcn which have an independent variable t.
Observation 1: I use .^. This is since the function will take an argument an evaluate it and this argument can also be a vector. In this particulat case I want pointwise evaluation. This is not really necessary when using fsolve but it is good practice if f is not a matrix equation, since "vectorization" is often used in matlab.
Observation 2: notice that even if a changes its value the function does not change. This is since matlab passes the value of a variable when defining a function and not the variable itself. A c programmer would say that a variable is passed by its value and not by a pointer. This means that fcn is really defined as fcn = #(x) 2*t.^3+t;. Using the variable a is just a conveniance (constants can may also be complicated to find, but when found they are just a value).
Armed with this knowledge, you should be able to tackle the problem in front of you. Also, the recursive call to flow in your function will eventuallt cause a crash. When you write a function that calls itself like this you must have a stopping criterium, something to tell the program when to stop. As it is now, flow will call ifself in the last row, like z = fzero(#flow,f_initial) for 500 times and then crash. Alos it is possible as well to define functions with zero inputs:
function plancksConstant = h()
plancksConstant = 6.62606957e−34;
Where the call h or h() will return Plancks constant.
Good luck!
I have a Matlab project in which I need to make a GUI that receives two mathematical functions from the user. I then need to find their intersection point, and then plot the two functions.
So, I have several questions:
Do you know of any algorithm I can use to find the intersection point? Of course I prefer one to which I can already find a Matlab code for in the internet. Also, I prefer it wouldn't be the Newton-Raphson method.
I should point out I'm not allowed to use built in Matlab functions.
I'm having trouble plotting the functions. What I basically did is this:
fun_f = get(handles.Function_f,'string');
fun_g = get(handles.Function_g,'string');
cla % To clear axes when plotting new functions
ezplot(fun_f);
hold on
ezplot(fun_g);
axis ([-20 20 -10 10]);
The problem is that sometimes, the axes limits do not allow me to see the other function. This will happen, if, for example, I will have one function as log10(x) and the other as y=1, the y=1 will not be shown.
I have already tried using all the axis commands but to no avail. If I set the limits myself, the functions only exist in certain limits. I have no idea why.
3 . How do I display numbers in a static text? Or better yet, string with numbers?
I want to display something like x0 = [root1]; x1 = [root2]. The only solution I found was turning the roots I found into strings but I prefer not to.
As for the equation solver, this is the code I have so far. I know it is very amateurish but it seemed like the most "intuitive" way. Also keep in mind it is very very not finished (for example, it will show me only two solutions, I'm not so sure how to display multiple roots in one static text as they are strings, hence question #3).
function [Sol] = SolveEquation(handles)
fun_f = get(handles.Function_f,'string');
fun_g = get(handles.Function_g,'string');
f = inline(fun_f);
g = inline(fun_g);
i = 1;
Sol = 0;
for x = -10:0.1:10;
if (g(x) - f(x)) >= 0 && (g(x) - f(x)) < 0.01
Sol(i) = x;
i = i + 1;
end
end
solution1 = num2str(Sol(1));
solution2 = num2str(Sol(2));
set(handles.roots1,'string',solution1);
set(handles.roots2,'string',solution2);
The if condition is because the subtraction will never give me an absolute zero, and this seems to somewhat solve it, though it's really not perfect, sometimes it will give me more than two very similar solutions (e.g 1.9 and 2).
The range of x is arbitrary, chosen by me.
I know this is a long question, so I really appreciate your patience.
Thank you very much in advance!
Question 1
I think this is a more robust method for finding the roots given data at discrete points. Looking for when the difference between the functions changes sign, which corresponds to them crossing over.
S=sign(g(x)-f(x));
h=find(diff(S)~=0)
Sol=x(h);
If you can evaluate the function wherever you want there are more methods you can use, but it depends on the size of the domain and the accuracy you want as to what is best. For example, if you don't need a great deal of accurac, your f and g functions are simple to calculate, and you can't or don't want to use derivatives, you can get a more accurate root using the same idea as the first code snippet, but do it iteratively:
G=inline('sin(x)');
F=inline('1');
g=vectorize(G);
f=vectorize(F);
tol=1e-9;
tic()
x = -2*pi:.001:pi;
S=sign(g(x)-f(x));
h=find(diff(S)~=0); % Find where two lines cross over
Sol=zeros(size(h));
Err=zeros(size(h));
if ~isempty(h) % There are some cross-over points
for i=1:length(h) % For each point, improve the approximation
xN=x(h(i):h(i)+1);
err=1;
while(abs(err)>tol) % Iteratively improve aproximation
S=sign(g(xN)-f(xN));
hF=find(diff(S)~=0);
xN=xN(hF:hF+1);
[~,I]=min(abs(f(xN)-g(xN)));
xG=xN(I);
err=f(xG)-g(xG);
xN=linspace(xN(1),xN(2),15);
end
Sol(i)=xG;
Err(i)=f(xG)-g(xG);
end
else % No crossover points - lines could meet at tangents
[h,I]=findpeaks(-abs(g(x)-f(x)));
Sol=x(I(abs(f(x(I))-g(x(I)))<1e-5));
Err=f(Sol)-g(Sol)
end
% We also have to check each endpoint
if abs(f(x(end))-g(x(end)))<tol && abs(Sol(end)-x(end))>1e-12
Sol=[Sol x(end)];
Err=[Err g(x(end))-f(x(end))];
end
if abs(f(x(1))-g(x(1)))<tol && abs(Sol(1)-x(1))>1e-12
Sol=[x(1) Sol];
Err=[g(x(1))-f(x(1)) Err];
end
toc()
Sol
Err
This will "zoom" in to the region around each suspected root, and iteratively improve the accuracy. You can tweak the parameters to see whether they give better behaviour (the tolerance tol, the 15, number of new points to generate, could be higher probably).
Question 2
You would probably be best off avoiding ezplot, and using plot, which gives you greater control. You can vectorise inline functions so that you can evaluate them like anonymous functions, as I did in the previous code snippet, using
f=inline('x^2')
F=vectorize(f)
F(1:5)
and this should make plotting much easier:
plot(x,f(x),'b',Sol,f(Sol),'ro',x,g(x),'k',Sol,G(Sol),'ro')
Question 3
I'm not sure why you don't want to display your roots as strings, what's wrong with this:
text(xPos,yPos,['x0=' num2str(Sol(1))]);
In my project I need a function which returns the index of the largest element of a given vector. Just like max. For more than one entry with the same maximum value (which occurs frequently) the function should choose one randomly. Unlike max.
The function is a subfunction in a MATLAB Function Block in Simulink. And the whole Simulink model is compiled.
My basic idea was:
function ind = findOpt(vector)
index_max = find(vector == max(vector));
random = randi([1,length(index_max)],1);
ind = index_max(random);
end
But I got problems with the comparison in find and with randi.
I found out about safe comparison here: Problem using the find function in MATLAB. Also I found a way to replace randi([1,imax],1): Implement 'randi' using 'rand' in MATLAB.
My Code now looks like this:
function ind = findOpt(vector)
tolerance = 0.00001;
index_max = find(abs(vector - max(vector)) < tolerance);
random = ceil(length(index_max)*rand(1));
ind = index_max(random);
end
Still doesn't work. I understand that the length of index_max is unclear and causes problems. But I can not think of any way to know it before. Any ideas how to solve this?
Also, I'm shocked that ceil doesn't work when the code gets executed?? In debug mode there is no change to the input visible.
I thought about creating an array like: index_max = abs(vector - max(vector)) < tolerance; But not sure how that could help. Also, it doesn't solve my problem with the random selection.
Hopefully somebody has more ideas or at least could give me some hints!
I am using MATLAB R2012b (32bit) on a Windows7-64bit PC, with the Lcc-win32 C 2.4.1 compiler.
Edit:
Vector usually is of size 5x1 and contains values between -2000 and zero which are of type double, e.g. vector = [-1000 -1200 -1000 -1100 -1550]'. But I think such a simple function should work with any kind of input vector.
The call of length(index_max) causes an system error in MATLAB and forces me to shut it down. I guess this is due to the strange return I get from find. For a vector with all the same values the return from find is something like [1.000 2.000 1.000 2.000 0.000]' which doesn't make any sense to me at all.
function v= findOpt(v)
if isempty(v)
return;
end
v = find((max(v) - v) < 0.00001);
v = v(ceil(rand(1)*end));
end
I was indeed overloading, just like user664303 suggested! Since I can not use objects in my project, I wanted an function that behaves similar, so I wrote:
function varargout = table(mode, varargin)
persistent table;
if isempty(table) && ~strcmp(mode,'writeTable')
error(...)
end
switch mode
case 'getValue'
...
case 'writeTable'
table = ...
...
end
end
Wanted to avoid passing the dimensions for the table in every call and thought it would be enough if the first call initializes the Table with mode='writeTable'. Looks like this caused my problem.
No problems after changing to:
if isempty(table)
table = zeros(dim1,dim2,...)
end
this is likely a few lines of code but I cannot figure it out...
I need to perform a convolution operation of experimental data with an analytical function which is singular at the beginning of the integration range. So if I use "conv" it won't work...
I have two experimental vectors, phi(time) and time
then want to calculate T(t)
T = \int_{0}^{t}\phi \left ( t-\tau \right )\frac{\textup{d}\tau}{\tau ^{1/2}}
So the the singularity at tau = 0 makes "conv" not work. I've been fiddling with anonymous functions and using quadgk to handle the singularity and the like, but can't make anything work. I would have bet anything that this is a solved problem but can't find anything like it. Any help is very much appreciated.
EDIT: solved it this way:
function T = TCalc(time, power)
warning off MATLAB:quadgk:NonFiniteValue %suppress warning at the zero point
T=zeros(size(time));
for par = 1:numel(time)
T(par) = s1(time(par));
end
function y = r1(t)
y = interp1q(time,power,t');%notice that t is transposed! to make interp1q work.
y = y';
end
function y = s1(tfin)
y = quadgk(#(t) (r1(tfin-t))./t.^(0.5),0,tfin,'RelTol',1.e-2);
end
end
I guess, the inelegant trick here is to make Matlab think the experimental data is an analytical function by passing through interp1, and using quadgk to tolerate a singularity at the limit of integration. About one second for 1000-point vectors. Ignoring or setting the tau = 0 point to a small value doesn't work, it still perceives a singularity and the result is wrong.
'Outliers.m' is called from a higher level .m file. The variables are all defined in the higher level file, and set as globals for access by Outliers.m. The purpose of the code is to identify outliers using Chauvenets Criterion, and for this, I have to calculate the integral of the guassian distribution, using the Integral function and function handles. The code works and gives sensible values when I enter specific variables as a test, but I cannot get it to work in a loop. My data set is comprised of 7 individual samples, each 1x30, all of which need to be analyzed. I have had various errors, read through the guidance on Integral and function handles, but cannot seem to find the solution...Any help or guidance would be very much appreciated.... Here is my code:
n = 7
for x = 1:n
for y = 1:30
z(x,y) = abs((cc(x,y) - mastercc(1,y))/masterccstd(1,y));
xmax(x,y) = mastercc(1,y)+z(x,y)*masterccstd(1,y);
xmin(x,y) = mastercc(1,y)-z(x,y)*masterccstd(1,y);
p(x,y) = 1/(masterccstd(1,y)*(sqrt(2*pi)));
fun(x,y)= #(x,y,z) (exp(-1/2)*z(x,y).^2);
q(x,y) = integral(fun(x,y),xmin(x,y),xmax(x,y),'ArrayValued',true);
pq(x,y) = p(x,y)*q(x,y); % probability
value(x,y) = n*(1/pq(x,y));
count(x,y) = logical(value(x,y) <0.5);
badbins(x)=sum(count(x,:));
end
end
It seems like your error is caused by an invald function definition.
If you try it like this it should work:
fun = #(x,y,z) (exp(-1/2)*z(x,y).^2)
Now it can be called like this for example:
fun(1,2,magic(4))
Solution to the loop problem, courtesy Andrei Bobrov via Matlab Central, link below:
http://www.mathworks.com/matlabcentral/answers/103958#comment_177000
NB: Please note the code is not complete for the purpose I explained in the problem description, but it does solve the Loop error.