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!
Related
I first defined functions for dy/dt=y and dy/dt=t:
function dy=d(y):
dy=y
end
function ddy=dd(t):
ddy=t
end
And then I used ode45, respectively:
[t,y]=ode45('d',[1 10],1)
[t,y]=ode45('dd',[1 10],1)
which returns the following error: Error using d
Too many input arguments.
My question is:
Where did I go wrong?
How does Matlab know whether y or t is the independent variable? When I define the first function, it could be reasonably interpreted as dt/dy=y instead of dy/dt=y. Is there a built-in convention for defining functions?
First things first: the docs on ode45 are on the mathworks website, or you can get them from the console by entering help ode45.
The function you pass in needs to take two variables, y then t. As you noticed, with just one it would be impossible to distinguish a function of only y from a function of only t. The first argument has to be the independent, the second is the dependent.
Try defining your function as dy = d(t, y) and ddy = dd(t, y) with the same bodies.
one other note, while using a string representing the function name should work, you can use #d and #dd to reference the functions directly.
I have a statement in my MATLAB program:
f = #(A)DistanceGauss(A,x_axis,Y_target,Y_initial,numOf,modus);
I understood that f is defined as the function handle to the function distancegauss which contains the parameters/arg list present inside the parentheses.
What does the variable A in #(A) do? Does it have any importance? While browsing I found that the variables within parentheses after # would be the input arguments for an anonymous function..
Can anyone explain what does that A do? Will this handle work even without that A after the # symbol ? Because it is already present as an argument to be passed after the function name.
Your code will create an anonymous function f which accepts one input A. In particular f will call the function DistanceGauss(A,x_axis,Y_target,Y_initial,numOf,modus); where the value of A is whatever you input with f(A) and the other inputs must already exist in your workspace and will be passed to the function. Note: if the other variables don't exist you should get an error when calling f.
Now a reasonable question is why would you want to do this you could just call DistanceGauss(A,x_axis,Y_target,Y_initial,numOf,modus); directly with whatever values you want, without having to worry about whether some of them exist.
There are two main reasons I can think of why you would do this (I'm sure there are others). Firstly for simplicity where your other inputs don't change and you don't want to have to keep retyping them or have users accidentally change them.
The other reason where you would want this is when optimizing/minimizing a function, for example with fminsearch. The matlab optimization functions will vary all inputs. If you want only vary some of them you can use this sort of syntax to reduce the number of input variables.
As to what A actually is in your case this will depend on what it does in DistanceGauss, which is not a standard MATLAB function so I suggest you look at the code for that.
"f(A)" or "f of A" or "The function of A" here has the handle "f"
DistanceGauss() here is another function that was defined elsewhere in your code.
You would set x_axis, Y_target, Y_initial, numOf, & modus before creating the function f. These arguments would stay the same for Function f, even if you try and set them again later.
'A' though, is different. You don't set it before you make the function. You can later operate on all values of A, such as if you plot the function or get the integral of the function. In that case, it would be performing the DistanceGauss function on every value of 'A', (though we can't see here what DistanceGauss function does. )
Anonymous function should be defined such as:
sqr = #(x) x.^2;
in which x shows the variable of the the function and there is no name for it (it is called anonymous!).
Although you can do something like this:
c = 10;
mygrid = #(x,y) ndgrid((-x:x/c:x),(-y:y/c:y));
[x,y] = mygrid(pi,2*pi);
in which you are modifying an existing function ndgrid to make a new anonymous function.
In your case also:
f = #(A)DistanceGauss(A,x_axis,Y_target,Y_initial,numOf,modus);
This is a new anonymous function by modifying the function DistanceGauss that you may want to have a single variable A.
If you remove (A) from the code, then f would be a handle to the existing function DistanceGauss:
f = #DistanceGauss;
Now you can evaluate the function simply by using the handle:
f(A,x_axis,...)
The following code won't work, but this is the idea I'm trying to get at.
c = #(x)constraints;
%this is where I would initialize sum as 0 but not sure how...
for i = 1:length(c)
sum = #(x)(sum(x) + (min(c(x)(i),0))^2);
end
penFunc = #(x)(funcHandle(x) + sig*sum(x));
where constraints and funcHandle are functions of x. This entire code would iterate for a sequence of sig's.
Obviously c(x)(i) isn't functional. I'm trying to write the function where the minimum of c(x) at i (c(x) is a vector) or 0 is taken and then squared.
I know I could calculate c(x) and then analyze it at each i, but I eventually want to pass penFunc as a handle to another function which calculates the minimum of penFunc, so I need to keep it as a function.
I confess I don't understand entirely what you're trying to achieve, but it appears you're trying to create a function handle of an anonymous function with a changing value sum that you precompute. MATLAB anonymous functions do allow you to do this.
It appears there might be some confusion with anonymous functions here. To start with, the line:
c = #(x)constraints;
is probably supposed to be something else, unless you really want c to be a function handle. The # at the start of the line declares a new anonymous function, when I think you just want to call the existing function constraints. It appears you really want c to be an array of constraints coming from the constraints function, in which case I think you mean to say
c = constraints(x);
Then we get to the sum, which I can't tell if you want as a vector or as a single sum. To start with, let's not name it 'sum', since that's the name of a built-in MATLAB function. Let's call it 'sumval'. If it's just a single value, then it's easy (it's easy both ways, but let's do this.) Start before the for loop with sumval=0; to initialize it, then the loop would be:
sumval = 0;
for i = 1:length(c)
sumval = sumval + (min(c(i),0))^2);
end
All four lines could be vectorized if you like to:
c(c>0) = 0; %Replace all positive values with 0
sumval = sum(c.^2); % Use .^ to do a element by element square.
The last line is obviously where you make your actual function handle, and I'm still not quite sure what is desired here. If sig is a function, then perhaps you really meant to have:
penFunc = #(x)(funcHandle(x) + sig*sumval);
But I'm not sure. If you wanted sum to be a vector, then how we specified it here wouldn't work.
Notice that it is indeed fine to have penFunc be an anonymous function with a variable within it (namely sumval), but it will continue to use the value of sumval that existed at the time of the function handle declaration.
So really the issues are A) the creation of c, which I don't think you meant to be a function handle, and B) the initialization of sum, which should probably be sumval (to not interact with MATLAB's own function), and which probably shouldn't declare a new function handle.
How do call a script to a function and vica versa in Matlab/Octave?
function mean_DNA_Microarray = Calc_mean_DNA_Microarray(M)
M = DNA_Microarray
mean_DNA_Microarray = M - ones(5,25)*mean(M(:,25))
end
The response is
error: invalid call to script
C:\Users\Nacho\Documents\Matlab\DNA_Microarray.m error: called from:
error: C:\Users\Nacho\Documents\Matlab\Calc_mean_DNA_Microarray.m at
line 3, column 3
Now this will work if I call DNA_Microarray a function, but the problem requires that it remain as a script.
First of all, you are not defining your function correctly, as the function does not know what M is (unless it is a global vairable, but I doubt so).
In ANY programming language, you need to tell a function which variables it is going to work with. This is not Matlab specific. In Matlab you will do it so:
function mean_DNA_Microarray = Calc_mean_DNA_Microarray(M) % Look! we are telling him what M is!
mean_DNA_Microarray = M - ones(5,25)*mean(M(:,25))
end
Then you want to all the function from somewhere else you would need to just type its name and pass in the arguments, in this case what inside the function is going to be called M
clear;
clc;
% Test code
Mnameoutofthefunction=rand(100,100);
DNAmean = DNA_Microarray(Mnameoutofthefunction); % here we are calling it!
Remember to save the function as functionname.m , in your case DNA_Microarray.m , else Matlab wont know which one it is.
But I HIGHLY recommend you to read a book about Matlab or just about programming in general, as it seems like you could benefit from some basic introduction.
Following #am304 suggestion, here you can find nice tutorials:
http://www.mathworks.co.uk/academia/student_center/tutorials/
EDIT What you want to do is create a function as follows:
function mean_DNA_Microarray = Calc_mean_DNA_Microarray(M) % Look! we are telling him what M is!
mean_DNA_Microarray = M - ones(5,25)*mean(M(:,25))
end
And then, inside your function DNA_Microarray call Calc_mean_DNA_Microarray with the input M
I have the following code:
syms t x;
e=symfun(x-t,[x,t]);
In the problem I want to solve x is a function of t but I only know its value at the given t,so I modeled it here as a variable.I want to differentiate e with respect to time without "losing" x,so that I can then substitute it with x'(t) which is known to me.
In another question of mine here,someone suggested that I write the following:
e=symfun(exp(t)-t,[t]);
and after the differentiation check if I can substitute exp(t) with the value of x'(t).
Is this possible?Is there any other neater way?
I'm really not sure I understand what you're asking (and I didn't understand your other question either), but here's an attempt.
Since, x is a function of time, let's make that explicit by making it what the help and documentation for symfun calls an "abstract" or "arbitrary" symbolic function, i.e., one without a definition. In Matlab R2014b:
syms t x(t);
e = symfun(x-t,t)
which returns
e(t) =
x(t) - t
Taking the derivative of the symfun function e with respect to time:
edot = diff(e,t)
returns
edot(t) =
D(x)(t) - 1
the expression for edot(t) is a function of the derivative of x with respect to time:
xdot = diff(x,t)
which is the abstract symfun:
xdot(t) =
D(x)(t)
Now, I think you want to be able to substitute a specific value for xdot (xdot_given) into e(t) for t at t_given. You should be able to do this just using subs, e.g., something like this:
sums t_given xdot_given;
edot_t_given = subs(edot,{t,xdot},{t_given, xdot_given});
You may not need to substitute t if the only parts of edot that are a function of time are the xdot parts.