I am trying to find use to Newton-Raphson method to find the roots. It does this by making a guess and then improving the guess after each iteration until you get one of the zeros.
Because the Newton-Raphson method quickly finds the zeros, it gives me a small error immediately and after two or three iterations max it should fail to meet the conditions of the while loop. However, the problem is that when I remove the semi-colon after "error" in my loop, I start getting fractions that should break the while loop, but its like Matlab doesn't know that 123/8328423 is less than 1. It continues to run until I manually force the program to stop running.
How do I fix this? I tried format long, format longe, and using double both in the command window, in the scrip file, and somewhere in the loop.
Thank you in advance for any tips, suggestions, or advice that may help!!
A = [1,2,-4;2,-2,-2;-4,-2,1;];
format longe
% syms x y z
% P = x^4 + 3*x^2*y^2-z^3+y+1;
% feval(symengine,'degree',P,x)
syms x
B = mateigenvalue(A);
f(x) = simplify(matdet(B));
x0 = 1;
error = 10;
while(error > .01)
x1 = x0 - f(x0)/(27*(x0)-3*(x0)^2);
error = abs(((f(x0)-f(x1))/f(x0))*100)
x0 = x1;
end
x0 = double(x0)
I reckon the main problem is with error.
It starts as double but inside the while-loop it turns into a symbolic variable and you can't easily compare symbolic variables with scalar values (the .01 in the while-loop condition).
Check in your workspace if error is symbolic (or type class(error) and check if sym is returned). I guess it is symbolic because a fraction is returned (123/8328423), as instead Matlab treats double values with decimals, not fractions.
If so, try doing (inside the while-loop) a conversion for error that is, under the line
error = abs(((f(x0)-f(x1))/f(x0))*100);
try putting
error=double(error);
So error will be temporarily converted in double and you can easily compare its value with .01 to check the while-loop condition.
Also, it is bad practice to call a variable error since error() is a built-in function in Matlab. By naming a variable error you cannot use the error() function. Same story goes for other built-in functions.
Related
I am working on my thesis and running in some programming problems in Matlab. I am trying to implement the ''golden Bisection Method'' to speed up my code. To this end, I've consulted the build in function FZERO.
So I am determining the difference between two vectors which are both (1x20).
Difference = Clmax_dist-cl_vec;
Clmax_dist comes from a semi-empirical method and cl_vec comes from the excecution of an external AVL.exe file.
Essentially, this difference depends only on one single variable AOA because the Clmax_dist vector is a constant. Hence, I am constantly feeding a new AOA value to the AVL.exe to obtain a new cl_vec and compare this again to the constant Clmax_dist.
I am iterating this until one of the element in the vector becomes either zero or negative. My loop stops and reveals the final AOA. This is a time consuming method and I wanted to use FZERO to speed this up.
However, the FZERO documentation reveals that it only works on function which has a scalar as input. Hence, my question is: How can I use FZERO with a function which has a vector as an output. Or do i need to do something totally different?
I've tried the following:
[Difference] = obj.DATCOMSPANLOADING(AOA);
fun=#obj.DATCOMSPANLOADING;
AOA_init = [1 20];
AOA_root = fzero(fun,AOA_init,'iter');
this gave me the following error:
Operands to the || and && operators must be convertible to logical scalar values.
Error in fzero (line 423)
while fb ~= 0 && a ~= b
Error in CleanCLmax/run (line 11)
AOA_root = fzero(fun,AOA_init,'iter');
Error in InitiatorController/moduleRunner (line 11)
ModuleHandle.run;
Error in InitiatorController/runModule (line 95)
obj.moduleRunner(ModuleHandle);
Error in RunSteps (line 7)
C.runModule('CleanCLmax');
The DATCOMSPANDLOADING function contains the following:
function [Difference] = DATCOMSPANLOADING(obj,AOA)
[Input]= obj.CLmaxInput; % Creates Input structure and airfoil list
obj.writeAirfoils(Input); % Creates airfoil coordinate files in AVL directory
[Clmax_dist,YClmax,Cla_mainsections] = obj.Clmax_spanwise(Input); % Creates spanwise section CLmax with ESDU method
[CLa] = obj.WingLiftCurveSlope(Input,Cla_mainsections); % Wing lift curve slope
[Yle_wing,cl_vec] = obj.AVLspanloading(Input,CLa,AOA); % Creates spanloading with AVL
Difference = Clmax_dist-cl_vec;
end
If I need to elaborate further, feel free to ask. And of course, Thank you very much.
fzero indeed only works on scalars. However, you can turn your criterion into a scalar: You are interested in AOA where any of the elements in the vector becomes zero, in which case you rewrite your objective function to return two output arguments: minDifference, which is min(Difference), and Difference. The first output, minDifference is the minimum of the difference, i.e. what fzero should try to optimize (from your question, I'm assuming all values start positive). The second output you'd use to inspect your difference vector in the end.
Essentially I want to have fminsearch run over a variety or parameters.
So I have the following snipet of code running:
%Setting up the changeable WIRX parameters:
L = 0.15; %Length along the electrodes in meters
I = 3000; %Current in amps
%Running the fminsearch:
TeNe = fminsearch(#(params) TeNe(params,L,I),[5,1.5e21],optimset('MaxFunEvals', 100000,'MaxIter', 100000));
What I want to do is be able to run this in a for loop with an array of values for L and I. However what i noticed is that I cannot even run this piece of code twice in a row with out getting the error:
Subscript indices must either be real positive integers or logicals.
Any insight would be much appreciated!
I assume that TeNe is a function which you call with the following inputs: (params,L,I).
However, the output of fminsearch is also assigned to TeNe.
That is why, after the first loop iteration you get the error you see. L has been set to 0.15, however, this makes no sense as an index to an array called TeNe - which you end up with after running fminsearch.
Consider changing the name of the output variable.
I wrote a small program in MATLAB to compute the Shapley value
using the multi-linear extension of a TU game. However, I run
into trouble with the Symbolic Math Toolbox of MATLAB. In
the program I have to integrate a set of functions to get the
Shapley value. However, inside a MATLAB program I cannot use
the int() command
Error using sym/subsindex (line 663) Ivalid indexing or function definition. When defining a function, ensure that the body of the function is a SYM object. When indexing, the input must be numeric, logical or ':'.
Error in ShapleyValueML (line 65)shv(jj)=int(dfy,0,1)
as a consequence I have to use integral() instead. In this case, I
need to transcribe the set of expressions into MATLAB function handle
with matlabFunction(). However, on all Linux machines (MATLAB R2014a) on
which I have access this command does not work (see the discussion below).
As a workaround, the MATLAB program returns the set of functions
into the current workspace, there the Shapley value can be computed
using the int() command.
To make the discussion more concrete, let us consider this small
MATLAB program first.
function [shv,F,dfm]=ShapleyValueML(v)
N=length(v);
[~, n]=log2(N);
S=1:N;
int=0:-1:1-n;
mat=(rem(floor(S(:)*pow2(int)),2)==1);
cmat=(rem(floor(S(:)*pow2(int)),2)==0);
x=sym('x',[1 n]);
mx=1-x;
y = sym('y');
vy=ones(1,n)*y;
F=0;
shv=zeros(1,n);
dfm=cell(1,n);
for ss=1:N
pd1=x(mat(ss,:));
pd2=mx(cmat(ss,:));
pd=prod(pd1)*prod(pd2)*v(ss);
F=pd+F;
end
F=expand(F);
for jj=1:n
dF=diff(F,x(jj));
dfy=subs(dF,x,vy);
%% Does not work!! MATLAB bug???
% mf=matlabFunction(dfy);
% shv(jj)=integral(mf,0,1);
%%
%% The best would be to use:
%%
% shv(jj)=int(dfy,0,1)
%% but it cannot be used inside a program.
dfm{jj}=dfy;
end
end
The commented parts are the parts that do not work inside
the program, but are needed to compute the Shapley value
with that program, which is its purpose. I tested this program
up to 12 players, and I was able to successfully calculate the
Shapley value by a two step procedure. Hence, the above program
specifies correctly the considered problem. To get a better
understanding of this two step procedure and of the functionality
of the above program, let us focus on a three person game.
The values of the coalitions are given by the following data array
>> v = [0,0,90,0,100,120,220];
Notice that coalitions are ordered in accordance with their unique
integer representations. The game is defined, we can now evaluate
the multi-linear extension and the set of partial derivatives with
the above program, but not the Shapley value.
>> [shv,F,dfm]=ShapleyValueML(v);
Integration of the set of partial derivatives runs over the diagonal
of the unit-cube, but then we can set the variables from [x1,x2,x3]
to [y,y,y], and integration runs from 0 to 1.
>> for k=1:3, shv(k)=int(dfm{k},0,1);end;
The solution of the integration is the Shapley value given by:
>> shv
shv =
65 75 80
Checking that this is indeed the Shapley value can be accomplished
with a potential function approach implemented in
>> sh_v=ShapleyValue(v)
sh_v =
65 75 80
that ships with my MATLAB Game Theory Toolbox MatTuGames from
http://www.mathworks.com/matlabcentral/fileexchange/35933-mattugames
Instead of integrating with int() one can also use integral(),
but then the contents like
>> dfm{1}
ans =
- 90*y^2 + 190*y
must be rewritten with matlabFunction() into a function handle. As I
have mentioned above this does not work under Linux
(MATLAB R2013a,R2013b,R2014a). To see this let us try to reproduce
the example
>> syms x y
>> r = sqrt(x^2 + y^2);
from the documentation at the URL:
http://www.mathworks.de/de/help/symbolic/generate-matlab-functions.html?searchHighlight=matlabFunction
This should give
ht =
#(x,y)tanh(sqrt(x.^2+y.^2))
but I get
>> ht = matlabFunction(tanh(r))
Cell contents reference from a non-cell array object.
Error in vectorize (line 15)
c = cells{i};
Error in sym/matlabFunction>mup2mat (line 319)
res = vectorize(res(2:end-1)); % remove quotes
Error in sym/matlabFunction>mup2matcell (line 304)
r = mup2mat(c{1});
Error in sym/matlabFunction (line 123)
body = mup2matcell(funs);
Here comes now my question: Exists there an alternative procedure to
get from
>> dfm{1}
ans =
- 90*y^2 + 190*y
a function handle
>> df=#(y) (- 90.*y.^2 + 190.*y)
df =
#(y)(-90.*y.^2+190.*y)
to integrate it by
>> integral(df,0,1)
ans =
65
Or to put it differently. Is there an alternative method available to
change multiplication * to element-wise multiplication .*, and the
power operation ^ to element-wise power.^?
Of course, any suggestions of improvement for the above MATLAB program
are highly appreciated.
I think I know what the problem is; Towards the beginning of ShapleyValueML function, you have a variable named int which shadows the builtin integration function:
...
int=0:-1:1-n; %# <-- problem!
...
shv(jj)=int(dfy,0,1)
...
That explains the error coming from sym/subsindex, you were using a symbolic object as an index into the numeric array int.
Change the variable name to something else, and the commented code runs fine (the symbolic integration)! Simple as that :)
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 recently started learning MatLab, and wrote the following script today as part of my practice to see how we can generate a vector:
x = [];
n = 4;
for i = i:n
x = [x,i^2];
end
x
When I run this script I get what I expect, namely the following vector:
x = 0 1 4 9 16
However, if I run the script a second time right afterwards I only get the following output:
x = 16
What is the reason for this? How come I only get the last vector entry as output the second time I run the script, and not the vector in its entirety? If anyone can explain this to me, I would greatly appreciate it.
Beginning with a fresh workspace, i will simply be the complex number 1i (as in x^2=-1). I imagine you got this warning on the first run:
Warning: Colon operands must be real scalars.
So the for statement basically loops over for i = real(1i):4. Note that real(1i)=0.
When you rerun the script again with the variables already initialized (assuming you didn't clear the workspace), i will refer to a variable containing the last value of 4, shadowing the builtin function i with the same name, and the for-loop executes:
x=[];
for i=4:4
x = [x, i^2]
end
which iterates only one time, thus you end up with x=16
you forget to initialize i.
after first execution i is 4 and remains 4.
then you initialize x as an empty vector but because i is 4 the loop runs only once.
clear your workspace and inspect it before and after first execution.
Is it possibly a simple typo?
for i = i:n
and should actually mean
for i = 1:n
as i is (probably) uninitialized in the first run, and therefore 0, it works just fine.
The second time, i is still n (=4), and only runs once.
Also, as a performance-tip: in every iteration of your loop you increase the size of your vector, the more efficient (and more matlaboid) way would be to create the vector with the basevalues first, for example with
x = 1:n
and then square each value by
x = x^2
In Matlab, using vector-operations (or matrix-operations on higher dimensions) should be prefered over iterative loop approaches, as it gives matlab the opportunity to do optimised operations. It is also often more readable that way.