I am trying to run fsolve to resolve a dissociation equation but I keep getting this error. I tracked it down to the x(1)^(1/2) term (removing the square root yields no error) but I can't find a way to solve the proper equation I need. The code is below.
T = 2000
Kp = exp(-deltaG/(Ru*T))
function [f]=func(x)
f(1) = 2-x(1)*4 / (3*x(1) - 1)*(x(1))^(1/2) - Kp
endfunction
x0 = [1]
[x,f_x] = fsolve(x0,func)
EDIT: More requested info
The error is
!--error 98 variable returned by scilab argument function is
incorrect
Ru is the gas constant, 8.315.
DeltaG is -135643.
Kp is 3.489e-3.
This is a book example, x should yield 0.3334.
What kind of solved this problem was that I updated scilab to version 6.0.1 from 5.5. The problem is that depending on the initial guess x0 the values of x get really absurd and x0 has to be so close to the real answer that it defeats the purpose of the calculation.
Also I don't have access to Maple, my other alternative would be MATLAB
Using symbolic calculus software like xcas or Maple one can solve your equation symbolycally
There are 3 solutions:
s1=((1/4)*t1+(1/4)*(Kp-2)^2/t1-(1/4)*Kp+1/2)^2
s2=(-(1/8)*t1-(1/8)*(Kp-2)^2/t1-(1/4)*Kp+1/2+(1/2*%i)*sqrt(3)*((1/4)*t1-(1/4)*(Kp-2)^2/t1))^2
s3=(-(1/8)*t1-(1/8)*(Kp-2)^2/t1-(1/4)*Kp+1/2-(1/2*%i)*sqrt(3)*((1/4)*t1-(1/4)*(Kp-2)^2/t1))^2
where
t=sqrt(-Kp*(Kp-4)*(Kp-2)^2)
t1=(-(Kp-2)*(Kp-2*sqrt(2))*(Kp+2*sqrt(2))+4*t)^(1/3);
depending on Kp values some solutions can be complex.
Related
I'm trying to solve the following ODE:
where R(T) is defined as:
This is my not so great attempt at using Octave:
1;
function xdot = f(t, T)
xdot = 987 * ( 0.0000696 * ( 1 + 0.0038 * ( T(t) - 25 ))) - ( 0.0168 * (T(t)-25 )) - (( 3.25 * 10 ^ (-13))) * ((T(t))^4 - (25^4));
endfunction
[x, istate, msg] = lsode( "f", 100, (t=linspace(0,3600,1000)'));
T_ref and T_infinity_sign are the same constant.
Why isn't my code correct?
If you type
debug_on_error(1)
on your octave session, and then run your code, you will see that the "f" part is called as expected, but then it fails inside lsode with the following error:
error: T(100): out of bound 1 (dimensions are 1x1)
error: called from
f at line 4 column 8
If you look at the documentation of lsode, it says it expects a function f whose first argument is a state vector x, and the second is a scalar, corresponding to time t at which that state vector occurs; f is expected to output the differential dx/dt at time t for state vector x.
From your code it seems that you are reversing the order of the arguments (and their meanings).
Therefore, when you passed T as a second argument to your function, lsode treats it like a scalar, so when you then try to evaluate T(t), it fails with an 'out_of_bounds_ error.
My advice would be, read the documentation of lsode and have a look at its examples carefully, and start playing with it to understand how it works.
PS. The debug_on_error directive launches the debugger if an error occurs during code execution. To exit the debugger type dbquit (or if you're using the GUI, click on the 'Quit Debugging Mode' button at the top right of the octave editor). If you don't know how to use the octave debugger, I recommend you spend some time to learn it, it is a very useful tool.
error: 'y' undefined near line 8 column 12
error: called from computeCost at line 8 column 3
Here is my code:
1;
function J = computeCost(X, y, theta)
%COMPUTECOST Compute cost for linear regression
% J = COMPUTECOST(X, y, theta) computes the cost of using theta as the
% parameter for linear regression to fit the data points in X and y
% Initialize some useful values
m = length(y); % number of training examples
% You need to return the following variables correctly
J = 0;
% ====================== YOUR CODE HERE ======================
% Instructions: Compute the cost of a particular choice of theta
% You should set J to the cost.
J = sum(( X * theta - y ) .^2 )/( 2 * m );
% =========================================================================
end
I am guessing it's an error from Coursera ML course assignment. I think you are trying to run the file which contains the implementation of function computeCost(X, y, theta), not the file which calls the computeCost(,,) function with values of X, y, theta. This is why you are getting the error as you aren't providing y.
Run the file which is calling computeCost() function, not the file which contains the implementation of computeCost() function.
That is:
For Week2 Assignment 1: Run ex1.m file
For Week3 Assignment 2: Run ex2.m file
There are two things happening here. First you are defining your function dynamically as opposed to in its own file; not sure why you would prefer that.
Second, after having defined this computeCost function, you are calling it from a context where you did not pass a y argument (or presumably, you didn't pass any arguments to it, and y happens to be the first one detected as missing inside the function).
Since this is a cost function and your code looks suspiciously like code from Andrew Ng's Machine Learning course on Coursera, I am going to go out on a limb here and guess that you called computeCost from something else that was supposed to use it as a cost function to be optimised, e.g. fminunc. Typically functions like fminunc expect a function handle as an argument, but they expect a very specific function handle too. If you look at the help of fminunc, it states that:
FCN should accept a vector (array) defining the unknown variables,
and return the objective function value, optionally with gradient.
Therefore, if you want to pass a function that is to be computed with three arguments, you need to "wrap" it into your own handle, which you can define on the spot, e.g. #(x) computeCost(x, y, t) (assuming 'y' and 't' exist already).
So, I'm guessing that instead of calling fminunc like so: fminunc( #(x) computeCost(x, y, t),
you probably called it like so: fminunc( #computeCost )
or even like so: fminunc( computeCost ) (which evaluates the function first, rather than pass a function handle as an argument).
Basically, go back to the code given to you by coursera, or read the notes carefully. You're calling things the wrong way.
Actually, you are trying to run a function and you can't run it until you provide the desired no. of parameters. Doing so, you may encounter the following error:
computeCost error: 'y' undefined near line 7 column 12
error: called from computeCost at line 7 column 3
As you see, here I'm calling this function without passing any argument.
SOLUTION:
You can test your code by running 'ex1' script. After that submit your work by calling 'submit' script.
Using Mathematica I need to evaluate the integral of a function. Since it is taking the program too much to compute it, would it be possible to use parallel computation to shorten the time needed? If so, how can I do it?
I uploaded a picture of the integrand function:
I need to integrate it with respect to (x3, y3, x, y) all of them ranging in a certain interval (x3 and y3 from 0 to 1) (x and y from 0 to 100). The parameters (a,b,c...,o) are preventing the NIntegrate function to work. Any suggestions?
If you evaluate this
expr=E^((-(x-y)^4-(x3-y3)^4)/10^4)*
(f x+e x^2+(m+n x)x3-f y-e y^2-(m+n y)y3)*
((378(x-y)^2(f x+e x^2+(m+n x)x3-f y-e y^2-(m+n y)y3))/
(Pi(1/40+Sqrt[((x-y)^2+(x3-y3)^2)^3]))+
(378(x-y)(x3-y3)(h x+g x^2+(o+p x)x3-h y-g y^2-(o+p y)y3))/
(Pi(1/40+Sqrt[((x-y)^2+(x3-y3)^2)^3])))+
(h x+g x^2+(o+p x)x3-h y-g y^2-(o +p y) y3)*
((378(x-y)(x3-y3)(f x+e x^2+(m+n x)x3-f y-e y^2-(m+n y)y3))/
(Pi(1/40+Sqrt[((x-y)^2+(x3-y3)^2)^3]))+
(378 (x3 - y3)^2 (h x + g x^2 + (o + p x)x3-h y-g y^2-(o+p y)y3))/
(Pi(1/40+Sqrt[((x-y)^2+(x3-y3)^2)^3])));
list=List ## Expand[expr]
then you will get a list of 484 expressions, each very similar in form to this
(378*f*h*x^3*x3)/(Pi*(1/40+Sqrt[(x^2+x3^2-2*x*y+y^2-2*x3*y3+y3^2)^3]))
Notice that you can then use NIntegrate in this way
f*h*NIntegrate[(378*x^3*x3)/(Pi*(1/40+Sqrt[(x^2+x3^2-2*x*y+y^2-2*x3*y3+y3^2)^3])),
{x,0,100},{y,0,100},{x3,0,1},{y3,0,1}]
but it gives warnings and errors about the convergence and accuracy, almost certainly due to your fractional powers in the denominator.
If you can find a way to pull out the scalar multipliers which are independent of x,y,x3,y3 and then perform that integration without warnings and errors and get an accurate result which isn't infinity then you could perhaps perform these integrals in parallel and total the results.
Some of the integrands are scalar multiples of others and if you combine similar integrands then you can reduce this down to 300 unique integrands.
I doubt this is going to lead to an acceptable solution for you.
Please check all this very carefully to make certain that no mistakes have been made.
EDIT
Since the variables that are independent of the integration appear to be easily separated from the dependent variables in the problem posed above, I think this will allow parallel NIntegrate
independentvars[z_] := (z/(z//.{e->1, f->1, g->1, h->1, m->1, n->1, o->1, p->1}))*
NIntegrate[(z//.{e->1, f->1, g->1, h->1, m->1, n->1, o->1, p->1}),
{x, 0, 100}, {y, 0, 100}, {x3, 0, 1}, {y3, 0, 1}]
Total[ParallelMap[independentvars, list]]
As I mentioned previously, the fractional powers in the denominator result in a flood of warnings and errors about convergence failing.
You can test this with the following much simpler example
expr = f x + f g x3 + o^2 x x3;
list = List ## Expand[expr];
Total[ParallelMap[independentvars, list]]
which instantly returns
500000. f + 5000. f g + 250000. o^2
This is a very primitive method of pulling independent symbolic variables outside an NIntegrate. This gives absolutely no warning if one of the integrands is not in a form where this primitive attempt at extraction is not appropriate or fails.
There may be a far better method that someone else has written out there somewhere. If someone could show a far better method of doing this then I would appreciate it.
It might be nice if Wolfram would consider incorporating something like this into NIntegrate itself.
I wrote this code in octave:
syms z;
f=z-2;
fsolve("f",0.)
Then this gives the error
#f: no function and no method found.
Also using fsolve(#f,0) gives the same error
When I write code as:
syms z;
f=z-2;
fsolve(f,0.)
Then this gives the error
ind2sub: subscript indices must be either positive integers less than 2^31 or logicals.
Please explain to me how to actually use fsolve.
% syms z; % Not needed, actually slows down the code
f=#(z)(z-2);
fsolve(f,0.)
You're missing the # symbol, which is a function handle. This tells Octave that f is not a variable, but actually is a(n anonymous) function, in this case of z, which is the first argument.
You'll probably want to have z to be a regular variable, because making it symbolic turns MATLAB from a speeding race car to a drudging farm vehicle. Unless there's a specific reason to have z symbolic (I cant think of any in case of usage with fsolve)' it's best to avoid symbolic maths.
I'm all new to Matlab and I'm supposed to use this function to find all 3 zero spots.
f.m (my file where the function can be found)
function fval = f(x)
% FVAL = F(X), compute the value of a test function in x
fval = exp(-x) - exp(-2*x) + 0.05*x - 0.25;
So obviously I write "type f" to read my function but then I try to do like fzero ('f', 0) and I get the ans 0.4347 and I assume that's 1 of my 3 zero spots but how to find the other 2?
From fzero documentation
x = fzero(fun,x0) tries to find a zero of fun near x0, if x0 is a scalar. fun is a function handle. The value x returned by fzero is near a point where fun changes sign, or NaN if the search fails. In this case, the search terminates when the search interval is expanded until an Inf, NaN, or complex value is found.
So it can't find all zeros by itself, only one! Which one depends on your inputted x0.
Here's an example of how to find some more zeros, if you know the interval. However it just repeatedly calls fzero for different points in the interval (and then still can miss a zero if your discretization is to coarse), a more clever technique will obviously be faster:
http://www.mathworks.nl/support/solutions/en/data/1-19BT9/index.html?product=ML&solution=1-19BT9
As you can see in the documentation and the example above, the proper way for calling fzero is with a function handle (#fun), so in your case:
zero1 = fzero(#f, 0);
From this info you can also see that the actual roots are at 0.434738, 1.47755 and 4.84368. So if you call fzero with 0.4, 1.5 and 4.8 you probably get those values out of it (convergence of fzero depends on which algorithm it uses and what function you feed it).
Just to complement Gunther Struyf's answer: there's a nice function on the file exchange by Stephen Morris called FindRealRoots. This function finds an approximation to all roots of any function on any interval.
It works by approximating the function with a Chebyshev polynomial, and then compute the roots of that polynomial. This obviously only works well with continuous, smooth and otherwise well-behaved functions, but the function you give seems to have those qualities.
You would use this something like so:
%# find approximate roots
R = FindRealRoots(#f, -1, 10, 100);
%# refine all roots thus found
for ii = 1:numel(R)
R(ii) = fzero(#f, R(ii)); end