How do I easiest solve an equation=0 with a function as a parameter?
My function with one input variable is called potd(angle), with one output variable, potNRGderiv. I tried:
syms x
solve(potd(x))
This gave me error: Undefined function 'sind' for input arguments of type 'sym'.
Have you got any ideas? Thanks in advance.
solve is the wrong avenue here, unless your function can be rewritten as a simple equation. solve uses muPAD functions which is why you can do solve(sin(x)) but not solve(sind(x)). You can, of course, just do the conversion yourself.
If your function is more complicated or you'd rather not rewrite it, look into fsolve:
x = fsolve(#myfun,x0)
Where x0 is your initial guess - i.e. myfun(x0) is close to 0 - and myfun is a function which takes x and returns a single output. Depending on what your function does, you may have to adjust the options using optimoptions (tolerance, step size, etc) to get a good result.
Related
I have the following function that I wish to solve using fzero:
f = lambda* exp(lambda^2)* erfc(lambda) - frac {C (T_m - T_i)}/{L_f*sqrt(pi)}
Here, C, T_m, T_i, and L_f are all input by the user.
On trying to solve using fzero, MATLAB gives the following error.
Undefined function or variable 'X'.
(where X are the variables stated above)
This error is understandable. But is there a way around it? How do I solve this?
This is answered to the best of my understanding after reading your question as it's not really clear what you are exactly trying and what you want exactly.
Posting the exact lines of code helps a big deal in understanding(as clean as possible, remove clutter). If then the output that matlab gives is added it becomes a whole lot easier to make sure we answer your question properly and it allows us to try it out. Usually it's a good idea to give some example values for data that is to be entered by the user anyway.
First of to make it a function it either needs a handle.
Or if you have it saved it as a matlab file you generally do not want other inputs in your m file then the variable.
So,
function [out]=yourfun(in)
constants=your values; %you can set a input or inputdlg to get a value from the user
out= something something, your lambda thingy probably; %this is the equation/function you're solving for
end
Now since that is not all that convenient I suggest the following
%declare or get your constants here, above the function makes it easier
syms lambda
f = lambda* exp(lambda^2)* erfc(lambda) - frac {C (T_m - T_i)}/{L_f*sqrt(pi)};
hf=matlabFunction(f); %this way matlab automatically converts it to a function handle, alternatively put #(lambda) in front
fzero(hf,x0)
Also this matlab page might help you as well ;)
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 am trying to run code similar to the following, I replaced the function I had with one much smaller, to provide a minimum working example:
clear
syms k m
n=2;
symsum(symsum(k*m,m,0,min(k,n-k)),k,0,n)
I receive the following error message:
"Error using sym/min (line 86)
Input arguments must be convertible to floating-point numbers."
I think this means that the min function cannot be used with symbolic arguments. However, I was hoping that MATLAB would be substituting in actual numbers through its iterations of k=0:n.
Is there a way to get this to work? Any help much appreciated. So far I the most relevant page I found was here, but I am somewhat hesitant as I find it difficult to understand what this function does.
EDIT following #horchler, I messed around putting it in various places to try and make it work, and this one did:
clear
syms k m
n=2;
symsum(symsum(k*m,m,0,feval(symengine, 'min', k,n-k)),k,0,n)
Because I do not really understand this feval function, I was curious to whether there was a better, perhaps more commonly-used solution. Although it is a different function, there are many pieces online advising against the eval function, for example. I thought perhaps this one may also carry issues.
I agree that Matlab should be able to solve this as you expect, even though the documentation is clear that it won't.
Why the issue occurs
The problem is due the inner symbolic summation, and the min function itself, being evaluated first:
symsum(k*m,m,0,min(k,n-k))
In this case, the input arguments to sym/min are not "convertible to floating-point numbers" as k is a symbolic variable. It is only after you wrap the above in another symbolic summation that k becomes clearly defined and could conceivably be reduced to numbers, but the inner expression has already generated an error so it's too late.
I think that it's a poor choice for sym/min to return an error. Rather, it should just return itself. This is what the sym/int function does when it can't evaluate an integral symbolically or numerically. MuPAD (see below) and Mathematica 10 also do something like this as well for their min functions.
About the workaround
This directly calls a MuPAD's min function. Calling MuPAD functions from Matlab is discussed in more detail in this article from The MathWorks.
If you like, you can wrap it in a function or an anonymous function to make calling it cleaner, e.g.:
symmin = #(x,y)feval(symengine,'min',x,y);
Then, you code would simply be:
syms k m
n = 2;
symsum(symsum(k*m,m,0,symmin(k,n-k)),k,0,n)
If you look at the code for sym/min in the Symbolic Math toolbox (type edit sym/min in your Command Window), you'll see that it's based on a different function: symobj::maxmin. I don't know why it doesn't just call MuPAD's min, other than performance reasons perhaps. You might consider filing a service request with The MathWorks to ask about this issue.
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'm working on an assignment for a class of mine and I'm supposed to write a code using a program of my choice (I've chosen Matlab) to solve the Bessel function differential equation using the 4th order Runge-Kutta method. For reference the Bessel function DE is:
x^2*(J_n)''+x*(J_n)'+(x^2-n^2)*J_n=0.
I'm able to separate this into two coupled first order DEs by:
(J_n)'=Z_n and
(Z_n)'+(1/x)*Z_n+[(x^2-n^2)/x^2]*J_n=0.
I have no experience with Matlab nor any other programming language before this assignment. I know Matlab has the 'ode45' command but I'm supposed to write the code myself, not rely on Matlab's commands. So far I've been working on the n=0 case for the Bessel function but I keep getting an error when I try and plot the function. The current error I have says: "Undefined function or method 'J' for input arguments of type 'double'." But I don't know how to fix this error nor if my code is even correct. Could someone tell me where I've gone wrong or what is the correct way to write this code?
h=0.01; %step size
J_0(1)=1; %initial condition for J_0
Z_0(1)=1; %initial condition for Z_0-This value should be zero
%but Matlab gives me an error. To fix this, I input
%Z_0(1)-1 to use the correct value for Z_0(1).
x(1)=0.001; %first value of x
dZ(Z_0,J_0)=(-1/x)*(Z_0-1)-J_0;
for i=[1:1:10]
dZ1=(-1/x)*(Z_0-1)-J_0;
dJ1=(Z_0(1)-1)*h;
dZ2=(-1/x)*(Z_0-1+0.5*h)-(J_0+0.5*h*dJ1);
dJ2=((Z_0(1)-1)+dZ1)*h;
dZ3=(-1/x)*(Z_0-1+0.5*h)-(J_0+0.5*h*dJ2);
dJ3=((Z_0(1)-1)+dZ1+dZ2)*h;
dZ4=(-1/x)*(Z_0-1+h)-(J_0+h*dJ3);
dJ4=((Z_0(1)-1)+dZ1+dZ2+dZ3)*h;
J(i+1)=J(i)+(h/6)*(dJ1+2*dJ2+2*dJ3+dJ4);
end
plot(J_0);
Thanks in advance for any help
Your problem is on the line:
J(i+1)=J(i)+(h/6)*(dJ1+2*dJ2+2*dJ3+dJ4);
In the right-hand side of your assignment operator you use the variable J that is never set before i is taking the value 1. Looks like a typo to me (should it be J_0 instead?)
Also, don't forget your index i when computing your dJ and dZ stuff in the for loop.