I'm trying to define a function h as a bessel function in the form:
h:=x-> (int(cos(x*sin(y)), y = 0 .. pi))/pi;
and then finding a value for x=0 with:
h(0);
But it's returning h(0) straight back at me.
I've also tried rephrasing the equation and defining the function in different ways and calling it with the subs command.
I'm fairly new to maple so is there something blaringly obvious I'm doing wrong here?
Related
I am trying to create a plot in MATLAB by iterating over values of a constant (A) with respect to a system of equations. My code is pasted below:
function F=Func1(X, A)
%X(1) is c
%X(2) is h
%X(3) is lambda
%A is technology (some constant)
%a is alpha
a=1;
F(1)=1/X(1)-X(3)
F(2)=X(3)*(X(1)-A*X(2)^a)
F(3)=-1/(24-X(2))-X(3)*A*a*X(2)^(a-1)
clear, clc
init_guess=[0,0,0]
for countA=1:0.01:10
result(countA,:)=[countA,fsolve(#Func1, init_guess)]
end
display('The Loop Has Ended')
display(result)
%plot(result(:,1),result(:,2))
The first set of code, I am defining the set of equations I am attempting to solve. In the second set of lines, I am trying to write a loop which iterates through the values of A that I want, [1,10] with an increment of 0.01. Right now, I am just trying to get my loop code to work, but I keep on getting this error:
"Failure in initial objective function evaluation. FSOLVE cannot continue."
I am not sure why this is the case. From what I understand, this is the result of my initial guess matrix not being the right size, but I believe it should be of size 3, as I am solving for three variables. Additionally, I'm fairly confident there's nothing wrong with how I'm referencing my Func1 code in my Loop code.
Any advice you all could provide would be sincerely appreciated. I've only been working on MATLAB for a few days, so if this is a rather trivial fix, pardon my ignorance. Thanks.
Couple of issues with your code:
1/X(1) in function Func1 is prone to the division by zero miscalculations. I would change it to 1/(X(1)+eps).
If countA is incrementing by 0.01, it cannot be used as an index for result. Perhaps introduce an index ind to increment.
I have included your constant A within the function to clarify what optimization variables are.
Here is the updated code. I don't know if the results are reasonable or not:
init_guess=[0,0,0];
ind = 1;
for countA=1:0.01:10
result(ind,:)=[countA, fsolve(#(X) Func1(X,countA), init_guess)];
ind = ind+1;
end
display('The Loop Has Ended')
display(result)
%plot(result(:,1),result(:,2))
function F=Func1(X,A)
%X(1) is c
%X(2) is h
%X(3) is lambda
%A is technology (some constant)
%a is alpha
a=1;
F(1)=1/(X(1)+eps)-X(3);
F(2)=X(3)*(X(1)-A*X(2)^a);
F(3)=-1/(24-X(2))-X(3)*A*a*X(2)^(a-1);
end
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'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!
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.
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.