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.
Related
I'm newbie to Matlab
I have an assignment :
Legendre polynomial Pn(x), n=0,1,2,. . . The recursive formula of is
Write recursive sub-functions and non-recursive sub-functions separately to find the value of the Legendre polynomial function
This is my code :
function P =Legendre(n,x)
syms x;
n = input('n=');
if n==0
P=1;
elseif n==1
P=x;
elseif n>=2
P=((2*n-1)/n)*x*Legendre(n-1)-((n-1)/n)*Legendre(n-2);
end
end
But I get an error message:
Unrecognized function or variable 'Legendre'.
Error in ti4 (line 9)
P=((2*n-1)/n)*x*Legendre(n-1)-((n-1)/n)*Legendre(n-2);
Sorry for the stupid question. Can anyone help me? Thank u so much
A few things are probably going on here.
File name needs to match function name (for the primary function)
In your case, the filename needs to be Legendre.m.
Symbolic toolbox OR do you want an answer
for most uses of this function, I would leave two full inputs, just as you have them. Bur I would remove the first two lines completely.
As it is, the first two lines will break your inputs. The value for n is reset by the input function. I'm actually not sure what happens when you declare an existing variable x, to a sym.
Input consistency
You are setting up a function with two inputs, named n and x. But when you maek your recursive calls you only pass in one variable. The easiest thing to do here is simply keep passing n in as the first input.
(Right now, you are trying to pass in x in the recursive calls, but it will be interpreted as n.)
I have a problem that needs to use ode45 in order to find the solution, but one of the state variables is defined using the derivative of another function.
The code I have for this is:
`[~,s]=ode45(#eom,time,IC,options);
function sdot=eom(~,s)
mu=0.01215510;
C=3;
x=s(1);
y=s(2);
gamma=s(3);
O=0/2*(x^2+y^2)+(1-mu)/sqrt((x+mu)^2+y^2)+mu/sqrt((x+mu-1)^2+y^2);
m=sqrt(2*O-C);
sdot=[m*cos(gamma);m*sin(gamma);(Oy*cos(gamma)-Ox*sin(gamma))/sqrt(2*O-C)-2];
sdot=sdot';
end`
Where Ox=dO/dx, and Oy=dO/dy.
However, I have tried putting it into a separate function file and several other methods, but I cannot figure out how to take the derivative of O with respect to x and y, and then transfer those x and y values back into the state defined variables.
In the separate function, I defined Od=O(mu,d), where d would determine what variable the derivative would be taken with respect to. Within this function, I defined it as Od(a,b)=diff(O,d,1). I had thought that I could input this function into my state function where I could define it as Od(s(1),s(2)), but this only returning error messages.
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,...)
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 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.