I am trying to evaluate the function in t = 1 in Matlab.
How can I get an answer to the following code ?
result = dsolve('D2y-23*Dy+22*y=sin(t)','y(0)=0','Dy(0)=0');
disp(result);
disp(fnval(result, 1));
The first answer is: exp(22*t)/10185 + (23*cos(t))/970 - exp(t)/42 + (21*sin(t))/970.
But when I am trying to find the evaluation of the function in the point t =1, the progrem is throwing some exception. Maybe the 'fnval' function is not suitable for here.
You can use subs.
result = dsolve('D2y-23*Dy+22*y=sin(t)','y(0)=0','Dy(0)=0');
t = 1;
subs(result)
ans =
3.5198e+005
You could also do it using eval, which is similar to your initial approach:
eval(result)
after assigning a value to t.
You can evaluate the function for a number range, using eval together with vectorize:
t = -0.1:0.01:0.1;
plot(t,eval(vectorize(result)))
This gives:
Related
I would like to have short-hand form about many parameters which I just need to keep fixed in Matlab 2016a because I need them in many places, causing many errors in managing them separately.
Code where the signal is 15x60x3 in dimensions
signal( 1:1 + windowWidth/4, 1:1 + windowWidth,: );
Its pseudocode
videoParams = 1:1 + windowWidth/4, 1:1 + windowWidth,: ;
signal( videoParams );
where you cannot write videoParams as string but should I think write ":" as string and everything else as integers.
There should be some way to do the pseudocode.
Output of 1:size(signal,3) is 3 so it gives 1:3. I do not get it how this would replace : in the pseudocode.
Extension for horcler's code as function
function videoParams = fix(k, windowWidth)
videoParams = {k:k + windowWidth/4, k:k + windowWidth};
end
Test call signal( fix(1,windowWidth){:}, : ) but still unsuccessful giving the error
()-indexing must appear last in an index expression.
so I am not sure if such a function is possible.
How can you make such a int-string-int input for the matrix?
This can be accomplished via comma-separated lists:
signal = rand(15,60,3); % Create random data
windowWidth = 2;
videoParams = {1:1+windowWidth/4, 1:1+windowWidth, 1:size(signal,3)};
Then use the comma-separated list as such:
signal(videoParams{:})
which is equivalent to
signal(1:1+windowWidth/4, 1:1+windowWidth, 1:size(signal,3))
or
signal(1:1+windowWidth/4, 1:1+windowWidth, :)
The colon operator by itself is shorthand for the entirety of a dimension. However, it is only applicable in a direct context. The following is meaningless (and invalid code) as the enclosing cell has no defined size for its third element:
videoParams = {1:1+windowWidth/4, 1:1+windowWidth, :};
To work around this, you could of course use:
videoParams = {1:1+windowWidth/4, 1:1+windowWidth};
signal(videoParams{:},:)
Could you please help me with the following issue: I have the following function handle:
r1 = #(lambda) b + lambda*(r - b); % r and b are vectors of return data
I want to find the optimal lambdas that set me a mean function to zero, for a given set of powers within that function. What I tried to do and didn't work, as it returns me an error for undefined operators for input arguments of type 'function_handle' is:
lambda0 = 0.3;
for a = 2:10 %power coefficient
S1(a) = fzero(mean((r - b)*r1.^(1/(a - 1))),lambda0);
end
Any suggestion as to how to go about this problem is highly appreciated! Thank you in advance.
fzero accepts a function handle as the first input. As you currently have it, you're trying to pass a statement as the first input. This statement can't even be properly evaluated because you are trying to perform numerical operations on a function handle (more on this in a bit).
You need to instead do something like this where we create a new function handle that evaluates the original function handle and performs the other operations you need.
S1(a) = fzero(#(lambda)mean((r - b)*r1(lambda).^(1/(a - 1))),lambda0);
Further Explanation
Performing operations on a function handle is not the same as performing them on the result.
So for example, if we had a function handle:
func = #(x)2*x;
If we evaluation this, by calling it with an input value for x
func(2)
4
This works as we would expect. If now we really want the value (2*x)^2, we could try to write it the way that you wrote your statement in your question
func2 = func^2;
We will get an error!
Undefined operator '^' for input arguments of type 'function_handle'.
This does not work because MATLAB attempts to apply the ^ operation to the function handle itself and not the value of the evaluated function handle.
Instead, we would need to create a new function handle that essentially wraps the other one and performs any additional options:
func2 = #(x)func(x)^2;
func2(2)
16
Bringing it Full-Circle
So if we go back to your question, you defined your anonymous function r1 like this.
r1 = #(lambda) b + lambda*(r - b); % r and b are vectors of return data
This all looks great. You have one input argument and you reference r and b from the parent workspace.
Now when you call fzero you try to perform operations on this function handle in hopes of creating a new function handle.
mean((r - b)*r1.^(1/(a - 1)))
Like we just showed this will result in a very similar error
Undefined operator .^ for input arguments of type 'function_handle'
So we need to wrap this into a new function.
newfunc = #(lambda)mean((r - b)*r1(lambda).^(1 / (a - 1)));
Now we can safely pass this to fzero.
result = fzero(newfunc, lambda0);
I'm having some trouble with local functions within my code so I've pasted a simple example below:
function [avg,testvar] = test(x) %Warning
n = length(x);
avg = mymean(x,n);
end
function [a,testvar] = mymean(v,n)
a = sum(v)/n;
testvar=123;
end
One can probably see what I'm attempting; to pass testvar out of the local functions. However Matlab returns the warning:
"The function return value 'testvar' might be unset"
with respect to the line I've commented "%Warning".
What's the best way of getting around this?
You need to specify the value of the second output of test(). Otherwise how can MATLAB know what its value is supposed to be? It doesn't know the second output of mymean() should be routed to the second output of test(). Perhaps this will solve your problem.
function [avg,testvar] = test(x) %Warning
n = length(x);
[avg, testvar] = mymean(x,n);
end
function [a,testvar] = mymean(v,n)
a = sum(v)/n;
testvar=123;
end
The variables between brackets after function are the output variables.
In your first function, you did not assign any value to testvar hence the warning. If you add testvar = 123; in the first function, the warning goes away. Or you can remove testvar from the output variables, leaving:
function avg = test(x)
I want to call a function in Matlab using another one, which has the same number of inputs and outputs. In fact, those inputs and outputs have the same name.
Example:
function [a,b] = gettwo(matrix,string,varargin)
[a,b] = getone(matrix,string,varargin{:});
end
This code produces the following error:
Error in getone(line 3)
aux = 'matrix(varargin{:})';
Output argument "b" (and maybe others) not assigned during
call to "C:\Users\baister\Documents\MATLAB\soft\getone.m>getone".
Error in results (line 4)
[a,b] = getone(matrix,string,varargin{:});
How should I wrap getone?
(The definitive function will have more lines than those shown in this post.)
Thanks.
The general wrapping for variable number of outputs should work like this:
function [varargout] = gettwo(matrix,string,varargin)
[varargout{1:nargout}] = getone(matrix,string,varargin{:});
end
You'll get the same error as above though, in case you do
[a,b] = gettwo(...);
and getone returns only 1 argument.
temp(i,1) = rand(1)*(pb(1,num).pos(i,1) - pw(1,num).pos(i,1));
This line gives the following error:
Error using ==> minus
Not enough input arguments.
The following are the definitions of pb and pw.
pw=struct('fitness',[],'pos',{});
pb=struct('fitness',[],'pos',{});
pos is a 2 x 1 array.
When tracking down errors like this, I break the problem up into smaller bits. Especially when the logic isn't readily apparent. Not only does it provide a path that can be used to step through your function using the debugger, but it also makes it more readable.
I've taken liberty with the intermediate variable names.
thisPb = pb(1,num);
thisPw = pw(1,num);
initialPos= pw.pos(i,1);
finalPos = pb.pos(i,1);
whos initialPos finalPos
temp(i,1) = rand(1) * (finalPos - initialPos);
The line with whos will print out the values. Make sure that finalPos and initialPos are both numbers.
One way that you can get this error is when num is an empty matrix.
The expression
>> s(x).a
can return a variable number of outputs, depending on the size of x.
If x = [1,2,3] for example, it will return three values (as long as s has at least three elements).
If x = [] on the other hand, then s(x).a will return no outputs, so the expression
>> disp(s(x).a)
will give you a Not enough input arguments error, which is almost certainly what you're seeing. You should check that num is not empty.
Are you sure, that all values are really initialised? Try to check this before your codeline.
disp(pb(1,num).pos(i,1))
disp(pw(1,num).pos(i,1))
temp(i,1) = rand(1)*(pb(1,num).pos(i,1) - pw(1,num).pos(i,1));