I try to make friends physics package and matrixes. But I am faced with difficulties. To be more specific, to consider a code:
with(Physics)
Setup(mathematicalnotation = true)
Coordinates(X)
Setup(tensors = A[mu](X))
PDEtools:-declare(A(X), p0(X), pm(X), pp(X), pt(X), U(X), m5(X))
And then I create a matrix:
pt := proc (X) options operator, arrow; Matrix(2, 2, [[p0(X), sqrt(2)*pp(X)], [sqrt(2)*pm(X), -p0(X)]]) end proc;
NULL;
It displays as
Why do p0, pp, pm become function only one variable x1?
Interestingly that maple understands
the next matrix
Why is only one variable?
In cases like this one, it is frequently useful to consult the documentation. In the help page for Coordinates, you read "The Coordinates command also sets an alias for a capital letter X to represent a list of coordinates ..." So: you type X and that automatically expands to x1, x2, x3, x4, while this sequence of variables is displayed as a single X. BUT, when you define a procedure, as in pt := proc(X) .... end proc, the X you see there is just a dummy representing (only one) procedure's parameter.
That answers your question: you see p0, pp, pm as functions of only one variable because pt is defined as a function of only one variable: when you execute pt(X) only the first element in the sequence, that is x1, is used, the other three are discarded. From context I understand that what you want to define is pt := proc(x1, x2, x3, x4) .... end proc. Then use it as pt(X) and you will see p0, pp, pm as functions of the four variables, which in turn are displayed as the single capital X.
Edgardo S. Cheb-Terrab
Physics, Differential Equations and Mathematical Functions, Maplesoft.
Related
I've been trying to plot a graph of function f(x) on interval 0,6. For some reason maple doesn't plot the graph when I use 'f(x)' as an argument. It works fine when I substitute assigned function as an argument. What can be the reason? How can I plot it using 'f(x)' as an argument?
My code is as follows and the error is on the pictures:
mystep:=proc(x,a,b,c)
if x<a
then
b;
else
c;
end if:
end proc:
f(x):=mystep(x,3,-2,7);
plot('f(x)', x=0..6);
enter image description here
Your syntax for creation of the operator (that you hope to assign to f) is incorrect for 1D plaintext input.
Here is the usual syntax for creation of such an operator, and assignment to name f.
restart;
mystep:=proc(x,a,b,c)
if x<a then b; else c; end if:
end proc:
f:=x->mystep(x,3,-2,7);
These should now all work as you expect.
plot('f(x)', x=0..6);
plot(f, 0..6);
plot(x->mystep(x,3,-2,7), 0..6);
In recent versions of Maple the syntax that you used can work in (only) 2D Input mode. But it is a poor syntax to use, since is easily confused with the 1D syntax for assigning into the remember table of an operator (and with even more confusion ensuing). You should avoid ambiguous syntax.
The type of function you have is piecewise-defined function (see this wikipedia page https://en.wikipedia.org/wiki/Piecewise). And in Maple there is already a command for defining this type of a function, piecewise, see its help page for a complete explanation of how to use it. Now for your mysetup, you have a condition x < a, and a value for when this happens, b, and a value for otherwise, c. So you want piecewise( x < a, b, c ). I think it is better to just use this command, but if it is really necessary to define a new procedure, then your mysetup becomes like the following.
mystep := proc(x, a, b, c)
return( piecewise( x < a, b, c ) ):
end proc:
Now for your plot you can use either
plot( piecewise( x < 3, -2, 7 ), x = 0..6 );
or
f(x) := mystep(x, 3, -2, 7);
plot( f(x), x = 0..6);
I have this simple procedure in Maple that I want to plot.
test:=proc(n)
local i,t;
for i from 1 to n do
t:=1;
od;
return t;
end:
The procedure itself works fine.
> test(19)
1
When I want to plot I get the following Error:
> plot(test(x),x=1..10)
Error, (in test) final value in for loop must be numeric or character
Please help
Maple's usual evaluation model is that arguments passed to commands are evaluated up front, prior to the computation done within the body of the command's own procedure.
So if you pass test(x) to the plot command then Maple will evaluate that argument test(x) up front, with x being simply a symbolic name.
It's only later in the construction of the plot that the plot command would substitute actual numeric values for that x name.
So, the argument test(x) is evaluated up front. But let's see what happens when we try such an up front evaluation of test(x).
test:=proc(n)
local i,t;
for i from 1 to n do
t:=1;
od;
return t;
end:
test(x);
Error, (in test) final value in for loop
must be numeric or character
We can see that your test procedure is not set up to receive a non-numeric, symbolic name such as x for its own argument.
In other words, the problem lies in what you are passing to the plot command.
This kind of problem is sometimes called "premature evaluation". It's a common Maple usage mistake. There are a few ways to avoid the problem.
One way is to utilize the so-called "operator form" calling sequence of the plot command.
plot(test, 1..10);
Another way is to delay the evaluation of test(x). The following use of so-called unevalution quotes (aka single right ticks, ie. apostrophes) delays the evaluation of test(x). That prevents test(x) from being evaluated until the internal plotting routines substitute the symbolic name x with actual numeric values.
plot('test(x)', x=1..10);
Another technique is to rewrite test so that any call to it will return unevaluated unless its argument is numeric.
test:=proc(n)
local i,t;
if not type(n,numeric) then
return 'procname'(args);
end if;
for i from 1 to n do
t:=1;
od;
return t;
end:
# no longer produces an error
test(x);
test(x)
# the passed argument is numeric
test(19);
1
plot(test(x), x=1..10);
I won't bother showing the actual plots here, as your example produces just the plot of the constant 1 (one).
#acer already talked about the technical problem, but your case may actually have a mathematical problem. Your function has Natural numbers as its domain, i.e. the set of positive integers {1, 2, 3, 4, 5, ...} Not the set of real numbers! How do you interpret doing a for-loop for until a real number for example Pi or sqrt(2) to 5/2? Why am I talking about real numbers? Because in your plot line you used plot( text(x), x = 1..10 ). The x=1..10 in plot is standing for x over the real interval (1, 10), not the integer set {1, 2, ..., 10}! There are two ways to make it meaningful.
Did you mean a function with integer domain? Then your plot should be a set of points. In that case you want a points-plot, you can use plots:-pointplot or adding the option style=point in plot. See their help pages for more details. Here is the simplest edit to your plot-line (keeping the part defininf test the same as in your post).
plot( [ seq( [n, test(n)], n = 1..10 ) ], style = point );
And the plot result is the following.
In your function you want the for loop to be done until an integer in relation with a real number, such as its floor? This is what the for-loop in Maple be default does. See the following example.
t := 0:
for i from 1 by 1 to 5/2 do
t := t + 1:
end do;
As you can see Maple do two steps, one for i=1 and one for i=2, it is treated literally as for i from 1 by 1 to floor(5/2) do, and this default behavior is not for every real number, if you replace 5/2 by sqrt(2) or Pi, Maple instead raise an error message for you. But anyway the fix that #acer provided for your code, plots the function "x -> test(floor(x))" in your case when x comes from rational numbers (and float numbers ^_^). If you change your code instead of returning constant number, you will see it in your plot as well. For example let's try the following.
test := proc(n)
local i,t:
if not type(n, numeric) then
return 'procname'(args):
end if:
t := 0:
for i from 1 to n do
t := t + 1:
end do:
return(t):
end:
plot(test(x), x=1..10);
Here is the plot.
It is indeed "x -> test(floor(x))".
I've calculated coefficients an, bn (100, T=2*pi) in c++ and checked that they are correct using few sources. Now i try to generate Fourier series graph for given example function in Scilab:
(x+2)*abs(cos(2*x*(x-pi/6)))
M=csvRead(filename, ";", [], 'double')
n=size(M,1)
for i = 1:n
A(i)=M(i)
B(i)=M(i + n)
end
function series=solution(x)
series=A(1)/2;
for i = 2:n
series=series+(A(i)*cos(i*x)+B(i)*sin(i*x));
end
endfunction
function series=solution2(x)
series=(x+2).*abs(cos(2.*x.*(x-%pi/6)));
endfunction
x = -%pi:%pi/100:%pi
plot2d(x, solution(x), 3)
x2 = -%pi:%pi/100:%pi
plot2d(x2, solution2(x2), 4)
Here is the result:
It clearly looks that tendency is ok but the beginning and the end of the period are wrong (reversed?). Do you see any issues in Scilab code? What could cause the problem - values in sin/cos in function solution(x)? Should i provide an, bn values and check for miscalculation there?
I don't know how did you calculated your A & B coefficients, but I assume that you used the usual notations to get the first line of the below formula:
Thus n starts from 1. Since Scilab starts vector indexing from 1, you correctly made your loop from 2, but forgot to compensate for this "offset".
Your function should be something like this:
function series=solution(x)
series=A(1)/2;
for i = 2:n
series=series+(A(i)*cos((i-1)*x)+B(i)*sin((i-1)*x));
end
endfunction
Since you didn't provided A & B, I can not check the result.
Additional note: Syntactically more correct if you explicitly define all input variables in a function, like this:
function series=solution(x,A,B)
This way you may be sure that your input is not changed somewhere else in the code.
How solve a system of ordinary differential equation ..an initial value problem ....with parameters dependent on time or independent variable?
say the equation I have
Dy(1)/dt=a(t)*y(1)+b(t)*y(2);
Dy(2)/dt=-a(t)*y(3)+b(t)*y(1);
Dy(3)/dt=a(t)*y(2);
where a(t) is a vector and b(t) =c*a(t); where the value of a and b are changing with time not in monotone way and each time step.
I tried to solve this using this post....but when I applied the same principle ...I got the error message
"Error using griddedInterpolant The point coordinates are not
sequenced in strict monotonic order."
Can someone please help me out?
Please read until the end to see whether the first part or second part of the answer is relevant to you:
Part 1:
First create an .m file with a function that describe your calculation and functions that will give a and b. For example: create a file called fun_name.m that will contain the following code:
function Dy = fun_name(t,y)
Dy=[ a(t)*y(1)+b(t)*y(2); ...
-a(t)*y(3)+b(t)*y(1); ...
a(t)*y(2)] ;
end
function fa=a(t);
fa=cos(t); % or place whatever you want to place for a(t)..
end
function fb=b(t);
fb=sin(t); % or place whatever you want to place for b(t)..
end
Then use a second file with the following code:
t_values=linspace(0,10,101); % the time vector you want to use, or use tspan type vector, [0 10]
initial_cond=[1 ; 0 ; 0];
[tv,Yv]=ode45('fun_name',t_values,initial_cond);
plot(tv,Yv(:,1),'+',tv,Yv(:,2),'x',tv,Yv(:,3),'o');
legend('y1','y2','y3');
Of course for the fun_name.m case I wrote you need not use sub functions for a(t) and b(t), you can just use the explicit functional form in Dy if that is possible (like cos(t) etc).
Part 2: If a(t) , b(t) are just vectors of numbers you happen to have that cannot be expressed as a function of t (as in part 1), then you'll need to have also a time vector for which each of them happens, this can be of course the same time you'll use for the ODE, but it need not be, as long as an interpolation will work. I'll treat the general case, when they have different time spans or resolutions. Then you can do something of the following, create the fun_name.m file:
function Dy = fun_name(t, y, at, a, bt, b)
a = interp1(at, a, t); % Interpolate the data set (at, a) at times t
b = interp1(at, b, t); % Interpolate the data set (bt, b) at times t
Dy=[ a*y(1)+b*y(2); ...
-a*y(3)+b*y(1); ...
a*y(2)] ;
In order to use it, see the following script:
%generate bogus `a` ad `b` function vectors with different time vectors `at` and `bt`
at= linspace(-1, 11, 74); % Generate t for a in a generic case where their time span and sampling can be different
bt= linspace(-3, 33, 122); % Generate t for b
a=rand(numel(at,1));
b=rand(numel(bt,1));
% or use those you have, but you also need to pass their time info...
t_values=linspace(0,10,101); % the time vector you want to use
initial_cond=[1 ; 0 ; 0];
[tv,Yv]= ode45(#(t,y) fun_name(t, y, at, a, bt, b), t_values, initial_cond); %
plot(tv,Yv(:,1),'+',tv,Yv(:,2),'x',tv,Yv(:,3),'o');
legend('y1','y2','y3');
I am new to Matlab. I was reading this code snippet, but in some parts (marked with asterisks) I don't understand what it means, so if anybody could help would be very much appreciated
function [A1nmb] = moran(initsize, popsize)
% MORAN generates a trajectory of a Moran type process
% which gives the number of genes of allelic type A1 in a population
% of haploid individuals that can exist in either type A1 or type A2.
% The population size is popsize and the initial number of type A1
% individuals os initsize.
% Inputs: initsize - initial number of A1 genes
% popsize - the total population size (preserved)
if (nargin==0)
initsize=10;
popsize=30;
end
A1nmb=zeros(1,popsize);
A1nmb(1)=initsize;
**lambda = inline('(x-1).*(1-(x-1)./N)', 'x', 'N');
mu = inline('(x-1).*(1-(x-1)./N)', 'x', 'N');**
x=initsize;
i=1;
while (x>1 & x<popsize+1)
if (lambda(x,popsize)/(lambda(x,popsize)+mu(x,popsize))>rand)
x=x+1;
A1nmb(i)=x;
else
x=x-1;
A1nmb(i)=x;
end;
i=i+1;
end;
nmbsteps=length(A1nmb);
***rate = lambda(A1nmb(1:nmbsteps-1),popsize) ...
+mu(A1nmb(1:nmbsteps-1),popsize);***
**jumptimes=cumsum(-log(rand(1,nmbsteps-1))./rate);**
jumptimes=[0 jumptimes];
stairs(jumptimes,A1nmb);
axis([0 jumptimes(nmbsteps) 0 popsize+1]);
The first line you marked
lambda = inline('(x-1).*(1-(x-1)./N)', 'x', 'N');
creates something called an inline function. It is equivalent to defining a mathematical function. Example:
y = inline('x^2')
would allow you to do
>> y(2)
4
This immediately explains the second line you marked.
rate = lambda(A1nmb(1:nmbsteps-1),popsize) ...
+mu(A1nmb(1:nmbsteps-1),popsize);
will compute the value of the function lambda(x,N) at x = A1nmb(1:nmbsteps-1) and N = popsize.
I will say immediately here that you should take a look at anonymous functions, a different format used to accomplish the same as inline. Only, anonymous functions are generally better supported, and usually a lot faster than inline functions.
Then, for the final line,
jumptimes = cumsum(-log(rand(1,nmbsteps-1))./rate);
is a nested command. rand will create a matrix containing pseudorandom numbers, log is the natural logarithm ("ln"), and cumsum creates a new matrix, where all the elements in the new matrix are the cumulative sum of the elements in the input matrix.
You will find the commands doc and help very useful. Try typing
doc cumsum
or
help inline
on the Matlab command prompt. Try that again with the commands forming the previous statement.
As a general word of advice: spend an insane lot of time reading through the documentation. Really, for each new command you encounter, read about it and play with it in a sandbox until you feel you understand it. Matlab only becomes powerful if you know all its commands, and there are a lot to get to know.
It defines an inline function object. For example this
lambda = inline('(x-1).*(1-(x-1)./N)', 'x', 'N')
defines lambda as a function with 2 variables. When you call lambda(A,n) Matlab simply expands the function you define in the first string. Thus lambda(A,n) using the variables you provide in the function call. lambda(A,n) would will evaluate to:
(A-1).*(1-(A-1)./n)
it just expands the function using the parameters you supply. Take a look at this link for more specific details http://www.mathworks.co.uk/help/techdoc/ref/inline.html
The cumsum function just returns the cumulative sum of a matrix along a particular dimension. Say we call cumsum on a vector X, then the value at element i in the result is equal to the sum of elements in X from index 1 to i. For example X = [1 2 1 3] we would get
AA = cumsum(X);
we would have
AA = [1 3 5 8]
See this link for more details and examples http://www.mathworks.co.uk/help/techdoc/ref/cumsum.html