I'm using maple for differentiation eguation. And I have a problem. I want to express the d/dt(alpha(t)) variable held constant from this equation (a part for example):
-2*(diff(alpha(t), t))*sin(beta(t))*(diff(beta(t), t))*cos(psi(t))*
cos(theta(t))-2*(diff(alpha(t), t))*cos(beta(t))*sin(psi(t))*(diff(psi(t),t))*
cos(theta(t))-2*(diff(alpha(t), t))*cos(beta(t))*cos(psi(t))*sin(theta(t))*
(diff(theta(t), t))-2*(diff(beta(t), t))*sin(alpha(t))^2*(diff(alpha(t),t))*
cos(beta(t))*sin(psi(t)).
Any help is appropriate. Thanks
You can do this with a substitution. For example, let's assume that the large output involving derivatives was produced by running some code that I'll abbreviate as 'mycode;'. Then you can do this:
output := mycode;
new_output := subs(diff(alpha(t), t) = v,output);
Then, in new_output, instance of the symbol diff(alpha(t), t) will be replaced by the symbol v, and then you can use a function like coeff to strip the coefficients of v. This way, you can figure out what the trig-polynomial representation of output is.
Related
I'm wondering if it's possible to define a natural variable n in TI-Nspire CAS. For example I'd like to write:
You can't define your own natural variables. However, Nspire has the following special variables you can use:
#n0...#n255: Restricted to natural numbers
#c0...#c255: Restricted to real numbers
You can replace the original variables with them by hand or for convience just put |x=#n0 and y=#n1 at the end of line.
Example: You are calculating fourier coefficients and know that variable k will only get real numbers from Σ operation. Replacing k with #n1 will simpilfy the function.
Picture
(Calculator needs to be in RAD mode if you want to try)
The answer is no. Variables in NSpire store a value. A variable has no type. Solve might return #n1 in a result to indicate an arbitrary natural number, but you can tell solve to look for integer solutions only.
I have successfully used annotation(derivative) in Modelica functions. Now I have reached a point where I think I need to use zeroDerivative or noDerivative, but from the specification I just do not understand what is the difference, and when to use what.
https://specification.modelica.org/v3.4/Ch12.html#declaring-derivatives-of-functions
It seems zeroDerivative is for time-constant parameters??
Does somebody have a simple example?
Use zeroDerivative to refer to inputs that are non-varying, i.e. parameters or constant values.
Use noDerivative for signals that do not have a derivative value. For example if an input signal comes from an external function.
The important case for noDerivative is when the input is "redundant".
As an example consider the computation of density for some media in MSL:
The density computation is found in Modelica.Media.R134a.R134a_ph.density_ph (note this does not contain any derivative in itself):
algorithm
d := rho_props_ph(
p,
h,
derivsOf_ph(
p,
h,
getPhase_ph(p, h)));
where the top function called is:
function rho_props_ph
"Density as function of pressure and specific enthalpy"
extends Modelica.Icons.Function;
input SI.Pressure p "Pressure";
input SI.SpecificEnthalpy h "Specific enthalpy";
input Common.InverseDerivatives_rhoT derivs
"Record for the calculation of rho_ph_der";
output SI.Density d "Density";
algorithm
d := derivs.rho;
annotation (
derivative(noDerivative=derivs) = rho_ph_der ...);
end rho_props_ph;
So the derivs-argument is sort of redundant and is given by p and h; and we don't need to differentiate it again. If you send in a derivs-argument that isn't given in this way may give unpredictable result, but describing this in detail would be too complicated. (There was some idea of noDerivative=something - but even just specifying it turned out to be too complicated.)
For zeroDerivative the corresponding requirement is that the arguments have zero derivative; that is straightforward to verify and if non-zero we cannot use the specific derivative (it is possible to specify multiple derivatives and use another derivative one for that case).
I'm using MATLAB 2012b.
I want to get d²/dxdy of a simple function:
f(x,y) = (x-1)² + 2y²
The documentation states that I can use syms and diff as in the following example:
> syms x y
> diff(x*sin(x*y), x, y)
ans =
2*x*cos(x*y) - x^2*y*sin(x*y)
But doing the same I got the wrong answer:
> syms x y
> f = (x-1)^2 + 2*y^2;
> diff(f,x,y)
ans =
4*y
The answer is right if I use diff like this:
diff(diff(f,x),y)
Well, it's not a problem for me to use it this way, but nevertheless why is the first variant not working? Is it a version issue?
The actual documentation from R2010a:
diff(expr) differentiates a symbolic expression expr with respect to its free variable as determined by symvar.
diff(expr, v) and diff(expr, sym('v')) differentiate expr with respect to v.
diff(expr, n) differentiates expr n times. n is a positive integer.
diff(expr, v, n) and diff(expr, n, v) differentiate expr with respect to v n times.
So, the command diff(f,x,y) is the last case. It would be equal to differentiating f w.r.t. x, y times, or w.r.t y, x times.
For some reason I don't quite understand, you don't get a warning or error, but one of the syms variables gets interpreted as n = 1, and then the differentiation is carried out. In this case, what diff seems to do is basically diff(f, y, 1).
In any case, it seems that the behavior changed from version to version, because in the documentation you link to (R2016b), there is an additional case:
diff(F,var1,...,varN) differentiates F with respect to the variables var1,...,varN
So I suspect you're running into a version issue.
If you want to differentiate twice, both w.r.t x and y, your second attempt is indeed the correct and most portable way to do that:
diff( diff(f,x), y )
or equivalently
diff( diff(f,y), x )
NB
I checked the R2010a code for symbolic/symbolic/#sym/diff.m and indeed, n is defaulted to 1 and only changed if one of the input variables is a double, and the variable to differentiate over is set equal to the last syms variable in the argument list. The multiple syms variable call is not supported, nor detected and error-trapped.
Syms is only creating symbolic variables.
The first code you execute is only a single derivative. The second code you provided differentiates two times. So I think you forgot to differentiate a second time in the first piece of code you provided.
I am also wondering what answer you expect? If you want 4*y as answer, than you should use
diff(f,y)
and not
diff(f,x,y)
Performing the second derivative is giving me zero?
diff(diff(f,x),y)
If you want 4 as answer than you have to do following:
diff(diff(f,y),y)
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.
i have two problems in mathematica and want to do them in matlab:
measure := RandomReal[] - 0.5
m = 10000;
data = Table[measure, {m}];
fig1 = ListPlot[data, PlotStyle -> {PointSize[0.015]}]
Histogram[data]
matlab:
measure =# (m) rand(1,m)-0.5
m=10000;
for i=1:m
data(:,i)=measure(:,i);
end
figure(1)
plot(data,'b.','MarkerSize',0.015)
figure(2)
hist(data)
And it gives me :
??? The following error occurred
converting from function_handle to
double: Error using ==> double
If i do :
measure =rand()-0.5
m=10000;
data=rand(1,m)-0.5
then, i get the right results in plot1 but in plot 2 the y=axis is wrong.
Also, if i have this in mathematica :
steps[m_] := Table[2 RandomInteger[] - 1, {m}]
steps[20]
Walk1D[n_] := FoldList[Plus, 0, steps[n]]
LastPoint1D[n_] := Fold[Plus, 0, steps[n]]
ListPlot[Walk1D[10^4]]
I did this :
steps = # (m) 2*randint(1,m,2)-1;
steps(20)
Walk1D =# (n) cumsum(0:steps(n)) --> this is ok i think
LastPointold1D= # (n) cumsum(0:steps(n))
LastPoint1D= # (n) LastPointold1D(end)-->but here i now i must take the last "folding"
Walk1D(10)
LastPoint1D(10000)
plot(Walk1D(10000),'b')
and i get an empty matrix and no plot..
Since #Itamar essentially answered your first question, here is a comment on the second one. You did it almost right. You need to define
Walk1D = # (n) cumsum(steps(n));
since cumsum is a direct analog of FoldList[Plus,0,your-list]. Then, the plot in your code works fine. Also, notice that, either in your Mathematica or Matlab code, it is not necessary to define LastPoint1D separately - in both cases, it is the last point of your generated list (vector) steps.
EDIT:
Expanding a bit on LastPoint1D: my guess is that you want it to be a last point of the walk computed by Walk1D. Therefore, it would IMO make sense to just make it a function of a generated walk (vector), that returns its last point. For example:
lastPoint1D = #(walk) (walk(end));
Then, you use it as:
walk = Walk1D(10000);
lastPoint1D(walk)
HTH
You have a few errors/mistakes translating your code to Matlab:
If I am not wrong, the line data = Table[measure, {m}]; creates m copies of measure, which in your case will create a random vector of size (1,m). If that is true, in Matlab it would simply be data = measure(m);
The function you define gets a single argument m, therefor it makes no sense using a matrix notation (the :) when calling it.
Just as a side-note, if you insert data into a matrix inside a for loop, it will run much faster if you allocate the matrix in advance, otherwise Matlab will re-allocate memory to resize the matrix in each iteration. You do this by data = zeros(1,m);.
What do you mean by "in plot 2 the y=axis is wrong"? What do you expect it to be?
EDIT
Regarding your 2nd question, it would be easier to help you if you describe in words what you want to achieve, rather than trying to read your (error producing) code. One thing which is clearly wrong is using expression like 0:steps(n), since you use m:n with two scalars m and n to produce a vector, but steps(n) produces a vector, not a scalar. You probably get an empty matrix since the first value in the vector returned by steps(n) might be -1, and 0:-1 produces an empty vector.