MATLAB: Piecewise for use in anonymous functions

piecewise

Hello,

clc,clear all,close all;
opengl('save', 'software')
syms beta theta r ;
B = 10;
H = 10;
C = 10;
L1 = 4;
Ri = B/2 + L1;
Rf = C + B/2 + L1;
Rbeta0 = Ri;
theta1 = atan((H/2)/(B/2+L1));
theta2 = pi - atan((H/2)/(B/2-L1));
Rt0 = piecewise((0<=theta) & (theta<theta1),(B/2+L1)/cos(theta),(theta1<=theta) & (theta<theta2),(H/2)/sin(theta),(theta2<=theta) & (theta<=pi),(B/2-L1)/cos(pi-theta));
Rmaxtb0 = (Rt0*Rbeta0)/Ri;
Rmaxtb00 = matlabFunction(Rmaxtb0)
Vmbeta0 = (2*(pi^2)*((B+(2*L1))^2)) / ((((2*L1*pi)+(pi*B))^2));
Vbeta0 = Vmbeta0 * (1 - ((r^2) / (Rmaxtb0^2)));
Vbeta00 = Vbeta0 * r;
Vbeta000 = matlabFunction(Vbeta00,'Vars',[theta r]);
Wsigmat = 2 * integral2(Vbeta000,0,pi,0,Rmaxtb00,'AbsTol',1e-5, 'RelTol',1e-5)

Error using symengine

Unable to generate code for piecewise for use in anonymous functions.

As you can see in my piecewise function there is 'theta' in the limits which is defined as 'syms' and is not defined prior

and I think it couldn't be written as if/else!! (I don't know if it can)

Is there any way to define a piecewise function as an anonymous function?

or any way to use an alternative for piecewise function??

thanks in advance for any help

Best Answer

Rmaxtb00 = matlabFunction(Rmaxtb0, 'File', 'Rmaxtb00.m', 'optimize', false);

The generated function in the .m file will use if/elseif to code the piecewise()

However, the generated if/elseif code will not be vectorized, so when you use it with integral() you need to use 'ArrayValued', true so that integral() does not pass in a vector of values to be evaluated.

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MATLAB: Is it so hard for MATLAB to solve the following symbolic equation

This seems to work

syms rp alpha beta theta psi z
A=cos(alpha)-cos(beta);
B=sin(alpha)-sin(beta);
C=(cos(beta)-1)/tan(beta)-(cos(alpha)-1)/tan(alpha);
R=A*(cos(theta)-cos(psi))+B*cos(theta)*sin(psi);
S=A*sin(theta)*sin(psi)-B*cos(theta)+C*cos(psi);
phi=atan(-R/S);
x=-rp*(cos(theta)*cos(phi)+sin(psi)*sin(theta)*sin(phi))*cos(alpha)-rp*(-cos(theta)*sin(phi)+sin(psi)*sin(theta)*cos(phi))*sin(alpha)+(rp/tan(alpha))*(cos(psi)*sin(phi)*(cos(alpha)-1)+cos(psi)*cos(phi)*sin(alpha));
y=-cos(psi)*sin(phi)*rp;
p=[x;y;z];
T=Rot('y',theta)*Rot('x',psi)*Rot('z',phi);
a1=T*[rp;0;0];
a2=T*[rp*cos(alpha);rp*sin(alpha);0];
a3=T*[rp*cos(beta);rp*sin(beta);0];  
Con1=(p+a1)'*[0;1;0];
Con2=(p+a2)'*[-sin(alpha);cos(alpha);0];
Con3=(p+a3)'*[-sin(beta);cos(beta);0];
disp([Con1; Con2; Con3])
function output=Rot(axis,angle)
syms x y z
if axis==x
    output=[1,0,0;0,cos(angle),-sin(angle);0,sin(angle),cos(angle)];
end
if axis==y
    output=[cos(angle),0,sin(angle);0,1,0;-sin(angle),0,cos(angle)];
end
if axis==z
    output=[cos(angle),-sin(angle),0;sin(angle),cos(angle),0;0,0,1];
end
end

MATLAB: AbsTol and RelTol in vpaintegral

Internally, vpaintegral() always works with tolerances. It has default values that it uses. It is not the case that providing a tolerance as an option results in the code proceeding differently -- no

if relative_tolerance_was_specified
    do some extra work
end

So you should not worry about only providing one instead of the other to have it do less work: it is always going to be doing the work, just with default values if you did not specify the option.

Typically reltol is more important in figuring out when to stop. Typically abstol is there as a fall-back for places where the relative values get small. You know that your final result is roughly 1000, but there could still be ranges of your integral where the values are (say) near 1e-15, and over that stretch vpaintegral() might be doing a lot of fine work to try to be more precise about the small numbers, but you in your knowledge of the overall range of results might say that it is pointless to be precise down below 1e-15 about a value that is effectively going to contribute nothing to the overall result, so you might want to put in an absolute tolerance of (say) 1e-5.

Absolute tolerance of 1e-5 does not mean that it will persue the integration until it is certain that the result over the entire integration is within 1e-5 of the "correct" value: instead it means that values less than 1e-5 would more or less filtered out.

There are some cases where the absolute values might get quite small, and yet the contribution could be very important. For example the integral of 1/x, as x goes to infinity, is infinite, ln(infinity), and yet the contributions up around realmax are only about 5e-309 each, with the total contribution by realmax only being about 709 -- the contributions beyond realmax are all negligibly small from the perspective of IEEE 754 double precision, and yet in theory they contribute an infinite amount out to infinity.

Therefore, it is not always appropriate to use an absolute tolerance: you have to know enough about the function to be able to establish that the integral of the missed values will not matter to your accuracy.

In the meantime, I recommand that you use matlabFunction() to turn the vpaintegral() with symbolic upper bound into an anonymous function that calls integral()

I was about to suggest you consider using the matlabFunction() options 'File' and 'optimize', but then I remembered that I have open cases against bugs in the optimization function that can result in the optimized function being quite wrong. It might still be worth using the 'File' option, but if you do, then explicitly set 'optimize' to false.

Best Answer

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MATLAB: Is it so hard for MATLAB to solve the following symbolic equation

MATLAB: AbsTol and RelTol in vpaintegral

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