I'm trying to use the symbolic math toolbox in Matlab 2012b to get symbolic solutions to a set of equations. My code looks like:
syms CPitch CRoll CYaw CX CY CZ
syms VPitch VRoll VZ
syms MPitch MRoll MZ
Eqs = [ MPitch == cos(VRoll)*CPitch + cos(CYaw)*VPitch;
MRoll == cos(CYaw)*VRoll + sin(CYaw)*VPitch + CRoll;
MZ == CZ*cos(VPitch)*sin(VRoll) + CY*sin(VPitch)*sin(VRoll) ];
solve(Eqs, {VRoll, VPitch, VZ})
I get back errors that look like:
Error in solve>tochar (line 289)
vc = char(v);
Error in solve>getEqns (line 254)
vc = tochar(v);
Error in solve (line 150)
[eqns,vars,options] = getEqns(varargin{:});
As far as I can tell everything matches how they do it in their examples. I'm not sure what it's complaining about.
Related
I am trying to solve the equation below for array d. I have used the snippet below:
channel_size = 9e-3;
d = [11e-3, 12e-3];
sigma = 0.49;
ee = 727/9806.65;
alpha = d-channel_size;
sym('p',[1 2])
for i = 1:2
eqn = alpha == (4.3^(1/7))*(p^(3/11))*(((1-(sigma^2))/ee)^(7/5))/(d.^(1/6))
S = solve(eqn, p)*0.015;
vpa(S/13e-12)
end
In fact, I should get two number corresponding to d(1) and d(2), but it does not work and this error appears:
Error using mupadmex
Error in MuPAD command: Operands are invalid. [linalg::matlinsolve]
Error in sym/privBinaryOp (line 1693)
Csym = mupadmex(op,args{1}.s, args{2}.s, varargin{:});
Error in sym/mrdivide (line 232)
X = privBinaryOp(A, B, 'symobj::mrdivide');
Declaring (d.^(1/6)) is wrong and instead (d(i)^(1/6)) should be used. In addition, as alpha = d-channel_size, in the equation, I should also declare alpha(i) instead of simple alpha.
>> syms x v(x) w(x);
>> eq1 = 2*v + 3*w == 4;
>> eq2 = 5*v + 4*w == 3;
>> sol = solve([eq1,eq2],[v,w])
I tried to implement this code in MATLAB, but error flashes out as "The second argument must be a vector of symbolic variables." I have tried similar things in Python using SymPy, but never such error comes. How to correct out this?
Have a look at the help file for the multivariate case of solve and the example in the help file
openExample('symbolic/SolveMultivariateEquationsAndAssignOutputsToStructureExample')
Applied to your problem
syms v w;
eq1 = [2*v + 3*w == 4;5*v + 4*w == 3];
sol = solve(eq1)
sol.v
sol.w
but if you just want to solve for v w then you could use e.g.
[2 3;5 4]\[4;3]
I've defined a function in Matlab like this:
clg = 2*pi*(alpha-alphai+0.5*A(1));
where A is an array which depends on alpha.
I want to evaluate clg for aplha=0.53.
I've tried subs(clg, alpha, 0.53), but It gives me a bunch of weird errors:
Error using sym/subs>normalize (line 210)
Entries in second argument must be scalar.
Error in sym/subs>mupadsubs (line 136)
[X2,Y2,symX,symY] = normalize(X,Y); %#ok
Error in sym/subs (line 124)
G = mupadsubs(F,X,Y);
Error in integral (line 45)
subs(clg, alpha, 0.53)
Any idea on how to achieve that?
Thanks!
I have no problem about your question.
>> syms x x1 x2
>> clg = 2*pi*(x-x1+0.5*x2);
>> subs(clg,x,0.53)
ans =
2*pi*(x2/2 - x1 + 53/100)
I am trying to reproduce the problem, but this works:
syms x
%A is a symbolic matrix that depends on x
A = inv([10*x, 2*x, 3*x;
4*x, 10, 6*x;
7*x, 8*x, 10*x])
%y is a symbolic expression defined such that
%it depends on A, but collapses to an expression of x
y = x + 0.5*A(2,2)
%subs returns a symbolic (in this case fraction) evaluation:
subs(y, x, 3)
%eval returns a numeric evaluation:
x = 3
eval(y)
(I have encountered the same error message as you in my own code, but have not yet hunted down its source.)
I want to calculate the part-expectation of log-normal distribution via:
m = 1;
v = 2;
mu = log((m^2)/sqrt(v+m^2));
sigma = sqrt(log(v/(m^2)+1));
syms x;
d = x*lognpdf(x,mu,sigma);
int(d, x, 0, 10);
However, MATLAB says:
Error using symfun>validateArgNames (line 211) Second input must be a
scalar or vector of unique symbolic variables.
Error in symfun (line 45)
y.vars = validateArgNames(inputs);
Error in sym/subsasgn (line 771)
C = symfun(B,[inds{:}]);
Error in lognpdf (line 36) x(x <= 0) = Inf;
Error in untitled (line 7) d = x*lognpdf(x,mu,sigma);
I even tried to just calculate the integral of the pdf by:
m = 1;
v = 2;
mu = log((m^2)/sqrt(v+m^2));
sigma = sqrt(log(v/(m^2)+1));
syms x;
d = lognpdf(x,mu,sigma);
int(d, x, 0, 10);
But there are still errors, and MATLAB says:
Error using symfun>validateArgNames (line 211) Second input must be a
scalar or vector of unique symbolic variables.
Error in symfun (line 45)
y.vars = validateArgNames(inputs);
Error in sym/subsasgn (line 771)
C = symfun(B,[inds{:}]);
Error in lognpdf (line 36) x(x <= 0) = Inf;
Error in untitled (line 7) d = lognpdf(x,mu,sigma);
I really don't know what happened. Should the integral of the pdf be the cdf?
Similar to an answer several months ago, the Statistics Toolbox doesn't support the Symbolic Toolbox currently.
Therefore, you can proceed by hard coding the PDF itself and integrating it:
d = exp(-(log(x)-mu)^2/(2*sigma^2))/(x*sigma*sqrt(2*pi));
int(d, x, 0, 10);
Or you can use the logncdf function, which may be cleaner.
I got 2 questions here.
1. If I have y = a*x^2 + 5. What function can make it into y = a.*x.^2 +5. As you seen, dot was inserted.
It's easy, but kinda diffcult to describe, but please have patience with me. Thank you so much.
First, let me make a very simple example of my problem.
If I want to calucate Y = F(x=1)+ 2.^2, and I know F(x=1) = a+b,(a and b are syms). This means Y = a + b + 4. Problem is here,
Matlab give me error if I write down as.
F = function( .... ); <==== output of function is F(X=1), and = F(x=1) = a+b
Y = integral2( F + 2.^2, .. , .. ,..)
However, if I just copy the output of F as
Y = integral2( a+b + 2.^2, .. , .. ,..)
Now it works!!!
Ok. Please allow me to talk about my code here. I am trying to find a double interation by using integral2. One part of my equcation(which is Y) is from another int output(which is F). Matlab will give ERROR for code below:
clear all;
a=4;
la1=1/(pi*500^2); la2= la1*5;
p1=25; p2=p1/25;
sgma2=10^(-11);
index=1;
g=2./a;
syms r u1 u2
powe= -2 ;
seta= 10^powe;
xNor = ( (u2./u1).^(a./2) + 1 ).^(2./a);
x = (xNor).^(0.5) * seta^(-1/a);
fun1 = r./(1+ r.^a );
out1 = int(fun1, x, Inf) ; %== This is my F in my example
q=pi.*(la1.*p1.^(2./a)+la2.*p2.^(2./a));
yi = #(u2,u1) exp(-u2.*(1+2.*...
( out1 )./... %=== out1 is the problem here.
( (( (u2./u1).^(a./2) + 1 ).^(2./a)).*seta.^(-2./a)))).*...
exp(-sgma2.*q.^(-a./2).* seta.*u2.^(a./2)./...
((( (u2./u1).^(a./2) + 1 ).^(2./a)).^(a./2)) );
maxF =#(u2) u2;
out2 = integral2(yi,0,Inf,0 ,maxF) % == this is Y in my previous example.
However, since I know the out1 = pi/4 - atan(10*(u2^2/u1^2 + 1)^(1/2))/2 (no dot,1/2, not 1./2). Instead of writing down out1, I will just type the equaction and add dot in the
yi = #(u2,u1) exp(-u2.*(1+2.*...
( pi./4 - atan(10.*(u2.^2./u1.^2 + 1).^(1./2))./2 )./... %=== not "out1"
( (( (u2./u1).^(a./2) + 1 ).^(2./a)).*seta.^(-2./a)))).*...
exp(-sgma2.*q.^(-a./2).* seta.*u2.^(a./2)./...
((( (u2./u1).^(a./2) + 1 ).^(2./a)).^(a./2)) );
Now the code is working!!!! The final output is = 0.9957.
Dear friends, I already spend a long time on this, but I still can not find out the problem. Could you please take a deeper look for me. Please copy the code to you matlab and test. Thank you so much.
Below is the error given by matlab, if I just use "out1" in yi = #(u2,u1) ......
Error using integralCalc/finalInputChecks (line 511)
Input function must return 'double' or 'single' values. Found 'sym'.
Error in integralCalc/iterateScalarValued (line 315)
finalInputChecks(x,fx);
Error in integralCalc/vadapt (line 133)
[q,errbnd] = iterateScalarValued(u,tinterval,pathlen);
Error in integralCalc (line 76)
[q,errbnd] = vadapt(#AtoBInvTransform,interval);
Error in
integral2Calc>#(xi,y1i,y2i)integralCalc(#(y)fun(xi*ones(size(y)),y),y1i,y2i,opstruct.integralOptions)
(line 18)
innerintegral = #(x)arrayfun(#(xi,y1i,y2i)integralCalc( ...
Error in
integral2Calc>#(x)arrayfun(#(xi,y1i,y2i)integralCalc(#(y)fun(xi*ones(size(y)),y),y1i,y2i,opstruct.integralOptions),x,ymin(x),ymax(x))
(line 18)
innerintegral = #(x)arrayfun(#(xi,y1i,y2i)integralCalc( ...
Error in integralCalc/iterateScalarValued (line 314)
fx = FUN(t);
Error in integralCalc/vadapt (line 133)
[q,errbnd] = iterateScalarValued(u,tinterval,pathlen);
Error in integralCalc (line 84)
[q,errbnd] = vadapt(#AToInfInvTransform,interval);
Error in integral2Calc>integral2i (line 21)
[q,errbnd] = integralCalc(innerintegral,xmin,xmax,opstruct.integralOptions);
Error in integral2Calc (line 8)
[q,errbnd] = integral2i(fun,xmin,xmax,ymin,ymax,optionstruct);
Error in integral2 (line 107)
Q = integral2Calc(fun,xmin,xmax,yminfun,ymaxfun,opstruct);
Error in ref7_equ11n2 (line 129)
out2 = integral2(yi,0,Inf,0 ,maxF)
The documentation of the function integral2 states:
integral2 - Numerically evaluate double integral
You won't be able to use integral2 with symbolic expressions, unless you convert them to function handles using matlabfunction.