problem with solve y sym

2017 10 20
1 个回答
更新时间：2017 10 23
12 次查看（30 天）

0 个投票

Hi, I am trying to solve a system of 11 equations with symbolic variables:
S=[q1 - q2.*(0.11.*x1aux) - q4.*(sumxxaux+0.17.*x2aux) - fc1aux + vpasaj1aux - lf1 - v1.*Delayxaux - q2*0.11.*x1aux - x1aux.*v1.*exp(0.3487*sumq0aux/6).*0.3487./6==0;
q6 - q7.*(0.17.*x2aux) - q4.*(sumxxaux+0.11.*x1aux) - fc2aux + vpasaj2aux - lf2 - v2.*Delayxaux - q7*0.17.*x2aux - x2aux.*v2.*exp(0.3487*sumq0aux/6).*0.3487./6==0;
q4/q2-dif2==0];
P= [q51 - q31.*xaux(1,1) - q4.*(0.11.*x1aux+0.17.*x2aux+sumxxaux-xaux(1,1)) - fcaux(1,1) + vpasajaux(1,1) - lfaux(1,1) - vaux(1,1).*Delayxaux - q31.*xaux(1,1) - xaux(1,1).*vaux(1,1).*exp(0.3487*sumq0aux/6).*0.3487./6==0;
q52 - q32.*xaux(1,2) - q4.*(0.11.*x1aux+0.17.*x2aux+sumxxaux-xaux(1,2)) - fcaux(1,2) + vpasajaux(1,2) - lfaux(1,2) - vaux(1,2).*Delayxaux - q32.*xaux(1,2) - xaux(1,2).*vaux(1,2).*exp(0.3487*sumq0aux/6).*0.3487./6==0;
q53 - q33.*xaux(1,3) - q4.*(0.11.*x1aux+0.17.*x2aux+sumxxaux-xaux(1,3)) - fcaux(1,3) + vpasajaux(1,3) - lfaux(1,3) - vaux(1,3).*Delayxaux - q33.*xaux(1,3) - xaux(1,3).*vaux(1,3).*exp(0.3487*sumq0aux/6).*0.3487./6==0];
R= [(-(3*q4^4 - 2*q4^3*q31 - 2*q4^3*q32 - 2*q4^3*q33 - 2*q4^3*q7 + q4^2*q7*q31 + q4^2*q7*q32 + q4^2*q7*q33 + q4^2*q31*q32 + q4^2*q31*q33 + q4^2*q32*q33 - q7*q31*q32*q33)/(3*q2*q4^4 + 3*q4^4*q7 + 3*q4^4*q31 + 3*q4^4*q32 + 3*q4^4*q33 - 4*q4^5 - 2*q2*q4^3*q7 - 2*q2*q4^3*q31 - 2*q2*q4^3*q32 - 2*q2*q4^3*q33 - 2*q4^3*q7*q31 - 2*q4^3*q7*q32 - 2*q4^3*q7*q33 - 2*q4^3*q31*q32 - 2*q4^3*q31*q33 - 2*q4^3*q32*q33 + q2*q4^2*q7*q31 + q2*q4^2*q7*q32 + q2*q4^2*q7*q33 + q2*q4^2*q31*q32 + q2*q4^2*q31*q33 + q2*q4^2*q32*q33 + q4^2*q7*q31*q32 + q4^2*q7*q31*q33 + q4^2*q7*q32*q33 + q4^2*q31*q32*q33 - q2*q7*q31*q32*q33))-(eta.*(0.11.*x1aux))./(q1 - q2.*(0.11.*x1aux) - q4.*(sumxxaux+0.17.*x2aux))==0;
(-(3*q4^4 - 2*q4^3*q31 - 2*q4^3*q32 - 2*q4^3*q33 - 2*q2*q4^3 + q2*q4^2*q31 + q2*q4^2*q32 + q2*q4^2*q33 + q4^2*q31*q32 + q4^2*q31*q33 + q4^2*q32*q33 - q2*q31*q32*q33)/(3*q2*q4^4 + 3*q4^4*q7 + 3*q4^4*q31 + 3*q4^4*q32 + 3*q4^4*q33 - 4*q4^5 - 2*q2*q4^3*q7 - 2*q2*q4^3*q31 - 2*q2*q4^3*q32 - 2*q2*q4^3*q33 - 2*q4^3*q7*q31 - 2*q4^3*q7*q32 - 2*q4^3*q7*q33 - 2*q4^3*q31*q32 - 2*q4^3*q31*q33 - 2*q4^3*q32*q33 + q2*q4^2*q7*q31 + q2*q4^2*q7*q32 + q2*q4^2*q7*q33 + q2*q4^2*q31*q32 + q2*q4^2*q31*q33 + q2*q4^2*q32*q33 + q4^2*q7*q31*q32 + q4^2*q7*q31*q33 + q4^2*q7*q32*q33 + q4^2*q31*q32*q33 - q2*q7*q31*q32*q33))-(eta.*(0.17.*x2aux))./(q6 - q7.*(0.17.*x2aux) - q4.*(0.11.*x1aux+sumxxaux))==0];
G= [(-(3*q4^4 - 2*q4^3*q7 - 2*q4^3*q32 - 2*q4^3*q33 - 2*q2*q4^3 + q2*q4^2*q7 + q2*q4^2*q32 + q2*q4^2*q33 + q4^2*q7*q32 + q4^2*q7*q33 + q4^2*q32*q33 - q2*q7*q32*q33)/(3*q2*q4^4 + 3*q4^4*q7 + 3*q4^4*q31 + 3*q4^4*q32 + 3*q4^4*q33 - 4*q4^5 - 2*q2*q4^3*q7 - 2*q2*q4^3*q31 - 2*q2*q4^3*q32 - 2*q2*q4^3*q33 - 2*q4^3*q7*q31 - 2*q4^3*q7*q32 - 2*q4^3*q7*q33 - 2*q4^3*q31*q32 - 2*q4^3*q31*q33 - 2*q4^3*q32*q33 + q2*q4^2*q7*q31 + q2*q4^2*q7*q32 + q2*q4^2*q7*q33 + q2*q4^2*q31*q32 + q2*q4^2*q31*q33 + q2*q4^2*q32*q33 + q4^2*q7*q31*q32 + q4^2*q7*q31*q33 + q4^2*q7*q32*q33 + q4^2*q31*q32*q33 - q2*q7*q31*q32*q33))-(eta.*(xaux(1,1)))./(q51 - q31.*xaux(1,1) - q4.*(0.11.*x1aux+0.17.*x2aux+sumxxaux-xaux(1,1)))==0;
(-(3*q4^4 - 2*q4^3*q7 - 2*q4^3*q31 - 2*q4^3*q33 - 2*q2*q4^3 + q2*q4^2*q7 + q2*q4^2*q31 + q2*q4^2*q33 + q4^2*q7*q31 + q4^2*q7*q33 + q4^2*q31*q33 - q2*q7*q31*q33)/(3*q2*q4^4 + 3*q4^4*q7 + 3*q4^4*q31 + 3*q4^4*q32 + 3*q4^4*q33 - 4*q4^5 - 2*q2*q4^3*q7 - 2*q2*q4^3*q31 - 2*q2*q4^3*q32 - 2*q2*q4^3*q33 - 2*q4^3*q7*q31 - 2*q4^3*q7*q32 - 2*q4^3*q7*q33 - 2*q4^3*q31*q32 - 2*q4^3*q31*q33 - 2*q4^3*q32*q33 + q2*q4^2*q7*q31 + q2*q4^2*q7*q32 + q2*q4^2*q7*q33 + q2*q4^2*q31*q32 + q2*q4^2*q31*q33 + q2*q4^2*q32*q33 + q4^2*q7*q31*q32 + q4^2*q7*q31*q33 + q4^2*q7*q32*q33 + q4^2*q31*q32*q33 - q2*q7*q31*q32*q33))-(eta.*(xaux(1,2)))./(q52 - q32.*xaux(1,2) - q4.*(0.11.*x1aux+0.17.*x2aux+sumxxaux-xaux(1,2)))==0;
(-(3*q4^4 - 2*q4^3*q7 - 2*q4^3*q31 - 2*q4^3*q32 - 2*q2*q4^3 + q2*q4^2*q7 + q2*q4^2*q31 + q2*q4^2*q32 + q4^2*q7*q31 + q4^2*q7*q32 + q4^2*q31*q32 - q2*q7*q31*q32)/(3*q2*q4^4 + 3*q4^4*q7 + 3*q4^4*q31 + 3*q4^4*q32 + 3*q4^4*q33 - 4*q4^5 - 2*q2*q4^3*q7 - 2*q2*q4^3*q31 - 2*q2*q4^3*q32 - 2*q2*q4^3*q33 - 2*q4^3*q7*q31 - 2*q4^3*q7*q32 - 2*q4^3*q7*q33 - 2*q4^3*q31*q32 - 2*q4^3*q31*q33 - 2*q4^3*q32*q33 + q2*q4^2*q7*q31 + q2*q4^2*q7*q32 + q2*q4^2*q7*q33 + q2*q4^2*q31*q32 + q2*q4^2*q31*q33 + q2*q4^2*q32*q33 + q4^2*q7*q31*q32 + q4^2*q7*q31*q33 + q4^2*q7*q32*q33 + q4^2*q31*q32*q33 - q2*q7*q31*q32*q33))-(eta.*(xaux(1,3)))./(q53 - q33.*xaux(1,3) - q4.*(0.11.*x1aux+0.17.*x2aux+sumxxaux-xaux(1,3)))==0];
D=[S;R;P;G];
[t11,t22,t331,t332,t333,t44,t551,t552,t553,t66,t77,param,cond]=solve(D,q1,q2,q31,q32,q33,q4,q51,q52,q53,q6,q7,'ReturnConditions',true,'Real',true);
But I obtain this answer:
Error using mupadengine/feval (line 163)
The row index is out of range.
Error in solve (line 369)
varargout{i} = transpose(eng.feval('map', solutions, '_index', i));
I don't know why! please help me!
Thanks

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qi (with i=1,2,3..) and dif2 are symbolic variables, the others variables are parameters.
Birdman
Birdman 2017-10-20
Firstly, you have to organize this code. Please use the code format. Then it will be possible to understand.
Sorry, but where do I put the code format? In the matrix D?
Steven Lord
Steven Lord 2017-10-20
Edit your question here on Answers. Select all of your code and press the "{}Code" button just above the edit box. That will format it nicely, so it's easier for people to read.
Walter Roberson
Walter Roberson 2017-10-20
编辑：Walter Roberson 2017-10-20
It looks to me as if the code could be tested if it were preceded with
[Delayxaux, eta, fc1aux, fc2aux, lf1, lf2, sumxxaux, sumq0aux, v1, v2, vpasaj1aux, vpasaj2aux, x1aux, x2aux] = deal(rand, rand, rand, rand, rand, rand, rand, rand, rand, rand, rand, rand, rand, rand);
fcaux = rand(1,3);
lfaux = rand(1,3);
vaux = rand(1,3);
vpasajaux = rand(1,3);
xaux = rand(1,3);
syms dif2 q1 q2 q4 q6 q7 q31 q32 q33 q51 q52 q53
You appear to be solving 11 equations for 11 out of the 12 variables, so you would expect most of the outputs to be in terms of dif2
Can any constraints be put on the symbols? Such as real valued, or non-negative?
Yes, this is the code: syms q1 positive
syms q2 positive
syms q31 positive
syms q32 positive
syms q33 positive
syms q4 positive
syms q51 positive
syms q52 positive
syms q53 positive
syms q6 positive %a united
syms q7 positive %b united
syms dif2

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回答（1 个）

Walter Roberson
Walter Roberson 2017-10-20

0 个投票

With the above initialization, in R2017b, it works hard on it for a while and then says "Warning: Unable to find explicit solution." and returns empty symbols.
This does not surprise me. If you
collect(D, q4)
then you will see that you have q4^5 in some of the equations. There is no general closed form solution to equations of degree 5 and it is not possible for MATLAB to look for numeric approximations because of the free variable dif2 .

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Thanks! I use R2015b, can be the error because my matlab is older? How I can fix the problem?
Thanks again!
Walter Roberson
Walter Roberson 2017-10-21
In R2017b, I also stepped through solving equations one by one. I was able to get most of the variables, but not all:
Q4 = solve(D(1),q4)
D2 = subs(D,q4,Q4)
Q1 = solve(D2(2),q1)
D3 = subs(D2,q1,Q1)
Q2 = solve(D3(3),q2)
D4 = subs(D3,q2,Q2)
Q31 = solve(D4(6),q31)
D5 = subs(D4,q31,Q31)
Q32 = solve(D5(7),q32)
D6 = subs(D5,q32,Q32)
Q33 = solve(D6(8),q33)
D7 = subs(D6,q33,Q33)
Q51 = solve(D7(9),q51)
D8 = subs(D7,q51,Q51)
%this next takes a while but does finish
Q52 = solve(D8(11),q52)
But it gets really really really bogged down in the following
D9 = subs(D8,q52,Q52)
I had to kill it when it got past 22 gigabytes
The variables remaining are q53, q6, and q7. Some of your equations involve division of terms, so solving for q6 and q7 would be especially difficult.
Walter Roberson
Walter Roberson 2017-10-21
编辑：Walter Roberson 2017-10-21
Cleaner version
[Delayxaux, eta, fc1aux, fc2aux, lf1, lf2, sumxxaux, sumq0aux, v1, v2, vpasaj1aux, vpasaj2aux, x1aux, x2aux] = deal(rand, rand, rand, rand, rand, rand, rand, rand, rand, rand, rand, rand, rand, rand);
fcaux = rand(1,3);
lfaux = rand(1,3);
vaux = rand(1,3);
vpasajaux = rand(1,3);
xaux = rand(1,3);
syms dif2
syms q1 q2 q4 q6 q7 q31 q32 q33 q51 q52 q53 positive
S=[q1 - q2.*(0.11.*x1aux) - q4.*(sumxxaux+0.17.*x2aux) - fc1aux + vpasaj1aux - lf1 - v1.*Delayxaux - q2*0.11.*x1aux - x1aux.*v1.*exp(0.3487*sumq0aux/6).*0.3487./6==0;
q6 - q7.*(0.17.*x2aux) - q4.*(sumxxaux+0.11.*x1aux) - fc2aux + vpasaj2aux - lf2 - v2.*Delayxaux - q7*0.17.*x2aux - x2aux.*v2.*exp(0.3487*sumq0aux/6).*0.3487./6==0;
q4/q2-dif2==0];
P= [q51 - q31.*xaux(1,1) - q4.*(0.11.*x1aux+0.17.*x2aux+sumxxaux-xaux(1,1)) - fcaux(1,1) + vpasajaux(1,1) - lfaux(1,1) - vaux(1,1).*Delayxaux - q31.*xaux(1,1) - xaux(1,1).*vaux(1,1).*exp(0.3487*sumq0aux/6).*0.3487./6==0;
q52 - q32.*xaux(1,2) - q4.*(0.11.*x1aux+0.17.*x2aux+sumxxaux-xaux(1,2)) - fcaux(1,2) + vpasajaux(1,2) - lfaux(1,2) - vaux(1,2).*Delayxaux - q32.*xaux(1,2) - xaux(1,2).*vaux(1,2).*exp(0.3487*sumq0aux/6).*0.3487./6==0;
q53 - q33.*xaux(1,3) - q4.*(0.11.*x1aux+0.17.*x2aux+sumxxaux-xaux(1,3)) - fcaux(1,3) + vpasajaux(1,3) - lfaux(1,3) - vaux(1,3).*Delayxaux - q33.*xaux(1,3) - xaux(1,3).*vaux(1,3).*exp(0.3487*sumq0aux/6).*0.3487./6==0];
R= [(-(3*q4^4 - 2*q4^3*q31 - 2*q4^3*q32 - 2*q4^3*q33 - 2*q4^3*q7 + q4^2*q7*q31 + q4^2*q7*q32 + q4^2*q7*q33 + q4^2*q31*q32 + q4^2*q31*q33 + q4^2*q32*q33 - q7*q31*q32*q33)/(3*q2*q4^4 + 3*q4^4*q7 + 3*q4^4*q31 + 3*q4^4*q32 + 3*q4^4*q33 - 4*q4^5 - 2*q2*q4^3*q7 - 2*q2*q4^3*q31 - 2*q2*q4^3*q32 - 2*q2*q4^3*q33 - 2*q4^3*q7*q31 - 2*q4^3*q7*q32 - 2*q4^3*q7*q33 - 2*q4^3*q31*q32 - 2*q4^3*q31*q33 - 2*q4^3*q32*q33 + q2*q4^2*q7*q31 + q2*q4^2*q7*q32 + q2*q4^2*q7*q33 + q2*q4^2*q31*q32 + q2*q4^2*q31*q33 + q2*q4^2*q32*q33 + q4^2*q7*q31*q32 + q4^2*q7*q31*q33 + q4^2*q7*q32*q33 + q4^2*q31*q32*q33 - q2*q7*q31*q32*q33))-(eta.*(0.11.*x1aux))./(q1 - q2.*(0.11.*x1aux) - q4.*(sumxxaux+0.17.*x2aux))==0;
(-(3*q4^4 - 2*q4^3*q31 - 2*q4^3*q32 - 2*q4^3*q33 - 2*q2*q4^3 + q2*q4^2*q31 + q2*q4^2*q32 + q2*q4^2*q33 + q4^2*q31*q32 + q4^2*q31*q33 + q4^2*q32*q33 - q2*q31*q32*q33)/(3*q2*q4^4 + 3*q4^4*q7 + 3*q4^4*q31 + 3*q4^4*q32 + 3*q4^4*q33 - 4*q4^5 - 2*q2*q4^3*q7 - 2*q2*q4^3*q31 - 2*q2*q4^3*q32 - 2*q2*q4^3*q33 - 2*q4^3*q7*q31 - 2*q4^3*q7*q32 - 2*q4^3*q7*q33 - 2*q4^3*q31*q32 - 2*q4^3*q31*q33 - 2*q4^3*q32*q33 + q2*q4^2*q7*q31 + q2*q4^2*q7*q32 + q2*q4^2*q7*q33 + q2*q4^2*q31*q32 + q2*q4^2*q31*q33 + q2*q4^2*q32*q33 + q4^2*q7*q31*q32 + q4^2*q7*q31*q33 + q4^2*q7*q32*q33 + q4^2*q31*q32*q33 - q2*q7*q31*q32*q33))-(eta.*(0.17.*x2aux))./(q6 - q7.*(0.17.*x2aux) - q4.*(0.11.*x1aux+sumxxaux))==0];
G= [(-(3*q4^4 - 2*q4^3*q7 - 2*q4^3*q32 - 2*q4^3*q33 - 2*q2*q4^3 + q2*q4^2*q7 + q2*q4^2*q32 + q2*q4^2*q33 + q4^2*q7*q32 + q4^2*q7*q33 + q4^2*q32*q33 - q2*q7*q32*q33)/(3*q2*q4^4 + 3*q4^4*q7 + 3*q4^4*q31 + 3*q4^4*q32 + 3*q4^4*q33 - 4*q4^5 - 2*q2*q4^3*q7 - 2*q2*q4^3*q31 - 2*q2*q4^3*q32 - 2*q2*q4^3*q33 - 2*q4^3*q7*q31 - 2*q4^3*q7*q32 - 2*q4^3*q7*q33 - 2*q4^3*q31*q32 - 2*q4^3*q31*q33 - 2*q4^3*q32*q33 + q2*q4^2*q7*q31 + q2*q4^2*q7*q32 + q2*q4^2*q7*q33 + q2*q4^2*q31*q32 + q2*q4^2*q31*q33 + q2*q4^2*q32*q33 + q4^2*q7*q31*q32 + q4^2*q7*q31*q33 + q4^2*q7*q32*q33 + q4^2*q31*q32*q33 - q2*q7*q31*q32*q33))-(eta.*(xaux(1,1)))./(q51 - q31.*xaux(1,1) - q4.*(0.11.*x1aux+0.17.*x2aux+sumxxaux-xaux(1,1)))==0;
(-(3*q4^4 - 2*q4^3*q7 - 2*q4^3*q31 - 2*q4^3*q33 - 2*q2*q4^3 + q2*q4^2*q7 + q2*q4^2*q31 + q2*q4^2*q33 + q4^2*q7*q31 + q4^2*q7*q33 + q4^2*q31*q33 - q2*q7*q31*q33)/(3*q2*q4^4 + 3*q4^4*q7 + 3*q4^4*q31 + 3*q4^4*q32 + 3*q4^4*q33 - 4*q4^5 - 2*q2*q4^3*q7 - 2*q2*q4^3*q31 - 2*q2*q4^3*q32 - 2*q2*q4^3*q33 - 2*q4^3*q7*q31 - 2*q4^3*q7*q32 - 2*q4^3*q7*q33 - 2*q4^3*q31*q32 - 2*q4^3*q31*q33 - 2*q4^3*q32*q33 + q2*q4^2*q7*q31 + q2*q4^2*q7*q32 + q2*q4^2*q7*q33 + q2*q4^2*q31*q32 + q2*q4^2*q31*q33 + q2*q4^2*q32*q33 + q4^2*q7*q31*q32 + q4^2*q7*q31*q33 + q4^2*q7*q32*q33 + q4^2*q31*q32*q33 - q2*q7*q31*q32*q33))-(eta.*(xaux(1,2)))./(q52 - q32.*xaux(1,2) - q4.*(0.11.*x1aux+0.17.*x2aux+sumxxaux-xaux(1,2)))==0;
(-(3*q4^4 - 2*q4^3*q7 - 2*q4^3*q31 - 2*q4^3*q32 - 2*q2*q4^3 + q2*q4^2*q7 + q2*q4^2*q31 + q2*q4^2*q32 + q4^2*q7*q31 + q4^2*q7*q32 + q4^2*q31*q32 - q2*q7*q31*q32)/(3*q2*q4^4 + 3*q4^4*q7 + 3*q4^4*q31 + 3*q4^4*q32 + 3*q4^4*q33 - 4*q4^5 - 2*q2*q4^3*q7 - 2*q2*q4^3*q31 - 2*q2*q4^3*q32 - 2*q2*q4^3*q33 - 2*q4^3*q7*q31 - 2*q4^3*q7*q32 - 2*q4^3*q7*q33 - 2*q4^3*q31*q32 - 2*q4^3*q31*q33 - 2*q4^3*q32*q33 + q2*q4^2*q7*q31 + q2*q4^2*q7*q32 + q2*q4^2*q7*q33 + q2*q4^2*q31*q32 + q2*q4^2*q31*q33 + q2*q4^2*q32*q33 + q4^2*q7*q31*q32 + q4^2*q7*q31*q33 + q4^2*q7*q32*q33 + q4^2*q31*q32*q33 - q2*q7*q31*q32*q33))-(eta.*(xaux(1,3)))./(q53 - q33.*xaux(1,3) - q4.*(0.11.*x1aux+0.17.*x2aux+sumxxaux-xaux(1,3)))==0];
D=[S;R;P;G];
%[t11,t22,t331,t332,t333,t44,t551,t552,t553,t66,t77,param,cond]=solve(D,q1,q2,q31,q32,q33,q4,q51,q52,q53,q6,q7,'ReturnConditions',true,'Real',true);
%sol = solve(D,q1,q2,q31,q32,q33,q4,q51,q52,q53,q6,q7,'ReturnConditions',true);
Q4 = solve(D(1), q4, 'ReturnConditions', true)
D2 = subs(D, q4, Q4.q4)
Q1 = solve(D2(2), q1, 'ReturnConditions', true)
D3 = subs(D2, q1, Q1.q1)
Q2 = solve(D3(3), q2, 'ReturnConditions', true)
D4 = subs(D3, q2, Q2.q2)
Q31 = solve(D4(6), q31, 'ReturnConditions', true)
D5 = subs(D4, q31, Q31.q31)
Q32 = solve(D5(7), q32, 'ReturnConditions', true)
D6 = subs(D5, q32, Q32.q32)
Q33 = solve(D6(8), q33, 'ReturnConditions', true)
D7 = subs(D6, q33, Q33.q33)
Q51 = solve(D7(9), q51, 'ReturnConditions', true)
D8 = subs(D7, q51, Q51.q51)
%this next takes a while but does finish
Q52 = solve(D8(11), q52, 'ReturnConditions', true)
keyboard %do not accidentally continue to the next one!
%gets really really bogged down in the following
D9 = subs(D8, q52, Q52.q52(1))
Thanks! I think I have to change the problem. Or, Do you think another solution? Like a command to approximate equations or simplify to the solve comand? I tried with simplify D but nothing change.
Thanks!
Walter Roberson
Walter Roberson 2017-10-23
Your R and your G both contain the ratio of quartics (polynomials of degree 4.) Just knowing that, we can predict that the solutions are going to be at least degree 7, or degree 8, or degree 16. You have 5 like that, and MATLAB needs to explore all of the possibilities for all of them. That is going to be slow and memory intensive.
Perhaps you can taylor with respect to q4 -- but to what order?
I extracted the division subexpression from the first R entry, and tossed in some random values for some of the variables
t2 = (3*q4^4 - 2*q4^3*q31 - 2*q4^3*q32 - 2*q4^3*q33 - 2*q4^3*q7 + q4^2*q7*q31 + q4^2*q7*q32 + q4^2*q7*q33 + q4^2*q31*q32 + q4^2*q31*q33 + q4^2*q32*q33 - q7*q31*q32*q33)/(3*q2*q4^4 + 3*q4^4*q7 + 3*q4^4*q31 + 3*q4^4*q32 + 3*q4^4*q33 - 4*q4^5 - 2*q2*q4^3*q7 - 2*q2*q4^3*q31 - 2*q2*q4^3*q32 - 2*q2*q4^3*q33 - 2*q4^3*q7*q31 - 2*q4^3*q7*q32 - 2*q4^3*q7*q33 - 2*q4^3*q31*q32 - 2*q4^3*q31*q33 - 2*q4^3*q32*q33 + q2*q4^2*q7*q31 + q2*q4^2*q7*q32 + q2*q4^2*q7*q33 + q2*q4^2*q31*q32 + q2*q4^2*q31*q33 + q2*q4^2*q32*q33 + q4^2*q7*q31*q32 + q4^2*q7*q31*q33 + q4^2*q7*q32*q33 + q4^2*q31*q32*q33 - q2*q7*q31*q32*q33)
t3 = subs(t2,[q2,q32,q33,q7],[2,32,33,7])
fsurf(t3,[-100 100 -80 40])
The result has at least 4 curved lines of discontinuity, indicating that if you cannot promise that the variables will stay in the "safe" ranges, that there cannot be any finite polynomial approximation.
I don't know... perhaps you could do something with continued fractions.

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