Solving a large equation

I have a large equation of the form:

8.5567=10.75-158.34*y+2112.07*y^2-12211.07*y^3-29.44*x+2185.91*x*y-35690.08*x*y^2+187137.42*x*y^3+129.16*x^2-9697.93*x^2*y+173834.73*x^2*y^2-935832.82*x^2*y^3-174.03*x^3+13191.71*x^3*y-245488.55*x^3*y^2+1384550.57*x^3*y^3

For x ranging from 0.15 to 0.39 I want to develop the value(s) of y.

Any ideas on how to do this would be greatly appreciated.

Thank you.

Answer 1

[Roger Stafford]

You can use 'roots'. For each value of x you can compute the corresponding value of each of the four coefficients of the powers of y. For example the powers of y^3 can be collected to form a single coefficient which would be:

   c3(x) = -12211.07+187137.42*x-935832.82*x^2+1384550.57*x^3

Use a for-loop:

   x = linspace(0.15,0.39,n); % You choose n (= 25?)
   y = zeros(n,3);
   for k = 1:n
     y(k,:) = roots([c3(x(k)),c2(x(k)),c1(x(k)),c0(x(k))]).'; % Get three y roots each trip
   end

This gives you an n x 3 array of the corresponding possible values of y, some of which may be complex-valued, depending on the numbers you are using.

You may wish to reorder the rows in y you obtain this way in order to preserve continuity in the three branches. That would be an extra step.

Answer 2

[MJTHDSN]

Dear Matlabers,

I have a similar question but a little bit confusing. Let`s assume we have 6 equations as below:

EQ1:a{{(L^2)*(Z^2)+(L^2)*(M^2)-2*L*(Z^2)+(Z^2)}}=(L^2)*(T^2)-2*L*(T^2)+ (T^2)-(2*L*T*B)+(T*B)+(B^2)

EQ2: b{{(L^2)*(Z^2)+(L^2)*(M^2)-2*L*(Z^2)+(Z^2)}}=(L^2)*(T^2)+(2*L*T*B)+(B^2)

EQ3: c{{(L^2)*(Z^2)+(L^2)*(M^2)-2*L*(Z^2)+(Z^2)}}=(T^2)+(2*T*B)+(B^2)

EQ4:d{{(L^2)*(Z^2)+(L^2)*(M^2)-2*L*(Z^2)+(Z^2)}}=(L^2)*(T^2)-2*L*(T^2)+ (T^2)-(2*L*T*B)-(T*B)+(B^2)

EQ5: e{{(L^2)*(Z^2)+(L^2)*(M^2)-2*L*(Z^2)+(Z^2)}}=(L^2)*(T^2)-(2*L*T*B)+(B^2)

EQ6: f{{(L^2)*(Z^2)+(L^2)*(M^2)-2*L*(Z^2)+(Z^2)}}=(T^2)-(2*T*B)+(B^2)

in the equations above a,b,c,d,e and f are the numerical known values (0.543 for example). So we have 6 equations with 5 unknowns as L, Z, M, T and B.

Can you please give me cues how to solve the equations to find these unknowns using MATLAB.

Best Regards,

[Walter Roberson]

Generally speaking if you have more equations than unknown, there will typically not be a solution. If you are certain that there is a solution then the implication would tend to be that some of the equations are not independent of each other.

[Walter Roberson]

With all of those ^2, these equations become analytically intractable quite quickly. The solution for the second variable will involve the root of an equation of order 4. If you use the placeholder form of the 4th root then you lose pretty much all ability to continue analysis. If you ask for the explicit value simplified then it takes a long time to do -- the equations become long enough to become impractical do proceed with.

You can use vpasolve() on any 5 of the 6 equations, and then substitute the values into the 6th equation to cross-check. You should expect most of the solutions to be complex valued. vpasolve() is probably only going to find one solution for the 5-equation subset but there should probably be about 32 of them.

[Adithya Valavi]

If you did know how to slove such equations then plz help me to

[Walter Roberson]

If you post your equations, we could try to solve them.

[Walter Roberson]

syms B L M T Z
Q = @(v) sym(v);
a = sqrt(Q(2)); b = Q(3); c = Q(-5); d = 1/Q(7); e = Q(11); f = sqrt(Q(13));

EQ1 = a*(((L^2)*(Z^2)+(L^2)*(M^2)-2*L*(Z^2)+(Z^2)))==(L^2)*(T^2)-2*L*(T^2)+ (T^2)-(2*L*T*B)+(T*B)+(B^2);
EQ2 = b*(((L^2)*(Z^2)+(L^2)*(M^2)-2*L*(Z^2)+(Z^2)))==(L^2)*(T^2)+(2*L*T*B)+(B^2);
EQ3 = c*(((L^2)*(Z^2)+(L^2)*(M^2)-2*L*(Z^2)+(Z^2)))==(T^2)+(2*T*B)+(B^2);
EQ4 = d*(((L^2)*(Z^2)+(L^2)*(M^2)-2*L*(Z^2)+(Z^2)))==(L^2)*(T^2)-2*L*(T^2)+ (T^2)-(2*L*T*B)-(T*B)+(B^2);
EQ5 = e*(((L^2)*(Z^2)+(L^2)*(M^2)-2*L*(Z^2)+(Z^2)))==(L^2)*(T^2)-(2*L*T*B)+(B^2);
EQ6 = f*(((L^2)*(Z^2)+(L^2)*(M^2)-2*L*(Z^2)+(Z^2)))==(T^2)-(2*T*B)+(B^2);
EQN = [EQ1, EQ2, EQ3, EQ4, EQ5, EQ6];
sol = vpasolve(EQN, [B L M T Z]);
temp = struct2cell(sol);
disp(vertcat(temp{:}));

The result turns out to be all-zero.

Answer 3

[Adithya Valavi]

Did you know how to slove these equations then plz tell us tooo