Solving a large equation

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Nicholas on 2 Sep 2014
Commented: Walter Roberson on 25 Mar 2020
I have a large equation of the form:
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.

Answers (3)

Roger Stafford
Roger Stafford on 2 Sep 2014
Edited: Roger Stafford on 2 Sep 2014
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
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.

MJTHDSN on 12 Apr 2018
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
Walter Roberson on 25 Mar 2020
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);
The result turns out to be all-zero.

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Adithya Valavi
Adithya Valavi on 24 Mar 2020
Did you know how to slove these equations then plz tell us tooo

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