Solve equation without symbolic math toolbox

Hello ,

I'm trying to solve this equation :

0 = (U/r.^3)* sqrt(-2+((3*r.^3)*(Bx/U)))-B

U is a constant

Bx and B are column matrix

I would have as a result a value of r for each value of Bx and B

is it possible to do this without the Symbolic math toolbox ?

Thank you

[mylene]

Move B to the right:

Square both sides:

Let :

Multiply by and rearrange:

[Walter Roberson]

@mylene

This appears to be the same technique described at https://www.mathworks.com/matlabcentral/answers/122212-solve-equation-without-symbolic-math-toolbox#comment_3343869

Answer 1

[Chris C]

You can run this with two functions. The first function I have designated as myMain.m and it is written as..

r0 = rand(3,3);
fsolve(@myfun,r0)

The second function myfun.m is called by the first function by passing initial guesses of r as r0. I altered the way your equation above was written and wrote the function as...

function F = myfun(r)
Bx = rand(3,1);
B = rand(3,1);
U = rand(1);

F = (U)*sqrt(-2+((3*r.^3)*(Bx/U)))-r.^3*B;
end

Change the matrices to whatever numbers you need and it should work.

Good luck.

[Benoit]

Hello,

Thank you for reactivity, I've tried you're solution. Im new to fsolve function so maybe I am missing something but your answer gives me just one solution.

[Benoit]

Ok , i've got it.

Thank you

[David Goodmanson]

Am I missing something here? It looks like the OP from awhile ago now was looking for the solution of

(U/r^3)*sqrt(-2 + 3*r^3*Bx/U) - B = 0

for pairs (Bx, B). Since r only comes in as r^3, set r^3 = a. Square both sides and rearrange to obtain

-2 + 3*a*Bx/U - (B/U)^2*a^2 = 0

Solve this with

a = roots([-(B/U)^2  3*(Bx/U) - 2])

then

 r = a^(1/3).

Since the equation was squared, you have to go back and determine which of the two roots satisfy the original equation and discard the other one. In exchange for that step there is no guessing around with a solver for what size the numerical solution might be.

Presumably if a is real then the real solution for r is the one that is wanted. If a is complex, then that is the way it is.

Answer 2

[Sudesh]

can solve this numerically without the Symbolic Math Toolbox. Since you want a value of rrr for each pair of BxBxBx and BBB, you can use numerical root-finding methods like fzero in MATLAB. Here’s a step-by-step guide:

Your equation is:

You want to solve for rrr, given UUU, BxBxBx (a column vector), and BBB (a column vector of the same size).

MATLAB Approach:

U = 1; % Example value
Bx = [0.5; 1.2; 0.8]; % Example column vector
B = [0.1; 0.2; 0.15]; % Example column vector

r_values = zeros(size(Bx)); % Preallocate

for k = 1:length(Bx)
    func = @(r) (U./r.^3).*sqrt(-2 + (3*r.^3)*(Bx(k)/U)) - B(k);
    % Provide an initial guess for r, say 0.5
    r_guess = 0.5;
    % Solve numerically
    r_values(k) = fzero(func, r_guess);
end

disp(r_values)

Notes:

1. Initial guess: fzero requires a starting point. You might need to adjust it depending on the expected range of r.
2. Vectorization: Each r depends on a pair (Bx(k), B(k)), so a for loop is appropriate.

Answer 3

[mylene]

% Example data U = 1; Bx = [1.5; 2.0; 2.5]; % Column vector B = [1.1; 1.2; 1.3]; % Column vector

% Define quadratic coefficients: c2*a^2 + c1*a + c0 = 0 c2 = B.^2; c1 = -3 * U * Bx; c0 = 2 * U^2;

% Solving for r for each row r_results = zeros(length(B), 2); % Preallocate for two possible roots

for i = 1:length(B) % Find roots of the quadratic for a (which is r^3) a_roots = roots([c2(i), c1(i), c0(i)]);

    % Convert back to r
    r_results(i, :) = a_roots.^(1/3);
end