Solve (a*B) + (c*D) = E without the Symbolic Toolbox

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Michael Garvin
Michael Garvin el 25 de Sept. de 2020
Comentada: Star Strider el 28 de Sept. de 2020
Solve (a*B) + (c*D) = E without the Symbolic Toolbox
where, B, D, & E are all known.
If the Symbolic Toolbox was available it would looke like this:
syms a c
eqn = ((a*B) + (c*D)) / E == 1;
x = solve( eqn );
Any help would be greatly appreciated.
(Available toolboxes include: Image Processing, Signal Processing, & Statistical and Machine Learning

Respuesta aceptada

Star Strider
Star Strider el 25 de Sept. de 2020
This would seem to be homework, and for homework we only give guidance and hints.
I would set it up as an implicit equation (so it equals 0), and use fsolve. To do this, ‘a’ and ‘c’ would have to be parameterized as ‘p(1)’ and ‘p(2)’, and you would have to code it as an anonymous function. .
  10 comentarios
Michael Garvin
Michael Garvin el 28 de Sept. de 2020
I'm needing to find a single ‘A’ & ‘C’ that best fits ‘B’, ‘D’, and ‘E’. I think the ‘\’will work, as described above by Star Strider, but I will definitely look at Ivo Houtzagar's link. Thank you.
Star Strider
Star Strider el 28 de Sept. de 2020
Experiment with something like this:
p = [B(:) D(:)] \ E(:);
a = p(1)
c = p(2)
If I understand correctly what you are doing, that should work.
To also get statistics with the parameter estimates, use the regress or fitlm functions, depending on what you want to do.

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Más respuestas (3)

Walter Roberson
Walter Roberson el 25 de Sept. de 2020
((a*B) + (c*D)) / E == 1
((a*B) + (c*D)) == 1 * E
a*B + c*D == E
a*B == E - c*D
a == (E-c*D) / B
a == E/B - D/B * c
a == (-D/B) * c + (E/B)
Parameterized:
c = t
a = (-D/B) * t + (E/B)
You have one equation in two variables; you are not going to be able to solve for both variables simultaneously.

Ivo Houtzager
Ivo Houtzager el 25 de Sept. de 2020
Editada: Ivo Houtzager el 25 de Sept. de 2020
A = E*pinv([B; D]);
a = A(1);
c = A(2);

Steven Lord
Steven Lord el 26 de Sept. de 2020
This is a generalization of Cleve's simplest impossible problem. Cleve's has B = 1/2, D = 1/2, E = 3.

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