Solving equation in terms of symbols

In summary, the conversation discussed solving an equation in terms of x, y, z, and m, with the desired answer being a function of x, y, and z. The equation was given in MATLAB code and was found to have four solutions for m. The conversation also mentioned using Mathematica to solve the equation, but it produced a large output and may have been due to using an older version. The conversation concluded with the success of finding the desired solution using a different version of MATLAB.
  • #1
quin
50
0
Hi friends
I want to solve an equation which is function of x,y,z,m but i want to find answer like this m=... i want my answer(m) to be a function of x, y, z
but when i solve it in MATLAB or mathematica none of them give me the answer
I know that my equation must have 4 answer for "m" in terms of x , y , z

in code of MATLAB my equation is:

-36 + m* (-64 - 24 *m + m^3) - 16* (2 + m)^2 *cos(x) + 8* cos(2*x) - 28* cos(x - y) + 2 *cos(2 *(x - y)) - 64 *cos(y) - 8* (2* m *(4 + m) + (1 + m)* (7 + m)* cos(x)) *cos(y) + 8 *cos(2*y) - 28* cos(x + y) + 2 *cos(2 *(x + y)) - 64 *cos(z) - 8* (2* m* (4 + m) + (14 + m *(8 + m) + 2 *cos(2*x)) *cos(y) + cos(x) *(14 + m *(8 + m) + 8 *(3 + m) *cos(y) + 2 *cos(2*y))) *cos(z) + 8 *(cos(x) - cos(y))^2 *cos(2*z)==0



i want to solve it and find "m" which is in terms of x,y,z

thank for your guidence
 
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  • #2
But surely you can do it by hand?

$$
-36 + m(-64 - 24 m + m^3) - 16(2 + m)^2 \cos(x) + 8\cos(2x) - 28\cos(x - y)
+ 2\cos(2(x - y)) - 64\cos(y)\\ - 8(2m(4 + m) + (1 + m)(7 + m)\cos(x))\cos(y) + 8\cos(2y) - 28\cos(x + y) + 2\cos(2(x + y)) - 64\cos(z)\\ - 8(2m(4 + m) + (14 + m(8 + m) + 2\cos(2x))\cos(y) + (14 + m(8 + m) + 8(3 + m)\cos(y) + 2\cos(2y))\cos(x))\cos(z)\\ + 8(\cos(x) - \cos(y))^2\cos(2z)=0$$... is basically finding the roots of a quartic in m:
http://en.wikipedia.org/wiki/Quartic_function

... expand the brackets and group the terms.
I know it's painful: use a whiteboard or a large window.
 
  • #3
Mathematica finds the four exact solutions of this in a few seconds.

m/.Solve[-36 + m*(-64 - 24*m + m^3) - 16*(2 + m)^2*Cos[x] + 8*Cos[2*x] - 28*Cos[x - y] + 2*Cos[2*(x - y)] - 64*Cos[y] - 8*(2*m*(4 + m) + (1 + m)*(7 + m)*Cos[x])*Cos[y] + 8*Cos[2*y] - 28*Cos[x + y] + 2*Cos[2*(x + y)] - 64*Cos[z] - 8*(2*m*(4 + m) + (14 + m*(8 + m) + 2*Cos[2*x])*Cos[y] + Cos[x]*(14 + m*(8 + m) + 8*(3 + m)*Cos[y] + 2*Cos[2*y]))*Cos[z] + 8*(Cos[x] - Cos[y])^2*Cos[2*z] == 0, m]

Each is huge and contains a forest of Cos[], but they are the exact solutions.

Just curious, what did you give Mathematica that did not find the solutions?
 
  • #4
Bill Simpson said:
Mathematica finds the four exact solutions of this in a few seconds.

m/.Solve[-36 + m*(-64 - 24*m + m^3) - 16*(2 + m)^2*Cos[x] + 8*Cos[2*x] - 28*Cos[x - y] + 2*Cos[2*(x - y)] - 64*Cos[y] - 8*(2*m*(4 + m) + (1 + m)*(7 + m)*Cos[x])*Cos[y] + 8*Cos[2*y] - 28*Cos[x + y] + 2*Cos[2*(x + y)] - 64*Cos[z] - 8*(2*m*(4 + m) + (14 + m*(8 + m) + 2*Cos[2*x])*Cos[y] + Cos[x]*(14 + m*(8 + m) + 8*(3 + m)*Cos[y] + 2*Cos[2*y]))*Cos[z] + 8*(Cos[x] - Cos[y])^2*Cos[2*z] == 0, m]

Each is huge and contains a forest of Cos[], but they are the exact solutions.

Just curious, what did you give Mathematica that did not find the solutions?

thanks for your help
I used "solve[]" too
and also now I test the sentence that you write above but mathematica said:
"a very larg output was generated"
maybe its reason is that the version of my mathematica is older than yours.

however finally I could reach to my result with other version of Matlab and it quickly gave answer

thank you friend
 
  • #5
quin said:
I test the sentence that you write above but mathematica said:
"a very large output was generated"
maybe its reason is that the version of my mathematica is older than yours.

I have never seen that error before. What version are you using?

$Version

will tell you this.
 
  • #6
Mathematica 7

thanks for your help
 

What is meant by solving an equation in terms of symbols?

Solving an equation in terms of symbols means finding the value of a variable in an equation without using any numerical values. This is done by manipulating the equation using algebraic techniques to isolate the variable on one side of the equation.

Why is it important to solve equations in terms of symbols?

Solving equations in terms of symbols allows us to understand and manipulate mathematical relationships without being limited by specific numerical values. This is especially useful in science and engineering, where variables often represent physical quantities that can vary in different situations.

What are the common techniques used to solve equations in terms of symbols?

The most common techniques used to solve equations in terms of symbols include simplifying expressions, combining like terms, using the distributive property, and isolating the variable by performing inverse operations.

What is the difference between solving equations in terms of symbols and solving for a specific value?

Solving equations in terms of symbols involves finding the general solution for a variable, while solving for a specific value involves finding the numerical value of a variable in a specific situation. Solving in terms of symbols is more general and can be applied to multiple situations, while solving for a specific value is more specific and only applies to one situation.

How do you know if your solution for an equation in terms of symbols is correct?

To check if a solution for an equation in terms of symbols is correct, you can substitute the solution back into the original equation and see if it satisfies the equation. The solution should make the equation true and not result in any contradictions.

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