How to reduce an equation

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Naveen Ramesh
Naveen Ramesh on 8 May 2013
Hi guys,
I am facing a problem in matlab here..
I have an equation that contains the unknown quantity on the right side of the eqaution also.
But somehow, matlab doesnt simplify the equation and i get no result.
I really dont know how to simplify this eqation.
this is my equation
a = (2.s_t)/((((2*s_1)/((a*cos(alpha))-(g*sin(alpha - xi)) - mu((a*sin(alpha))+g*cos(alpha - xi))))^0.5) + (((2*s_1)/((a*cos(beta))+(g*sin(beta+xi))))^0.5))
If anybody could give me a solution, i d be grateful
Thanks once again... !!!!
  3 Comments
Naveen Ramesh
Naveen Ramesh on 10 May 2013
yes.... It is '' 2*s_t''.. and yes... i am trying to solve for 'a'.......

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Answers (2)

Youssef  Khmou
Youssef Khmou on 8 May 2013
hi,
You mean algebraic simplification? you need to use the cos(a+b)/sin(a+b) properties ,
if you mean numerical simplification then break it into parts :
A=2*s_t;
B=(((2*s_1)/((a*cos(alpha))-(g*sin(alpha - xi)) - mu((a*sin(alpha))+g*cos(alpha - xi))))^0.5);
C= (((2*s_1)/((a*cos(beta))+(g*sin(beta+xi))))^0.5);
a=A./(B+C);

Walter Roberson
Walter Roberson on 8 May 2013
If you examine the structure of the equation, and substitute constants for some expressions that are independent of "a", and then bring the denominator of the right side up, you can arrive at a structure of the form
a*(sqrt(s_l/(a*c1+c2))+sqrt(s_l/(a*c3+c4))) = 2*a_l
This can be solve()'d for a, and the answer will be of the form
(s_l-T^2*c2)/(T^2*c1)
where T is
RootOf((-c3*c2^3+c4*c1*c2^2)*z^6 + (4*c4*a_l*c1^2*c2-4*c3*a_l*c1*c2^2)*z^5 + (3*c3*s_l*c2^2-2*c4*c1*s_l*c2-4*c3*a_l^2*c1^2*c2+4*c4*a_l^2*c1^3-s_l*c1*c2^2)*z^4 + (-4*c4*a_l*c1^2*s_l+8*c3*a_l*c1*s_l*c2)*z^3 + (2*s_l^2*c1*c2+4*c3*a_l^2*c1^2*s_l-3*c3*s_l^2*c2+c4*c1*s_l^2)*z^2 - 4*c3*a_l*_Z*c1*s_l^2-s_l^3*c1+c3*s_l^3, z)
Here, Rootof(f(z),z) means the set of values of z such that f(z) is 0 -- i.e., the roots of the 6th order polynomial.
Unfortunately for you, there is no general analytic solution for the roots of a 6th order polynomial.
It is not impossible that substituting back from the original expression into the constants here would happen to give you coefficients that simplified sufficiently to reduce the polynomial degree down to 4, but I highly doubt that to be the case here.
Still, as it is the square of the root of a 6th order that is needed, possibly the expression could be transformed into the roots of a 3rd order; whether that is possible is out of my depths.

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