Hi guys, I have 2 polynomials, one with known coefficients and one with unknown coefficients. Is there any way in Matlab that I can find the unknown coefficients by equating of coefficients of the like power? For instance, P1=2*s^2+1 and P2=a*s^2+b*s+c, to make P1=P2 we will have a=2,b=0 and c=1. Thanks in advance for the help. Cheers,
MATLAB: How Can I find the Coefficients of a Polynomials by equating Coefficients Method
equating coefficientsMATLABpolynomials
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You DON'T have TWO equations. There is only one you have written.
Given two variables and one equation, there will generally be an infinite set of solutions, a curve in the (k1,k2) plane. Since s is symbolic, the shape of that curve will be a function of s. As it turns out, this is a trivial problem, doable with paper and pencil.
First, note that the s^2 term goes away here. So the problem reduces to:
3*k1 + 3*k2 + (3*s)/2 - 2*k1*k2 + 2*k1*s - 1 = 0Factor out k1 from the terms it appears in and isolate k1 on the left hand side.
k1*(3 - 2*k2 + 2*s) = 1 - 3*k2 - 3*s/2So as long as we do not have
3 - 2*k2 + 2*s = 0then we can solve for k1 as a function of k2 and s. If that denominator is zero, then we have a divide by zero, so a singularity of some sort.
k1 = (1 - 3*k2 - 3*s/2)/(3 - 2*k2 + 2*s)You state that you KNOW a solution. I am sorry, but you do not know a solution, except for a specific value of s!
Lets see what happens when k2 = 13/18.
k1 = @(k2) (1 - 3*k2 - 3*s/2)/(3 - 2*k2 + 2*s);syms sk1(13/18)ans = -((3*s)/2 + 7/6)/(2*s + 14/9)See that we get a function of s still. If s = 0, then indeed we get your answer.
subs(k1(13/18),s,0)ans = -3/4The solution is as I said, a 1-manifold in the (k1,k2) plane, that is parameterized in its shape by s. If you want to put a name on the general curve, I think hyperbola would apply. Since you seem to think that s=0, here is the solution locus for that value.
ezplot(subs(k1,s,0))grid onylabel k_1
Finally, the point where we saw a divide by zero corresponds to a singularity, thus the location of the asymptote (again, it is a function of s.)
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