How Can I find the Coefficients of a Polynomials by equating Coefficients Method?

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,

Answer 1

[Image Analyst]

If you have a bunch of (s,P1) pairs (at least 3) then you can train a polynomial with polyfit (does a least squares) and it will tell you the best values of a, b, and c to fit your training data. If that what you mean? If not, explain why not.

Answer 2

[John D'Errico]

Not too difficult, and I am sure there are various better solutions.

syms s a b c
P1=2*s^2+1;
P2=a*s^2+b*s+c;

C = coeffs(P1 - P2,s);

solve(C(1),c)
ans =
1

solve(C(2),b)
ans =
0

solve(C(3),a)
ans =
2

[Ethen]

Thanks John, your answer is exactly the answer to the question, but I probably should have asked the question in this way: I have 5 polynomials A,B,C,R,S. A,B,C have known coefficients and R&S have unknown coefficients. I have the following equation AR+BS=C. Now I want to find the coefficients. I know that I can use the same approach, but I don't exactly know what combination of unknown and known coefficients I have in AR+BS. What I want to do is to solve the Diaphontine Equation. Once again thanks. Cheers

[John D'Errico]

Why do people never actually state anything like their actual problem? It is Diophantine? Note the spelling. :) So there are integer coefficients here to worry about? Regardless, it simply becomes a set of independent linear Diophantine equations. Extract the coefficients using a similar scheme to what I showed, then solve the equations as appropriate. I'm not sure I see the issue here. Perhaps you can give a better example that shows what is the real problem.

[Ethen]

Sorry about both spelling and not mentioning the real problem. Ok, assume that A=s^2+s-2, B=s^2+s and you have R=s+a, S=bs+c. now a,b and c are the coefficients that need to be found. these values are 3 output of a function and are going to be sent to another function to be used there. So, I can't extract the coefficient every time that A or B or both change. I want to write a code that does the trick without me being involve. thanks.

[John D'Errico]

I was just harassing you about the spelling. But is it truly a Diophantine problem, thus only integer solutions are admitted?

[John D'Errico]

And in your more complete example, what is C, if we are to find solutions for A*R + B*S = C?

For example, if I set C to zero, then no solutions exist at all. Try it.

syms s a b c
A = s^2 + s - 2;
B = s^2 + s;
R = s + a;
S = b*s + c;

A*R + B*S
ans =
(a + s)*(s^2 + s - 2) + (s^2 + s)*(c + b*s)

expand(A*R + B*S)
ans =
a*s - 2*s - 2*a + c*s + a*s^2 + b*s^2 + b*s^3 + c*s^2 + s^2 + s^3

scoeffs = coeffs(expand(A*R + B*S),s)
scoeffs =
[ -2*a, a + c - 2, a + b + c + 1, b + 1]

Here we have a linear system of 4 equations in 3 unknowns if we assume the right hand side © was zero, then careful inspection shows us there is no solution at all. Solve would tell us that too. In general, as you have set it up, most of the time there will be no exact solution to the system you have posed, since it is over-determined.

[Ethen]

C is a polynomial with known coefficients. the degree of c is equal to degree of AR. as long as A and B don't ave a common factor, the equation should have a solution.

[John D'Errico]

Are you positive of that claim?

C = 1 + 2*s + 3*s^2 + 4*s^3;
expand(A*R + B*S - C)
ans =
a*s - 4*s - 2*a + c*s + a*s^2 + b*s^2 + b*s^3 + c*s^2 - 2*s^2 - 3*s^3 - 1

scoeffs = coeffs(expand(A*R + B*S - C),s)
scoeffs =
[ - 2*a - 1, a + c - 4, a + b + c - 2, b - 3]

As we see, a linear system of 4 equations in 3 unknowns, where each of those coefficients must be zero.

I don't even need to use the symbolic TB at this point.

- We must have b = 3 (from the 4th coefficient)
- Likewise, a = -0.5 (from the first)
- So then we know that c = 4 - a = 4.5.

But then a + b + c = 3 - 0.5 + 4.5 = 7, which contradicts what we know from the third coefficient, that the sum must be 2.

So, NO. Sorry, but A and B had no common factor, yet there exists no solution.

Again, the result of this operation is a linear system with 4 equations in 3 unknowns. Linear algebra tells us that there will often be no solution. For SOME SPECIFIC values of C, there may be a solution.

If A,B,C were all given as simple integers, then your statement is correct, we could find integer solutions R and S under the conditions you pose. It seems not to hold for polynomials though.

[Ethen]

Yes I'm sure.I guess I have pointed out by the examples that C and AR are monic.

[John D'Errico]

NO!!!!! This is the very first time you made that claim. In fact, you have NEVER yet even given an example for C. Please be serious. Don't lie about having made a claim that never was, especially when the written record is plain to see.

Regardless, let me try to show you that again, you are wrong. Here is a rather arbitrary C.

C = 4 + 3*s + 2*s^2 + s^3;
scoeffs = coeffs(expand(A*R + B*S - C),s)
scoeffs =
  [ - 2*a - 4, a + c - 5, a + b + c - 1, b]

Clearly, by equating coefficients, b must be zero.

That tells us that both

a + c = 5

and

a + c = 1

You don't think there is some inconsistency here? Maybe it only works on alternate tuesdays.

[Ethen]

John, here is not a place that I wanna lie. in fact I never lie. what I meant was that if you look at A and R, both are monic. Consequently, AR would be monic. this is what I meant. I have a question and looking for the answer to that question, so by lying I just fool myself. I guess we are off the track here. Anyway thanks

[John D'Errico]

The point is, there is NO solution to the problem you have posed for most such problems (which contradicts your claims.) I've even shown how to compute a solution IF one exists, despite what seems to be the moving target of your claims.

Suppose I start with your example, but lets look at whether C can ever be a monic polynomial, as you claim?

syms s a b c
A=s^2+s-2;
B=s^2+s;
R=s+a;
S=b*s+c;

C = expand(A*R + B*S)
C =
  a*s - 2*s - 2*a + c*s + a*s^2 + b*s^2 + b*s^3 + c*s^2 + s^2 + s^3

Ccoeffs = coeffs(C,s)
Ccoeffs =
  [ -2*a, a + c - 2, a + b + c + 1, b + 1]

Hmm. Apparently C is never monic UNLESS b = 0. So if a monic polynomial C is somehow important to you, then the problem reduces to a simpler one.

The point is, I think you don't really know what you want here, or what must happen for a solution to exist. You have just been making random claims, then changing them when I've disproved your claims.

Anyway, lets try a simple version of your problem that actually DOES have a solution. There are some rare cases it might seem. Here I'll arbitrarily construct a problem where

a = 2
b = 0
c = 3

Therefore we would have...

R0 = s + 2;
S0 = 3;

C0 = expand(A*R0 + B*S0)
C0 =
  s^3 + 6*s^2 + 3*s - 4

So, for this simple case, can we recover those coefficient for a,b,c? The problem is now well posed.

scoeffs = coeffs(expand(A*R + B*S - C0),s)
abc = solve(scoeffs)

Warning: 4 equations in 3 variables. 
> In /Applications/MATLAB_R2014a.app/toolbox/symbolic/symbolic/symengine.p>symengine at 56
  In mupadengine.mupadengine>mupadengine.evalin at 97
  In mupadengine.mupadengine>mupadengine.feval at 150
  In solve at 170 
abc = 
    a: [1x1 sym]
    b: [1x1 sym]
    c: [1x1 sym]

abc.a
ans =
2

abc.b
ans =
0

abc.c
ans =
3

So IF a solution exists, what I've showed you is how to solve it (and I showed you that long ago.) But my guess is that soon you will have changed the problem. For example, you called this a Diophantine problem before. Did you really mean that?

[Ethen]

Thanks