# Math Help - solving for 3 unknows, how difficult can that be....

1. ## solving for 3 unknows, how difficult can that be....

I have a problem related to a topic in electronics (filter design) which I have been studying.

Basically, I'm considering 3 unknown variables $a_1, a_2, b_1$ and one constant $c$.

From two different starting points I can workout 3 equations which should be enough to solve for $a_1, a_2, b_1$. However, I'm having difficulties trying to establish/proof the equivalence of the 2 sets of 3 equations.

for instance, from starting point 1, I obtain the following 3 equations

$\\p_1 = \frac{1}{b_1} + a_1 = 3 \frac{c^3+1}{c^3-1} \\ p_2 = \frac{a_1}{b_1} + a_2 = 3 \\ p_3 = \frac{a_2}{b_1} \qquad = \frac{c^3+1}{c^3-1}$

and from starting point 2, I obtain the following 3 equations

$\\ 2(1+q_1+q_2+q_3) &=& c(q_1-q_2-3(q_3-1)) \\ \qquad(1+q_1+q_2+q_3) &=& c^3(q_2-q_1+1-q_3) \\ 2(1+q_1+q_2+q_3) &=& c^2(3(q_3+1)-q_2-q_1)$

where

$\\q_1 = a_1+b_1 \\ q_2 = a_2+a_1b_1 \\ q_3 = a_2b_1$

The two sets of 3 equations are equivalent in the sense that they lead to the same solution for the 3 unknows $a_1, a_2, b_1$ given any $c$, I have verified this numerically.
My problem is that I'm having problems showing this. At first sight it seems that it might be easy to show this, however, after having spend considerable time on this I have started to worry that it might not be so easy.

Any comments or feedback on this problem is highly appreciated.

2. ## Re: solving for 3 unknows, how difficult can that be....

Hi niaren!

Have you tried solving your equations with wolframalpha?
The exact solution of your first set equations can for instance be found like this:
{\frac{1}{z} + x = 3 \frac{c^3+1}{c^3-1}, \frac{x}{z} + y = 3, \frac{y}{z} = \frac{c^3+1}{c^3-1}} - Wolfram|Alpha

(Click on the button "Exact forms" to get the exact solution.)

3. ## Re: solving for 3 unknows, how difficult can that be....

Hi ILikeSerena,

Thank you very much for your feedback. I now have an exact solution

$\\a1 = 2\frac{c^2-1}{c^2+c+1} \\ a2 = \frac{-c+1+c^2}{c^2+c+1} \\ b1 = \frac{c-1}{c+1}$

It turns out (not surprisingly) that there are 3 solutions to both set of equations. However, what surprises me is that only one solution is common to both sets of equations, which is real. That must mean that the two sets of equations are in fact not equal (in the sense of have equal solutions) as I have been trying to proof for quite some time now. They are only equal in the sense of having an equal real solution!

As I tried to convey in the first post I'm not only interested in the solution(s) of the equations but my primary objective is to show in a simpler way that the two sets of equations are equal in the sense of having an equal real solution. I hope this can be achieved without having to derive exact solutions to set of equations and them compare them....

4. ## Re: solving for 3 unknows, how difficult can that be....

Well... your first set of equations contains divisions, while your second set does not.
That has impact on the solutions, since you're not allowed to divide by zero.

Typically when you solve a set of equations, you have to check your solutions in the original equation to check that you have not introduced extra solutions.
Extra solutions are typically introduced when you multiply a denominator away, square the equation, or apply a log rule.

I haven't checked (yet) if this is your problem though.

5. ## Re: solving for 3 unknows, how difficult can that be....

Hi ILikeSerena,

Sorry, I wasn't clear. I don't have a problem with the solutions.
What I would like to show is that the two equations have a common real solution. And I would like to find out if it is possible to show that in a simpler way than having to obtain exact solutions and compare them.

6. ## Re: solving for 3 unknows, how difficult can that be....

I guess I still don't understand.
This is starting to look a bit esoteric.
You're in electronic filter design aren't you?
What you seem to be doing is setting arbitrary solving conditions to yourself (that I don't quite get) and trying to find an exact mathematical proof within those conditions.
Are you?

To check for a common real solution you might simply fill in that real solution and see if it fits both sets of equations...
Or am I misunderstanding yet again?

7. ## Re: solving for 3 unknows, how difficult can that be....

For instance, one idea I have tried is to express the q's in terms of p's but I haven't been successful in doing that. I don't think that is possible.

8. ## Re: solving for 3 unknows, how difficult can that be....

Sorry, ILikeSerena,

You're right. I'm actually looking at a bigger problem. I'm trying to proof that a special class of filters can be expressed (decomposed) as an algebraic sum of two simpler filters. Because I'm not a mathematician, I'm now analyzing a simple case of a 3. order filter and hope to find some pattern that I can generalize to all order filters. For a N'th order filter I can find two sets of equations with N unknowns and N equations. What I have shown in the first post is the two sets of equations the case N=3.