Consider the sum of two factorized polynomials:
k1(s+a1)...(s+am) + k2(s+b1)...(s+bn)
is there some nice theorem telling us something about the roots of this sum? A theorem that tells us all the roots algebraically would be grand slam.
k1(s+a1)...(s+am) + k2(s+b1)...(s+bn)
is there some nice theorem telling us something about the roots of this sum? A theorem that tells us all the roots algebraically would be grand slam.
Christer said:
Yes, it is called MATLAB 😉
Alternatively, there is a cool freeware program called Maxima that might be able to help out.
http://wxmaxima.sourceforge.net/wiki/index.php/Main_Page
Re: Re: Roots of sum of factorized polynomials?
I actually used to have Maxima installed, but haven't yet reinstalled since my last sytem reinstall. I admit I didn't think of that possibility, but I must say I question whether either Maxima or Matlab could present a general algebraic solution unless there already are some useful theorems they can apply. Oh well, maybe I should get Maxima up and running and see if can do it.
BWRX said:
I actually used to have Maxima installed, but haven't yet reinstalled since my last sytem reinstall. I admit I didn't think of that possibility, but I must say I question whether either Maxima or Matlab could present a general algebraic solution unless there already are some useful theorems they can apply. Oh well, maybe I should get Maxima up and running and see if can do it.
Cal Weldon said:
I'm afraid the white bear should have found itself 10km to the west of where it started unless it walks another 10km to the east...🙂
jacco vermeulen said:
What do you mean I don't know? 🙂
I don't know of any theorems of the kind I was asking for, but hope there might be something of the kind, despite me not knowing it. 🙂
I have tried Maxima now, and it seems not possible to do what I was asking, or at least I can't see any way of doing it. It seems not to like constant symbols, but want numerical values for them, in order to even accept the polynomial for finding roots. And for the general problem that I stated, I am not sure there is even a way to enter the problem unless we fix m and n.
I am affraid, that if order is higher than 5 algebraic solutions may not exist. Numerical solutions are possible.
Consider this: If there are any roots,
0 = k1(s+a1)...(s+am) + k2(s+b1)...(s+bn)
then
k1(s+a1)...(s+am) = -k2(s+b1)...(s+bn)
Assume m<n
Then if s = -ai then left side =0 and therefore one of the bi = -ai
we can then write
k1(s+a1)...(s+am) = k2(s-a1)...(s-am) - k2(s+b[m+1])...(s+bn)
Don't know if that gets us anywhere...
davce
0 = k1(s+a1)...(s+am) + k2(s+b1)...(s+bn)
then
k1(s+a1)...(s+am) = -k2(s+b1)...(s+bn)
Assume m<n
Then if s = -ai then left side =0 and therefore one of the bi = -ai
we can then write
k1(s+a1)...(s+am) = k2(s-a1)...(s-am) - k2(s+b[m+1])...(s+bn)
Don't know if that gets us anywhere...
davce
Lets consider m=n=1
then from above (if i didn't screw that up)
k1(s+a1)+k2(s-a1)=0
(k1+k2)s +(k1-k2)a1 =0
s = (k2-k1)a1 / (k1+k2)
dave
then from above (if i didn't screw that up)
k1(s+a1)+k2(s-a1)=0
(k1+k2)s +(k1-k2)a1 =0
s = (k2-k1)a1 / (k1+k2)
dave
Lets consider m=1 n=2
then
k1(s+a1)+k2(s-a1)(s+b2)=0
k1s+k1a1+k2(s^2+(b2-a1)s - a1b2)=0
k2s^2 + (k1+k2(b2-a1))s +a1(k1-k2b2) =0
There is a formula for that.
dave
then
k1(s+a1)+k2(s-a1)(s+b2)=0
k1s+k1a1+k2(s^2+(b2-a1)s - a1b2)=0
k2s^2 + (k1+k2(b2-a1))s +a1(k1-k2b2) =0
There is a formula for that.
dave
darkfenriz said:
If it were about finding roots from scratch for a polynomial, yes most probably, and the general solutions are very messy already at degree 3. However, since we already have the roots (or assume we have them) for the two polynomials constituting the sum, I was hoping there might be some interesting relationship between the roots of these original polynomials and the roots of the new polyniomial (the sum).
Re: Re: Roots of sum of factorized polynomials?
Very clever Dave, but I am afraid it is not of much help. While we can rewrite the equation as
k1(s+a1)...(s+am) = -k2(s+b1)...(s+bn)
there is no guarantee that either side is 0 for any root of the sum. In fact it is easy to find counterexamples. Consider eg. the the two one-degree polynomials (s+1) and (s-3). Their sum (s+1)+(s-3) has the root s = 1, but neither of the original polynomials is zero for this value of s.
planet10 said:
Very clever Dave, but I am afraid it is not of much help. While we can rewrite the equation as
k1(s+a1)...(s+am) = -k2(s+b1)...(s+bn)
there is no guarantee that either side is 0 for any root of the sum. In fact it is easy to find counterexamples. Consider eg. the the two one-degree polynomials (s+1) and (s-3). Their sum (s+1)+(s-3) has the root s = 1, but neither of the original polynomials is zero for this value of s.
I don't think any such formula exists.
If you add polynomials, factorized or not, you are finding new roots from scratch.
Check this example:
P1(x)=x^2+ 2x +1
P2(x)=x^2+2x - 3
you know the roots of both, but if you subtact them P1(x)-P2(x) there are no longer any roots.
Just an example...
If you add polynomials, factorized or not, you are finding new roots from scratch.
Check this example:
P1(x)=x^2+ 2x +1
P2(x)=x^2+2x - 3
you know the roots of both, but if you subtact them P1(x)-P2(x) there are no longer any roots.
Just an example...
Blues said:
The bear is back where he started and the only way that can happen is if he started at one of the poles. The bear is white.
Darkfenriz,
Yes, you are right there need not be any roots at all. But that doesn't rule out useful theoreoms about the roots, when they exist. Thetheorems just have to be conditional or partial, covering some cases. Besides, even though I stated the problem as a general one, my interest in this is for cases where the two polynomials express the poles and zeroes respectively of an impedance/transfer function etc. which puts certain restrictions on the problem. I don't think something like your example can happen, for instance.
Yes, you are right there need not be any roots at all. But that doesn't rule out useful theoreoms about the roots, when they exist. Thetheorems just have to be conditional or partial, covering some cases. Besides, even though I stated the problem as a general one, my interest in this is for cases where the two polynomials express the poles and zeroes respectively of an impedance/transfer function etc. which puts certain restrictions on the problem. I don't think something like your example can happen, for instance.
you could take a geometric pole/zero view
seperate and then divide through to get
A(s)/B(s) = -1
then you could plot the | A(s)/B(s) | = 1 contours
and the Arg( A(s)/B(s) ) = pi isolines
their intersections are the roots of the summed polynomials, less the common roots which cancel in the division - of course the common roots are roots of the final sum too
computation wize this may not be a win but it should help in visualizing the relation of the roots of the sum to the starting polynomial's individual roots
there are likely lots of conformal mapping/compelx analysis ideas that come in to play that I've mostly never learned or forgot
seperate and then divide through to get
A(s)/B(s) = -1
then you could plot the | A(s)/B(s) | = 1 contours
and the Arg( A(s)/B(s) ) = pi isolines
their intersections are the roots of the summed polynomials, less the common roots which cancel in the division - of course the common roots are roots of the final sum too
computation wize this may not be a win but it should help in visualizing the relation of the roots of the sum to the starting polynomial's individual roots
there are likely lots of conformal mapping/compelx analysis ideas that come in to play that I've mostly never learned or forgot
And there came the Nyquist diagrams and complex analysis. 
Oh well, I suppose I asked for it. Maybe I should read up a bit on that and see if I understand more now than 25 years ago. I am not sure how much it helps though, unless I have actual polynomials, but maybe experimenting with various polynomials gives some insight.
Thanks everybody for your efforts to help. Maybe there isn't much general knowledge of the kind I had hoped for. Meanwhile, I think I have started to get some better understanding of the problem where this came up, so maybe I can do without an answer to my question. In case anybody wonders, the question came up when considering a feedback loop containing an unknown impedance and what one can know about or do to guarantee stability.
Oh well, I suppose I asked for it. Maybe I should read up a bit on that and see if I understand more now than 25 years ago. I am not sure how much it helps though, unless I have actual polynomials, but maybe experimenting with various polynomials gives some insight.
Thanks everybody for your efforts to help. Maybe there isn't much general knowledge of the kind I had hoped for. Meanwhile, I think I have started to get some better understanding of the problem where this came up, so maybe I can do without an answer to my question. In case anybody wonders, the question came up when considering a feedback loop containing an unknown impedance and what one can know about or do to guarantee stability.
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