Factorize reciprocal polynomial 4th-order
I try to factorize any polynomial like :
$x^4 + a.x^3 + b.x^2 + a.x + 1$ with $ a, b inBbbR$
into :
$(x^2 + c.x + d)(x^2 + e.x + f)$ with $ c, d, e, f inBbbR$
I also want $c(a, b)$, $d(a, b)$, $e(a, b)$, $f(a, b)$ to be continous, so I can be smooth at runtime when changing $(a, b)$.
The only way I succeeded for now is to compute all the roots and regroup them by conjugate pairs. But it's tricky because roots can be paired in many way, there are also cases where there are many solutions (when all roots are real), and I noticed roots may swap for specific values of $(a, b)$.
I would like to now if there is a simpler known method for this typical polynom.
Many thanks !
polynomials
asked Nov 12 at 10:15
user2443456
424
add a comment |
I try to factorize any polynomial like :
$x^4 + a.x^3 + b.x^2 + a.x + 1$ with $ a, b inBbbR$
into :
$(x^2 + c.x + d)(x^2 + e.x + f)$ with $ c, d, e, f inBbbR$
I also want $c(a, b)$, $d(a, b)$, $e(a, b)$, $f(a, b)$ to be continous, so I can be smooth at runtime when changing $(a, b)$.
The only way I succeeded for now is to compute all the roots and regroup them by conjugate pairs. But it's tricky because roots can be paired in many way, there are also cases where there are many solutions (when all roots are real), and I noticed roots may swap for specific values of $(a, b)$.
I would like to now if there is a simpler known method for this typical polynom.
Many thanks !
polynomials
asked Nov 12 at 10:15
user2443456
424
Please use MathJax to type the equations. This way they will be much more understandable.
– Matti P.
Nov 12 at 10:20
Thanks, edited.
– user2443456
Nov 12 at 10:35
Well, just opening the parentheses in the second equation, we get $$ beginarrayrcc e + c &=& a \ f+d+ce &=& b \ cf + de &=& a \ df &=& 1 endarray $$
– Matti P.
Nov 12 at 10:42
add a comment |
I try to factorize any polynomial like :
$x^4 + a.x^3 + b.x^2 + a.x + 1$ with $ a, b inBbbR$
into :
$(x^2 + c.x + d)(x^2 + e.x + f)$ with $ c, d, e, f inBbbR$
I also want $c(a, b)$, $d(a, b)$, $e(a, b)$, $f(a, b)$ to be continous, so I can be smooth at runtime when changing $(a, b)$.
The only way I succeeded for now is to compute all the roots and regroup them by conjugate pairs. But it's tricky because roots can be paired in many way, there are also cases where there are many solutions (when all roots are real), and I noticed roots may swap for specific values of $(a, b)$.
I would like to now if there is a simpler known method for this typical polynom.
Many thanks !
polynomials
asked Nov 12 at 10:15
user2443456
424
I try to factorize any polynomial like :
$x^4 + a.x^3 + b.x^2 + a.x + 1$ with $ a, b inBbbR$
into :
$(x^2 + c.x + d)(x^2 + e.x + f)$ with $ c, d, e, f inBbbR$
I also want $c(a, b)$, $d(a, b)$, $e(a, b)$, $f(a, b)$ to be continous, so I can be smooth at runtime when changing $(a, b)$.
The only way I succeeded for now is to compute all the roots and regroup them by conjugate pairs. But it's tricky because roots can be paired in many way, there are also cases where there are many solutions (when all roots are real), and I noticed roots may swap for specific values of $(a, b)$.
I would like to now if there is a simpler known method for this typical polynom.
Many thanks !
polynomials
polynomials
asked Nov 12 at 10:15
user2443456
424
asked Nov 12 at 10:15
user2443456
424
edited Nov 12 at 10:32
asked Nov 12 at 10:15
user2443456
424
asked Nov 12 at 10:15
user2443456
424
asked Nov 12 at 10:15
user2443456
424
424
Please use MathJax to type the equations. This way they will be much more understandable.
– Matti P.
Nov 12 at 10:20
Thanks, edited.
– user2443456
Nov 12 at 10:35
Well, just opening the parentheses in the second equation, we get $$ beginarrayrcc e + c &=& a \ f+d+ce &=& b \ cf + de &=& a \ df &=& 1 endarray $$
– Matti P.
Nov 12 at 10:42
add a comment |
Please use MathJax to type the equations. This way they will be much more understandable.
– Matti P.
Nov 12 at 10:20
Thanks, edited.
– user2443456
Nov 12 at 10:35
Well, just opening the parentheses in the second equation, we get $$ beginarrayrcc e + c &=& a \ f+d+ce &=& b \ cf + de &=& a \ df &=& 1 endarray $$
– Matti P.
Nov 12 at 10:42
Please use MathJax to type the equations. This way they will be much more understandable.
– Matti P.
Nov 12 at 10:20
Please use MathJax to type the equations. This way they will be much more understandable.
– Matti P.
Nov 12 at 10:20
Thanks, edited.
– user2443456
Nov 12 at 10:35
Thanks, edited.
– user2443456
Nov 12 at 10:35
Well, just opening the parentheses in the second equation, we get $$ beginarrayrcc e + c &=& a \ f+d+ce &=& b \ cf + de &=& a \ df &=& 1 endarray $$
– Matti P.
Nov 12 at 10:42
Well, just opening the parentheses in the second equation, we get $$ beginarrayrcc e + c &=& a \ f+d+ce &=& b \ cf + de &=& a \ df &=& 1 endarray $$
– Matti P.
Nov 12 at 10:42
add a comment |
2 Answers
2
active
oldest
votes
As you pointed out yourself, the polynomial is palindromic (=equal to its own reciprocal). This means that $1/alpha$ is a zero whenever $alpha$ is.
In the case of four real roots this suggests using factors like $(x+alpha)(x+1/alpha)$. Such a factor has constant term $=1$, so you will have $d=f=1$.
This reduces your system of equations to $c+e=a$, $2+ce=b$. Meaning that $c$ and $e$ are the roots of the quadratic
$$
(T-c)(T-e)=T^2-(c+e)T+ce=T^2-aT+(b-2).
$$
The roots of this quadratic are real whenever we have the inequality
$$
a^2>4(b-2),
$$
when
$$
c,e=fracapmsqrta^2-4b+82.
$$
If this equality is not satisfied then you can proceed as follows.
- Solve for $c$ and $e$ anyway (quadratic formula).
- Find the zeros of $x^2+cx+1$ and $x^2+ex+1$ (quadratic formula again).
- Match the complex zeros into conjugate pairs (one from each of the above pairs of solutions).
- Build the factors.
Of course, when you have four real zeros you have many options with the factorization. The one I suggested is just easy to find, and the coefficients $c,e(,d,f)$ are easy to write down as functions of $a$ and $b$.
answered Nov 12 at 11:03
Jyrki Lahtonen
108k12166367
1
Thanks, Also, for the case where the equality is not satisfied, you can just calculate one root and get the 3 others with conjugate/inversion of this one.
– user2443456
Nov 13 at 11:11
add a comment |
Write $$x^4+a x^3+b x^2+a x+1-left(x^2+c x+dright) left(x^2+e x+fright)=0$$ Expand and group terms to get
$$(1-d f)+x (a-c f-d e)+x^2 (b-c e-d-f)+x^3 (a-c-e)=0$$ Set each coefficient equal to $0$ and you should get four nasty equations in $c,d,e,f$
answered Nov 12 at 10:44
Claude Leibovici
118k1157132
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
As you pointed out yourself, the polynomial is palindromic (=equal to its own reciprocal). This means that $1/alpha$ is a zero whenever $alpha$ is.
In the case of four real roots this suggests using factors like $(x+alpha)(x+1/alpha)$. Such a factor has constant term $=1$, so you will have $d=f=1$.
This reduces your system of equations to $c+e=a$, $2+ce=b$. Meaning that $c$ and $e$ are the roots of the quadratic
$$
(T-c)(T-e)=T^2-(c+e)T+ce=T^2-aT+(b-2).
$$
The roots of this quadratic are real whenever we have the inequality
$$
a^2>4(b-2),
$$
when
$$
c,e=fracapmsqrta^2-4b+82.
$$
If this equality is not satisfied then you can proceed as follows.
- Solve for $c$ and $e$ anyway (quadratic formula).
- Find the zeros of $x^2+cx+1$ and $x^2+ex+1$ (quadratic formula again).
- Match the complex zeros into conjugate pairs (one from each of the above pairs of solutions).
- Build the factors.
Of course, when you have four real zeros you have many options with the factorization. The one I suggested is just easy to find, and the coefficients $c,e(,d,f)$ are easy to write down as functions of $a$ and $b$.
answered Nov 12 at 11:03
Jyrki Lahtonen
108k12166367
1
Thanks, Also, for the case where the equality is not satisfied, you can just calculate one root and get the 3 others with conjugate/inversion of this one.
– user2443456
Nov 13 at 11:11
add a comment |
As you pointed out yourself, the polynomial is palindromic (=equal to its own reciprocal). This means that $1/alpha$ is a zero whenever $alpha$ is.
In the case of four real roots this suggests using factors like $(x+alpha)(x+1/alpha)$. Such a factor has constant term $=1$, so you will have $d=f=1$.
This reduces your system of equations to $c+e=a$, $2+ce=b$. Meaning that $c$ and $e$ are the roots of the quadratic
$$
(T-c)(T-e)=T^2-(c+e)T+ce=T^2-aT+(b-2).
$$
The roots of this quadratic are real whenever we have the inequality
$$
a^2>4(b-2),
$$
when
$$
c,e=fracapmsqrta^2-4b+82.
$$
If this equality is not satisfied then you can proceed as follows.
- Solve for $c$ and $e$ anyway (quadratic formula).
- Find the zeros of $x^2+cx+1$ and $x^2+ex+1$ (quadratic formula again).
- Match the complex zeros into conjugate pairs (one from each of the above pairs of solutions).
- Build the factors.
Of course, when you have four real zeros you have many options with the factorization. The one I suggested is just easy to find, and the coefficients $c,e(,d,f)$ are easy to write down as functions of $a$ and $b$.
answered Nov 12 at 11:03
Jyrki Lahtonen
108k12166367
1
Thanks, Also, for the case where the equality is not satisfied, you can just calculate one root and get the 3 others with conjugate/inversion of this one.
– user2443456
Nov 13 at 11:11
add a comment |
As you pointed out yourself, the polynomial is palindromic (=equal to its own reciprocal). This means that $1/alpha$ is a zero whenever $alpha$ is.
In the case of four real roots this suggests using factors like $(x+alpha)(x+1/alpha)$. Such a factor has constant term $=1$, so you will have $d=f=1$.
This reduces your system of equations to $c+e=a$, $2+ce=b$. Meaning that $c$ and $e$ are the roots of the quadratic
$$
(T-c)(T-e)=T^2-(c+e)T+ce=T^2-aT+(b-2).
$$
The roots of this quadratic are real whenever we have the inequality
$$
a^2>4(b-2),
$$
when
$$
c,e=fracapmsqrta^2-4b+82.
$$
If this equality is not satisfied then you can proceed as follows.
- Solve for $c$ and $e$ anyway (quadratic formula).
- Find the zeros of $x^2+cx+1$ and $x^2+ex+1$ (quadratic formula again).
- Match the complex zeros into conjugate pairs (one from each of the above pairs of solutions).
- Build the factors.
Of course, when you have four real zeros you have many options with the factorization. The one I suggested is just easy to find, and the coefficients $c,e(,d,f)$ are easy to write down as functions of $a$ and $b$.
answered Nov 12 at 11:03
Jyrki Lahtonen
108k12166367
As you pointed out yourself, the polynomial is palindromic (=equal to its own reciprocal). This means that $1/alpha$ is a zero whenever $alpha$ is.
In the case of four real roots this suggests using factors like $(x+alpha)(x+1/alpha)$. Such a factor has constant term $=1$, so you will have $d=f=1$.
This reduces your system of equations to $c+e=a$, $2+ce=b$. Meaning that $c$ and $e$ are the roots of the quadratic
$$
(T-c)(T-e)=T^2-(c+e)T+ce=T^2-aT+(b-2).
$$
The roots of this quadratic are real whenever we have the inequality
$$
a^2>4(b-2),
$$
when
$$
c,e=fracapmsqrta^2-4b+82.
$$
If this equality is not satisfied then you can proceed as follows.
- Solve for $c$ and $e$ anyway (quadratic formula).
- Find the zeros of $x^2+cx+1$ and $x^2+ex+1$ (quadratic formula again).
- Match the complex zeros into conjugate pairs (one from each of the above pairs of solutions).
- Build the factors.
Of course, when you have four real zeros you have many options with the factorization. The one I suggested is just easy to find, and the coefficients $c,e(,d,f)$ are easy to write down as functions of $a$ and $b$.
answered Nov 12 at 11:03
Jyrki Lahtonen
108k12166367
edited Nov 15 at 5:16
answered Nov 12 at 11:03
Jyrki Lahtonen
108k12166367
answered Nov 12 at 11:03
Jyrki Lahtonen
108k12166367
answered Nov 12 at 11:03
Jyrki Lahtonen
108k12166367
108k12166367
1
Thanks, Also, for the case where the equality is not satisfied, you can just calculate one root and get the 3 others with conjugate/inversion of this one.
– user2443456
Nov 13 at 11:11
add a comment |
1
Thanks, Also, for the case where the equality is not satisfied, you can just calculate one root and get the 3 others with conjugate/inversion of this one.
– user2443456
Nov 13 at 11:11
1
1
Thanks, Also, for the case where the equality is not satisfied, you can just calculate one root and get the 3 others with conjugate/inversion of this one.
– user2443456
Nov 13 at 11:11
Thanks, Also, for the case where the equality is not satisfied, you can just calculate one root and get the 3 others with conjugate/inversion of this one.
– user2443456
Nov 13 at 11:11
add a comment |
Write $$x^4+a x^3+b x^2+a x+1-left(x^2+c x+dright) left(x^2+e x+fright)=0$$ Expand and group terms to get
$$(1-d f)+x (a-c f-d e)+x^2 (b-c e-d-f)+x^3 (a-c-e)=0$$ Set each coefficient equal to $0$ and you should get four nasty equations in $c,d,e,f$
answered Nov 12 at 10:44
Claude Leibovici
118k1157132
add a comment |
Write $$x^4+a x^3+b x^2+a x+1-left(x^2+c x+dright) left(x^2+e x+fright)=0$$ Expand and group terms to get
$$(1-d f)+x (a-c f-d e)+x^2 (b-c e-d-f)+x^3 (a-c-e)=0$$ Set each coefficient equal to $0$ and you should get four nasty equations in $c,d,e,f$
answered Nov 12 at 10:44
Claude Leibovici
118k1157132
add a comment |
Write $$x^4+a x^3+b x^2+a x+1-left(x^2+c x+dright) left(x^2+e x+fright)=0$$ Expand and group terms to get
$$(1-d f)+x (a-c f-d e)+x^2 (b-c e-d-f)+x^3 (a-c-e)=0$$ Set each coefficient equal to $0$ and you should get four nasty equations in $c,d,e,f$
answered Nov 12 at 10:44
Claude Leibovici
118k1157132
Write $$x^4+a x^3+b x^2+a x+1-left(x^2+c x+dright) left(x^2+e x+fright)=0$$ Expand and group terms to get
$$(1-d f)+x (a-c f-d e)+x^2 (b-c e-d-f)+x^3 (a-c-e)=0$$ Set each coefficient equal to $0$ and you should get four nasty equations in $c,d,e,f$
answered Nov 12 at 10:44
Claude Leibovici
118k1157132
answered Nov 12 at 10:44
Claude Leibovici
118k1157132
answered Nov 12 at 10:44
Claude Leibovici
118k1157132
answered Nov 12 at 10:44
Claude Leibovici
118k1157132
118k1157132
add a comment |
add a comment |
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Please use MathJax to type the equations. This way they will be much more understandable.
– Matti P.
Nov 12 at 10:20
Thanks, edited.
– user2443456
Nov 12 at 10:35
Well, just opening the parentheses in the second equation, we get $$ beginarrayrcc e + c &=& a \ f+d+ce &=& b \ cf + de &=& a \ df &=& 1 endarray $$
– Matti P.
Nov 12 at 10:42