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Analytical solution to polynomial system

0

$begingroup$

I have a polynomial system with three equations and three unknowns i wish to solve analytically. I can obtain a numerical solution easily but for my project i need to find a analytical solution. The system is
$$(fracsqrt32x-frac12y+a)^2+z^2=r_1^2$$
$$(-fracsqrt32x-frac12y+a)^2+z^2=r_2^2$$
$$(y+a)^2+z^2=r_3^2$$

where $a$ and the $r_i$ are constants.

I think the problem can be stated as the intersection between three elliptical cylinders. But I don't know if that helps or how to proceed from there.

systems-of-equations quadratics

share|cite|improve this question

edited Apr 3 at 4:49

Harry Peter

5,53411439

asked Mar 23 at 17:38

GuywithaproblemGuywithaproblem

31

$endgroup$

add a comment | 

0

$begingroup$

I have a polynomial system with three equations and three unknowns i wish to solve analytically. I can obtain a numerical solution easily but for my project i need to find a analytical solution. The system is
$$(fracsqrt32x-frac12y+a)^2+z^2=r_1^2$$
$$(-fracsqrt32x-frac12y+a)^2+z^2=r_2^2$$
$$(y+a)^2+z^2=r_3^2$$

where $a$ and the $r_i$ are constants.

I think the problem can be stated as the intersection between three elliptical cylinders. But I don't know if that helps or how to proceed from there.

systems-of-equations quadratics

share|cite|improve this question

edited Apr 3 at 4:49

Harry Peter

5,53411439

asked Mar 23 at 17:38

GuywithaproblemGuywithaproblem

31

$endgroup$

add a comment | 

$begingroup$

I have a polynomial system with three equations and three unknowns i wish to solve analytically. I can obtain a numerical solution easily but for my project i need to find a analytical solution. The system is
$$(fracsqrt32x-frac12y+a)^2+z^2=r_1^2$$
$$(-fracsqrt32x-frac12y+a)^2+z^2=r_2^2$$
$$(y+a)^2+z^2=r_3^2$$

where $a$ and the $r_i$ are constants.

I think the problem can be stated as the intersection between three elliptical cylinders. But I don't know if that helps or how to proceed from there.

systems-of-equations quadratics

share|cite|improve this question

edited Apr 3 at 4:49

Harry Peter

5,53411439

asked Mar 23 at 17:38

GuywithaproblemGuywithaproblem

31

$endgroup$

share|cite|improve this question

edited Apr 3 at 4:49

Harry Peter

5,53411439

asked Mar 23 at 17:38

GuywithaproblemGuywithaproblem

31

share|cite|improve this question

edited Apr 3 at 4:49

Harry Peter

5,53411439

asked Mar 23 at 17:38

GuywithaproblemGuywithaproblem

31

edited Apr 3 at 4:49

Harry Peter

5,53411439

edited Apr 3 at 4:49

Harry Peter

5,53411439

asked Mar 23 at 17:38

GuywithaproblemGuywithaproblem

31

asked Mar 23 at 17:38

GuywithaproblemGuywithaproblem

31

2 Answers
2

active

oldest

votes

0

$begingroup$

Let us change variables defining
$$X=fracsqrt 32 x qquadqquad Y=frac 12 y -aqquad qquad Z=z^2$$ to make
$$(X-Y)^2+Z=r_1^2tag 1$$
$$(X+Y)^2+Z=r_2^2tag 2$$
$$(2Y+a)^2+Z=r_3^2tag 3$$

Subtracting $(1)$ for $(2)$ gives
$$X=fracr_2^2-r_1^24 Ytag 4$$ and $(3)$ gives
$$Z=r_3^2-(2Y+a)^2tag 5$$ All of that makes
$$48 Y^4+64 a Y^3+8 left(2 a^2+r_1^2+r_2^2-2 r_3^2right)Y^2-(r_1^4+r_2^4-2 r_1^2
r_2^2)=0tag 6$$ that you can solve with radicals (have a look here and have fun !).

You will have at most four real roots but, more than likely, some of them will be discarded because of $(5)$ ($Z$ must be positive).

When solved, go back to $(x,y,z)$.

share|cite|improve this answer

answered Mar 25 at 8:08

Claude LeiboviciClaude Leibovici

125k1158135

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you for the derivation. Now i know how to tackle the problem.
  $endgroup$
  – Guywithaproblem
  Apr 6 at 13:00
  

  
  
  
  

add a comment | 

0

$begingroup$

Hint: With the last equation you will get
$$left(fracsqrt32x-frac12y+aright)^2+r_3^2-(y+a)^2=r_1^2$$
$$left(-fracsqrt32x-frac12y+aright)^2+r_3^2-(y+a)^2=r_2^2$$
Subtracting both equations you will get
$$sqrt3r_1^2-sqrt3r_2^2-6xa+3xy=0$$

share|cite|improve this answer

edited Mar 23 at 18:15

answered Mar 23 at 17:52

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

78.9k42867

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you to all downvoters!!!!!!!!!!!!!!!!!!
  $endgroup$
  – Dr. Sonnhard Graubner
  Mar 23 at 17:58
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You are missing the $x$ inside the first set of parentheses in your first two lines.
  $endgroup$
  – KReiser
  Mar 23 at 18:12
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you for your hint! Just corrected!
  $endgroup$
  – Dr. Sonnhard Graubner
  Mar 23 at 18:13
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Your conclusion is still wrong. Maybe the downvoters had a point?
  $endgroup$
  – KReiser
  Mar 23 at 18:13
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  It looks like your subsequent edit fixed it. And that's an awfully aggressive tone to have here.
  $endgroup$
  – KReiser
  Mar 23 at 18:18
  

  
  
  
  

 | 
show 4 more comments

Your Answer

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0

$begingroup$

Let us change variables defining
$$X=fracsqrt 32 x qquadqquad Y=frac 12 y -aqquad qquad Z=z^2$$ to make
$$(X-Y)^2+Z=r_1^2tag 1$$
$$(X+Y)^2+Z=r_2^2tag 2$$
$$(2Y+a)^2+Z=r_3^2tag 3$$

Subtracting $(1)$ for $(2)$ gives
$$X=fracr_2^2-r_1^24 Ytag 4$$ and $(3)$ gives
$$Z=r_3^2-(2Y+a)^2tag 5$$ All of that makes
$$48 Y^4+64 a Y^3+8 left(2 a^2+r_1^2+r_2^2-2 r_3^2right)Y^2-(r_1^4+r_2^4-2 r_1^2
r_2^2)=0tag 6$$ that you can solve with radicals (have a look here and have fun !).

You will have at most four real roots but, more than likely, some of them will be discarded because of $(5)$ ($Z$ must be positive).

When solved, go back to $(x,y,z)$.

share|cite|improve this answer

answered Mar 25 at 8:08

Claude LeiboviciClaude Leibovici

125k1158135

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you for the derivation. Now i know how to tackle the problem.
  $endgroup$
  – Guywithaproblem
  Apr 6 at 13:00
  

  
  
  
  

add a comment | 

0

$begingroup$

Let us change variables defining
$$X=fracsqrt 32 x qquadqquad Y=frac 12 y -aqquad qquad Z=z^2$$ to make
$$(X-Y)^2+Z=r_1^2tag 1$$
$$(X+Y)^2+Z=r_2^2tag 2$$
$$(2Y+a)^2+Z=r_3^2tag 3$$

Subtracting $(1)$ for $(2)$ gives
$$X=fracr_2^2-r_1^24 Ytag 4$$ and $(3)$ gives
$$Z=r_3^2-(2Y+a)^2tag 5$$ All of that makes
$$48 Y^4+64 a Y^3+8 left(2 a^2+r_1^2+r_2^2-2 r_3^2right)Y^2-(r_1^4+r_2^4-2 r_1^2
r_2^2)=0tag 6$$ that you can solve with radicals (have a look here and have fun !).

You will have at most four real roots but, more than likely, some of them will be discarded because of $(5)$ ($Z$ must be positive).

When solved, go back to $(x,y,z)$.

share|cite|improve this answer

answered Mar 25 at 8:08

Claude LeiboviciClaude Leibovici

125k1158135

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you for the derivation. Now i know how to tackle the problem.
  $endgroup$
  – Guywithaproblem
  Apr 6 at 13:00
  

  
  
  
  

add a comment | 

$begingroup$

Let us change variables defining
$$X=fracsqrt 32 x qquadqquad Y=frac 12 y -aqquad qquad Z=z^2$$ to make
$$(X-Y)^2+Z=r_1^2tag 1$$
$$(X+Y)^2+Z=r_2^2tag 2$$
$$(2Y+a)^2+Z=r_3^2tag 3$$

Subtracting $(1)$ for $(2)$ gives
$$X=fracr_2^2-r_1^24 Ytag 4$$ and $(3)$ gives
$$Z=r_3^2-(2Y+a)^2tag 5$$ All of that makes
$$48 Y^4+64 a Y^3+8 left(2 a^2+r_1^2+r_2^2-2 r_3^2right)Y^2-(r_1^4+r_2^4-2 r_1^2
r_2^2)=0tag 6$$ that you can solve with radicals (have a look here and have fun !).

You will have at most four real roots but, more than likely, some of them will be discarded because of $(5)$ ($Z$ must be positive).

When solved, go back to $(x,y,z)$.

share|cite|improve this answer

answered Mar 25 at 8:08

Claude LeiboviciClaude Leibovici

125k1158135

$endgroup$

share|cite|improve this answer

answered Mar 25 at 8:08

Claude LeiboviciClaude Leibovici

125k1158135

answered Mar 25 at 8:08

Claude LeiboviciClaude Leibovici

125k1158135

answered Mar 25 at 8:08

Claude LeiboviciClaude Leibovici

125k1158135

0

$begingroup$

Hint: With the last equation you will get
$$left(fracsqrt32x-frac12y+aright)^2+r_3^2-(y+a)^2=r_1^2$$
$$left(-fracsqrt32x-frac12y+aright)^2+r_3^2-(y+a)^2=r_2^2$$
Subtracting both equations you will get
$$sqrt3r_1^2-sqrt3r_2^2-6xa+3xy=0$$

share|cite|improve this answer

edited Mar 23 at 18:15

answered Mar 23 at 17:52

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

78.9k42867

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you to all downvoters!!!!!!!!!!!!!!!!!!
  $endgroup$
  – Dr. Sonnhard Graubner
  Mar 23 at 17:58
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You are missing the $x$ inside the first set of parentheses in your first two lines.
  $endgroup$
  – KReiser
  Mar 23 at 18:12
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you for your hint! Just corrected!
  $endgroup$
  – Dr. Sonnhard Graubner
  Mar 23 at 18:13
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Your conclusion is still wrong. Maybe the downvoters had a point?
  $endgroup$
  – KReiser
  Mar 23 at 18:13
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  It looks like your subsequent edit fixed it. And that's an awfully aggressive tone to have here.
  $endgroup$
  – KReiser
  Mar 23 at 18:18
  

  
  
  
  

 | 
show 4 more comments

0

$begingroup$

Hint: With the last equation you will get
$$left(fracsqrt32x-frac12y+aright)^2+r_3^2-(y+a)^2=r_1^2$$
$$left(-fracsqrt32x-frac12y+aright)^2+r_3^2-(y+a)^2=r_2^2$$
Subtracting both equations you will get
$$sqrt3r_1^2-sqrt3r_2^2-6xa+3xy=0$$

share|cite|improve this answer

edited Mar 23 at 18:15

answered Mar 23 at 17:52

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

78.9k42867

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you to all downvoters!!!!!!!!!!!!!!!!!!
  $endgroup$
  – Dr. Sonnhard Graubner
  Mar 23 at 17:58
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You are missing the $x$ inside the first set of parentheses in your first two lines.
  $endgroup$
  – KReiser
  Mar 23 at 18:12
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you for your hint! Just corrected!
  $endgroup$
  – Dr. Sonnhard Graubner
  Mar 23 at 18:13
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Your conclusion is still wrong. Maybe the downvoters had a point?
  $endgroup$
  – KReiser
  Mar 23 at 18:13
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  It looks like your subsequent edit fixed it. And that's an awfully aggressive tone to have here.
  $endgroup$
  – KReiser
  Mar 23 at 18:18
  

  
  
  
  

 | 
show 4 more comments

$begingroup$

Hint: With the last equation you will get
$$left(fracsqrt32x-frac12y+aright)^2+r_3^2-(y+a)^2=r_1^2$$
$$left(-fracsqrt32x-frac12y+aright)^2+r_3^2-(y+a)^2=r_2^2$$
Subtracting both equations you will get
$$sqrt3r_1^2-sqrt3r_2^2-6xa+3xy=0$$

share|cite|improve this answer

edited Mar 23 at 18:15

answered Mar 23 at 17:52

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

78.9k42867

$endgroup$

share|cite|improve this answer

edited Mar 23 at 18:15

answered Mar 23 at 17:52

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

78.9k42867

answered Mar 23 at 17:52

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

78.9k42867

answered Mar 23 at 17:52

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

78.9k42867