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Hyper-simultaneous equation

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0

1

$begingroup$

I am new here and of course I would like to be specific. I've found a general solution to the following system:

If $a$, $b$ and $c$ are given such that:
$$
x+y+z=a
$$
$$
x²+y²+z²=b
$$
$$
xⁿ+yⁿ+zⁿ=c
$$

Where $n=8$.

$x$, $y$ and $z$ could be devised using the method I have derived.

I just want to basically ask if this kind of thing is publishable. Thanks for your reply in anticipation.

polynomials

share|cite|improve this question

edited Mar 13 at 17:12

Lateef Suaib

asked Mar 13 at 16:30

Lateef SuaibLateef Suaib

12

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Publishable it might as well be. Unless there is something super special about the method itself, I would not expect to publish in a journal, but rather on a conference paper. Note that MathOverflow (instead of MathStackExchange) might be more suitable for research-level mathematics.
  $endgroup$
  – Mefitico
  Mar 13 at 16:41
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks for this sir. I will take that into account. But pls, how do I switch to mathoverflow?
  $endgroup$
  – Lateef Suaib
  Mar 13 at 17:08
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Here you go!
  $endgroup$
  – Mefitico
  Mar 13 at 17:15
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks sir! I really appreciate your response.
  $endgroup$
  – Lateef Suaib
  Mar 13 at 18:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  For me to not be too delusional about my method, please, looking at the system, can it be easily solved without much stress?
  $endgroup$
  – Lateef Suaib
  Mar 13 at 18:36
  

  
  
  
  

add a comment | 

0

1

$begingroup$

I am new here and of course I would like to be specific. I've found a general solution to the following system:

If $a$, $b$ and $c$ are given such that:
$$
x+y+z=a
$$
$$
x²+y²+z²=b
$$
$$
xⁿ+yⁿ+zⁿ=c
$$

Where $n=8$.

$x$, $y$ and $z$ could be devised using the method I have derived.

I just want to basically ask if this kind of thing is publishable. Thanks for your reply in anticipation.

polynomials

share|cite|improve this question

edited Mar 13 at 17:12

Lateef Suaib

asked Mar 13 at 16:30

Lateef SuaibLateef Suaib

12

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Publishable it might as well be. Unless there is something super special about the method itself, I would not expect to publish in a journal, but rather on a conference paper. Note that MathOverflow (instead of MathStackExchange) might be more suitable for research-level mathematics.
  $endgroup$
  – Mefitico
  Mar 13 at 16:41
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks for this sir. I will take that into account. But pls, how do I switch to mathoverflow?
  $endgroup$
  – Lateef Suaib
  Mar 13 at 17:08
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Here you go!
  $endgroup$
  – Mefitico
  Mar 13 at 17:15
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks sir! I really appreciate your response.
  $endgroup$
  – Lateef Suaib
  Mar 13 at 18:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  For me to not be too delusional about my method, please, looking at the system, can it be easily solved without much stress?
  $endgroup$
  – Lateef Suaib
  Mar 13 at 18:36
  

  
  
  
  

add a comment | 

$begingroup$

I am new here and of course I would like to be specific. I've found a general solution to the following system:

If $a$, $b$ and $c$ are given such that:
$$
x+y+z=a
$$
$$
x²+y²+z²=b
$$
$$
xⁿ+yⁿ+zⁿ=c
$$

Where $n=8$.

$x$, $y$ and $z$ could be devised using the method I have derived.

I just want to basically ask if this kind of thing is publishable. Thanks for your reply in anticipation.

polynomials

share|cite|improve this question

edited Mar 13 at 17:12

Lateef Suaib

asked Mar 13 at 16:30

Lateef SuaibLateef Suaib

12

$endgroup$

share|cite|improve this question

edited Mar 13 at 17:12

Lateef Suaib

asked Mar 13 at 16:30

Lateef SuaibLateef Suaib

12

share|cite|improve this question

edited Mar 13 at 17:12

Lateef Suaib

asked Mar 13 at 16:30

Lateef SuaibLateef Suaib

12

asked Mar 13 at 16:30

Lateef SuaibLateef Suaib

12

asked Mar 13 at 16:30

Lateef SuaibLateef Suaib

12

2 Answers
2

active

oldest

votes

0

$begingroup$

I'm doubtful that this would be publishable unless the method is particularly neat or generalizable; it's not hard to find a method using matrices. Explicitly:

If a 3 by 3 matrix $A$ has eigenvalues $x, y, z$, then $A^8$ has eigenvalues $x^8, y^8, z^8$. Therefore, $tr(A^8) = x^8 + y^8 + z^8$. We can choose $A$ to be the companion matrix for a polynomial $x^3 - l x^2 - mx - n$. Then

$$x + y + z = tr(A) = l$$

$$x^2 + y^2 + z^2 = tr(A^2) = l^2 + 2m$$

$$x^8 + y^8 + z^8 = tr(A^8) = l^8 + 8 l^6 m + 20 l^4 m^2 + 16 l^2 m^3 + 2 m^4 + 8 l^5 n + 32 l^3 m n + 24 l m^2 n + 12 l^2 n^2 + 8 m n^2$$

So we set $l = a, m = frac{b - a^2}{2}$, and solve the third equation to find $n$ (it's quadratic in $n$, so the solution won't be too difficult). Then $x, y, z$ are the eigenvalues of $A$, which can be determined by solving the above cubic.

share|cite|improve this answer

answered Mar 13 at 19:43

user44191user44191

1032

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  To reduce the eighth powers to third powers only requires solving quadratic equation and i have confirmed that if the problem is reduced to that level, the new problem which involves linear, quadratic and cubic equations can be solved without much stress. If that is the case, do u still think I have something interesting at hand?
  $endgroup$
  – Lateef Suaib
  Mar 16 at 8:44
  

  
  
  
  

add a comment | 

0

$begingroup$

See Newton-Girard's formulas for a more complete explanation of this technique.

Let:
$$
S_n =sum_{k=1}^l x_i ^n
$$
$$
sigma_1 = S_1
$$
$$
sigma_2 = sum_{ineq j} x_i x_j
$$
$$
sigma_3 = sum_{ineq j neq k} x_i x_j x_k
$$

Then:
$$
S_2= sigma_1^2-2sigma_2
$$

$$
S_n = sigma_1 S_{n-1}-sigma_2S_{n-2}+sigma_3 S_{n-3}-cdots
$$

Thus in your case, we have:
$$
a=sigma_1
$$
$$
b= S_2
$$
$$
c = S_8
$$

Note that if we have $sigma_1$, $sigma_2$ and $sigma_3$, then $x$, $y$ and $z$ are the roots of the equation:

$$
lambda^3-sigma_1 lambda^2+sigma_2 lambda-sigma_3=0
$$

Then $sigma_2$ can be computed almost directly:

$$
sigma_2 = frac{a^2-b}{2}
$$

It remains to find $sigma_3$, but it can be computed be recurring the formula until $S_8$ is written as a function of $sigma_1$, $sigma_2$ and $S_2$. It will likely yield an equation similar to the one @user44191 presented.

Basically, if you have an analytic method that does not incur into solving a third or higher degree polynomial, you might have something interesting in hand.

share|cite|improve this answer

answered Mar 14 at 13:28

MefiticoMefitico

1,111218

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  To reduce the eighth powers to third powers only requires quadrac
  $endgroup$
  – Lateef Suaib
  Mar 16 at 8:37
  

  
  
  
  

add a comment | 

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0

$begingroup$

I'm doubtful that this would be publishable unless the method is particularly neat or generalizable; it's not hard to find a method using matrices. Explicitly:

If a 3 by 3 matrix $A$ has eigenvalues $x, y, z$, then $A^8$ has eigenvalues $x^8, y^8, z^8$. Therefore, $tr(A^8) = x^8 + y^8 + z^8$. We can choose $A$ to be the companion matrix for a polynomial $x^3 - l x^2 - mx - n$. Then

$$x + y + z = tr(A) = l$$

$$x^2 + y^2 + z^2 = tr(A^2) = l^2 + 2m$$

$$x^8 + y^8 + z^8 = tr(A^8) = l^8 + 8 l^6 m + 20 l^4 m^2 + 16 l^2 m^3 + 2 m^4 + 8 l^5 n + 32 l^3 m n + 24 l m^2 n + 12 l^2 n^2 + 8 m n^2$$

So we set $l = a, m = frac{b - a^2}{2}$, and solve the third equation to find $n$ (it's quadratic in $n$, so the solution won't be too difficult). Then $x, y, z$ are the eigenvalues of $A$, which can be determined by solving the above cubic.

share|cite|improve this answer

answered Mar 13 at 19:43

user44191user44191

1032

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  To reduce the eighth powers to third powers only requires solving quadratic equation and i have confirmed that if the problem is reduced to that level, the new problem which involves linear, quadratic and cubic equations can be solved without much stress. If that is the case, do u still think I have something interesting at hand?
  $endgroup$
  – Lateef Suaib
  Mar 16 at 8:44
  

  
  
  
  

add a comment | 

0

$begingroup$

I'm doubtful that this would be publishable unless the method is particularly neat or generalizable; it's not hard to find a method using matrices. Explicitly:

If a 3 by 3 matrix $A$ has eigenvalues $x, y, z$, then $A^8$ has eigenvalues $x^8, y^8, z^8$. Therefore, $tr(A^8) = x^8 + y^8 + z^8$. We can choose $A$ to be the companion matrix for a polynomial $x^3 - l x^2 - mx - n$. Then

$$x + y + z = tr(A) = l$$

$$x^2 + y^2 + z^2 = tr(A^2) = l^2 + 2m$$

$$x^8 + y^8 + z^8 = tr(A^8) = l^8 + 8 l^6 m + 20 l^4 m^2 + 16 l^2 m^3 + 2 m^4 + 8 l^5 n + 32 l^3 m n + 24 l m^2 n + 12 l^2 n^2 + 8 m n^2$$

So we set $l = a, m = frac{b - a^2}{2}$, and solve the third equation to find $n$ (it's quadratic in $n$, so the solution won't be too difficult). Then $x, y, z$ are the eigenvalues of $A$, which can be determined by solving the above cubic.

share|cite|improve this answer

answered Mar 13 at 19:43

user44191user44191

1032

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  To reduce the eighth powers to third powers only requires solving quadratic equation and i have confirmed that if the problem is reduced to that level, the new problem which involves linear, quadratic and cubic equations can be solved without much stress. If that is the case, do u still think I have something interesting at hand?
  $endgroup$
  – Lateef Suaib
  Mar 16 at 8:44
  

  
  
  
  

add a comment | 

$begingroup$

I'm doubtful that this would be publishable unless the method is particularly neat or generalizable; it's not hard to find a method using matrices. Explicitly:

If a 3 by 3 matrix $A$ has eigenvalues $x, y, z$, then $A^8$ has eigenvalues $x^8, y^8, z^8$. Therefore, $tr(A^8) = x^8 + y^8 + z^8$. We can choose $A$ to be the companion matrix for a polynomial $x^3 - l x^2 - mx - n$. Then

$$x + y + z = tr(A) = l$$

$$x^2 + y^2 + z^2 = tr(A^2) = l^2 + 2m$$

$$x^8 + y^8 + z^8 = tr(A^8) = l^8 + 8 l^6 m + 20 l^4 m^2 + 16 l^2 m^3 + 2 m^4 + 8 l^5 n + 32 l^3 m n + 24 l m^2 n + 12 l^2 n^2 + 8 m n^2$$

So we set $l = a, m = frac{b - a^2}{2}$, and solve the third equation to find $n$ (it's quadratic in $n$, so the solution won't be too difficult). Then $x, y, z$ are the eigenvalues of $A$, which can be determined by solving the above cubic.

share|cite|improve this answer

answered Mar 13 at 19:43

user44191user44191

1032

$endgroup$

share|cite|improve this answer

answered Mar 13 at 19:43

user44191user44191

1032

answered Mar 13 at 19:43

user44191user44191

1032

answered Mar 13 at 19:43

user44191user44191

1032

0

$begingroup$

See Newton-Girard's formulas for a more complete explanation of this technique.

Let:
$$
S_n =sum_{k=1}^l x_i ^n
$$
$$
sigma_1 = S_1
$$
$$
sigma_2 = sum_{ineq j} x_i x_j
$$
$$
sigma_3 = sum_{ineq j neq k} x_i x_j x_k
$$

Then:
$$
S_2= sigma_1^2-2sigma_2
$$

$$
S_n = sigma_1 S_{n-1}-sigma_2S_{n-2}+sigma_3 S_{n-3}-cdots
$$

Thus in your case, we have:
$$
a=sigma_1
$$
$$
b= S_2
$$
$$
c = S_8
$$

Note that if we have $sigma_1$, $sigma_2$ and $sigma_3$, then $x$, $y$ and $z$ are the roots of the equation:

$$
lambda^3-sigma_1 lambda^2+sigma_2 lambda-sigma_3=0
$$

Then $sigma_2$ can be computed almost directly:

$$
sigma_2 = frac{a^2-b}{2}
$$

It remains to find $sigma_3$, but it can be computed be recurring the formula until $S_8$ is written as a function of $sigma_1$, $sigma_2$ and $S_2$. It will likely yield an equation similar to the one @user44191 presented.

Basically, if you have an analytic method that does not incur into solving a third or higher degree polynomial, you might have something interesting in hand.

share|cite|improve this answer

answered Mar 14 at 13:28

MefiticoMefitico

1,111218

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  To reduce the eighth powers to third powers only requires quadrac
  $endgroup$
  – Lateef Suaib
  Mar 16 at 8:37
  

  
  
  
  

add a comment | 

0

$begingroup$

See Newton-Girard's formulas for a more complete explanation of this technique.

Let:
$$
S_n =sum_{k=1}^l x_i ^n
$$
$$
sigma_1 = S_1
$$
$$
sigma_2 = sum_{ineq j} x_i x_j
$$
$$
sigma_3 = sum_{ineq j neq k} x_i x_j x_k
$$

Then:
$$
S_2= sigma_1^2-2sigma_2
$$

$$
S_n = sigma_1 S_{n-1}-sigma_2S_{n-2}+sigma_3 S_{n-3}-cdots
$$

Thus in your case, we have:
$$
a=sigma_1
$$
$$
b= S_2
$$
$$
c = S_8
$$

Note that if we have $sigma_1$, $sigma_2$ and $sigma_3$, then $x$, $y$ and $z$ are the roots of the equation:

$$
lambda^3-sigma_1 lambda^2+sigma_2 lambda-sigma_3=0
$$

Then $sigma_2$ can be computed almost directly:

$$
sigma_2 = frac{a^2-b}{2}
$$

It remains to find $sigma_3$, but it can be computed be recurring the formula until $S_8$ is written as a function of $sigma_1$, $sigma_2$ and $S_2$. It will likely yield an equation similar to the one @user44191 presented.

Basically, if you have an analytic method that does not incur into solving a third or higher degree polynomial, you might have something interesting in hand.

share|cite|improve this answer

answered Mar 14 at 13:28

MefiticoMefitico

1,111218

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  To reduce the eighth powers to third powers only requires quadrac
  $endgroup$
  – Lateef Suaib
  Mar 16 at 8:37
  

  
  
  
  

add a comment | 

$begingroup$

See Newton-Girard's formulas for a more complete explanation of this technique.

Let:
$$
S_n =sum_{k=1}^l x_i ^n
$$
$$
sigma_1 = S_1
$$
$$
sigma_2 = sum_{ineq j} x_i x_j
$$
$$
sigma_3 = sum_{ineq j neq k} x_i x_j x_k
$$

Then:
$$
S_2= sigma_1^2-2sigma_2
$$

$$
S_n = sigma_1 S_{n-1}-sigma_2S_{n-2}+sigma_3 S_{n-3}-cdots
$$

Thus in your case, we have:
$$
a=sigma_1
$$
$$
b= S_2
$$
$$
c = S_8
$$

Note that if we have $sigma_1$, $sigma_2$ and $sigma_3$, then $x$, $y$ and $z$ are the roots of the equation:

$$
lambda^3-sigma_1 lambda^2+sigma_2 lambda-sigma_3=0
$$

Then $sigma_2$ can be computed almost directly:

$$
sigma_2 = frac{a^2-b}{2}
$$

It remains to find $sigma_3$, but it can be computed be recurring the formula until $S_8$ is written as a function of $sigma_1$, $sigma_2$ and $S_2$. It will likely yield an equation similar to the one @user44191 presented.

Basically, if you have an analytic method that does not incur into solving a third or higher degree polynomial, you might have something interesting in hand.

share|cite|improve this answer

answered Mar 14 at 13:28

MefiticoMefitico

1,111218

$endgroup$

share|cite|improve this answer

answered Mar 14 at 13:28

MefiticoMefitico

1,111218

answered Mar 14 at 13:28

MefiticoMefitico

1,111218

answered Mar 14 at 13:28

MefiticoMefitico

1,111218