Non-linear system of 3 equations
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I have a system of 3 equations with the unknown $(x,y,z)$ to solve in the space $mathbb{R}^2 times mathbb{R}_+^{*}$ :
$$m=x-y \ v=x^2(z-1) \gamma=x^3(z^3+3z-2)$$
with $m in mathbb{R}$, $v>0$ , $gamma in mathbb{R}$ and $z>1$
Numerically, I easily find the solution with a solver, but I cannot figure out whether a closed-form formula of the solution exists.
Thanks.
real-analysis linear-algebra systems-of-equations
asked Nov 21 at 15:01
Canardini
2,4651519
add a comment |
up vote
0
down vote
favorite
I have a system of 3 equations with the unknown $(x,y,z)$ to solve in the space $mathbb{R}^2 times mathbb{R}_+^{*}$ :
$$m=x-y \ v=x^2(z-1) \gamma=x^3(z^3+3z-2)$$
with $m in mathbb{R}$, $v>0$ , $gamma in mathbb{R}$ and $z>1$
Numerically, I easily find the solution with a solver, but I cannot figure out whether a closed-form formula of the solution exists.
Thanks.
real-analysis linear-algebra systems-of-equations
asked Nov 21 at 15:01
Canardini
2,4651519
1
It looks like you can treat the two equations in $x,z$ as a separate pair and solve those, then finish by solving the first. Have you attempted to factor $z^3+3z-2$?
– abiessu
Nov 21 at 15:08
1
Actually, $z>1$, I tried but I end up solving $$gamma=(z^3+3z-2)*(frac{v}{z-1})^{frac{3}{2}}$$
– Canardini
Nov 21 at 15:31
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I have a system of 3 equations with the unknown $(x,y,z)$ to solve in the space $mathbb{R}^2 times mathbb{R}_+^{*}$ :
$$m=x-y \ v=x^2(z-1) \gamma=x^3(z^3+3z-2)$$
with $m in mathbb{R}$, $v>0$ , $gamma in mathbb{R}$ and $z>1$
Numerically, I easily find the solution with a solver, but I cannot figure out whether a closed-form formula of the solution exists.
Thanks.
real-analysis linear-algebra systems-of-equations
asked Nov 21 at 15:01
Canardini
2,4651519
I have a system of 3 equations with the unknown $(x,y,z)$ to solve in the space $mathbb{R}^2 times mathbb{R}_+^{*}$ :
$$m=x-y \ v=x^2(z-1) \gamma=x^3(z^3+3z-2)$$
with $m in mathbb{R}$, $v>0$ , $gamma in mathbb{R}$ and $z>1$
Numerically, I easily find the solution with a solver, but I cannot figure out whether a closed-form formula of the solution exists.
Thanks.
real-analysis linear-algebra systems-of-equations
real-analysis linear-algebra systems-of-equations
asked Nov 21 at 15:01
Canardini
2,4651519
asked Nov 21 at 15:01
Canardini
2,4651519
edited Nov 21 at 15:30
asked Nov 21 at 15:01
Canardini
2,4651519
asked Nov 21 at 15:01
Canardini
2,4651519
asked Nov 21 at 15:01
Canardini
2,4651519
2,4651519
1
It looks like you can treat the two equations in $x,z$ as a separate pair and solve those, then finish by solving the first. Have you attempted to factor $z^3+3z-2$?
– abiessu
Nov 21 at 15:08
1
Actually, $z>1$, I tried but I end up solving $$gamma=(z^3+3z-2)*(frac{v}{z-1})^{frac{3}{2}}$$
– Canardini
Nov 21 at 15:31
add a comment |
1
It looks like you can treat the two equations in $x,z$ as a separate pair and solve those, then finish by solving the first. Have you attempted to factor $z^3+3z-2$?
– abiessu
Nov 21 at 15:08
1
Actually, $z>1$, I tried but I end up solving $$gamma=(z^3+3z-2)*(frac{v}{z-1})^{frac{3}{2}}$$
– Canardini
Nov 21 at 15:31
1
1
It looks like you can treat the two equations in $x,z$ as a separate pair and solve those, then finish by solving the first. Have you attempted to factor $z^3+3z-2$?
– abiessu
Nov 21 at 15:08
It looks like you can treat the two equations in $x,z$ as a separate pair and solve those, then finish by solving the first. Have you attempted to factor $z^3+3z-2$?
– abiessu
Nov 21 at 15:08
1
1
Actually, $z>1$, I tried but I end up solving $$gamma=(z^3+3z-2)*(frac{v}{z-1})^{frac{3}{2}}$$
– Canardini
Nov 21 at 15:31
Actually, $z>1$, I tried but I end up solving $$gamma=(z^3+3z-2)*(frac{v}{z-1})^{frac{3}{2}}$$
– Canardini
Nov 21 at 15:31
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
down vote
A standard way to solve the last two equations, would be to square the last and cube the other to solve for $z$ (expect when $x=0$, but in this case $y=-m$ and $z$ can be anything):
$$
v³(z³+3z-2)²-gamma^2(z-1)³ = 0
$$
This gives a sixth order polynomial that you still have to check for extraneous solutions (as we have squared and lost the sign of $gamma$ and $x$). But than, using the second, we get:
$$
x = pmsqrt{frac{v}{z-1}}
$$
Which also requires $z>1$ for a solution, and finally using the first one:
$$
y = x - m
$$
answered Nov 21 at 15:32
Eduardo Elael
1865
Yes, exactly, but how do I solve the first equation ? I feel anyhow I need to you use a numerical solver .
– Canardini
Nov 21 at 15:35
It is not possible if you cannot reduce the degree of the polynomial with some cleaver trick, which does not seems to be possible as your coefficients depends on the parameters.
– Eduardo Elael
Nov 21 at 15:50
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
A standard way to solve the last two equations, would be to square the last and cube the other to solve for $z$ (expect when $x=0$, but in this case $y=-m$ and $z$ can be anything):
$$
v³(z³+3z-2)²-gamma^2(z-1)³ = 0
$$
This gives a sixth order polynomial that you still have to check for extraneous solutions (as we have squared and lost the sign of $gamma$ and $x$). But than, using the second, we get:
$$
x = pmsqrt{frac{v}{z-1}}
$$
Which also requires $z>1$ for a solution, and finally using the first one:
$$
y = x - m
$$
answered Nov 21 at 15:32
Eduardo Elael
1865
Yes, exactly, but how do I solve the first equation ? I feel anyhow I need to you use a numerical solver .
– Canardini
Nov 21 at 15:35
It is not possible if you cannot reduce the degree of the polynomial with some cleaver trick, which does not seems to be possible as your coefficients depends on the parameters.
– Eduardo Elael
Nov 21 at 15:50
add a comment |
up vote
1
down vote
A standard way to solve the last two equations, would be to square the last and cube the other to solve for $z$ (expect when $x=0$, but in this case $y=-m$ and $z$ can be anything):
$$
v³(z³+3z-2)²-gamma^2(z-1)³ = 0
$$
This gives a sixth order polynomial that you still have to check for extraneous solutions (as we have squared and lost the sign of $gamma$ and $x$). But than, using the second, we get:
$$
x = pmsqrt{frac{v}{z-1}}
$$
Which also requires $z>1$ for a solution, and finally using the first one:
$$
y = x - m
$$
answered Nov 21 at 15:32
Eduardo Elael
1865
Yes, exactly, but how do I solve the first equation ? I feel anyhow I need to you use a numerical solver .
– Canardini
Nov 21 at 15:35
It is not possible if you cannot reduce the degree of the polynomial with some cleaver trick, which does not seems to be possible as your coefficients depends on the parameters.
– Eduardo Elael
Nov 21 at 15:50
add a comment |
up vote
1
down vote
up vote
1
down vote
A standard way to solve the last two equations, would be to square the last and cube the other to solve for $z$ (expect when $x=0$, but in this case $y=-m$ and $z$ can be anything):
$$
v³(z³+3z-2)²-gamma^2(z-1)³ = 0
$$
This gives a sixth order polynomial that you still have to check for extraneous solutions (as we have squared and lost the sign of $gamma$ and $x$). But than, using the second, we get:
$$
x = pmsqrt{frac{v}{z-1}}
$$
Which also requires $z>1$ for a solution, and finally using the first one:
$$
y = x - m
$$
answered Nov 21 at 15:32
Eduardo Elael
1865
A standard way to solve the last two equations, would be to square the last and cube the other to solve for $z$ (expect when $x=0$, but in this case $y=-m$ and $z$ can be anything):
$$
v³(z³+3z-2)²-gamma^2(z-1)³ = 0
$$
This gives a sixth order polynomial that you still have to check for extraneous solutions (as we have squared and lost the sign of $gamma$ and $x$). But than, using the second, we get:
$$
x = pmsqrt{frac{v}{z-1}}
$$
Which also requires $z>1$ for a solution, and finally using the first one:
$$
y = x - m
$$
answered Nov 21 at 15:32
Eduardo Elael
1865
answered Nov 21 at 15:32
Eduardo Elael
1865
answered Nov 21 at 15:32
Eduardo Elael
1865
answered Nov 21 at 15:32
Eduardo Elael
1865
1865
Yes, exactly, but how do I solve the first equation ? I feel anyhow I need to you use a numerical solver .
– Canardini
Nov 21 at 15:35
It is not possible if you cannot reduce the degree of the polynomial with some cleaver trick, which does not seems to be possible as your coefficients depends on the parameters.
– Eduardo Elael
Nov 21 at 15:50
add a comment |
Yes, exactly, but how do I solve the first equation ? I feel anyhow I need to you use a numerical solver .
– Canardini
Nov 21 at 15:35
It is not possible if you cannot reduce the degree of the polynomial with some cleaver trick, which does not seems to be possible as your coefficients depends on the parameters.
– Eduardo Elael
Nov 21 at 15:50
Yes, exactly, but how do I solve the first equation ? I feel anyhow I need to you use a numerical solver .
– Canardini
Nov 21 at 15:35
Yes, exactly, but how do I solve the first equation ? I feel anyhow I need to you use a numerical solver .
– Canardini
Nov 21 at 15:35
It is not possible if you cannot reduce the degree of the polynomial with some cleaver trick, which does not seems to be possible as your coefficients depends on the parameters.
– Eduardo Elael
Nov 21 at 15:50
It is not possible if you cannot reduce the degree of the polynomial with some cleaver trick, which does not seems to be possible as your coefficients depends on the parameters.
– Eduardo Elael
Nov 21 at 15:50
add a comment |
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1
It looks like you can treat the two equations in $x,z$ as a separate pair and solve those, then finish by solving the first. Have you attempted to factor $z^3+3z-2$?
– abiessu
Nov 21 at 15:08
1
Actually, $z>1$, I tried but I end up solving $$gamma=(z^3+3z-2)*(frac{v}{z-1})^{frac{3}{2}}$$
– Canardini
Nov 21 at 15:31