Solve System of Equations: 2z+z2x=x, 2y+y2z=z, 2x+x2y=y

[anemone]

Solve for real solutions of the following system:

$2z+z^2x=x$

$2y+y^2z=z$

$2x+x^2y=y$

I know, this isn't a very hard problem, thus, we should compete in terms of how many minutes (or hours) that we used to come up with a good solution. I said good but not smart because to me:

$\text{all workable solutions, be them short or long}=\text{good solutions}$ :p

 

[mente oscura]

Solve for real solutions of the following system:

$2z+z^2x=x$

$2y+y^2z=z$

$2x+x^2y=y$

I know, this isn't a very hard problem, thus, we should compete in terms of how many minutes (or hours) that we used to come up with a good solution. I said good but not smart because to me:

$\text{all workable solutions, be them short or long}=\text{good solutions}$ :p

Hello.

Since you ask for a solution, and I'm so lazy.(Sun)

Spoiler

$$x=y=z=0$$

(Whew)

Regards.

 

[anemone]

Hello.

Since you ask for a solution, and I'm so lazy.(Sun)

Regards.

There are several ways to overcome laziness:

1.	Figure out the real issue.
2.	Focus on the actual problem.
3.	Get organized.
4.	Tell yourself you can do it.
5.	Taking action: Start, Take your time, Give yourself pep talks, Ask for help/hint when you need it, Compliment yourself every step of the way, Staying motivated, Know that the work is worth it, Stay on track, and most importantly, NEVER GIVE UP!

(Tongueout)(Sun)

 

[anemone]

Spoiler

$$x=y=z=0$$

I forgot to mention that there are more answers to the problem, mente!(Emo)

 

[Opalg]

Solve for real solutions of the following system:

$2z+z^2x=x$

$2y+y^2z=z$

$2x+x^2y=y$

[sp]If $x=\pm1$ then the third equation becomes $\pm2 + y = y$, which clearly has no solution. So we may assume that $x\ne\pm1$, and similarly for the other variables.

Then we can write the equations as $$x = \frac{2z}{1-z^2}, \qquad z = \frac{2y}{1-y^2}, \qquad y = \frac{2x}{1-x^2}.$$ Hence each of $x,y,z$ belongs to a cycle of length 3 in the discrete dynamical system given by the function $f(x) = \dfrac{2x}{1-x^2}.$ So we need to calculate $$f(f(x)) = \frac{\frac{4x}{1-x^2}}{1 - \Bigl(\frac{2x}{1-x^2}\Bigr)^{\!2}} = \frac{4x(1-x^2)}{1-6x^2+x^4},$$ and $$x = f(f(f(x))) = \frac{\frac{8x(1-x^2)}{1-6x^2+x^4}}{1 - \Bigl(\frac{4x(1-x^2)}{1-6x^2+x^4}\Bigr)^{\!2}} = \frac{8x(1-x^2)(1-6x^2+x^4)}{(1-6x^2+x^4)^2 - 16x^2(1-x^2)^2}.$$ That gives the equation $8x(1-x^2)(1-6x^2+x^4) = x\bigl((1-6x^2+x^4)^2 - 16x^2(1-x^2)^2\bigr)$. One solution is $x=0$. After dividing through by $x$ you can multiply out and rearrange the equation as $x^8 - 20x^6 + 14x^4 + 28x^2 - 7 = 0$. The left side factorises as $(x^2+1)(x^6 - 21x^4 + 35x^2 - 7).$ Ignoring the first factor and writing $w=x^2$, we have to solve the cubic equation $w^3 - 21w^2 + 35w - 7 = 0.$ At this point, I turned (for the first and I hope last time in my life) to Wolfram Alpha, which gave the solutions as $19.19567\ldots$, $1.572417\ldots$, $0.231914\ldots$. Taking square roots, we get the solutions for $x$, namely $\pm4.381$, $\pm1.254$, $\pm0.482$ (to 3 decimal places). Together with $0$, that gives seven solutions for $x$, and in each case $y$ and $z$ then come from the equations $y=f(x)$, $z=f(y)$.[/sp]

I know, this isn't a very hard problem, thus, we should compete in terms of how many minutes (or hours) that we used to come up with a good solution.

Not very hard? Maybe I'm being dumb (which happens often enough), but it took me quite a long time to see how to do it and then to plough through the calculations.

 

[mathbalarka]

I don't really see how this is an easy problem. Easy in the sense that all can be done by simple substitution, right, but I don't think any smart solution can be derived. A few pointers might help someone come up with a smart one (Opalg's is not really smart, but indeed that is what I'd do) :

The simultaneous equation there satisfies cyclic relationship, i.e., they all are left invariant by the three cycles $(1 2 3)$ and $(1 3 2)$. The conjugates are thus formed by all the 3 2-cycles. Indeed, the Galois group of the cubic on Opalg's solution is actually cyclic.

My approach now would be two sum all the original equations up as well as doing the same for the conjugates. Now sum all these up to get a 3-varaible equation in terms of $x + y + z$, $xy + yz + xz$ and $xyz$, which can be derived by breaking up the symmetric powersum polynomials in terms of elementary symmetric polynomials, which are just these.

The other 2 equations attached to these can be derived the same way, considering the other elementary symmetric polynomials instead. The breakdown is that all the equations are of cubic degree, but in the case, original one way quadratic degree. I am not sure if that'd help at all, but thinking of what resolvent do, I think it might be of some use. Perhaps someone can go through all these and derive a "smart" solution?

 

[kaliprasad]

[sp]If $x=\pm1$ then the third equation becomes $\pm2 + y = y$, which clearly has no solution. So we may assume that $x\ne\pm1$, and similarly for the other variables.

Then we can write the equations as $$x = \frac{2z}{1-z^2}, \qquad z = \frac{2y}{1-y^2}, \qquad y = \frac{2x}{1-x^2}.$$ Hence each of $x,y,z$ belongs to a cycle of length 3 in the discrete dynamical system given by the function $f(x) = \dfrac{2x}{1-x^2}.$ So we need to calculate $$f(f(x)) = \frac{\frac{4x}{1-x^2}}{1 - \Bigl(\frac{2x}{1-x^2}\Bigr)^{\!2}} = \frac{4x(1-x^2)}{1-6x^2+x^4},$$ and $$x = f(f(f(x))) = \frac{\frac{8x(1-x^2)}{1-6x^2+x^4}}{1 - \Bigl(\frac{4x(1-x^2)}{1-6x^2+x^4}\Bigr)^{\!2}} = \frac{8x(1-x^2)(1-6x^2+x^4)}{(1-6x^2+x^4)^2 - 16x^2(1-x^2)^2}.$$ That gives the equation $8x(1-x^2)(1-6x^2+x^4) = x\bigl((1-6x^2+x^4)^2 - 16x^2(1-x^2)^2\bigr)$. One solution is $x=0$. After dividing through by $x$ you can multiply out and rearrange the equation as $x^8 - 20x^6 + 14x^4 + 28x^2 - 7 = 0$. The left side factorises as $(x^2+1)(x^6 - 21x^4 + 35x^2 - 7).$ Ignoring the first factor and writing $w=x^2$, we have to solve the cubic equation $w^3 - 21w^2 + 35w - 7 = 0.$ At this point, I turned (for the first and I hope last time in my life) to Wolfram Alpha, which gave the solutions as $19.19567\ldots$, $1.572417\ldots$, $0.231914\ldots$. Taking square roots, we get the solutions for $x$, namely $\pm4.381$, $\pm1.254$, $\pm0.482$ (to 3 decimal places). Together with $0$, that gives seven solutions for $x$, and in each case $y$ and $z$ then come from the equations $y=f(x)$, $z=f(y)$.[/sp]Not very hard? Maybe I'm being dumb (which happens often enough), but it took me quite a long time to see how to do it and then to plough through the calculations.

Spoiler

Opalg has found

x = 2z/(1-$z^2$)

y = 2y/(1-$y^2$)

z = 2x/(1-$x^2$)letting z = tan a we get x = tan 2a, y = tan 4a, z = tan 8a

or tan 8a = tan a

so 8a = npi + a ( n= 0 to 6)

or 7a = npi

so solution set

z= tan (npi/7), x = tan (2npi/7), y = tan (4npi/7)

for n = 0 to 6

.

 

[mathbalarka]

(Speechless) চমৎকার!

Oops, spilled some of my native language at the look of that. (Wasntme)

 

[Opalg]

Spoiler

Opalg has found

x = 2z/(1-$z^2$)

y = 2y/(1-$y^2$)

z = 2x/(1-$x^2$)letting z = tan a we get x = tan 2a, y = tan 4a, z = tan 8a

or tan 8a = tan a

so 8a = npi + a ( n= 0 to 6)

or 7a = npi

so solution set

z= tan (npi/7), x = tan (2npi/7), y = tan (4npi/7)

for n = 0 to 6

.

Excellent! (Bow) (why didn't I think of that?).

 

[anemone]

Not very hard? Maybe I'm being dumb (which happens often enough), but it took me quite a long time to see how to do it and then to plough through the calculations.

I think I've put my foot in my mouth by saying that problem is easy, because I thought everyone would think of solving the given equation using trigonometric approach, after they rewritten any of the given equations to make either $x$, $y$ or $z$ the subject.

I hope no one will take it amiss that I am trying to brag here because I am not. I am just a silly girl who loves to solve some challenging problems.

Last but not least, thanks for participating, Opalg!:)

Spoiler

Opalg has found

x = 2z/(1-$z^2$)

y = 2y/(1-$y^2$)

z = 2x/(1-$x^2$)letting z = tan a we get x = tan 2a, y = tan 4a, z = tan 8a

or tan 8a = tan a

so 8a = npi + a ( n= 0 to 6)

or 7a = npi

so solution set

z= tan (npi/7), x = tan (2npi/7), y = tan (4npi/7)

for n = 0 to 6

.

Congrats, kaliprasad! My approach is the same as you and thanks for participating!

 

[MarkFL]

I think I've put my foot in my mouth by saying that problem is easy, because I thought everyone would think of solving the given equation using trigonometric approach, after they rewritten any of the given equations to make either $x$, $y$ or $z$ the subject.

I played with the problem a bit soon after you posted, and that clever insight never occurred to me. Well done to both you and kaliprasad for seeing it. (Clapping)

I hope no one will take it amiss that I am trying to brag here because I am not. I am just a silly girl who loves to solve some challenging problems.

Anyone who sees you as either a braggart or a "silly girl" is simply not aware of your humble demeanor, your affinity for challenging problems and your dedication here to make MHB a better place, both as a member of the staff and as someone who tirelessly shares interesting problems with us. (Sun) Your dedication is admirable. (Yes)