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Changed to a useful title. Added reference from the comments.
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YCor
YCor
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How I can prove this conjecture if it's not open? Boundedness of solutions of a difference equation

Is there someone who can show me how I can prove this conjecture? Or at least show me how to do the first implication ?

Conjecture: Assume $\alpha,\beta, \lambda \in [0,\infty)$. Then every positive solution of the difference equation : $$z_{n+1}=\frac{\alpha+z_{n}\beta +z_{n-1}\lambda}{z_{n-2}},\quad n=0,1,\ldots$$ is bounded if and only if $\beta=\lambda$

Any help is very welcome. Thank you for any comments or any replies.

Edit: as mentioned in the comments, this is conjecture 8 in this paper by Ladas, Lugo and Palladino

How I can prove this conjecture if it's not open?

Is there someone who can show me how I can prove this conjecture? Or at least show me how to do the first implication ?

Conjecture: Assume $\alpha,\beta, \lambda \in [0,\infty)$. Then every positive solution of the difference equation : $$z_{n+1}=\frac{\alpha+z_{n}\beta +z_{n-1}\lambda}{z_{n-2}},\quad n=0,1,\ldots$$ is bounded if and only if $\beta=\lambda$

Any help is very welcome. Thank you for any comments or any replies.

Boundedness of solutions of a difference equation

Is there someone who can show me how I can prove this conjecture? Or at least show me how to do the first implication ?

Conjecture: Assume $\alpha,\beta, \lambda \in [0,\infty)$. Then every positive solution of the difference equation : $$z_{n+1}=\frac{\alpha+z_{n}\beta +z_{n-1}\lambda}{z_{n-2}},\quad n=0,1,\ldots$$ is bounded if and only if $\beta=\lambda$

Any help is very welcome. Thank you for any comments or any replies.

Edit: as mentioned in the comments, this is conjecture 8 in this paper by Ladas, Lugo and Palladino

As long as there was a recent bump, let's get rid of all those nasty typos, the annoying "how to proof", etc.
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Arturo Magidin
Arturo Magidin
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How I can proofprove this conjecture if it's not open?

Is there someone who can show me how I can proofprove this conjecture? Or at least show me how i can doing

to do the first implication ?

conjecture :Assume $\alpha,\beta , \lambda \in [0,\infty) $ Then every positive solution of

the difference equation :

$z_{n+1}=\frac{\alpha+z_{n}\beta +z_{n-1}\lambda}{z_{n-2}}$,$n=0,1,....$

is bounded if and only if : $\beta=\lambda$

Conjecture: Assume $\alpha,\beta, \lambda \in [0,\infty)$. Then every positive solution of the difference equation : $$z_{n+1}=\frac{\alpha+z_{n}\beta +z_{n-1}\lambda}{z_{n-2}},\quad n=0,1,\ldots$$ is bounded if and only if $\beta=\lambda$

anyAny help is very welcom , thankwelcome. Thank you for any comments or any replies  .

How I can proof this conjecture if it's not open?

Is there someone show me how I can proof this conjecture at least show me how i can doing

the first implication ?

conjecture :Assume $\alpha,\beta , \lambda \in [0,\infty) $ Then every positive solution of

the difference equation :

$z_{n+1}=\frac{\alpha+z_{n}\beta +z_{n-1}\lambda}{z_{n-2}}$,$n=0,1,....$

is bounded if and only if : $\beta=\lambda$

any help is very welcom , thank you for any comments or any replies  .

How I can prove this conjecture if it's not open?

Is there someone who can show me how I can prove this conjecture? Or at least show me how to do the first implication ?

Conjecture: Assume $\alpha,\beta, \lambda \in [0,\infty)$. Then every positive solution of the difference equation : $$z_{n+1}=\frac{\alpha+z_{n}\beta +z_{n-1}\lambda}{z_{n-2}},\quad n=0,1,\ldots$$ is bounded if and only if $\beta=\lambda$

Any help is very welcome. Thank you for any comments or any replies.

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salimmath15
salimmath15
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How I can proof this conjecture if it's not open?

Is there someone show me how I can proof this conjecture at least show me how i can doing

the first implication ?

conjecture :Assume $\alpha,\beta , \lambda \in [0,\infty) $ Then every positive solution of

the difference equation :

$z_{n+1}=\frac{\alpha+z_{n}\beta +z_{n-1}\lambda}{z_{n-2}}$,$n=0,1,....$

is bounded if and only if : $\beta=\lambda$

any help is very welcom , thank you for any comments or any replies .