For what values of the parameter $c$ the difference equation have periodic (2-cycle) solutions











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I have difference equation:
$y(n+1)=y(n)^2+c$



At what value of parametr $c$ the equation have a 2-cycle solution?



In this example $f(y)=y^2+c$



To solve the problem, I wanted to solve this equation: $y=f(f(y))$ because solution of the equation is periodic 2-cycle.



i got the roots of this equation:



$y_1=frac{1}2(-sqrt{-4c-3}-1)$



$y_2=frac{1}2(sqrt{-4c-3}-1)$



$y_3=frac{1}2(-sqrt{1-4c}+1)$



$y_4=frac{1}2(sqrt{1-4c}+1)$



but I don't know what to do next. How to find the necessary parameter values using the roots?










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  • This is a difference equation not a differential equation.
    – Anurag A
    Nov 18 at 18:28










  • thank you, i fixed.
    – bvl
    Nov 18 at 18:31















up vote
0
down vote

favorite












I have difference equation:
$y(n+1)=y(n)^2+c$



At what value of parametr $c$ the equation have a 2-cycle solution?



In this example $f(y)=y^2+c$



To solve the problem, I wanted to solve this equation: $y=f(f(y))$ because solution of the equation is periodic 2-cycle.



i got the roots of this equation:



$y_1=frac{1}2(-sqrt{-4c-3}-1)$



$y_2=frac{1}2(sqrt{-4c-3}-1)$



$y_3=frac{1}2(-sqrt{1-4c}+1)$



$y_4=frac{1}2(sqrt{1-4c}+1)$



but I don't know what to do next. How to find the necessary parameter values using the roots?










share|cite|improve this question
























  • This is a difference equation not a differential equation.
    – Anurag A
    Nov 18 at 18:28










  • thank you, i fixed.
    – bvl
    Nov 18 at 18:31













up vote
0
down vote

favorite









up vote
0
down vote

favorite











I have difference equation:
$y(n+1)=y(n)^2+c$



At what value of parametr $c$ the equation have a 2-cycle solution?



In this example $f(y)=y^2+c$



To solve the problem, I wanted to solve this equation: $y=f(f(y))$ because solution of the equation is periodic 2-cycle.



i got the roots of this equation:



$y_1=frac{1}2(-sqrt{-4c-3}-1)$



$y_2=frac{1}2(sqrt{-4c-3}-1)$



$y_3=frac{1}2(-sqrt{1-4c}+1)$



$y_4=frac{1}2(sqrt{1-4c}+1)$



but I don't know what to do next. How to find the necessary parameter values using the roots?










share|cite|improve this question















I have difference equation:
$y(n+1)=y(n)^2+c$



At what value of parametr $c$ the equation have a 2-cycle solution?



In this example $f(y)=y^2+c$



To solve the problem, I wanted to solve this equation: $y=f(f(y))$ because solution of the equation is periodic 2-cycle.



i got the roots of this equation:



$y_1=frac{1}2(-sqrt{-4c-3}-1)$



$y_2=frac{1}2(sqrt{-4c-3}-1)$



$y_3=frac{1}2(-sqrt{1-4c}+1)$



$y_4=frac{1}2(sqrt{1-4c}+1)$



but I don't know what to do next. How to find the necessary parameter values using the roots?







differential-equations






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share|cite|improve this question













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share|cite|improve this question








edited Nov 18 at 20:48

























asked Nov 18 at 18:21









bvl

72




72












  • This is a difference equation not a differential equation.
    – Anurag A
    Nov 18 at 18:28










  • thank you, i fixed.
    – bvl
    Nov 18 at 18:31


















  • This is a difference equation not a differential equation.
    – Anurag A
    Nov 18 at 18:28










  • thank you, i fixed.
    – bvl
    Nov 18 at 18:31
















This is a difference equation not a differential equation.
– Anurag A
Nov 18 at 18:28




This is a difference equation not a differential equation.
– Anurag A
Nov 18 at 18:28












thank you, i fixed.
– bvl
Nov 18 at 18:31




thank you, i fixed.
– bvl
Nov 18 at 18:31















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