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?
differential-equations
add a comment |
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?
differential-equations
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
add a comment |
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?
differential-equations
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
differential-equations
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
add a comment |
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
add a comment |
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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