Transition from one difference equation to another
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I have initial equation: $y(n+1)=y^2(n)+C$, $Cleqfrac{1}4$
How I can show that if $C=frac{a}2+frac{a^2}4$ then by using replacement $y(n)=alpha x(n)+beta$
I will get the equation $x(n+1)=ax(n)(1-x(n))$
differential-equations differential
asked Nov 18 at 13:52
bvl
123
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0
down vote
favorite
I have initial equation: $y(n+1)=y^2(n)+C$, $Cleqfrac{1}4$
How I can show that if $C=frac{a}2+frac{a^2}4$ then by using replacement $y(n)=alpha x(n)+beta$
I will get the equation $x(n+1)=ax(n)(1-x(n))$
differential-equations differential
asked Nov 18 at 13:52
bvl
123
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I have initial equation: $y(n+1)=y^2(n)+C$, $Cleqfrac{1}4$
How I can show that if $C=frac{a}2+frac{a^2}4$ then by using replacement $y(n)=alpha x(n)+beta$
I will get the equation $x(n+1)=ax(n)(1-x(n))$
differential-equations differential
asked Nov 18 at 13:52
bvl
123
I have initial equation: $y(n+1)=y^2(n)+C$, $Cleqfrac{1}4$
How I can show that if $C=frac{a}2+frac{a^2}4$ then by using replacement $y(n)=alpha x(n)+beta$
I will get the equation $x(n+1)=ax(n)(1-x(n))$
differential-equations differential
differential-equations differential
asked Nov 18 at 13:52
bvl
123
asked Nov 18 at 13:52
bvl
123
asked Nov 18 at 13:52
bvl
123
asked Nov 18 at 13:52
bvl
123
asked Nov 18 at 13:52
bvl
123
123
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
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0
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accepted
Using $C=a/2+a^2/4$ and $y=alpha x+beta$ in the equation leads to
$$
alpha x(n+1) + beta = (alpha^2 x^2(n) + 2alphabeta x(n) + beta^2) + frac{a}{2} + frac{a^2}{4}
$$
or
$$
x(n+1) = x(n)(alpha x(n) + 2beta) + frac{1}{alpha} left(beta(beta-1) + frac{a}{2} + frac{a^2}{4}right).
$$
The term in brackets vanishes if
$$
beta^2 - beta + frac{a}{2} + frac{a^2}{4} = 0,
$$
leading to
$$
beta = frac{1 pm i(a+1)}{2}.
$$
If we choose $alpha = -2beta$, we have
$$
x(n+1) = 2beta x(n)(1 - x(n)),
$$
which is the logistic equation, but with a complex coefficient. If we rather use $C=a/2-a^2/4$, we can choose $2beta = a$ and $alpha=-1$, leading then to
$$
x(n+1) = a x(n)(1- x(n)).
$$
answered Nov 18 at 14:57
rafa11111
985417
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
Using $C=a/2+a^2/4$ and $y=alpha x+beta$ in the equation leads to
$$
alpha x(n+1) + beta = (alpha^2 x^2(n) + 2alphabeta x(n) + beta^2) + frac{a}{2} + frac{a^2}{4}
$$
or
$$
x(n+1) = x(n)(alpha x(n) + 2beta) + frac{1}{alpha} left(beta(beta-1) + frac{a}{2} + frac{a^2}{4}right).
$$
The term in brackets vanishes if
$$
beta^2 - beta + frac{a}{2} + frac{a^2}{4} = 0,
$$
leading to
$$
beta = frac{1 pm i(a+1)}{2}.
$$
If we choose $alpha = -2beta$, we have
$$
x(n+1) = 2beta x(n)(1 - x(n)),
$$
which is the logistic equation, but with a complex coefficient. If we rather use $C=a/2-a^2/4$, we can choose $2beta = a$ and $alpha=-1$, leading then to
$$
x(n+1) = a x(n)(1- x(n)).
$$
answered Nov 18 at 14:57
rafa11111
985417
add a comment |
up vote
0
down vote
accepted
Using $C=a/2+a^2/4$ and $y=alpha x+beta$ in the equation leads to
$$
alpha x(n+1) + beta = (alpha^2 x^2(n) + 2alphabeta x(n) + beta^2) + frac{a}{2} + frac{a^2}{4}
$$
or
$$
x(n+1) = x(n)(alpha x(n) + 2beta) + frac{1}{alpha} left(beta(beta-1) + frac{a}{2} + frac{a^2}{4}right).
$$
The term in brackets vanishes if
$$
beta^2 - beta + frac{a}{2} + frac{a^2}{4} = 0,
$$
leading to
$$
beta = frac{1 pm i(a+1)}{2}.
$$
If we choose $alpha = -2beta$, we have
$$
x(n+1) = 2beta x(n)(1 - x(n)),
$$
which is the logistic equation, but with a complex coefficient. If we rather use $C=a/2-a^2/4$, we can choose $2beta = a$ and $alpha=-1$, leading then to
$$
x(n+1) = a x(n)(1- x(n)).
$$
answered Nov 18 at 14:57
rafa11111
985417
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
Using $C=a/2+a^2/4$ and $y=alpha x+beta$ in the equation leads to
$$
alpha x(n+1) + beta = (alpha^2 x^2(n) + 2alphabeta x(n) + beta^2) + frac{a}{2} + frac{a^2}{4}
$$
or
$$
x(n+1) = x(n)(alpha x(n) + 2beta) + frac{1}{alpha} left(beta(beta-1) + frac{a}{2} + frac{a^2}{4}right).
$$
The term in brackets vanishes if
$$
beta^2 - beta + frac{a}{2} + frac{a^2}{4} = 0,
$$
leading to
$$
beta = frac{1 pm i(a+1)}{2}.
$$
If we choose $alpha = -2beta$, we have
$$
x(n+1) = 2beta x(n)(1 - x(n)),
$$
which is the logistic equation, but with a complex coefficient. If we rather use $C=a/2-a^2/4$, we can choose $2beta = a$ and $alpha=-1$, leading then to
$$
x(n+1) = a x(n)(1- x(n)).
$$
answered Nov 18 at 14:57
rafa11111
985417
Using $C=a/2+a^2/4$ and $y=alpha x+beta$ in the equation leads to
$$
alpha x(n+1) + beta = (alpha^2 x^2(n) + 2alphabeta x(n) + beta^2) + frac{a}{2} + frac{a^2}{4}
$$
or
$$
x(n+1) = x(n)(alpha x(n) + 2beta) + frac{1}{alpha} left(beta(beta-1) + frac{a}{2} + frac{a^2}{4}right).
$$
The term in brackets vanishes if
$$
beta^2 - beta + frac{a}{2} + frac{a^2}{4} = 0,
$$
leading to
$$
beta = frac{1 pm i(a+1)}{2}.
$$
If we choose $alpha = -2beta$, we have
$$
x(n+1) = 2beta x(n)(1 - x(n)),
$$
which is the logistic equation, but with a complex coefficient. If we rather use $C=a/2-a^2/4$, we can choose $2beta = a$ and $alpha=-1$, leading then to
$$
x(n+1) = a x(n)(1- x(n)).
$$
answered Nov 18 at 14:57
rafa11111
985417
answered Nov 18 at 14:57
rafa11111
985417
answered Nov 18 at 14:57
rafa11111
985417
answered Nov 18 at 14:57
rafa11111
985417
985417
add a comment |
add a comment |
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