Mixing
Explore for convergence the following recurrence: $x_{n+1} = (1-x_n)^2, x_1 = {1over 2}, ninBbb N$
$begingroup$
Explore for convergence the following recurrence:
$$
x_{n+1} = (1-x_n)^2\
x_1 = {1over 2}\
ninBbb N
$$
To show a sequence is convergent it suffices to show it is bounded and monotonic. Obviously the the sequence is bounded by $1$ and $0$:
$$
0 le x_n le 1
$$
Check whether the sequence is monotonic:
$$
{1over 2} = x_1 > x_2 = {1over 4}
$$
Put $P(n): x_n > x_{n+1}forall ninBbb N$. $P(1)$ is true. Assume $P(n)$ is true. Then:
$$
x_{n+1} < x_n\
-x_{n+1} > -x_n \
1 - x_{n+1} > 1 - x_n \
(1 - x_{n+1})^2 > (1 - x_n)^2 \
x_{n+2} > x_{n+1}
$$
showing the induction hypotheses doesn't hold. However if we put it the following way:
$$
x_{n+1} < x_n \
x_{k+1} - 1 < x_n - 1 \
(x_{k+1} - 1)^2 < (x_n - 1)^2 \
(1 - x_{k+1})^2 < (1 - x_n)^2 \
x_{n+2} < x_{n+1}
$$
which completes the induction.
Clearly that is impossible. I believe the mistake is in the squaring step. Apparently there is a veil before my eyes preventing me from spotting it. What went wrong with the two cases?
calculus sequences-and-series limits proof-verification recurrence-relations
asked Jan 27 at 12:03
romanroman
2,39321225
$endgroup$
add a comment |
$begingroup$
Explore for convergence the following recurrence:
$$
x_{n+1} = (1-x_n)^2\
x_1 = {1over 2}\
ninBbb N
$$
To show a sequence is convergent it suffices to show it is bounded and monotonic. Obviously the the sequence is bounded by $1$ and $0$:
$$
0 le x_n le 1
$$
Check whether the sequence is monotonic:
$$
{1over 2} = x_1 > x_2 = {1over 4}
$$
Put $P(n): x_n > x_{n+1}forall ninBbb N$. $P(1)$ is true. Assume $P(n)$ is true. Then:
$$
x_{n+1} < x_n\
-x_{n+1} > -x_n \
1 - x_{n+1} > 1 - x_n \
(1 - x_{n+1})^2 > (1 - x_n)^2 \
x_{n+2} > x_{n+1}
$$
showing the induction hypotheses doesn't hold. However if we put it the following way:
$$
x_{n+1} < x_n \
x_{k+1} - 1 < x_n - 1 \
(x_{k+1} - 1)^2 < (x_n - 1)^2 \
(1 - x_{k+1})^2 < (1 - x_n)^2 \
x_{n+2} < x_{n+1}
$$
which completes the induction.
Clearly that is impossible. I believe the mistake is in the squaring step. Apparently there is a veil before my eyes preventing me from spotting it. What went wrong with the two cases?
calculus sequences-and-series limits proof-verification recurrence-relations
asked Jan 27 at 12:03
romanroman
2,39321225
$endgroup$
add a comment |
$begingroup$
Explore for convergence the following recurrence:
$$
x_{n+1} = (1-x_n)^2\
x_1 = {1over 2}\
ninBbb N
$$
To show a sequence is convergent it suffices to show it is bounded and monotonic. Obviously the the sequence is bounded by $1$ and $0$:
$$
0 le x_n le 1
$$
Check whether the sequence is monotonic:
$$
{1over 2} = x_1 > x_2 = {1over 4}
$$
Put $P(n): x_n > x_{n+1}forall ninBbb N$. $P(1)$ is true. Assume $P(n)$ is true. Then:
$$
x_{n+1} < x_n\
-x_{n+1} > -x_n \
1 - x_{n+1} > 1 - x_n \
(1 - x_{n+1})^2 > (1 - x_n)^2 \
x_{n+2} > x_{n+1}
$$
showing the induction hypotheses doesn't hold. However if we put it the following way:
$$
x_{n+1} < x_n \
x_{k+1} - 1 < x_n - 1 \
(x_{k+1} - 1)^2 < (x_n - 1)^2 \
(1 - x_{k+1})^2 < (1 - x_n)^2 \
x_{n+2} < x_{n+1}
$$
which completes the induction.
Clearly that is impossible. I believe the mistake is in the squaring step. Apparently there is a veil before my eyes preventing me from spotting it. What went wrong with the two cases?
calculus sequences-and-series limits proof-verification recurrence-relations
asked Jan 27 at 12:03
romanroman
2,39321225
$endgroup$
Explore for convergence the following recurrence:
$$
x_{n+1} = (1-x_n)^2\
x_1 = {1over 2}\
ninBbb N
$$
To show a sequence is convergent it suffices to show it is bounded and monotonic. Obviously the the sequence is bounded by $1$ and $0$:
$$
0 le x_n le 1
$$
Check whether the sequence is monotonic:
$$
{1over 2} = x_1 > x_2 = {1over 4}
$$
Put $P(n): x_n > x_{n+1}forall ninBbb N$. $P(1)$ is true. Assume $P(n)$ is true. Then:
$$
x_{n+1} < x_n\
-x_{n+1} > -x_n \
1 - x_{n+1} > 1 - x_n \
(1 - x_{n+1})^2 > (1 - x_n)^2 \
x_{n+2} > x_{n+1}
$$
showing the induction hypotheses doesn't hold. However if we put it the following way:
$$
x_{n+1} < x_n \
x_{k+1} - 1 < x_n - 1 \
(x_{k+1} - 1)^2 < (x_n - 1)^2 \
(1 - x_{k+1})^2 < (1 - x_n)^2 \
x_{n+2} < x_{n+1}
$$
which completes the induction.
Clearly that is impossible. I believe the mistake is in the squaring step. Apparently there is a veil before my eyes preventing me from spotting it. What went wrong with the two cases?
calculus sequences-and-series limits proof-verification recurrence-relations
calculus sequences-and-series limits proof-verification recurrence-relations
asked Jan 27 at 12:03
romanroman
2,39321225
asked Jan 27 at 12:03
romanroman
2,39321225
edited Jan 27 at 12:13
roman
asked Jan 27 at 12:03
romanroman
2,39321225
asked Jan 27 at 12:03
romanroman
2,39321225
asked Jan 27 at 12:03
romanroman
2,39321225
2,39321225
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
You can not do $x_{n+1} - 1 < x_n - 1 \
implies (x_{n+1} - 1)^2 < (x_n - 1)^2 $
because both LHS and RHS of inequalities is negative.
For eg. $-1 > -2$ but $1^2 < 2^2$
answered Jan 27 at 12:10
ab123ab123
1,799423
$endgroup$
$begingroup$
Thank you. Seems like the limit doesn't exist after all.
$endgroup$
– roman
Jan 27 at 12:16
$begingroup$
@roman this is not a correct conclusion. Only because your method doesn't work doesn't mean that the assumption is wrong.
$endgroup$
– Nathanael Skrepek
Jan 27 at 12:22
$begingroup$
@NathanaelSkrepek You are right, i meant after some further inspection there are two limit points for $x_{2k}$ and $x_{2k+1}$, which shows the sequence is divergent.
$endgroup$
– roman
Jan 27 at 12:25
$begingroup$
@roman Yes, the sequence starts alternating between a value close to $0$ and a value close to $1$
$endgroup$
– ab123
Jan 27 at 12:27
add a comment |
$begingroup$
If you look at the sequences $newcommand{N}{mathbb{N}}(a_n)_{ninN} :=(x_{2n})_{ninN}$ and $(b_n)_{ninN}:=(x_{2n-1})_{ninN}$ then you can show that both of them will converge to different limits. We start with using the recursion twice
$$
x_{n+2} = (1-x_{n+1})^2 = (1-(1-x_n)^2)^2 = dots = x_n^4 - 4x_n^3 + 4x_n^2
$$
We define the polynomial function $p(x) := x^4 - 4x^3 + 4x^2$. some curve sketching yields that this function is monoton increasing for $xin (0,1)$. The recursion for $a_n$ and $b_n$ is given by
$$
a_{n+1} = a_n^4 - 4a_n^3 + 4a_n^2, quad a_1 = frac{1}{4} \
b_{n+1} = b_n^4 - 4b_n^3 + 4b_n^2, quad b_1 = frac{1}{2}
$$
As you already said $0<x_n<1$. Clearly, this is also true for $a_n$ and $b_n$.
It is straight forward to check that $b_2 > b_1$. Hence, $(b_n)_{ninN}$ is convergent. To evaluate the limit point you have to solve $x = p(x)$. It is easy to see that $0$ and $1$ are solutions. By polynomial division you can find the other solution by solving a quadratic equation. You will realize that only the solution $1$ can be the limit of $(b_n)_{ninN}$.
Since $a_n = (1-b_n)^2$, the limit of $(a_n)_{ninN}$ is $0$. Therefore, $(x_n)_{nin N}$ cannot converge.
answered Jan 27 at 12:53
Nathanael SkrepekNathanael Skrepek
1,7111615
$endgroup$
add a comment |
$begingroup$
Just to have a complete answer, from $x_{n+1}=f(x_n)$, where $f(x)=(1-x)^2$, we have $f'(x)=-2(1-x)$. Or $f(x)$ is descending on $[0,1]$. Sequence is bounded indeed $0leq x_n leq 1$, can be show using induction. Let's compute a few values
$$x_1=frac{1}{2} > x_2=frac{1}{4} > x_4=frac{49}{256}$$
$$x_1=frac{1}{2}< x_3=frac{9}{16} < x_5=frac{42849}{65536}$$
and because the function $f(x)$ is descending
$$f(x_2)leq f(x_4) iff x_3 leq x_5 \
f(x_3) geq f(x_5) iff x_4 geq x_6 \
f(x_4) leq f(x_6) iff x_5 leq x_7 \
...$$
By induction, we have
$$x_1>x_2>x_4geq x_6 geq ... geq x_{2k} geq ...$$
$$x_1<x_3<x_5leq x_7 leq ... leq x_{2k+1} leq ...$$
i.e. we have a decreasing subsequence $left(x_{2k}right)_{kinmathbb{N}}$ and increasing one $left(x_{2k+1}right)_{kinmathbb{N}}$ both bounded, so both these subsequences have limits
$$frac{1}{4}=x_2 geq limlimits_{krightarrowinfty}x_{2k}$$
$$frac{9}{16}=x_3 leq limlimits_{krightarrowinfty}x_{2k+1}$$
and these limits are different. As a result, the original sequence is diverging.
answered Jan 27 at 13:06
rtybasertybase
11.5k31534
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
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$begingroup$
You can not do $x_{n+1} - 1 < x_n - 1 \
implies (x_{n+1} - 1)^2 < (x_n - 1)^2 $
because both LHS and RHS of inequalities is negative.
For eg. $-1 > -2$ but $1^2 < 2^2$
answered Jan 27 at 12:10
ab123ab123
1,799423
$endgroup$
$begingroup$
Thank you. Seems like the limit doesn't exist after all.
$endgroup$
– roman
Jan 27 at 12:16
$begingroup$
@roman this is not a correct conclusion. Only because your method doesn't work doesn't mean that the assumption is wrong.
$endgroup$
– Nathanael Skrepek
Jan 27 at 12:22
$begingroup$
@NathanaelSkrepek You are right, i meant after some further inspection there are two limit points for $x_{2k}$ and $x_{2k+1}$, which shows the sequence is divergent.
$endgroup$
– roman
Jan 27 at 12:25
$begingroup$
@roman Yes, the sequence starts alternating between a value close to $0$ and a value close to $1$
$endgroup$
– ab123
Jan 27 at 12:27
add a comment |
$begingroup$
You can not do $x_{n+1} - 1 < x_n - 1 \
implies (x_{n+1} - 1)^2 < (x_n - 1)^2 $
because both LHS and RHS of inequalities is negative.
For eg. $-1 > -2$ but $1^2 < 2^2$
answered Jan 27 at 12:10
ab123ab123
1,799423
$endgroup$
$begingroup$
Thank you. Seems like the limit doesn't exist after all.
$endgroup$
– roman
Jan 27 at 12:16
$begingroup$
@roman this is not a correct conclusion. Only because your method doesn't work doesn't mean that the assumption is wrong.
$endgroup$
– Nathanael Skrepek
Jan 27 at 12:22
$begingroup$
@NathanaelSkrepek You are right, i meant after some further inspection there are two limit points for $x_{2k}$ and $x_{2k+1}$, which shows the sequence is divergent.
$endgroup$
– roman
Jan 27 at 12:25
$begingroup$
@roman Yes, the sequence starts alternating between a value close to $0$ and a value close to $1$
$endgroup$
– ab123
Jan 27 at 12:27
add a comment |
$begingroup$
You can not do $x_{n+1} - 1 < x_n - 1 \
implies (x_{n+1} - 1)^2 < (x_n - 1)^2 $
because both LHS and RHS of inequalities is negative.
For eg. $-1 > -2$ but $1^2 < 2^2$
answered Jan 27 at 12:10
ab123ab123
1,799423
$endgroup$
You can not do $x_{n+1} - 1 < x_n - 1 \
implies (x_{n+1} - 1)^2 < (x_n - 1)^2 $
because both LHS and RHS of inequalities is negative.
For eg. $-1 > -2$ but $1^2 < 2^2$
answered Jan 27 at 12:10
ab123ab123
1,799423
answered Jan 27 at 12:10
ab123ab123
1,799423
answered Jan 27 at 12:10
ab123ab123
1,799423
answered Jan 27 at 12:10
ab123ab123
1,799423
1,799423
$begingroup$
Thank you. Seems like the limit doesn't exist after all.
$endgroup$
– roman
Jan 27 at 12:16
$begingroup$
@roman this is not a correct conclusion. Only because your method doesn't work doesn't mean that the assumption is wrong.
$endgroup$
– Nathanael Skrepek
Jan 27 at 12:22
$begingroup$
@NathanaelSkrepek You are right, i meant after some further inspection there are two limit points for $x_{2k}$ and $x_{2k+1}$, which shows the sequence is divergent.
$endgroup$
– roman
Jan 27 at 12:25
$begingroup$
@roman Yes, the sequence starts alternating between a value close to $0$ and a value close to $1$
$endgroup$
– ab123
Jan 27 at 12:27
add a comment |
$begingroup$
Thank you. Seems like the limit doesn't exist after all.
$endgroup$
– roman
Jan 27 at 12:16
$begingroup$
@roman this is not a correct conclusion. Only because your method doesn't work doesn't mean that the assumption is wrong.
$endgroup$
– Nathanael Skrepek
Jan 27 at 12:22
$begingroup$
@NathanaelSkrepek You are right, i meant after some further inspection there are two limit points for $x_{2k}$ and $x_{2k+1}$, which shows the sequence is divergent.
$endgroup$
– roman
Jan 27 at 12:25
$begingroup$
@roman Yes, the sequence starts alternating between a value close to $0$ and a value close to $1$
$endgroup$
– ab123
Jan 27 at 12:27
$begingroup$
Thank you. Seems like the limit doesn't exist after all.
$endgroup$
– roman
Jan 27 at 12:16
$begingroup$
Thank you. Seems like the limit doesn't exist after all.
$endgroup$
– roman
Jan 27 at 12:16
$begingroup$
@roman this is not a correct conclusion. Only because your method doesn't work doesn't mean that the assumption is wrong.
$endgroup$
– Nathanael Skrepek
Jan 27 at 12:22
$begingroup$
@roman this is not a correct conclusion. Only because your method doesn't work doesn't mean that the assumption is wrong.
$endgroup$
– Nathanael Skrepek
Jan 27 at 12:22
$begingroup$
@NathanaelSkrepek You are right, i meant after some further inspection there are two limit points for $x_{2k}$ and $x_{2k+1}$, which shows the sequence is divergent.
$endgroup$
– roman
Jan 27 at 12:25
$begingroup$
@NathanaelSkrepek You are right, i meant after some further inspection there are two limit points for $x_{2k}$ and $x_{2k+1}$, which shows the sequence is divergent.
$endgroup$
– roman
Jan 27 at 12:25
$begingroup$
@roman Yes, the sequence starts alternating between a value close to $0$ and a value close to $1$
$endgroup$
– ab123
Jan 27 at 12:27
$begingroup$
@roman Yes, the sequence starts alternating between a value close to $0$ and a value close to $1$
$endgroup$
– ab123
Jan 27 at 12:27
add a comment |
$begingroup$
If you look at the sequences $newcommand{N}{mathbb{N}}(a_n)_{ninN} :=(x_{2n})_{ninN}$ and $(b_n)_{ninN}:=(x_{2n-1})_{ninN}$ then you can show that both of them will converge to different limits. We start with using the recursion twice
$$
x_{n+2} = (1-x_{n+1})^2 = (1-(1-x_n)^2)^2 = dots = x_n^4 - 4x_n^3 + 4x_n^2
$$
We define the polynomial function $p(x) := x^4 - 4x^3 + 4x^2$. some curve sketching yields that this function is monoton increasing for $xin (0,1)$. The recursion for $a_n$ and $b_n$ is given by
$$
a_{n+1} = a_n^4 - 4a_n^3 + 4a_n^2, quad a_1 = frac{1}{4} \
b_{n+1} = b_n^4 - 4b_n^3 + 4b_n^2, quad b_1 = frac{1}{2}
$$
As you already said $0<x_n<1$. Clearly, this is also true for $a_n$ and $b_n$.
It is straight forward to check that $b_2 > b_1$. Hence, $(b_n)_{ninN}$ is convergent. To evaluate the limit point you have to solve $x = p(x)$. It is easy to see that $0$ and $1$ are solutions. By polynomial division you can find the other solution by solving a quadratic equation. You will realize that only the solution $1$ can be the limit of $(b_n)_{ninN}$.
Since $a_n = (1-b_n)^2$, the limit of $(a_n)_{ninN}$ is $0$. Therefore, $(x_n)_{nin N}$ cannot converge.
answered Jan 27 at 12:53
Nathanael SkrepekNathanael Skrepek
1,7111615
$endgroup$
add a comment |
$begingroup$
If you look at the sequences $newcommand{N}{mathbb{N}}(a_n)_{ninN} :=(x_{2n})_{ninN}$ and $(b_n)_{ninN}:=(x_{2n-1})_{ninN}$ then you can show that both of them will converge to different limits. We start with using the recursion twice
$$
x_{n+2} = (1-x_{n+1})^2 = (1-(1-x_n)^2)^2 = dots = x_n^4 - 4x_n^3 + 4x_n^2
$$
We define the polynomial function $p(x) := x^4 - 4x^3 + 4x^2$. some curve sketching yields that this function is monoton increasing for $xin (0,1)$. The recursion for $a_n$ and $b_n$ is given by
$$
a_{n+1} = a_n^4 - 4a_n^3 + 4a_n^2, quad a_1 = frac{1}{4} \
b_{n+1} = b_n^4 - 4b_n^3 + 4b_n^2, quad b_1 = frac{1}{2}
$$
As you already said $0<x_n<1$. Clearly, this is also true for $a_n$ and $b_n$.
It is straight forward to check that $b_2 > b_1$. Hence, $(b_n)_{ninN}$ is convergent. To evaluate the limit point you have to solve $x = p(x)$. It is easy to see that $0$ and $1$ are solutions. By polynomial division you can find the other solution by solving a quadratic equation. You will realize that only the solution $1$ can be the limit of $(b_n)_{ninN}$.
Since $a_n = (1-b_n)^2$, the limit of $(a_n)_{ninN}$ is $0$. Therefore, $(x_n)_{nin N}$ cannot converge.
answered Jan 27 at 12:53
Nathanael SkrepekNathanael Skrepek
1,7111615
$endgroup$
add a comment |
$begingroup$
If you look at the sequences $newcommand{N}{mathbb{N}}(a_n)_{ninN} :=(x_{2n})_{ninN}$ and $(b_n)_{ninN}:=(x_{2n-1})_{ninN}$ then you can show that both of them will converge to different limits. We start with using the recursion twice
$$
x_{n+2} = (1-x_{n+1})^2 = (1-(1-x_n)^2)^2 = dots = x_n^4 - 4x_n^3 + 4x_n^2
$$
We define the polynomial function $p(x) := x^4 - 4x^3 + 4x^2$. some curve sketching yields that this function is monoton increasing for $xin (0,1)$. The recursion for $a_n$ and $b_n$ is given by
$$
a_{n+1} = a_n^4 - 4a_n^3 + 4a_n^2, quad a_1 = frac{1}{4} \
b_{n+1} = b_n^4 - 4b_n^3 + 4b_n^2, quad b_1 = frac{1}{2}
$$
As you already said $0<x_n<1$. Clearly, this is also true for $a_n$ and $b_n$.
It is straight forward to check that $b_2 > b_1$. Hence, $(b_n)_{ninN}$ is convergent. To evaluate the limit point you have to solve $x = p(x)$. It is easy to see that $0$ and $1$ are solutions. By polynomial division you can find the other solution by solving a quadratic equation. You will realize that only the solution $1$ can be the limit of $(b_n)_{ninN}$.
Since $a_n = (1-b_n)^2$, the limit of $(a_n)_{ninN}$ is $0$. Therefore, $(x_n)_{nin N}$ cannot converge.
answered Jan 27 at 12:53
Nathanael SkrepekNathanael Skrepek
1,7111615
$endgroup$
If you look at the sequences $newcommand{N}{mathbb{N}}(a_n)_{ninN} :=(x_{2n})_{ninN}$ and $(b_n)_{ninN}:=(x_{2n-1})_{ninN}$ then you can show that both of them will converge to different limits. We start with using the recursion twice
$$
x_{n+2} = (1-x_{n+1})^2 = (1-(1-x_n)^2)^2 = dots = x_n^4 - 4x_n^3 + 4x_n^2
$$
We define the polynomial function $p(x) := x^4 - 4x^3 + 4x^2$. some curve sketching yields that this function is monoton increasing for $xin (0,1)$. The recursion for $a_n$ and $b_n$ is given by
$$
a_{n+1} = a_n^4 - 4a_n^3 + 4a_n^2, quad a_1 = frac{1}{4} \
b_{n+1} = b_n^4 - 4b_n^3 + 4b_n^2, quad b_1 = frac{1}{2}
$$
As you already said $0<x_n<1$. Clearly, this is also true for $a_n$ and $b_n$.
It is straight forward to check that $b_2 > b_1$. Hence, $(b_n)_{ninN}$ is convergent. To evaluate the limit point you have to solve $x = p(x)$. It is easy to see that $0$ and $1$ are solutions. By polynomial division you can find the other solution by solving a quadratic equation. You will realize that only the solution $1$ can be the limit of $(b_n)_{ninN}$.
Since $a_n = (1-b_n)^2$, the limit of $(a_n)_{ninN}$ is $0$. Therefore, $(x_n)_{nin N}$ cannot converge.
answered Jan 27 at 12:53
Nathanael SkrepekNathanael Skrepek
1,7111615
answered Jan 27 at 12:53
Nathanael SkrepekNathanael Skrepek
1,7111615
answered Jan 27 at 12:53
Nathanael SkrepekNathanael Skrepek
1,7111615
answered Jan 27 at 12:53
Nathanael SkrepekNathanael Skrepek
1,7111615
1,7111615
add a comment |
add a comment |
$begingroup$
Just to have a complete answer, from $x_{n+1}=f(x_n)$, where $f(x)=(1-x)^2$, we have $f'(x)=-2(1-x)$. Or $f(x)$ is descending on $[0,1]$. Sequence is bounded indeed $0leq x_n leq 1$, can be show using induction. Let's compute a few values
$$x_1=frac{1}{2} > x_2=frac{1}{4} > x_4=frac{49}{256}$$
$$x_1=frac{1}{2}< x_3=frac{9}{16} < x_5=frac{42849}{65536}$$
and because the function $f(x)$ is descending
$$f(x_2)leq f(x_4) iff x_3 leq x_5 \
f(x_3) geq f(x_5) iff x_4 geq x_6 \
f(x_4) leq f(x_6) iff x_5 leq x_7 \
...$$
By induction, we have
$$x_1>x_2>x_4geq x_6 geq ... geq x_{2k} geq ...$$
$$x_1<x_3<x_5leq x_7 leq ... leq x_{2k+1} leq ...$$
i.e. we have a decreasing subsequence $left(x_{2k}right)_{kinmathbb{N}}$ and increasing one $left(x_{2k+1}right)_{kinmathbb{N}}$ both bounded, so both these subsequences have limits
$$frac{1}{4}=x_2 geq limlimits_{krightarrowinfty}x_{2k}$$
$$frac{9}{16}=x_3 leq limlimits_{krightarrowinfty}x_{2k+1}$$
and these limits are different. As a result, the original sequence is diverging.
answered Jan 27 at 13:06
rtybasertybase
11.5k31534
$endgroup$
add a comment |
$begingroup$
Just to have a complete answer, from $x_{n+1}=f(x_n)$, where $f(x)=(1-x)^2$, we have $f'(x)=-2(1-x)$. Or $f(x)$ is descending on $[0,1]$. Sequence is bounded indeed $0leq x_n leq 1$, can be show using induction. Let's compute a few values
$$x_1=frac{1}{2} > x_2=frac{1}{4} > x_4=frac{49}{256}$$
$$x_1=frac{1}{2}< x_3=frac{9}{16} < x_5=frac{42849}{65536}$$
and because the function $f(x)$ is descending
$$f(x_2)leq f(x_4) iff x_3 leq x_5 \
f(x_3) geq f(x_5) iff x_4 geq x_6 \
f(x_4) leq f(x_6) iff x_5 leq x_7 \
...$$
By induction, we have
$$x_1>x_2>x_4geq x_6 geq ... geq x_{2k} geq ...$$
$$x_1<x_3<x_5leq x_7 leq ... leq x_{2k+1} leq ...$$
i.e. we have a decreasing subsequence $left(x_{2k}right)_{kinmathbb{N}}$ and increasing one $left(x_{2k+1}right)_{kinmathbb{N}}$ both bounded, so both these subsequences have limits
$$frac{1}{4}=x_2 geq limlimits_{krightarrowinfty}x_{2k}$$
$$frac{9}{16}=x_3 leq limlimits_{krightarrowinfty}x_{2k+1}$$
and these limits are different. As a result, the original sequence is diverging.
answered Jan 27 at 13:06
rtybasertybase
11.5k31534
$endgroup$
add a comment |
$begingroup$
Just to have a complete answer, from $x_{n+1}=f(x_n)$, where $f(x)=(1-x)^2$, we have $f'(x)=-2(1-x)$. Or $f(x)$ is descending on $[0,1]$. Sequence is bounded indeed $0leq x_n leq 1$, can be show using induction. Let's compute a few values
$$x_1=frac{1}{2} > x_2=frac{1}{4} > x_4=frac{49}{256}$$
$$x_1=frac{1}{2}< x_3=frac{9}{16} < x_5=frac{42849}{65536}$$
and because the function $f(x)$ is descending
$$f(x_2)leq f(x_4) iff x_3 leq x_5 \
f(x_3) geq f(x_5) iff x_4 geq x_6 \
f(x_4) leq f(x_6) iff x_5 leq x_7 \
...$$
By induction, we have
$$x_1>x_2>x_4geq x_6 geq ... geq x_{2k} geq ...$$
$$x_1<x_3<x_5leq x_7 leq ... leq x_{2k+1} leq ...$$
i.e. we have a decreasing subsequence $left(x_{2k}right)_{kinmathbb{N}}$ and increasing one $left(x_{2k+1}right)_{kinmathbb{N}}$ both bounded, so both these subsequences have limits
$$frac{1}{4}=x_2 geq limlimits_{krightarrowinfty}x_{2k}$$
$$frac{9}{16}=x_3 leq limlimits_{krightarrowinfty}x_{2k+1}$$
and these limits are different. As a result, the original sequence is diverging.
answered Jan 27 at 13:06
rtybasertybase
11.5k31534
$endgroup$
Just to have a complete answer, from $x_{n+1}=f(x_n)$, where $f(x)=(1-x)^2$, we have $f'(x)=-2(1-x)$. Or $f(x)$ is descending on $[0,1]$. Sequence is bounded indeed $0leq x_n leq 1$, can be show using induction. Let's compute a few values
$$x_1=frac{1}{2} > x_2=frac{1}{4} > x_4=frac{49}{256}$$
$$x_1=frac{1}{2}< x_3=frac{9}{16} < x_5=frac{42849}{65536}$$
and because the function $f(x)$ is descending
$$f(x_2)leq f(x_4) iff x_3 leq x_5 \
f(x_3) geq f(x_5) iff x_4 geq x_6 \
f(x_4) leq f(x_6) iff x_5 leq x_7 \
...$$
By induction, we have
$$x_1>x_2>x_4geq x_6 geq ... geq x_{2k} geq ...$$
$$x_1<x_3<x_5leq x_7 leq ... leq x_{2k+1} leq ...$$
i.e. we have a decreasing subsequence $left(x_{2k}right)_{kinmathbb{N}}$ and increasing one $left(x_{2k+1}right)_{kinmathbb{N}}$ both bounded, so both these subsequences have limits
$$frac{1}{4}=x_2 geq limlimits_{krightarrowinfty}x_{2k}$$
$$frac{9}{16}=x_3 leq limlimits_{krightarrowinfty}x_{2k+1}$$
and these limits are different. As a result, the original sequence is diverging.
answered Jan 27 at 13:06
rtybasertybase
11.5k31534
answered Jan 27 at 13:06
rtybasertybase
11.5k31534
answered Jan 27 at 13:06
rtybasertybase
11.5k31534
answered Jan 27 at 13:06
rtybasertybase
11.5k31534
11.5k31534
add a comment |
add a comment |
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