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Continuous function with 2 attractive fixed points [closed]

0

1

$begingroup$

I stumbled upon a question and I can't seem to find the answer. Here it is:

Suppose $f$ a continuous function (from $Bbb R$ to $Bbb R$) with 2 attractive fixed points. Let's call them $a$ and $b$. The basin of attraction of $a$ is $(-infty,p)$ and the basin of attraction of $b$ is $(p, infty)$ for $p$ in $Bbb R$. What can you say about the point $p$? My intuition would be to say that the function $f$ is not derivable in $p$, but I don't really know why.

Also, what would be an example of a function that has these properties?

Thank you in advance for your answers!

dynamical-systems

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edited Apr 2 at 17:39

Daniele Tampieri

2,79221023

asked Apr 2 at 16:54

A. SmithA. Smith

52

$endgroup$

closed as off-topic by Thomas Shelby, Eevee Trainer, max_zorn, José Carlos Santos, Trần Thúc Minh Trí Apr 3 at 11:52

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Thomas Shelby, Eevee Trainer, max_zorn, José Carlos Santos, Trần Thúc Minh Trí

If this question can be reworded to fit the rules in the help center, please edit the question.

add a comment | 

0

1

$begingroup$

I stumbled upon a question and I can't seem to find the answer. Here it is:

Suppose $f$ a continuous function (from $Bbb R$ to $Bbb R$) with 2 attractive fixed points. Let's call them $a$ and $b$. The basin of attraction of $a$ is $(-infty,p)$ and the basin of attraction of $b$ is $(p, infty)$ for $p$ in $Bbb R$. What can you say about the point $p$? My intuition would be to say that the function $f$ is not derivable in $p$, but I don't really know why.

Also, what would be an example of a function that has these properties?

Thank you in advance for your answers!

dynamical-systems

share|cite|improve this question

edited Apr 2 at 17:39

Daniele Tampieri

2,79221023

asked Apr 2 at 16:54

A. SmithA. Smith

52

$endgroup$

closed as off-topic by Thomas Shelby, Eevee Trainer, max_zorn, José Carlos Santos, Trần Thúc Minh Trí Apr 3 at 11:52

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Thomas Shelby, Eevee Trainer, max_zorn, José Carlos Santos, Trần Thúc Minh Trí

If this question can be reworded to fit the rules in the help center, please edit the question.

add a comment | 

$begingroup$

I stumbled upon a question and I can't seem to find the answer. Here it is:

Suppose $f$ a continuous function (from $Bbb R$ to $Bbb R$) with 2 attractive fixed points. Let's call them $a$ and $b$. The basin of attraction of $a$ is $(-infty,p)$ and the basin of attraction of $b$ is $(p, infty)$ for $p$ in $Bbb R$. What can you say about the point $p$? My intuition would be to say that the function $f$ is not derivable in $p$, but I don't really know why.

Also, what would be an example of a function that has these properties?

Thank you in advance for your answers!

dynamical-systems

share|cite|improve this question

edited Apr 2 at 17:39

Daniele Tampieri

2,79221023

asked Apr 2 at 16:54

A. SmithA. Smith

52

$endgroup$

share|cite|improve this question

edited Apr 2 at 17:39

Daniele Tampieri

2,79221023

asked Apr 2 at 16:54

A. SmithA. Smith

52

share|cite|improve this question

edited Apr 2 at 17:39

Daniele Tampieri

2,79221023

asked Apr 2 at 16:54

A. SmithA. Smith

52

edited Apr 2 at 17:39

Daniele Tampieri

2,79221023

edited Apr 2 at 17:39

Daniele Tampieri

2,79221023

asked Apr 2 at 16:54

A. SmithA. Smith

52

asked Apr 2 at 16:54

A. SmithA. Smith

52

2 Answers
2

active

oldest

votes

1

$begingroup$

If $f(p)>p$, then there is a neighborhood of $p$ on which $f(x)>p$. Therefore, that neighborhood would belong to the attractive basin of $b$. This is a contradiction, since all points $<p$ belong to the attractive basin of $a$.

Similarly, if $f(p)<p$ then a neighborhood of $p$ would belong to the attractive basin of $a$, which also cannot be since all points $>p$ must belong to the basin of $b$.

Therefore, $f(p)=p$, i.e. $p$ is also a fixed point.

The information given doesn't tell you anything about the differentiability at $p$. You were given the example of $frac4piarctan(x)$ with $p=0$, which is differentiable at $p$. You can bend it a bit to produce an example that is not differentiable at $p$:

$$f(x)=begincasesfrac4piarctan(x),&xgeq0\frac4(1+epsilon)piarctanleft(x/(1+epsilon)right),&x>0endcases$$

share|cite|improve this answer

edited Apr 2 at 19:10

answered Apr 2 at 18:48

user647486user647486

917111

$endgroup$

add a comment | 

0

$begingroup$

$$ f(x) = frac4pi ; arctan x $$

share|cite|improve this answer

answered Apr 2 at 17:50

Will JagyWill Jagy

105k5103202

$endgroup$

add a comment | 

1

$begingroup$

If $f(p)>p$, then there is a neighborhood of $p$ on which $f(x)>p$. Therefore, that neighborhood would belong to the attractive basin of $b$. This is a contradiction, since all points $<p$ belong to the attractive basin of $a$.

Similarly, if $f(p)<p$ then a neighborhood of $p$ would belong to the attractive basin of $a$, which also cannot be since all points $>p$ must belong to the basin of $b$.

Therefore, $f(p)=p$, i.e. $p$ is also a fixed point.

The information given doesn't tell you anything about the differentiability at $p$. You were given the example of $frac4piarctan(x)$ with $p=0$, which is differentiable at $p$. You can bend it a bit to produce an example that is not differentiable at $p$:

$$f(x)=begincasesfrac4piarctan(x),&xgeq0\frac4(1+epsilon)piarctanleft(x/(1+epsilon)right),&x>0endcases$$

share|cite|improve this answer

edited Apr 2 at 19:10

answered Apr 2 at 18:48

user647486user647486

917111

$endgroup$

add a comment | 

1

$begingroup$

If $f(p)>p$, then there is a neighborhood of $p$ on which $f(x)>p$. Therefore, that neighborhood would belong to the attractive basin of $b$. This is a contradiction, since all points $<p$ belong to the attractive basin of $a$.

Similarly, if $f(p)<p$ then a neighborhood of $p$ would belong to the attractive basin of $a$, which also cannot be since all points $>p$ must belong to the basin of $b$.

Therefore, $f(p)=p$, i.e. $p$ is also a fixed point.

The information given doesn't tell you anything about the differentiability at $p$. You were given the example of $frac4piarctan(x)$ with $p=0$, which is differentiable at $p$. You can bend it a bit to produce an example that is not differentiable at $p$:

$$f(x)=begincasesfrac4piarctan(x),&xgeq0\frac4(1+epsilon)piarctanleft(x/(1+epsilon)right),&x>0endcases$$

share|cite|improve this answer

edited Apr 2 at 19:10

answered Apr 2 at 18:48

user647486user647486

917111

$endgroup$

add a comment | 

$begingroup$

If $f(p)>p$, then there is a neighborhood of $p$ on which $f(x)>p$. Therefore, that neighborhood would belong to the attractive basin of $b$. This is a contradiction, since all points $<p$ belong to the attractive basin of $a$.

Similarly, if $f(p)<p$ then a neighborhood of $p$ would belong to the attractive basin of $a$, which also cannot be since all points $>p$ must belong to the basin of $b$.

Therefore, $f(p)=p$, i.e. $p$ is also a fixed point.

The information given doesn't tell you anything about the differentiability at $p$. You were given the example of $frac4piarctan(x)$ with $p=0$, which is differentiable at $p$. You can bend it a bit to produce an example that is not differentiable at $p$:

$$f(x)=begincasesfrac4piarctan(x),&xgeq0\frac4(1+epsilon)piarctanleft(x/(1+epsilon)right),&x>0endcases$$

share|cite|improve this answer

edited Apr 2 at 19:10

answered Apr 2 at 18:48

user647486user647486

917111

$endgroup$

share|cite|improve this answer

edited Apr 2 at 19:10

answered Apr 2 at 18:48

user647486user647486

917111

answered Apr 2 at 18:48

user647486user647486

917111

answered Apr 2 at 18:48

user647486user647486

917111

0

$begingroup$

$$ f(x) = frac4pi ; arctan x $$

share|cite|improve this answer

answered Apr 2 at 17:50

Will JagyWill Jagy

105k5103202

$endgroup$

add a comment | 

0

$begingroup$

$$ f(x) = frac4pi ; arctan x $$

share|cite|improve this answer

answered Apr 2 at 17:50

Will JagyWill Jagy

105k5103202

$endgroup$

add a comment | 

$begingroup$

$$ f(x) = frac4pi ; arctan x $$

share|cite|improve this answer

answered Apr 2 at 17:50

Will JagyWill Jagy

105k5103202

$endgroup$

share|cite|improve this answer

answered Apr 2 at 17:50

Will JagyWill Jagy

105k5103202

answered Apr 2 at 17:50

Will JagyWill Jagy

105k5103202

answered Apr 2 at 17:50

Will JagyWill Jagy

105k5103202