How can $ln(x+2)$ have a fixed point in $(-2,-1]$?
$begingroup$
I have to determine the fixed points of $ln(x+2)$. So as a first step I plotted $ln(x+2)-x$ and found that it does have two fixed points. One between $[0, infty]$, which is perfectly fine and one in $(-2,-1]$.
The fixed point in the first interval makes sense to me since it fulfills all requirements for the Banach fixed point theorem:
$(mathbb{R},d)$ where d is the Euclidean metric is a metric space- The derivative in the metric is $|f'(x)| leq frac{1}{2} < 1$, thus it is bounded and Lipschitz-continuous, thus it is a contraction.
But in the interval $(-2,-1]$, its derivative is unbounded (isn't it?) or at least $L > 1$, thus it is not Lipschitz-continous in that interval.
Can someone explain to me how it is then possible that $ln(x+2)$ does have a fixed point in that interval, if one of the conditions is clearly not fulfilled?
Or am I missing something/made a mistake?
logarithms fixed-point-theorems lipschitz-functions
edited Jan 2 at 15:45
Clement C.
51k34093
asked Jan 2 at 15:30
MLKMLK
80112
$endgroup$
add a comment |
$begingroup$
I have to determine the fixed points of $ln(x+2)$. So as a first step I plotted $ln(x+2)-x$ and found that it does have two fixed points. One between $[0, infty]$, which is perfectly fine and one in $(-2,-1]$.
The fixed point in the first interval makes sense to me since it fulfills all requirements for the Banach fixed point theorem:
$(mathbb{R},d)$ where d is the Euclidean metric is a metric space- The derivative in the metric is $|f'(x)| leq frac{1}{2} < 1$, thus it is bounded and Lipschitz-continuous, thus it is a contraction.
But in the interval $(-2,-1]$, its derivative is unbounded (isn't it?) or at least $L > 1$, thus it is not Lipschitz-continous in that interval.
Can someone explain to me how it is then possible that $ln(x+2)$ does have a fixed point in that interval, if one of the conditions is clearly not fulfilled?
Or am I missing something/made a mistake?
logarithms fixed-point-theorems lipschitz-functions
edited Jan 2 at 15:45
Clement C.
51k34093
asked Jan 2 at 15:30
MLKMLK
80112
$endgroup$
1
$begingroup$
A fixed point doesn't mean an attractor. It just means $x_0$ such that $f(x_0)=x_0$.
$endgroup$
– Clement C.
Jan 2 at 15:40
add a comment |
$begingroup$
I have to determine the fixed points of $ln(x+2)$. So as a first step I plotted $ln(x+2)-x$ and found that it does have two fixed points. One between $[0, infty]$, which is perfectly fine and one in $(-2,-1]$.
The fixed point in the first interval makes sense to me since it fulfills all requirements for the Banach fixed point theorem:
$(mathbb{R},d)$ where d is the Euclidean metric is a metric space- The derivative in the metric is $|f'(x)| leq frac{1}{2} < 1$, thus it is bounded and Lipschitz-continuous, thus it is a contraction.
But in the interval $(-2,-1]$, its derivative is unbounded (isn't it?) or at least $L > 1$, thus it is not Lipschitz-continous in that interval.
Can someone explain to me how it is then possible that $ln(x+2)$ does have a fixed point in that interval, if one of the conditions is clearly not fulfilled?
Or am I missing something/made a mistake?
logarithms fixed-point-theorems lipschitz-functions
edited Jan 2 at 15:45
Clement C.
51k34093
asked Jan 2 at 15:30
MLKMLK
80112
$endgroup$
I have to determine the fixed points of $ln(x+2)$. So as a first step I plotted $ln(x+2)-x$ and found that it does have two fixed points. One between $[0, infty]$, which is perfectly fine and one in $(-2,-1]$.
The fixed point in the first interval makes sense to me since it fulfills all requirements for the Banach fixed point theorem:
$(mathbb{R},d)$ where d is the Euclidean metric is a metric space- The derivative in the metric is $|f'(x)| leq frac{1}{2} < 1$, thus it is bounded and Lipschitz-continuous, thus it is a contraction.
But in the interval $(-2,-1]$, its derivative is unbounded (isn't it?) or at least $L > 1$, thus it is not Lipschitz-continous in that interval.
Can someone explain to me how it is then possible that $ln(x+2)$ does have a fixed point in that interval, if one of the conditions is clearly not fulfilled?
Or am I missing something/made a mistake?
logarithms fixed-point-theorems lipschitz-functions
logarithms fixed-point-theorems lipschitz-functions
edited Jan 2 at 15:45
Clement C.
51k34093
asked Jan 2 at 15:30
MLKMLK
80112
edited Jan 2 at 15:45
Clement C.
51k34093
asked Jan 2 at 15:30
MLKMLK
80112
edited Jan 2 at 15:45
Clement C.
51k34093
edited Jan 2 at 15:45
Clement C.
51k34093
edited Jan 2 at 15:45
Clement C.
51k34093
51k34093
asked Jan 2 at 15:30
MLKMLK
80112
asked Jan 2 at 15:30
MLKMLK
80112
asked Jan 2 at 15:30
MLKMLK
80112
80112
1
$begingroup$
A fixed point doesn't mean an attractor. It just means $x_0$ such that $f(x_0)=x_0$.
$endgroup$
– Clement C.
Jan 2 at 15:40
add a comment |
1
$begingroup$
A fixed point doesn't mean an attractor. It just means $x_0$ such that $f(x_0)=x_0$.
$endgroup$
– Clement C.
Jan 2 at 15:40
1
1
$begingroup$
A fixed point doesn't mean an attractor. It just means $x_0$ such that $f(x_0)=x_0$.
$endgroup$
– Clement C.
Jan 2 at 15:40
$begingroup$
A fixed point doesn't mean an attractor. It just means $x_0$ such that $f(x_0)=x_0$.
$endgroup$
– Clement C.
Jan 2 at 15:40
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The Banach Fixed Point theorem gives a sufficient condition for the existence (and uniqueness) of a fixed point, but that condition is by no means necessary.
edited Jan 2 at 15:45
Namaste
1
answered Jan 2 at 15:40
Julián AguirreJulián Aguirre
69.5k24297
$endgroup$
$begingroup$
Maybe worth addressing the fact that the OP seems to assume fixed points have to be attractors. While a fixed point literally just means "point mapped to itself." (I.e., your answer points that out in a rather implicit way.)
$endgroup$
– Clement C.
Jan 2 at 15:45
$begingroup$
Thank you for the clarification. I really thought that every fixpoint needs to fulfill the conditions
$endgroup$
– MLK
Jan 2 at 16:01
add a comment |
$begingroup$
Your confusion is the following one:
- If a map satisfies the conditions of the Banach fix point theorem, it has a fix point.
- However the converse is not true. A map may have a fix point without satisfying the Banach fix point hypothesis.
This is the case here. $g(x) = ln(x+2)-x$ is continuous in the interval $(-2,1]$ and $g(1)>0$ while $limlimits_{x to -2^+} g(x)=-infty$. Hence $g$ vanishes in that interval, which means that $ln(x+2)$ has a fix point (even if the hypothesis of Banach fix point theorem are not met).
answered Jan 2 at 15:41
mathcounterexamples.netmathcounterexamples.net
27k22158
$endgroup$
$begingroup$
Thank you very much for the clarification :)
$endgroup$
– MLK
Jan 2 at 16:03
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The Banach Fixed Point theorem gives a sufficient condition for the existence (and uniqueness) of a fixed point, but that condition is by no means necessary.
edited Jan 2 at 15:45
Namaste
1
answered Jan 2 at 15:40
Julián AguirreJulián Aguirre
69.5k24297
$endgroup$
$begingroup$
Maybe worth addressing the fact that the OP seems to assume fixed points have to be attractors. While a fixed point literally just means "point mapped to itself." (I.e., your answer points that out in a rather implicit way.)
$endgroup$
– Clement C.
Jan 2 at 15:45
$begingroup$
Thank you for the clarification. I really thought that every fixpoint needs to fulfill the conditions
$endgroup$
– MLK
Jan 2 at 16:01
add a comment |
$begingroup$
The Banach Fixed Point theorem gives a sufficient condition for the existence (and uniqueness) of a fixed point, but that condition is by no means necessary.
edited Jan 2 at 15:45
Namaste
1
answered Jan 2 at 15:40
Julián AguirreJulián Aguirre
69.5k24297
$endgroup$
$begingroup$
Maybe worth addressing the fact that the OP seems to assume fixed points have to be attractors. While a fixed point literally just means "point mapped to itself." (I.e., your answer points that out in a rather implicit way.)
$endgroup$
– Clement C.
Jan 2 at 15:45
$begingroup$
Thank you for the clarification. I really thought that every fixpoint needs to fulfill the conditions
$endgroup$
– MLK
Jan 2 at 16:01
add a comment |
$begingroup$
The Banach Fixed Point theorem gives a sufficient condition for the existence (and uniqueness) of a fixed point, but that condition is by no means necessary.
edited Jan 2 at 15:45
Namaste
1
answered Jan 2 at 15:40
Julián AguirreJulián Aguirre
69.5k24297
$endgroup$
The Banach Fixed Point theorem gives a sufficient condition for the existence (and uniqueness) of a fixed point, but that condition is by no means necessary.
edited Jan 2 at 15:45
Namaste
1
answered Jan 2 at 15:40
Julián AguirreJulián Aguirre
69.5k24297
edited Jan 2 at 15:45
Namaste
1
edited Jan 2 at 15:45
Namaste
1
edited Jan 2 at 15:45
Namaste
1
1
answered Jan 2 at 15:40
Julián AguirreJulián Aguirre
69.5k24297
answered Jan 2 at 15:40
Julián AguirreJulián Aguirre
69.5k24297
answered Jan 2 at 15:40
Julián AguirreJulián Aguirre
69.5k24297
69.5k24297
$begingroup$
Maybe worth addressing the fact that the OP seems to assume fixed points have to be attractors. While a fixed point literally just means "point mapped to itself." (I.e., your answer points that out in a rather implicit way.)
$endgroup$
– Clement C.
Jan 2 at 15:45
$begingroup$
Thank you for the clarification. I really thought that every fixpoint needs to fulfill the conditions
$endgroup$
– MLK
Jan 2 at 16:01
add a comment |
$begingroup$
Maybe worth addressing the fact that the OP seems to assume fixed points have to be attractors. While a fixed point literally just means "point mapped to itself." (I.e., your answer points that out in a rather implicit way.)
$endgroup$
– Clement C.
Jan 2 at 15:45
$begingroup$
Thank you for the clarification. I really thought that every fixpoint needs to fulfill the conditions
$endgroup$
– MLK
Jan 2 at 16:01
$begingroup$
Maybe worth addressing the fact that the OP seems to assume fixed points have to be attractors. While a fixed point literally just means "point mapped to itself." (I.e., your answer points that out in a rather implicit way.)
$endgroup$
– Clement C.
Jan 2 at 15:45
$begingroup$
Maybe worth addressing the fact that the OP seems to assume fixed points have to be attractors. While a fixed point literally just means "point mapped to itself." (I.e., your answer points that out in a rather implicit way.)
$endgroup$
– Clement C.
Jan 2 at 15:45
$begingroup$
Thank you for the clarification. I really thought that every fixpoint needs to fulfill the conditions
$endgroup$
– MLK
Jan 2 at 16:01
$begingroup$
Thank you for the clarification. I really thought that every fixpoint needs to fulfill the conditions
$endgroup$
– MLK
Jan 2 at 16:01
add a comment |
$begingroup$
Your confusion is the following one:
- If a map satisfies the conditions of the Banach fix point theorem, it has a fix point.
- However the converse is not true. A map may have a fix point without satisfying the Banach fix point hypothesis.
This is the case here. $g(x) = ln(x+2)-x$ is continuous in the interval $(-2,1]$ and $g(1)>0$ while $limlimits_{x to -2^+} g(x)=-infty$. Hence $g$ vanishes in that interval, which means that $ln(x+2)$ has a fix point (even if the hypothesis of Banach fix point theorem are not met).
answered Jan 2 at 15:41
mathcounterexamples.netmathcounterexamples.net
27k22158
$endgroup$
$begingroup$
Thank you very much for the clarification :)
$endgroup$
– MLK
Jan 2 at 16:03
add a comment |
$begingroup$
Your confusion is the following one:
- If a map satisfies the conditions of the Banach fix point theorem, it has a fix point.
- However the converse is not true. A map may have a fix point without satisfying the Banach fix point hypothesis.
This is the case here. $g(x) = ln(x+2)-x$ is continuous in the interval $(-2,1]$ and $g(1)>0$ while $limlimits_{x to -2^+} g(x)=-infty$. Hence $g$ vanishes in that interval, which means that $ln(x+2)$ has a fix point (even if the hypothesis of Banach fix point theorem are not met).
answered Jan 2 at 15:41
mathcounterexamples.netmathcounterexamples.net
27k22158
$endgroup$
$begingroup$
Thank you very much for the clarification :)
$endgroup$
– MLK
Jan 2 at 16:03
add a comment |
$begingroup$
Your confusion is the following one:
- If a map satisfies the conditions of the Banach fix point theorem, it has a fix point.
- However the converse is not true. A map may have a fix point without satisfying the Banach fix point hypothesis.
This is the case here. $g(x) = ln(x+2)-x$ is continuous in the interval $(-2,1]$ and $g(1)>0$ while $limlimits_{x to -2^+} g(x)=-infty$. Hence $g$ vanishes in that interval, which means that $ln(x+2)$ has a fix point (even if the hypothesis of Banach fix point theorem are not met).
answered Jan 2 at 15:41
mathcounterexamples.netmathcounterexamples.net
27k22158
$endgroup$
Your confusion is the following one:
- If a map satisfies the conditions of the Banach fix point theorem, it has a fix point.
- However the converse is not true. A map may have a fix point without satisfying the Banach fix point hypothesis.
This is the case here. $g(x) = ln(x+2)-x$ is continuous in the interval $(-2,1]$ and $g(1)>0$ while $limlimits_{x to -2^+} g(x)=-infty$. Hence $g$ vanishes in that interval, which means that $ln(x+2)$ has a fix point (even if the hypothesis of Banach fix point theorem are not met).
answered Jan 2 at 15:41
mathcounterexamples.netmathcounterexamples.net
27k22158
edited Jan 2 at 15:47
answered Jan 2 at 15:41
mathcounterexamples.netmathcounterexamples.net
27k22158
answered Jan 2 at 15:41
mathcounterexamples.netmathcounterexamples.net
27k22158
answered Jan 2 at 15:41
mathcounterexamples.netmathcounterexamples.net
27k22158
27k22158
$begingroup$
Thank you very much for the clarification :)
$endgroup$
– MLK
Jan 2 at 16:03
add a comment |
$begingroup$
Thank you very much for the clarification :)
$endgroup$
– MLK
Jan 2 at 16:03
$begingroup$
Thank you very much for the clarification :)
$endgroup$
– MLK
Jan 2 at 16:03
$begingroup$
Thank you very much for the clarification :)
$endgroup$
– MLK
Jan 2 at 16:03
add a comment |
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$begingroup$
A fixed point doesn't mean an attractor. It just means $x_0$ such that $f(x_0)=x_0$.
$endgroup$
– Clement C.
Jan 2 at 15:40