Mixing
Convergence of slowly decreasing sequences
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Let $A$ be a large number, say $A=100$, and $a$ be a small number, say $a=0.01$.
Let $x_1=1$ and define $x_n+1-x_n=-a exp(-A/x_n)$. Is it true that $x_nto 0$?
We know that $x_n$ is strictly decreasing. But I do not know how to show that $x_n>0$ for all $n$, since mathematical induction fails here.
Note that $x=0$ is a fixed point for the iteration.
sequences-and-series analysis
asked Jul 26 at 4:21
Dong Li
702413
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up vote
4
down vote
favorite
Let $A$ be a large number, say $A=100$, and $a$ be a small number, say $a=0.01$.
Let $x_1=1$ and define $x_n+1-x_n=-a exp(-A/x_n)$. Is it true that $x_nto 0$?
We know that $x_n$ is strictly decreasing. But I do not know how to show that $x_n>0$ for all $n$, since mathematical induction fails here.
Note that $x=0$ is a fixed point for the iteration.
sequences-and-series analysis
asked Jul 26 at 4:21
Dong Li
702413
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
Let $A$ be a large number, say $A=100$, and $a$ be a small number, say $a=0.01$.
Let $x_1=1$ and define $x_n+1-x_n=-a exp(-A/x_n)$. Is it true that $x_nto 0$?
We know that $x_n$ is strictly decreasing. But I do not know how to show that $x_n>0$ for all $n$, since mathematical induction fails here.
Note that $x=0$ is a fixed point for the iteration.
sequences-and-series analysis
asked Jul 26 at 4:21
Dong Li
702413
Let $A$ be a large number, say $A=100$, and $a$ be a small number, say $a=0.01$.
Let $x_1=1$ and define $x_n+1-x_n=-a exp(-A/x_n)$. Is it true that $x_nto 0$?
We know that $x_n$ is strictly decreasing. But I do not know how to show that $x_n>0$ for all $n$, since mathematical induction fails here.
Note that $x=0$ is a fixed point for the iteration.
sequences-and-series analysis
asked Jul 26 at 4:21
Dong Li
702413
asked Jul 26 at 4:21
Dong Li
702413
asked Jul 26 at 4:21
Dong Li
702413
asked Jul 26 at 4:21
Dong Li
702413
702413
add a comment |Â
add a comment |Â
2 Answers
2
active
oldest
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up vote
1
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accepted
Hint: Â $,f(x) = x - a e^-A/x gt x - e^-1/x gt 0,$ on $,Bbb R^+,$ for $,0 lt a lt 1 lt A,$. Therefore $,x_n gt 0,$ $,implies x_n+1=f(x_n) gt 0,$.
answered Jul 26 at 5:06
dxiv
53.9k64796
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1
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The dynamical system $(mathbbR^+,h)$, where $h(x) = x - a e^-fracAx$ is well-defined if and only if $ h: mathbbR^+ rightarrow mathbbR^+$, i.e. if for $x>0$, $h(x)>0$.
You can show that this inequality is equivalent to $frace^Aa x e^frac1x > 1$. The LHS function takes on a minimum at $x = 1$, where the value is $frace^A+1a$. Hence you need a and A to satisfy the condition $frace^A+1a > 1$.
Further, if A is positive, you have a monotone decreasing sequence $x_n = h^n(1)$ that is bounded below. In that case $x_n rightarrow 0$.
The values you specified satisfy these conditions.
answered Jul 26 at 5:14
ertl
445110
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Hint: Â $,f(x) = x - a e^-A/x gt x - e^-1/x gt 0,$ on $,Bbb R^+,$ for $,0 lt a lt 1 lt A,$. Therefore $,x_n gt 0,$ $,implies x_n+1=f(x_n) gt 0,$.
answered Jul 26 at 5:06
dxiv
53.9k64796
add a comment |Â
up vote
1
down vote
accepted
Hint: Â $,f(x) = x - a e^-A/x gt x - e^-1/x gt 0,$ on $,Bbb R^+,$ for $,0 lt a lt 1 lt A,$. Therefore $,x_n gt 0,$ $,implies x_n+1=f(x_n) gt 0,$.
answered Jul 26 at 5:06
dxiv
53.9k64796
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Hint: Â $,f(x) = x - a e^-A/x gt x - e^-1/x gt 0,$ on $,Bbb R^+,$ for $,0 lt a lt 1 lt A,$. Therefore $,x_n gt 0,$ $,implies x_n+1=f(x_n) gt 0,$.
answered Jul 26 at 5:06
dxiv
53.9k64796
Hint: Â $,f(x) = x - a e^-A/x gt x - e^-1/x gt 0,$ on $,Bbb R^+,$ for $,0 lt a lt 1 lt A,$. Therefore $,x_n gt 0,$ $,implies x_n+1=f(x_n) gt 0,$.
answered Jul 26 at 5:06
dxiv
53.9k64796
answered Jul 26 at 5:06
dxiv
53.9k64796
answered Jul 26 at 5:06
dxiv
53.9k64796
answered Jul 26 at 5:06
dxiv
53.9k64796
53.9k64796
add a comment |Â
add a comment |Â
up vote
1
down vote
The dynamical system $(mathbbR^+,h)$, where $h(x) = x - a e^-fracAx$ is well-defined if and only if $ h: mathbbR^+ rightarrow mathbbR^+$, i.e. if for $x>0$, $h(x)>0$.
You can show that this inequality is equivalent to $frace^Aa x e^frac1x > 1$. The LHS function takes on a minimum at $x = 1$, where the value is $frace^A+1a$. Hence you need a and A to satisfy the condition $frace^A+1a > 1$.
Further, if A is positive, you have a monotone decreasing sequence $x_n = h^n(1)$ that is bounded below. In that case $x_n rightarrow 0$.
The values you specified satisfy these conditions.
answered Jul 26 at 5:14
ertl
445110
add a comment |Â
up vote
1
down vote
The dynamical system $(mathbbR^+,h)$, where $h(x) = x - a e^-fracAx$ is well-defined if and only if $ h: mathbbR^+ rightarrow mathbbR^+$, i.e. if for $x>0$, $h(x)>0$.
You can show that this inequality is equivalent to $frace^Aa x e^frac1x > 1$. The LHS function takes on a minimum at $x = 1$, where the value is $frace^A+1a$. Hence you need a and A to satisfy the condition $frace^A+1a > 1$.
Further, if A is positive, you have a monotone decreasing sequence $x_n = h^n(1)$ that is bounded below. In that case $x_n rightarrow 0$.
The values you specified satisfy these conditions.
answered Jul 26 at 5:14
ertl
445110
add a comment |Â
up vote
1
down vote
up vote
1
down vote
The dynamical system $(mathbbR^+,h)$, where $h(x) = x - a e^-fracAx$ is well-defined if and only if $ h: mathbbR^+ rightarrow mathbbR^+$, i.e. if for $x>0$, $h(x)>0$.
You can show that this inequality is equivalent to $frace^Aa x e^frac1x > 1$. The LHS function takes on a minimum at $x = 1$, where the value is $frace^A+1a$. Hence you need a and A to satisfy the condition $frace^A+1a > 1$.
Further, if A is positive, you have a monotone decreasing sequence $x_n = h^n(1)$ that is bounded below. In that case $x_n rightarrow 0$.
The values you specified satisfy these conditions.
answered Jul 26 at 5:14
ertl
445110
The dynamical system $(mathbbR^+,h)$, where $h(x) = x - a e^-fracAx$ is well-defined if and only if $ h: mathbbR^+ rightarrow mathbbR^+$, i.e. if for $x>0$, $h(x)>0$.
You can show that this inequality is equivalent to $frace^Aa x e^frac1x > 1$. The LHS function takes on a minimum at $x = 1$, where the value is $frace^A+1a$. Hence you need a and A to satisfy the condition $frace^A+1a > 1$.
Further, if A is positive, you have a monotone decreasing sequence $x_n = h^n(1)$ that is bounded below. In that case $x_n rightarrow 0$.
The values you specified satisfy these conditions.
answered Jul 26 at 5:14
ertl
445110
answered Jul 26 at 5:14
ertl
445110
answered Jul 26 at 5:14
ertl
445110
answered Jul 26 at 5:14
ertl
445110
445110
add a comment |Â
add a comment |Â
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