Mixing
Find limit of $a_n = ln (1 + a_{n-1})$
Sequence $a_n$ is defined as $a_n = ln (1 + a_{n-1})$, where $a_0 > -1$. Find all $a_0$ for which the sequence converges.
Using arithmetic properties of limits, I've found that if the sequence converges, then it converges to $0$, but I don't know how to prove it and find $a_0$.
calculus limits
asked Nov 20 '18 at 23:24
user4201961
634411
add a comment |
Sequence $a_n$ is defined as $a_n = ln (1 + a_{n-1})$, where $a_0 > -1$. Find all $a_0$ for which the sequence converges.
Using arithmetic properties of limits, I've found that if the sequence converges, then it converges to $0$, but I don't know how to prove it and find $a_0$.
calculus limits
asked Nov 20 '18 at 23:24
user4201961
634411
add a comment |
Sequence $a_n$ is defined as $a_n = ln (1 + a_{n-1})$, where $a_0 > -1$. Find all $a_0$ for which the sequence converges.
Using arithmetic properties of limits, I've found that if the sequence converges, then it converges to $0$, but I don't know how to prove it and find $a_0$.
calculus limits
asked Nov 20 '18 at 23:24
user4201961
634411
Sequence $a_n$ is defined as $a_n = ln (1 + a_{n-1})$, where $a_0 > -1$. Find all $a_0$ for which the sequence converges.
Using arithmetic properties of limits, I've found that if the sequence converges, then it converges to $0$, but I don't know how to prove it and find $a_0$.
calculus limits
calculus limits
asked Nov 20 '18 at 23:24
user4201961
634411
asked Nov 20 '18 at 23:24
user4201961
634411
asked Nov 20 '18 at 23:24
user4201961
634411
asked Nov 20 '18 at 23:24
user4201961
634411
asked Nov 20 '18 at 23:24
user4201961
634411
634411
add a comment |
add a comment |
4 Answers
4
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oldest
votes
You are correct in finding the limit in case of existence to be $0$
Notice that you are finding the fixed point of the function $$f(x)=ln (1+x)$$
The derivative of your function is $$f'(x) = frac {1}{1+x}$$
As you notice the derivative is positive for the domain of your function which is $x>-1$
We have $frac {1}{1+x} <1$ for $x>0$ and $frac {1}{1+x} >1$ for $-1<x<0$
Thuerefore the sequence will converge to $0$ for $x_0>0$ and diverge for $x_0<0$
answered Nov 20 '18 at 23:49
Mohammad Riazi-Kermani
41k42059
When $x_0<0$, $(a_n)$ is not a sequence.
– hamam_Abdallah
Nov 21 '18 at 11:02
add a comment |
hint
for $x>-1$
$$ln(1+x)le x$$
the sequence is decreasing.
if $-1<a_0<0$ the sequence is not well defined
if $a_0=0$ it is constant.
if $a_0>0$ the sequence is decreasing and positive.
the unique fixed point is zero . the sequence goes to zero.
answered Nov 20 '18 at 23:46
hamam_Abdallah
38k21634
add a comment |
Note that we must have $a_n > 1$ for all $n$, else $a_{n+1}$ is not even well defined (I'll assume this silently throughout).
We note that $a_n geq a_{n+1}$ if and only if $e^{a_n}geq 1 + a_n$. Given the power series expansion of $e^x$, we have $$e^{a_n} - (1 + a_n) = sumlimits_{i=2}^{infty}frac{a_n^i}{i!},$$ which is clearly non-negative for all $a_n$. Thus, the sequence is decreasing, and hence monotonic, for any $a_0$.
Now, decreasing sequences converge if and only if they are bounded below, and given what you already know about what it must converge to, you essentially already know the answer, so we just need to show that the sequence is bounded below for $a_0 geq 0$, but not for $a_0 < 0$.
We'll do those in order: if some $a_n geq 0$, then $a_{n+1} = ln(1 + a_n) geq ln(1) = 0$, so indeed the sequence is bounded below by $0$ for $a_0 geq 0$, and so converges for all such values.
On the other hand, if $a_0 < 0$, then, since our sequence is decreasing, it cannot converge to $0$ (it ever gets within a distance $|a_0|$ of $0$), so, by the argument that you already have, cannot converge.
Thus, our sequence converges exactly for $a_0 geq 0$.
answered Nov 20 '18 at 23:49
user3482749
2,672414
add a comment |
HINT
We can show that as $a_0 >0$
$a_n$ is decreasing
$a_n$ is bounded below
answered Nov 20 '18 at 23:37
gimusi
1
add a comment |
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4 Answers
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4 Answers
4
active
oldest
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votes
You are correct in finding the limit in case of existence to be $0$
Notice that you are finding the fixed point of the function $$f(x)=ln (1+x)$$
The derivative of your function is $$f'(x) = frac {1}{1+x}$$
As you notice the derivative is positive for the domain of your function which is $x>-1$
We have $frac {1}{1+x} <1$ for $x>0$ and $frac {1}{1+x} >1$ for $-1<x<0$
Thuerefore the sequence will converge to $0$ for $x_0>0$ and diverge for $x_0<0$
answered Nov 20 '18 at 23:49
Mohammad Riazi-Kermani
41k42059
When $x_0<0$, $(a_n)$ is not a sequence.
– hamam_Abdallah
Nov 21 '18 at 11:02
add a comment |
You are correct in finding the limit in case of existence to be $0$
Notice that you are finding the fixed point of the function $$f(x)=ln (1+x)$$
The derivative of your function is $$f'(x) = frac {1}{1+x}$$
As you notice the derivative is positive for the domain of your function which is $x>-1$
We have $frac {1}{1+x} <1$ for $x>0$ and $frac {1}{1+x} >1$ for $-1<x<0$
Thuerefore the sequence will converge to $0$ for $x_0>0$ and diverge for $x_0<0$
answered Nov 20 '18 at 23:49
Mohammad Riazi-Kermani
41k42059
When $x_0<0$, $(a_n)$ is not a sequence.
– hamam_Abdallah
Nov 21 '18 at 11:02
add a comment |
You are correct in finding the limit in case of existence to be $0$
Notice that you are finding the fixed point of the function $$f(x)=ln (1+x)$$
The derivative of your function is $$f'(x) = frac {1}{1+x}$$
As you notice the derivative is positive for the domain of your function which is $x>-1$
We have $frac {1}{1+x} <1$ for $x>0$ and $frac {1}{1+x} >1$ for $-1<x<0$
Thuerefore the sequence will converge to $0$ for $x_0>0$ and diverge for $x_0<0$
answered Nov 20 '18 at 23:49
Mohammad Riazi-Kermani
41k42059
You are correct in finding the limit in case of existence to be $0$
Notice that you are finding the fixed point of the function $$f(x)=ln (1+x)$$
The derivative of your function is $$f'(x) = frac {1}{1+x}$$
As you notice the derivative is positive for the domain of your function which is $x>-1$
We have $frac {1}{1+x} <1$ for $x>0$ and $frac {1}{1+x} >1$ for $-1<x<0$
Thuerefore the sequence will converge to $0$ for $x_0>0$ and diverge for $x_0<0$
answered Nov 20 '18 at 23:49
Mohammad Riazi-Kermani
41k42059
answered Nov 20 '18 at 23:49
Mohammad Riazi-Kermani
41k42059
answered Nov 20 '18 at 23:49
Mohammad Riazi-Kermani
41k42059
answered Nov 20 '18 at 23:49
Mohammad Riazi-Kermani
41k42059
41k42059
When $x_0<0$, $(a_n)$ is not a sequence.
– hamam_Abdallah
Nov 21 '18 at 11:02
add a comment |
When $x_0<0$, $(a_n)$ is not a sequence.
– hamam_Abdallah
Nov 21 '18 at 11:02
When $x_0<0$, $(a_n)$ is not a sequence.
– hamam_Abdallah
Nov 21 '18 at 11:02
When $x_0<0$, $(a_n)$ is not a sequence.
– hamam_Abdallah
Nov 21 '18 at 11:02
add a comment |
hint
for $x>-1$
$$ln(1+x)le x$$
the sequence is decreasing.
if $-1<a_0<0$ the sequence is not well defined
if $a_0=0$ it is constant.
if $a_0>0$ the sequence is decreasing and positive.
the unique fixed point is zero . the sequence goes to zero.
answered Nov 20 '18 at 23:46
hamam_Abdallah
38k21634
add a comment |
hint
for $x>-1$
$$ln(1+x)le x$$
the sequence is decreasing.
if $-1<a_0<0$ the sequence is not well defined
if $a_0=0$ it is constant.
if $a_0>0$ the sequence is decreasing and positive.
the unique fixed point is zero . the sequence goes to zero.
answered Nov 20 '18 at 23:46
hamam_Abdallah
38k21634
add a comment |
hint
for $x>-1$
$$ln(1+x)le x$$
the sequence is decreasing.
if $-1<a_0<0$ the sequence is not well defined
if $a_0=0$ it is constant.
if $a_0>0$ the sequence is decreasing and positive.
the unique fixed point is zero . the sequence goes to zero.
answered Nov 20 '18 at 23:46
hamam_Abdallah
38k21634
hint
for $x>-1$
$$ln(1+x)le x$$
the sequence is decreasing.
if $-1<a_0<0$ the sequence is not well defined
if $a_0=0$ it is constant.
if $a_0>0$ the sequence is decreasing and positive.
the unique fixed point is zero . the sequence goes to zero.
answered Nov 20 '18 at 23:46
hamam_Abdallah
38k21634
answered Nov 20 '18 at 23:46
hamam_Abdallah
38k21634
answered Nov 20 '18 at 23:46
hamam_Abdallah
38k21634
answered Nov 20 '18 at 23:46
hamam_Abdallah
38k21634
38k21634
add a comment |
add a comment |
Note that we must have $a_n > 1$ for all $n$, else $a_{n+1}$ is not even well defined (I'll assume this silently throughout).
We note that $a_n geq a_{n+1}$ if and only if $e^{a_n}geq 1 + a_n$. Given the power series expansion of $e^x$, we have $$e^{a_n} - (1 + a_n) = sumlimits_{i=2}^{infty}frac{a_n^i}{i!},$$ which is clearly non-negative for all $a_n$. Thus, the sequence is decreasing, and hence monotonic, for any $a_0$.
Now, decreasing sequences converge if and only if they are bounded below, and given what you already know about what it must converge to, you essentially already know the answer, so we just need to show that the sequence is bounded below for $a_0 geq 0$, but not for $a_0 < 0$.
We'll do those in order: if some $a_n geq 0$, then $a_{n+1} = ln(1 + a_n) geq ln(1) = 0$, so indeed the sequence is bounded below by $0$ for $a_0 geq 0$, and so converges for all such values.
On the other hand, if $a_0 < 0$, then, since our sequence is decreasing, it cannot converge to $0$ (it ever gets within a distance $|a_0|$ of $0$), so, by the argument that you already have, cannot converge.
Thus, our sequence converges exactly for $a_0 geq 0$.
answered Nov 20 '18 at 23:49
user3482749
2,672414
add a comment |
Note that we must have $a_n > 1$ for all $n$, else $a_{n+1}$ is not even well defined (I'll assume this silently throughout).
We note that $a_n geq a_{n+1}$ if and only if $e^{a_n}geq 1 + a_n$. Given the power series expansion of $e^x$, we have $$e^{a_n} - (1 + a_n) = sumlimits_{i=2}^{infty}frac{a_n^i}{i!},$$ which is clearly non-negative for all $a_n$. Thus, the sequence is decreasing, and hence monotonic, for any $a_0$.
Now, decreasing sequences converge if and only if they are bounded below, and given what you already know about what it must converge to, you essentially already know the answer, so we just need to show that the sequence is bounded below for $a_0 geq 0$, but not for $a_0 < 0$.
We'll do those in order: if some $a_n geq 0$, then $a_{n+1} = ln(1 + a_n) geq ln(1) = 0$, so indeed the sequence is bounded below by $0$ for $a_0 geq 0$, and so converges for all such values.
On the other hand, if $a_0 < 0$, then, since our sequence is decreasing, it cannot converge to $0$ (it ever gets within a distance $|a_0|$ of $0$), so, by the argument that you already have, cannot converge.
Thus, our sequence converges exactly for $a_0 geq 0$.
answered Nov 20 '18 at 23:49
user3482749
2,672414
add a comment |
Note that we must have $a_n > 1$ for all $n$, else $a_{n+1}$ is not even well defined (I'll assume this silently throughout).
We note that $a_n geq a_{n+1}$ if and only if $e^{a_n}geq 1 + a_n$. Given the power series expansion of $e^x$, we have $$e^{a_n} - (1 + a_n) = sumlimits_{i=2}^{infty}frac{a_n^i}{i!},$$ which is clearly non-negative for all $a_n$. Thus, the sequence is decreasing, and hence monotonic, for any $a_0$.
Now, decreasing sequences converge if and only if they are bounded below, and given what you already know about what it must converge to, you essentially already know the answer, so we just need to show that the sequence is bounded below for $a_0 geq 0$, but not for $a_0 < 0$.
We'll do those in order: if some $a_n geq 0$, then $a_{n+1} = ln(1 + a_n) geq ln(1) = 0$, so indeed the sequence is bounded below by $0$ for $a_0 geq 0$, and so converges for all such values.
On the other hand, if $a_0 < 0$, then, since our sequence is decreasing, it cannot converge to $0$ (it ever gets within a distance $|a_0|$ of $0$), so, by the argument that you already have, cannot converge.
Thus, our sequence converges exactly for $a_0 geq 0$.
answered Nov 20 '18 at 23:49
user3482749
2,672414
Note that we must have $a_n > 1$ for all $n$, else $a_{n+1}$ is not even well defined (I'll assume this silently throughout).
We note that $a_n geq a_{n+1}$ if and only if $e^{a_n}geq 1 + a_n$. Given the power series expansion of $e^x$, we have $$e^{a_n} - (1 + a_n) = sumlimits_{i=2}^{infty}frac{a_n^i}{i!},$$ which is clearly non-negative for all $a_n$. Thus, the sequence is decreasing, and hence monotonic, for any $a_0$.
Now, decreasing sequences converge if and only if they are bounded below, and given what you already know about what it must converge to, you essentially already know the answer, so we just need to show that the sequence is bounded below for $a_0 geq 0$, but not for $a_0 < 0$.
We'll do those in order: if some $a_n geq 0$, then $a_{n+1} = ln(1 + a_n) geq ln(1) = 0$, so indeed the sequence is bounded below by $0$ for $a_0 geq 0$, and so converges for all such values.
On the other hand, if $a_0 < 0$, then, since our sequence is decreasing, it cannot converge to $0$ (it ever gets within a distance $|a_0|$ of $0$), so, by the argument that you already have, cannot converge.
Thus, our sequence converges exactly for $a_0 geq 0$.
answered Nov 20 '18 at 23:49
user3482749
2,672414
answered Nov 20 '18 at 23:49
user3482749
2,672414
answered Nov 20 '18 at 23:49
user3482749
2,672414
answered Nov 20 '18 at 23:49
user3482749
2,672414
2,672414
add a comment |
add a comment |
HINT
We can show that as $a_0 >0$
$a_n$ is decreasing
$a_n$ is bounded below
answered Nov 20 '18 at 23:37
gimusi
1
add a comment |
HINT
We can show that as $a_0 >0$
$a_n$ is decreasing
$a_n$ is bounded below
answered Nov 20 '18 at 23:37
gimusi
1
add a comment |
HINT
We can show that as $a_0 >0$
$a_n$ is decreasing
$a_n$ is bounded below
answered Nov 20 '18 at 23:37
gimusi
1
HINT
We can show that as $a_0 >0$
$a_n$ is decreasing
$a_n$ is bounded below
answered Nov 20 '18 at 23:37
gimusi
1
edited Nov 20 '18 at 23:48
answered Nov 20 '18 at 23:37
gimusi
1
answered Nov 20 '18 at 23:37
gimusi
1
answered Nov 20 '18 at 23:37
gimusi
1
1
add a comment |
add a comment |
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