Understanding Rate/Speed of Convergence of a sequence.
$begingroup$
Consider the following text taken from this link.
What does it say?
As far as I can get:
If $p_n$ (i.e. $p_0, p_1, p_2, ...$) is a sequence, $p_n$'s point of convergence is $p$, $lambda$ and $alpha$ (where $alpha < 0$) are constants, and $$lim_{n rightarrow infty} frac{left| p_{n+1} - p right|}{left|p_n - pright|} = lambda $$ (i.e. if $n$ goes to infinity, then, the ratio of "difference of the $n$-th $p$ and $p$", and the "$n+1$-th $p$ and $p$" would be $lambda$)
[$text{What does it mean though?}$],
then we can say that Rate of Convergence is ... ????
Can anyone help me to complete the rest of the text in pain English?
convergence nonlinear-system
asked Dec 3 '18 at 18:25
user366312user366312
591317
$endgroup$
add a comment |
$begingroup$
Consider the following text taken from this link.
What does it say?
As far as I can get:
If $p_n$ (i.e. $p_0, p_1, p_2, ...$) is a sequence, $p_n$'s point of convergence is $p$, $lambda$ and $alpha$ (where $alpha < 0$) are constants, and $$lim_{n rightarrow infty} frac{left| p_{n+1} - p right|}{left|p_n - pright|} = lambda $$ (i.e. if $n$ goes to infinity, then, the ratio of "difference of the $n$-th $p$ and $p$", and the "$n+1$-th $p$ and $p$" would be $lambda$)
[$text{What does it mean though?}$],
then we can say that Rate of Convergence is ... ????
Can anyone help me to complete the rest of the text in pain English?
convergence nonlinear-system
asked Dec 3 '18 at 18:25
user366312user366312
591317
$endgroup$
add a comment |
$begingroup$
Consider the following text taken from this link.
What does it say?
As far as I can get:
If $p_n$ (i.e. $p_0, p_1, p_2, ...$) is a sequence, $p_n$'s point of convergence is $p$, $lambda$ and $alpha$ (where $alpha < 0$) are constants, and $$lim_{n rightarrow infty} frac{left| p_{n+1} - p right|}{left|p_n - pright|} = lambda $$ (i.e. if $n$ goes to infinity, then, the ratio of "difference of the $n$-th $p$ and $p$", and the "$n+1$-th $p$ and $p$" would be $lambda$)
[$text{What does it mean though?}$],
then we can say that Rate of Convergence is ... ????
Can anyone help me to complete the rest of the text in pain English?
convergence nonlinear-system
asked Dec 3 '18 at 18:25
user366312user366312
591317
$endgroup$
Consider the following text taken from this link.
What does it say?
As far as I can get:
If $p_n$ (i.e. $p_0, p_1, p_2, ...$) is a sequence, $p_n$'s point of convergence is $p$, $lambda$ and $alpha$ (where $alpha < 0$) are constants, and $$lim_{n rightarrow infty} frac{left| p_{n+1} - p right|}{left|p_n - pright|} = lambda $$ (i.e. if $n$ goes to infinity, then, the ratio of "difference of the $n$-th $p$ and $p$", and the "$n+1$-th $p$ and $p$" would be $lambda$)
[$text{What does it mean though?}$],
then we can say that Rate of Convergence is ... ????
Can anyone help me to complete the rest of the text in pain English?
convergence nonlinear-system
convergence nonlinear-system
asked Dec 3 '18 at 18:25
user366312user366312
591317
asked Dec 3 '18 at 18:25
user366312user366312
591317
asked Dec 3 '18 at 18:25
user366312user366312
591317
asked Dec 3 '18 at 18:25
user366312user366312
591317
asked Dec 3 '18 at 18:25
user366312user366312
591317
591317
add a comment |
add a comment |
1 Answer
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$begingroup$
You've got the $alpha$ inequality wrong (it's $alpha>0$) and you're missing an $alpha$ in the denominator, that should read $|p_n-p|^alpha$. Perhaps it will help you to write
$$lim_{n to infty}frac{|p_{n+1}-p|}{|p_{n}-p|^alpha}=lim_{n to infty}frac{|frac{p_{n+1}}{p}-1|}{|frac{p_n}{p}-1|^alpha}.$$
To me, this means that the rate of change of the series tends to stabilize as $n$ grows.
answered Dec 3 '18 at 18:49
PatricioPatricio
3076
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1 Answer
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1 Answer
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votes
$begingroup$
You've got the $alpha$ inequality wrong (it's $alpha>0$) and you're missing an $alpha$ in the denominator, that should read $|p_n-p|^alpha$. Perhaps it will help you to write
$$lim_{n to infty}frac{|p_{n+1}-p|}{|p_{n}-p|^alpha}=lim_{n to infty}frac{|frac{p_{n+1}}{p}-1|}{|frac{p_n}{p}-1|^alpha}.$$
To me, this means that the rate of change of the series tends to stabilize as $n$ grows.
answered Dec 3 '18 at 18:49
PatricioPatricio
3076
$endgroup$
add a comment |
$begingroup$
You've got the $alpha$ inequality wrong (it's $alpha>0$) and you're missing an $alpha$ in the denominator, that should read $|p_n-p|^alpha$. Perhaps it will help you to write
$$lim_{n to infty}frac{|p_{n+1}-p|}{|p_{n}-p|^alpha}=lim_{n to infty}frac{|frac{p_{n+1}}{p}-1|}{|frac{p_n}{p}-1|^alpha}.$$
To me, this means that the rate of change of the series tends to stabilize as $n$ grows.
answered Dec 3 '18 at 18:49
PatricioPatricio
3076
$endgroup$
add a comment |
$begingroup$
You've got the $alpha$ inequality wrong (it's $alpha>0$) and you're missing an $alpha$ in the denominator, that should read $|p_n-p|^alpha$. Perhaps it will help you to write
$$lim_{n to infty}frac{|p_{n+1}-p|}{|p_{n}-p|^alpha}=lim_{n to infty}frac{|frac{p_{n+1}}{p}-1|}{|frac{p_n}{p}-1|^alpha}.$$
To me, this means that the rate of change of the series tends to stabilize as $n$ grows.
answered Dec 3 '18 at 18:49
PatricioPatricio
3076
$endgroup$
You've got the $alpha$ inequality wrong (it's $alpha>0$) and you're missing an $alpha$ in the denominator, that should read $|p_n-p|^alpha$. Perhaps it will help you to write
$$lim_{n to infty}frac{|p_{n+1}-p|}{|p_{n}-p|^alpha}=lim_{n to infty}frac{|frac{p_{n+1}}{p}-1|}{|frac{p_n}{p}-1|^alpha}.$$
To me, this means that the rate of change of the series tends to stabilize as $n$ grows.
answered Dec 3 '18 at 18:49
PatricioPatricio
3076
answered Dec 3 '18 at 18:49
PatricioPatricio
3076
answered Dec 3 '18 at 18:49
PatricioPatricio
3076
answered Dec 3 '18 at 18:49
PatricioPatricio
3076
3076
add a comment |
add a comment |
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