Convergence of Shanks Transformation
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If sequence $A_n rightarrow 0$ as $n rightarrow infty$, and the Shanks Transformation of $A_n$ defined as $$Sleft(A_nright)=frac{A_{n+1}A_{n-1}-A_n^2}{A_{n+1}+A_{n-1}-2A_n}$$ also converges, prove $Sleft(A_nright)$ converges to zero.
It's problem 8-1 in Carl M. Bender's book, Advanced Mathematical Methods for Scientists and Engineers. I tried the definition of limit but couldn't finish the proof.
sequences-and-series limits convergence
edited Jan 20 at 22:43
Davide Giraudo
128k17156268
asked Jan 17 at 6:11
BruceBruce
114
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add a comment |
$begingroup$
If sequence $A_n rightarrow 0$ as $n rightarrow infty$, and the Shanks Transformation of $A_n$ defined as $$Sleft(A_nright)=frac{A_{n+1}A_{n-1}-A_n^2}{A_{n+1}+A_{n-1}-2A_n}$$ also converges, prove $Sleft(A_nright)$ converges to zero.
It's problem 8-1 in Carl M. Bender's book, Advanced Mathematical Methods for Scientists and Engineers. I tried the definition of limit but couldn't finish the proof.
sequences-and-series limits convergence
edited Jan 20 at 22:43
Davide Giraudo
128k17156268
asked Jan 17 at 6:11
BruceBruce
114
$endgroup$
add a comment |
$begingroup$
If sequence $A_n rightarrow 0$ as $n rightarrow infty$, and the Shanks Transformation of $A_n$ defined as $$Sleft(A_nright)=frac{A_{n+1}A_{n-1}-A_n^2}{A_{n+1}+A_{n-1}-2A_n}$$ also converges, prove $Sleft(A_nright)$ converges to zero.
It's problem 8-1 in Carl M. Bender's book, Advanced Mathematical Methods for Scientists and Engineers. I tried the definition of limit but couldn't finish the proof.
sequences-and-series limits convergence
edited Jan 20 at 22:43
Davide Giraudo
128k17156268
asked Jan 17 at 6:11
BruceBruce
114
$endgroup$
If sequence $A_n rightarrow 0$ as $n rightarrow infty$, and the Shanks Transformation of $A_n$ defined as $$Sleft(A_nright)=frac{A_{n+1}A_{n-1}-A_n^2}{A_{n+1}+A_{n-1}-2A_n}$$ also converges, prove $Sleft(A_nright)$ converges to zero.
It's problem 8-1 in Carl M. Bender's book, Advanced Mathematical Methods for Scientists and Engineers. I tried the definition of limit but couldn't finish the proof.
sequences-and-series limits convergence
sequences-and-series limits convergence
edited Jan 20 at 22:43
Davide Giraudo
128k17156268
asked Jan 17 at 6:11
BruceBruce
114
edited Jan 20 at 22:43
Davide Giraudo
128k17156268
asked Jan 17 at 6:11
BruceBruce
114
edited Jan 20 at 22:43
Davide Giraudo
128k17156268
edited Jan 20 at 22:43
Davide Giraudo
128k17156268
edited Jan 20 at 22:43
Davide Giraudo
128k17156268
128k17156268
asked Jan 17 at 6:11
BruceBruce
114
asked Jan 17 at 6:11
BruceBruce
114
asked Jan 17 at 6:11
BruceBruce
114
114
add a comment |
add a comment |
1 Answer
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Equivalently $S(A_n)=A_{n-1}-frac{(Delta A_{n-1})^2}{Delta^2 A_{n-1}}$ with the forward difference operator $Delta u_n:=u_{n+1}-u_n$. (Note the problem statement should have included $Delta^2 A_nne 0$.) Thus $$frac{S(A_n)}{A_{n-1}}=1-frac{Delta A_{n-1}}{A_{n-1}}cdotfrac{Delta A_{n-1}}{Delta^2A_{n-1}}.$$By Stolz–Cesàro this has limit $1-1=0$, so $S_n$ has limit $0times 0=0$.
answered Jan 21 at 6:53
J.G.J.G.
33.3k23252
$endgroup$
$begingroup$
Thanks. I haven't heart about Stolz–Cesàro theorem before but wiki says that it can be viewed as a l'Hôpital's rule for sequences, which makes this easier to grasp. However, the statement of Stolz–Cesàro theorem says that the sequence in the denominator is strickly monotonic and divergence, while in our case, the sequence $A_{n-1}$ is neither monotonic nor divergence. So why can we apply this theorem?
$endgroup$
– Bruce
Feb 2 at 14:49
$begingroup$
@Bruce You can get divergence by working with sequences such as $1/A_n$.
$endgroup$
– J.G.
Feb 2 at 15:07
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Equivalently $S(A_n)=A_{n-1}-frac{(Delta A_{n-1})^2}{Delta^2 A_{n-1}}$ with the forward difference operator $Delta u_n:=u_{n+1}-u_n$. (Note the problem statement should have included $Delta^2 A_nne 0$.) Thus $$frac{S(A_n)}{A_{n-1}}=1-frac{Delta A_{n-1}}{A_{n-1}}cdotfrac{Delta A_{n-1}}{Delta^2A_{n-1}}.$$By Stolz–Cesàro this has limit $1-1=0$, so $S_n$ has limit $0times 0=0$.
answered Jan 21 at 6:53
J.G.J.G.
33.3k23252
$endgroup$
$begingroup$
Thanks. I haven't heart about Stolz–Cesàro theorem before but wiki says that it can be viewed as a l'Hôpital's rule for sequences, which makes this easier to grasp. However, the statement of Stolz–Cesàro theorem says that the sequence in the denominator is strickly monotonic and divergence, while in our case, the sequence $A_{n-1}$ is neither monotonic nor divergence. So why can we apply this theorem?
$endgroup$
– Bruce
Feb 2 at 14:49
$begingroup$
@Bruce You can get divergence by working with sequences such as $1/A_n$.
$endgroup$
– J.G.
Feb 2 at 15:07
add a comment |
$begingroup$
Equivalently $S(A_n)=A_{n-1}-frac{(Delta A_{n-1})^2}{Delta^2 A_{n-1}}$ with the forward difference operator $Delta u_n:=u_{n+1}-u_n$. (Note the problem statement should have included $Delta^2 A_nne 0$.) Thus $$frac{S(A_n)}{A_{n-1}}=1-frac{Delta A_{n-1}}{A_{n-1}}cdotfrac{Delta A_{n-1}}{Delta^2A_{n-1}}.$$By Stolz–Cesàro this has limit $1-1=0$, so $S_n$ has limit $0times 0=0$.
answered Jan 21 at 6:53
J.G.J.G.
33.3k23252
$endgroup$
$begingroup$
Thanks. I haven't heart about Stolz–Cesàro theorem before but wiki says that it can be viewed as a l'Hôpital's rule for sequences, which makes this easier to grasp. However, the statement of Stolz–Cesàro theorem says that the sequence in the denominator is strickly monotonic and divergence, while in our case, the sequence $A_{n-1}$ is neither monotonic nor divergence. So why can we apply this theorem?
$endgroup$
– Bruce
Feb 2 at 14:49
$begingroup$
@Bruce You can get divergence by working with sequences such as $1/A_n$.
$endgroup$
– J.G.
Feb 2 at 15:07
add a comment |
$begingroup$
Equivalently $S(A_n)=A_{n-1}-frac{(Delta A_{n-1})^2}{Delta^2 A_{n-1}}$ with the forward difference operator $Delta u_n:=u_{n+1}-u_n$. (Note the problem statement should have included $Delta^2 A_nne 0$.) Thus $$frac{S(A_n)}{A_{n-1}}=1-frac{Delta A_{n-1}}{A_{n-1}}cdotfrac{Delta A_{n-1}}{Delta^2A_{n-1}}.$$By Stolz–Cesàro this has limit $1-1=0$, so $S_n$ has limit $0times 0=0$.
answered Jan 21 at 6:53
J.G.J.G.
33.3k23252
$endgroup$
Equivalently $S(A_n)=A_{n-1}-frac{(Delta A_{n-1})^2}{Delta^2 A_{n-1}}$ with the forward difference operator $Delta u_n:=u_{n+1}-u_n$. (Note the problem statement should have included $Delta^2 A_nne 0$.) Thus $$frac{S(A_n)}{A_{n-1}}=1-frac{Delta A_{n-1}}{A_{n-1}}cdotfrac{Delta A_{n-1}}{Delta^2A_{n-1}}.$$By Stolz–Cesàro this has limit $1-1=0$, so $S_n$ has limit $0times 0=0$.
answered Jan 21 at 6:53
J.G.J.G.
33.3k23252
answered Jan 21 at 6:53
J.G.J.G.
33.3k23252
answered Jan 21 at 6:53
J.G.J.G.
33.3k23252
answered Jan 21 at 6:53
J.G.J.G.
33.3k23252
33.3k23252
$begingroup$
Thanks. I haven't heart about Stolz–Cesàro theorem before but wiki says that it can be viewed as a l'Hôpital's rule for sequences, which makes this easier to grasp. However, the statement of Stolz–Cesàro theorem says that the sequence in the denominator is strickly monotonic and divergence, while in our case, the sequence $A_{n-1}$ is neither monotonic nor divergence. So why can we apply this theorem?
$endgroup$
– Bruce
Feb 2 at 14:49
$begingroup$
@Bruce You can get divergence by working with sequences such as $1/A_n$.
$endgroup$
– J.G.
Feb 2 at 15:07
add a comment |
$begingroup$
Thanks. I haven't heart about Stolz–Cesàro theorem before but wiki says that it can be viewed as a l'Hôpital's rule for sequences, which makes this easier to grasp. However, the statement of Stolz–Cesàro theorem says that the sequence in the denominator is strickly monotonic and divergence, while in our case, the sequence $A_{n-1}$ is neither monotonic nor divergence. So why can we apply this theorem?
$endgroup$
– Bruce
Feb 2 at 14:49
$begingroup$
@Bruce You can get divergence by working with sequences such as $1/A_n$.
$endgroup$
– J.G.
Feb 2 at 15:07
$begingroup$
Thanks. I haven't heart about Stolz–Cesàro theorem before but wiki says that it can be viewed as a l'Hôpital's rule for sequences, which makes this easier to grasp. However, the statement of Stolz–Cesàro theorem says that the sequence in the denominator is strickly monotonic and divergence, while in our case, the sequence $A_{n-1}$ is neither monotonic nor divergence. So why can we apply this theorem?
$endgroup$
– Bruce
Feb 2 at 14:49
$begingroup$
Thanks. I haven't heart about Stolz–Cesàro theorem before but wiki says that it can be viewed as a l'Hôpital's rule for sequences, which makes this easier to grasp. However, the statement of Stolz–Cesàro theorem says that the sequence in the denominator is strickly monotonic and divergence, while in our case, the sequence $A_{n-1}$ is neither monotonic nor divergence. So why can we apply this theorem?
$endgroup$
– Bruce
Feb 2 at 14:49
$begingroup$
@Bruce You can get divergence by working with sequences such as $1/A_n$.
$endgroup$
– J.G.
Feb 2 at 15:07
$begingroup$
@Bruce You can get divergence by working with sequences such as $1/A_n$.
$endgroup$
– J.G.
Feb 2 at 15:07
add a comment |
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