Convergence acceleration of successions with logarithms
$begingroup$
I have a numerical question regarding acceleration of a succession.
A preliminary: suppose that I have a succession $a_g$ that, for high $g$, asymptotically goes as
$$
a_g=s_0+frac{s_1}g+frac{s_2}{g^2}+...=sum_{k=0}^infty frac{s_k}{g^k}.
$$
I am interested in computing the coefficients $s_k$, but in particular I am interested in computing the leading coefficient $s_0$ (that would also be the limit of the succession as $gtoinfty$). A way to accelerate this convergence is given by the Richardson transform: by defining the succession
$$
a_g^{(N)}=sum_{n=0}^N(-1)^{n+N}frac{(g+n)^N}{n!(N-n)!}a_{g+n},
$$
it can be proven that $a_g^{(N)}$ goes to the same limit as $a_g$, but the succession is accelerated as the asymptotic behavior is
$$
a_g^{(N)}simeq s_0+sum_{k=N+1}^infty frac{d_k}{g^k}.
$$
This works nicely and gives very good numerical results in the examples I've used.
The problem is that I'm now working with more general successions, of the form
$$
b_g=sum_{t=0}^Tleft(sum_{k=0}^infty frac{s_{(k,t)}}{g^k}right)frac{1}{(log g)^t}
$$
$T$ is a finite integer, the logarithm powers are finite. The coefficients $s_{(k,t)}$ are arbitrary coefficients, in principle not related to one another. As an example, the succession for $T=2$ would be
$$
sum_{k=0}^inftyfrac{s_{(k,0)}}{g^k}+frac{1}{log g}sum_{k=0}^inftyfrac{s_{(k,1)}}{g^k}+frac{1}{(log g)^2}sum_{k=0}^inftyfrac{s_{(k,2)}}{g^k}.
$$
Now convergence is way slower: as an example, for $T=1$, terms with no $1/g$ powers attached ($k=0$) are
$$
s_{(0,0)}+frac{s_{(0,1)}}{log g}.
$$
If I want to compute $s_{(0,0)}$ by computing $b_g$ for high $g$, I get very slow convergence, as $(log g)^{-1}$ goes to zero very slowly. The standard Richardson transform does not really work here.
The question: is there a generalization of this Richardson transform to successions like the $b_g$ succession?
Thanks everybody!
P.s.: this is a crossposting from https://mathoverflow.net/questions/316816/convergence-acceleration-of-successions-with-logarithms
numerical-methods convergence-acceleration
asked Dec 7 '18 at 11:48
Salvatore BaldinoSalvatore Baldino
210212
$endgroup$
add a comment |
$begingroup$
I have a numerical question regarding acceleration of a succession.
A preliminary: suppose that I have a succession $a_g$ that, for high $g$, asymptotically goes as
$$
a_g=s_0+frac{s_1}g+frac{s_2}{g^2}+...=sum_{k=0}^infty frac{s_k}{g^k}.
$$
I am interested in computing the coefficients $s_k$, but in particular I am interested in computing the leading coefficient $s_0$ (that would also be the limit of the succession as $gtoinfty$). A way to accelerate this convergence is given by the Richardson transform: by defining the succession
$$
a_g^{(N)}=sum_{n=0}^N(-1)^{n+N}frac{(g+n)^N}{n!(N-n)!}a_{g+n},
$$
it can be proven that $a_g^{(N)}$ goes to the same limit as $a_g$, but the succession is accelerated as the asymptotic behavior is
$$
a_g^{(N)}simeq s_0+sum_{k=N+1}^infty frac{d_k}{g^k}.
$$
This works nicely and gives very good numerical results in the examples I've used.
The problem is that I'm now working with more general successions, of the form
$$
b_g=sum_{t=0}^Tleft(sum_{k=0}^infty frac{s_{(k,t)}}{g^k}right)frac{1}{(log g)^t}
$$
$T$ is a finite integer, the logarithm powers are finite. The coefficients $s_{(k,t)}$ are arbitrary coefficients, in principle not related to one another. As an example, the succession for $T=2$ would be
$$
sum_{k=0}^inftyfrac{s_{(k,0)}}{g^k}+frac{1}{log g}sum_{k=0}^inftyfrac{s_{(k,1)}}{g^k}+frac{1}{(log g)^2}sum_{k=0}^inftyfrac{s_{(k,2)}}{g^k}.
$$
Now convergence is way slower: as an example, for $T=1$, terms with no $1/g$ powers attached ($k=0$) are
$$
s_{(0,0)}+frac{s_{(0,1)}}{log g}.
$$
If I want to compute $s_{(0,0)}$ by computing $b_g$ for high $g$, I get very slow convergence, as $(log g)^{-1}$ goes to zero very slowly. The standard Richardson transform does not really work here.
The question: is there a generalization of this Richardson transform to successions like the $b_g$ succession?
Thanks everybody!
P.s.: this is a crossposting from https://mathoverflow.net/questions/316816/convergence-acceleration-of-successions-with-logarithms
numerical-methods convergence-acceleration
asked Dec 7 '18 at 11:48
Salvatore BaldinoSalvatore Baldino
210212
$endgroup$
add a comment |
$begingroup$
I have a numerical question regarding acceleration of a succession.
A preliminary: suppose that I have a succession $a_g$ that, for high $g$, asymptotically goes as
$$
a_g=s_0+frac{s_1}g+frac{s_2}{g^2}+...=sum_{k=0}^infty frac{s_k}{g^k}.
$$
I am interested in computing the coefficients $s_k$, but in particular I am interested in computing the leading coefficient $s_0$ (that would also be the limit of the succession as $gtoinfty$). A way to accelerate this convergence is given by the Richardson transform: by defining the succession
$$
a_g^{(N)}=sum_{n=0}^N(-1)^{n+N}frac{(g+n)^N}{n!(N-n)!}a_{g+n},
$$
it can be proven that $a_g^{(N)}$ goes to the same limit as $a_g$, but the succession is accelerated as the asymptotic behavior is
$$
a_g^{(N)}simeq s_0+sum_{k=N+1}^infty frac{d_k}{g^k}.
$$
This works nicely and gives very good numerical results in the examples I've used.
The problem is that I'm now working with more general successions, of the form
$$
b_g=sum_{t=0}^Tleft(sum_{k=0}^infty frac{s_{(k,t)}}{g^k}right)frac{1}{(log g)^t}
$$
$T$ is a finite integer, the logarithm powers are finite. The coefficients $s_{(k,t)}$ are arbitrary coefficients, in principle not related to one another. As an example, the succession for $T=2$ would be
$$
sum_{k=0}^inftyfrac{s_{(k,0)}}{g^k}+frac{1}{log g}sum_{k=0}^inftyfrac{s_{(k,1)}}{g^k}+frac{1}{(log g)^2}sum_{k=0}^inftyfrac{s_{(k,2)}}{g^k}.
$$
Now convergence is way slower: as an example, for $T=1$, terms with no $1/g$ powers attached ($k=0$) are
$$
s_{(0,0)}+frac{s_{(0,1)}}{log g}.
$$
If I want to compute $s_{(0,0)}$ by computing $b_g$ for high $g$, I get very slow convergence, as $(log g)^{-1}$ goes to zero very slowly. The standard Richardson transform does not really work here.
The question: is there a generalization of this Richardson transform to successions like the $b_g$ succession?
Thanks everybody!
P.s.: this is a crossposting from https://mathoverflow.net/questions/316816/convergence-acceleration-of-successions-with-logarithms
numerical-methods convergence-acceleration
asked Dec 7 '18 at 11:48
Salvatore BaldinoSalvatore Baldino
210212
$endgroup$
I have a numerical question regarding acceleration of a succession.
A preliminary: suppose that I have a succession $a_g$ that, for high $g$, asymptotically goes as
$$
a_g=s_0+frac{s_1}g+frac{s_2}{g^2}+...=sum_{k=0}^infty frac{s_k}{g^k}.
$$
I am interested in computing the coefficients $s_k$, but in particular I am interested in computing the leading coefficient $s_0$ (that would also be the limit of the succession as $gtoinfty$). A way to accelerate this convergence is given by the Richardson transform: by defining the succession
$$
a_g^{(N)}=sum_{n=0}^N(-1)^{n+N}frac{(g+n)^N}{n!(N-n)!}a_{g+n},
$$
it can be proven that $a_g^{(N)}$ goes to the same limit as $a_g$, but the succession is accelerated as the asymptotic behavior is
$$
a_g^{(N)}simeq s_0+sum_{k=N+1}^infty frac{d_k}{g^k}.
$$
This works nicely and gives very good numerical results in the examples I've used.
The problem is that I'm now working with more general successions, of the form
$$
b_g=sum_{t=0}^Tleft(sum_{k=0}^infty frac{s_{(k,t)}}{g^k}right)frac{1}{(log g)^t}
$$
$T$ is a finite integer, the logarithm powers are finite. The coefficients $s_{(k,t)}$ are arbitrary coefficients, in principle not related to one another. As an example, the succession for $T=2$ would be
$$
sum_{k=0}^inftyfrac{s_{(k,0)}}{g^k}+frac{1}{log g}sum_{k=0}^inftyfrac{s_{(k,1)}}{g^k}+frac{1}{(log g)^2}sum_{k=0}^inftyfrac{s_{(k,2)}}{g^k}.
$$
Now convergence is way slower: as an example, for $T=1$, terms with no $1/g$ powers attached ($k=0$) are
$$
s_{(0,0)}+frac{s_{(0,1)}}{log g}.
$$
If I want to compute $s_{(0,0)}$ by computing $b_g$ for high $g$, I get very slow convergence, as $(log g)^{-1}$ goes to zero very slowly. The standard Richardson transform does not really work here.
The question: is there a generalization of this Richardson transform to successions like the $b_g$ succession?
Thanks everybody!
P.s.: this is a crossposting from https://mathoverflow.net/questions/316816/convergence-acceleration-of-successions-with-logarithms
numerical-methods convergence-acceleration
numerical-methods convergence-acceleration
asked Dec 7 '18 at 11:48
Salvatore BaldinoSalvatore Baldino
210212
asked Dec 7 '18 at 11:48
Salvatore BaldinoSalvatore Baldino
210212
asked Dec 7 '18 at 11:48
Salvatore BaldinoSalvatore Baldino
210212
asked Dec 7 '18 at 11:48
Salvatore BaldinoSalvatore Baldino
210212
asked Dec 7 '18 at 11:48
Salvatore BaldinoSalvatore Baldino
210212
210212
add a comment |
add a comment |
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