Approximation for confluent hypergeometric function 1F1(-aX,-X,-1)
$begingroup$
This question concerns Kummers confluent Hypergeometric function $_1F_1$, sometimes denoted $M$. Recall that
$$
M(a;b;z) = {_1F_1 (a;b;z)} = sum_{n=0}^{infty} frac{(-1)^n}{n!}frac{(a)_n}{(b)_n}
$$
I am seeking to approximate $M(-lfloor alpha Xrfloor;-X;-1)$ when $X$ is a large positive real and $alpha in [0,1]$, perhaps by way of a series in $alpha$.
Numerically, I find that:
$$
M(-lfloor alpha Xrfloor;-X;-1) approx 1 - left( frac{e-1}{e}+frac{pi}{10}right)alpha + left( frac{pi}{10}right) alpha^2
$$
is within 0.4% of the true value when $Xgg100$ and $alpha in [0,1]$. However, I see no combinatorial interpretation, or way to derive this from the hypergeometric series, or general approximation to higher orders.
power-series approximation hypergeometric-function
asked May 22 '17 at 10:52
Bilal KhanBilal Khan
215
$endgroup$
add a comment |
$begingroup$
This question concerns Kummers confluent Hypergeometric function $_1F_1$, sometimes denoted $M$. Recall that
$$
M(a;b;z) = {_1F_1 (a;b;z)} = sum_{n=0}^{infty} frac{(-1)^n}{n!}frac{(a)_n}{(b)_n}
$$
I am seeking to approximate $M(-lfloor alpha Xrfloor;-X;-1)$ when $X$ is a large positive real and $alpha in [0,1]$, perhaps by way of a series in $alpha$.
Numerically, I find that:
$$
M(-lfloor alpha Xrfloor;-X;-1) approx 1 - left( frac{e-1}{e}+frac{pi}{10}right)alpha + left( frac{pi}{10}right) alpha^2
$$
is within 0.4% of the true value when $Xgg100$ and $alpha in [0,1]$. However, I see no combinatorial interpretation, or way to derive this from the hypergeometric series, or general approximation to higher orders.
power-series approximation hypergeometric-function
asked May 22 '17 at 10:52
Bilal KhanBilal Khan
215
$endgroup$
add a comment |
$begingroup$
This question concerns Kummers confluent Hypergeometric function $_1F_1$, sometimes denoted $M$. Recall that
$$
M(a;b;z) = {_1F_1 (a;b;z)} = sum_{n=0}^{infty} frac{(-1)^n}{n!}frac{(a)_n}{(b)_n}
$$
I am seeking to approximate $M(-lfloor alpha Xrfloor;-X;-1)$ when $X$ is a large positive real and $alpha in [0,1]$, perhaps by way of a series in $alpha$.
Numerically, I find that:
$$
M(-lfloor alpha Xrfloor;-X;-1) approx 1 - left( frac{e-1}{e}+frac{pi}{10}right)alpha + left( frac{pi}{10}right) alpha^2
$$
is within 0.4% of the true value when $Xgg100$ and $alpha in [0,1]$. However, I see no combinatorial interpretation, or way to derive this from the hypergeometric series, or general approximation to higher orders.
power-series approximation hypergeometric-function
asked May 22 '17 at 10:52
Bilal KhanBilal Khan
215
$endgroup$
This question concerns Kummers confluent Hypergeometric function $_1F_1$, sometimes denoted $M$. Recall that
$$
M(a;b;z) = {_1F_1 (a;b;z)} = sum_{n=0}^{infty} frac{(-1)^n}{n!}frac{(a)_n}{(b)_n}
$$
I am seeking to approximate $M(-lfloor alpha Xrfloor;-X;-1)$ when $X$ is a large positive real and $alpha in [0,1]$, perhaps by way of a series in $alpha$.
Numerically, I find that:
$$
M(-lfloor alpha Xrfloor;-X;-1) approx 1 - left( frac{e-1}{e}+frac{pi}{10}right)alpha + left( frac{pi}{10}right) alpha^2
$$
is within 0.4% of the true value when $Xgg100$ and $alpha in [0,1]$. However, I see no combinatorial interpretation, or way to derive this from the hypergeometric series, or general approximation to higher orders.
power-series approximation hypergeometric-function
power-series approximation hypergeometric-function
asked May 22 '17 at 10:52
Bilal KhanBilal Khan
215
asked May 22 '17 at 10:52
Bilal KhanBilal Khan
215
edited May 22 '17 at 11:10
Bilal Khan
asked May 22 '17 at 10:52
Bilal KhanBilal Khan
215
asked May 22 '17 at 10:52
Bilal KhanBilal Khan
215
asked May 22 '17 at 10:52
Bilal KhanBilal Khan
215
215
add a comment |
add a comment |
1 Answer
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$begingroup$
Since
$${_1hspace{-1.5px}F_1}(-lflooralpha Xrfloor;-X;-1)=
sum_{k=0}^{infty}frac{(-lflooralpha Xrfloor)_k}{(-X)_k}frac{(-1)^k}{k!}=
\sum_{k=0}^{infty}
frac{(lflooralpha Xrfloor)!,(X-k)!}{X!,(lflooralpha Xrfloor-k)!}
frac{(-1)^k}{k!}$$
and
$$lim_{X to infty}
frac{(alpha X-{alpha X})!,(X-k)!}{X!,(alpha X-{alpha X}-k)!}=
alpha^k,$$
taking the element-wise limit gives
$$lim_{X to infty}
{_1hspace{-1.5px}F_1}(-lflooralpha Xrfloor;-X;-1)=e^{-alpha}.$$
answered Jan 20 '18 at 15:16
MaximMaxim
5,5981219
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add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Since
$${_1hspace{-1.5px}F_1}(-lflooralpha Xrfloor;-X;-1)=
sum_{k=0}^{infty}frac{(-lflooralpha Xrfloor)_k}{(-X)_k}frac{(-1)^k}{k!}=
\sum_{k=0}^{infty}
frac{(lflooralpha Xrfloor)!,(X-k)!}{X!,(lflooralpha Xrfloor-k)!}
frac{(-1)^k}{k!}$$
and
$$lim_{X to infty}
frac{(alpha X-{alpha X})!,(X-k)!}{X!,(alpha X-{alpha X}-k)!}=
alpha^k,$$
taking the element-wise limit gives
$$lim_{X to infty}
{_1hspace{-1.5px}F_1}(-lflooralpha Xrfloor;-X;-1)=e^{-alpha}.$$
answered Jan 20 '18 at 15:16
MaximMaxim
5,5981219
$endgroup$
add a comment |
$begingroup$
Since
$${_1hspace{-1.5px}F_1}(-lflooralpha Xrfloor;-X;-1)=
sum_{k=0}^{infty}frac{(-lflooralpha Xrfloor)_k}{(-X)_k}frac{(-1)^k}{k!}=
\sum_{k=0}^{infty}
frac{(lflooralpha Xrfloor)!,(X-k)!}{X!,(lflooralpha Xrfloor-k)!}
frac{(-1)^k}{k!}$$
and
$$lim_{X to infty}
frac{(alpha X-{alpha X})!,(X-k)!}{X!,(alpha X-{alpha X}-k)!}=
alpha^k,$$
taking the element-wise limit gives
$$lim_{X to infty}
{_1hspace{-1.5px}F_1}(-lflooralpha Xrfloor;-X;-1)=e^{-alpha}.$$
answered Jan 20 '18 at 15:16
MaximMaxim
5,5981219
$endgroup$
add a comment |
$begingroup$
Since
$${_1hspace{-1.5px}F_1}(-lflooralpha Xrfloor;-X;-1)=
sum_{k=0}^{infty}frac{(-lflooralpha Xrfloor)_k}{(-X)_k}frac{(-1)^k}{k!}=
\sum_{k=0}^{infty}
frac{(lflooralpha Xrfloor)!,(X-k)!}{X!,(lflooralpha Xrfloor-k)!}
frac{(-1)^k}{k!}$$
and
$$lim_{X to infty}
frac{(alpha X-{alpha X})!,(X-k)!}{X!,(alpha X-{alpha X}-k)!}=
alpha^k,$$
taking the element-wise limit gives
$$lim_{X to infty}
{_1hspace{-1.5px}F_1}(-lflooralpha Xrfloor;-X;-1)=e^{-alpha}.$$
answered Jan 20 '18 at 15:16
MaximMaxim
5,5981219
$endgroup$
Since
$${_1hspace{-1.5px}F_1}(-lflooralpha Xrfloor;-X;-1)=
sum_{k=0}^{infty}frac{(-lflooralpha Xrfloor)_k}{(-X)_k}frac{(-1)^k}{k!}=
\sum_{k=0}^{infty}
frac{(lflooralpha Xrfloor)!,(X-k)!}{X!,(lflooralpha Xrfloor-k)!}
frac{(-1)^k}{k!}$$
and
$$lim_{X to infty}
frac{(alpha X-{alpha X})!,(X-k)!}{X!,(alpha X-{alpha X}-k)!}=
alpha^k,$$
taking the element-wise limit gives
$$lim_{X to infty}
{_1hspace{-1.5px}F_1}(-lflooralpha Xrfloor;-X;-1)=e^{-alpha}.$$
answered Jan 20 '18 at 15:16
MaximMaxim
5,5981219
edited Jan 7 at 2:58
answered Jan 20 '18 at 15:16
MaximMaxim
5,5981219
answered Jan 20 '18 at 15:16
MaximMaxim
5,5981219
answered Jan 20 '18 at 15:16
MaximMaxim
5,5981219
5,5981219
add a comment |
add a comment |
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