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Behavior of the Gaussian Hypergeometric function when one of its arguments approaches $0$ or $1$

Hypergeometric function argument simplificationAlgebraic operations on hypergeometric functionsApproximating hypergeometric function F(1,1+a,2+a,z) for z->1Waring's Problem and the floor function - solving a recurrence relation by handComputing the hypergometric function $_1F_2$ in special casesExpressing the hypergeometric function as an infinite sum of squaresCan we establish a relation between the $F$ distribution and Binomial Distribution?Continuity of hypergeometric function in the argumentsMellin transform of a Gaussian Hypergeometric Function with negative $x$-argumentRewriting Appell's Hypergeometric Function $F_1$ in terms of Gauss' Hypergeometric Function $_2F_1$

1

$begingroup$

For two positive integers $a,b$, denote by $_2F_1(a,1-b;a+1;z)$ the Gaussian Hypergeometric function whose first three parameters are fixed at $a,1-b$ and $a+1$, respectively. such function is linked to the regularized incomplete Beta function (namely the CDF of a Beta distribution) by the relation:
$$
B(z;a,b)=frac{z^a}{a} {}_2F_1(a,1-b;a+1;z).
$$
I was wondering whether there exist functions $f,, g$, such that
$$
f(z)sim {}_2F_1(a,1-b;a+1;z) quad (zdownarrow 0),\
g(z)sim {}_2F_1(a,1-b;a+1;z) quad (zuparrow 1).
$$
In particular, I would be interested in the case $a,bgeq 1$.

calculus limits statistics asymptotics hypergeometric-function

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edited Mar 13 at 16:22

Jack London

asked Mar 13 at 16:15

Jack LondonJack London

34018

$endgroup$

add a comment | 

1

$begingroup$

For two positive integers $a,b$, denote by $_2F_1(a,1-b;a+1;z)$ the Gaussian Hypergeometric function whose first three parameters are fixed at $a,1-b$ and $a+1$, respectively. such function is linked to the regularized incomplete Beta function (namely the CDF of a Beta distribution) by the relation:
$$
B(z;a,b)=frac{z^a}{a} {}_2F_1(a,1-b;a+1;z).
$$
I was wondering whether there exist functions $f,, g$, such that
$$
f(z)sim {}_2F_1(a,1-b;a+1;z) quad (zdownarrow 0),\
g(z)sim {}_2F_1(a,1-b;a+1;z) quad (zuparrow 1).
$$
In particular, I would be interested in the case $a,bgeq 1$.

calculus limits statistics asymptotics hypergeometric-function

share|cite|improve this question

edited Mar 13 at 16:22

Jack London

asked Mar 13 at 16:15

Jack LondonJack London

34018

$endgroup$

add a comment | 

$begingroup$

For two positive integers $a,b$, denote by $_2F_1(a,1-b;a+1;z)$ the Gaussian Hypergeometric function whose first three parameters are fixed at $a,1-b$ and $a+1$, respectively. such function is linked to the regularized incomplete Beta function (namely the CDF of a Beta distribution) by the relation:
$$
B(z;a,b)=frac{z^a}{a} {}_2F_1(a,1-b;a+1;z).
$$
I was wondering whether there exist functions $f,, g$, such that
$$
f(z)sim {}_2F_1(a,1-b;a+1;z) quad (zdownarrow 0),\
g(z)sim {}_2F_1(a,1-b;a+1;z) quad (zuparrow 1).
$$
In particular, I would be interested in the case $a,bgeq 1$.

calculus limits statistics asymptotics hypergeometric-function

share|cite|improve this question

edited Mar 13 at 16:22

Jack London

asked Mar 13 at 16:15

Jack LondonJack London

34018

$endgroup$

share|cite|improve this question

edited Mar 13 at 16:22

Jack London

asked Mar 13 at 16:15

Jack LondonJack London

34018

share|cite|improve this question

edited Mar 13 at 16:22

Jack London

asked Mar 13 at 16:15

Jack LondonJack London

34018

asked Mar 13 at 16:15

Jack LondonJack London

34018

asked Mar 13 at 16:15

Jack LondonJack London

34018

1 Answer
1

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oldest

votes

0

$begingroup$

We have
$$
eqalign{
& F(z,a,b) = {}_2F_{,1} left( {left. {matrix{
{a,;1 - b} cr
{a + 1} cr
} ;} right|;z} right) = cr
& = sumlimits_{0, le ,k} {{{a^{,overline {,k,} } left( {1 - b} right)^{,overline {,k,} } } over {left( {a + 1} right)^{,overline {,k,} } }}
{{z^{,k} } over {k!}}} = cr
& = asumlimits_{0, le ,k} {{{left( {1 - b} right)^{,overline {,k,} } } over {left( {a + k} right)}}{{z^{,k} } over {k!}}} = cr
& = asumlimits_{0, le ,k} {{{left( { - 1} right)^{,k} left( {b - 1} right)^{,underline {,k,} } } over {left( {a + k} right)}}{{z^{,k} } over {k!}}} cr
& = asumlimits_{0, le ,k} {left( { - 1} right)^{,k} left( matrix{
b - 1 cr
k cr} right){{z^{,k} } over {left( {a + k} right)}}} cr}
$$

Then, for the first limit you have
$$
mathop {lim }limits_{z, to ,0 + } F(z,a,b) = 1
$$

Continuing the development
$$
eqalign{
& F(z,a,b) = asumlimits_{0, le ,k} {left( { - 1} right)^{,k} left( matrix{
b - 1 cr
k cr} right){{z^{,k} } over {left( {a + k} right)}}} = cr
& = a,z^{, - ,a} sumlimits_{0, le ,k} {left( { - 1} right)^{,k} left( matrix{
b - 1 cr
k cr} right){{z^{,a + k} } over {left( {a + k} right)}}} = cr
& = a,z^{, - ,a} sumlimits_{0, le ,k} {left( { - 1} right)^{,k} left( matrix{
b - 1 cr
k cr} right)int_{t = 0}^z {t^{,a + k - 1} } dt} = cr
& = a,z^{, - ,a} int_{t = 0}^z {t^{,a - 1} sumlimits_{0, le ,k} {left( { - 1} right)^{,k} left( matrix{
b - 1 cr
k cr} right)} left( { - t} right)^{,k} } dt = cr
& = a,z^{, - ,a} int_{t = 0}^z {t^{,a - 1} left( {1 - t} right)^{,b - 1} } dt cr}
$$
which gives the relation with the Incomplete Beta function you already know.
$$
F(z,a,b) = a,z^{, - ,a} ,{rm B}(z,;a,b)quad left| {;0 < a,b} right.
$$

From here we get
$$
eqalign{
& mathop {lim }limits_{z, to ,1 - } F(z,a,b) = aint_{t = 0}^1 {t^{,a - 1} left( {1 - t} right)^{,b - 1} } dt = cr
& = a,{rm B}(a,b) cr}
$$
and since (re. for instance to this)
$$
{rm B}(z,;a,b) = {{z^{,,a} } over a}left( {1 + O(z)} right)
$$
also
$$
mathop {lim }limits_{z, to ,0 + } left( {F(z,a,b) = a,z^{, - ,a} ,{rm B}(z,;a,b)} right) = 1quad left| {;0 < a,b} right.
$$
and you can use the above expression for the whole domain of your interest.

share|cite|improve this answer

answered Mar 13 at 17:49

G CabG Cab

20.4k31341

$endgroup$

add a comment | 

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0

$begingroup$

We have
$$
eqalign{
& F(z,a,b) = {}_2F_{,1} left( {left. {matrix{
{a,;1 - b} cr
{a + 1} cr
} ;} right|;z} right) = cr
& = sumlimits_{0, le ,k} {{{a^{,overline {,k,} } left( {1 - b} right)^{,overline {,k,} } } over {left( {a + 1} right)^{,overline {,k,} } }}
{{z^{,k} } over {k!}}} = cr
& = asumlimits_{0, le ,k} {{{left( {1 - b} right)^{,overline {,k,} } } over {left( {a + k} right)}}{{z^{,k} } over {k!}}} = cr
& = asumlimits_{0, le ,k} {{{left( { - 1} right)^{,k} left( {b - 1} right)^{,underline {,k,} } } over {left( {a + k} right)}}{{z^{,k} } over {k!}}} cr
& = asumlimits_{0, le ,k} {left( { - 1} right)^{,k} left( matrix{
b - 1 cr
k cr} right){{z^{,k} } over {left( {a + k} right)}}} cr}
$$

Then, for the first limit you have
$$
mathop {lim }limits_{z, to ,0 + } F(z,a,b) = 1
$$

Continuing the development
$$
eqalign{
& F(z,a,b) = asumlimits_{0, le ,k} {left( { - 1} right)^{,k} left( matrix{
b - 1 cr
k cr} right){{z^{,k} } over {left( {a + k} right)}}} = cr
& = a,z^{, - ,a} sumlimits_{0, le ,k} {left( { - 1} right)^{,k} left( matrix{
b - 1 cr
k cr} right){{z^{,a + k} } over {left( {a + k} right)}}} = cr
& = a,z^{, - ,a} sumlimits_{0, le ,k} {left( { - 1} right)^{,k} left( matrix{
b - 1 cr
k cr} right)int_{t = 0}^z {t^{,a + k - 1} } dt} = cr
& = a,z^{, - ,a} int_{t = 0}^z {t^{,a - 1} sumlimits_{0, le ,k} {left( { - 1} right)^{,k} left( matrix{
b - 1 cr
k cr} right)} left( { - t} right)^{,k} } dt = cr
& = a,z^{, - ,a} int_{t = 0}^z {t^{,a - 1} left( {1 - t} right)^{,b - 1} } dt cr}
$$
which gives the relation with the Incomplete Beta function you already know.
$$
F(z,a,b) = a,z^{, - ,a} ,{rm B}(z,;a,b)quad left| {;0 < a,b} right.
$$

From here we get
$$
eqalign{
& mathop {lim }limits_{z, to ,1 - } F(z,a,b) = aint_{t = 0}^1 {t^{,a - 1} left( {1 - t} right)^{,b - 1} } dt = cr
& = a,{rm B}(a,b) cr}
$$
and since (re. for instance to this)
$$
{rm B}(z,;a,b) = {{z^{,,a} } over a}left( {1 + O(z)} right)
$$
also
$$
mathop {lim }limits_{z, to ,0 + } left( {F(z,a,b) = a,z^{, - ,a} ,{rm B}(z,;a,b)} right) = 1quad left| {;0 < a,b} right.
$$
and you can use the above expression for the whole domain of your interest.

share|cite|improve this answer

answered Mar 13 at 17:49

G CabG Cab

20.4k31341

$endgroup$

add a comment | 

0

$begingroup$

We have
$$
eqalign{
& F(z,a,b) = {}_2F_{,1} left( {left. {matrix{
{a,;1 - b} cr
{a + 1} cr
} ;} right|;z} right) = cr
& = sumlimits_{0, le ,k} {{{a^{,overline {,k,} } left( {1 - b} right)^{,overline {,k,} } } over {left( {a + 1} right)^{,overline {,k,} } }}
{{z^{,k} } over {k!}}} = cr
& = asumlimits_{0, le ,k} {{{left( {1 - b} right)^{,overline {,k,} } } over {left( {a + k} right)}}{{z^{,k} } over {k!}}} = cr
& = asumlimits_{0, le ,k} {{{left( { - 1} right)^{,k} left( {b - 1} right)^{,underline {,k,} } } over {left( {a + k} right)}}{{z^{,k} } over {k!}}} cr
& = asumlimits_{0, le ,k} {left( { - 1} right)^{,k} left( matrix{
b - 1 cr
k cr} right){{z^{,k} } over {left( {a + k} right)}}} cr}
$$

Then, for the first limit you have
$$
mathop {lim }limits_{z, to ,0 + } F(z,a,b) = 1
$$

Continuing the development
$$
eqalign{
& F(z,a,b) = asumlimits_{0, le ,k} {left( { - 1} right)^{,k} left( matrix{
b - 1 cr
k cr} right){{z^{,k} } over {left( {a + k} right)}}} = cr
& = a,z^{, - ,a} sumlimits_{0, le ,k} {left( { - 1} right)^{,k} left( matrix{
b - 1 cr
k cr} right){{z^{,a + k} } over {left( {a + k} right)}}} = cr
& = a,z^{, - ,a} sumlimits_{0, le ,k} {left( { - 1} right)^{,k} left( matrix{
b - 1 cr
k cr} right)int_{t = 0}^z {t^{,a + k - 1} } dt} = cr
& = a,z^{, - ,a} int_{t = 0}^z {t^{,a - 1} sumlimits_{0, le ,k} {left( { - 1} right)^{,k} left( matrix{
b - 1 cr
k cr} right)} left( { - t} right)^{,k} } dt = cr
& = a,z^{, - ,a} int_{t = 0}^z {t^{,a - 1} left( {1 - t} right)^{,b - 1} } dt cr}
$$
which gives the relation with the Incomplete Beta function you already know.
$$
F(z,a,b) = a,z^{, - ,a} ,{rm B}(z,;a,b)quad left| {;0 < a,b} right.
$$

From here we get
$$
eqalign{
& mathop {lim }limits_{z, to ,1 - } F(z,a,b) = aint_{t = 0}^1 {t^{,a - 1} left( {1 - t} right)^{,b - 1} } dt = cr
& = a,{rm B}(a,b) cr}
$$
and since (re. for instance to this)
$$
{rm B}(z,;a,b) = {{z^{,,a} } over a}left( {1 + O(z)} right)
$$
also
$$
mathop {lim }limits_{z, to ,0 + } left( {F(z,a,b) = a,z^{, - ,a} ,{rm B}(z,;a,b)} right) = 1quad left| {;0 < a,b} right.
$$
and you can use the above expression for the whole domain of your interest.

share|cite|improve this answer

answered Mar 13 at 17:49

G CabG Cab

20.4k31341

$endgroup$

add a comment | 

$begingroup$

We have
$$
eqalign{
& F(z,a,b) = {}_2F_{,1} left( {left. {matrix{
{a,;1 - b} cr
{a + 1} cr
} ;} right|;z} right) = cr
& = sumlimits_{0, le ,k} {{{a^{,overline {,k,} } left( {1 - b} right)^{,overline {,k,} } } over {left( {a + 1} right)^{,overline {,k,} } }}
{{z^{,k} } over {k!}}} = cr
& = asumlimits_{0, le ,k} {{{left( {1 - b} right)^{,overline {,k,} } } over {left( {a + k} right)}}{{z^{,k} } over {k!}}} = cr
& = asumlimits_{0, le ,k} {{{left( { - 1} right)^{,k} left( {b - 1} right)^{,underline {,k,} } } over {left( {a + k} right)}}{{z^{,k} } over {k!}}} cr
& = asumlimits_{0, le ,k} {left( { - 1} right)^{,k} left( matrix{
b - 1 cr
k cr} right){{z^{,k} } over {left( {a + k} right)}}} cr}
$$

Then, for the first limit you have
$$
mathop {lim }limits_{z, to ,0 + } F(z,a,b) = 1
$$

Continuing the development
$$
eqalign{
& F(z,a,b) = asumlimits_{0, le ,k} {left( { - 1} right)^{,k} left( matrix{
b - 1 cr
k cr} right){{z^{,k} } over {left( {a + k} right)}}} = cr
& = a,z^{, - ,a} sumlimits_{0, le ,k} {left( { - 1} right)^{,k} left( matrix{
b - 1 cr
k cr} right){{z^{,a + k} } over {left( {a + k} right)}}} = cr
& = a,z^{, - ,a} sumlimits_{0, le ,k} {left( { - 1} right)^{,k} left( matrix{
b - 1 cr
k cr} right)int_{t = 0}^z {t^{,a + k - 1} } dt} = cr
& = a,z^{, - ,a} int_{t = 0}^z {t^{,a - 1} sumlimits_{0, le ,k} {left( { - 1} right)^{,k} left( matrix{
b - 1 cr
k cr} right)} left( { - t} right)^{,k} } dt = cr
& = a,z^{, - ,a} int_{t = 0}^z {t^{,a - 1} left( {1 - t} right)^{,b - 1} } dt cr}
$$
which gives the relation with the Incomplete Beta function you already know.
$$
F(z,a,b) = a,z^{, - ,a} ,{rm B}(z,;a,b)quad left| {;0 < a,b} right.
$$

From here we get
$$
eqalign{
& mathop {lim }limits_{z, to ,1 - } F(z,a,b) = aint_{t = 0}^1 {t^{,a - 1} left( {1 - t} right)^{,b - 1} } dt = cr
& = a,{rm B}(a,b) cr}
$$
and since (re. for instance to this)
$$
{rm B}(z,;a,b) = {{z^{,,a} } over a}left( {1 + O(z)} right)
$$
also
$$
mathop {lim }limits_{z, to ,0 + } left( {F(z,a,b) = a,z^{, - ,a} ,{rm B}(z,;a,b)} right) = 1quad left| {;0 < a,b} right.
$$
and you can use the above expression for the whole domain of your interest.

share|cite|improve this answer

answered Mar 13 at 17:49

G CabG Cab

20.4k31341

$endgroup$

share|cite|improve this answer

answered Mar 13 at 17:49

G CabG Cab

20.4k31341

answered Mar 13 at 17:49

G CabG Cab

20.4k31341

answered Mar 13 at 17:49

G CabG Cab

20.4k31341