Analytic continuation of the incomplete beta function
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Is there a rigorous proof for the analytic continuation of the incomplete beta function $B(x;a,b)$ for all values of $a$ and $b$? The incomplete beta function normally restricts the values of $a,b$ as $a>0$, $b>0$. So I would like to extend these values to negative but I cannot find a good and comprehensive reference for this.
There is a given expression in wikipedia for the analytic continuation of beta function but I cannot see one for the incomplete one. By the way, the integral that I am working on is:
$$rho=frac{b_0}{1-q}int_0^{1-(b_0/x)^{1-q}}v^{-1/2}(1-v)^{frac{1}{q-1}-1}dv=frac{b_0}{1-q}Beta(1-(b_0/x)^{1-q};frac{1}{2},frac{1}{q-1})$$
And I am working on $-infty<q<1$. Thanks for the help.
beta-function analytic-continuation
edited Nov 21 at 14:47
Klangen
1,37711231
asked Aug 21 at 3:40
user583893
264
add a comment |
up vote
0
down vote
favorite
Is there a rigorous proof for the analytic continuation of the incomplete beta function $B(x;a,b)$ for all values of $a$ and $b$? The incomplete beta function normally restricts the values of $a,b$ as $a>0$, $b>0$. So I would like to extend these values to negative but I cannot find a good and comprehensive reference for this.
There is a given expression in wikipedia for the analytic continuation of beta function but I cannot see one for the incomplete one. By the way, the integral that I am working on is:
$$rho=frac{b_0}{1-q}int_0^{1-(b_0/x)^{1-q}}v^{-1/2}(1-v)^{frac{1}{q-1}-1}dv=frac{b_0}{1-q}Beta(1-(b_0/x)^{1-q};frac{1}{2},frac{1}{q-1})$$
And I am working on $-infty<q<1$. Thanks for the help.
beta-function analytic-continuation
edited Nov 21 at 14:47
Klangen
1,37711231
asked Aug 21 at 3:40
user583893
264
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Is there a rigorous proof for the analytic continuation of the incomplete beta function $B(x;a,b)$ for all values of $a$ and $b$? The incomplete beta function normally restricts the values of $a,b$ as $a>0$, $b>0$. So I would like to extend these values to negative but I cannot find a good and comprehensive reference for this.
There is a given expression in wikipedia for the analytic continuation of beta function but I cannot see one for the incomplete one. By the way, the integral that I am working on is:
$$rho=frac{b_0}{1-q}int_0^{1-(b_0/x)^{1-q}}v^{-1/2}(1-v)^{frac{1}{q-1}-1}dv=frac{b_0}{1-q}Beta(1-(b_0/x)^{1-q};frac{1}{2},frac{1}{q-1})$$
And I am working on $-infty<q<1$. Thanks for the help.
beta-function analytic-continuation
edited Nov 21 at 14:47
Klangen
1,37711231
asked Aug 21 at 3:40
user583893
264
Is there a rigorous proof for the analytic continuation of the incomplete beta function $B(x;a,b)$ for all values of $a$ and $b$? The incomplete beta function normally restricts the values of $a,b$ as $a>0$, $b>0$. So I would like to extend these values to negative but I cannot find a good and comprehensive reference for this.
There is a given expression in wikipedia for the analytic continuation of beta function but I cannot see one for the incomplete one. By the way, the integral that I am working on is:
$$rho=frac{b_0}{1-q}int_0^{1-(b_0/x)^{1-q}}v^{-1/2}(1-v)^{frac{1}{q-1}-1}dv=frac{b_0}{1-q}Beta(1-(b_0/x)^{1-q};frac{1}{2},frac{1}{q-1})$$
And I am working on $-infty<q<1$. Thanks for the help.
beta-function analytic-continuation
beta-function analytic-continuation
edited Nov 21 at 14:47
Klangen
1,37711231
asked Aug 21 at 3:40
user583893
264
edited Nov 21 at 14:47
Klangen
1,37711231
asked Aug 21 at 3:40
user583893
264
edited Nov 21 at 14:47
Klangen
1,37711231
edited Nov 21 at 14:47
Klangen
1,37711231
edited Nov 21 at 14:47
Klangen
1,37711231
1,37711231
asked Aug 21 at 3:40
user583893
264
asked Aug 21 at 3:40
user583893
264
asked Aug 21 at 3:40
user583893
264
264
add a comment |
add a comment |
1 Answer
1
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oldest
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1
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There is the general complete description with the regularized Gauss hypergeometric function ${{_2}tilde{F}_{1}}(a,b,c,x)$, see 1 and 2: $$
B_x(a,b)= Gamma(a),x^a ;{{_2}tilde{F}_{1}}(a,1-b,a+1,x), qquad -a notin mathbb{N}
$$
When $a le 0$ or $b le 0$, the Gauss hypergeometric function ${{_2}F_{1}}$ function can be used: If $a neq 0$ is no negative integer, the result is (3)
$$
B_x(a,b)= frac{x^a}{a},{{_2}F_{1}}(a,1-b,a+1,x), qquad -a notin mathbb{N},
$$
else if $b neq 0$ is no negative integer, then (4):
$$
B_x(a,b)= B(a,b)- frac{(1-x)^b x^a}{b};{{_2}F_{1}}(1,a+b,b+1,1-x), qquad -b notin mathbb{N}.
$$
answered Aug 21 at 8:16
gammatester
16.6k21631
Hi. Thanks for the answer. I don't actually understand this. If the parameter $b$ takes a value of zero or any negative value, I can use the last equation? Is that right?
– user583893
Aug 21 at 9:44
No, you can use it if $b ne -1, -2, dots$ otherwise $ c= b+1$ becomes zero, and the the Gauss function is undefined. The regularized function $${{_2}tilde{F}_{1}}(a,b,c,x) = frac{1}{Gamma(c)} {{_2}F_{1}}(a,b,c,x)$$ can handle this because together with $Gamma(c)$ you get a finite limit if $c$ approaches a negative integer or zero.
– gammatester
Aug 21 at 9:56
What i need is what works for negative values of b. I already tried in the mathematica, and it gave me the correct answer.
– user583893
Aug 21 at 10:36
What i need is an extension of incomplete beta function that works for negative values of the parameter $b$. But i tried feeding up negative values of $b$ to the last equation and it gave me the right answers. Now I am confused since the last equation must only work for positive $b$'s.
– user583893
Aug 21 at 10:40
No it does work for negative non-integer $b$. If $b$ is a negative integer and $a$ is not you can use the second formula, otherwise use the regularized function. But note the hint given at functions.wolfram.com/GammaBetaErf/Beta3/02/02. Using my own implementation I get for example $B_{0.5}(-1.5, -2.5)= 6.4$ (verified on Wolfram function evaluation).
– gammatester
Aug 21 at 10:56
|
show 4 more comments
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1 Answer
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1 Answer
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votes
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down vote
accepted
There is the general complete description with the regularized Gauss hypergeometric function ${{_2}tilde{F}_{1}}(a,b,c,x)$, see 1 and 2: $$
B_x(a,b)= Gamma(a),x^a ;{{_2}tilde{F}_{1}}(a,1-b,a+1,x), qquad -a notin mathbb{N}
$$
When $a le 0$ or $b le 0$, the Gauss hypergeometric function ${{_2}F_{1}}$ function can be used: If $a neq 0$ is no negative integer, the result is (3)
$$
B_x(a,b)= frac{x^a}{a},{{_2}F_{1}}(a,1-b,a+1,x), qquad -a notin mathbb{N},
$$
else if $b neq 0$ is no negative integer, then (4):
$$
B_x(a,b)= B(a,b)- frac{(1-x)^b x^a}{b};{{_2}F_{1}}(1,a+b,b+1,1-x), qquad -b notin mathbb{N}.
$$
answered Aug 21 at 8:16
gammatester
16.6k21631
Hi. Thanks for the answer. I don't actually understand this. If the parameter $b$ takes a value of zero or any negative value, I can use the last equation? Is that right?
– user583893
Aug 21 at 9:44
No, you can use it if $b ne -1, -2, dots$ otherwise $ c= b+1$ becomes zero, and the the Gauss function is undefined. The regularized function $${{_2}tilde{F}_{1}}(a,b,c,x) = frac{1}{Gamma(c)} {{_2}F_{1}}(a,b,c,x)$$ can handle this because together with $Gamma(c)$ you get a finite limit if $c$ approaches a negative integer or zero.
– gammatester
Aug 21 at 9:56
What i need is what works for negative values of b. I already tried in the mathematica, and it gave me the correct answer.
– user583893
Aug 21 at 10:36
What i need is an extension of incomplete beta function that works for negative values of the parameter $b$. But i tried feeding up negative values of $b$ to the last equation and it gave me the right answers. Now I am confused since the last equation must only work for positive $b$'s.
– user583893
Aug 21 at 10:40
No it does work for negative non-integer $b$. If $b$ is a negative integer and $a$ is not you can use the second formula, otherwise use the regularized function. But note the hint given at functions.wolfram.com/GammaBetaErf/Beta3/02/02. Using my own implementation I get for example $B_{0.5}(-1.5, -2.5)= 6.4$ (verified on Wolfram function evaluation).
– gammatester
Aug 21 at 10:56
|
show 4 more comments
up vote
1
down vote
accepted
There is the general complete description with the regularized Gauss hypergeometric function ${{_2}tilde{F}_{1}}(a,b,c,x)$, see 1 and 2: $$
B_x(a,b)= Gamma(a),x^a ;{{_2}tilde{F}_{1}}(a,1-b,a+1,x), qquad -a notin mathbb{N}
$$
When $a le 0$ or $b le 0$, the Gauss hypergeometric function ${{_2}F_{1}}$ function can be used: If $a neq 0$ is no negative integer, the result is (3)
$$
B_x(a,b)= frac{x^a}{a},{{_2}F_{1}}(a,1-b,a+1,x), qquad -a notin mathbb{N},
$$
else if $b neq 0$ is no negative integer, then (4):
$$
B_x(a,b)= B(a,b)- frac{(1-x)^b x^a}{b};{{_2}F_{1}}(1,a+b,b+1,1-x), qquad -b notin mathbb{N}.
$$
answered Aug 21 at 8:16
gammatester
16.6k21631
Hi. Thanks for the answer. I don't actually understand this. If the parameter $b$ takes a value of zero or any negative value, I can use the last equation? Is that right?
– user583893
Aug 21 at 9:44
No, you can use it if $b ne -1, -2, dots$ otherwise $ c= b+1$ becomes zero, and the the Gauss function is undefined. The regularized function $${{_2}tilde{F}_{1}}(a,b,c,x) = frac{1}{Gamma(c)} {{_2}F_{1}}(a,b,c,x)$$ can handle this because together with $Gamma(c)$ you get a finite limit if $c$ approaches a negative integer or zero.
– gammatester
Aug 21 at 9:56
What i need is what works for negative values of b. I already tried in the mathematica, and it gave me the correct answer.
– user583893
Aug 21 at 10:36
What i need is an extension of incomplete beta function that works for negative values of the parameter $b$. But i tried feeding up negative values of $b$ to the last equation and it gave me the right answers. Now I am confused since the last equation must only work for positive $b$'s.
– user583893
Aug 21 at 10:40
No it does work for negative non-integer $b$. If $b$ is a negative integer and $a$ is not you can use the second formula, otherwise use the regularized function. But note the hint given at functions.wolfram.com/GammaBetaErf/Beta3/02/02. Using my own implementation I get for example $B_{0.5}(-1.5, -2.5)= 6.4$ (verified on Wolfram function evaluation).
– gammatester
Aug 21 at 10:56
|
show 4 more comments
up vote
1
down vote
accepted
up vote
1
down vote
accepted
There is the general complete description with the regularized Gauss hypergeometric function ${{_2}tilde{F}_{1}}(a,b,c,x)$, see 1 and 2: $$
B_x(a,b)= Gamma(a),x^a ;{{_2}tilde{F}_{1}}(a,1-b,a+1,x), qquad -a notin mathbb{N}
$$
When $a le 0$ or $b le 0$, the Gauss hypergeometric function ${{_2}F_{1}}$ function can be used: If $a neq 0$ is no negative integer, the result is (3)
$$
B_x(a,b)= frac{x^a}{a},{{_2}F_{1}}(a,1-b,a+1,x), qquad -a notin mathbb{N},
$$
else if $b neq 0$ is no negative integer, then (4):
$$
B_x(a,b)= B(a,b)- frac{(1-x)^b x^a}{b};{{_2}F_{1}}(1,a+b,b+1,1-x), qquad -b notin mathbb{N}.
$$
answered Aug 21 at 8:16
gammatester
16.6k21631
There is the general complete description with the regularized Gauss hypergeometric function ${{_2}tilde{F}_{1}}(a,b,c,x)$, see 1 and 2: $$
B_x(a,b)= Gamma(a),x^a ;{{_2}tilde{F}_{1}}(a,1-b,a+1,x), qquad -a notin mathbb{N}
$$
When $a le 0$ or $b le 0$, the Gauss hypergeometric function ${{_2}F_{1}}$ function can be used: If $a neq 0$ is no negative integer, the result is (3)
$$
B_x(a,b)= frac{x^a}{a},{{_2}F_{1}}(a,1-b,a+1,x), qquad -a notin mathbb{N},
$$
else if $b neq 0$ is no negative integer, then (4):
$$
B_x(a,b)= B(a,b)- frac{(1-x)^b x^a}{b};{{_2}F_{1}}(1,a+b,b+1,1-x), qquad -b notin mathbb{N}.
$$
answered Aug 21 at 8:16
gammatester
16.6k21631
answered Aug 21 at 8:16
gammatester
16.6k21631
answered Aug 21 at 8:16
gammatester
16.6k21631
answered Aug 21 at 8:16
gammatester
16.6k21631
16.6k21631
Hi. Thanks for the answer. I don't actually understand this. If the parameter $b$ takes a value of zero or any negative value, I can use the last equation? Is that right?
– user583893
Aug 21 at 9:44
No, you can use it if $b ne -1, -2, dots$ otherwise $ c= b+1$ becomes zero, and the the Gauss function is undefined. The regularized function $${{_2}tilde{F}_{1}}(a,b,c,x) = frac{1}{Gamma(c)} {{_2}F_{1}}(a,b,c,x)$$ can handle this because together with $Gamma(c)$ you get a finite limit if $c$ approaches a negative integer or zero.
– gammatester
Aug 21 at 9:56
What i need is what works for negative values of b. I already tried in the mathematica, and it gave me the correct answer.
– user583893
Aug 21 at 10:36
What i need is an extension of incomplete beta function that works for negative values of the parameter $b$. But i tried feeding up negative values of $b$ to the last equation and it gave me the right answers. Now I am confused since the last equation must only work for positive $b$'s.
– user583893
Aug 21 at 10:40
No it does work for negative non-integer $b$. If $b$ is a negative integer and $a$ is not you can use the second formula, otherwise use the regularized function. But note the hint given at functions.wolfram.com/GammaBetaErf/Beta3/02/02. Using my own implementation I get for example $B_{0.5}(-1.5, -2.5)= 6.4$ (verified on Wolfram function evaluation).
– gammatester
Aug 21 at 10:56
|
show 4 more comments
Hi. Thanks for the answer. I don't actually understand this. If the parameter $b$ takes a value of zero or any negative value, I can use the last equation? Is that right?
– user583893
Aug 21 at 9:44
No, you can use it if $b ne -1, -2, dots$ otherwise $ c= b+1$ becomes zero, and the the Gauss function is undefined. The regularized function $${{_2}tilde{F}_{1}}(a,b,c,x) = frac{1}{Gamma(c)} {{_2}F_{1}}(a,b,c,x)$$ can handle this because together with $Gamma(c)$ you get a finite limit if $c$ approaches a negative integer or zero.
– gammatester
Aug 21 at 9:56
What i need is what works for negative values of b. I already tried in the mathematica, and it gave me the correct answer.
– user583893
Aug 21 at 10:36
What i need is an extension of incomplete beta function that works for negative values of the parameter $b$. But i tried feeding up negative values of $b$ to the last equation and it gave me the right answers. Now I am confused since the last equation must only work for positive $b$'s.
– user583893
Aug 21 at 10:40
No it does work for negative non-integer $b$. If $b$ is a negative integer and $a$ is not you can use the second formula, otherwise use the regularized function. But note the hint given at functions.wolfram.com/GammaBetaErf/Beta3/02/02. Using my own implementation I get for example $B_{0.5}(-1.5, -2.5)= 6.4$ (verified on Wolfram function evaluation).
– gammatester
Aug 21 at 10:56
Hi. Thanks for the answer. I don't actually understand this. If the parameter $b$ takes a value of zero or any negative value, I can use the last equation? Is that right?
– user583893
Aug 21 at 9:44
Hi. Thanks for the answer. I don't actually understand this. If the parameter $b$ takes a value of zero or any negative value, I can use the last equation? Is that right?
– user583893
Aug 21 at 9:44
No, you can use it if $b ne -1, -2, dots$ otherwise $ c= b+1$ becomes zero, and the the Gauss function is undefined. The regularized function $${{_2}tilde{F}_{1}}(a,b,c,x) = frac{1}{Gamma(c)} {{_2}F_{1}}(a,b,c,x)$$ can handle this because together with $Gamma(c)$ you get a finite limit if $c$ approaches a negative integer or zero.
– gammatester
Aug 21 at 9:56
No, you can use it if $b ne -1, -2, dots$ otherwise $ c= b+1$ becomes zero, and the the Gauss function is undefined. The regularized function $${{_2}tilde{F}_{1}}(a,b,c,x) = frac{1}{Gamma(c)} {{_2}F_{1}}(a,b,c,x)$$ can handle this because together with $Gamma(c)$ you get a finite limit if $c$ approaches a negative integer or zero.
– gammatester
Aug 21 at 9:56
What i need is what works for negative values of b. I already tried in the mathematica, and it gave me the correct answer.
– user583893
Aug 21 at 10:36
What i need is what works for negative values of b. I already tried in the mathematica, and it gave me the correct answer.
– user583893
Aug 21 at 10:36
What i need is an extension of incomplete beta function that works for negative values of the parameter $b$. But i tried feeding up negative values of $b$ to the last equation and it gave me the right answers. Now I am confused since the last equation must only work for positive $b$'s.
– user583893
Aug 21 at 10:40
What i need is an extension of incomplete beta function that works for negative values of the parameter $b$. But i tried feeding up negative values of $b$ to the last equation and it gave me the right answers. Now I am confused since the last equation must only work for positive $b$'s.
– user583893
Aug 21 at 10:40
No it does work for negative non-integer $b$. If $b$ is a negative integer and $a$ is not you can use the second formula, otherwise use the regularized function. But note the hint given at functions.wolfram.com/GammaBetaErf/Beta3/02/02. Using my own implementation I get for example $B_{0.5}(-1.5, -2.5)= 6.4$ (verified on Wolfram function evaluation).
– gammatester
Aug 21 at 10:56
No it does work for negative non-integer $b$. If $b$ is a negative integer and $a$ is not you can use the second formula, otherwise use the regularized function. But note the hint given at functions.wolfram.com/GammaBetaErf/Beta3/02/02. Using my own implementation I get for example $B_{0.5}(-1.5, -2.5)= 6.4$ (verified on Wolfram function evaluation).
– gammatester
Aug 21 at 10:56
|
show 4 more comments
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