Mixing
Evaluating product of Upper Incomplete Gamma functions
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I have checked several posts but couldn't find the equivalent of $Gamma(m,a) cdot Gamma(m,b)$, where '$cdot$' means multiplication. I suspect that it can be solved by applying the equivalent of Gamma function $(n-1)!e^{-x}sum limits_{k=0}^{m}dfrac{x^k}{k!}$ but then there will be two summations with same limits which I have no clue how to solve
Any suggestions?
summation gamma-function gamma-distribution
edited 2 days ago
David G. Stork
8,97621232
asked 2 days ago
hakkunamattata
454
add a comment |
up vote
2
down vote
favorite
I have checked several posts but couldn't find the equivalent of $Gamma(m,a) cdot Gamma(m,b)$, where '$cdot$' means multiplication. I suspect that it can be solved by applying the equivalent of Gamma function $(n-1)!e^{-x}sum limits_{k=0}^{m}dfrac{x^k}{k!}$ but then there will be two summations with same limits which I have no clue how to solve
Any suggestions?
summation gamma-function gamma-distribution
edited 2 days ago
David G. Stork
8,97621232
asked 2 days ago
hakkunamattata
454
What do you understand by $;Gamma(m,a);$ ? That's not the usual Gamma Function...
– DonAntonio
2 days ago
Its the upper incomplete Gamma function
– hakkunamattata
2 days ago
I think it'd be a rather good idea to explicitly say that, and not only "Gamma Functions", which can mislead.
– DonAntonio
2 days ago
Upper or lower incomplete? "*" means multiplication (or something else)? The formula you state only applies when the first argument is a positive integer; is $m$ a positive integer?
– Eric Towers
2 days ago
yes m is positive, I have changed the title. Thanks for suggestion
– hakkunamattata
2 days ago
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I have checked several posts but couldn't find the equivalent of $Gamma(m,a) cdot Gamma(m,b)$, where '$cdot$' means multiplication. I suspect that it can be solved by applying the equivalent of Gamma function $(n-1)!e^{-x}sum limits_{k=0}^{m}dfrac{x^k}{k!}$ but then there will be two summations with same limits which I have no clue how to solve
Any suggestions?
summation gamma-function gamma-distribution
edited 2 days ago
David G. Stork
8,97621232
asked 2 days ago
hakkunamattata
454
I have checked several posts but couldn't find the equivalent of $Gamma(m,a) cdot Gamma(m,b)$, where '$cdot$' means multiplication. I suspect that it can be solved by applying the equivalent of Gamma function $(n-1)!e^{-x}sum limits_{k=0}^{m}dfrac{x^k}{k!}$ but then there will be two summations with same limits which I have no clue how to solve
Any suggestions?
summation gamma-function gamma-distribution
summation gamma-function gamma-distribution
edited 2 days ago
David G. Stork
8,97621232
asked 2 days ago
hakkunamattata
454
edited 2 days ago
David G. Stork
8,97621232
asked 2 days ago
hakkunamattata
454
edited 2 days ago
David G. Stork
8,97621232
edited 2 days ago
David G. Stork
8,97621232
edited 2 days ago
David G. Stork
8,97621232
8,97621232
asked 2 days ago
hakkunamattata
454
asked 2 days ago
hakkunamattata
454
asked 2 days ago
hakkunamattata
454
454
What do you understand by $;Gamma(m,a);$ ? That's not the usual Gamma Function...
– DonAntonio
2 days ago
Its the upper incomplete Gamma function
– hakkunamattata
2 days ago
I think it'd be a rather good idea to explicitly say that, and not only "Gamma Functions", which can mislead.
– DonAntonio
2 days ago
Upper or lower incomplete? "*" means multiplication (or something else)? The formula you state only applies when the first argument is a positive integer; is $m$ a positive integer?
– Eric Towers
2 days ago
yes m is positive, I have changed the title. Thanks for suggestion
– hakkunamattata
2 days ago
add a comment |
What do you understand by $;Gamma(m,a);$ ? That's not the usual Gamma Function...
– DonAntonio
2 days ago
Its the upper incomplete Gamma function
– hakkunamattata
2 days ago
I think it'd be a rather good idea to explicitly say that, and not only "Gamma Functions", which can mislead.
– DonAntonio
2 days ago
Upper or lower incomplete? "*" means multiplication (or something else)? The formula you state only applies when the first argument is a positive integer; is $m$ a positive integer?
– Eric Towers
2 days ago
yes m is positive, I have changed the title. Thanks for suggestion
– hakkunamattata
2 days ago
What do you understand by $;Gamma(m,a);$ ? That's not the usual Gamma Function...
– DonAntonio
2 days ago
What do you understand by $;Gamma(m,a);$ ? That's not the usual Gamma Function...
– DonAntonio
2 days ago
Its the upper incomplete Gamma function
– hakkunamattata
2 days ago
Its the upper incomplete Gamma function
– hakkunamattata
2 days ago
I think it'd be a rather good idea to explicitly say that, and not only "Gamma Functions", which can mislead.
– DonAntonio
2 days ago
I think it'd be a rather good idea to explicitly say that, and not only "Gamma Functions", which can mislead.
– DonAntonio
2 days ago
Upper or lower incomplete? "*" means multiplication (or something else)? The formula you state only applies when the first argument is a positive integer; is $m$ a positive integer?
– Eric Towers
2 days ago
Upper or lower incomplete? "*" means multiplication (or something else)? The formula you state only applies when the first argument is a positive integer; is $m$ a positive integer?
– Eric Towers
2 days ago
yes m is positive, I have changed the title. Thanks for suggestion
– hakkunamattata
2 days ago
yes m is positive, I have changed the title. Thanks for suggestion
– hakkunamattata
2 days ago
add a comment |
1 Answer
1
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up vote
0
down vote
We have that
$$
eqalign{
& Gamma (m,a)Gamma (m,b) = Gamma (m)^{,2} Q(m,a)Q(m,b) = cr
& = Gamma (m)^{,2} e^{, - left( {a + b} right)} sumlimits_{k = 0}^{m - 1} {{{a^{,k} } over {k!}}}
sumlimits_{j = 0}^{m - 1} {{{b^{,j} } over {j!}}} cr}
$$
The sum is over a square in $k,j$ and , also with the help of the following scheme,
we can re-write it as
$$
eqalign{
& sumlimits_{k = 0}^{m - 1} {{{a^{,k} } over {k!}}} sumlimits_{j = 0}^{m - 1} {{{b^{,j} } over {j!}}}
= sumlimits_{k = 0}^{m - 1} {sumlimits_{j = 0}^{m - 1} {{{a^{,k} b^{,j} } over {k!j!}}} } = cr
& = sumlimits_{s = 0}^{2m - 1} {sumlimits_{k = 0}^s {{{a^{,k} b^{,s - k} } over {k!left( {s - k} right)!}}} }
- sumlimits_{s = 0}^{m - 1} {sumlimits_{k = 0}^s {{{a^{,m + k} b^{,s - k} } over {left( {m + k} right)!left( {s - k} right)!}}} }
- sumlimits_{s = 0}^{m - 1} {sumlimits_{k = 0}^s {{{a^{,s - k} b^{,m + k} } over {left( {m + k} right)!left( {s - k} right)!}}} } = cr
& = sumlimits_{s = 0}^{2m - 1} {{{left( {a + b} right)^{,s} } over {s!}}}
- a^{,m} sumlimits_{s = 0}^{m - 1} {sumlimits_{k = 0}^s {{{a^{,k} b^{,s - k} } over {left( {m + k} right)!left( {s - k} right)!}}} }
- b^{,m} sumlimits_{s = 0}^{m - 1} {sumlimits_{k = 0}^s {{{a^{,s - k} b^{,k} } over {left( {m + k} right)!left( {s - k} right)!}}} } cr}
$$
note the summation extends to $m-1$, not to $m$.
The formula above can be managed in various other ways, but I cannot see
a way of getting rid of the $m+k$ at denominator.
answered 2 days ago
G Cab
16.9k31237
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
We have that
$$
eqalign{
& Gamma (m,a)Gamma (m,b) = Gamma (m)^{,2} Q(m,a)Q(m,b) = cr
& = Gamma (m)^{,2} e^{, - left( {a + b} right)} sumlimits_{k = 0}^{m - 1} {{{a^{,k} } over {k!}}}
sumlimits_{j = 0}^{m - 1} {{{b^{,j} } over {j!}}} cr}
$$
The sum is over a square in $k,j$ and , also with the help of the following scheme,
we can re-write it as
$$
eqalign{
& sumlimits_{k = 0}^{m - 1} {{{a^{,k} } over {k!}}} sumlimits_{j = 0}^{m - 1} {{{b^{,j} } over {j!}}}
= sumlimits_{k = 0}^{m - 1} {sumlimits_{j = 0}^{m - 1} {{{a^{,k} b^{,j} } over {k!j!}}} } = cr
& = sumlimits_{s = 0}^{2m - 1} {sumlimits_{k = 0}^s {{{a^{,k} b^{,s - k} } over {k!left( {s - k} right)!}}} }
- sumlimits_{s = 0}^{m - 1} {sumlimits_{k = 0}^s {{{a^{,m + k} b^{,s - k} } over {left( {m + k} right)!left( {s - k} right)!}}} }
- sumlimits_{s = 0}^{m - 1} {sumlimits_{k = 0}^s {{{a^{,s - k} b^{,m + k} } over {left( {m + k} right)!left( {s - k} right)!}}} } = cr
& = sumlimits_{s = 0}^{2m - 1} {{{left( {a + b} right)^{,s} } over {s!}}}
- a^{,m} sumlimits_{s = 0}^{m - 1} {sumlimits_{k = 0}^s {{{a^{,k} b^{,s - k} } over {left( {m + k} right)!left( {s - k} right)!}}} }
- b^{,m} sumlimits_{s = 0}^{m - 1} {sumlimits_{k = 0}^s {{{a^{,s - k} b^{,k} } over {left( {m + k} right)!left( {s - k} right)!}}} } cr}
$$
note the summation extends to $m-1$, not to $m$.
The formula above can be managed in various other ways, but I cannot see
a way of getting rid of the $m+k$ at denominator.
answered 2 days ago
G Cab
16.9k31237
add a comment |
up vote
0
down vote
We have that
$$
eqalign{
& Gamma (m,a)Gamma (m,b) = Gamma (m)^{,2} Q(m,a)Q(m,b) = cr
& = Gamma (m)^{,2} e^{, - left( {a + b} right)} sumlimits_{k = 0}^{m - 1} {{{a^{,k} } over {k!}}}
sumlimits_{j = 0}^{m - 1} {{{b^{,j} } over {j!}}} cr}
$$
The sum is over a square in $k,j$ and , also with the help of the following scheme,
we can re-write it as
$$
eqalign{
& sumlimits_{k = 0}^{m - 1} {{{a^{,k} } over {k!}}} sumlimits_{j = 0}^{m - 1} {{{b^{,j} } over {j!}}}
= sumlimits_{k = 0}^{m - 1} {sumlimits_{j = 0}^{m - 1} {{{a^{,k} b^{,j} } over {k!j!}}} } = cr
& = sumlimits_{s = 0}^{2m - 1} {sumlimits_{k = 0}^s {{{a^{,k} b^{,s - k} } over {k!left( {s - k} right)!}}} }
- sumlimits_{s = 0}^{m - 1} {sumlimits_{k = 0}^s {{{a^{,m + k} b^{,s - k} } over {left( {m + k} right)!left( {s - k} right)!}}} }
- sumlimits_{s = 0}^{m - 1} {sumlimits_{k = 0}^s {{{a^{,s - k} b^{,m + k} } over {left( {m + k} right)!left( {s - k} right)!}}} } = cr
& = sumlimits_{s = 0}^{2m - 1} {{{left( {a + b} right)^{,s} } over {s!}}}
- a^{,m} sumlimits_{s = 0}^{m - 1} {sumlimits_{k = 0}^s {{{a^{,k} b^{,s - k} } over {left( {m + k} right)!left( {s - k} right)!}}} }
- b^{,m} sumlimits_{s = 0}^{m - 1} {sumlimits_{k = 0}^s {{{a^{,s - k} b^{,k} } over {left( {m + k} right)!left( {s - k} right)!}}} } cr}
$$
note the summation extends to $m-1$, not to $m$.
The formula above can be managed in various other ways, but I cannot see
a way of getting rid of the $m+k$ at denominator.
answered 2 days ago
G Cab
16.9k31237
add a comment |
up vote
0
down vote
up vote
0
down vote
We have that
$$
eqalign{
& Gamma (m,a)Gamma (m,b) = Gamma (m)^{,2} Q(m,a)Q(m,b) = cr
& = Gamma (m)^{,2} e^{, - left( {a + b} right)} sumlimits_{k = 0}^{m - 1} {{{a^{,k} } over {k!}}}
sumlimits_{j = 0}^{m - 1} {{{b^{,j} } over {j!}}} cr}
$$
The sum is over a square in $k,j$ and , also with the help of the following scheme,
we can re-write it as
$$
eqalign{
& sumlimits_{k = 0}^{m - 1} {{{a^{,k} } over {k!}}} sumlimits_{j = 0}^{m - 1} {{{b^{,j} } over {j!}}}
= sumlimits_{k = 0}^{m - 1} {sumlimits_{j = 0}^{m - 1} {{{a^{,k} b^{,j} } over {k!j!}}} } = cr
& = sumlimits_{s = 0}^{2m - 1} {sumlimits_{k = 0}^s {{{a^{,k} b^{,s - k} } over {k!left( {s - k} right)!}}} }
- sumlimits_{s = 0}^{m - 1} {sumlimits_{k = 0}^s {{{a^{,m + k} b^{,s - k} } over {left( {m + k} right)!left( {s - k} right)!}}} }
- sumlimits_{s = 0}^{m - 1} {sumlimits_{k = 0}^s {{{a^{,s - k} b^{,m + k} } over {left( {m + k} right)!left( {s - k} right)!}}} } = cr
& = sumlimits_{s = 0}^{2m - 1} {{{left( {a + b} right)^{,s} } over {s!}}}
- a^{,m} sumlimits_{s = 0}^{m - 1} {sumlimits_{k = 0}^s {{{a^{,k} b^{,s - k} } over {left( {m + k} right)!left( {s - k} right)!}}} }
- b^{,m} sumlimits_{s = 0}^{m - 1} {sumlimits_{k = 0}^s {{{a^{,s - k} b^{,k} } over {left( {m + k} right)!left( {s - k} right)!}}} } cr}
$$
note the summation extends to $m-1$, not to $m$.
The formula above can be managed in various other ways, but I cannot see
a way of getting rid of the $m+k$ at denominator.
answered 2 days ago
G Cab
16.9k31237
We have that
$$
eqalign{
& Gamma (m,a)Gamma (m,b) = Gamma (m)^{,2} Q(m,a)Q(m,b) = cr
& = Gamma (m)^{,2} e^{, - left( {a + b} right)} sumlimits_{k = 0}^{m - 1} {{{a^{,k} } over {k!}}}
sumlimits_{j = 0}^{m - 1} {{{b^{,j} } over {j!}}} cr}
$$
The sum is over a square in $k,j$ and , also with the help of the following scheme,
we can re-write it as
$$
eqalign{
& sumlimits_{k = 0}^{m - 1} {{{a^{,k} } over {k!}}} sumlimits_{j = 0}^{m - 1} {{{b^{,j} } over {j!}}}
= sumlimits_{k = 0}^{m - 1} {sumlimits_{j = 0}^{m - 1} {{{a^{,k} b^{,j} } over {k!j!}}} } = cr
& = sumlimits_{s = 0}^{2m - 1} {sumlimits_{k = 0}^s {{{a^{,k} b^{,s - k} } over {k!left( {s - k} right)!}}} }
- sumlimits_{s = 0}^{m - 1} {sumlimits_{k = 0}^s {{{a^{,m + k} b^{,s - k} } over {left( {m + k} right)!left( {s - k} right)!}}} }
- sumlimits_{s = 0}^{m - 1} {sumlimits_{k = 0}^s {{{a^{,s - k} b^{,m + k} } over {left( {m + k} right)!left( {s - k} right)!}}} } = cr
& = sumlimits_{s = 0}^{2m - 1} {{{left( {a + b} right)^{,s} } over {s!}}}
- a^{,m} sumlimits_{s = 0}^{m - 1} {sumlimits_{k = 0}^s {{{a^{,k} b^{,s - k} } over {left( {m + k} right)!left( {s - k} right)!}}} }
- b^{,m} sumlimits_{s = 0}^{m - 1} {sumlimits_{k = 0}^s {{{a^{,s - k} b^{,k} } over {left( {m + k} right)!left( {s - k} right)!}}} } cr}
$$
note the summation extends to $m-1$, not to $m$.
The formula above can be managed in various other ways, but I cannot see
a way of getting rid of the $m+k$ at denominator.
answered 2 days ago
G Cab
16.9k31237
edited yesterday
answered 2 days ago
G Cab
16.9k31237
answered 2 days ago
G Cab
16.9k31237
answered 2 days ago
G Cab
16.9k31237
16.9k31237
add a comment |
add a comment |
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What do you understand by $;Gamma(m,a);$ ? That's not the usual Gamma Function...
– DonAntonio
2 days ago
Its the upper incomplete Gamma function
– hakkunamattata
2 days ago
I think it'd be a rather good idea to explicitly say that, and not only "Gamma Functions", which can mislead.
– DonAntonio
2 days ago
Upper or lower incomplete? "*" means multiplication (or something else)? The formula you state only applies when the first argument is a positive integer; is $m$ a positive integer?
– Eric Towers
2 days ago
yes m is positive, I have changed the title. Thanks for suggestion
– hakkunamattata
2 days ago