Mixing
entire function of gamma function
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$$frac{Gamma(az+b)}{Gamma(cz+d)Gamma(mz+n)}$$
my question is as follows:
what can we say about if the function is entire function or not? under which conditions this function is en entire function.
complex-analysis
asked yesterday
efe kağan
133
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$$frac{Gamma(az+b)}{Gamma(cz+d)Gamma(mz+n)}$$
my question is as follows:
what can we say about if the function is entire function or not? under which conditions this function is en entire function.
complex-analysis
asked yesterday
efe kağan
133
2
The Gamma function has no zeros, so the singularities come from the numerator, i.e. $az+b =0,-1,-2,ldots$. Provided, of course, that neither $cz+d$ nor $mz+n$ are in ${0,-1,-2,ldots}$.
– Richard Martin
yesterday
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
$$frac{Gamma(az+b)}{Gamma(cz+d)Gamma(mz+n)}$$
my question is as follows:
what can we say about if the function is entire function or not? under which conditions this function is en entire function.
complex-analysis
asked yesterday
efe kağan
133
$$frac{Gamma(az+b)}{Gamma(cz+d)Gamma(mz+n)}$$
my question is as follows:
what can we say about if the function is entire function or not? under which conditions this function is en entire function.
complex-analysis
complex-analysis
asked yesterday
efe kağan
133
asked yesterday
efe kağan
133
asked yesterday
efe kağan
133
asked yesterday
efe kağan
133
asked yesterday
efe kağan
133
133
2
The Gamma function has no zeros, so the singularities come from the numerator, i.e. $az+b =0,-1,-2,ldots$. Provided, of course, that neither $cz+d$ nor $mz+n$ are in ${0,-1,-2,ldots}$.
– Richard Martin
yesterday
add a comment |
2
The Gamma function has no zeros, so the singularities come from the numerator, i.e. $az+b =0,-1,-2,ldots$. Provided, of course, that neither $cz+d$ nor $mz+n$ are in ${0,-1,-2,ldots}$.
– Richard Martin
yesterday
2
2
The Gamma function has no zeros, so the singularities come from the numerator, i.e. $az+b =0,-1,-2,ldots$. Provided, of course, that neither $cz+d$ nor $mz+n$ are in ${0,-1,-2,ldots}$.
– Richard Martin
yesterday
The Gamma function has no zeros, so the singularities come from the numerator, i.e. $az+b =0,-1,-2,ldots$. Provided, of course, that neither $cz+d$ nor $mz+n$ are in ${0,-1,-2,ldots}$.
– Richard Martin
yesterday
add a comment |
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The Gamma function has no zeros, so the singularities come from the numerator, i.e. $az+b =0,-1,-2,ldots$. Provided, of course, that neither $cz+d$ nor $mz+n$ are in ${0,-1,-2,ldots}$.
– Richard Martin
yesterday