hypergeometric function with special Pochhammer symbol
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What is the relationship between these two hypergeometric functions? Can the following function be written as another function of some hypergeometric functions ?
$$1F1(a+b,2a,x)$$ and $$1F1(a+b,a,x)$$ Can I convert 'a' to '2a'?
hypergeometric-function
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show 6 more comments
up vote
0
down vote
favorite
What is the relationship between these two hypergeometric functions? Can the following function be written as another function of some hypergeometric functions ?
$$1F1(a+b,2a,x)$$ and $$1F1(a+b,a,x)$$ Can I convert 'a' to '2a'?
hypergeometric-function
I haven't an answer. But have a look at the interesting document url{fuw.edu.pl/~derezins/hyper-published.pdf} around paragraph (4.9). Maybe Kummer's relationship is a step...
– Jean Marie
2 days ago
@ Jean Marie:Unfortunately, I did not see anything in this field.
– Dana
2 days ago
Have you had a look at {en.wikipedia.org/wiki/…} ?
– Jean Marie
2 days ago
Another source again {functions.wolfram.com/HypergeometricFunctions/Hypergeometric1F1/…}
– Jean Marie
2 days ago
I did not find anything related.
– Dana
2 days ago
|
show 6 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
What is the relationship between these two hypergeometric functions? Can the following function be written as another function of some hypergeometric functions ?
$$1F1(a+b,2a,x)$$ and $$1F1(a+b,a,x)$$ Can I convert 'a' to '2a'?
hypergeometric-function
What is the relationship between these two hypergeometric functions? Can the following function be written as another function of some hypergeometric functions ?
$$1F1(a+b,2a,x)$$ and $$1F1(a+b,a,x)$$ Can I convert 'a' to '2a'?
hypergeometric-function
hypergeometric-function
edited 18 hours ago
Jean Marie
28k41848
28k41848
asked 2 days ago
Dana
84
84
I haven't an answer. But have a look at the interesting document url{fuw.edu.pl/~derezins/hyper-published.pdf} around paragraph (4.9). Maybe Kummer's relationship is a step...
– Jean Marie
2 days ago
@ Jean Marie:Unfortunately, I did not see anything in this field.
– Dana
2 days ago
Have you had a look at {en.wikipedia.org/wiki/…} ?
– Jean Marie
2 days ago
Another source again {functions.wolfram.com/HypergeometricFunctions/Hypergeometric1F1/…}
– Jean Marie
2 days ago
I did not find anything related.
– Dana
2 days ago
|
show 6 more comments
I haven't an answer. But have a look at the interesting document url{fuw.edu.pl/~derezins/hyper-published.pdf} around paragraph (4.9). Maybe Kummer's relationship is a step...
– Jean Marie
2 days ago
@ Jean Marie:Unfortunately, I did not see anything in this field.
– Dana
2 days ago
Have you had a look at {en.wikipedia.org/wiki/…} ?
– Jean Marie
2 days ago
Another source again {functions.wolfram.com/HypergeometricFunctions/Hypergeometric1F1/…}
– Jean Marie
2 days ago
I did not find anything related.
– Dana
2 days ago
I haven't an answer. But have a look at the interesting document url{fuw.edu.pl/~derezins/hyper-published.pdf} around paragraph (4.9). Maybe Kummer's relationship is a step...
– Jean Marie
2 days ago
I haven't an answer. But have a look at the interesting document url{fuw.edu.pl/~derezins/hyper-published.pdf} around paragraph (4.9). Maybe Kummer's relationship is a step...
– Jean Marie
2 days ago
@ Jean Marie:Unfortunately, I did not see anything in this field.
– Dana
2 days ago
@ Jean Marie:Unfortunately, I did not see anything in this field.
– Dana
2 days ago
Have you had a look at {en.wikipedia.org/wiki/…} ?
– Jean Marie
2 days ago
Have you had a look at {en.wikipedia.org/wiki/…} ?
– Jean Marie
2 days ago
Another source again {functions.wolfram.com/HypergeometricFunctions/Hypergeometric1F1/…}
– Jean Marie
2 days ago
Another source again {functions.wolfram.com/HypergeometricFunctions/Hypergeometric1F1/…}
– Jean Marie
2 days ago
I did not find anything related.
– Dana
2 days ago
I did not find anything related.
– Dana
2 days ago
|
show 6 more comments
1 Answer
1
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We have that
$$
{}_1F_1 (a + b,2a,x) = sumlimits_{0, le ,k} {{{left( {a + b} right)^{,overline {,k,} } } over {left( {2a} right)^{,overline {,k,} } }}{{x^{,k} } over {k!}}}
$$
where the exponent overlined is a way to indicate the Rising Factorial or Pochammer symbol.
Now, the denominator equals
$$
eqalign{
& left( {2a} right)^{,overline {,k,} } = prodlimits_{0, le ,j, le ,k - 1} {left( {2a + j} right)} = cr
& = prodlimits_{0, le ,2j, le ,leftlfloor {{{k - 1} over 2}} rightrfloor = leftlceil {{k over 2}} rightrceil - 1} {left( {2a + 2j} right)}
prodlimits_{0, le ,2j, le ,leftlfloor {{k over 2}} rightrfloor - 1} {left( {2a + 1 + 2j} right)} = cr
& = 2^{,k} prodlimits_{0, le ,2j, le ,leftlfloor {{{k - 1} over 2}} rightrfloor = leftlceil {{k over 2}} rightrceil - 1} {left( {a + j} right)}
prodlimits_{0, le ,2j, le ,leftlfloor {{k over 2}} rightrfloor - 1} {left( {a + 1/2 + j} right)} = cr
& = 2^{,k} a^{,overline {,leftlceil {{k over 2}} rightrceil ,} } left( {a + 1/2} right)^{,overline {,leftlfloor {{k over 2}} rightrfloor ,} } cr}
$$
or, alternatively
$$
left( {2a} right)^{,overline {,k,} } = left( {2a} right)^{,overline {,k + a - a,} } = left( {2a} right)^{,overline {, - a,} } left( a right)^{,overline {,k + a,} }
$$
In both versions it is not possible to simply relate the above results with $a^{,overline {,k ,} }$.
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
$$
{}_1F_1 (a + b,2a,x) = sumlimits_{0, le ,k} {{{left( {a + b} right)^{,overline {,k,} } } over {left( {2a} right)^{,overline {,k,} } }}{{x^{,k} } over {k!}}}
$$
where the exponent overlined is a way to indicate the Rising Factorial or Pochammer symbol.
Now, the denominator equals
$$
eqalign{
& left( {2a} right)^{,overline {,k,} } = prodlimits_{0, le ,j, le ,k - 1} {left( {2a + j} right)} = cr
& = prodlimits_{0, le ,2j, le ,leftlfloor {{{k - 1} over 2}} rightrfloor = leftlceil {{k over 2}} rightrceil - 1} {left( {2a + 2j} right)}
prodlimits_{0, le ,2j, le ,leftlfloor {{k over 2}} rightrfloor - 1} {left( {2a + 1 + 2j} right)} = cr
& = 2^{,k} prodlimits_{0, le ,2j, le ,leftlfloor {{{k - 1} over 2}} rightrfloor = leftlceil {{k over 2}} rightrceil - 1} {left( {a + j} right)}
prodlimits_{0, le ,2j, le ,leftlfloor {{k over 2}} rightrfloor - 1} {left( {a + 1/2 + j} right)} = cr
& = 2^{,k} a^{,overline {,leftlceil {{k over 2}} rightrceil ,} } left( {a + 1/2} right)^{,overline {,leftlfloor {{k over 2}} rightrfloor ,} } cr}
$$
or, alternatively
$$
left( {2a} right)^{,overline {,k,} } = left( {2a} right)^{,overline {,k + a - a,} } = left( {2a} right)^{,overline {, - a,} } left( a right)^{,overline {,k + a,} }
$$
In both versions it is not possible to simply relate the above results with $a^{,overline {,k ,} }$.
add a comment |
up vote
0
down vote
We have that
$$
{}_1F_1 (a + b,2a,x) = sumlimits_{0, le ,k} {{{left( {a + b} right)^{,overline {,k,} } } over {left( {2a} right)^{,overline {,k,} } }}{{x^{,k} } over {k!}}}
$$
where the exponent overlined is a way to indicate the Rising Factorial or Pochammer symbol.
Now, the denominator equals
$$
eqalign{
& left( {2a} right)^{,overline {,k,} } = prodlimits_{0, le ,j, le ,k - 1} {left( {2a + j} right)} = cr
& = prodlimits_{0, le ,2j, le ,leftlfloor {{{k - 1} over 2}} rightrfloor = leftlceil {{k over 2}} rightrceil - 1} {left( {2a + 2j} right)}
prodlimits_{0, le ,2j, le ,leftlfloor {{k over 2}} rightrfloor - 1} {left( {2a + 1 + 2j} right)} = cr
& = 2^{,k} prodlimits_{0, le ,2j, le ,leftlfloor {{{k - 1} over 2}} rightrfloor = leftlceil {{k over 2}} rightrceil - 1} {left( {a + j} right)}
prodlimits_{0, le ,2j, le ,leftlfloor {{k over 2}} rightrfloor - 1} {left( {a + 1/2 + j} right)} = cr
& = 2^{,k} a^{,overline {,leftlceil {{k over 2}} rightrceil ,} } left( {a + 1/2} right)^{,overline {,leftlfloor {{k over 2}} rightrfloor ,} } cr}
$$
or, alternatively
$$
left( {2a} right)^{,overline {,k,} } = left( {2a} right)^{,overline {,k + a - a,} } = left( {2a} right)^{,overline {, - a,} } left( a right)^{,overline {,k + a,} }
$$
In both versions it is not possible to simply relate the above results with $a^{,overline {,k ,} }$.
add a comment |
up vote
0
down vote
up vote
0
down vote
We have that
$$
{}_1F_1 (a + b,2a,x) = sumlimits_{0, le ,k} {{{left( {a + b} right)^{,overline {,k,} } } over {left( {2a} right)^{,overline {,k,} } }}{{x^{,k} } over {k!}}}
$$
where the exponent overlined is a way to indicate the Rising Factorial or Pochammer symbol.
Now, the denominator equals
$$
eqalign{
& left( {2a} right)^{,overline {,k,} } = prodlimits_{0, le ,j, le ,k - 1} {left( {2a + j} right)} = cr
& = prodlimits_{0, le ,2j, le ,leftlfloor {{{k - 1} over 2}} rightrfloor = leftlceil {{k over 2}} rightrceil - 1} {left( {2a + 2j} right)}
prodlimits_{0, le ,2j, le ,leftlfloor {{k over 2}} rightrfloor - 1} {left( {2a + 1 + 2j} right)} = cr
& = 2^{,k} prodlimits_{0, le ,2j, le ,leftlfloor {{{k - 1} over 2}} rightrfloor = leftlceil {{k over 2}} rightrceil - 1} {left( {a + j} right)}
prodlimits_{0, le ,2j, le ,leftlfloor {{k over 2}} rightrfloor - 1} {left( {a + 1/2 + j} right)} = cr
& = 2^{,k} a^{,overline {,leftlceil {{k over 2}} rightrceil ,} } left( {a + 1/2} right)^{,overline {,leftlfloor {{k over 2}} rightrfloor ,} } cr}
$$
or, alternatively
$$
left( {2a} right)^{,overline {,k,} } = left( {2a} right)^{,overline {,k + a - a,} } = left( {2a} right)^{,overline {, - a,} } left( a right)^{,overline {,k + a,} }
$$
In both versions it is not possible to simply relate the above results with $a^{,overline {,k ,} }$.
We have that
$$
{}_1F_1 (a + b,2a,x) = sumlimits_{0, le ,k} {{{left( {a + b} right)^{,overline {,k,} } } over {left( {2a} right)^{,overline {,k,} } }}{{x^{,k} } over {k!}}}
$$
where the exponent overlined is a way to indicate the Rising Factorial or Pochammer symbol.
Now, the denominator equals
$$
eqalign{
& left( {2a} right)^{,overline {,k,} } = prodlimits_{0, le ,j, le ,k - 1} {left( {2a + j} right)} = cr
& = prodlimits_{0, le ,2j, le ,leftlfloor {{{k - 1} over 2}} rightrfloor = leftlceil {{k over 2}} rightrceil - 1} {left( {2a + 2j} right)}
prodlimits_{0, le ,2j, le ,leftlfloor {{k over 2}} rightrfloor - 1} {left( {2a + 1 + 2j} right)} = cr
& = 2^{,k} prodlimits_{0, le ,2j, le ,leftlfloor {{{k - 1} over 2}} rightrfloor = leftlceil {{k over 2}} rightrceil - 1} {left( {a + j} right)}
prodlimits_{0, le ,2j, le ,leftlfloor {{k over 2}} rightrfloor - 1} {left( {a + 1/2 + j} right)} = cr
& = 2^{,k} a^{,overline {,leftlceil {{k over 2}} rightrceil ,} } left( {a + 1/2} right)^{,overline {,leftlfloor {{k over 2}} rightrfloor ,} } cr}
$$
or, alternatively
$$
left( {2a} right)^{,overline {,k,} } = left( {2a} right)^{,overline {,k + a - a,} } = left( {2a} right)^{,overline {, - a,} } left( a right)^{,overline {,k + a,} }
$$
In both versions it is not possible to simply relate the above results with $a^{,overline {,k ,} }$.
answered 17 hours ago
G Cab
16.8k31237
16.8k31237
add a comment |
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I haven't an answer. But have a look at the interesting document url{fuw.edu.pl/~derezins/hyper-published.pdf} around paragraph (4.9). Maybe Kummer's relationship is a step...
– Jean Marie
2 days ago
@ Jean Marie:Unfortunately, I did not see anything in this field.
– Dana
2 days ago
Have you had a look at {en.wikipedia.org/wiki/…} ?
– Jean Marie
2 days ago
Another source again {functions.wolfram.com/HypergeometricFunctions/Hypergeometric1F1/…}
– Jean Marie
2 days ago
I did not find anything related.
– Dana
2 days ago