Timeline for Why can't Mathematica evaluate the limit of this hypergeometric function? Is there any way to find the answer?
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
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| Nov 11, 2021 at 6:00 | history | tweeted | twitter.com/StackMma/status/1458675887088578562 | ||
| Nov 10, 2021 at 19:51 | answer | added | Carl Woll | timeline score: 2 | |
| Nov 10, 2021 at 17:37 | history | became hot network question | |||
| Nov 10, 2021 at 16:21 | vote | accept | Kheeyal | ||
| Nov 10, 2021 at 10:35 | answer | added | Ulrich Neumann | timeline score: 4 | |
| Nov 10, 2021 at 10:19 | answer | added | Roman | timeline score: 8 | |
| Nov 10, 2021 at 9:56 | comment | added | user64494 | And such a general formula may not exist. | |
| Nov 10, 2021 at 9:48 | comment | added | user64494 |
Table[Limit[f, r -> Infinity, Assumptions -> d > 1], {d, 2, 7}] results in {\[Infinity], \[Infinity], -(2/9) 2^(3/4) Sqrt[\[Pi]] (9 Gamma[-(3/4)] + 4 Gamma[1/4]) Gamma[5/4], -((\[Pi]^( 3/2) (8 Gamma[-(2/3)] Gamma[7/6] + 25 Gamma[1/3] Gamma[7/6] - 11 Gamma[1/6] Gamma[4/3]))/(4 2^(1/3))), -(2/225) 2^(5/8) \[Pi]^( 3/2) (150 Gamma[-(5/8)] Gamma[9/8] + 513 Gamma[3/8] Gamma[9/8] - 155 Gamma[1/8] Gamma[11/8]), -((\[Pi]^( 5/2) (18 Gamma[-(3/5)] Gamma[11/10] + 65 Gamma[2/5] Gamma[11/10] - 15 Gamma[1/10] Gamma[7/5]))/(18 2^(2/5)))}, so general formula is unclear.
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| Nov 10, 2021 at 9:37 | comment | added | user64494 |
Limit[f, r -> Infinity, Assumptions -> d > 1] results in ConditionalExpression[\[Infinity], d < 3/2] and both Limit[f, r -> Infinity, Assumptions -> d > 3/2 && d \[Element] Integers]andLimit[f, r -> Infinity, Assumptions -> d > 3/2] return the input.
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| Nov 10, 2021 at 8:34 | history | edited | user64494 |
edited tags
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| Nov 10, 2021 at 5:32 | history | asked | Kheeyal | CC BY-SA 4.0 |