Timeline for How to solve recurrence relation with 2 variables？
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
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| May 10, 2023 at 2:01 | comment | added | Max Alekseyev | A closed expression for $g(n,m)$: $$g(n,m) = (n+m+1)\binom{n+m}n \int_0^1 (1+(\alpha-1)t)^n (1+(\beta-1)t)^m {\rm d}t,$$ | |
| May 9, 2023 at 4:28 | vote | accept | Lili Si | ||
| May 5, 2023 at 5:12 | history | edited | Notamathematician | CC BY-SA 4.0 |
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| May 3, 2023 at 21:16 | comment | added | Peter Mueller | @Notamathematician Your answer is not "useless" as you guessed the correct form. Note that writing the OP in terms of $g(n,m)=\binom{n+m}{m}f(n,m)$ results in a simpler recursion ($\frac{\alpha n}{n+m}$ replaced with $\alpha$, and the other factor with $\beta$) which should also allow for a generating function proof. | |
| May 3, 2023 at 8:51 | comment | added | Notamathematician | @PeterTaylor, thak you for comment! Could you write your own answer including the proof? In that case, I could delete my useless answer based on the experimental result. | |
| May 3, 2023 at 8:46 | comment | added | Peter Taylor | The obvious way to prove it would be to show that the identity holds at the boundary conditions and is preserved by the recurrence. Since the terms can be grouped trivially by the exponents of $\alpha$ and $\beta$ this should be easy. | |
| May 3, 2023 at 8:04 | history | edited | Notamathematician | CC BY-SA 4.0 |
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| May 3, 2023 at 7:44 | history | undeleted | Notamathematician | ||
| May 3, 2023 at 7:44 | history | edited | Notamathematician | CC BY-SA 4.0 |
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| May 3, 2023 at 7:16 | history | deleted | Notamathematician | via Vote | |
| May 3, 2023 at 6:42 | history | answered | Notamathematician | CC BY-SA 4.0 |