Timeline for Limit of recursion relation
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
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| Jan 26, 2023 at 13:11 | comment | added | José María Grau Ribas | Perhaps the following proposal would have been better $$ F_n(0)=3 \textrm{ and }F_n(k)=\frac{1}{k^2}+\frac{\left(1-\frac{1}{n}\right) (k+n+k/n)}{k+n+2} F_n(k-1), \textrm{ for } 0< k\leq n $$ | |
| Jan 26, 2023 at 13:04 | history | edited | Carlo Beenakker | CC BY-SA 4.0 |
added 215 characters in body
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| Jan 26, 2023 at 12:45 | history | edited | Carlo Beenakker | CC BY-SA 4.0 |
added 233 characters in body
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| Jan 26, 2023 at 12:39 | comment | added | Carlo Beenakker | I will add the Mathematica commands (I use version 13.2, perhaps they also work on earlier versions) | |
| Jan 26, 2023 at 12:30 | history | edited | Carlo Beenakker | CC BY-SA 4.0 |
added 44 characters in body
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| Jan 26, 2023 at 12:27 | comment | added | José María Grau Ribas | Excuse me. Mathematica 11.0 solve the recursion but does not calculate that limit | |
| Jan 26, 2023 at 11:53 | history | answered | Carlo Beenakker | CC BY-SA 4.0 |