Timeline for Shannon's communication paper and finite differences
Current License: CC BY-SA 2.5
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 21, 2011 at 17:16 | comment | added | Did | @darij No problem. Since your 2nd interpretation and Shannon's equation are unrelated, no surprise that the floor function (plus constant or not) never solves the latter. Of course you already know this, so I will stop here. :-) | |
| Jan 21, 2011 at 14:10 | comment | added | darij grinberg | @Didier: Sorry, it's only now that I have noticed your first comment to my post. It seems that you have confused my 1st interpretation with the 2nd one. The 2nd one allows arbitrarily long pauses, so I can transmit any number of symbols from 0 to 13 within a timespan of 13.37 seconds. So, you are right that it's not the floor function: it's the floor function + 1 ;) | |
| Jan 21, 2011 at 6:48 | comment | added | Did | @darij Bis repetita: if $n=1$, then $N$ is not the floor function. | |
| Jan 21, 2011 at 4:30 | comment | added | Anthony Quas | Is it obvious in Morse code that the length of a dash is an exact integer (or rational) multiple of the length of a dot? | |
| Jan 20, 2011 at 20:48 | comment | added | darij grinberg | "Not necessarily the same" doesn't mean "not necessarily commensurable". So you are going with the 2nd interpretation. Just read what I wrote about it: If you had just one symbol with length $1$, then your $N(t)$ would be the floor of $t$, contradicting the asymptotics. | |
| Jan 20, 2011 at 20:21 | comment | added | Anthony Quas | I interpret $N(t)$ as the number of sequences that can be transmitted in time $t$ or less. Thanks for the commensurate -> commensurable. I fixed this in the question. I don't see any evidence that Shannon wants the $t_i$'s to be integers. Indeed when he introduces them he says "Each of the symbols $S_i$ is assumed to have a duration in time $t_i$ seconds (not necessarily the same for different $S_i$, for example the dots and dashes in telegraphy)." | |
| Jan 20, 2011 at 20:17 | comment | added | Did | @darij When $n=1$ and $t_1=1$, $N$ is any $1$-periodic function (and not the floor function) hence $N$ is bounded as soon as it is bounded on any interval of length $1$. | |
| Jan 20, 2011 at 19:40 | history | answered | darij grinberg | CC BY-SA 2.5 |