Timeline for How do I define the following piecewise function?
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
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| S Feb 2, 2015 at 4:27 | history | suggested | Lola | CC BY-SA 3.0 |
This is what I originally wrote. I corrected the typos.
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| Feb 2, 2015 at 2:54 | comment | added | Lola | So the function of n I wanted is f[n,\floor{n/5}] (using the definition of f[n,t] given above). | |
| Feb 2, 2015 at 2:51 | comment | added | Lola | I wanted to have an efficient way of defining this function. Something that would work when the range is much larger or the conditional more complicated. | |
| Feb 2, 2015 at 2:47 | review | Suggested edits | |||
| S Feb 2, 2015 at 4:27 | |||||
| Feb 2, 2015 at 2:45 | comment | added | Lola | No, f as I wrote it, is a function of n alone. In order to define it, I used the auxiliary variable t. In other words, what I wrote means: f[n]=0 if 0<=n<2, f[n]=1 if 3<=n<=4, f[n]=4 if 5<=n<=7, f[n]=5 if 8<=n<=9, f[n]=8 if 10<=n<=12, etc | |
| Feb 1, 2015 at 22:01 | comment | added | mgamer | f depends on two variables. | |
| Feb 1, 2015 at 21:32 | vote | accept | Lola | ||
| Feb 2, 2015 at 15:34 | |||||
| Feb 1, 2015 at 21:31 | comment | added | Lola | I guess you can solve for t in terms of n and plug it in the function f[n,t], so I guess this will do it in this particular example. However, I am still curious on how would one define a function of n without solving for t. This will allow for a more complicated condition instead of 5t<=n<=5t+3... | |
| Feb 1, 2015 at 21:12 | comment | added | Lola | Thanks! This is useful. The only thing is that I would really need a function f of n alone, without keeping track of the t. | |
| Feb 1, 2015 at 21:07 | vote | accept | Lola | ||
| Feb 1, 2015 at 21:09 | |||||
| Feb 1, 2015 at 18:49 | history | answered | k_v | CC BY-SA 3.0 |