Timeline for Finding solutions to $f'(x) = f(x + k)$
Current License: CC BY-SA 2.5
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 27, 2010 at 4:37 | history | edited | Hashem sazegar | CC BY-SA 2.5 |
added 27 characters in body
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| Sep 27, 2010 at 4:35 | comment | added | Hashem sazegar | no problem if you omit k in the first line so you can reach in no.1 to $k<1$ and second one to $k\ge 1$ | |
| Sep 26, 2010 at 21:00 | comment | added | Yemon Choi | Hashem: $k$ is given at the start, and $a$ must be found. You cannot say "if $a$ does this then $k$ will do this" - that is not a proof. | |
| Sep 26, 2010 at 10:21 | comment | added | Hashem sazegar | $a$ is larger than 1 and less than almost 1.76 so $k$ is almost >=1 | |
| Sep 26, 2010 at 10:13 | comment | added | Hashem sazegar | since $k\ge 1$ and $0<x<1 so $x+k>1$ then $g(x+k)=-1$,from this results we have $f(x+k)=-a^{-x-k}$,so $a^{-k}=log{a}$,hence $k=-loglog{a}/log{a}$ | |
| Sep 26, 2010 at 10:02 | comment | added | Yemon Choi | It appears to be a pair of Ansätze for the problem. However, it seems to be lacking a proof that given a particular $k$ one can actually find such an $a$. This would require rather more care than is being demonstrated here. | |
| Sep 26, 2010 at 9:49 | comment | added | S. Carnahan♦ | I'm going to delete your old answer. I have fixed the LaTeX in your new answer, but I still have no idea what it means. | |
| Sep 26, 2010 at 9:45 | history | edited | S. Carnahan♦ | CC BY-SA 2.5 |
fixed LaTeX
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| Sep 26, 2010 at 8:47 | history | edited | Hashem sazegar | CC BY-SA 2.5 |
added 12 characters in body
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| Sep 26, 2010 at 8:32 | history | edited | Hashem sazegar | CC BY-SA 2.5 |
deleted 7 characters in body; deleted 7 characters in body
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| Sep 26, 2010 at 8:23 | history | answered | Hashem sazegar | CC BY-SA 2.5 |