Timeline for How does the function g(x) behave as x tends to 1?
Current License: CC BY-SA 3.0
17 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 15, 2017 at 21:14 | comment | added | H. H. Rugh | Don't know if this is of interest to you, but if you instead require $g:(0,1]\rightarrow {\Bbb R}$ continuous then there is a solution that behaves like $g(x)=\frac12 - \frac13 \ln(x) + O((\ln x)^2)$ as $x\rightarrow 1$ but this solution does not extend continuously to $x=0$ | |
| Mar 15, 2017 at 11:00 | answer | added | Todd Trimble | timeline score: 9 | |
| Mar 15, 2017 at 10:45 | history | reopened |
Fedor Petrov Neil Strickland Todd Trimble |
||
| Mar 15, 2017 at 10:37 | comment | added | Todd Trimble | Wait... this problem seems very familiar, and with a somewhat surprising solution. (My memory could be deceiving me, but for some reason Noam Elkies comes to mind as having discussed this (?!)) | |
| Mar 15, 2017 at 10:06 | review | Reopen votes | |||
| Mar 15, 2017 at 10:45 | |||||
| Mar 15, 2017 at 9:51 | comment | added | Fedor Petrov | I think, it is quite a normal and not trivial question. | |
| Mar 15, 2017 at 8:53 | comment | added | user44191 | @NeilStrickland Ah, of course, was so focused on 1 that I forgot about 0. The periodic function I derived my homogeneous solution must converge as it goes to $\infty$, so it must be constant, and the only solution is $g(x) = 0$. | |
| Mar 15, 2017 at 8:39 | history | edited | Neil Strickland | CC BY-SA 3.0 |
edited title
|
| Mar 15, 2017 at 8:38 | comment | added | Neil Strickland | @user44191 Your functions are not continuous at $x=0$. For example, if $g(x)=\cos(\pi\log_2(-\ln(x)))$ then we can consider $a_n=\exp(-2^n)$ and $a_n\to 0$ with $g(a_n)=(-1)^n$. If $g(x)=-g(x^2)$ then $g(x)=g(x^{2^{2k}})$ and $x^{2^{2k}}\to 0$, so if $g$ is continuous then $g(x)=g(0)$. Also, the identity $g(x)=-g(x^2)$ gives $g(0)=0$. | |
| Mar 15, 2017 at 8:35 | history | closed |
Michael Albanese Steven Landsburg Stefan Waldmann Carlo Beenakker R.P. |
Not suitable for this site | |
| Mar 15, 2017 at 8:18 | comment | added | user44191 | @NeilStrickland That can't be the only solution, given that the homogeneous form $g(x^2) = -g(x)$ has nontrivial solutions, of the form $\sum_n c_n cos((2n + 1) \pi log_2(-ln(x)) + d_n sin((2n + 1) \pi log_2(-ln(x))$. | |
| Mar 15, 2017 at 7:52 | history | edited | Neil Strickland | CC BY-SA 3.0 |
Spelling
|
| Mar 15, 2017 at 7:44 | comment | added | Neil Strickland | Obviously, if there is a limit, then it has to be 1/2. One can prove that $g(x)=\sum_k(-1)^kx^{2^k}$. I think that this is nondecreasing and tends to $1/2$ as $x\to 1$, but it does not seem to be trivial to prove that. | |
| Mar 15, 2017 at 5:39 | comment | added | Eugene | Emm Tends to 0.5? | |
| Mar 15, 2017 at 2:25 | review | Close votes | |||
| Mar 15, 2017 at 8:35 | |||||
| Mar 15, 2017 at 2:09 | review | First posts | |||
| Mar 15, 2017 at 3:29 | |||||
| Mar 15, 2017 at 2:07 | history | asked | Sarah Terrep | CC BY-SA 3.0 |