Timeline for Is the following function decreasing on $(0,1)$?
Current License: CC BY-SA 3.0
22 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 24, 2012 at 18:15 | vote | accept | Malik Younsi | ||
| Jul 24, 2012 at 18:15 | history | bounty ended | Malik Younsi | ||
| Jul 24, 2012 at 17:59 | comment | added | Malik Younsi | @juan : I was confused, but I get it now. Thank you! | |
| Jul 24, 2012 at 15:48 | comment | added | juan | We have $f(e^{-\pi/x}) = g(x) \vartheta_4^2(e^{-\pi x})$. The factor $\vartheta_4^2(e^{-\pi x})$ is almost constant $=1$. Therefore the character of $f(e^{-\pi/x}) $ will be that of $g(x)$. Of course this need careful bounds. Those of $\vartheta_4^2(e^{-\pi x})$ are easily obtained from the expression given for the $\vartheta$ functions at the start of my answer. Those of $g(x)$ must be elementary. For example the error in $\vartheta_4^2(e^{-\pi x})\approx1$ is of the order of $C e^{-\pi x}$. This will work for $x\gg1$. | |
| Jul 24, 2012 at 14:47 | comment | added | Malik Younsi | @juan The function $g(x)=\frac{x}{2} \sinh{\frac{\pi}{2x}}$ is decreasing on $(0,\infty)$. However, I'm not sure I get why this implies that the function $f(e^{-\pi/x})$ is decreasing. Could you explain a little? | |
| Jul 23, 2012 at 20:29 | comment | added | Malik Younsi | @juan +1, very nice! | |
| Jul 23, 2012 at 16:17 | history | edited | GH from MO | CC BY-SA 3.0 |
deleted 10 characters in body; deleted 19 characters in body
|
| Jul 23, 2012 at 16:14 | history | edited | juan | CC BY-SA 3.0 |
edited body; added 2 characters in body
|
| Jul 23, 2012 at 15:55 | comment | added | juan | @Malik Younsi I have added some ideas to finish the argument. I have proved that the function decrease for 0<q< 0.29... I have some problems with the TeX. May somebody edit this? | |
| Jul 23, 2012 at 15:53 | history | edited | juan | CC BY-SA 3.0 |
I add some ideas to finish the argument
|
| Jul 23, 2012 at 13:04 | history | edited | juan | CC BY-SA 3.0 |
deleted 946 characters in body
|
| Jul 23, 2012 at 8:20 | history | edited | juan | CC BY-SA 3.0 |
I complete my solution; added 14 characters in body
|
| Jul 23, 2012 at 2:17 | comment | added | Malik Younsi | Rewriting the function in terms of theta functions is a good idea, even though it's not clear to me that the last function in your answer is a decreasing function of $q$...! | |
| Jul 23, 2012 at 1:48 | comment | added | GH from MO | I see: $\vartheta_3(q)/\vartheta_4(q)$ is increasing as it has positive Taylor coefficients, hence $\vartheta_2^2(q)/\vartheta_3^2(q)=(1-\vartheta_4^4(q)/\vartheta_3^4(q))^{1/2}$ is also increasing. | |
| Jul 23, 2012 at 1:07 | comment | added | GH from MO | Why is the function of $q$ in (*) increasing on $(0,1)$? | |
| Jul 23, 2012 at 1:06 | history | edited | GH from MO | CC BY-SA 3.0 |
edited body
|
| Jul 22, 2012 at 20:41 | history | edited | Ryan Reich | CC BY-SA 3.0 |
define \Z
|
| Jul 22, 2012 at 19:13 | history | edited | juan | CC BY-SA 3.0 |
deleted 2 characters in body
|
| Jul 22, 2012 at 17:47 | comment | added | juan | I have changed the problem in other I think more tractable. But I see that still there is something missing. | |
| Jul 22, 2012 at 17:45 | history | edited | juan | CC BY-SA 3.0 |
added 29 characters in body; edited body; added 2 characters in body
|
| Jul 22, 2012 at 17:31 | comment | added | juan | I have a little confusion with increasing decreasing, but this must be the solution | |
| Jul 22, 2012 at 17:24 | history | answered | juan | CC BY-SA 3.0 |