Timeline for Reflection of light from function graph
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 2, 2019 at 15:21 | comment | added | Hans | +1. Neglected to say that this is a wonderful solution. But do you agree with my minor criticism above, Robert? | |
| Oct 1, 2019 at 22:47 | comment | added | Hans | Your statement "In order for $x_n \to \infty$ with $\alpha_n$ increasing but staying below $\pi/2$, we would certainly need this to go to $0$." is not true in general, since we can have a sub-index-sequence $n_i$ where $x_{n_i+1}-x_{n_i}$ summing over $i$ to a finite positive number (fast) and $\dfrac{\alpha_{n_i+1}-\alpha_{n_i}}{x_{n_i+1} - x_{n_i}}=a$ for some positive $a$. However, we can conclude that $\liminf\limits_{n\to\infty}\dfrac{\alpha_{n+1}-\alpha_n}{x_{n+1} - x_n}=0$. It is impossible for the right hand side of the last equation which gives the desired result. Do you agree? | |
| Feb 23, 2015 at 6:10 | comment | added | user64494 | @ Robert Israel : I find that approach constructive. How about the general case? | |
| Feb 23, 2015 at 0:12 | comment | added | Joseph O'Rourke | Nice calculation! So, just to be clear, "that certainly won't happen" implies that $f(x)=e^{-x}$ has the property that $x_n$ stays bounded---Correct? | |
| Feb 22, 2015 at 23:13 | history | answered | Robert Israel | CC BY-SA 3.0 |