Lonely Runner Conjecture- II

The Lonely Runner conjecture states that each runner is lonely at some point in time. Let the speeds of the runners be $\{a_1,a_2,\dots,a_k\}$ , and let us prove “loneliness” for the runner with speed $a_e$ .

As we know, the distance between $a_i$ and $a_e$ is given by the formula $|a_e-a_i|t$ for $t\leq \frac{1}{2|a_e-a_i|}$ and $1-|a_e-a_i|t$ for $t\geq \frac{1}{2|a_e-a_i|}$ , where $t$ stands for time. Also, $speed\times time=distance$ .

$|a_e-a_i|t\geq \frac{1}{k}$ therefore $t\geq \frac{1}{k|a_e-a_i|}$ when $k>2$ . Also, $1-|a_e-a_i|t\geq \frac{1}{k}$ therefore $t\leq \left (1-\frac{1}{k}\right )\left(\frac{1}{|a_e-a_i|}\right)$ . Finally, observing that this is a periodic process with a period of $\frac{1}{|a_e-a_i|}$ , we come upon the conclusion that $a_e$ is lonely as compared to $a_i$ when $t\in \left[\frac{1}{k|a_e-a_i|}+\frac{n_i}{|a_e-a_i|}, \left (1-\frac{1}{k}\right )\left(\frac{1}{|a_e-a_i|}\right)\frac{n_i}{|a_e-a_i|}\right]$ , where $n_i\in\Bbb{N}$ .

Now we prove the loneliess of $a_e$ with respect to every other runner. This is equivalent to the statement

$\bigcap_{i\in A} \left[\frac{1}{k|a_e-a_i|}+\frac{n_i}{|a_e-a_i|}, \left (1-\frac{1}{k}\right )\left(\frac{1}{|a_e-a_i|}\right)\frac{n_i}{|a_e-a_i|}\right]$ , where $A=\{1,2,3\dots,k\}\setminus\{e\}$ . Also, note that all $n_i$ can take different integral values.

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Hello! My name is Ayush Khaitan, and I'm a graduate student in Mathematics. I am always excited about talking to people about their research. Please please set up a meeting with me if you feel that I might have an interesting perspective to offer- https://calendly.com/ayushkhaitan/meeting-with-ayush View more posts

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