If Achilles runs D=2048 meters in T=491.52 seconds, then his average pace for the full 2048 meters is 491.52/2048 = .24 seconds/meter. Let P(d) denote Achilles's *average* pace for the run between 0 and distance d, so P(2048)=P(D)=0.24 s/m. Thus Achilles's average pace for the whole 2048 meters is P(2048)=P(D), and his average pace for the first half is P(1024)=P(D/2). But Achilles's average pace for the first half is 5 secs/500m faster than his average pace for the second half, so his average pace for the first half is 2.5 secs/500m faster than his average pace for the full 2048 meters. i.e., P(D/2)=P(D)-2.5/500=P(D)-0.005. But this is true for any distance 0<d<D=2048, so P(d/2)=P(d)-2.5/500=P(d)-0.005. Let S=2.5/500=0.005, so P(d/2)=P(d)-S. Then let k=log2(d), so that d=2^k. Then P(d/2)=P(d)-S = P(2^(k-1))=P(2^k)-S. Now let Q(k)=P(2^k), so that Q(k-1)=Q(k)-S. Using the methods of Wikipedia's difference equations, https://en.wikipedia.org/wiki/Linear_difference_equation we find that Q(k)=2*Q(k-1)-Q(k-2) so that Q(k)=c1+c2*k. Inserting this into Q(k-1)=Q(k)-S, we find that Q(k)=c1+S*k Since k=log2(d), we have Q(log2(d))=c1+S*log2(d), i.e., Q(log2(d))=P(2^log2(d))=P(d)=c1+S*log2(d) So, indeed, c1 is an arbitrary constant (the pace of the turtle) which we could set to 0 for Galilean relativity. Thus we have the form of the solution; we need to calculate constant c1 for our particular example of Achilles's run. But P(2048)=0.24 s/m = c1+S*log2(2048) = c1+(2.5/500)*11, so c1 = 0.185. Now since S=2.5/500=0.005, P(d) = S*log2(d)+0.185 = 0.005*log2(d)+0.185. Checking, P(2048) = 0.005*11+0.185 = 0.24; check! P(1024) = 0.005*10+0.185 = 0.235 = P(2048) - 0.005; check! Note that we have assiduously avoided the issue of what exactly happens at d=0! We have also assiduously avoided discussing Achilles's *speed*; by focusing solely on his *pace*, we have avoided a number of land mines. Q: How to calculate Achilles's *instantaneous pace* from his *average pace* ? At 12:37 PM 5/26/2017, Henry Baker wrote:

[A modern Zeno problem.]

Achilles runs a total distance D=2^k meters in a total time T seconds. For example, Achilles might run 2048 meters in 491.52 seconds (= 8:11.52). Achilles wasn't terribly fast by today's world record standards. :-)

Achilles's average *pace* (=1/speed) for the first half (=D/2) is 5 seconds/500m (= 0.01 seconds/meter) *faster* than his average pace for the second half.

Achilles's average pace for the first quarter (=D/4) is 0.01 seconds/meter faster than his average pace for the second quarter.

and so forth.

Achilles's motion is a smooth analytic function of time and/or distance, excepting perhaps the origin.

What is the functional form of his motion -- e.g., distance as a function of time, time as a function of distance, speed as a function of time, speed as a function of distance, pace as a function of time, pace as a function of distance -- any one of these forms should be useful.

If I haven't made a mistake, Achilles's absolute *average pace* over the entire run shouldn't matter; clearly if the average pace for the first half differs by 5 seconds/500m from the average pace for the second half, then we could simply set the average pace for the entire run to zero (i.e., Newton relativity), so he may end up going backwards part of the time (i.e., relative to a tortoise moving at constant pace = Achilles's average pace).

But this still leaves the problem of the functional form of his speed.