Math on the Run

A link to the article: http://arxiv.org/pdf/1507.00871v1.pdf

Very cool. Thanks.

super cool. I had always just assumed the intermediate value theorem proof worked all the time -- there is always some mile within the race that you run at average pace. Turns out it only works if the race is an exactly integer number of miles (true for 3 miles, not necessarily for 5k).

Yes, I think it is really amazing that if you run a 5k at 6:00 average pace, there is not necessarily any mile at exactly 6:00 pace!

By the way, I became interested in this problem because of this LetsRun thread about the fact that Kenenisa Bekele has not run a sub-4:00 mile:

http://www.letsrun.com/forum/flat_read.php?thread=4464001

Someone pointed out that he ran 4:50 for 2,000m and then someone else carefully constructed a speed function for him where he would run no mile in sub-4:00. I was astonished that this was possible, and decided to look into it more closely. Hence, the paper!

- Diana

Neat!

However, for Dara's assertion that, during a 4 mile run at 15:00 a mile, the Intermediate Value Theorem holds true - Is this because If you divide the total distance by the desired split distance, it is an even integer?

So the theorem would hold that Komen ran at least one mile segment at 3:59.2 during his 2-mile WR (average pace), because it can be broken into 2 mile segments. But you could not say the relative thing about his 3K WR (That somewhere in there is a 3:56.35 mile), because it is not evenly divisible?

Suppose that Keteiny ran really fast for the first and last 9.1km and really slow for the middle 2.9 km, then her total time for the race would still be 65.50 but her time for each 12km subinterval would still be much slower than Huddle's.

Mindblowingly amazing stuff eh?

Lenny Leonard wrote:

However, for Dara's assertion that, during a 4 mile run at 15:00 a mile, the Intermediate Value Theorem holds true - Is this because If you divide the total distance by the desired split distance, it is an even integer?

Yes! Exactly: you can only apply the IVT argument when the race distance is an integer multiple of the split distance.

So the theorem would hold that Komen ran at least one mile segment at 3:59.2 during his 2-mile WR (average pace), because it can be broken into 2 mile segments. But you could not say the relative thing about his 3K WR (That somewhere in there is a 3:56.35 mile), because it is not evenly divisible?

Yes! He could have run the first and last 0.86 miles fast, and the middle 0.14 miles very slow, and his time for 3K (1.86 mi) would be the same, but there would be no sub-4 mile in there because they are all too slow. You can make this construction because all of the 1-mile subintervals overlap the slow part.

A couple of silly questions that I have been pondering:

All of the examples that I have seen where no segment pace is as fast or faster than the average pace are symmetrical in nature and have exactly 2 types of partitions. Are there any such distributions that are asymmetric and/or have varying partitions?

If Keitany's HM was EXACTLY duplicated and put back to back (or in an infinite loop), then would there have to be a 12 km segment faster than Huddle's (spanning the 2 runs, if necessary)?

Maths are hard wrote:

All of the examples that I have seen where no segment pace is as fast or faster than the average pace are symmetrical in nature and have exactly 2 types of partitions. Are there any such distributions that are asymmetric and/or have varying partitions?

Sure. We give an example where Keitany runs 27:00 for the first and last 9.1km and 11:50 for the middle 2.9km, so the whole 21.1km is 65:50 as required, but every 12km subset is 38:50, slower than Huddle's 37:49. But within that 9.1km, Keitany could have wildly erratic paces, as long as her total time comes out to 27:00; her pace need not be symmetric.

The idea behind this example is that she runs fast at the beginning and end, and slow in the middle, and every 12km includes that slow middle part. Any example has to be something like this, so it has to be "symmetric" in that sense. You could change the 9.1 and 2.9 a little bit to break the symmetry. (But then it wouldn't be as elegant of an example!)

Awesome, thanks for sharing. I am thinking if her run was 'modular' (e.g. could take the last 4k and the first 8k) then we could always find a 12k section of at least average pace? But if the run was extended at the same average, not necessarily. That is pretty weird.

Cool stuff, and I like that you used Keitany and Huddle for this example!

This type of discussion is one of the reasons some parts of Letsrun are still worth it :)

Also, where is the elegant proof on the Bekele thread? All I see is a bunch of the standard Letsrun name-calling haha.

oyrtikhjgfg wrote:

Also, where is the elegant proof on the Bekele thread? All I see is a bunch of the standard Letsrun name-calling haha.

Ah, I linked to the wrong thread - whoops! It was actually on the Ritzenhein sub-4:00 thread (though both contain lively mathematical discussion!). Here is the post:

and here's how the discussion went down:

otter wrote: Bekele ran a 7:26 3000. That averages under a 4 minute mile for the entire race.

liouifsdj wrote: Yes, but were there 1760 yards that were consecutively run in under 4 minutes? Otherwise the answer is no, Bekele has never gone sub-4.

goalposts wrote: How would someone run 3000m in 7:26 without ever breaking sub 4-mile pace for a mile portion of the race? Explain that.

cookieswithmandms: For an easy example, imagine he ran 1 second for the first 200m, 3:00 for the next 3 laps, 59 seconds for the middle 200m, 3:00 for the next 3 laps, then 1 second for the final 200m. No matter how you slice that, no 1609m is faster than 4:00. That gives a 7:01 3k. Adding time would not create any faster intermediate splits. Realistically, say 25, 60, 60, 60, 36, 60, 60, 60, 25, run evenly for each split.

This result has applied to me in real life! My 3k pr is 8:43, which I ran back in college. At the time my mile pr was 4:43, slower than the average pace of my 3k (4:41/mile). Moreover, the overall pattern of the splits - went out hard, slower in the middle, closed hard - makes it possible that every mile-long subset of the race was slower than the race average. It's very interesting that I can be sure I broke my 1500m pr during this 3k based on IVT, but that I can't be sure about my mile!

Two more comments. Well, three - first I want to join the others above in saying how super cool this is. Dr. Davis, thank you so much not only for writing that neat paper, but also talking with us about it on letsrun!

Second, if you are having trouble imagining what the speed function would have to look like when the split distance is less than half of the race distance, you should definitely check out the math paper (http://arxiv.org/pdf/1507.00871v1.pdf). Specifically Figure 3 at the top of page 5, and the associated text on the preceding pages if necessary. The key is 'decreasing sawtooth with 'period' less than the split length'...but a picture is worth a thousand words.

This relates to another thought I had about this - IVT says that your instantaneous pace equals your race average pace at some point. But in the limit of shorter and shorter split lengths, don't those finite average paces become instantaneous paces, so that 'in the limit' IVT applies once again? I understand that as the intervals get short, the pace function becomes arbitrarily 'spiky'. But I'm curious about how one would talk rigorously about this limit that I'm trying to gesture towards, and how IVT arguments would behave.

And finally (sorry, can't count!) I am curious about other cool math/running things that people have stumbled across on the internetz. Here is my contribution:

http://barnyard.syr.edu/articles/articles.shtml

This is a page with links to some great short articles about running, many of which are math related. Fans of Dr. Davis's paper will probably like some of these articles.

Davis's counterexample fails not because 12k isn't a divisor of HM, but merely because it's greater than half a half marathon.

first let's find the rates per kilometer: 65:50/HM = 3950s/21.1k= 187.2 s/k

9.1k segment pace = 27x60s/9.1k = 178.0 s/k

middle 2.9k segment pace = 710s/2.9k = 244.8 s/k

Let L less than 21.1 be the length of the chosen segment, and s the portion of L done at the slow segment pace. The average pace over L is given by the function

p (s) = (244.8s + 178.0(L-s))/L

= 244.8s/L + 178.0 - 178s/L

= 66.8s/L + 178

p = 187.2 where 66.8s/L + 178 = 187.2

66.8s/L = 9.2

s/L = .138

A 12km chunk must always include the entire slow segment in which case s/L = 2.9/9.1 = .318 > .138. But take a 6km chunk - still not an integer segment of a HM, but all you have to do is pick

s = .828 km

L - s = 5.172km

to check:

(.828 x 244.8 + 178 x 5.172)/6 = 187.2 s/k

more generally, the example fails only where L is so great that

(L-9.1)/L > .138

L - .138L > 9.1

.862L > 9.1

L > 10.55k = half a half mary

Only a segment so long that it must include more than 1.45k of the slow section cannot be at 187.2 s/k. I suspect that it is true in general that any segment less than half the distance will work however you slice it up. In which case it is sufficient - but not necessary - that the segment be an integer divisor in length. 1/3 the distance, 1/4 the distance, all are less than half the distance, but so is race distance x 1/e. I will think about that some other time if someone doesn't find and post a proof first.

and only now do I notice the link to the actual paper... it seemed at first like they were relying on only the example in the SA article.

But I would still like to see an example where non-divisor segments less than half the distance fail.

The same basic premise was found here in 2004, you should consider citing this paper:

http://condor.depaul.edu/mash/gash1609.pdf

This is linked to on the barnyard site as well.

Bad Wigins wrote:

But I would still like to see an example where non-divisor segments less than half the distance fail.

Here is a realistic one you can do in your turkey trot 5k or your track:

1km 4:00

333 1/3 m 1:40

1km 4:00

333 1/3 m 1:40

1km 4:00

333 1/3 m 1:40

1km 4:00

total distance 5km, pace 21/5 = 4.2 minutes/k = 4:12/k

pace per any 1.333km segment = 5:40/1.3333 = 17/3 x 3/4 = 17/4 = 4.25 minutes = 4:15/k

this wouldn't work with two intervals that added up to an integer because you'd have to include the same number of each for an integer total distance. It's the extra 1k that throws the overall average off of the 1.333k average.