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Catch-Up

Zeno's paradox

Long ago

Mathematics should then be studied, that it puts the mind right.

Mikhail Lomonosov

Mathematics gives the most reliable rules: the one who follows them is not in danger for the deception of the senses.

Leonard Euler

Long ago

Zeno's long-known paradox about how swift-footed Achilles was catching up a tortoise. He was catching up, catching up and ...
...until now, many people cannot solve the task that was formulated by philosopher Zeno of Elea (c. 490–430 BC).

Let's look at the "problem":

One day, a leisurely tortoise got a journey. Sometime later, when the tortoise had traveled some distance, swift-footed Achilles ran after it. Now the discussion is offered:

1. While Achilles runs the distance already covered by the tortoise, some time will pass
2. During this time the tortoise will cover some another distance.

We get the same situation as at the very beginning. And the reasoning (p. 1, 2) can be repeated an infinite number of times. So, we conclude: Achilles will never catch up with the tortoise.

Someone start to argue: everyone understands that this is nonsense. Achilles will catch up and overrun. It is enough to take the difference in speeds and calculate, a task for schoolchildren ...
That's right. But Zenon's task was, of course, not about "calculating".

The question is: how to resolve the contradiction (the conclusion that Achilles will never catch up with the tortoise) remaining within the model proposed by Zeno. Or to show that Zeno's model is incorrect.

Look in details

Sure, when Achilles is just starting to run, the tortoise is already somewhere ahead. And, following to Zeno, we get an infinitely repeating set of positions, in each of which Achilles is (at least slightly) behind the tortoise.

Everything seems to be normal, no tricks.

What about the conclusion? What does it mean: "Achilles will never catch up with the tortoise", or: "Achilles is always behind the tortoise"?

It means that there is no such moment in time when Achilles will catch up with the tortoise. Or - that there is no such place (point on the way) where Achilles will catch up with the tortoise.

This is clearly not true! And these conclusions do not follow from the reasoning at all!

Indeed, Achilles at every stage is following the tortoise. And number of such situations is infinite. Reasoning is only about this! However, an infinite number of situations does not mean that the elapsed time (or the length passed by Achilles) to catch up with the tortoise are infinite!

In each situation (when Achilles reaches to the place where the tortoise was before), the distance between the tortoise and Achilles becomes smaller. The segment of the path between them gradually "shrinks" to a point. Remember that any segment consists of an infinite number of points. And this surprises no one. In the case of Achilles and the tortoise, an infinite number of decreasing segments add up to a segment of finite length! That is, Achilles will catch up with the tortoise after running a finite distance. And the fact that this distance can be represented as an infinite sum of infinitely decreasing segments is not a paradox, it is easy! It's just a model choice.

The same can be said about time. The time spent by Achilles on each (decreasing) segment of the path becomes less and less, shrinking to zero. So, that infinite sum of decreasing "spans" of time forms a finite value. And here the solution to the paradox lies.

As soon as the talk turns to distance (or time), reasoning is no longer enough, measurements become necessary.

Concrete

And now let's calculate (remaining within the framework of the model proposed by Zeno). Let's sum up the segments of the path that Achilles runs while catching up with the tortoise.

If at the very beginning the distance to the tortoise is L0, Achilles will run it (at a speed of VA) in the time

During this time, the tortoise will cover (at a speed VT) the distance

Achilles will run this distance in time

During this time, the tortoise will cover the distance

and so on...

Let's write down the distances:
Starting distance is L0 , then

That is, each next distance is equal to the previous one, multiplied by VT / VА.

If, for example, the speed of Achilles is twice the speed of the tortoise, then

and then the distances from the tortoise to Achilles are equal:

Suppose the initial distance from Achilles to the tortoise is 100m. The next segment of the path L1 will be 50 meters, then - 25 meters, ... the tenth segment is less than 10 cm, and the twentieth is less than 0.1 mm - about the thickness of a hair.

(It is so if the speed of Achilles is twice the speed of the turtle. If Achilles runs faster, then the segments of the path will decrease faster.)

Now let's write down the sum of all the segments that Achilles runs, catching up with the tortoise:

The sum in brackets has an infinite number of terms, but this sum is a finite number. And it's not hard to find it. Let's denote

Take a closer look at the terms, recall, and note that each next is half the previous one. Now rewrite:

What is written in brackets is a sum of infinite number of terms. It starts with 1 and each next term is half the previous one. But we denoted such an infinite sum (the one that is now in brackets) with the letter S. Let's substitute S for the sum in brackets and get:

Now everything is quite simple:

And finally: S = 2.

So the infinite sum is

This means that Achilles will catch up with the tortoise when he runs a distance twice as large as L0 (what separated him from the tortoise at the beginning).

Of course, this is true if the speed of Achilles is twice that of a tortoise. Fr another ratio of the speeds of Achilles and the tortoise the distance to the meeting will be different. More on this below.

Similar calculations can be done about the time of movement.

By the way

There is a wonderful illustration of how an infinite sum can be equal to a finite number. Let's draw several segments one after another. Every next is twice less than previous.

Then let's draw square (of corresponding size) on each segment.

Now let's draw a straight line (and this will be a straight line!) through the upper right corners of the squares.

Look carefully and think: this line will intersect the horizontal base of all the segments just there (at that point) where the sum of an infinite number of infinitely decreasing segments converges (!)

Of course, if the segments (and squares) do not decrease by half at each step (Ln = 1/2 . Ln-1), but, for example, by a factor of three (Ln = 1/3 . Ln-1), then the inclined "enveloping" straight line will go steeper and cross the horizontal closer to the beginning. Depending on the ratio of the lengths of the segments, the slope of the "envelope" will be less or more, and the point of intersection with the horizontal will move to the left or to the right (the sum of the lengths of all segments will be less or more). The main thing is that the ratio of the length of each next segment to the length of the previous one should be less than 1.

Common view

Returning to the distance that Achilles ran while catching up with the turtle. If the ratio of speeds (tortoise and Achilles) denote

, then the sum of all segments of the path is:

Let's denote

then

Thus

and finally

This is the expression for the sum of a convergent geometric progression.

So, if the ratio of the speed of the turtle to the speed of Achilles is equal to q, then Achilles will catch up with the turtle by running the distance

And the time it takes Achilles to catch up with the tortoise

Which, generally speaking, is understandable without Zenon's tricks 😉.

Naturally, all arguments are valid for cases where q < 1.
In other cases (q = 1 or q > 1) Achilles runs no faster than a tortoise, which means that Achilles will never catch up with a tortoise. 😉

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