# Did calculus solve Zeno's paradoxes [duplicate]

Let's start form two versions of Dichotomy paradox

Version 1: Anyone can't walk though any distance without walk though half of it and so on.

solution 0 (?) ... which is false I think, that's not what the statement talking about

For any distance d we have

``````d/4 + d/8 + ... + d/(2^n)+... = d/2
``````

That we can walk though half of d

Therefore we can walk though any distance d

solution 1

Let f : T -> V be a function with domain time(T) and codomain velocity(V)

Suppose we started from 0

For any distance d that we want to walk though have:

Exists 0,c_0 in T if we take integral from 0 to c_0 on f(x) it equal to d

Alse exists 0,c_1 in T s.t. take integral from 0 to c_1 on f(x) have d/2^1

...(Keep doing this to n)

Alse exists 0,c_n in T s.t. take integral from 0 to c_n on f(x) have d/2^n

...(Keep doing this to infinite that as n->inf, c_n=d/2^n=0)

That will have, exists 0 in T s.t. take integral from 0 to 0 on f(x) have 0

In another word ... Zeno states that in order to move any distance d, what we have to do first is just walk though 0 infinite many times, but infinite many 0 distance add together will still be 0.

But is this just the wrong way to do limit.

Or is there something else I missed

Zeno's arrow paradox: ...(the mistake is similar to Version 1 of Dichotomy paradox)

Version 2: Even anyone can walk though half of any distance, he still have to walk though the half of the rest and so on.

solution 1 (Calculus--James Stewart pg.6 The sum of a series)

For any distance d we have

``````d/2 + d/4 + ... + d/(2^n)+... = d
``````

Suppose we can indeed walk though all of those sumed distances in a finite time

That implies that we can walk though any distance d

solution 2

Let f : T -> D be a function with domain time(T) and codomain distance(D)

For any people who walk with constant velocity v

If it's the case that range(f)=[0,d), for some d in R

It's either time being defined differently or space being defined differently

For example we define Zeno's time as the following:

Let c,t be some real number then

``````Zeno's time(T) = t iff Normal time = (L-(1/c)^t)/v where c > 1
``````

Have f(t) = (L-(1/c)^t)

That with finite Zeno's time no one can reach d.

Similarly we can also construct a Zeno's space s.t. range(f)=[0,d)

Basicly, this solution want to say: paradoxes caused by using different definitions.

Achilles and the tortoise: ... (This is similar to second version of Dichotomy paradox)

In both versions, I think solution 1s are fine, are they correct?

And please tell me if you know where I can find some formal proof in calculus which shows the mistakes of Zeno's paradoxes

Any help or suggestion would be appreciated.

• No, calculus only gives a technical workaround, see Papa-Grimaldi, Why Mathematical Solutions of Zeno’s Paradoxes Miss The Point. Sep 6 '19 at 4:32
• try posting this to the mathematics stack exchange. Sep 6 '19 at 6:15
• The basic calculus solution to it is the theorem that an infinite series of terms may have a finite sum, provided suitable properties of the terms of the series. See Convergent series. Sep 6 '19 at 6:28
• This has many good answers here: math.stackexchange.com/questions/813880/…
– jhch
Sep 6 '19 at 15:52
• @Conifold the paper relies on an idea of "motion" which is completely at odds with physics. Mathematical solutions miss the point Zeno was trying to make, but that doesn't mean Zeno was correct.
– Era
Sep 6 '19 at 17:18