On 5/31/2022 3:15 PM, mitchr...@gmail.com wrote:
> On Monday, May 30, 2022 at 9:54:09 PM UTC-7, zelos...@gmail.com wrote:
>> fredag 27 maj 2022 kl. 19:38:33 UTC+2 skrev mitchr...@gmail.com:
>>> On Friday, May 27, 2022 at 10:23:29 AM UTC-7, Ross A. Finlayson wrote:
>>>> On Thursday, May 26, 2022 at 2:17:50 PM UTC-7, sergi o wrote:
>>>>> On 5/26/2022 3:47 PM, mitchr...@gmail.com wrote:
>>>>>> On Thursday, May 26, 2022 at 1:37:42 PM UTC-7, Chris M. Thomasson wrote:
>>>>>>> On 5/26/2022 1:25 PM, Michael Moroney wrote:
>>>>>>>> On 5/25/2022 11:49 PM, Chris M. Thomasson wrote:
>>>>>>>>> On 5/25/2022 7:21 PM, Dan joyce wrote:
>>>>>>>>>> On Monday, May 23, 2022 at 7:11:22 PM UTC-4, Chris M. Thomasson wrote:
>>>>>>>>>>> On 5/23/2022 1:25 PM, Michael Moroney wrote:
>>>>>>>>>>>> Just because in theory an infinite number of steps is required doesn't
>>>>>>>>>>>> mean the limit cannot be reached. Consider Zeno's Paradox where
>>>>>>>>>>>> Achilles
>>>>>>>>>>>> races a tortoise with a head start. Each time Achilles reaches a point
>>>>>>>>>>>> where the tortoise was, the tortoise advances somewhat. When Achilles
>>>>>>>>>>>> reaches that point, the tortoise advances more. And so on for an
>>>>>>>>>>>> infinite number of steps. Yet Achilles catches up to the tortoise and
>>>>>>>>>>>> passes it and wins the race, despite taking an infinite number of
>>>>>>>>>>>> steps
>>>>>>>>>>>> to catch up to the tortoise.
>>>>>>>>>>> If Achilles strictly plays by the tortoises rules on a step-by-step
>>>>>>>>>>> basis:
>>>>>>>>
>>>>>>>> What are "the tortoise's rules"? The only rules are the tortoise gets a
>>>>>>>> head start and both it and Achilles run as fast as they can to the
>>>>>>>> finish line, and whoever does so first, wins.
>>>>>>>>>>>
>>>>>>>>>>> step 1: tortoise moves one meter; Achilles moves one meter. The
>>>>>>>>>>> tortoise
>>>>>>>>>>> is ahead because of the head start.
>>>>>>>>>>>
>>>>>>>>>>> step 2: tortoise moves one meter; Achilles moves one meter. The
>>>>>>>>>>> tortoise
>>>>>>>>>>> is still ahead because of the head start.
>>>>>>>>>>>
>>>>>>>>>>> on and on. The turtle will cross the finish line before Achilles.
>>>>>>>>>>
>>>>>>>>>> The turtle will never cross the finish line but will always be ahead
>>>>>>>>>> of Achilles.
>>>>>>>>>
>>>>>>>>> Yes. True. It gets infinitely closer and closer to the finish line.
>>>>>>>>
>>>>>>>> That is not Zeno's Paradox. The tortoise gets a head start, at point
>>>>>>>> A[1]. The race starts. When Achilles reaches A[1], the tortoise has
>>>>>>>> moved ahead somewhat, to what we call A[2]. When Achilles reaches A[2].
>>>>>>>> the tortoise has reached A[3], at A[3] the tortoise is at A[4] and so
>>>>>>>> forth.
>>>>>>>>
>>>>>>>> Since Achilles is faster than the tortoise, the distances A[1], A[2],
>>>>>>>> [A3], ... get smaller and smaller, since the time it takes Achilles to
>>>>>>>> run from the start to A[1] equals the time it takes the slower tortoise
>>>>>>>> to run from A[1] to A[2], and so on.
>>>>>>>>
>>>>>>>> The paradox is, no matter how big n gets, A[n] (Achilles' position) is
>>>>>>>> always behind A[n+1] (the tortoise's position), even as n approaches
>>>>>>>> infinity. So Achilles can never beat the tortoise, right? But, as long
>>>>>>>> as the head start isn't _too_ large, in real life, Achilles passes the
>>>>>>>> tortoise and wins, just as you'd expect. So what's wrong with this?
>>>>>>>>
>>>>>>>> As I said, just because there's an infinite limit, it doesn't mean the
>>>>>>>> limit is absolute. In this case, the total time passed also reaches a
>>>>>>>> limit (at n=infinity) but that time limit isn't infinite, so what
>>>>>>>> happens after the "limit" on time passes? As always, time marches on...
>>>>>>>> At that point Achilles passes the tortoise and remains ahead for the
>>>>>>>> rest of the race, and the infinite series no longer applies.
>>>>>>>>>
>>>>>>>>>
>>>>>>>>>>> Now, if Achilles tells the tortoise to f-off and just starts
>>>>>>>>>>> running, he
>>>>>>>>>>> will quickly pass the tortoise...
>>>>>>>>
>>>>>>>> In real life, yes, but in Zeno's Paradox, no.
>>>>>>>>>>>
>>>>>>>>>>> ;^)
>>>>>>>>>
>>>>>>>>
>>>>>>>> Extra credit: Given the speeds of Achilles S1 and the tortoise S2
>>>>>>>> (S1>S2), and the head start distance A1, how long does it take for
>>>>>>>> Achilles to pass the tortoise? :-)
>>>>>>>>
>>>>>>> I did some equations on this a while back:
>>>>>>>
>>>>>>> https://groups.google.com/g/sci.math/c/UKBgW2IOZkI/m/6tr-_qY-3DgJ
>>>>>>>
>>>>>>> Here are my comments:
>>>>>>>
>>>>>>> Iirc, scale was speed:
>>>>>>> ____________________________
>>>>>>> [...]
>>>>>>> Ahhhh, now this is a direct formula:
>>>>>>>
>>>>>>> n = iteration count
>>>>>>> d = distance
>>>>>>> s = scale
>>>>>>>
>>>>>>> r_[n] = (d / s^n) * (s^n - (s-1)^n)
>>>>>>>
>>>>>>>
>>>>>>> just might work for finding the total distance
>>>>>>> traveled at a given iteration count of the following
>>>>>>> iterated equation:
>>>>>>>
>>>>>>> r_[n+1] = r_[n] + (d - r_[n]) / s
>>>>>>>
>>>>>>>
>>>>>>>
>>>>>>> Here is the sequence for d = 10 and s = 4 using the
>>>>>>> iterative formula:
>>>>>>> __________________________________
>>>>>>> r_[0] = 0
>>>>>>> r_[1] = 0 + (10 - 0) / 4 = 2.5
>>>>>>> r_[2] = 2.5 + (10 - 2.5) / 4 = 4.375
>>>>>>> r_[3] = 4.375 + (10 - 4.375) / 4 = 5.78125
>>>>>>> r_[4] = 5.78125 + (10 - 5.78125) / 4 = 6.8359375
>>>>>>> __________________________________
>>>>>>>
>>>>>>>
>>>>>>> And here is the sequence for d = 10 and s = 4 using
>>>>>>> the direct formula:
>>>>>>> __________________________________
>>>>>>> r_[0] = 10 / 1 * 0 = 0
>>>>>>> r_[1] = 10 / 4 * 1 = 2.5
>>>>>>> r_[2] = 10 / 16 * 7 = 4.375
>>>>>>> r_[3] = 10 / 64 * 37 = 5.78125
>>>>>>> r_[4] = 10 / 256 * 175 = 6.8359375
>>>>>>> __________________________________
>>>>>>>
>>>>>>>
>>>>>>> As you can see, they are identical!
>>>>>>>
>>>>>>> Humm...
>>>>>>> ____________________________
>>>>>>>
>>>>>>>
>>>>>>> Here is another post:
>>>>>>>
>>>>>>> https://groups.google.com/g/sci.math/c/UKBgW2IOZkI/m/ysjxQWu9URMJ
>>>>>>> ____________________________
>>>>>>> I think I found a way to find the handicap of a
>>>>>>> runner in an infinite race on a finite track...
>>>>>>>
>>>>>>> How about something like:
>>>>>>>
>>>>>>>
>>>>>>> Let:
>>>>>>>
>>>>>>> d = total distance in track
>>>>>>> s = scale, which relates to speed
>>>>>>> n = integer iteration count, which relates to time
>>>>>>> r_h = a runners starting handicap
>>>>>>>
>>>>>>>
>>>>>>>
>>>>>>> Here is the iterative equation for finding the
>>>>>>> distance a runner is down the track that I posted
>>>>>>> up thread:
>>>>>>>
>>>>>>> r_[n + 1] = r_[n] + (d - r_[n]) / s
>>>>>>>
>>>>>>>
>>>>>>> The handicap of the runner is equal to r_[0]
>>>>>>> because n = 0 is the starting position of every
>>>>>>> runner.
>>>>>>>
>>>>>>> The goal is to find the handicap of a runner with
>>>>>>> a given distance, iteration count, total distance
>>>>>>> of the track, and a scale or speed. AFAICT, the
>>>>>>> following formula solves for the handicap of a
>>>>>>> runner using that information:
>>>>>>>
>>>>>>>
>>>>>>> r_h = ((s-1) / s)^(-n) * ( (d * (s-1)^n * s^(-n) - d + r)
>>>>>>>
>>>>>>>
>>>>>>>
>>>>>>> Here is output of a racer using the iterative equation
>>>>>>> with the following attributes:
>>>>>>>
>>>>>>> d = 10
>>>>>>> s = 4
>>>>>>> r_h = 6.8
>>>>>>> _______________________________________
>>>>>>> r_[0] = 6.8
>>>>>>> r_[1] = 6.8 + (10 - 6.8) / 4 = 7.6
>>>>>>> r_[2] = 7.6 + (10 - 7.6) / 4 = 8.2
>>>>>>> r_[3] = 8.2 + (10 - 8.2) / 4 = 8.65
>>>>>>> r_[4] = 8.65 + (10 - 8.65) / 4 = 8.9875
>>>>>>> _______________________________________
>>>>>>>
>>>>>>>
>>>>>>>
>>>>>>> As we can see this runner has a head start of 6.8 out
>>>>>>> of 10. Also, in the third frame, the runner r_[2] has
>>>>>>> traveled 8.2 out of a possible 10.0.
>>>>>>>
>>>>>>> Given that information alone, we can plug it all into
>>>>>>> the formula for finding the handicap, and get:
>>>>>>>
>>>>>>>
>>>>>>> r_h = ((4-1) / 4)^(-2) * ((10 * (4-1)^2 * 4^(-2) - 10 + 8.2) = 6.8
>>>>>>>
>>>>>>>
>>>>>>>
>>>>>>> Bingo! We now know that the handicap for the runner
>>>>>>> is 6.8 at n = 0 by information reaped in a later moment
>>>>>>> in time when n = 2... Three frames later.
>>>>>>>
>>>>>>>
>>>>>>> Is this Kosher?!?!
>>>>>>>
>>>>>>>
>>>>>>>
>>>>>>> :^o
>>>>>>>
>>>>>>> ____________________________
>>>>>>
>>>>>> If you add zero to .999 repeating you still get .999 repeating.
>>>>>> Add the infinitely small and you get 1 instead.
>>>>> .999 repeating = 1.000 repeating anyway
>>>> Mitch, for that ".999... is add infinitesimal", just first
>>>> have it that "1 minus infinitesimal, is, .999..., lesser".
>>> .999 is lesser than one by the infinitely small not zero.
>>>
>>> Mitchell Raemsch
>>>>
>>>> Then though it's always that "the .999..., lesser, is
>>>> only on its way to zero, least or none", because there
>>>> are two kinds of relations: related motion and lattice
>>>> relations, that the field defines lattice relations while
>>>> the infinitesimals is only part of a "range" or "course".
>>>>
>>>> I.e., the infinitesimal changes between 1.0 and 0.0,
>>>> going through each .aaa... as far as it could be measured,
>>>> are instead of that "this .333... times 3 = .999... = 1", that
>>>> this "1 minus .000...1" is writing out a notation, where
>>>> the ...1's "sum their differences, to zero", while the numbers,
>>>> "round up".
>>>>
>>>> So, when someone writes ".999, ..., repeating", is mostly
>>>> reflecting the notion that the notation after numbers introducing
>>>> the "..." or over-bar or the usual way of indicating the
>>>> repeating part for any rational number, basically works from
>>>> the field of course that _all_ and _only_ rational numbers,
>>>> end with a repeating terminus.
>>>>
>>>> Then there's only that
>>>>
>>>> 000... <- 0
>>>> 000...
>>>>
>>>> 011...
>>>> 011... <- 1/2
>>>> 100...
>>>>
>>>> 111...
>>>> 111... <- 1
>>>>
>>>> Notice the bounds are only at the ends,
>>>> and each column is half 1's and half 0's.
>>>>
>>>> It's easier to reduce the discussion to [0,1] instead of
>>>> involving all the real numbers.
>> There are no infinitesimals in real numbers. The real numbers are archimedian. I have told you this
>
> How do you know they are more real than the Calculus fundamental infinitesimal?
> .999 repeating is not the same quantity as the first integer.
> Add zero to .999 repeating and you get .999 repeating.
>
> Mitchell Raemsch

how do you know you actually have an infinitesimal ?