On 5/27/2022 12:38 PM, mitchr...@gmail.com wrote:
> 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

if you add an infinitesimal to 1 you get 1.000... repeating

>>
>> 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.