Zeno's paradox and pi.

[djoyce099]

I know the argument of 1 =.999999999999... and
the counter-argument of 1< .9999999999... as JG and
a few others propose.

Putting it into a pure mathematics prospective using
pi as an example and employing Zeno's paradox.
Does this make the case for JG's argument?

We have two racing elements (A) and (B).
(B) runs twice as fast as (A) so (A) is given a head
start of half the distance to the finish line.
In this case the course distance = pi kilometers.
(A) starts at pi/2 = 1.57079632679489661923... kilometers.
(B) starts at pi = 3.14159265358979323846... kilometers.

Now in Zeno's paradox --
As each leg of the race the distance of 1/2 the remaining
distance too pi is assigned a new value for (A) where (B)
is assigned the previous value of (A) because (B) runs
twice as fast as (A) and is closing the gap at each leg
of the race, but never eliminating that gap no matter
how small that gap becomes.

There is also a time element for each leg that is also halved at
each leg of the race. Otherwise, the race would end in a tie.

Length of racecourse = 3.14159265358979323846... kilometers.
Starting position for (A) = 1.57079632679489661923... kilometers.
(A) having half the distance advantage.
Starting position for (B) = 3.14159265358979323846... kilometers.
(B) runs twice as fast as (A) so is starting the full distance of
pi kilometers.
The first leg of the race ---
(A) is @ position on the course = 2.35619449019234492884... kilometers.
(B) is @ position on the course = 1.57079632679489661923... kilometers.
The leg(2) of the race ---
(A) is @ position on the course = 2.74889357189106908365... kilometers.
(B) is @ position on the course = 2.35619449019234492884... kilometers.
..
The leg(6) of the race ---
(A) is @ position on the course = 3.11704896098362297878...kilometers.
Where the first decimal expansion of (1) in pi shows up.
(B) is @ position on the course = 3.09250526837745271911...kilometers.

And so on as each integer of PI in (A) increases to its true value
for that digit expansion but always < pi for the next digit.
In the case of trailing (9's)where the course is 1 kilometer long, it is
pi kilometers long.
Will the one-kilometer course for (A) and (B) have an average converging
rate as the pi kilometer course?


Another interesting aspect of trailing 9's after the decimal point is --

2^4-1 =15
1/(1/15 +1) =0.9375
2^7-1 =127
1/(1/127 +1) =0.9921875
2^10-1 =1023
1/(1/1023 +1) =0.9990234375
2^14-1 =16383
1/(1/16383 +1) =0.99993896484375
2^17-1 =131071
1/(1/131071 +1) =0.99999237060546875
2^20-1 =1048575
1/(1/1048575 +1) =0.99999904632568359375
2^24-1 =16777215
1/(1/16777215 +1) =0.999999940395355224609375
2^27-1 =134217727
1/(1/134217727 +1) =0.999999992549419403076171875
Adding a new (9) at each point.
..
2^n-1 =n1
1/(1/n1 +1)=.99999999999999999999999999999999...-->00
Where (n) is not countable

2^ n where n= 4,7,10,14,17,20,24,27,30,34..(-1 for each).
gap = 3,3,4,3,3,4,3,3,4,3,3,4...
Will this gap pattern stay the same --->00?

Dan

[Mitch Raemsch]

You can't reach the end of the transcendental of Pi.

[djoyce099]

> > Where (n1) is not countable

> >
> > 2^ n where n= 4,7,10,14,17,20,24,27,30,34..(-1 for each).
> > gap = 3,3,4,3,3,4,3,3,4,3,3,4...
> > Will this gap pattern stay the same --->00?
> >
> > Dan
>
> You can't reach the end of the transcendental of Pi.

That is the point, where (n1)is not countable.

[Mike Terry]

In both cases the convergence rate is exponential (considered as a
function of the leg number), the remaining distance halving at each
successive leg.

There is an explanation behind the apparent pattern - it is due to 2^10
= 1024, which above, but not far off 10^3.

Note that your n'th entry is just (1 - 2^-n). So the 2^10 entry causes
the sequence to break (just) above (1 - 10^-3). Then the next two
entries follow the observed pattern, which is down to the closeness of
the 2^10 to 10^3.

Then the 2^20 entry = (2^10)^2 is again just above (but close to) 10^6
and so on. But notice that the relative closeness with 2^20 is less
than with 2^10. It's still close enough to maintain the pattern for the
next few entries, but 2^30 will have an even larger relative
discrepancy, and continuing the sequence the discrepancies grow and the
pattern breaks down. It should be easy to program this and find where
the pattern breaks, but I haven't done this. (I think the "why" is more
interesting than the "when, exactly".)

In the long run, on average we expect it would take log_2 (10) steps to
get another 9 digit, and log_2 (10) = ln(10)/ln(2) = 3.321928.... The
average for your observed pattern would be exactly 10/3 = 3.3333333...
which is too high. (i.e. the 4's in your pattern are slightly too
frequent for the long run)

Regards,
Mike.

[Mitch Raemsch]

Are you half way to Pi?

[Dan Christensen]

On Saturday, March 7, 2020 at 9:05:13 PM UTC-5, djoyce099 wrote:

>
> Now in Zeno's paradox --
> As each leg of the race the distance of 1/2 the remaining
> distance too pi is assigned a new value for (A) where (B)
> is assigned the previous value of (A) because (B) runs
> twice as fast as (A) and is closing the gap at each leg
> of the race, but never eliminating that gap no matter
> how small that gap becomes.
>

IMHO decomposing the race between a slower runner (The Tortoise) and a faster runner (Achilles) into ever decreasing intervals like this turned out to be an intellectual dead end that held back European science for perhaps 2 thousand years. The ancient Greek philosophers of Zeno's day had no notion or any interest in how to measure the speed of an object.

It took the Italian scientist Galileo, some 2 thousand years later, to define the speed of an object with his simple formula s=d/t. Today, even school children can use it to calculate precisely when and where Achilles would over take The Tortoise (assuming constant speeds). There was never really any paradox, just stubborn ignorance.


Another Dan

Download my DC Proof 2.0 freeware at http://www.dcproof.com
Visit my Math Blog at http://www.dcproof.wordpress.com

[Chris M. Thomasson]

pi/2

[Chris M. Thomasson]

Generally, we define a nice precision level for pi, suitable for the
needs of the task at hand.

[FromTheRafters]

djoyce099 pretended :

> I know the argument of 1 =.999999999999... and
> the counter-argument of 1< .9999999999... as JG and
> a few others propose.

Nobody thinks that one is less than .999... .

> Putting it into a pure mathematics prospective using
> pi as an example and employing Zeno's paradox.
> Does this make the case for JG's argument?

No, because of the construction of decimal representations.

[...]

[Eram semper recta]

On Saturday, 7 March 2020 21:05:13 UTC-5, djoyce099 wrote:
> I know the argument of 1 =.999999999999... and
> the counter-argument of 1< .9999999999... as JG and
> a few others propose.

I have never said that 1 < .999...

You cannot compare a number (1) to a sequence (0.999...) because they are not the same thing.

I reject Euler's definition S (0.999...) = Lim S (1).

>
> Putting it into a pure mathematics prospective using
> pi as an example and employing Zeno's paradox.

There is no paradox in Zeno's story. If you try to traverse a unit distance in the following way: 1/2 + 1/4 + 1/8 + ..., then you will NEVER complete the traversal.

There is NOTHING paradoxical about that claim - it is pure logic. Whoever called Zeno's philosophical discussion (because that's all it was!) a paradox was an absolute moron who is as ignorant as most of you today. The Greeks liked to debate and philosophise with one another - each trying to outdo the other.

One can traverse a unit distance because one NEVER traverses that distance in the way claimed by Zeno.

1/2 + 1/2
1/4 + 1/4 + 1/2
3/4 + 1/4

All these are FINITE traversals and there are many ways to complete the traversal - all of them FINITE.

The fact that one passes all the invisible markers or points or road signs DOES NOT mean that those distances are all added up as per Zeno's anecdote.

> Does this make the case for JG's argument?

No. LOGIC and common sense only.

[Mostowski Collapse]

.999... is not a sequence, its a limit of
a sequence. Whats wrong with you?

[Eram semper recta]

On Sunday, March 8, 2020 at 9:55:37 AM UTC-4, Jan Burse aka Mostowski Collapse wrote:
> .999... is not <blirp, derp, burp...>

SHUT UP MORON.

[Mostowski Collapse]

More nonsense by John Gabbermonkey.

> You cannot compare a number (1) to a sequence
> (0.999...) because they are not the same thing.

- JG, 08.03.2020

He still doesn't get what 0.999... means.
He is sure the greatest, the greatest dumbo.

[Me]

Closely followed by Mückenheim:

"0.999... is an infinite sequence of terms of the form 9/10^n added and written in an abbreviated form by the usual notation where the 10^n are indicated by the position of the digit." (WM)

[djoyce099]

Thanks, Mike.

Just by noticing the slight increase over each 4th gap the pattern,
as you stated it, will break.

Dan

[djoyce099]

The pattern gap --3,3,4,3,3,4,3,3,4,3,3,4... brakes @ 2^784-1 where
2^780-1 has 234 trailing 9's and 2^784-1 has 236 trailing 9's

[Mitch Raemsch]

Nobody could ever count to infinity...

[Dan Christensen]

On Sunday, March 8, 2020 at 8:23:28 AM UTC-4, Eram semper recta wrote:
> On Saturday, 7 March 2020 21:05:13 UTC-5, djoyce099 wrote:
> > I know the argument of 1 =.999999999999... and
> > the counter-argument of 1< .9999999999... as JG and
> > a few others propose.
>
> I have never said that 1 < .999...
>
> You cannot compare a number (1) to a sequence (0.999...) because they are not the same thing.
>
> I reject Euler's definition S (0.999...) = Lim S (1).
>
> >
> > Putting it into a pure mathematics prospective using
> > pi as an example and employing Zeno's paradox.
>
> There is no paradox in Zeno's story. If you try to traverse a unit distance in the following way: 1/2 + 1/4 + 1/8 + ..., then you will NEVER complete the traversal.
>
> There is NOTHING paradoxical about that claim - it is pure logic. Whoever called Zeno's philosophical discussion (because that's all it was!) a paradox was an absolute moron who is as ignorant as most of you today. The Greeks liked to debate and philosophise with one another - each trying to outdo the other.
>

Their "philosophizing" may have held back the development of European science by as much as 2 thousand years. It took the Italian Galileo, to sort out this silliness with his simple s=d/t some 2 thousand years later.

[Chris M. Thomasson]

On 3/8/2020 12:47 AM, FromTheRafters wrote:
> djoyce099 pretended :
>> I know the argument of 1 =.999999999999... and the counter-argument of
>> 1< .9999999999... as JG and
>> a few others propose.
>
> Nobody thinks that one is less than .999...   .

There are two ways to think about it. One is the limit, .999... = 1 for
sure, is one... The other is building it up step-by-step. Well, in this
case, there is no iteration where 1 equals .999... unless we use an
epsilon to say, close enough, and halt iteration.

[Chris M. Thomasson]

On 3/8/2020 12:47 AM, FromTheRafters wrote:

> djoyce099 pretended :
>> I know the argument of 1 =.999999999999... and the counter-argument of
>> 1< .9999999999... as JG and
>> a few others propose.
>
> Nobody thinks that one is less than .999...

No. Using step-by-step iterative method .999... is always less than one.

[Chris M. Thomasson]

On 3/8/2020 5:23 AM, Eram semper recta wrote:
> On Saturday, 7 March 2020 21:05:13 UTC-5, djoyce099 wrote:
>> I know the argument of 1 =.999999999999... and
>> the counter-argument of 1< .9999999999... as JG and
>> a few others propose.
>
> I have never said that 1 < .999...

[...]

Using the iterative step-by-step method, .999... is always less than one.

[FromTheRafters]

It happens that Chris M. Thomasson formulated :

> On 3/8/2020 12:47 AM, FromTheRafters wrote:
>> djoyce099 pretended :
>>> I know the argument of 1 =.999999999999... and the counter-argument of 1<
>>> .9999999999... as JG and
>>> a few others propose.
>>
>> Nobody thinks that one is less than .999...   .
>
> There are two ways to think about it. One is the limit, .999... = 1 for sure,
> is one... The other is building it up step-by-step. Well, in this case, there
> is no iteration where 1 equals .999... unless we use an epsilon to say, close
> enough, and halt iteration.

Still doesn't make one *less then* .999... though.

[FromTheRafters]

FromTheRafters brought next idea :

Should read *less than* not *less then*.

[djoyce099]

On Sunday, March 8, 2020 at 8:23:28 AM UTC-4, Eram semper recta wrote:

Well, that was really stupid of me I just got my arrow going the wrong way.
Not --the counter-argument of 1< .9999999999...
But --the counter-argument of 1> .9999999999...

Dan

[Chris M. Thomasson]

On 3/8/2020 2:39 PM, Chris M. Thomasson wrote:
> On 3/8/2020 12:47 AM, FromTheRafters wrote:
>> djoyce099 pretended :
>>> I know the argument of 1 =.999999999999... and the counter-argument
>>> of 1< .9999999999... as JG and
>>> a few others propose.
>>
>> Nobody thinks that one is less than .999...
>
> No. Using step-by-step iterative method .999... is always less than one.

I meant to respond to djoyce099 sorry.

[Zelos Malum]

Den söndag 8 mars 2020 kl. 03:21:13 UTC+1 skrev Mitch Raemsch:
> On Saturday, March 7, 2020 at 6:05:13 PM UTC-8, djoyce099 wrote:

> > I know the argument of 1 =.999999999999... and
> > the counter-argument of 1< .9999999999... as JG and
> > a few others propose.
> >

> > Putting it into a pure mathematics prospective using
> > pi as an example and employing Zeno's paradox.
> > Does this make the case for JG's argument?
> >

> You can't reach the end of the transcendental of Pi.

and it doesn't matter

[FromTheRafters]

Chris M. Thomasson wrote on 3/9/2020 :
> On 3/8/2020 2:39 PM, Chris M. Thomasson wrote:
>> On 3/8/2020 12:47 AM, FromTheRafters wrote:
>>> djoyce099 pretended :
>>>> I know the argument of 1 =.999999999999... and the counter-argument of 1<
>>>> .9999999999... as JG and
>>>> a few others propose.
>>>
>>> Nobody thinks that one is less than .999...
>>
>> No. Using step-by-step iterative method .999... is always less than one.
>
> I meant to respond to djoyce099 sorry.

He used "<" where he should have used ">" up there. Both are wrong
because "=" is the correct relation.

[FromTheRafters]

djoyce099 has brought this to us :

Much better, though both are equally wrong.

[Eram semper recta]

To compare 0.999... with 1 is WRONG whichever comparative relation you use: <, =, >

0.999... is a series/sequence. It is NOT a number.

1 is a very well formed number.

[djoyce099]

On Saturday, March 7, 2020 at 11:41:01 PM UTC-5, Dan Christensen wrote:

> On Saturday, March 7, 2020 at 9:05:13 PM UTC-5, djoyce099 wrote:
>
> >
> > Now in Zeno's paradox --
> > As each leg of the race the distance of 1/2 the remaining
> > distance too pi is assigned a new value for (A) where (B)
> > is assigned the previous value of (A) because (B) runs
> > twice as fast as (A) and is closing the gap at each leg
> > of the race, but never eliminating that gap no matter
> > how small that gap becomes.
> >
>

> IMHO decomposing the race between a slower runner (The Tortoise) and a faster runner (Achilles) into ever decreasing intervals like this turned out to be an intellectual dead end that held back European science for perhaps 2 thousand years. The ancient Greek philosophers of Zeno's day had no notion or any interest in how to measure the speed of an object.
>
> It took the Italian scientist Galileo, some 2 thousand years later, to define the speed of an object with his simple formula s=d/t. Today, even school children can use it to calculate precisely when and where Achilles would over take The Tortoise (assuming constant speeds). There was never really any paradox, just stubborn ignorance.
>
>
> Another Dan

>
> Download my DC Proof 2.0 freeware at http://www.dcproof.com
> Visit my Math Blog at http://www.dcproof.wordpress.com

The scenario I propose was the speed of runner (B) is twice as
fast as a runner (A) but runner (A) starting line is 1/2 the
distance to the finish line and runner (B) is the full distance to the finish line. Assuming that this is a flying mile and each (A)
and (B)are approaching their starting position at the same time
and their speed, the race will be a tie at the finish line.
I just used pi as that distance of the course but in that case,
taking each leg of the race (B) will never overtake (A) within
the length of the course.

Dan

[Eram semper recta]

It matters a great deal you stupid fuck!

You are very fast to repeat the gibberish you have been brainwashed to believe is true: "If they belong to the same equivalence class, then they are equal".

The chief attribute of equivalence classes is that they ALL have the same limit. That their component-wise differences tend to null is hardly remarkable because it is a consequence of the fact that they have the same *LIMIT* - the very thing you just claimed "doesn't matter".

In any case, the idea that two sequences belonging to the same equivalence class makes them equal is lame. It's like saying:

A man and a baboon both have penises, therefore they are equivalent.

In your case, you are equivalent to a baboon intellectually because your IQs are the same.

It **matters** very much what is this "limit". So much so, that through S = Lim S, you still drivel out nonsense like 0.333... = 1/3.

[djoyce099]

> IMHO decomposing the race between a slower runner (The >Tortoise) and a faster runner (Achilles) into ever decreasing >intervals like this turned out to be an intellectual dead end >that held back European science for perhaps 2 thousand years. >The ancient Greek philosophers of Zeno's day had no notion or >any interest in how to measure the speed of an object.
>
> It took the Italian scientist Galileo, some 2 thousand years >later, to define the speed of an object with his simple formula >s=d/t. Today, even school children can use it to calculate >precisely when and where Achilles would over take The Tortoise >(assuming constant speeds). There was never really any paradox, >just stubborn ignorance.
>
>
> Another Dan
>
> Download my DC Proof 2.0 freeware at http://www.dcproof.com
> Visit my Math Blog at http://www.dcproof.wordpress.com

---------------------------------------------------------------

The scenario I propose was the speed of runner (B) is twice as
fast as a runner (A) but runner (A) starting line is 1/2 the
distance to the finish line and runner (B) is the full distance to the finish line. Assuming that this is a flying mile and each

(A)and (B)are approaching their starting position at the same

time and their speed, the race will be a tie at the finish line.

I just used pi as that distance of the course but in that case,
taking each leg of the race (B) will never overtake (A) within
the length of the course.

Dan

---------------------------------------------------------------
Can anyone refute the above logic?