
ZENO'S PARADOX
An athlete can jump infinitely, but each of his jumps are the half of the previous one...
Starting with a jump of half meter, how many jumps it will take to reach 1 meter?
And while we all know the answer, I would like to know why in reality it is not true!
Skipper: We'll fix it.
Alex: Fix it? How you gonna fix this?
Skipper: Grit, spit and a whole lotta duct tape.





But it is true: try it.
You can no longer reach your coffee...
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Sigh[^]... I'm so happy that I do not drink coffee... fortunately the tea still in my reach...
Skipper: We'll fix it.
Alex: Fix it? How you gonna fix this?
Skipper: Grit, spit and a whole lotta duct tape.





Only briefly: it's all quantum. And as Heisenberg states "I AM the danger" you cannot know the position and speed of anything at the same time. Therefore, if you measure the position of the athlete at any point in the movement, his speed cannot be determined and he has probably arrived somewhere else already.
Alternatively:
Quote: Heat crackled off the white walls of the tavern but, Teppic thought, how different it was from the Old Kingdom. There even the heat was old; the air was musty and lifeless, it pressed like a vice, you felt it was made of boiled centuries. Here it was leavened by the breeze from the sea. It was edged with salt crystals. It carried exciting hints of wine; more than a hint in fact, because Xeno was already on his second amphora. This was the kind of place where things rolled up their sleeves and started.
“But I still don’t understand about the tortoise,” he said, with some difficulty. He’d just taken his first mouthful of Ephebian wine, and it had apparently varnished the back of his throat.
“’S quite simple,” said Xeno. “Look, let’s say this olive stone is the arrow and this, and this—” he cast around aimlessly—“and this stunned seagull is the tortoise, right? Now, when you fire the arrow it goes from here to the seag—the tortoise, am I right?”
“I suppose so, but—”
“But, by this time, the seagu—the tortoise has moved on a bit, hasn’t he? Am I right?”
“I suppose so,” said Teppic, helplessly. Xeno gave him a look of triumph.
“So the arrow has to go a bit further, doesn’t it, to where the tortoise is now. Meanwhile the tortoise has flow—moved on, not much, I’ll grant you, but it doesn’t have to be much. Am I right? So the arrow has a bit further to go, but the point is that by the time it gets to where the tortoise is now the tortoise isn’t there. So, if the tortoise keeps moving, the arrow will never hit it. It’ll keep getting closer and closer but never hit it. QED.”
“Are you right?” said Teppic automatically.
“No,” said Ibid coldly. “There’s a dozen tortoise kebabs to prove him wrong. The trouble with my friend here is that he doesn’t know the difference between a postulate and a metaphor of human existence. Or a hole in the ground.”
“It didn’t hit it yesterday,” snapped Xeno.
“Yes, I was watching. You hardly pulled the string back. I saw you,” said Ibid. The Late Great Sir PTerry, Pyramids
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I am currently* rereading from Colour of Magic to Shepherd's Crown in order. The book I am on funnily enough is Pyramids; at the bit just after the King has died and Ptepic is on the Brass Bridge.
Bloody seagulls!
veni bibi saltavi





Wanna buy a grilled tortoise? Onna stick?
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A more difficult and interesting puzzle would be:
What is the smallest first jump the athlete must make to eventually get to 1 meter?
Cheers,
Mick

It doesn't matter how often or hard you fall on your arse, eventually you'll roll over and land on your feet.





Do you mean in finite jumps?
That's not a problem  1 meter...
Skipper: We'll fix it.
Alex: Fix it? How you gonna fix this?
Skipper: Grit, spit and a whole lotta duct tape.





No, that's the jump for minimum number of jumps. My question is what is the smallest first jump that will eventually get to a distance of 1 meter (e.g. 1st jump = 2/3rd meter + 2nd jump 1/3 meter = 1 meter). We know that 1/2 meter 1st jump never gets there  each jump gets to half the remaining distance. What is the minimum 1st jump distance that will get there (with unlimited number of jumps allowed).
Cheers,
Mick

It doesn't matter how often or hard you fall on your arse, eventually you'll roll over and land on your feet.





I see what do you mean... However it is the one and only solution  starting with the 2/3 of the requested distance...
See this:
1 = x/n^{0} + x/2^{1} + x/2^{2} + ... + x/2^{n}
It is easy (relatively) to prove that for that there is only one solution x = 2/3...
The trick is to start with 1 part of the distance. In that case the distance made by the athlete after n jumps can be written like this (the initial jumps is 1/base):
(2^{n1} + 2^{n2} + 2^{n3} + ... + n^{0}) / base * 2^{n1}
And that number never reaches 1...
Skipper: We'll fix it.
Alex: Fix it? How you gonna fix this?
Skipper: Grit, spit and a whole lotta duct tape.
modified 6Feb17 8:48am.





No  actually, the smallest distance > 1/2 meter will eventually get there. Even .50000000(with a million zeros)1 will. The interesting thing is that it is impossible to express that number, but eventually, the fractional part, however small, will cover at least the remaining distance to the full meter (it may be not be exactly the meter, but marginally greater than).
change your equation to
1 <= x/2^{0} + x/2^{1} + x/2^{2} + ... + x/2^{n}
Also, just try it with the calculator with x = .51, and then x= .501, and you will see what I mean  they only take a few jumps each.
Cheers,
Mick

It doesn't matter how often or hard you fall on your arse, eventually you'll roll over and land on your feet.





The problem is that .51 and .50000....00001 are not real numbers...
Zeno (and me too) are people of the real world... And that's the exact problem... Computation with real numbers gets you nowhere, but in reality it is not true!!!
Skipper: We'll fix it.
Alex: Fix it? How you gonna fix this?
Skipper: Grit, spit and a whole lotta duct tape.





That's not true.# Steps Needed      Starting Distance From 0  1  2/3  2  4/7  3  8/15  n  2^{n}/2^{n+1}1 
However, anything greater than half will get above 1, but never on 1, except for the fractions shown.
CP bug? Table not appearing immediately after text.
modified 6Feb17 8:42am.





What's not true? You just proved, that you can not get to 1 meter... or above or below...
Skipper: We'll fix it.
Alex: Fix it? How you gonna fix this?
Skipper: Grit, spit and a whole lotta duct tape.





2/3 + 1/3 = 1
4/7 + 2/7 + 1/7 = 1
8/15 + 4/15 + 2/15 + 1/15 = 1
All the fractional distances that I showed you lead to 1 in n number of steps.





You right about the first part  I was a bit hasty on that...
Skipper: We'll fix it.
Alex: Fix it? How you gonna fix this?
Skipper: Grit, spit and a whole lotta duct tape.





Updated the text for the exceptions.





Actually, wrote a small program to demonstrate:
namespace TestJumps {
class Program {
static void Main(string[] args) {
string s;
do {
double d;
Console.Write("Input test value: ");
s = Console.ReadLine();
if (double.TryParse(s, out d) && d > .5D) {
int i = 0;
double total = d;
while (total < 1D) {
d /= 2;
total += d;
i++;
}
Console.WriteLine("Total {0} reached in {1} steps", total, i);
}
} while (!string.IsNullOrEmpty(s));
}
}
}
Results:
Input test value: .51
Total 1.0040625 reached in 5 steps
Input test value: .501
Total 1.00004296875 reached in 8 steps
Input test value: .500000001
Total 1.00000000013735 reached in 28 steps
Cheers,
Mick

It doesn't matter how often or hard you fall on your arse, eventually you'll roll over and land on your feet.





And you proved what I told...
If your initial number is in the form of 1/n you are getting close to 1 from below, but never reaching...
If your initial number is in the form of 1/n + 1/10^n you are getting close to 1 from above, but never reaching...
The only way to get to 1  exactly  is starting with 2/3 as first jump...
Skipper: We'll fix it.
Alex: Fix it? How you gonna fix this?
Skipper: Grit, spit and a whole lotta duct tape.





What about 8/15 (0.53 recurring) meters? 4 jumps required.
8/15 + 8/30 + 8/60 + 8/120
= 64/120 + 32/120 + 16/120 + 8/120
= 120/120
= 1
Cheers,
Mick

It doesn't matter how often or hard you fall on your arse, eventually you'll roll over and land on your feet.





Yes. On the first part I was too hasty (and made some mistakes)... There are more than one solutions if you are not starting with 1/n...
Now that can be fascinating to find out a relation between the numerator, the denominator and the number of steps...
Skipper: We'll fix it.
Alex: Fix it? How you gonna fix this?
Skipper: Grit, spit and a whole lotta duct tape.





The relationship is very simple:
For a divisor that may be represented in binary as 1...1, the number of steps is the number of bits in the divisor:
1  1 step
3  2 steps
7  3 steps
15  4 steps
...
If the ratio between step sizes is 'q', use numbers that may be represented as 1...1 in base q.
If you have an important point to make, don't try to be subtle or clever. Use a pile driver. Hit the point once. Then come back and hit it again. Then hit it a third time  a tremendous whack.
Winston Churchill





Although I am not good at mathematics, I am good at arithmetic, and (possibly because of my programming experience), I do tend to notice patterns. I think the clue to solving that question would be to take note of the expression that normalises the fractions: Each numerator becomes 2 ^ (totalSteps  stepsTaken), and the denominator is (2 ^ totalSteps)  1. I think (but cannot prove) that the resulting numerator is 2 ^ (totalSteps  1).
resulting in
x = (2^(n  1)) / ((2 ^ n)  1)
where n is the number of jumps needed. This works for both the 2/3 and 8/15 we've already explored. As I'm not sure how to prove it, cannot guarantee I'm right for all cases.
Cheers,
Mick

It doesn't matter how often or hard you fall on your arse, eventually you'll roll over and land on your feet.





Simple: one Planck length[^]
Mind you, he'd have to jump a whole load of times to get to a meter!
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1 jumps
... doing a negative number of jumps is just as easy as doing an infinite number of them.
Sin tack ear lol
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