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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
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It doesn't matter how often or hard you fall on your arse, eventually you'll roll over and land on your feet.
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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.
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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
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It doesn't matter how often or hard you fall on your arse, eventually you'll roll over and land on your feet.
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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/n0 + x/21 + x/22 + ... + x/2n
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):
(2n-1 + 2n-2 + 2n-3 + ... + n0) / base * 2n-1
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 6-Feb-17 8:48am.
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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/20 + x/21 + x/22 + ... + x/2n
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.
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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.
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That's not true.| # Steps Needed - - - - | Starting Distance From 0 | | 1 | 2/3 | | 2 | 4/7 | | 3 | 8/15 | | n | 2n/2n+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 6-Feb-17 8:42am.
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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.
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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.
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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.
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Updated the text for the exceptions.
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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.
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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.
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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.
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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.
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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
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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.
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Simple: one Planck length[^]
Mind you, he'd have to jump a whole load of times to get to a meter! 
Bad command or file name. Bad, bad command! Sit! Stay! Staaaay...
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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
Pressing the any key may be continuate
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So all he has to do is turn round, and jump backwards?
Bad command or file name. Bad, bad command! Sit! Stay! Staaaay...
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Kornfeld Eliyahu Peter wrote: I would like to know why in reality it is not true!
Because he can't jump less than the Planck length. 
Marc
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Planck length is a far too complicated definition of a simple thing... to my taste...
When I got these kind of question in school I used to answer: The shoe size... because of the shoes size...
Skipper: We'll fix it.
Alex: Fix it? How you gonna fix this?
Skipper: Grit, spit and a whole lotta duct tape.
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Depending on the length of his feet, it should take between six and seven jumps before his toes cross the line.
I wanna be a eunuchs developer! Pass me a bread knife!
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Publish rip-off books. (5)
Slogans aren't solutions.
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