

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!
Bad command or file name. Bad, bad command! Sit! Stay! Staaaay...





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





So all he has to do is turn round, and jump backwards?
Bad command or file name. Bad, bad command! Sit! Stay! Staaaay...






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





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.





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!





Publish ripoff books. (5)
Slogans aren't solutions.





Print?
ripoff  pri
books  NT (New Testament)
Andy B





Yes, bang on!
Slogans aren't solutions.





Damn Great! Looking forward to it
Andy B





I know that feeling!
Slogans aren't solutions.






Rajesh R Subramanian wrote: I've been to Dehradun and Mussoorie Me too, but that was a long time ago.





Falling whilst taking a selfie is one of the biggest causes of selfie related deaths worldwide.
Not in India though, apparently there is more risk of being hit by a train whilst taking a selfie, I'm not sure how a train gets into a 50 foot ravine though.





55378008 wrote: Falling whilst taking a selfie is one of the biggest causes of selfie related deaths worldwide. The second biggest is taking a selfie with me in the frame.
I wanna be a eunuchs developer! Pass me a bread knife!





Yes, but isn't a large knife also in the frame?
Mongo: Mongo only pawn... in game of life.





You've given me an idea for an app.
Peter Wasser
"The whole problem with the world is that fools and fanatics are always so certain of themselves, and wiser people so full of doubts."  Bertrand Russell





"Ultimate Selfie"?
I wanna be a eunuchs developer! Pass me a bread knife!





Its best we don't discuss it on a public forum, somebody will steal the idea.
I just had an idea for an App too, but I would need to go to the Trumpbox to share it.





Maybe the other one pushed him down.
Maybe I need to stop reading Sherlock Holmes adventures.





CPallini wrote: Maybe I need to stop reading Sherlock Holmes adventures. Never, else you'd end up liking the two TV farts "Sherlock" and "Elementary".
It would have been painful for the other one to trace back the path on the same set of footprints.
CALL APOGEE, SAY AARDWOLF
GCS d s/++ a C++++ U+++ P L E W++ N++ o+ K w+++ O? M V? PS+ PE Y+ PGP t++ 5? X R++ tv b+ DI+++ D++ G e++>+++ h ++>+++ y+++* Weapons extension: ma k++ F+2 X
If you think 'goto' is evil, try writing an Assembly program without JMP.  TNCaver
"Go ahead, make my day"





Nae danger.





CPallini wrote: Nae danger. Ah, so you're more the James Bond type.
I wanna be a eunuchs developer! Pass me a bread knife!






Pictures or it didn't happen
Tom





Taken by a ghost photographer?
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





NEW SOCIAL MEDIA IDEA!
www.SnuffSelfies.com
One caveat . . . membership fees must be paid in advance.
Ravings en masse^ 

"The difference between genius and stupidity is that genius has its limits."  Albert Einstein  "If you are searching for perfection in others, then you seek disappointment. If you are seek perfection in yourself, then you will find failure."  Balboos HaGadol Mar 2010 





I'm looking for a cracking phrase for everything to do with the application except the application itself. So beyond the software development there is all the environment stuff, databases, connectivity, etc, etc, etc.
I need a short phrase to cover that. Any ideas?
veni bibi saltavi







That's the bunny! My brain was farting and it wouldn't tell me. Thanks.
veni bibi saltavi





You're welcome. I'll raise a virtual Gin to celebrate





Nagy Vilmos wrote: My brain was farting I thought I smelled something.
There are two kinds of people in the world: those who can extrapolate from incomplete data.
There are only 10 types of people in the world, those who understand binary and those who don't.



