dj_ivocha on 24/11/2008 at 09:54
Quote Posted by Taffer_Boy_Elvis
...car going 60 miles per hour, when it's 60 miles from the destination, and
slowing by 1 mph each mile it gets closer to the destination.
Quote Posted by demagogue
Think about the 60 mile distance divided evenly into 1/60 parts, each one 1 mile = 1 minute of travel @ 60 mph. As you slow down, a progressively bigger chunk is taken out of each 1/60 section that you don't travel in 1 minute (0/60, 1/60, 2/60...).
Why do you keep talking about 1-minute chunks? :weird: Nowhere in the problem's definition is any fixed time mentioned.
Car goes 60mph.
Car travels 1mile (in 1 minute).
Car goes 59mph.
Car travels 1mile (in 1.016949153 minutes).
Car goes 58mph.
...
...
Car goes 1mph.
Car travels 1 mile (in 60 minutes).
Car stops.
Car has traveled 60 miles in
how much time?
As I understand it, the bolded part is the real question to that problem.
hopper on 24/11/2008 at 10:40
It takes Σ[n=0->59] 1 / (60-n) hours, or 1/60 + 1/59 + ... + 1/2 +1 hours.
dj_ivocha on 24/11/2008 at 11:45
Indeed. Or in other words, about 4 hours and 40 minutes. :idea:
demagogue on 24/11/2008 at 14:58
Quote Posted by dj_ivocha
Why do you keep talking about 1-minute chunks? :weird: Nowhere in the problem's definition is any fixed time mentioned.
I wasn't talking about 1-minute chunks, but 1-mile chunks ("60 mile distance divided evenly" into "1-mile" chunks; you quoted it yourself).
It's just, if you think about 1-minute periods inside each 1-mile chunks, then you only have to keep track of the
extra time over a minute you travel for every mile. You know from the beginning that you travel 59/60 of the 2nd mile (@59mph) in a minute. So you only need to compute the extra 1/60 mil at 59 mph. You know you travel 58/60 of the 3rd mile in a minute, so you only need to compute the extra 2/60 @58mph, etc. i.e., every time, 1 minute +1/60mi@59mph, +2/60mi@58mph, +3/60mi@57mph... We're all (TBE, me, you, hopper) posting the same correct answers ... 1mile@59mph=1.0169 min, 1mile@58mph=1.0344 min ... 1mi@1mph=60min.
It's just a mental shortcut ... Anyway, it was for me at the time. I liked the 1+2+3 ... part of it, and that 1 hour of the answer was taken care of automatically, right from the start (each 1-minute bit for each 1-mile chunk, 60 times; 1:59 hours if you include the easy last mile). So all you need to do is add the increasing bits of extra time for each 1-mile chunk.
If you look at the bold answer I finally gave, it's the same answer that hopper just posted from a different route.
So don't knock it if it leads to the same right answer; it's math(s). :p
Also: variety spice of life, etc...
Quote Posted by dj_ivocha
Car has traveled 60 miles in
how much time?
As I understand it, the bolded part is the real question to that problem.
That's why my answer is "1 hour + Σ[0->59] n / (60-n) minutes" ;)
Aaanyway, I also wanted to see the answer as a fraction, not a big decimal... But I can only think of forcing a common denominator by just multiplying each time by (59!/n)/(59!/n); then every denominator will be 59!. But as that's so massive, doesn't really help much.
Muzman on 24/11/2008 at 15:13
Quote Posted by d0om
Actually the bouncing ball one has an infinite number of bounces in a finite length of time.
If it halves in height each bounce it will halve in time each bounce too.
1 + 1/2 + 1/4 + 1/8 + 1/16 + ... =2.
So therefore it stops after 2 times the initial bounce time, after an infinite number of bounces.
There isn't any particular name for this; the numbers never quite converging until you (I dunno what the right word is) abstract it with time rather than step through the iterations, is there?
There probably isn't, I would expect it would be fairly well known if there was one since it's pretty old and rudimentry stuff it seems. I just have some vague memory of some cool label some teacher associated wth it (could have been something obvious like 'the convergence problem', 'the horizon problem' I guess. Might stick with 'The bouncing ball problem'.)
hopper on 24/11/2008 at 16:10
Quote Posted by demagogue
That's why my answer is "1 hour + Σ[0->59] n / (60-n) minutes"
Which is wrong.
It probably gives you the right answer if you go exactly 60 miles - I haven't tested it - but if you reduce the number of one-mile legs, the answer will be one minute off for each deleted leg, to the point where, if you go only one leg, according to your formula, that first leg will take 1h. It's like the proverbial clock standing still.
Take the formula v = s*t, where v = distance, s = speed and t = time, so for each 1-mile leg we get the time
--> t = v/s
where you only need to adjust the speed for each leg, since v is constant.
or, in sum,
Σt = Σ[n=1->x] v/sn = v*Σ[n=1->x] 1/sn.
(x being the number of legs you want to play this game for.)
Now, starting with the last mile and going backwards and setting v=1 (leg length of 1 mile) and x=60 (number of legs) and sn=n (i.e., the speed of each leg equals that leg's position in the row), we get:
t = Σ[n=1->60] 1/n
Which is the same as in my last post, in even simpler form. And the beauty of it is, it works with any number of legs :cheeky:
demagogue on 24/11/2008 at 16:24
Quote Posted by hopper
Which is wrong. ... the answer will be one minute off for each deleted leg
Well, the extra minute for each leg is covered in the opening 60 minutes (60 minutes + the sum)...
[Edit: oh, ok, if you're thinking about deleting legs then ... it's better written as
"
Σ[0->59] 1+ [n / (60-n)] minutes"
so you put those 60 minutes inside the sum, and it can handle the change in # legs.
Edit2. I don't know much physics, so maybe I don't know what you mean by "deleting a leg" ... Do you mean only going 59 miles? Mine still gets the right answer. Or do you mean measuring the full 60 miles with fewer legs, 59 or even 1?? But wouldn't "1 hour for that first leg" still be the right answer if you "go only one leg" of 60 miles? Or do you want it to apply to any arbitrary length? That I don't know... But I wasn't trying to apply it to any arbitrary length. I was trying to answer TBE's problem. It's not
wrong wrong if it gives the right answer for the problem! Maybe if you're a physics student; but this wasn't physics for me. It was just math. And the math is right!
Do mine leg for leg:
n=0. 1 minute + 0/60 = 1 minute
1. 1 minute + 1/59 min = 1.0169 min's
2. 1 minute + 2/58 min = 1.0344 min's
...
59. 1 minute + 59/1 min's = 60 minutes
Now let's do yours leg for leg:
n=0. 1 / 60 hours = .0166 hours *60 = 1 minute
1. 1/59 hours = 0.0169 hours * 60 = 1.1069 min's
2. 1/58 hours = 0.01724 hours * 60 = 1.0344 min's
...
59. 1/1 hours = 1 hour *60 = 60 minutes
It's the same answer every leg.
I grant that your solution is much more elegant, esp in the simpler form. I saw that immediately! But I didn't think my answer was bad for a lawyer brushing up on his math for fun... :sweat:
It came up with the right answer for this problem...
Quote:
to the point where, if you go only one leg, according to your formula, that first leg will take 1h. It's like the proverbial clock standing still.
Well, don't forget what the sum is actually measuring. It's measuring the
extra time you go above 1 minute (or the baseline time) for each leg in a progressive deceleration. In the first leg, by definition you don't go above 1 minute (the baseline), so it's a meaningless nil, like you say. So it's right for what its job is, even for the blip you pointed out. I know it's not as elegant, but I thought there was a certain cleverness to it... It's different, you have to give it that. I actually suspect it's a derivation of the textbook answer you gave, and there's some relationship between them ... just looking at how each one was created (one measuring a speed over a whole leg (mile) which slows each round, the other taking a baseline v/s on the first leg then just repeating it and adding a v/s remainder that grows/slows each round. Two ways of measuring the same "time" per mile. Looking back, I guess I was thinking if I just repeated a baseline and dealt with a growing remainder I could stumble into some shortcut to give me a quicker answer than brute-force computing the time each round. But looking at it now, it didn't turn out simpler at all; quite the opposite.
hopper on 24/11/2008 at 17:22
Quote Posted by demagogue
I don't know much physics, so maybe I don't know what you mean by "deleting a leg" ... Do you mean only going 59 miles?
That's what I meant, yes. Your NEW AND IMPROVED FORMULA</advertising voice> seems to me to work better, too. ;)
ZylonBane on 24/11/2008 at 22:53
Quote Posted by Muzman
There isn't any particular name for this; the numbers never quite converging until you (I dunno what the right word is) abstract it with time rather than step through the iterations, is there?
There is, and I already told you what it is. It's called a convergent series.
