Pure maths, physics, logic (braingames.ru): non-trade-related brain games - page 187
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Oh, man...
Well, for example, like this: to the cat sequence 5,4,5,4,5,4 the mouse responds as follows: 4,5,4,5,4,5.
No, you don't understand. I was the one who suggested the mouse's way out of your solution:
Note: you won't find a similar rebuttal to the cat sequence 2,3,4,2,3,4. Don't even try (but you will anyway).
But I can already see for myself that it doesn't fit (on the last move the mouse is in 4 and the cat is there too).
1. I'm paraphrasing:
...
2. I swear by the tailrace.
3. No way. The exponent is not bounded at the top. This shit is definitely bounded.
1. OK, I hear you right. It doesn't matter when to dump everything into a single vessel - at once or gradually. The heat is not going anywhere.
2. That's overkill, but I'm inclined to think it does, too.
3. Well yes, not an exponent, but a number e. I'm talking about the marginal transition where the number of parts tends to infinity. Well, that's a big deal...
There in a finite case (i.e. when the number of parts N is finite) creeps out ( N/(N+1) )^N -> 1/e.
But I'm having trouble accurately calculating the finite case. It's a rather cumbersome expression. And in Excel it's easy to calculate, it's understandable.
P.S. I calculated it - for the infinite case. Data from your table:
Mathemat:
I get something murderously simple: in infinite crushing, the final temperature of all tea is
T - (T-t)/e = 95 - 90/e ~ 61.89085.
Here e is the great constant of the even greater comrade Leonhard Euler.
Slightly more than the last line of your file. Probably got the conversion wrong. Or you've got an error piled up somewhere.
Can you give the data of your calculations, say, for n=100 000?
P.S. I did it - for infinite case. The data is from your table:
I get something very simple: In the infinite crushing case, the final temperature of all tea is
T - (T-t)/e = 95 - 90/e ~ 61.89085.
Here e is the great constant of the even greater comrade Leonhard Euler.
Can you report your calculation data for, say, n=100,000?
Excel (VBA) is a terrible retard. I spent half an hour calculating, and then it overflowed somewhere at 32768
// actually in my prog, but it's boring to deal with data types, easier to rewrite in a normal language (: like mql :).
Here are the results for 32000: // well, I suggest that you don't count further.
Initial temperatures were 1 and 0 degrees, respectively, for clarity
see, what should be according to your formula (T=1, t=0): 1 - 1/e ~ (1 - 0.367879441171442) = 0.632120558828558
yep, looks like it all adds up. score.
// however, check out the left-hand column pair. it's converging nicely to a complete temperature interchange. isn't it miraculous? ;-)
Ahem. Yep. You've got a bad head... :)
Anyway. I was wondering about some more details to calculate, and to stretch my brain a bit more.
In particular, what happens if one drink splits infinitely, and the second only slightly: in 2 parts, in 3, etc.
An intuitive hypothesis was that the degree of the number e in Alexei's formula would correspond to the number of fractions of the second drink.
As a result I made a script in mql (don't bother with this slow Excel. brr...), at the same time I calculated Alexey's order (n1 = 100 000), and launched it for mulyon too, just for full satisfaction. So:
At n1 = 100,000 :
2014.06.14 12:10:05.508 TeaCoffee EURJPY,H1 : Result: t tea = 0.367881280559, t coffee = 0.632118719437
2014.06.14 12:10:05.508 TeaCoffee EURJPY,H1: Start: t tea = 1.000000, t coffee = 0.000000, n tea = 1, n coffee = 100000, v tea = 1.000000, v coffee = 1.000000
at n1 = 1 000 000 :
2014.06.14 12:11:00.218 TeaCoffee EURJPY,H1 : Result: t tea = 0.367879625141, t coffee = 0.632120374911
2014.06.14 12:11:00218 TeaCoffee EURJPY,H1: Start: t tea = 1.000000, t coffee = 0.000000, n tea = 1, n coffee = 1000000, v tea = 1.000000, v coffee = 1.000000
// the Mathemat formula should result in the following limit: endT = 1 - 1/e ~ (1 - 0.367879441171442) = 0.632120558828558
// which agrees perfectly with the result, now down to the sixth digit.
now let's check the "intuitive hypothesis":
When n tea = 1000000, n coffee = 2
2014.06.14 12:29:57.770 TeaCoffee EURJPY,H1: Result: t tea = 0.270670837135, t coffee = 0.729329162824
2014.06.14 12:29:57.770 TeaCoffee EURJPY,H1: Start: t tea = 1.000000, t coffee = 0.000000, n tea = 1000000, n coffee = 2, v tea = 1.000000, v coffee = 1.000000
According to the hypothesis it should be: endT = 1 - 1/(e ^2 ) ~ (1 - 0.135335283236613) = 0.864664716763387
Bummer, the hypothesis is not confirmed.
I have tried to build another hypothesis for the case [N1->∞, N2 = 2, 3, 4 ....] around Alexey's formula, but have not found anything yet.
Alexey, if there is still powder left, look please, what there should be obtained analytically.
Here are more results for some N2:
2014.06.14 12:47:24.782 TeaCoffee EURJPY,H1: Result: t tea = 0.224042143726, t coffee = 0.775957856295
2014.06.14 12:47:24782 TeaCoffee EURJPY,H1: Start: t tea = 1.000000, t coffee = 0.000000, n tea = 1000000, n coffee = 3, v tea = 1.000000, v coffee = 1.000000
2014.06.06.14 12:47:49.782 TeaCoffee EURJPY,H1: Result: t tea = 0.195367205557, t coffee = 0.804632794492
2014.06.14 12:47:49782 TeaCoffee EURJPY,H1: Start: t tea = 1.000000, t coffee = 0.000000, n tea = 1000000, n coffee = 4, v tea = 1.000000, v coffee = 1.000000
2014.06.06.14 12:54:39.154 TeaCoffee EURJPY,H1: Result: t tea = 0.175467808435, t coffee = 0.824532191564
2014.06.14 12:54:39154 TeaCoffee EURJPY,H1: Start: t tea = 1.000000, t coffee = 0.000000, n tea = 1000000, n coffee = 5, v tea = 1.000000, v coffee = 1.000000
2014.06.06.14 12:54:48.454 TeaCoffee EURJPY,H1: Result: t tea = 0.125110661269, t coffee = 0.874889338728
2014.06.14 12:54:48.454 TeaCoffee EURJPY,H1: Start: t tea = 1.000000, t coffee = 0.000000, n tea = 1000000, n coffee = 10, v tea = 1.000000, v coffee = 1.000000
------------
At the same time I calculated for a large number of parts (100,000) of both drinks:
2014.06.14 13:33:12.788 TeaCoffee EURJPY,H1: Result: t tea = 0.001784121886, t coffee = 0.998215878114
2014.06.14 13:33:12.788 TeaCoffee EURJPY,H1: Start: t tea = 1.000000, t coffee = 0.000000, n tea = 100000, n coffee = 100000, v tea = 1.000000, v coffee = 1.000000
It took a very long time (26 minutes) to count, so I would not recommend repeating this feat. However, you can see that the result clearly converges at infinity to a complete exchange of temperatures at the drinks.
Attached is a script, you can play around with it if you're interested. // It is in mql4, so it works in MT5, just rename it to .mq5.
By the way, the script can calculate the heat exchange at different initial volumes of drinks. I haven't played with it yet, I will try it now.
joo:
joo:
oh...
:) :) :)
In fact, you can still laugh here and at the same time refute the result.
There are (at least) two good reasons for this: (1) the clap has a duration in time, (2) the bird has mass.
It follows that (1) it takes a finite and not zero time to recognize the clap and (2) the bird accelerates not instantly but during a finite period of time.
And from this it follows with necessity, that the psyche will hear the fourth clap twice. With a corresponding tripling of the speed again.
But that's not all, the laughter can go on. Of course the clap, when played backwards, sounds quite different from the forward one. It is just an acoustic fact, which cannot be ignored. It is logical to assume that with proper intelligence the bird will also react to it differently - namely by tripling the speed. ;)
Then try to guess what happens after four claps, the bird will start to accelerate and decelerate crossing the propagation front of the fourth clap and will do it permanently (or until battery will run out or it will collapse from wild vibration and overloading). Its average speed will be fully consistent with the speed of sound (330m/s).
--
Well, it's already possible to have the last laugh and to finish. Or for perverts (like Mathemat, TheXpert, Avals, alsu and others) you can ask a question: what's the frequency of bird's movement around front of fourth flap spreading, if to define delay of recognition and acceleration/deceleration.
Here you may choose something arbitrary for certainty (1) duration of the clap (from the beginning till the moment when it is recognized exactly as a clap), let it be, say, 1 ms.
(2) the acceleration(deceleration) time to triple the speed. let it be say 100 ms.
Good luck! ;) ;)
Explain in human terms how you did it. I'll do some analysis, check it out. I don't believe it, it does look like a miracle.
You hinted a couple of pages ago, but in your own style, very briefly. I still don't understand what it is.
The gunpowder isn't very good, it's almost dry. Here's a screenshot of how I calculated the formula with e. Took me about three hours, got it on about the fifth try...
In short, tell me exactly what you were doing in the left pair of columns.
Then let's see what happens after four claps: the psyche will start accelerating and slowing down across the propagation front of the fourth clap, and will do so forever (or until the batteries run out or it collapses from wild vibration and overloading). Its average speed will, of course, fully match that of the clap, i.e. sound (330 m/s).
--
Well, that's it, we can laugh for the last time and finish. Or for perverts (like Mathemat, TheXpert, Avals, alsu and others) you can also ask a question: what frequency will the bird vibrate around the propagation front of the fourth clap, if you define recognition delay and acceleration/deceleration.
Here you may choose something arbitrary for certainty (1) duration of the clap (from the beginning till the moment when it is recognized exactly as a clap), let it be, say, 1 ms.
(2) the acceleration(deceleration) time to triple the speed. let it be say 100 ms.
Good luck! ;) ;)
So you tell that shit to the moderators of that resource. It's logical in principle, by the way.
At the beginning, I too was existentially depressed after all this pile of assumptions and neglects. But I coped with it: first I published inaccurate result, and then corrected it (the moderator hinted that there were inaccuracies and misspellings).
It is useless to argue with a moderator. The task has its own laws, which do not have to agree with the physical ones.
so at what speed will the bird fly into the depths of the universe?
/***********/
I.e. reacts to seven events - three baby claps (twice for each) and once for the fourth.