[Archive!] Pure mathematics, physics, chemistry, etc.: brain-training problems not related to trade in any way - page 375
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Да хватит уже истерить. Тем более в такой неподходящей ветке. У 5ки огромная куча проблем и без этого.
Неужели думаете, что кого-то тут больше слушать будут? Или скопом лучше получится?
Придумайте адекватный выход, и возможно к Вам прислушаются.
Dear Hpert.
Do you know the difference between hysterical and prankish?
And useful from unnecessary?
;)
Swetten, факториал - это, грубо говоря, дискретный аналог гамма-функции. Гамма-функцию как обобщение факториала на нецелые числа начал систематически изучать Эйлер. Где ее только нет. И не верю, что в аэродинамике она не встречается.
Вообще в огромном множестве интегралов, без которых физика не обходится, неожиданно выползает именно гамма-функция.
2 FreeLance: чтобы помочь Метаквотам с интерполяцией, надо хотя бы узнать, как она делается сейчас, - чтобы не тыкать пальцем в небо. Какая-нибудь реакция раработчиков терминала на Ваш вопль о помощи была?
There is no reaction. I think there's the old rake with missing quotes and the time axis.
But a calendar for the instrument has appeared in the heel. So a solution might show up.
There just aren't any mathematicians on the team there. :(
Regarding SCOPOM.
Rinat once said that he listens to the community.
Not individual "upstarts"...
:)
And about the rake of graphic synchronization almost the entire community has been browbeaten.
Isn't it?
Here's a good one:
An old maths professor has put six of the most primitive locks in the door of his flat, which can be opened with a nail file. But the professor, when he leaves for work, randomly closes only three of them, three locks remain open (assuming that the key is turned in the lock anyway, i.e. it is impossible to know whether the lock is closed or not).
How many variants will it take for a student with a failing grade to get to the flat to get his credit?
ONE option is to turn all six locks... The student does not know how many locks are open and how many are closed... Since the locks are primitive, they are opened in one direction (usually by turning away from the jamb) and even if the key is turned... turning all locks in this direction he will get three unlocked and three unlocked locks... that is, one turn for each lock... no question about closing the locks afterwards... :)
А чем ценен факториал?
You should read something like "Amusing Combinatorics".
What is the value of the factorial is best illustrated by an example. Suppose we have four letters - A, B, C and D. How many ways can we make a four-letter word out of them?
Combinatorics calls it permutations and states that the number of combinations to compute permutations = n! (reads "En-factorial"). So we have 4 letters and so the number of permutations of those letters would be = 4! = 4*3*2*1 = 24. So, by permutations of the letters A,B,C and d, we can make only 24 words. For example, the combinations would be:
ABWG GABV AGBV ABGV ... and so on.
In short, in combinatorics the factorial is an integral part of many formulas and facilitates the process of solving many combinatorial problems.
Once upon a time in the USSR after World War II, in order to wean young people off gambling, the state introduced elements of combinatorics and probability theory into the compulsory curriculum of mathematics. It was done for the youth to estimate their chances of winning at gambling and to understand that it is not worth it. I still have a textbook like that at home. Too bad it's not in the school curriculum now. That section of mathematics is the most interesting thing.
For those who are brain-dead, there is a more serious problem:
There are three equal chords AB, CD and PQ in the circle with centre O (see figure). Prove that MOK is half of angle BLD.
Two equal chords of a circle, intersecting, divide the intersection point in equal ratios. I do not give the proof, those who wish check it independently.
Then AM=MQ and PK=KD. Hence the equality on three sides of the following triangles follows:
AOM=QOM,
POK=DOK,
then MO and KO are the bisectors of angles AMQ and PKD respectively. Then for angles we have
OMK=1/2*AMK=1/2*(180-LMK)
OKM=1/2*DKM=1/2*(180-LKM)
OMK+OKM=1/2*(180-LKM)+1/2*(180-LMK)=180-(LMK+LKM)/2=180-BLD/2
MOK=180-(OMK+OKM)=BLD/2,
as needed to prove it:)
Great, alsu, just got to this thread.
2 Michelangelo: that's right, the student doesn't know how many locks are closed. Well then 63 options (not counting the obvious one, i.e. checking if the door is closed).
In short, the problem is incorrect from the very beginning. It does not say anything about what information the student has.
А вот, кстати, крокодил: он больше зелёный или больше длинный? :)
How old is that joke?
at least 20 years.