[Archive!] Pure mathematics, physics, chemistry, etc.: brain-training problems not related to trade in any way - page 351
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x+(1/y)=2-(y-z)^2
y+(1/z)=2-(x-y)^2
z+(1/x)=2-(z-x)^2
Решите систему уравнений для положительных x, y и z:
x+(1/y)=2-(y-z)^2
y+(1/z)=2-(x-y)^2
z+(1/x)=2-(z-x)^2
Man, you guys...
The decimal mahas still haven't been solved.
And I haven't heard a word from the procurator.
In the senate...
;)
And S is unlikely to come here, they are very busy.
А как с ними решать, с машками, avatara? Общепринятой всеми процедуры пока не предложено. Что-то конкретное все равно должно исходить из требований к реализациям при нецелых периодах. Я пока ничего окромя непрерывности и не вижу.
А S сюда вряд ли заглянет, сильно занятые оне.
Another clue.
Pure geometry.
Peter has - so he thinks, infallibly.
But if the last value (i-1) is greater than the one added (i) with a remainder and vice versa - it is less, the results must be different.
And he has the same.
;)
----
like series (timeseries) -
6 3 7 5
6 7 3 5...
periood the same 3,333
Решите систему уравнений для положительных x, y и z:
x+(1/y)=2-(y-z)^2
y+(1/z)=2-(x-y)^2
z+(1/x)=2-(z-x)^2
x+1/x +y+1/y+z+1/z =6-(y-z)^2-(x-y)^2-(z-x)^2
x+1/x >=2
6-(y-z)^2-(x-y)^2-(z-x)^2 >=6
x=y=z=1
Еще одна подсказка.
Чистая геометрия.
У Петра - так он считает, непогрешимо.
That's a strong one. Just an option. // I've already pestered Alexey in private with my doubts. What is "infallible" here...)))
But if the last value (i-1) is larger than the value added (i) with a remainder and vice versa - it's smaller, the results should be different.
And it is the same.
;)
----
like series (timeseries) -
6 3 7 5
6 7 3 5...
periood is the same 3,333
Explain this one. I don't quite understand.
А вот это поясните. Не вполне понимаю.
mashka is not just an average... eh? ;)
now calculate it for the first row 6 3 7 5
and for the second 6 7 3 5.
I assert (and can show:) that MA/*3.333*/(0) is different for these rows.
If no one is interested in this problem, solve the others next...
I'm already embarrassed.
What am I saying? A simple waving machine is invariant with respect to any reshuffle of prices involved in the calculation. In principle, the "fractal" should behave the same. No? OK, justify it then.
It's different for other mash-ups. For a linearly weighted one the invariance of the swing will be with respect to other movements of the settlement prices.
Ну можно тогда и совсем в дебри залезть, если смотреть на инварианты.
О чем толкую? Простая машка инвариантна относительно любой перестановки цен, участвующих в расчете. В принципе так же должна вести себя и "фрактальная". Нет? ОК, обоснуй тогда.
Для других машек все по-другому. Для линейно взвешенной инвариантность машки будет относительно других движений расчетных цен.
Not really. It's kind of like a... sliding average. used for time series. Imho. ;)