[Archive!] Pure mathematics, physics, chemistry, etc.: brain-training problems not related to trade in any way - page 285
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Сам такой! :)
Can you write a formula? You're better at it. I'm a little slow on the uptake.
Yeah, I had a similar idea about cutting out the pieces. That sounds about right. So, do I lay out the solution? Or shall we build our own formulas? Purely visually I flashed a three-storey expression, true for the different segments, but I decided not to look at it.
And still I gnaw at the thought that the function may exist at an angle other than 90.
пять баллов
The only thing wrong with angles is that it's a pi/6 when the hypotenuse is half the size of the hypotenuse, not the other. Your angle is about 26.565 degrees.
The forumula is lazy, but it will be of the following approximate form
y=0.5*x at x belonging to [some expression from 2k, the same expression from 2k+1)
y=2*x for x as [same as 2k+1, same as 2k+2], where k is an integer
Ну да, у меня похожая идейка была с вырезанием кусочков. Кажись, похоже на правду. Ну что, выкладываю решение? Или сами формулы соорудим? Чисто визуально промелькнуло трехэтажное выражение, справедливое для разных отрезков, но я решил не смотреть на него.
И все же гложет мысль, что функция может существовать и при другом угле, не 90.
I don't think it will be 90 at all. At least my idea won't work there - the chunks will overlap... although... there could be three cut branches on each side of the graph.
I guess so. As for the formula, I'm in favour of making it myself. It doesn't feel right to give up at the finish line.
:)
OK. But the explanation of the problems states that it is not the proof but the construction of the particular case that is difficult in this particular problem. Draw the formulas.
For the angle not equal to 90, already later, who will be interested. Let's say 60.
.. хотя.. может ведь и три порезанных ветки быть у графика с каждой стороны.
Наверно есть.
Something like this (-infinity < n < infinity, n is an integer; x0 is arbitrary positive):
y = 0.5*x if x0*(1/2)^(2n) < |x| <= x0*(1/2)^(2n+1)
y = -2*x if x0*(1/2)^(2n+1) < |x| <= x0*(1/2)^(2n)
y = 0 if x = 0
A similar construction, only 4-storey (or more storey?), might work out for 60 degrees.
But what are the lines themselves? The tangents of their slopes have to be calculated from the condition that the nearest continuum segments differ in length by a factor of 3. The lines in which the function itself falls are defined depending on the value modulo 3 taken by degree 1/3 at the right end of the segment.
But I'm not sure, to be honest.
https://www.mql5.com/ru/forum/123519/page282
Answer: this is the burner of a powerful electric cooker (you can look up the picture on Yandex for 'KE burner').
These cookers are used in cafés, canteens and educational establishments.
Most likely they are numbers that are described by some well-known formula like: Mersen, Markov, Fermat, Germain, Wilson, etc. Which means that it is not that difficult to solve a problem with a computer. If they were not described by such a formula, there would be no way of proving anything.