[Archive!] Pure mathematics, physics, chemistry, etc.: brain-training problems not related to trade in any way - page 280
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It could be even simpler:
31,
331,
3031,
30031,
300031.
True, with mutual simplicity you still have to check. But the law of education is simpler.
Next:
Find all alpha such that the sequence cos(alpha), cos(2*alpha), cos(4*alpha), cos(8*alpha), ..., cos(2^n*alpha), ... - are all negative. 451
I can still do it too:
91
991
9991
99991
999991
;)
I.e. you can also do this:
Find all such alpha that cos(alpha) = z(0)<0 and all z(n+1) = 2*z(n)^2 - 1 are negative. I hope that's clear?
Т.е. можно и так:
Найти все такие alpha, что cos(alpha) = z(0)<0, а все z(n+1) = 2*z(n)^2 - 1 отрицательны. Надеюсь, понятно?
The first formulation makes more sense to me. I'm off to find out.
A graphical construction would probably help. The parabola y=2*z^2 - 1 and the line y=z.
Obviously, the fixed point of the mapping z -> 2*z^2 - 1 is the intersection of these graphs.
We need a negative one. We solve the equation: 2*z^2 - z - 1 = 0, z<0.
This is z=-1/2, i.e. alpha = 2*Pi/3. This is one point.
Тут графическое построение поможет, наверно. Парабола y=2*z^2 - 1 и прямая y=z.
Очевидно, неподвижная точка отображения z -> 2*z^2 - 1 - пересечение этих графиков.
Нам нужна отрицательная. Решаем уравнение: 2*z^2 - z - 1 = 0, z<0. Это z=-1/2, т.е. alpha = 2*Pi/3. Это одна точка.
The rest of the solutions are obtained by "cloning" - multiplying by powers of two.
OK, so far, trivial. What about other solutions or proof that there are none? Yes, there are no other solutions, but the proof is non-trivial.
Next:
Так, пока тривиально. А как насчет других решений или доказательства, что их нет?
You saw it on the graph, didn't you?
Ты же на графике видел?
Well, I only pointed out a fixed point. We need more than that. It doesn't have to be that all these negative points are equal. Yeah, the solution proves there are no other points.
Yeah, I saw your congratulations. It's been a while since I've had black caviar...