[Archive!] Pure mathematics, physics, chemistry, etc.: brain-training problems not related to trade in any way - page 114
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It's not just x=0. It's all points x(n) = 1/((2n+0.5)*Pi). There's a countable set of them, and they have a limit point.
Yes, of course I blurted it out. I meant the points sin(x)=0. :-)
However, this countable set of admissible points does not satisfy the definition of the limit: "for any convergent sequence". Or in delta-epsilon language: "for any delta, there exists such an epsilon".
А как на счёт дробных степеней?
OK, you can do fractional, just not transcendental. If you can find the last four digits in them. Just don't talk about rounding errors encountered in real life.
How do you do this? It just says ln(2) (Maple 13)
with(Student[Calculus1]):
LimitTutor();
I type in the limit and click "All steps".
One more question. How can I change the default settings for plotting limits? When I do refresh the sheet, the chart's appearance changes. :(
Next: Prove that the degree of two cannot end in four identical digits.
There cannot be odd ones among them. Only 2,4,6 and 8 can be even. If it is 2, then we divide it by 2 and get 1111. If it is 6, then we obtain 3333 in the same way.
Note, too, that if you subtract an even number from an even number, you end up with an even number. That is, our number can be represented as (abc...xyz0000 + 4444) or (abc...xyz0000 + 8888).
1. If z is even, then (abc...xyz0000 + 4444)/2 = (klm...pqr0000 + 2222) and we come to option with 2.
If z is odd, then (abc...xyz0000 + 4444)/2 = (klm...pqr5000 + 2222) = klm...pqr7222 and we come to a similar variant c 111 at the end. So 4444 cannot be either.
2. If z is even, then (abc...xyz0000 + 8888)/2 = (klm...pqr0000 + 4444) and we arrive at variant with 4.
If z is odd, then (abc...xyz0000 + 8888)/2 = (klm...pqr5000 + 4444)
Continuing in the same way, (klm...pqr5000 + 4444)/2 = (def...ghjT500 + 2222), where T can be either even or odd.
Repeating this operation one more time we get 1 at the end of the number, which cannot be at the degree of two.
Any degree can. If those four digits = 0 :)
2^1,16=2,23457427614444000000
Спасибо, очень интересно. И очень странно, что без задания направления берет, хотя слева и справа не берет. Не должно быть так.
-1 я добавил сам, чтобы продемонстрировать функцию, которая в правой окрестности нуля имеет предельную точку в области определения (нуль), но сама область ее определения счетна. Т.е. функция не определена почти везде (термин "почти везде" вполне математический и означает "везде, кроме не более чем счетного множества" - конечно, если мы говорим об исходном множестве мощности континуум).
Загляни сюда, тут весь спор.
А преподам попробуй сначала дать первый предел, послушай, и, если будут считать, что существует, задай второй, с минус единицей. Обрати ихнее внимание на область определения второй функции.
Why be so radical about this function? Well, yes, it's a bit exotic, so what? The set of its values, although countable, is still infinite. After all, somewhere must the boundary between analog and discrete be drawn? Your function is the boundary - it exists only at the touch points of a modulated sine wave with some line.
Yurixx wrote (a) >>
Sergei, you have been the most hurt in this discussion. I understand that your "ego and conceit" have nothing to do with it, which makes it all the more interesting to ask why ?
And I also wonder why in the entire discussion you never once perceived or responded to a single physical argument. On the contrary, you have only been busy catching someone up on something.
And using the results of "voting" as an argument - waaaayyyyy.
And now, finally, the apotheosis - the switch to personalities.
Is it worth getting so worked up over nothing, Sergei?
I cited the vote as a reference to the argument and my observations and as a joke that there are slightly more of us :o) It was a joke - I thought you'd understand. There are several hundred pages of arguments about physical arguments, I just thought as if objectively - what are they arguing about. It's a shame it took me a while to figure it out.
I wasn't the first one to get personal, otherwise I wouldn't have reacted at all. But I am glad it ended so well.
to Mathemat.
I have a big request for Farnsworth and lea. Please check, if it's not a bummer, this limit on the same packages as before (Mathematica, Maple, MathCad - all three):
Not a bore at all. Mathematica - only after crashing (laptop crashed), saving data and stuff
PS: and MatCAD version M035 -ln(2).
But I'm glad it ended so well.
Ну зачем так радикально про эту функцию?
I see what you mean, alexeros. I didn't think of that right away :)
Except you don't have to write about 0.9999(9). It's still one. Infinite periodic fractions don't scare us.
2 Farnsworth: thank you, dear. I'm almost 100% convinced there's a limit.
Yurixx >> If it's 2, then we divide it by 2 and get 1111. If 6, we get 3333 in the same way.
It's a bit more vague: 92222/2 = 46111.
А 98888/8 = 12361. You're lucky, you still have one left at the end.
The funniest thing is that your reasoning seems like it should be correct for three identical digits, but it probably isn't. Looking for a counterargument.
Нечетных среди них быть не может. Четными могут быть только 2,4,6 и 8. Если 2, то делим число на 2 и получаем в конце 1111. Если 6, то аналогично получим 3333.