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  • Vivek

    Member
    July 28, 2021 at 6:10 pm
  • Vivek

    Member
    March 22, 2021 at 5:12 pm

    Nice, Artur 🙂

  • Vivek

    Member
    March 13, 2021 at 2:58 pm

    Fantastic, Artur. 🙂 You didn’t miss a thing. A minor typo though – first one is 8, not 4. So, 8,9,12,18,8p,12p

  • Vivek

    Member
    March 12, 2021 at 10:33 pm

    Finally, Artur 🙂 This is close to your idea….

  • Vivek

    Member
    March 10, 2021 at 11:01 am

    You are right about the existence of derivative. Even though the function is continuous, it may not be differentiable at corner/junction points. I think it can be avoided by taking one sided derivative, for the sake of this problem.

  • Vivek

    Member
    March 10, 2021 at 10:53 am

    Artur, For the assumption f(x) is differentiable, We only need that it is differentiable at right hand side and that I think is possible because of the way the function is defined. Like both f(x+h) and f(x) are going to exist.

    And for the second , I used the limit definition of right hand derivative. f'(x) is lim (f(x+h)-f(x))/h as h->0+ (right hand limit). Instead of h, We have 1/f(x) which also approaches 0 as f(x) goes to infinity.. wondering why we can’t use that?? Little bit confused here…

    Also we can prove that this limit indeed goes to -infinity by letting L>-a and getting contradiction.

    Anyway If this does not go to -infinity for any non decreasing f(x), Can we find f(x), such that L is not -infinity?

  • Vivek

    Member
    March 9, 2021 at 6:08 pm

    This is super confusing….

  • Vivek

    Member
    June 18, 2021 at 4:41 pm

    ha, Interesting Artur. Yes, I think I used the chinese remainder theorem. And the first such sequence is of length 17, I think. There are no such sequences of lesser length.

  • Vivek

    Member
    May 21, 2021 at 4:32 pm

    Hi Artur,

    We just need to find one pair (n,k) , preferably the least such pair ( as in dictionary order).

    The generalisation for all k larger than certain number is the interesting take !! (And That seems to be the case.)

    The original question was this :- Prove that in any 10 consecutive numbers, We can always find a number which is co-prime to all the other. Just extended it and searched for the sequence that first shows exception to this rule. It is the case with 11,12,13 and so on. But Soon we get the exception. And What is that first exceptional sequence?

    We can get it easily with programming, Artur 🙂 But, We can also get it analytically. It is the one good fun search !!

  • Vivek

    Member
    March 21, 2021 at 3:19 pm

    Yes, Artur 🙂

  • Vivek

    Member
    March 15, 2021 at 7:33 pm

    Hi Artur, Years back while we were learning about sorting, I thought of this game, like randomly adding certain powers like flipping two adjacent numbers, using functions. I used to play with mobile numbers and randomly generated numbers, with functions like f(5) meaning exchange 5th number with the next. Later programmed it in Android with buttons. Then , Tried this in html with the limited knowledge in html code. If you can improve it, It will be great , Artur 🙂 This one, I feel, is not that much appealing , like we have to zoom in, to get better view. Also there is no need for repeated numbers and the limit of 10 numbers , is due to the screen size. I think we can improve it a bit.

  • Vivek

    Member
    March 13, 2021 at 2:55 pm

    🙂 I have some minor confusions, Artur. Can we always find xn such that f(xn)= 2^n*T ? This may not be possible if f has jump discontinuities , right?

  • Vivek

    Member
    March 13, 2021 at 2:52 pm

    Thank you for the nice illustration, Artur 🙂

  • Vivek

    Member
    March 10, 2021 at 6:32 pm

    In the second part, I argued L>=0 is not possible. So, L<0. I think that is not completely true. L may not exist at all, oscillating between positive and negative values forever. Even in this case, The existence of negative value, proves our point.

    The problem is this contradicts with the earlier attempt in which I said L goes to -infinity for sure. So, Either that or this must be wrong. Now I understand there is some problem…

  • Vivek

    Member
    March 10, 2021 at 3:10 pm

    Confusing , Artur 🙁 I think I have to give some time gap to this problem. Can come to this freshly again later. I try to think about it. I too tried to get some counter example to better understand. Could not though. If you come across one, Please share it , Artur. Anyway the problem is already closed as you provided the solution. This is for the understanding of some concepts.

  • Vivek

    Member
    March 10, 2021 at 8:48 am

    **in the first image, That is for all x>M.

  • Vivek

    Member
    March 10, 2021 at 8:46 am

    You are right, Artur. We should be careful while dealing with infinities. It is certainly the case like when we have something like f(x+x^2)-2f(x). I thought we can approach in that way , if we have f(x+g(x))-2f(x) as g(x) goes to 0 as x->infinity. I tried to prove it from the scratch from the limit definition. But Not sure of it. I might be wrong. Anyway I tried the same in slightly different way. Let me know your opinion on this…

  • Vivek

    Member
    March 9, 2021 at 11:37 pm

    Artur, Based on the clue from your solution, to observe the tail end behaviour, I tried something. But not sure of its completeness. I might have missed something.

    Let G(x)=f(x+1/f(x))-2f(x)

    lim G(x) as x->infinity is

    lim f(x+1/f(x)) – 2 * lim f(x)

    Suppose f(x)->infinity, lim f(x+1/f(x))=lim f(x) as x->infinity

    So, lim G(x) = -lim f(x) <0 , as f(x)>0

    So, There must exist G(x)<0, that is f(x+1/f(x))<2f(x)

    Something incomplete here….??

    *This is exactly your solution as in part 1, but trying to extend that to part 2 as well…

  • Vivek

    Member
    March 9, 2021 at 6:00 pm

    Fantastic , Artur!! Good Job…Love it 🙂

  • Vivek

    Member
    March 9, 2021 at 5:59 pm

    James, There is a solution below. Yet Let us know if you find some other way to solve it.

Viewing 1 - 20 of 51 posts