Vivek
TeacherForum Replies Created
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Nice, Artur 🙂
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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
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Finally, Artur 🙂 This is close to your idea….
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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.
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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?
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This is super confusing….
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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.
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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 !!
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Yes, Artur 🙂
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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.
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🙂 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?
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Thank you for the nice illustration, Artur 🙂
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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…
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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.
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**in the first image, That is for all x>M.
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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…
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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…
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Fantastic , Artur!! Good Job…Love it 🙂
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James, There is a solution below. Yet Let us know if you find some other way to solve it.
