I'm learning real analysis for a short course in optimization, and this is the clearest explanation of lim sup that I've ever witnessed. You rock Dr Peyam!
I think the interesting sequences are ones like s_n = (1+1/n)cos n, where for all N, there's an n > N such that s_n > limsup s_n and s_n also doesn't converge. And limsup s_n isn't a term of the sequence, either.
You helped me think a little wiser about the limits and more... I don't subscribe to people easily but I did it in you just because of your love about maths(simple or more complicated,doesn't matter). Keep going. Hard work beats tallent!
My analysis prof constantly told us to "play with the soup" to prove things, etc. I can only ever think of lime soup. Also, we defined it as the inf of all the sups, rather than a limit of the sups. Seems to be the same thing. Do you have to mention the case where the sequence isn't bounded from below? Then the limsup would be -infinity.
My god you are a lifesaver. I came to grads school of statistics without proper math skills, and couldnt understand my professor going over limsup and liminf on day 1. You are perfect teacher
An excellent video, I have been struggling with understanding the intuition behind this concept; but you have answered all of my questions. I can't thank you enough!
@@josephhajj1570 I think the theorem demonstrates the existence of limit points in infinite bounded sets, not necessarily compact. In that case, I don’t know the proof either, but observe that infinity is not a sufficient condition: Z is infinite, although it doesn’t have any limit points. Hence, it seems that being bounded is another necessary condition.
To fully grasp the concept, realize that the sequence may exceed its limsup infinitely often, but only by amounts that approach zero. Such as s_n=(1+1/n)(-1)^n
Also of interest is that the limsup and liminf are the largest and smallest values that are limits of some subsequence. (Such limits can be proved to exist)
The lim in the limsup scared me (also the sup scared me), but the idea is straightforward. We define the limsup of a sequence s_1, s_2, ... , s_n, ... , as the limit of a sequence of sups, specifically, sup {s_1, s_2, ... }, sup {s_2, s_3, ...} , ..., sup { s_3, s_4, ... } , .. (this is made rigorous in the youtube video). The threshold-wall analogy is great. We can visualize the limsup as the sup of the graph of (n, s(n)) as you keep moving the wall to the right (so you cannot see initial terms to the left). Alternatively, fix the screen and move the points (n, s(n)) to the left, and take the sup of the terms remaining to the right (like a mario brothers game ). Moreover, at the risk of being handwavy, limsup n→∞ s_n = sup { s _∞ , s_∞ + 1, ... } . Note that the latter is only meant to be a heuristic, not a rigorous statement. Also note for the ultrafinitists, the symbol ∞ is intended to be merely suggestive of a very large number (in this context a very large integer) . This permits us to avoid the whole 'potential infinity' versus 'completed infinity' debate.
There is an alternate definition of lim sup: lim sup is defined is as the largest limit point (or accumulation point) of the sequence. Let's call this def 2 and yours def 1. I can prove def 1 implies def 2. But I am unable to prove def 2 implies def 1. Please help. Any relevant link is suffice. Thank you!
Great video! There seems to be one more option though besides the limit superior going to infinity or converging to a finite number: It could be negative infinity also.
what if the you have something like 1/x which has a pole at zero. if you define Sn=1/(n-10) for example at 10 it diverges but after that it goes down, so it's not bounded, but when n is large it's going to tend to 0. ?
Dr peyam, if a sequence is bounded, will it always be that the sup is larger before the lim sup before N? Is this why V is a decreasing sequence? What if it started off small before it started oscillating? The analogy if the class grades wouldn't apply then?
Hey Dr Peyam.. Could you please do a video which talks about lim sups and lim infs as the greatest and the least subsequential limits of a sequence? I understand both the definitions of lim sup separately (i.e 1) as the greatest subsequential limit and 2) in terms of helper sequences) but can't quite understand why these two characterizations are equal. :-/
What if a sequence was to increase and then oscillate, and the limsup is larger than any other element before N, does that mean that V(N) is still decreasing?
It’s still decreasing, since the larger N is, there are fewer values to compare to. The sup of 20 elements is always bigger than the sup of a subset with 10 elements
Mathematicians are always coming up with new definitions to sort out the singularities that former definitions cannot fix. Like cesaro sums, analytic continuity, lim sup and so on.
can you explain why the limsup is still decreasing when the sequence (for example, at 2:30) is monotonically increasing and is bounded above? I don't quite get that part when reading the textbook. I get the idea of the proof of why {a_n} (defined as sup{x_k : k greater equal to n}) is decreasing and {b_n} is increasing, but visually, what if the sequence is monotonically increasing and bounded above? Isn't the sup also increasing in that case? What is the inf? Thank you so much!
In the case of an increasing sequence the Sup would just be constant every time so it’s increasing in a degenerate case. The inf would be the value at N
What if you had a limsup such that it fit a function rather than a constant. Perhaps you could expand models of limsup with lim arbitrary function f(x)
Now, I'm no mathmatical wizard and my formal education in math is limited to the "need to know"-level of economists. Now my math professor in university was one of the few I bothered to listen to. But I do find the level of ignorance and interest disconcerting. I know the abysmal lack of comprehension for a fact. My farther did a course in pharmaco-kinetics shortly before retiring and I as a student was able to derive the fairly simple first order differential equations. The elimination of a drug is a straight line on single logarithmic paper. The problem were that neither his collegues nor the medical professor GOT IT! It simply flew into interstellar space over their pin-heads. There was a nobel laureate in economic that had a model of the price of shares based on the assumption that stock market quotes were normal distributed - probably because it was the only statistical distribution he knew. The result was as devastating as it was predictable: He ruined his business because in so far as the stock market quotes are statistically distributed they are NOT following a normal distribution. The Normal Distribution is an extreemly "slimtailed" distribution which does not take rare but very extreme occurences into consideration. Personally I have given up - I do not want to waste my life telling people that they are ignorant idiots, as they will never believe it - no matter what. So I have resigned myself to be the nasty disrespectfull person. There are no perks in being right.
It's a simple fix really. Just get your own Nobel prize because you can easily beat the chump who doesn't understand normal distributions and with the money open a more successful business which will never collapse in your lifetime. Only perks in being right are when you apply it to your own decisions.
@@drpeyam thank you Dr! Can you explain about how to use completeness axiom in finding the sup and inf of an interval? This analysis course is really struggling for me 😢
If S n is bounded Sup{S n | n> N} = 1 For all N Sup biggest value in sequence Lim sup Sn lim [Inf]= 1 Limsup[Inf] = sup (biggest value) converging in the long run Has to do with the helpers sequence Sequence decrease & non increasing Sometimes limit But the bright side Limsup[inf] always exists Lim Sup doesnt exist Infimum the smallest value Lim inf [Inf] {SN | n >N} Fact lim inf sequence fn Is Lim inf [Inf]=-lim sup [Inf] -Sn Recall inf S = - sup -S Long run Inf {Sn | n > N }= - sup{ - Sn | n >N } Taking lim of both sides Lim inf Sn[Inf] = - lim sup Sn[Inf]
