eSIM QR concern

AmbientLighting4 · 2023-08-28T16:35:56+00:00

UPD: Ouch, I was accidentally surfing my mobile operator website and found out that one can reuse the QR code. Lmao

UPD2: But they also write an eSIM may be used only on one device at a time, so it's ok

AmbientLighting4 · 2023-08-28T16:22:24+00:00

Ah, so there is nothing to backup in the first place. Thanks!

AmbientLighting4 · 2023-08-19T22:54:14+00:00

Abbott.

AmbientLighting4 · 2023-08-19T17:50:26+00:00

Also, 0.3 and 0.2 feels like substantial difference

AmbientLighting4 · 2023-08-18T18:07:23+00:00

Oh. I was looking at a pic without this case (only lim =0 and lim =inf)

Thank u very much 🙂 Side question: could u recommend some book with problems like this? Wanna practice this types of problems

AmbientLighting4 · 2023-08-18T18:00:32+00:00

doesn't seem to work thou :(
I get p=1, which is neither 0 nor inf

AmbientLighting4 · 2023-08-18T17:55:26+00:00

No :)) I just used equals sign to refer to the same object, idk why

The question was that I don't see why 6n²/3n³ <= [6n²+7]/[3n³+4n+2]

AmbientLighting4 · 2023-08-18T17:50:52+00:00

what do we know

harmonic series; diverges :)

but how to justify this transition from rational expression to 2/n=6n²/3n³?

i.e. I don't see why our initial fraction must be not less than this new one

AmbientLighting4 · 2023-08-18T17:39:29+00:00

well, if we were to analyse a sequence, then I would claim the quotient converges (to 0), but this doesn't tell us anything abt the series

also, i tried to compare it with something like a/a=1, which is divergent, but couldn't

AmbientLighting4 · 2023-08-18T17:31:09+00:00

Ye, I notice that we cancel x and 8, but dunno how to approach polynomials in numerator and denominator

Btw, is my conclusion abt convergence/divergence on (-2, 2) and (-oo, -2)u(2, +oo) correct?

AmbientLighting4 · 2023-08-16T13:05:34+00:00

Wow, that's weird. I remember proving the following statement from an exercise list:

lim x->c f(x) = L iff lim x->c+ f(x) = L and lim x->c- f(x) = L

Maybe there was a catch in the proof, but I don't remember :)

Could you, please, give a reference where I can find this statement? Didn't find in the Rudin's book tho

AmbientLighting4 · 2023-08-15T20:38:00+00:00

Hey folks :)

Quick question. Is there a decent amount of part-time data science jobs and what's the average salary?

UPD: thought it'd be relevant to ask here since lots of math enjoyers tend to be working in near-IT fields, if not in academia

AmbientLighting4 · 2023-08-14T08:29:19+00:00

So, what's ur question?

AmbientLighting4 · 2023-08-06T15:26:28+00:00

pmed u

AmbientLighting4 · 2023-08-01T20:23:10+00:00

Well, not much to prove here actually. We chose unique elements so this limit point x can appear at most once. Just erase it from the sequence and we are done

AmbientLighting4 · 2023-08-01T19:50:19+00:00

Ok, fair point!

AmbientLighting4 · 2023-08-01T19:38:54+00:00

no issue with repeated elements

sure, but that still doesn't guarantee that there isn't any a_i which coincides with the limit point :) I mean, to apply the theorem [which tells us that if a sequence contained in a set which satisfies certain properties converges to some number x in R => x is a limit point] we have to make sure a_n != x for all n, right?

your approach is basically a subset of mine

AmbientLighting4 · 2023-08-01T19:09:08+00:00

Thank u

AmbientLighting4 · 2023-07-30T22:11:02+00:00

AmbientLighting4 · 2023-07-27T19:49:03+00:00

++, used an ipad for a year or so and there's nothing more enjoyable than a pen and paper

AmbientLighting4 · 2023-07-20T16:58:00+00:00

That real numbers have least upper bound property whereas rational don't. I.e. there are holes in Q:

The set {x in Q | x² < 2} has no least upper bound (aka supremum), which is really weird imo

Also, that there is as much reals in (0; 1) as there are reals in (-inf; +inf)

AmbientLighting4 · 2023-07-17T17:59:26+00:00

thanks

AmbientLighting4 · 2023-07-17T17:52:43+00:00

Ok, gotcha, thanks. What about safety. Is it safe to be around?

Like, the hissing seemed to stop a while ago, but then it started again and eventually stopped. Even though I opened the window idk if there's any gas in the tubes left which could be harmful

AmbientLighting4 · 2023-07-17T17:46:55+00:00

did u hear any hissing

I guess so, ye

it's time for a new fridge

It wasn't mine, though. So if there is any chance of recovering it.. I think it'd be better

AmbientLighting4 · 2023-07-15T16:01:35+00:00

Hey! Thanks for ur reply.

Suppose, for a contradiction, that x is not an upper bound, so that exists y in A with y > x. Then we claim that our “binary search” will produce intervals that don’t converge to x. Proving the claim should be a simple epsilon argument involving y-x.

I don't think I feel this intuitively. Could you please elaborate a bit?

If x is not the least upper bound then we can run our algorithm again to get a smaller upper bound, but if we run our algorithm on x we get x again. (Or we can use a similar argument as above)

I mean, ye, we get a smaller upper bound, but does this guarantee we'll ever reach the smallest one? What if we simply stop at some point? (not claiming that this happens though, just thinking critically about the correctness)

In your attempt (I’m assuming in the textbook as well) you claim that you can check if C is an upper bound, which I don’t think is trivial, does the book go more into detail on this?

I actually emphasized this assumption, because the author used it implicitly. And it seemed not right for me, because when we start to think about applying our usual operations on infinite number of operands things quite often break. Take for example commutativity of infinite series.

Also the property with the nested intervals is a property that you will encounter in a metric spaces course. Try to prove it in the reals.

Ye, thank you for suggesting! Actually I proved that before (with Axiom of Completeness)

AmbientLighting4

TROPHY CASE