Mathematics - 2018-08-26 (page 1 of 3)

Daminark

00:18

@Symposium why those two subjects exactly? This is interesting

Oh shit they have Toby Gee. I dunno much about him but I think he's a serious person for sure

Symposium

@Daminark I have no idea why those two subjects only!

They have Fred Diamond too.

Daminark

Not sure who that guy is, though I picked up a book on modular forms by "Diamond and Shurman" so there's that

Symposium

I'm pretty sure that's him.

Mary Star

00:34

@user31415 How could we show that? Could you give me a hint?

user 2646

To make it sound even more impressive, they should've added "analytic" to number theory :P

Daminark

:0

loch

lsgnt is very good!

Jasper Loy

00:50

ANT can stand for both analytic number theory and algebraic number theory. =D

Symposium

@loch Do/did you go there? I searched 'lsgnt' in chat and I read you suggest to someone that they could perhaps sneak into the lectures given under LSGNT.

I may do that when I'm feeling like a rebel/adventurous.

loch

01:31

i studied in london so i've heard of lsgnt

i guess if you want to lean on the safe side you can always email people and ask if you can sit intheir lectures

uhoh

03:41

1

Q: Can Lissajous orbits have stable/unstable manifolds?

The question Did DSCOVR travel “along the stable manifold of it's future SE L1 Halo orbit” to get there? is specific to DSCOVR's trajectory from Earth to its primarily heliocentric orbit near Sun-Earth L1, which is a Lissajous oribit. Here I would just like to ask the general question: Can Lissa...

any thoughts? (see also meta)

Vrouvrou

05:59

@mercio sorry problem of internet

Asaf Karagila

08:25

@LeakyNun That's like saying that major problems in number theory have been decided because we have a proof of Fermat's Last Theorem.

Jasper Loy

08:38

@AsafKaragila Congrats on being a moderator. The last time we talked was years ago.

jmindel

Question... if I have a set of numbers, is it possible to use only addition, subtraction, and multiplication to get to any other number? And if so, how? (This might be better asked as an actual SE post, but wanted to see if anyone had any ideas.) Thanks!

Jasper Loy

Hello @MatsGranvik. I hope you are well.

@jmindel This is an interesting question to me. Interestingly, I have never thought about it.

Nick

08:54

math is awesome!

Jasper Loy

Hello @Nick. I hope you are well.

jmindel

@JasperLoy Nor had I until recently! It's an odd one.

I was hoping to use something like the fundamental theorem of arithmetic recursively, but it still doesn't give me enough constraints to determine which numbers should be applied as b, q, and r first in order to eventually create an a = bq + r where a = the desired number.

Jasper Loy

@jmindel Of course, if we can get 1, we can add 1 to get any number.

jmindel

Right. Unfortunately, 1 is not part of the set.

The numbers in this particular set are ASCII values--they tend to range from 32 to 120.

Jasper Loy

I mean if we can get 1 by doing operations, yeah.

jmindel

09:02

Oh, understood.

Interesting, then...

Jasper Loy

And in fact, that might be the way if there is a way.

jmindel

So I should try to get to the smallest value possible, and then repeatedly add it or add one of its multiples in order to get to the desired number?

That doesn't quite work as I'd hoped... but it's interesting.

Jasper Loy

Who knows? That's why we have a question. =D

usukidoll

Got a question

Jasper Loy

Hello @usukidoll. I hope you are well.

jmindel

09:04

I had originally hoped that I would only be able to use each number in the set once in the final calculation.

Which is unrealistic, but I'm still curious... and yes! Exciting to have a new question. :)

usukidoll

If F is a collection of subsets of the sample space. Could the converse of this statement be true? If A is in F then Omega/A is in F. Omega/A is also known as not A in probability and I'm wondering if the complement is in F then so is A?

On a phone so I can't do fancy F. So yeah I'm aware that regular F means field

Leaky Nun

@JasperLoy you're gonna need the planets to align to see him again

usukidoll

I could take a pic...

Jasper Loy

@usukidoll The complement of the complement is the set itself.

usukidoll

? Yeah I know but I'm asking if definition 1.3 is also true in reverse. I gotta request desktop version on my phone to upload

-_- latex save me

Jasper Loy

09:12

@usukidoll If you have the property that A is in F implies A complement is in F, then you have the property that A complement is in F implies A is in F, because the complement of A complement is A. I hope I did not misunderstand.

usukidoll

The collection F of subsets of the sample space $\omega$ is called an event space if

If $ A \in F$ then $\omega \backslash A \in F$

$\omega \backslash A = A^{c}$

Jasper Loy

Yes, we are taking complements with respect to the space omega, and A is an element of the collection F of subsets of omega.

usukidoll

Is the converse true? I think it is.

I mean if A is in the event space then not A is also in there so if not A is in F then A is also in F. I have to deal with A\B so I was thinking hat if not B is in the event space then so is B

Jasper Loy

A\B is just A intersect B complement.

usukidoll

Yeah I know

But if not B is in the event space is B also in the event space?

Jasper Loy

09:16

To answer your question, if A complement is in F, then A is in F, because the property implies that A which is A complement complement is in F.

Well B is just a subset of omega, right? Same thing. Replace A by B, just a dummy variable in this context.

usukidoll

Ah mmk. I was doing this earlier and I thought well... Consider $A \backslash B = A \cap B^{c}$. By taking the compliment I have $ (A \cap B^{c})^{c}$ which becomes $A^{c} \cup B$ but I'm wondering if that's it to prove it as an event space or os there more to it?

Oh gawd complement

Doing $(A^{c} \cup B)^{c}$ returns it back to where I began

Jasper Loy

What do you want to prove now exactly? As far as your original question goes, I think we answered it.

usukidoll

Wanted to prove that A\B is in the event space

Wonder if there is more needed or did whatever I text is enough. I got a strange feeling that's not enough idk

Jasper Loy

I think that's all you need.

I guess you are doing a course in probability/measure theory, where you are dealing with a probability space or measure space.

usukidoll

Yup

Also vector analysis but I finished that assignment already

Jasper Loy

09:27

Standard 4th year undergrad or 1st year grad course.

usukidoll

I'm a postbac...for the third semester now

Jasper Loy

A very good book I have come across that covers measure-theoretic probability and stochastic processes is Knowing the Odds by John Walsh, in the AMS GSM series.

And a very good book I have come across that covers measure theory and integration itself is Real Analysis by James Yeh.

Hello @Secret I hope you are well.

Nick

09:39

@JasperLoy And I hope you are well and recovered.

I like the new lighter shade of blue you have going

Jasper Loy

@Nick I am not, alas. But still trying.

Nick

@JasperLoy I actually had a phase for quite some time. I believed I was on some version of the Truman Show.

Secret

pretty fine except busy with PhD

Leaky Nun

10:06

@loch hi

what is the future of set theory?

loch

Hi @LeakyNun

I know nothing about foundations lol

Leaky Nun

I talked to my professor about it

and he sounds like set theory has no future

he strongly advises against me studying it

but hey, one of our mods completed a set-theory post-doc

2 hours ago, by Asaf Karagila

@LeakyNun That's like saying that major problems in number theory have been decided because we have a proof of Fermat's Last Theorem.

loch

Oh career wise i think the problem is its not very popular

Leaky Nun

so did one of our mods make a poor decision?

Secret

Set theory can have a future only if there is a way to find an uncountable physical infinity

Leaky Nun

10:14

@Secret i'm serious

Secret

as without that, we cannot motivate the uncountables beyond the set theoric framework and counterexamples

loch

I wouldnt say its a poor decision - many would say math phd is a poor decision because academia is hard and other jobs are easier

Secret

The problem I get from set theory (at least ZFC) is that too many of the uncountable infinities don't have applications outside of set theory other than to construct counterexamples

so even from a pure maths perspective set theory looks grim because its framework is too localised

loch

but even if i dont go into academia in the end i dont think its a poor decision? Although we’ll probbaly have to wait a few years later to find out :p

Jasper Loy

I don't know anything about academia but you should do what you love @LeakyNun.

user91500

10:16

@JasperLoy Hi jasper, I was looking for you several days ago!

Jasper Loy

@user91500 Hello, is anything the matter?

Secret

But again, I am not a set theorist. I will still have the optimism that fundemental research is very important, and that includes set theory, so don't give up on it

Too many governments nowadays not focus enough on fundemental research

loch

Probably objectively you can say there are more positions in many other fields compared to logic — and leave the judgement of whether the decision is good or not to the individual

user91500

@JasperLoy Not really, just for fun!

Leaky Nun

I am the individual

Jasper Loy

10:18

The nicest set theory book I know about is Thomas Jech's Set Theory, but before that you would need to read Thomas Jech and Karel Hrbacek's Introduction to Set Theory @LeakyNun.

loch

Then it depends on your priorities!

Leaky Nun

@JasperLoy thanks

Jasper Loy

Whichever field you choose it's gonna be hard anyway.

Be yourself and be original. Enough copycats.

Secret

I hate copycats, this reality is already filled with them

the last thing humanity should produce is more copycats that arrests innovation

And to counter copy cat, we need more people to take risk

so, do whatever you want to do, don't care what the Establishment said

loch

I think your prof’s point is just that if your dream is to work in math , youre more likely to be able to do that if you work in a field thats more popular/ has questions fhat more people care about

Secret

10:22

@loch Actually.. do you knew of any direct application of set theory? Most applications I heard from often only involve its framework to construct mathematical object, but not really making use of the objects themselves?

loch

I know very little about foundations lol so idk - probably also depends on what you mean by application

Secret

well, just theorems in pure maths will suffice, almost everything I knew in other maths disciplines that make use of set theory objects only ever make use of the axiom of choice, what about things like nonmeasurable sets, inaccessibles and so on?

I never saw things like $\omega_1$, $\aleph_{x > 2}$ etc. being used in anything other than counterexamples constructions

It really annoys me that something as awesome as nonmeasurable sets is not mentioned anywhere other than proving the banarch tarski paradox

loch

22

A: What practical applications does set theory have?

Set theory is an extremely convenient language for being able to rigorously define and manipulate various "completed infinities" - not only just infinite sets such as the natural numbers or real numbers, but much "larger" completed infinities, such as Stone-Cech compactifications, the hyperreals,...

Secret

Stone–Čech compactification

In the mathematical discipline of general topology, Stone–Čech compactification (or Čech–Stone compactification) is a technique for constructing a universal map from a topological space X to a compact Hausdorff space βX. The Stone–Čech compactification βX of a topological space X is the largest compact Hausdorff space "generated" by X, in the sense that any map from X to a compact Hausdorff space factors through βX (in a unique way). If X is a Tychonoff space then the map from X to its image in βX is a homeomorphism, so X can be thought of as a (dense) subspace of βX. For general topological...

hmm... that fell out of my radar, need to study more...

Leaky Nun

@KasmirKhaan til GIFTER and GIFter

@loch fair enough

Secret

10:35

I am attempting to apply very large cardinals to public key cryptography, but since these developments are in their rudimentary stages, it is hard to say how well these cryptosystems will fare (and hence I give a comment here instead of an answer). — 35093731895230467514051 May 9 '17 at 2:52

hmm...

Leaky Nun

plot twist: username is an encrypted message

Secret

17

Q: What are the most prominent uses of transfinite induction outside of set theory?

What are the most prominent uses of transfinite induction in fields of mathematics other than set theory? (Was it used in Cantor's investigations of trigonometric series?)

set-theory big-list ordinals applications transfinite-induction

Leaky Nun

also why are three Hongkongers talking in English

Secret

because it is supremely slower to type chinese on a computer at my end. Also that others may be interested in the conversation and they are almost always english speakers

citeseerx.ist.psu.edu/viewdoc/…

loch

TIL @Secret is from hong kong

Secret

10:39

> It is well known that some results about finite constructions can be obtained
only by using infinite concepts (Goodstein theorem for example). Many results in
number theory (about properties of integers) have been proved by using a natural
framework with infinite notions (for example the prime number theorem, with
holomorphic functions, even if it is possible to prove it only from the Peano
arithmetic).

ok, that sorts of consistent with a certain blog post by Terry Tao, that there exists correspondence between finite sentences and infinite sentences, so infinite objects can be a shortcut in the space of all mathematics to reach those otherwise hard to reach finitary theorems

I guess for now, that's the major route that uncountable and inaccessible sets can found application outside of set theory

Alessandro Codenotti

Who is the third Hongkonger?

Leaky Nun

@AlessandroCodenotti logic exercise :P

user 2646

10:57

i think Terrance Tao's family is from Hong Kong

Jasper Loy

Terence, not Terrance.

Hong Kong is part of China, and so is Taiwan, according to China. =)

Leaky Nun

>_>

Jasper Loy

11:13

China has over 1 billion people. Maybe more Chinese learn English as a second language than all native English speakers combined.

Easier for Chinese to learn English than English to learn Chinese.

But according to effectivelanguagelearning.com the hardest language for an English speaker to learn is Japanese.

usukidoll

I am so confused. I need a hint assuming that this image I took from my phone will upload. Just 1-11 . 1-12 isn't required and I already did 1-8 through 1-10

Jasper Loy

I hope it loads.

usukidoll

Ughhhhhhh. Hold on I'll get a URL out of this thing brb

user 2646

<_<

Mary Star

From 16 men and 12 omen we want to create a committee of 10 people, with at least 3 men and at least 5 women. In how many ways can we create such a committee (men and women are discrete) ?

For that do we have to calculate the number of ways without restrictions minus th enumber of ways that the number of men is less than 3 and th enumber of women less than 5?

usukidoll

11:20

@JasperLoy try this ibb.co/gUuHsp

12 omens? That's one scary committee

Uh oh

Gone 😭

Jasper Loy

@MaryStar Hello Mary. I see you are still a star. =D

Mary Star

@JasperLoy Yes I am :D

Jasper Loy

@usukidoll It is just a matter of writing out this set in terms of some unions, intersections, and complements. Just bothersome notation, but a simple exercise.

Alessandro Codenotti

@LeakyNun loch?

Leaky Nun

right

usukidoll

11:26

😥

Alessandro Codenotti

I see, I didn't know

usukidoll

Is that the family set index of union families?

Jasper Loy

@usukidoll Maybe try it out with a specific example first and then you will see the general pattern. Say m=3 and k=2 for example.

usukidoll

That's for all the A_1, A_2 things?

Jasper Loy

Yes. Taking the set operations on those subsets of omega.

usukidoll

11:30

I did the previous exercise but I don't see the link to it. It said m= 2 and k = 1 but where did that come from in #10? I just used parts of the event space definition

Avantgarde

@LeakyNun Look back at your question and try to answer it then.

Jasper Loy

@usukidoll If there are only two sets and you want it to belong to one of them, then that means A intersect B complement! Of course you also need B intersect A complement. Then take the union of the two.

Leaky Nun

@Avantgarde no u

Avantgarde

@LeakyNun I was replying to your question

"what will you be doing after 10 years?"

Leaky Nun

i was memeing

Avantgarde

11:34

:(

usukidoll

Oooohhhh

$A \cap B^{c} \cup B \cap A^{c}$

Jasper Loy

@LeakyNun The soprano in Puccini's La Bohemia sings the character Mimi.

usukidoll

$(A \cap B^{c} ) \cup (B \cap A^{c})$

Jasper Loy

@usukidoll Yes that represents the points belonging to exactly one of A and B, no more no less.

usukidoll

Right

Sooo... Do I take the complement of that again? How does that related to m=3 and k=2?

Jasper Loy

11:38

Well, that's an exercise for you.

Just draw some Venn diagrams to help you figure out if it helps.

usukidoll

If I take the complement it's $( A^{c} \cup B) \cup (B^{c} \cup A)$

user91500

@JasperLoy I have a question?

usukidoll

Taking the union is all of the points in A and B

Jasper Loy

@user91500 What is it? I can only answer personal questions, not math questions, because I have forgotten all the math I know. =D

usukidoll

D=

user91500

11:41

@JasperLoy Do you like the new theme for MSE?

Jasper Loy

@user91500 I see they moved some buttons to the left from the top, but other than that I don't see much change. It's fine for me, but I do prefer a more colourful them to replace the existing one, which is too white for me overall.

user91500

@JasperLoy So please downvote here math.meta.stackexchange.com/questions/28842/…

Jasper Loy

@user91500 I see. I prefer not to cast votes on those posts, because I would need to think of the ideal theme for the site for myself before judging each of the other possibilities individually, and I don't want to do that now.

user91500

@JasperLoy But you can't downvote just because you do not have enough reputations :)

Jasper Loy

@user91500 LOL. Yeah.

user91500

11:50

@JasperLoy I recall you later to downvote!

Jasper Loy

@usukidoll If there are three sets you can use say A, B, C. Think A intersect B intersect C complement...

@user91500 LOL. OK. I will see you around more in future hopefully.

user91500

@JasperLoy I will be here until 30 september 2018!

Jasper Loy

@user91500 I see. Would you like to email me?

user91500

@JasperLoy I think you will have enough reputations until that date!

Jasper Loy

@user91500 Oh OK. If you want to email me in future just let me know. Otherwise I won't tell you my new email. =D

Oskar Tegby

12:01

@TedShifrin What happens when we have that e.g. $M_n=1$ for $x_n=0$? How do we know how close to $0$ we have uniform convergence? We can check if $f_n(x)\to f(x)$ as $n\to\infty$ for $x\in(a,b)\backslash\{0\}$ where $0\in(a,b)$ to see that it is uniformly convergent for certain points, but how do we find the exact point where we go from having uniform convergence to not having it?

usukidoll

12:19

Sooo if I have three sets doesn't the symmetric difference formula change as well? Or are we just considering three sets A,B,C. Then if we take all of the intersections then $A \cap B \cap C$ which represents all of the points in A B and C. But of I take the complement that would mean $(A^{c} \cup B^{c} \cup C^{c})$ which all of the points that aren't in sets A B and C

I'm assuming that m must be the number of sets since the symmetric difference formula is only A and B but I still can't figure out what k is

Besides being a positive integer. So no zeros and no negative numbers

Jasper Loy

You should think of the set that the question asks for.

In this case, you want the points that are in exactly 2 out of the 3.

usukidoll

Which is I don't know where it is unless it's that A string that they mentioned

So wait maybe it's something like

Jasper Loy

Case 1: In A and B but not in C

usukidoll

A1, A2,...,

Jasper Loy

Case 2: In A and C but not in B

usukidoll

12:22

B1, B2,...,

Jasper Loy

Case 3: In B and C but not in A.

usukidoll

C1, C2,...

Jasper Loy

So the A, B, C is really your A1, A2, A3.

I just used A, B, C for convenience.

usukidoll

D= so could it be the symmetric difference of A+B, A+C, and B+C?

Jasper Loy

It's just the union of the three cases above!

usukidoll

12:24

The union of A+B, B+C, and A+C? One of the letters is gonna be left out in each case

Either no C's, no B's, or no A's

Jasper Loy

In Case 1, it is just A intersect B intersect C complement!

usukidoll

Which is $A^{c} \cup B^{c} \cup C^{c}$

The union of all points not in A,B, or C

Jasper Loy

@usukidoll No you don't need to think of that set. Just combine the three cases I wrote above.

usukidoll

All three cases combined together are the sets A,B, and C

Jasper Loy

If you know that the sets are closed under complements, countable intersections, and countable unions, you are done.

For any value of m and k, you can always write the required set in terms of these set operations.

usukidoll

12:33

Oh wait a sec. I have to use the definition in the book. That F is nonempty which means the event space contains the empty set and the whole set. So if there exists A,B,C in F then all of their complements also exist in F.

Jasper Loy

Hmm, maybe you misunderstood the question...

usukidoll

I don't understand what it's asking

I understood 1-8 through 1-10

That was the definition

Jasper Loy

OK, now there are m subsets of omega.

usukidoll

I took the complements and got them. The way 1-11 is written is getting me confused.

Jasper Loy

And we have 1 less than or equals k less than or equals m.

So far so good?

These subsets are of course labelled A1, ..., Am.

usukidoll

12:36

Yeah there in F

Jasper Loy

Now think hard about what the set containing precisely the points belonging to exactly k of these m sets means.

If we use A B C and set m=3 and k=2 for example.

usukidoll

Hmm.. 2 of these 3 sets

Jasper Loy

Then we want (A intersect B intersect C') union (A intersect B' intersect C) union (A' intersect B intersect C).

Now A, B, and C are in F. And since F is closed under complements, countable intersections, and countable unions, this set is also in F. QED.

usukidoll

Damn it. The words in the problem that was being asked tripped me up x.x

Jasper Loy

Yeah, sometimes you just need to really understand the question, just like sometimes you just need to really understand the paragraph in the textbook. =)

Math is about words more than it is about symbols. =)

Symbols are arbitrarily assigned, but words do mean something. =)

usukidoll

12:42

OMG that's what the m= 2 and k= 1 was because we had two sets A and B so A+B= A\B or B\A. That means for the two sets we either have points in A or points on B. Over here we have 3 sets with two points. Son of aaaaaaaaaa

Jasper Loy

Yyyyeeeaaaahhh

usukidoll

Fffffffffffffffffffff....that was a trick question type of thing . I thought is uses variations of the symmetric difference

Jasper Loy

I am glad you get it now.

The problem you describe is one of how to generalise lower-dimensional results to higher-dimensional analogues.

Sometimes we need some time to figure out how or what to generalise.

usukidoll

So the pattern will be like m= 2,3,4... and k=1,2,3 and it's like even though we have more sets and points it will still be closed under countable unions, countable intersections, and complements?

Jasper Loy

Yes, because that is the property of a probability space or measure space isn't it?

usukidoll

12:54

Yeah

Jasper Loy

As long as you have 1\leq k\leq m that question is asking you to show.

Now you just need to write it out using the proper notation.

usukidoll

So I have to produce a generalized version of this. Like it does belong to exactly k of the Airport

Omg

I meant Ai

XD

Jasper Loy

airport, lol, wtf

usukidoll

Autocorrect

Jasper Loy

That's why I turn off all spellcheck and autocorrect on all my hardware and software.

usukidoll

12:57

So does that mean do 26 sets like the letters of the alphabet 0.0

Jasper Loy

Yeah, just that 26 is only one number, not any number. You just need to write it out properly using things like 1,...,k,...,m you know

usukidoll

Yeah because I need an overall version not just a specific let's pick a number case

RockDock

I have a doubt , suppose in simplex method of LPP , in ratio test , one ratio i got is zero .Should i choose that corresponding row as a pivot row or we can choose only positive ratio ( IN MAXIMISATION PROBLEM ) ?

usukidoll

Let m be the number of sets and k be the number of points

Jasper Loy

Yeah and you wrote correctly, 26 letters of 1 alphabet, not 26 alphabets.

@usukidoll Nope, k is not the number of points. It is the number of sets also, lol.

usukidoll

13:00

F. So let k and m be the number of sets

Jasper Loy

You can have more than one element in each of those k sets ya know.

You have a big set omega.

And then you have m sets, and then you have k of these m sets.

usukidoll

And then we shall prove that it belongs to an event space

That it being these sets

Like...consider an Omega set. Suppose we have m sets and k of m sets???

Weeeeeeeeeeeeeeeeeeeeee

Jasper Loy

Hmm I think you can write down yourself everything now. Just take your time and be as precise as possible.

usukidoll

Oh wait ... Suppose we have m sets and k of m sets . We need to prove that it belongs to an event space

Jasper Loy

One way to write good proofs is to read good books and see how proofs are written. Read more books and you will be able to write better.

usukidoll

13:04

And then that string of A1,A2,...Am

Jasper Loy

This may sound superficial, but there are two steps to writing a proof clearly...

usukidoll

And then B1,B2,...,Bm and then C1,C2,...,Cm

Eiiaeiiwf

Jasper Loy

Step 1 is to be clear what the argument really is, mathematically.

And Step 2 is to express that argument in words as clearly as you can.

usukidoll

I'm trying to. Kinda sucks that when I was a community

Jasper Loy

You need to get Step 1 right first. And then go to Step 2.

usukidoll

13:06

college student I never took technical writing

So I'm using freelancer writing to get a proof together x.x

Jasper Loy

To take a technical writing class gives you some practice, but really you can train yourself by developing good habits of mind, patterns of thinking.

usukidoll

:/

I took Discrete Math and got the same grade as that proof writing course which was a B-

So I must be doing something on track. 🤔

Unless my uni really sucks and don't care about quality control

Jasper Loy

I think the best thing for you to do is to read good mathematical exposition and try to emulate it.

Lecture notes written by professors are not meant to be proper published works.

usukidoll

So if m = 4 and k = 3 ABCD... ABC,ACD,BCD no D, no B, no A

Jasper Loy

I am leaving the room now, byeeee.

usukidoll

13:13

Byyyyeee

Thanks for helping :)

Mancala

13:28

I only managed to show the existence of the fields, but I could not show the rest, that is, first order frame field and the others ... math.stackexchange.com/questions/2894490/…

Leaky Nun

@Mancala fieldn't

Mancala

13:42

@LeakyNun I am referring to the smooth vector fields

Leaky Nun

"field theory" is for abstract algebra

I've removed the tag for you

Mancala

Ah yes! Thanks

Oskar Tegby

13:54

How do I show that $f(s)=1/n^s$ is differentiable? Here, $n\in\Bbb{N}$.

Oskar Tegby

14:21

@TedShifrin I get stuck at $$\lim_{h\to0}\frac{1-n^h}{n^{s+h}}/h,$$ and don't arrive to $-n^{-s}\log n$ as I should do.

Holo

$m:=1/n\implies f(s)=m^s$?

Oskar Tegby

?

I don't need to compute the derivative. I just need to show that it exists. Namely, that the left and right limit are the same for all $s$.

Holo

So what is your definition of $a^x?$

@OskarTegby do you use the definition of $a^x=\exp(x\ln a):=\lim\left(1+\frac{x\ln a}{k}\right)^k$?

Oskar Tegby

14:37

I haven't this far. Maybe that's what I need to do.

Holo

There are other definitions to, but you have to have a definition to be able to do show that

Oskar Tegby

Yeah! That makes sense. Thanks, @Holo.

Holo

No problem @OskarTegby , but you can advance even without the definition using the fact that $a^{c+b}=a^ca^b$ and $a^0=1$ for $a\ne 0$: $$\lim_h \frac{m^{x+h}-m^x}{h}=\lim_h \frac{m^xm^h-m^x}{h}=\lim_h \frac{m^x(m^h-1)}{h}=m^x\lim_h \frac{m^h-m^0}{h}=m^x\lim_h \frac{m^{0+h}-m^0}{h}$$So you only need to show that $m^s$ is differentiable at $0$(Now using the definition)

tatan

I am doing definite integrals. But I am facing problems in breaking the limits while working with inverse trigonometric functions. Is there an easy intuitive way I can use to remember the variation of the inverse trig functions. For eg. tan^-1(2x/1-x^2) has different expressions for x<-1,x \in [-1,1] and x>+1. Is there an easy way to find that out?

Oskar Tegby

14:52

Okay, but dont we get $0/0$ when $h\to0$ in the above?

Do you mean the definition of Euler's number?

Holo

Both Euler's number and the definition of exponential

Oskar Tegby

Hm... Can you show me?

Holo

@OskarTegby yes we get $0/0$, so you need to go to the definition, using the definition I gave you can show that $1+x\ln a\le \exp(x\ln a) \le 1+x\ln a+(x\ln a)^2$, so you have $$\lim_h \frac{m^{0+h}-m^0}{h}=\lim_h \frac{\exp(h\ln m)-1}{h}$$Now using the inequality you have $$\lim_h \frac{1+h\ln m-1}{h}=\ln m\le \lim_h \frac{\exp(h\ln m)-1}{h}\le \lim_h \frac{1+h\ln m+(h\ln m)^2-1}{h}=\\\lim_h \frac{h(\ln m+h(\ln m)^2)}{h}=\lim_h (\ln m+h(\ln m)^2)=\ln m$$

So by the squeeze theorem $\lim_h \frac{m^{0+h}-m^0}{h}=\ln m$

The inequality holds only when $0<x\ln a<1$

Secret

15:43

5

Q: Ultrainfinitism, or a step beyond the transfinite

Cantor has, in the immortal words of D. Hilbert, given all of us a paradise (or perhaps, I would rather say, a great vacation spot), the TRANSFINITE. $\aleph_0, \aleph_1,\aleph_2\dots$ the lists goes on forever, into higher and higher ethereal realms. In his theological mind, Cantor thought ...

set-theory lo.logic large-cardinals foundations mathematical-philosophy

2-classes wtf

Fréchet filter

In mathematics, the Fréchet filter, also called the cofinite filter, on a set is a special subset of the set's power set. A member of this power set is in the Fréchet filter if and only if its complement in the power set is finite. This is of interest in topology, where filters originated, and relates to order and lattice theory because a set's power set is a partially ordered set (and more specifically, a lattice) under set inclusion. The Fréchet filter is named after the French mathematician Maurice Fréchet (1878-1973), who worked in topology. It is alternatively called a cofinite filter because...

Finite intersection property

In general topology, a branch of mathematics, a collection A of subsets of a set X is said to have the finite intersection property (FIP) if the intersection over any finite subcollection of A is nonempty. It has the strong finite intersection property (SFIP) if the intersection over any finite subcollection of A is infinite. A centered system of sets is a collection of sets with the finite intersection property. == Definition == Let X {\displaystyle X} be a set with A = ...

free ultrafilters are not bounded from below

it's turtles all the way down

Alessandro Codenotti

16:13

Hi @Paul

Paul Plummer

16:30

@AlessandroCodenotti Hi

Oskar Tegby

16:49

@Holo I don't see how the inequality follows from the definition.

We can of course add $(x\ln a)^2$ an obtain something larger, and the lower limit is just when $k=1$, but how do we know that the last inequality is true?

Holo

Can you see why $1+a\le \left(1+\frac an\right)^n$? @OskarTegby

Oskar Tegby

Not straight away.

Holo

It is Bernoulli's inequality

@OskarTegby look here math.stackexchange.com/questions/475309/…

Oskar Tegby

I see. Thank you!

I'm doing mathematics right now, but I'm stilling missing doing mathematics because I want to look into other parts of it. I wish more people had this problem.

Holo

17:05

@OskarTegby the other side goes like this: $$\left(1+\frac an\right)^n\le 1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+...+\frac{x^n}{n!}$$This can be proven using binomial theorem. from this you do the following:

Oskar Tegby

Cool! Thanks, Holo. I appreciate it a lot.

yasar

17:31

To estimate number of decimal digits in $2^n$ I am the formula $n/3$, but in big numbers it estimates too highly. Do you know better way to estimate it without using a calculator or computers?

According to math.stackexchange.com/questions/177973/… my estimation should hold. But n=1000 has 302 decimal digits, whereas formula estimates 333

mercio

according to the accepted answer your estimation should not hold

Leaky Nun

@yasar well log10(2) is more like 0.301 so ok

0.30102 iirc

Abdullah UYU

As the name implies it is an estimation.

yasar

@mercio For some reason, my brain interpreted 0.3 as one-third.

Sorry about that

mercio

1/3 is 0.333 so yeah 0.3 would be closer

Holo

17:45

Use $n\cdot0.30102999566398119521373889472449302676818988146210854131$

Leaky Nun

no u

yasar

@Holo No way I am going to calculate that without a calculator :D

Holo

@yasar so take as many digits you want from this. The more you take the more accurate it will be

yasar

@Holo I am going to go with 0.3

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