Mathematics - 2022-06-02

Xander Henderson

00:00

@D.C.theIII I should probably flag it for moderator attention.

Is there a moderator here who can protect me from the horrible depredations of @TedShifrin?!

Okay, I gots ta go make diners. Buh-bye.

robjohn

I'll alert the room owner.

Ted...

Ted Shifrin

00:15

You summoned?

DC, I think you’ve received more venom than anyone else.

Why did you bother changing your name for just a roman numeral?

leslie townes

00:38

it's a sponsored promotion of lesliecoin III. it's the third iteration of our cryptocurrency. the second was based on a fork of the original blockchain after a cybercriminal going by the name shed tifrin made off with most of our assets. the second was a switch from proof-of-work to proof-of-laziness, which we did in the name of environmental friendliness.

the third, rather. this is really deep crypto stuff, a lot of people wouldn't understand it.

Ted Shifrin

Some of us couldn’t care less.

leslie townes

we have a lot of promotions in store for june. june 22 is the month of lesliecoin.

Ted Shifrin

A day = a month?

leslie townes

whoops. june 2022.

D.C. the III

@TedShifrin the former chap was deviant, this individual is more refined and distinguished.....

leslie townes

00:47

or june '22.

or as some would have it june MMXXII.

Ted Shifrin

I see no evidence, DC

leslie townes

this year looks pretty cool in roman numerals. the 1980s were a mess.

Ted Shifrin

I only like palindromic :)

leslie townes

there's a bridge on the berkeley campus from 1914 or 1924 or something like that where they write IIII instead of IV because that was an alternate style and maybe regarded as more official at one time. i noticed that on one of my first visits there because it struck me as odd.

it went against everything i'd been told about roman numerals.

hippies with a chisel

D.C. the III

@leslietownes iconoclasts

copper.hat

03:53

I have never seen IIII. They goofed and did a McCarthy on it.

robjohn

04:19

Why Do Clocks and Watches Use the Roman Numeral IIII instead of IV?

EM4

04:48

no idea, why?

copper.hat

05:09

Wow, I must be wrong.

D.C. the III

@robjohn Watch shopping Robjohn?....I could see you as an Audemars type of guy....

copper.hat

05:24

i gave my son an 1965 omega seamaster for his prom. it belonged to my uncle.

D.C. the III

05:36

family heirloom

copper.hat

my uncle was a large animal vet, my aunt used to refer to it as the arse watch.

no, he didn't :-) but it is an amusing visual

D.C. the III

Lol....my mind didn't even venture to those dimensions.....I'm not sure I was the deviant in the chat before.....🤣

copper.hat

05:57

well, he was the bull in the bowler hat from time to time, but that of course would be the wrong label on my aunt's part

one potato two potato

09:54

Do students usually learn (linear) presentation of finite group in undergraduate algebra class?

Belu cat

12:16

Can the associative property of addition be proved I mean can a+(b+c)=(a+b)+c be proven?

Ofek

maybe with induction?

Xander Henderson

@Belucat From what set of axioms?

Or, perhaps, a better response would be "Either associativity must be assumed as an axiom, or it must be proved."

Belu cat

Do u mean it's an axiom

and has no proof

Xander Henderson

(Also, presumably, you mean associativity of the real numbers? or the natural numbers? There are algebraic structures for which associativity does not hold, e.g. the octonions).

Belu cat

Oh

I meant the real numbers

Xander Henderson

12:25

@Belucat No. That's not what I said.

Belu cat

oh

then?

Xander Henderson

The real numbers can be defined axiomatically. The associativity of addition is generally one of the assumptions of such an axiomatic definition.

On the other hand, there are other ways of defining the reals, for which associativity of addition must be proved.

What is your starting point?

What are you assuming?

Ofek

for example, you can also start with the set theory

and build numbers from sets, define operations, and prove stuff

Belu cat

I ain't assuming anything I am going through Calculus by Spivak where in the first chapter he states a+(b+c)=(a+b)+c and uses this to prove the associativity that holds for sum of 4, 5 numbers.

So I wanted to know how was (a+b)+c=a+(b+c) accepted?

Xander Henderson

@Belucat You are assuming something. You are assuming whatever definitions or assumptions Spivak is assuming.

Belu cat

12:28

Will you please be clear what Spivak is assuming?

Xander Henderson

Since Spivak is presenting calculus, he is likely defining the reals axiomatically. Hence he assumes that addition is associative. But I don't have a copy of Spivak handy, so I don't know what assumptions he is actually making.

Belu cat

He calls it a property

Oh Can you please present a proof for associativity?

Xander Henderson

@Belucat Starting from what axioms?

And in what context?

Belu cat

I ain't sure about it, will you please give a random proof (u can use any axioms)

Xander Henderson

Again, if you are reading a text on calculus or analysis, the associativity of addition on the reals is usually taken as an axiom. Hence there is no proof.

Belu cat

12:31

I am just learning mathematics

Ofek

@Belucat maybe he is relying on the assumption that the real numbers is a field? and then its a property

Belu cat

Oh

What do u mean by context?

OMG MATHS IS SOO GOOD..

He also says we simply agree a+b+c just as (a+b)+c=a+(b+c)

Xander Henderson

@Belucat Sure. Because if your goal is to study calculus, then you don't want to spend a lot of time building up a system of real numbers, since the goal is to actually do something with those numbers.

So, you introduce the reals axiomatically (i.e. you "simply agree" to the properties of the reals) and proceed from there.

Belu cat

Yeah But I really feel bad for not knowing addition properly!

Xander Henderson

I don't understand the complaint.

If you want to have a deeper understanding of these properties, take another class.

Belu cat

12:35

Of what?

Xander Henderson

Perhaps a course in naive set theory, or abstract algebra.

Belu cat

Any book you would like to refer?

Xander Henderson

Or a course in number systems.

Belu cat

I can't take any course right now!

Ofek

i just have proved inductively that if (a+1)+c=a+(1+c) then (a+b)+c=a+(b+c)

for natural numbers

Belu cat

12:37

Oh

Ofek

you can try it if you want, its not hard

linear_combinatori_probabi

Sorry for hijacking the thread, but I just want to say that my disability in making decisions just helped me saving money on choosing a new laptop. I go with my old PC :)

Ofek

XD

Belu cat

@Ofek Yep sure!

linear_combinatori_probabi

Choosing either brand and/or hardware specs doesn't help in calculating my combinatorics problems lol

Ofek

12:42

do you know set theory btw? if so, i can try write something to prove all the properties of real numbers from set theory.

tbh im kinda interested to do it

Belu cat

I don't know set theory, I am a school student btw

I am sorry

linear_combinatori_probabi

I remember that set theory made me upset

Ofek

up set

Belu cat

@Ofek unhappy?

Ofek

im just kidding about upset about sets

linear_combinatori_probabi

12:46

set = sad

Belu cat

😂😂

linear_combinatori_probabi

So instead I gave up choosing laptop and spent about $80 buying introduction to algorithm 4ed. I'm so smart

Xander Henderson

@linear_combinatori_probabi I have a first edition copy of text called "Introduction to Algorithms" which I apparently got for nine dollars.

Belu cat

Was that book soo much expensive?!

linear_combinatori_probabi

I'm so dumb

Xander Henderson

12:51

$80 is cheap thrills for modern textbooks.

New copies of introductory calculus texts can cost as much as $200.

Belu cat

😯😯

linear_combinatori_probabi

I once obsessed in choosing calculus text book like choosing a laptop

because they're exp as fu*k

Belu cat

pdfs are better!

Xander Henderson

@Belucat Strong disagree. I hate reading off of a screen.

Belu cat

You could take a printout

linear_combinatori_probabi

12:54

I once forked a Repo. on GitHub about pdfs and received an email, saying that I (and those other users that forked it) are violating some laws

Xander Henderson

@Belucat Sure, but the cost of printing out a book is not nothing. And hardcover books are actually bound.

Belu cat

@XanderHenderson I know it hurts

linear_combinatori_probabi

I read pdfs but I enjoy read textbook, two different worlds

Belu cat

I agree @linear_combinatori_probabi

linear_combinatori_probabi

paper books make me calm

Jam

13:42

For example, when E is a vector bundle a section of E is an element of the vector space $E_x$ lying over each point $x\in B$"

i dont get this sentence

sections of vector undles are functions from the base to the product

geocalc33

What is a function $f(x,y)$ s.t. holding x constant at $x=0$ yields $\Gamma(y-1)$, and holding y constant at $y=2$ yields the zeta function?

Steve

Hi, is there a name to this lemma? I am trying to find a proof for it https://imgur.com/a/BmXznfM
Thanks!

Xander Henderson

Yeah, I'm pretty sure that lemma is called "Robert".

His friends call him "Bob", but you should call him "Dr."

Alternative snarky remark: Sure. That lemma is named "1.2.7".

geocalc33

14:02

I'm looking for examples of $f(x,y)$ just in case nobody saw that coming

Xander Henderson

More seriously, that looks like the Chebyshev inequality to me (in the case where $X$ has mean zero).

geocalc33

game 1 tonight

Xander Henderson

@geocalc33 The World Series does not start until the Fall, hence your "Game 1" is irrelevant and meaningless. :P

geocalc33

14:21

haha. Rafa into the semi's at the French open

Xander Henderson

Is that, like, ice hockey or golf or something?

geocalc33

Tennis!

Xander Henderson

Tennis is the one with the net, right?

You play it in a bikini? on the beach?

Big, round, white ball?

geocalc33

Yes

fine

Xander Henderson

See, I know sportsballs!

Koro

14:39

@XanderHenderson: After marking close vote to a post, does a comment automatically get dropped below the post?

Xander Henderson

@Koro Not generally, no. But it depends on the reason you cite for closure.

Koro

the comment that "Does this answer your question?"

Xander Henderson

That comment is automatically left when you vote to close a question as a duplicate of another question.

Koro

Ah, I see.

Belu cat

14:57

Can I prove AM GM inequality from (a+b)^2?

Koro

yes.

Calvin Khor

@XanderHenderson 'defining' something axiomatically always leaves me wondering if I'm talking about the empty set

Belu cat

@Koro tq

Calvin Khor

@Steve The proof is $$\Bbb E(|X|^p) = \int_{\Omega} |X|^p d\Bbb P ≥ \int_{|X|\ge \epsilon} |X|^p d\Bbb P \ge \epsilon^p \int_{|X|\ge \epsilon} d \Bbb P = \epsilon^p \Bbb P(|X|\ge \epsilon).$$

Koro

@Belucat :-)

V.S.e.H.

15:01

hey all, I'm struggling a bit with problem 2.4.P13 from Horn and Johnson's Matrix Analysis. Below is one part of the exercise:

Suppose that A ∈ Mn and B ∈ Mm have no eigenvalues in common. Consider the linear transformations T1 , T2 : Mn,m → Mn,m defined by T1 (X ) = AX and T2 (X ) = XB. Show that T1 and T2 commute, and deduce from (2.4.8.1) that the eigenvalues of T = T1 − T2 are differences of eigenvalues of T1 and T2 .

how can T1 and T2 have eigenvalues/commute, when m \neq n

Calvin Khor

$T_1 T_2(X) = T_1 ( XB) = A(XB ) = AXB = (AX)B = T_2(AX)= T_2T_1(X)$

note that $A$ is never multiplied with $B$

V.S.e.H.

aaah, so what should be proven is $T_1 \circ T_2 (X) = T_2\circ T_1(X)$?

Xander Henderson

@CalvinKhor inorite?

There are rumours of some poor graduate student at my phd institution who managed to build up an entire theory for spaces which satisfied certain properties. During his defense, he was asked for an example. It turned out that the unit circle was the only example of such a space.

(Most of that story is probably apocryphal---seems too awful to be true---but it is the kind of thing which reminds one to always have a few examples handy).

Calvin Khor

i mean that to me sounds like the supervisor's fault tho @XanderHenderson

@V.S.e.H. yup. And asking for eigenvalues is like asking for the eigenvalues of any linear operator on a finite dimensional vector space (which in this case is a space of matrices)

Koro

0

Q: Evaluate the definite integral $\int_0^1 x^b b^x \,dx$.

Since, the question asks to evaluate $\int_0^1 x^b b^x\, dx$. Sharing my thought shots on the same: I multiplied and divided my integrand with $\log b^x$ and afterwards substituted $b^x = t$. So my Integrand reduced to $\int_1^b \frac{(\log t)^n}{(\log b)^{b+1}}\,dt$. This becomes an improper in...

definite-integrals

Calvin Khor

15:12

oh no its this integral again

I will pretend to be away

Koro

What is your opinion on close voting of this question?

@CalvinKhor :D

Calvin Khor

it looks correctly closed to me @Koro

by the way where is the integral from?

leslie townes

hm, is "2.4.8.1" the simultaneous upper-triangularizability of commuting operators? as calvin noted T_1 and T_2 commute even if A and B do not.

Koro

@CalvinKhor it seems to have come from one of the booklets of some coaching centre.

Calvin Khor

:) ty

@leslietownes you are purposely capitalising in a unique pattern!

Koro

15:17

@CalvinKhor Thanks. In the post, the OP asked for evaluating the integral (didn't say anything about finding a limit) but from the comments, it seems that they wanted to find limit only.

Calvin Khor

..oh. Then I'm not so sure, but OP voted for it himself (that's the community vote)

Koro

The limit question (which I asked here few days ago) immediately came to my mind when I saw that post and posted the link to that, which the OP said had helped them. So I think that the post getting closed is OK.

V.S.e.H.

@CalvinKhor yeah, that makes sense then, thanks!

Belu cat

15:31

Spivak mentions That the set of natural numbers must have a smallest number except it not being a null set, how can it be possible? if suppose natural numbers were like integers then it wouldn't it possible to find a least member? does he assume natural numbers starts with 1 and extends further in such a case why does he have to worry for it being a null set?

Xander Henderson

@Belucat I don't know. I don't have a copy of Spivak handy.

Belu cat

He claims set of natural numbers being a null set is the only possible condition where he couldn't find a least number

Do does he assume that the set of natural numbers doesn't have negative numbers in them?

in making such a claim?

Xander Henderson

Usually, the set of natural numbers is either (a) the set of positive integers or (b) the set of nonnegative integers.

Belu cat

cause he talks about the well ordering principle

@XanderHenderson yep

Xander Henderson

So Spivak is almost certainly building a system like that.

Belu cat

15:34

Oh so first he concentrates only on positive integers right?

Why does he worry for it to be a null set?

Xander Henderson

I don't know. I don't have a copy of that text.

leslie townes

it's likely just that usual corner case. he wants to say 'every subset of N has a least element.' which is true, except for the one subset of N that doesn't have any elements.

Belu cat

Oh

leslie townes

sometimes people separately distinguish the empty set for no reason (just because it makes people nervous and is a common corner case) but here you really gotta do it. you need elements to have a least element.

just a guess, like xander i do not have the book

Belu cat

Also wouldn't the normal induction do I mean if for a set A P(1) is true and P(k+1) must be proven true whenever p(K) is true

But he says there is another form of induction where P(k+1) would be proven true if and only if p(1) is true and p(K) is true and p(l) where l<k in natural numbers must also be true

Is that right?

Are these 2 really different?

@leslietownes Tq :)!

Xander Henderson

15:42

@Belucat This appears to be a statement of "strong induction" (or "complete induction"---either term should be Googlable). Weak and strong induction are mathematically equivalent.

Strong induction just gives you more hypotheses to work with (which can make things easier).

Belu cat

Hmm tq

Oh both are just the same right?

Xander Henderson

That depends on what you mean by "the same".

They are mathematically equivalent approaches, but they are not quite "the same".

Belu cat

How do they differ? @XanderHenderson

Xander Henderson

They differ in exactly the way you describe.

For weak induction, the induction hypothesis is that $P[k]$ holds. For strong induction, the induction hypothesis is that $P[\ell]$ holds for all $\ell \le k$. In either case, you then show that $P[k+1]$ holds.

Belu cat

in the so called " weak induction" we do consider all the natural number if k=1, k+1 would be 2 and soo on every number is taken into consideration and even for the strong induction we cover every number right?

How do they differ?

Xander Henderson

15:46

I don't understand.

Mathematical induction

Mathematical induction is a mathematical proof technique. It is essentially used to prove that a statement P(n) holds for every natural number n = 0, 1, 2, 3, ... ; that is, the overall statement is a sequence of infinitely many cases P(0), P(1), P(2), P(3), ... . Informal metaphors help to explain this technique, such as falling dominoes or climbing a ladder: Mathematical induction proves that we can climb as high as we like on a ladder, by proving that we can climb onto the bottom rung (the basis) and that from each rung we can climb up to the next one (the step). A proof by induction consists...

Belu cat

My analogy would be say few members standing in a line and a secret is passed and every one would pass it to the person next to him induction states if the (k+1)st person knows the secret it is certain for the kth person must know itand that does make sure everyone betweent the first and the kth person would know the secret and hoe does strong induction make the arguments more strong?

How does it differ?

Xander Henderson

Don't think about the analogy. Look at the definitions.

Belu cat

Oh

I just don't when to use the strong and weak inductions?

Xander Henderson

The usual process for proving a statement using induction is to first prove that $P[1]$ holds, then show that if $P[k]$ holds, then $P[k+1]$ holds.

Belu cat

yeah

Xander Henderson

15:51

For strong induction, first prove that $P[1]$ holds, then prove that if $P[\ell]$ holds for all $\ell \le k$, then $P[k+1]$ holds.

Strong induction assumes a stronger induction hypothesis.

Belu cat

Oh Ok

Xander Henderson

However, the two forms of induction are mathematically equivalent, which means that anything which can be proved by one can be proved by the other.

Belu cat

Yep

Yeah the definitions sound different but they equivalent!

Xander Henderson

The definitions are different. The two forms of induction are different. But they are mathematically equivalent.

Belu cat

Yeah I think I got it must think about the definitions!

@XanderHenderson Thank you soo much :)

robjohn

15:55

Let $Q[k]$ be that $P[j]$ is true for $1\le j\le k$...

Applying weak induction to $Q$ is the same as strong induction to $P$

Belu cat

yeah

robjohn

20:57

@Ted: it's quiet in here...

leslie townes

it was quiet in here.

Ted Shifrin

Ssshhhh …

robjohn

🤭

geocalc33

21:59

hello

I concocted a neat function $$ I(x,y)=\sum_{n=1}^{\infty}\bigg|\int_0^1 \frac{1}{t~(\log t)^y}\exp\bigg(\frac{n^x}{\log t}\bigg) ~dt~\bigg|$$

$I(0,y)=\Gamma(y-1)$

$I(x,n)$ for $n=2,3,4,...$ are zeta functions

but I'm not sure about a closed form for $I(x,y)$

geocalc33

22:16

And summing you get: $$ \sum_{k=1}^\infty I(x,k)=\sum_{k=1}^\infty \sum_{n=1}^\infty \bigg|\int_0^1\frac{1}{t~(\log t)^k}\exp\bigg(\frac{n^x}{\log t}\bigg) ~dt~\bigg|=\zeta(x)+\zeta(2x)+2\zeta(3x)\cdot\cdot~\cdot$$

which looks odly familiar to: $$\varphi(x)=\sum_{n=1}^\infty (e^{-n^{-x}}-1)=\sum_{n=1}^\infty \frac{(-1)^n}{n!}\zeta(nx)$$

Steve

23:16

@CalvinKhor I see, thanks!

Transcript for

$Mathematics$ Mathematics