Mathematics - 2023-11-18

Obliv

00:06

um is $\sum_{k=1}^{s+1}k\binom{s+1}{k}$ taking out the last term $=(s+1)\binom{s+1}{s+1}+\sum_{k=1}^{s}k\binom{s}{k}$

Ted Shifrin

No.

Obliv

Ok, so how do I take out the last sum term so I can get $\sum_{k=1}^{s}k\binom{s}{k}$?

Ted Shifrin

You can’t on a whim change $s+1$ to $s$.

Obliv

doesn't it just include 1 extra number if $s$ is a natural number

Ted Shifrin

have you tried writing out the formulas?

Obliv

00:09

doesn't that just complicate things? (if you mean the factorial form)

oh I am given a hint that $\binom{n}{k} = \binom{n-1}{k-1}+\binom{n-1}{k}$

Ted Shifrin

Yes, being correct may be more complicated than being wrong.

3

Obliv

$\sum_{k=1}^{s+1}k\binom{s+1}{k} = 1\binom{s+1}{1}+2\binom{s+1}{2}+...+(s+1)\binom{s+1}{s+1}$

if I take out that last term, I now have $\sum_{k=1}^{s}(...)$ something

but it's not $k\binom{s}{k}$?

ohhh

it's $k\binom{s+1}{k}$

$\sum_{k=1}^{s}k\binom{s+1}{k} = \sum_{k=1}^{s}k\binom{s}{k-1}+\sum_{k=1}^{s}k\binom{s}{k}$

but now i have another sum I must work with

mick

hi all

Jakobian

00:33

hey Mick

user 726941

00:48

welcome

Obliv

can I split apart $\sum_{k=0}^{s}(k+1)\binom{s}{k}$?

into $\sum_{k=0}^{s}k\binom{s}{k} + \sum_{k=0}^{s}\binom{s}{k}$

Obliv

01:18

I'm trying to prove $\sum_{k=1}^{n}k\binom{n}{k}=n2^{n-1}$ for all natural numbers

after doing the base case, I do the case for an arbitrary natural number $s$, so assumte true that $\sum_{k=1}^{s}k\binom{s}{k} =s2^{s-1}$

$\sum_{k=1}^{s+1}k\binom{s+1}{k}\stackrel{?}{=}(s+1)2^{s}$

since $\binom{n}{k}=\binom{n-1}{k-1}+\binom{n-1}{k}$

and we can take out a term so $=(s+1)\binom{s+1}{s+1}+\sum_{k=1}^{s}k\binom{s+1}{k}$

$=s+1+\sum_{k=1}^{s}k\binom{s}{k-1}+\sum_{k=1}^{s}k\binom{s}{k}$

$=s+1+s2^{s-1}+\sum_{k=1}^{s}k\binom{s}{k-1}$

Let's change the bounds so $\sum_{k=1}^{s}k\binom{s}{k-1}=\sum_{k=0}^{s}(k+1)\binom{s}{k}$

Which can be split into
$=s2^{s-1}+s+1+\sum_{k=0}^{s}k\binom{s}{k}+\sum_{k=0}^{s}\binom{s}{k}$

Taking out the $k=0$ terms, $\\=s2^{s-1}+s+2+\sum_{k=1}^{s}k\binom{s}{k}+\sum_{k=1}^{s}\binom{s}{k}$

We use the binomial theorem so that the last sum becomes $2^s -1$,$\\=s2^{s-1}+s+2+s2^{s-1}+2^s-1\\=s2^s+2^s+s+1=(s+1)(2^s)+s+1$

so clearly that's not $(s+1)(2^s)$ because it has an extra $s+1$ term. Idk where I went wrong.

Ted Shifrin

Check your summation limits when you changed variable.

Obliv

Oh ok I changed $\sum_{k=1}^{s}k\binom{s}{k-1}$ incorrectly

It looked ok for the first 2 terms :D but i didn't check the last for some reason..

I have an extra term

wait yeah the extra term is $(s+1)\binom{s+1}{s+1}$ I'm saved, thank you @TedShifrin

Ted Shifrin

01:36

I find it safer use a new letter. Let $j=k-1$ and write it all with $j$.

At the end, you can rename $j$ as $k$.

Obliv

I actually get even more confused with introducing new letters for bounds

Idk maybe it'll get less confusing as I get more experienced

I seriously get confused by daylight savings every year so maybe i'm dyslexic in some way

Obliv

01:55

does $A \setminus \varnothing = A$?

Ted Shifrin

02:16

@Obliv Think about substitutions in integrals.

Allie

02:57

Hey @TedShifrin

Oh he left

Ted Shifrin

Or not.

Obliv

dun dun dun

Allie

Hi ted

Its meow mix

How r u doing

Ted Shifrin

Hi Allie … Haven’t seen you in a while. You doing OK?

Allie

Yes, actually life has been not so bad lately

Specifically this semester

Ted Shifrin

03:07

good. How’s biochem going?

Allie

I switched to pure chem

Jakobian

@TedShifrin Here, I've answered that question with gauge integrals you've brought my attention to. Link to answer.

Allie

And im absolutely loving it

Ted Shifrin

great. I loved thermo and took two courses on it!

Allie

Ill be in physical chemistry next semester

Ted Shifrin

03:08

Great. You’ll love all the multivariable calculus in it.

Allie

But im really happy i started research and i really want to get accepted at a UC this summer for a SURF REU

Obliv

Ofc Ted loved thermo.

Ted Shifrin

neat, @Jakobian.

@Obliv Yes, my chem major friends hated me.

Obliv

My sister is having a hard time in her bio class because there were biochem topics. She briefly considered doing a "hard" science major but I think she didn't realize the commitment it takes.

I'm supportive either way, I really don't like to believe in the stigma that it's impossible to succeed unless you're a genius or whatever

Ted Shifrin

It’s tougher than majoring in psychology or elementary education, for sure .

Obliv

03:12

i think the world needs more scientists of all variety, so the bar should be lowered.

Ted Shifrin

No, no genius required, but knowing how to study effectively helps.

No, lowering bars is not the answer.

I categorically reject that.

Obliv

well lowered, but then throw a bunch more bars that increase in height

Ted Shifrin

But too many people want to be doctors for the wrong reasons. Sadly, the hassles with insurance in the US are chasing away good doctors..

Obliv

Eh, it is true though that as time goes on the standards increase just by virtue of more progress being made?

Ted Shifrin

No, I don’t think so. The academic standards in the US are very weak, cimpared to Europe and Asia.

Obliv

03:16

how about the standards of ivy leagues? More or less the same over the last few decades?

I've heard it's gone lower there too actually

Ted Shifrin

Covid wreaked all sorts of havoc.

Allie

Do u guys know any good introductory e&m texts tho

Ted Shifrin

Purcell

Obliv

@TedShifrin She actually wanted to be a doctor forever. I think numerous factors have led to her decision to change but I think healthcare workers get taken advantage of in the US?

purcell, jackson, griffiths?

Ted Shifrin

Part of the Berkeley physics series. We used it in my course at MIT.

Those are more advanced.

Allie

03:19

Ok sounds good. And that would be appropriate for someone comfortable with physics I and II and multivariable?

Obliv

physics II was calc based for you? @Allie

Ted Shifrin

Isn’t physics II E&M?

Obliv

It is.. but as I've found you need to take it again lol

I took AP physics in HS, Physics I, II & III in community, now I'm taking them again for my bachelors

Ted Shifrin

Griffiths is a junior/senior level course. Purcell is meant for physics II using calculus.

Allie

Yeah but we didnt get super into it

Obliv

03:21

not complaining though, because I need the extra runs

Oh ok, for some reason I thought griffiths was the intro book. I'm confusing it with my Physics II text

Ted Shifrin

Nooooo

Obliv

As long as you have a good understanding of vector products, vector dot products, and vector shenanigans in general I think you will do fine

Ted Shifrin

Stokes and Gauss are relevant.

Obliv

maybe looking over the very basic idea of what a differential equation is and partial derivatives might help

but i'm sure the course/txt will motivate the math

Is this a lazy proof

If $A,B,C$ and $D$ are sets, then $(A \times B)\cap(C\times D)=(A\cap C)\times(B \cap D)$
\\In setbuilder notation we can see that $(A \cap C)\times(B\cap D) = \{(a,b)\mid (a \in (A\cap C))\land (b \in (B\cap D))\}\\=\{(a,b)\mid(a\in A)\land(a\in C)\land(b\in B)\land(b \in D)\}\\=\{(a,b)\mid(a\in A)\land(b\in B)\land(a\in C)\land(b \in D)\}\\=\{(a,b)\mid(a\in A)\land(b\in B)\}\cup \{(a,b)\mid(a\in C)\land(b\in D)\}\\=(A\times B)\cap(C\times D)\}$

Maybe I should write why the set can be broken up into that union and why that implies the intersection

nvm i just copied it wrong, should be intersect not union

Ted Shifrin

03:40

Yes, intersection

Obliv

04:08

What's a good way to prove $|A+B|=|A|+|B|$ for finite sets

wow it's getting late, I copied that wrong as well

leslie townes

depends on your setup. if you're going from set theory axioms, probably induction.

Obliv

should be $|A\cup B| = |A|+|B|$

leslie townes

and your definition of 'disjoint union'

Obliv

I'm not sure if we went over the axioms tbh

Jakobian

If you're not sure then you didn't. Set theory axioms leave a huge mark in your brain

Obliv

04:15

what's the difference between principia mathematica and ZF axiomatic set theory

yeah idk what system we're even working in

i think it's just elementary set theory

Jakobian

Principia Mathematica

The Principia Mathematica (often abbreviated PM) is a three-volume work on the foundations of mathematics written by mathematician–philosophers Alfred North Whitehead and Bertrand Russell and published in 1910, 1912, and 1913. In 1925–1927, it appeared in a second edition with an important Introduction to the Second Edition, an Appendix A that replaced ✱9 and all-new Appendix B and Appendix C. PM was originally conceived as a sequel volume to Russell's 1903 The Principles of Mathematics, but as PM states, this became an unworkable suggestion for practical and philosophical reasons: "The present...

The difference is that nobody cares about principia mathematica axioms

@Obliv its called naive set theory, which is doing math without caring about axioms

Very common approach and usually doesn't obstruct anything

Ted Shifrin

@Obliv what if $A\cap B\ne\emptyset$?

Jakobian

Note that if someone is not doing naive set theory, they won't work with ZF but rather with ZFC. Unless you're a set theory pervert

one potato two potato

04:31

how about KFC?

Ted Shifrin

👎

Koro

Too oily

Obliv

Oh snap so that's not always true then @TedShifrin

so I can just disprove by counterexample

Obliv

04:49

@Jakobian why is this? are they flawed

Obliv

05:02

How would one determine the atomic assertions of a statement?

for example if $n$ is prime or $n+2$ is prime, then $n^2+2$ is prime or $n^2-2$ is prime. Does P(n)=n is prime and P(n+2)=n+2 is prime make sense

but I'd need like 4 atomic statements

i wonder if that's the right way..

yeah I guess that works

but maybe changing the letter each time is better so $A(n),B(n),...$

shintuku

11:55

@Obliv it was meant to set mathematics on a purely logical and absolute footing, but turns out you can't (see Godel's incompleteness theorems)

i think it takes a few hundred pages to prove 1+1=2

Jakobian

12:09

@Obliv it should be thought of as a historical curiosity

Shaun

12:30

I don't get why the following was closed.

-2

Q: Prove that there are $a,b,c\in G$ with $ab = c$

Let $p$ be a prime and let $G$ be a subgroup of $\mathbb{F}_p^\times$ of order divisible by 6. Prove (or disprove) that there exist $a,b,c\in G$ so that $ab=c$. I know that subgroups of $\mathbb{F}_p^\times$ are cyclic, and this follows from the nontrivial fact that there is a primitive root mo...

Jakobian

@onepotatotwopotato

Shaun

I understand it says why, but it has context and everything.

What's more is that my answer to it was downvoted.

-1

A: Prove that there are $a,b,c\in G$ with $ab = c$

Hints: All subgroups of a group are themselves groups with respect to the operation restricted to their underlying sets. Each group, and hence subgroup, is nonempty (because it has an identity) and closed with respect to its operation.

Jakobian

$Mathematics$ CURED

For feedback/discussion/requests of Close/Undelete/Reopen/Edit...

2

1

@Shaun

Shaun

I just posted there :)

0

A: Requests for Reopen & Undeletion Votes (volume 01/2022 - today)

Please reopen Prove that there are $a,b,c\in G$ with $ab = c$ It has context. The OP was overthinking the answer. Note: I answered it.

oneofvalts

I am reading Modern Projective Geometry and in a proposition, it gives $s: P(V \times K) \to V \mathop{\dot{\cup}} P(V)$.

And it says this is an isomorphism, that's OK. The part that I don't get is why the disjoint union is used there.

$V$ and $P(V)$ are not even subsets of the same set.

Jakobian

12:42

@oneofvalts coproduct?

I don't know projective geometry, but this looks like coproduct of whatever objects you are considering

For example, if $X, Y$ are sets then their coproduct $X\sqcup Y = \{0\}\times X \cup \{1\} \times Y$ can be thought of as a disjoint union of $X, Y$, even thought we can have $X\cap Y \neq \empty$. Its just that we can work with their copies that are disjoint

Similar construction can be done for, say, topological spaces

one potato two potato

it's worth being closed

I don't know if this is a correct sentence btw

Jakobian

It is worth of being closed, would be more correct I think? But not sure

oneofvalts

@Jakobian Well, I now realize using disjoint union there is not a problem, because it turns out my definition of disjoint union was terribly wrong. Thanks for your comments anyways.

Jakobian

No problem. :)

one potato two potato

12:58

I've seen several times that when the question is "stupid" or trivial enough, the answer post usually gets downvotes. This is related to the so-called 'gamification' of MSE discussed here before. The answerer posts their answer to get reputation essentially.

Jan 28 at 19:51, by copper.hat

@Shaun downvoting without reason is one of the most aggravating aspects of the site gamification.

this is a bit different but look who's tagged here

one potato two potato

14:04

kmill.github.io/knotfolio

enjoy

Jakobian

14:15

Pretty cool. Though knots aren't my thing

Koro

15:09

why is the first part i.e., measure of the set {x| $\lim_{|B|\to 0} |\frac {\int_B (f(y)-g(y) dy}{|B|}|>\lambda/2\}$ is $\le c_1 \|f-g\|_1$?

low footfall here today. what might have happened? $\ddot\frown$

Koro

15:32

I want to eat cooked okra but can't find them in this region. Even cooked cauliflower is not available here.

only biryani with potato is available.

0

Q: Proving that measure of a set is bounded by a norm

Suppose that $f\in L^1(\mathbb R^n)$. Fix an $\epsilon>0$. Let $g$ be in $C_c(\mathbb R^n)$ such that $\|f-g\|_1\lt \epsilon$. Consider the set $E=\{x\in \mathbb R^n: \lim_{r\to 0+}\frac 1{|B(x,r)|}|\int_{B(x,r)} f(y)- g(y) dy|\gt a/2\}$ for some $a>0$. $|X|$ denotes Lebesgue outer measure of the...

real-analysis measure-theory

in case anyone wants to take a look.

Mad

16:21

@TedShifrin hi

Ted Shifrin

Howdy

Mad

When we talk about Homomorphisms of a Lie algebra, we say its linear maps, that also respect the brackets.
When we talk about for example the space END(g) of all endomorphisms over a lie algebra g.
Do these also respect the bracket? i am not finding a clear definition.

Or is this the space of endomorphisms in the traditional sense, of linear maps from g to g as a vector space

Ted Shifrin

I would assume it’s in the category of Lie algebras, but authors may differ.

Mad

Alright

Thorgott

you can consider both the Lie algebra endomorphisms of $\mathfrak{g}$, which form a group and the linear endomorphisms of the underlying vector space of $\mathfrak{g}$, which is itself a Lie algebra with Lie bracket given by the commmutator

Mad

16:33

yes we are considering the second part you mentioned.

Thorgott

which one an author means should probably be clarified somewhere

personally, I would denote the former as $\mathrm{End}(\mathfrak{g})$ and the latter as $\mathfrak{gl}(\mathfrak{g})$, but others may differ

Ted Shifrin

I would concur.

Mad

So i am asking, because i am asked to prove, that a derivation which is a linear map D: g to g with the attribute D[x,y]=[Dx,y]+[x,Dy]that all the derivation Der(g) form a sub lie algebra of End(g)

So do i need to show, that for a Derivation, the following must apply? D[x,y]=[Dx,Dy]

This is why i am asking

Ted Shifrin

They say explicitly sub-Lie algebra. So they are thinking of Thor’s second option, it seems.

Thorgott

agreed

Mad

16:37

And in the second option, these "linear endomorphisms" are the ones one knows from typical linear algebra? thus they do not respect the lie bracket?

Thorgott

yes

Mad

Okay thank you.

Thorgott

the lie algebra gl(g) does not depend on the Lie bracket of g whatsoever

the subalgebra Der(g) however does depend on the Lie bracket of g

A. P.

Hi, is there any norm $\|\cdot\|$ in order that $L^{1}([0,1])\cap L^{2}([0,1])$ is not Banach? Thank you

Thorgott

any norm?

A. P.

16:46

What I meant is if there is any norm that I can define on the mentioned space, such that the space is not complete.

Jakobian

17:01

@A.P. $L^1$ norm, $L^2$ norm

Obvious choices

Thorgott

you can find an incomplete norm on any infinite-dimensional vector space

Jakobian

@Thorgott that too, but I think it might not be too useful, A.P. most likely is hiding context from us

Thorgott

yeah, that's why I questioned

one potato two potato

norm-minimizing

Jakobian

You could also take $L^p$ norm for any $1 < p < 2$

A. P.

17:11

The problem is that I am trying to find an example of a space constructed as the intersection of the spaces L^1([0,1]) and L^2([0,1]) that is not Banach. So, I have to define a norm in such a way that the space E = L^1 ∩ L^2 is not Banach with that norm. This would be the context. I also thought about defining ||.||_1 on E. Then I have to show that there exists a Cauchy sequence that does not converge. How can I do this?

@Jakobian I also don't see why it should be evident to only consider associating it with the norm of one of the two spaces. Any advice to improve intuition on this is appreciated.

Thorgott

surely you are not being asked for an arbitrary norm on the intersection

A. P.

Well, the task is to construct a space that is not Banach in the intersection, so I think the way is to propose a norm that makes this possible.

Jakobian

@A.P. $L^1\cap L^2$ with $L^1$ norm is dense in $L^1$ so it can't be complete

Same thing happens with $L^p$ norm for $1\leq p \leq 2$

Thorgott

I mean, usually you wanna equip such an intersection with the sum norm, no? Like when doing interpolation stuff.

Jakobian

Actually $L^2\subseteq L^1$ in this case

So you want to specifically take the $L^1$ norm

@Thorgott that would definitely make it Banach

Thorgott

17:27

that is true

idk, this problem just seems oddly posed to me

cause if you just want any incomplete norm, it hardly matters to pretend the space is something specific like the intersection of L^1 and L^2

Ted Shifrin

I think the question is fine.

A. P.

@Jakobian I know that L^1 is complete with the L^1 norm. Why L^1\cap L^2 with L^1 norm is dense? and why this implies that it can not be complete? I am sorry, I cannot follow that.

is there any reference for that?

Jakobian

@A.P. for example, any function in $L^1$ can be approximated by continuous functions

One way to do this is prove that bounded measurable functions can be approximated so

Actually just take bounded measurable functions directly

A. P.

ok, the intersection is dense in L^1, but why from here one can deduce that is not complete the intersection?

Jakobian

Because otherwise it'd be equal to L^1

It'd be closed

But something like $1/\sqrt{x}$ shows otherwise

A. P.

17:48

Okay, I think I've got it (sorry, I still don't navigate very easily through results that connect everything; in fact, I was thinking of trying to prove it's not complete by definition, but your approach seems more appropriate).

If L^1 ∩ L^2 is dense in L^1 (I have to verify this), then by definition cl(L^1 ∩ L^2) = L^1. Since L^1 is Banach with the L^1 norm, and as L^1 ∩ L^2 is a subset of L^1, then L^1 ∩ L^2 is not complete if and only if L^1 ∩ L^2 is not closed. We show that L^1 ∩ L^2 is not closed in L^1 because cl(L^1 ∩ L^2) != L^1 ∩ L^2, which is true because density gives us cl(L^1 ∩ L^2) = L^1, and this is not equal to L^1 ∩ L^2.

Jakobian

Yes

@A.P. to show that $L^1\cap L^2 = L^2$ with $L^1$ norm is dense in $L^1$, for any $f\in L^1$ consider truncate $f_n = \text{mid}(-n, f, n)$ of $f$. Then $f_n\in L^1\cap L^2$ and $\|f-f_n\|_1\to 0$

Also, you can use LaTeX in chat to read it here (easier on pc). See chat description

A. P.

Okay, so by definition, the work on density. Is there a different way to write f_n? I don't understand the notation. I tried to install LaTeX, but it didn't work. I'll reread the instructions.

f_n = \text{mid}(-n, f, n)

sunny

I'm reading about asymptotically stable system, i.e. where the solutions of $x'=Ax$ tend to $0$ as $t\to\infty$. For such a system, all eigenvalues of $A$ have negative real part. The matrix $e^{tA}$ contains as its columns the solutions. It is claimed in my exercise sheet that $\lVert e^{tA}\rVert\leq ce^{-\epsilon t}$ holds for some $\epsilon>0$ and a constant $c$ (for $t\geq0$). This is not very obvious to me.

I know that every element in $e^{tA}$ is a linear combination of terms of the form $t^je^{\lambda t}$, where $\lambda$ is an eigenvalue of $A$ and $j$ is less than the multiplicity of that eigenvalue. Moreover, I know of $\lVert e^{A}\rVert\leq e^{\lVert A\rVert}$, but I don't know if this is helpful.

Koro

18:08

how to find the range of \alpha?

i.e., the mapping $V: S(\mathbb R^n)\to \mathbb R$ defined as $V(\phi)= \int f \phi$ is continuous.

Jakobian

@A.P. middle value of $n, f(x), -n$

Lucky Chouhan

18:34

aayi aayi aa

A. P.

what is the definition of middle value of f,g,h? I cannot find a reference @Jakobian

Jakobian

@A.P. $\text{mid}(a, b, c) = b$ if $a\leq b\leq c$ for example

the value thats in between the two other values

A. P.

ok, the problem is the english again

so mid(a,b,c)=a if b<=a<=c

Jakobian

yes

A. P.

I am sorry. My question now feels very naive about the middle

Jakobian

18:38

no its okay, this isn't a standard definition, I'm just using it because I saw it recently and took liking to it

normally I'd write something like $\max(\min(n, f), -n)$

user123234

can someone help me here?

I want to use exercise b) for the part c) but I don't get it

for definition: $ad_X(Y)=[X,Y]=XY-YX$

@TedShifrin can you maybe help me?

I wanted to define $f(z)=e^{x+tz/(1-e^{-\alpha})}\frac{Z}{1-e^{-\alpha}}$

but I don't think this works

Mad

19:09

I see you are asking as well about Lie theory.

user123234

yes

Mad

cant help, i myself need help :D Actually i am facing problems right now proving a statement about the Adjunct function haha

user123234

19:33

no worries maybe someone else can help

Koro

no1 studies distribution here?

user 726941

№1

leslie townes

19:54

ted: at lunch, munchkin just announced "three threes and one gets you to ten" and "three threes and one more three makes twelve"

Ted Shifrin

Ah, she's progressing apace. Now wait for her to start instructing munchkin$'$.

Does she intuit that three fours is the same as four threes?

(Maybe start with two rather than four.)

@user123234 No, but I see that this exercise appears as a theorem on Wikipedia's entry on the Baker-Campbell-Hausdorff formula (of which this is a special case).

user 726941

20:11

leslie: ask her to replace "three" with "negative-three"

at supper

Ted Shifrin

That’s in another few years.

user 726941

How about "zero"

user123234

@TedShifrin Yes this then would be part d) where we need to show that $e^Xe^Y=\exp(X+\frac{\alpha}{1-e^{-\alpha}}Y)$

But I also don't see that I mean I only know $A(t)$ and $B(t)$ satisfy the same linear differential equation

this also is written as a hint, but what does this help me

Soumik Mukherjee

@user726941 I think $i$ should be introduced before $-1$. Natural numbers, addition, multiplication, numbers of the form $a+bi$, subtraction. In this order.

user123234

I wanted to use that since $A'=AY$ and $B'=BY$ then $A'/A=B'/B$ respectively $A'(t)B(t)=B'(t)A(t)$ for all $t$ then I thought picking $t=1$ but this does not help

Ted Shifrin

20:19

@user123234 It means they are probably the same, up to constant multiple.

@user123234 Careful cavalierly “dividing” matrices.

user123234

But I mean if I solve the general differential equation $f'=fY$ I get $f(t)=ce^{tY}$

Ted Shifrin

Only when $Y$ is constant.

user123234

okey, but I mean Y is a matrix

Ted Shifrin

Yes, and so?

user123234

What do you mean by constant

I mean I don't get it can I immediately say that since they solve the same linear DE they are equal?

Joe Shmo

20:27

Hi all

Jakobian

@Koro I've studied some in the past but I feel very not competent about it

@JoeShmo hello

Joe Shmo

whats going on with you these days? what are you working on?

Jakobian

I'm learning about Henstock-Kurzweil integrals

user123234

@TedShifrin If I assume $A(t)=cB(t)$ then I can take $t=1$ and get $e^Xe^Y=c\exp(X+\frac{\alpha}{1-e^{-\alpha}}Y)$ but how do I show $c=1$?

Jakobian

so far only on $[a, b]$, but I might try learning them on $\mathbb{R}^n$, or even on manifolds later

Koro

21:20

@Jakobian I came across this during Lebesgue differentiation theorem.

but I've not yet dived into HK integrals.

leslie townes

ugh, don't say it like you're going to. there is still hope. nobody needs to 'dive in' to HK integrals. jakobian has to do it as part of a court-ordered program.

Koro

gkorpal.github.io/files/lebesgue.pdf (section 1, last para).

user 726941

is he on probation doing math community work :P

Jakobian

I mean, yeah you don't have to follow in my steps

I'm not doing this because I know the way, I think we have very vastly different goals in mind

I'm doing what I might find fun and what might expand my knowledge, is all

user 726941

22:02

All in all your community work is appreciated 👍🏻

Jakobian

thanks

Summerday

22:32

does someone knows about mathematical finance

I have a question here:

0

Q: How can I price this option?

In the Black-Scholes model, I want to price the so called Butterfly option, where the payoff $P(x)$ is the following function: $P(x)=0$ if $0\leq x\leq 40$, $P(x)=x-40$ for $40\leq x\leq 60$, $P(x)=-x+80$ for $60\leq x\leq 80$ and $P(x)=0$ for $80\leq x\leq 100$ In the lecture they gave us the h...