Mathematics - 2024-06-28

Xander Henderson

01:19

Ugh... senile old man vs senile old man... I don't know if I can keep watching this...

Semiclassical

04:44

@XanderHenderson i'm glad i didn't

user123234

06:24

can someone help me here

0

A: How can I find the argument of this complex numbers under the preimage of $f(\omega)=\frac{\omega+z}{1+\overline{z}\omega}$?

Let $z=re^{it}$, $w = e^{it_1}$, and $$ W = \frac{w-z}{1-\overline z w} = e^{it_1} \frac{1-re^{i(t-t_1)}}{1-re^{-i(t-t_1)}}. $$ Note that the numerator and denominator are complex conjugates, and $$ \arg(1-re^{i(t-t_1)}) = -\arg(1-re^{-i(t-t_1)}) = \tan^{-1}\left(\frac{r\sin(t_1-t)}{1-r\cos(t_1-t...

Ryder Rude

07:40

Ignore It - Short Horror Film ( Award Winning )

►

Pizza

08:29

Question: but what does it mean when transparent avatar appear on online users ?

psie

08:43

From the Hölder inequality, if $\mu$ being a finite measure, and $f,g$ mapping from $E$ into $\mathbb R$, we have $$\lVert f\rVert_r\leq\mu(E)^{\frac1{r}-\frac1{r'}}\lVert f\rVert_{r'},\tag1$$for $1\leq r<r'<\infty$. Then my book states that; "...thus $L^q\subset L^p$ when $1\leq p<q\leq\infty$." Why can $q$ equal $\infty$?

You get $(1)$ from the Hölder inequality by putting $g=1$ and $f=|f|^r$ and $p=r'/r$.

Sine of the Time

08:58

@psie if $f\in L^{\infty}$ there exists $k>0$ s.t $|f|\le k$ a.e

psie

@SineoftheTime indeed, the norm is $$\lVert f\rVert_\infty =\inf\{C\in [0,\infty]:|f|\leq C,\mu \text{ a.e.}\},$$ so $|f|\leq \lVert f\rVert_\infty$. But I still do not see how $L^\infty\subset L^p$ for all $p\in[1,\infty)$ when $\mu$ is finite.

wait, I think I see now

Sine of the Time

you want to show $L^{\infty}\subseteq L^p$ right?

psie

right

Sine of the Time

so if $f \in L^{\infty} \implies f\in L^p$

$\int_E |f|^p \le \int_E k^p=k^p\mu(E)$

psie

ah! ok, I think I understand 👍

thanks

ephe

09:25

Let $M$ be a filtered module and $N$ be a submodule. Then is the following proof about every Cauchy sequence in $M/N$ being convergent true?

Let $(x_n+N)$ be a Cauchy sequence in $M/N$. We will construct a Cauchy sequence $(y_n)$ in $M$ such that $(x_n+N)=(y_n+N)$. Set $x_1=y_1$. Suppose $y_n$ has been chosen such that $y_n+N=x_n+N$. Note that since $(x_n+N)$ is Cauchy for all $k\in\mathbb{Z}$ we have $x_{n+1}-x_{n}+N\in (M_k+N)/N$ eventually. So $x_{n+1}-x_n+N=p+j+N$ for some $p\in M_k$ and $j\in N$. Thus $x_{n+1}-x_n-p-j\in N$. Then for some $u\in N$ we have $x_{n+1}-x_n+u\in M_k$ eventu…

Pizza

10:02

$y''+6y'+10y=2\sin x$

$y(t) = e^{-3t}(c_1\cos(t)+c_s\sin(t))$

Semiclassical

@Pizza presumably $\sin t$ intended?

Pizza

Can i use the indeterminate method, taking a form $y_p(x) = A\cos(x)+B\sin(x)$?

Semiclassical

should work, since it's a distinct solution set from $e^{(-3\pm i)t}$

Pizza

@Semiclassical sorry in was $c_2 \sin(t)$

i write $c_s$ ...

Semiclassical

no worries

Pizza

10:07

anyway

i did that before: $\lambda = \frac{-6 \pm \sqrt{6^2 - 4 \cdot 1 \cdot 10}}{2 \cdot 1} = \frac{-6 \pm \sqrt{36 - 40}}{2} = \frac{-6 \pm \sqrt{-4}}{2} = \frac{-6 \pm 2i}{2} = -3 \pm i$

$\alpha+i\beta = -3+i$
$\alpha-i\beta=-3-i$

$\alpha=-3, \quad \beta = 1$

Semiclassical

another idea here is to begin with $y''+6y'+10y=2e^{i t}$, then take the imaginary part at the end

Pizza

$y''+6y'+10y=0$
$\lambda^2+6\lambda+10=0$

Semiclassical

the advantage is that then it's clearly sufficient to take $y_p=Ae^{it}$

Pizza

complex substitution method?

Semiclassical

p much yeah

Pizza

10:11

i can try

$y_p' = Aie^{it}$
$y_p'' = -A e^{it}$

$-A e^{it} + 6A i e^{it} + 10A e^{it} = 2e^{it}.$
$(-A + 6Ai + 10A)e^{it} = 2e^{it}.$

question: is $e^{it}$ always 1?

Semiclassical

what?

Pizza

sorry i mean $|e^{it}|$

Semiclassical

ah. then yes

anyways, do you see how to proceed with the calculation you were making above (using $y_p=Ae^{i t}$)

Pizza

@Semiclassical but isnt $y_p = \frac{2}{9}e^{it}$ ?

Semiclassical

no

the $6i$ that shows up on the LHS cannot be disregarded

Pizza

10:25

$-A + 6Ai + 10A = 2$
$(10A - A) + 6Ai = 2$
$9A + 6Ai = 2$

Semiclassical

you have to solve for $A$, plug that into $Ae^{it}$, and only then take the imaginary part

you'll probably also want to rationalize $A$ first, so that the denominator is real

Pizza

$9A = 2 \quad \Rightarrow \quad A = \frac{2}{9}$
$6Ai = 0 \quad \Rightarrow \quad A = 0$

is wrong?

Semiclassical

yes

you're treating $A$ as though it's a real number

but there's no reason it should be here (and indeed can't be, since as your work shows no real $A$ would solve this)

Pizza

@Semiclassical can i write $A = a + bi$?

Semiclassical

sure, that would work

do it that way first and see what you get

Pizza

10:34

$9(a + bi) + 6i(a + bi) = 2$
$9a + 9bi + 6ai - 6b = 2$
$(9a - 6b) + (9b + 6a)i = 2$

$9a - 6b = 2$
$9b + 6a = 0$

now can i do this right?

Semiclassical

yes, so what do you get?

Pizza

$a = \frac{2}{13}.$
$b = -\frac{4}{39}$

Semiclassical

yep

that said, this way is slow. a much faster way is to proceed as you would in more familiar algebra

namely, $9A+6Ai=(9+6i)A=2\implies A=2/(9+6i)$. this can be rationalized by multiplying top/bottom by $9-6i$ to get $A=2(9-6i)/(81+36)=\frac{2}{13}(3-2i)$, which argrees with your result

Pizza

mm yes

$y_p = \left( \frac{2}{13} - \frac{4}{39}i \right) e^{it}$

Semiclassical

@Pizza right. now need to take imaginary part of that. neither term is hard to deal with but have to be careful with the sign on the second

oh. blah, i messed up my algebra slightly above

Pizza

10:41

$\text{Im}(y_p) = \text{Im} \left( \left( \frac{2}{13} - \frac{4}{39}i \right) (\cos t + i \sin t) \right).$

where

$\left( \frac{2}{13} - \frac{4}{39}i \right) (\cos t + i \sin t) = \frac{2}{13} \cos t + \frac{2}{13} i \sin t - \frac{4}{39}i \cos t - \frac{4}{39} \sin t$

Semiclassical

$A=2(9-6i)/(81+36)=2(3-2i)/(39)=\frac{2}{13}-\frac{4}{39}i$

Binky McSquigglebottom

But couldn't we have used the similarity solution method?

Semiclassical

@Pizza was off by 1/3 in my initial result for $A$

Pizza

👍

Semiclassical

@BinkyMcSquigglebottom which method do you mean? certainly one could have looked for $y_p=A\cos t+B\sin t$ directly

i just like the complex-valued method better here because the derivatives are tidier

@Pizza right, so what's the imaginary part of that?

Pizza

10:44

$y_p(t) = \frac{2}{13} \sin t - \frac{4}{39} \cos t$

Semiclassical

yep

which indeed works

there's a whole bunch of solution methods for equations like this tho

Pizza

$y(t) = e^{-3t}(c_1 \cos t + c_2 \sin t) + \frac{2}{13} \sin t - \frac{4}{39} \cos t$

Semiclassical

i'm partial to the Laplace transform personally, tho here the lack of initial conditions makes that less preferable

Pizza

wait

what do I do now?

Semiclassical

well, are you asked for the general solution or were you supplied with initial conditions on $y(t)$ as well?

if it's the former you're done

Jakobian

10:47

Interesting thing about $y_p(t) = Ae^{zt}$ is that its two-dimensional since $A$ is also a complex number here

Semiclassical

if you were given $y(0),y'(0)$ then you need to deduce $c_1,c_2$

@Jakobian yeah. so having two parameters vs one is only a superficial advantage

Pizza

Compute the integral of the following Cauchy problem: $$\begin{cases}y''+6y'+10y=2\sin x \\ y(0)=y'(0)=0\end{cases}$$

Semiclassical

ah. then yeah, need to actually find $c_1,c_2$

@Pizza based on this expression, what are $y(0)$ and $y'(0)$?

Pizza

okok

@Semiclassical mm i mean for y(0) i need to replace t with 0

so $t=0$

they are conditions

$y(0) = e^{-3 \cdot 0}(c_1 \cos 0 + c_2 \sin 0) + \frac{2}{13} \sin 0 - \frac{4}{39} \cos 0 = c_1 - \frac{4}{39} = 0$
$c_1 = \frac{4}{39}$

Semiclassical

yeah. that's the easy one

Pizza

10:52

now ...

derivative :,)

Semiclassical

$y'(0)$ is the more annoying one. not hard, just annoying

Binky McSquigglebottom

@Semiclassical using similarity transformations to reduce partial differential equations to ordinary differential equations by exploiting symmetries in the problem.

Semiclassical

@BinkyMcSquigglebottom ...this isn't a PDE in the first place

this is just a linear inhomogeneous ODE with constant coefficients and initial conditions

Pizza

$y'(0) = e^{-3 \cdot 0}(-c_1 \sin 0 + c_2 \cos 0 - 3c_1 \cos 0 - 3c_2 \sin 0) + \frac{2}{13} \cos 0 + \frac{4}{39} \sin 0.$
$y'(0) = -3c_1 + c_2 + \frac{2}{13} = 0$

wait

i need to replace c1

Semiclassical

yeah, but you already found it above

Pizza

10:59

$c_2 = \frac{2}{39}$

$y(t) = e^{-3t} \left( \frac{4}{39} \cos t + \frac{2}{39} \sin t \right) + \frac{2}{13} \sin t - \frac{4}{39} \cos t$

niceeee

Semiclassical

yeah

the nice thing about the Laplace transform approach is that you get rid of the initial conditions along the way

the bad things are 1) having to do partial fractions, and 2) having to do the transforms/inverse transforms

so for instance it's easy to get that the Laplace transform $Y(s)$ of $y(t)$ is $Y(s)=\frac{1}{s^2+6s+10}G(s)$ where $G(s)$ is the Laplace transform of $g(t)=2\sin t$

Pizza

Could the Lagrange method be used?

Semiclassical

Lagrange method?

Pizza

yes i found it online

Semiclassical

source?

Pizza

11:04

youmath.it/lezioni/analisi-due/equazioni-differenziali/…

Semiclassical

ah, variation of parameters

Pizza

@BinkyMcSquigglebottom do you mean that method?

youmath.it/lezioni/analisi-due/equazioni-differenziali/…

Semiclassical

tbh i forget about variation of parameters, mostly b/c i don't often do cases that aren't covered by the "guess the particular solution" method

Binky McSquigglebottom

@Pizza yes

Pizza

this is Wolfram solution

Semiclassical

11:11

looks right if you expand that out

Pizza

👍

Thanks for the help @Semiclassical :)

Semiclassical

np

Pizza

@Semiclassical Was this the best way to do it?

Semiclassical

best is subjective

but

there's three routes i see for problems like this

1) guess the solution up to coefficients and deduce these (this is what you did)
2) variation of parameters, which is effective but i can't remember how it works without looking it up,
3) integral transform methods, which are neat but rely on a lot of machinery

the nice thing about route 1 is that it allows you a pretty direct mechanical solution

the problem is that it's not very effective if the ODE you're dealing with isn't constant-coefficients

Binky McSquigglebottom

Can you check my method pls ?

@Pizza It should be this

Pizza

11:26

@Semiclassical sorry, i was afk

Semiclassical

Laplace transform methods are very cool, when they work

particularly in the context of circuits

variation of parameters has the advantage of not requiring as many assumptions

Pizza

i have a question

but why can I replace 2sin(t) with 2e^it

Semiclassical

suppose $z$ solves $z''+6z'+10z=2e^{it}$

if $y(t)$ is the imaginary part of $z(t)$, then taking the imaginary part of both sides gives $y''+6y'+10y=2 \sin t$

so solving the complex-valued problem automatically yields a solution to the imaginary part

it'd also work if we wanted to do $2\cos t$, just taking the real part instead

(this kind of method is ubiquitous in physics/engineering applications---it's why electrical engineers use complex numbers a lot in their education)

more discussion here: physics.stackexchange.com/a/643532/55641

Pizza

but so if I have a differential equation with trigonometric terms, can I immediately make this change of e^it?

Semiclassical

for the inhomogeneous part, yes

for instance, that's exactly the convention i remember from my graduate electrodynamics book

Pizza

11:41

last question

Semiclassical

after a certain point they'd only ever write the diff-eqs as $y''+...=e^{i\omega t}$

(tho you still had to be careful, mostly for interpreting expressions like $|y(t)|^2$ correctly)

Pizza

when you told me to write y = Ae^it , you told me that I was treating A as a real number, when it shouldn't be, so I wrote A = a+bi, would you have done that? or another way?

Semiclassical

that's what i was getting at with my comments here

namely, that you can instead solve as $9A+6iA=(9+6i)A=2\implies A=\frac{2}{9+6i}$

the only problem with this is the denominator, but that can be cleared by multiplying by $9-6i$ on top/bottom

Pizza

yes

Semiclassical

hence $A=\frac{2(9-6i)}{9^2+6^2}=\frac{18-12i}{117}=\frac{2}{13}-\frac{4}{13}i$

which what you get before

so just treat $A$ as a complex number, solve for it like you otherwise would and deduce real/imaginary parts from that

Pizza

11:48

so if it had been 2cos(t), something would have changed

@Semiclassical 👍

Semiclassical

yeah, tho not a lot

real part of $Ae^{it}$ would be $\frac{2}{13}\cos t+\frac{4}{13}\sin t$

in actual practice you'd usually stick to taking the real part, and thus obtain 2 sin(t) as the real part of $-2i e^{it}$

but bleh

it's important to get good at that kind of thing when you have to do it a lot and need to consistently say "the real answer is the real part of the complex-valued answer"

but here it's a one-off so you might as well just say it's the imaginary part

Pizza

okok i understand

Semiclassical

(there's also the ambiguity of whether you do $e^{it}$ or $e^{-it}$. gotta pick one and stick to it)

Pizza

but are there cases where it is better to write e^-it?

Xander Henderson

12:24

@Semiclassical Yeah, it was bad. I gave up after about an hour.

That was when they started comparing gold handicaps. Like children. "Mine's bigger than yours!" "Nuh uh! Mine's the biggest!"

Ugh...

Semiclassical

@Pizza the nice thing about $e^{-it}$ is that the real part of $ie^{-it}$ is $\sin t$

also, in physics it's typical to write plane waves in PDEs (especially quantum physics) as $e^{ikx-i\omega t}$

(you need to have $x$ and $t$ having opposite signs to describe a right-moving wave, so you have to make some choices along with that)

psie

13:17

> If $\phi$ is convex, for each point $(\alpha, \phi(\alpha))$, there exists an affine function $f_\alpha(x) = a_\alpha x + b_\alpha$ such that

> - the line $L_\alpha$ corresponding to $f_\alpha$ passes through $(\alpha, \phi(\alpha))$;

> - the graph $\phi$ lies above $L_\alpha$.

Why is this true?

Xander Henderson

@psie What have you done to try to convince yourself that it is true?

psie

The definition of a convex function: $\phi$ is convex if $\phi(tx+(1-t)y)\leq t\phi(x)+(1-t)\phi(y)$ for all $x,y\in\mathbb R$ and $t\in[0,1]$.

Sine of the Time

isn't it the geometric interpretation of convex function?

the graph $\phi$ lies above?

Peter

this should be downvoted. It needs more focus.

psie

ok, I will try to look at the definition more carefully

Xander Henderson

13:22

@psie You might want to focus on what kind of line could possibly intersect $\varphi$ at a single point (because it intersects at $(a,\varphi(a))$, and at no other point). There is a word for such a line, and a description of such a line from elementary calculus...

psie

ok 👍

Sine of the Time

$t\phi(x)+(1-t)\phi(y)$ is the parametric eq of the line through $\phi(x)$ and $\phi (y)$

ephe

Let $M$ and $N$ be two filtered $A$-modules with filtrations $(M_n)$ and $(N_n)$ and $u\colon M\to N$ be a morphism of modules. We can show that if $u(M_n)=N_n$ and $u$ is an isomorphism then $u$ is continuous. Indeed, let $O$ be an open subset of $N$. We will show that $u^{-1}(O)$ is open. Let $x\in u^{-1}(O)$. Then $u(x)\in O$ and since $O$ is open there exists $k\in\mathbb{Z}$ such that $u(x)+M_k\subset O$. Applying the inverse to both sides we attain $x+N_k\subset u^{-1}(O)$ which shows that the inverse image is open.

psie

@XanderHenderson you don't mean the tangent, do you?

user 85795

think about it

Xander Henderson

13:26

@psie Do I? Possibly. Think about it. Do some work on your own. Math is not a spectator sport.

user 85795

you will not remember it other wise

Pizza

14:20

@XanderHenderson Is there a way to see if a user has deleted my invitation to the room?

@Semiclassical 👍

Xander Henderson

14:40

@Pizza No.

Obliv

15:17

does "collection" have a precise meaning?

or can we just insert the relevant definition/primitive notion from whatever framework we're working from (NBG,ZFC,MK,etc)

Xander Henderson

@Obliv Not really, no. It is just a set.

But it is nice to have other words for "set", to distinguish different types of sets. In this case a topology is a set in which each of the elements is itself a set (specifically, a subset of the space for which the topology is defined).

Obliv

got it

Xander Henderson

As English, it is perhaps easier to parse "a topology is a collection of subsets..." than it is to parse "a topology is a set of subsets...".

But they have the same meaning.

You'll also see "family" used in this way, as in "a family of sets".

Semiclassical

i feel like i tend to see "family" used more in a context where there's some parameter for members of the family

e.g. a family of curves

Xander Henderson

@Semiclassical Yes, that is generally how I would use it, too.

But, ultimately, it is just a set whose elements are curves or sets or whatever.

Obliv

15:28

In this definition in munkres, an "m-tuple" of a set (where m is a positive integer) is a function mapping each and every number from $\{1,2,...,m\}$ to some element of $X$. In the cartesian product of a family of sets $A_m$, the m-tuples $(x_1,....,x_m)$ are mapping $\{1,...,m\}$ to $\{A_1,....,A_m\}$?

er rather all of $\{1,...,m\}$ goes to $A_1$ then map it all to $A_2$ and so on

Xander Henderson

Uh... no?

That doesn't look right.

Obliv

ok how about this: the function definition of m-tuple maps all the elements of a set of natural numbers up to $m$ to some set $X$ (kind of like an indeterminate in polynomial rings?)

i will break it down to see where I am going wrong

Xander Henderson

Per the definition, an $m$-tuple is a function $\iota$ which takes $\{1,\dotsc,m\}$ into some space. So Munkres defines a space $X$ as the union of several sets, and then says that the Cartesian product is the set of all $m$-tuples which satisfy $\iota(j) \in A_j\subseteq X$.

An $m$-tuple is a kind of function. A Cartesian product is a set of $m$-tuples which satisfy a particular property.

Obliv

the indexed family of sets $\{A_{\alpha}\}_{\alpha\in M}$ where I guess I'll just write $M = \{1,...,m\} \subset \mathbb{N}$ (I'm making this indexing set up), is now the target set of the previous definition (taking the place of $X$)? Or are we doing the mapping of tuples to each individual set $A_{\alpha}$

@XanderHenderson ah okay I see

Xander Henderson

Start in a simple case: take the Cartesian product of two sets, $A$ and $B$.

Obliv

15:36

So it's the latter, that $A_{\alpha} \subseteq X$

Xander Henderson

The set $A\times B$ consists of all ordered pairs ($2$-tuples) $(a,b)$ such that $a \in A$ and $b\in B$.

Obliv

what if $|A| \neq |B|$ though

Xander Henderson

Doesn't matter? Why should it matter?

$\{1,2,3\} \times \{a,b\} = \{ (1,a), (1,b), (2,a), (2,b), (3,a), (3,b)\}$.

I don't understand...

Obliv

Ohhhhhhhh

thanks

ephe

If I made some progress on my question and now have a more precise but different question about the topic, is it better to ask a new question or edit the original?

Semiclassical

15:43

new question usually

ephe

thanks

Xander Henderson

@ephe Has the question been answered?

ephe

@XanderHenderson no it hasnt

Thorgott

@ephe You can take $M=N$ as modules, but shift one of the filtrations by one, then take $u$ to be the identity

Xander Henderson

Then I would suggest either editing the original question, or deleting the original question and asking a new one (depending on how severe the changes are).

nickbros123

15:46

is it true in infinite dim case that invertible linear map maps bases to bases? non singularity insists that it maps linearly independent to linearly independent, but cant go beyond this

I* cant go beydond this

ephe

@XanderHenderson Well it is this question: math.stackexchange.com/questions/4938539/…. I was asking to make sense of the diagram first but now I get the diagram I just dont understand how the universal property of cokernel and kernel doesn't force the map between coim and im to be not just a bijection but also an isomorphism

@Thorgott I will try this. Thank you so much.

Thorgott

@nickbros123 define invertible

nickbros123

16:04

@Thorgott If $T:V \to W$, T is invertible if there exists $T^{-1}:W \to V$ such that $T \circ T^{-1} =I_W; T^{-1} \circ T =I_V$ where $I_V:V \to V$ and correspondingly for $W$

and if it exists, T^-1 would be linear if T is linear

Thorgott

ok, so take a basis of $V$ and look at its image under $T$, what do you need to check?

nickbros123

@Thorgott yeah, I got it; Since non singularity preserves linear independence, direct image of a basis would definitely span W since if $\beta \in W$, $\beta=T\alpha$ for some $\alpha$, and $\alpha$ would be a linear combo of the chosen bases, and $\beta$ would be a linear combo of vectors present in direct image of basis...

Jakobian

16:50

@Obliv for you its most likely a class

Since the book you're reading considers NBG

Although I'm not sure what the consensus is for NBG

And other of those axiomatic systems that consider proper classes as objects

People just work in ZFC until the class issues cause serious problems

A class has then an informal meaning as a proposition, and a set belongs to that class is the same as satisfying this proposition

Its more common, I find, to consider things like Grothendieck universes or other ways for which we don't have to consider classes as objects

The approach with Grothendieck universes is something people do in category theory for example

Another way in which you can do things is e.g. consider ZFC with global axiom of choice

user10478

I'm watching a lecture on regularization, and the professor seems to be making the assumption that solving a constrained optimization problem will always yield the feasible point which has the shortest distance to the minimum (or maximum as appropriate) of the objective function. This isn't entirely obvious to me.

Is this a general fact of differentiable objective functions (perhaps related to the fact that the steepest uphill/downhill direction is always perpendicular to the contour lines), or does it hinge on some special fact about regularization?

Jakobian

This approach is a little different because it adds a global choice function to your symbols, but it helps to consider it sometimes

Thorgott

17:10

@nickbros123 yup, that works

ephe

Let $F_A$ denote the category of filtered modules over a commutative unital filtered ring $A$ with morphisms being $A$-linear maps $u\colon M\to N$ such that $u(M_n)\subset N_n$. The book states that we have a canonical factorization:

$\mathrm{Ker}(u)\to M\to \mathrm{Coim}(u)\xrightarrow{\theta}\mathrm{Im}(u)\to N\to \mathrm{Coker}(u)$

where $\theta$ is a bijection.

But why is this $\theta$ only a bijection when it is the unique isomorphism induced by the universal properties of cokernel and kernel? My guess is that the kernel and cokernel are defined as they are in the category of modul…

Thorgott

@Jakobian in that context, surely not

2

he's just defining what a top. space is

@ephe yeah, you can use examples like the one I mentioned earlier to see this

ephe

@Thorgott Thank you!

Thorgott

no problem

Soumik Mukherjee

18:29

Can someone suggest a learning roadmap for algebraic k theory? My goal is to understand Bass-Quillen conjecture.

Jakobian

@Thorgott yes.

Binky McSquigglebottom

19:55

Guys guys guys there are 5 pizzas left in the h bar 🍕🍕🍕🍕🍕 hurry up

Claudio

guys, how would you determine the residue at infinity of $z\frac{\sqrt{z^2+1}}{(z-2)^2}$?

pardon, I was thinking about something else lol, I meant the Laurent series expansion at $z = \infty$ of that function

user 85795

20:11

@BinkyMcSquigglebottom Thanks for sharing :-)

psie

Suppose $\phi$ has left and right derivates such that $$\lim_{h \to 0^+} \frac{\phi(c)−\phi(c−h)}{h} \leq \lim_{h \to 0^+} \frac{\phi(c+h)−\phi(c)}{h}.\tag1$$ Let $a$ be a number between those limits. Is it true that $a(x-c)+\phi(c)\leq\phi(x)$?

Jakobian

@BinkyMcSquigglebottom watch out for I have an endless apetite and a bottomless stomach

psie

@psie the reason I am asking is because the function $l(x)=a(x-c)+\phi(c)$ is constructed in a proof of Jensen's inequality in Durret's book, and it also needs to satisfy $l(c)=\phi(c)$.

Binky McSquigglebottom

Nice guys!!!

Jakobian

@psie is it true for differentiable functions

Didn't you forget some assumptions

Jensen inequality has assumption of convexity

psie

20:18

true, let $\phi$ be convex

but a convex function might not be differentiable everywhere

Jakobian

I think it always has upper and lower derivatives or something similar like in your setting

Sine of the Time

@Claudio do you want to write explicitely the series or compute the first terms?

psie

for example, with $\phi(x)=|x|$ and $c=0$, we pick $a\in [-1,1]$, and $l(x)$ will then lie under the graph of the function $\phi(x)$ for all $x$, but I don't see how to show this for some arbitrary $\phi$

I think I see something now

nah, I probably don't

Claudio

20:40

@SineoftheTime why? Would the procedure differ?

The answer would be only the first terms since I only need the $c_{-1}$ coefficient

Sine of the Time

the standard method should work

let $w=1/z$ and expand around $0$

Binky McSquigglebottom

How do we do this?

Is there any topic I need to read about for this?

Somehow it strikes me as triangle inequality

leslie townes

its a consequence of the connection between the definition of the norm and the properties of an inner product (or whatever you call the 'beta' in your 'euclidean space')

Parallelogram law

In mathematics, the simplest form of the parallelogram law (also called the parallelogram identity) belongs to elementary geometry. It states that the sum of the squares of the lengths of the four sides of a parallelogram equals the sum of the squares of the lengths of the two diagonals. We use these notations for the sides: AB, BC, CD, DA. But since in Euclidean geometry a parallelogram necessarily has opposite sides equal, that is, AB = CD and BC = DA, the law can be stated as 2 ( A _ B ) 2...

don't let the fact that it "has a name" and part of a wikipedia page distract you from the fact the only thing you really "need" to read about for this are the relevant definitions, which connect ||u + v|| and ||u - v|| with ||u|| and ||v||

a general tip in that direction is that when working with things that involve both the norm and the inner product it is usually more handy to work with the square of the norm than the norm itself

because ||v||^2 = beta(v,v) is a simpler formula than the definition ||v|| := sqrt(beta(v,v)), it is sometimes easier to 'see' or algebraically manipulate facts about the squared norm

Binky McSquigglebottom

20:58

Thank you so much

leslie townes

👍

Binky McSquigglebottom

$\langle v, v \rangle = 2$

:)

Claudio

@SineoftheTime that's what I was trying to do. Perfect

and thanks

Claudio

21:57

I was thinking that maybe expanding $f(1/\zeta)$ around $\zeta = 0$ and then finding the $c_1$ coefficient does not give me $Res(f,\infty)$ but $-Res(f, \infty)$ since the transformation only maps the singularities outside my path $\gamma$ and does not take into account that I'm looking at the complementary path $\gamma^{-}$

Sine of the Time

22:10

@Claudio $\sqrt{1+z^2}=z\sqrt{1+1/z^2}=zg(z)$ so that $\frac{z^2g(z)}{z^2-4z+4}=g(z)\frac1{1-4/z+4/z^2}\sim 1+4/z$ applying the geometric series and considering that the coefficient of $1/z$ of $g(z)$ is $0$ since $g(z)\sim 1+\frac12\frac1{z^2}$

Claudio

22:27

i got 4 as well, but the result should be $-2\pi i \cdot Res(f,\infty) = 8i\pi$ and if I plug in 4 I get an extra minus sign

@SineoftheTime that's also a great way of computing it, definitely quicker, thanks again

Jakobian

23:37

@AlessandroCodenotti finished Gillman and Jerison together with all the exercises in it

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