Mathematics - 2017-08-20 (page 1 of 2)

Will Hunting

04:16

@BalarkaSen Seifert, not Siefert.

Dietmar and Deitmar are both mathematicians, though.

Secret

04:44

0

Q: Why cosine and sine functions are used while representing a signal or a wave?

Actually, in the mathematics sine and cosine functions are defined based on right angled triangle.But how the representation of a wave or signal will say based on these trigonometric functions ( we can't draw any right angled triangles in the mediums(air) ))then how can we say that?

electronics

Akiva Weinberger

05:20

@Secret They satisfy the right differential equations

At least for simple harmonic motion, you need the second derivative to be proportional to the negative of position

Secret

I wonder if it is possible to produce a sine wave from a linear combination of bessels

since the bessels do form an orthonormal basis

So theoretically, the inner products $\langle J_{\nu}(x), sin(x)\rangle$ will give the coefficients in the series combination

Fourier–Bessel series

In mathematics, Fourier–Bessel series is a particular kind of generalized Fourier series (an infinite series expansion on a finite interval) based on Bessel functions. Fourier–Bessel series are used in the solution to partial differential equations, particularly in cylindrical coordinate systems. == Definition == The Fourier–Bessel series of a function f(x) with a domain of [0,b] f : [ 0 , b ] → R {\displaystyle f:[0,b]\rightarrow \mathbb {R} } is the notation of ...

that's sounds like overkill

Anyway, for an ODE with a linear differential operator, if $a_i$ are solutions, then so is any convergent series combinations of $a_i$

If the differential operator is nonlinear (because it contains nonlinear terms like $(-)^2, yD(-),$ etc.), then it might be possible for a solution to be some series combination of functions, but the individual functions are not solutions

Actually I am wrong, while for linear differential operators, "if $a_i(x)$ is a solution $\implies$ $\sum_{i}a_i$ are solutions", the converse is not necessary true.

One counterexample being sines and cosines don't solve the bessel equation. It will take a bessel function (which can be constructed from sin cos by fourier series) to solve te bessel equation, despite both it and the wave equation are linear

EyesOnBud

05:41

Hi, I have a math question

LOL

(87.2057<(-(((((sin(sin(((|((({a}+210.161)--271.942)<(({b}*{b})<=-293.63))|<(49.87^exp({a}[41]))[33])-256.239)[36][35]))=tan((-85.8608*((-104.479^sin((atan({a})/{b}[87]))[69][30][93])-((ln(|(((-{b})*{b}[46])|(ln(|{b}|)>172.746))|)[83]!=tan(((-99.2256-(({a}<=-382.473)/({a}<=-240.893)))>=(94.5361*(-(-395.226/{a}))))))=-372.904)))))=(367.384/((((217.513>(pow((-313.092&cos({a}))[15],124.239)>=(375.489|exp(((-79.3096^{b})/-17.7684))))[81])>(((sin({b})>|{a}|)<-54.43)[26][50][65]-266.508))|-304.039)!=224.515)))%2.36137)[25]&((-((asin(ln(|pow(((-342.912<=({b}[24]--185.244))+50.6944),(|{b}[65][30]|+…

Secret

that seemed to be a bunch of inequalities

user147690

06:28

@EyesOnBud The answer is approximately $13.37\pi$ (actually compute this)

Secret

07:39

What operator is [] in the above expression?

Anonymous

08:07

0

Q: How to prove L'Hospital's rule using epsilon-delta method?

For epsilon-delta proofs, basically we need to find a $\delta$ such that $|F(x)-L|<\epsilon$ whenever, $0<|x-a|<\delta$ (for a small positive number $\epsilon$). To prove L'Hospital's rule (for when numerator and denominator function both tend to $0$ as $x\rightarrow a^{+}$) let us assume $F(x)...

real-analysis limits epsilon-delta

Anonymous

Can someone help me with this ? ^

Danu

09:51

10

A: Classification of $SU(2)$ principal fibre bundles over four-dimensional manifolds

Isomorphism classes of principal $SU(2)$-bundles $P\to X$ over a closed 4-manifold $X$ are classified by $H^4(X;\mathbb{Z})$. One assigns to $P$ the second Chern class (a.k.a. Euler class) of the associated $\mathbb{C}^2$-bundle. (In the oriented case, the number $c_2(P)[X]$ is also called the in...

How come we can trivialize $X$ in the complement of a ball, over an arbitrary connected 4-manifold?

Ah, because it has to be trivial over an open manifold. Never mind! :D

Balarka Sen

11:05

@Danu Hm, that's a cute fact. I don't know how to get from [X, BSU(2)] to [X, K(Z, 4)] immediately.

Ah maybe K(Z, 4) is obtained from attaching higher cells to BSU(2) or something so that there is an inclusion BSU(2) --> K(Z, 4).

Secret

I have no idea how to calculate the cardinality of the neighborhood of $\omega_2$ under the order topology

Alessandro Codenotti

11:21

@Danu Do you have a simple example of an even dimensional real manifold which isn't a complex manifold? @Balarka said I should ask you

Danu

@AlessandroCodenotti $S^4$ lel

Fun fact, $S^4$ is not even almost complex.

Balarka Sen

how does one prove that

huh nice

Danu

The almost complex? Wu classes IIRC

Balarka Sen

shit

Danu

characteristic classes stuff

becuase complex implies restrictions on e.g. divisibility

especially Hirzebruch-Riemann-Roch is powerful in this direction

Balarka Sen

11:24

I see

Do you have to invoke big things to come up with examples like that?

I told Alessandro about F_n not being a fundamental group of any compact complex surface thing

so eg (S^1 x S^3) # (S^1 x S^3) is an example

Danu

I think it's probably not a difficult thing to prove that $S^4$ is not complex

Yeah, you just use Pontryagin numbers

Calculate $p_1[S^4]$. Now calculate it assuming $S^4$ is complex.

But proving it is not almost complex is more difficult.

Balarka Sen

What restrictions does being complex have on Pontryagin numbers? (I am not very familiar with those characteristic classes)

Danu

The Pontryagin classes of a complex bundle have a simple expression in terms of Chern classes

Balarka Sen

ah

Danu

Recall that $p_j(V)=(-1)^jc_{2j}(V\otimes_{\mathbb R} \Bbb C)$

Now if $V$ is complex then $V\otimes_{\Bbb R}\Bbb C=V\oplus\bar V$

Balarka Sen

11:27

Right. Ah.

Danu

The Pontryagin classes are then $p_j(V)=c_j^2(V)-2c_{j-1}(V)c_{j+1}(V)+\dots$

(easy exercise)

Balarka Sen

By using the direct sum formula for Chern classes, I expect. Gotcha.

Danu

So if $S^4$ were complex then $p_1[S^4]=\pm 2 c_2[S^4]=\pm 4$ (I forget the sign)

But $S^4$ bounds, hence $p_1[S^4]=0$

Balarka Sen

Nice! Thanks a lot.

Alessandro Codenotti

@Danu well that's surely easier than I expected! I suppose showing it's not a complex manifold is not easy at all?

Danu

11:30

See the above.

You need characteristic classes

Characteristic classes are basically the best thing ever :)

Alessandro Codenotti

I have no idea what does are, but thanks for the example!

Danu

The best

A beautiful combination of geometry and topology

with extremely powerful applications

@BalarkaSen Yeah apparently

idk

Also, someone on Physics claimed that principal $G$-bundles over four-manifolds are classified by second Chern class for any compact, simple Lie group $G$

This just has to be horribly wrong. Any easy counterexamples @BalarkaSen?

Balarka Sen

I don't really know enough about Chern classes to help you with that.

Secret

some fb stuff

Danu

11:47

I don't even know how to make sense out of "second Chern class" for principal bundles unless they're frame bundles of a vector bundle :P

I guess that for the compact simple Lie groups you can associate a vector bundle to them via the defining representations

But the result is not oging to be a complex vector bundle in general so it makes no sense to talk of $c_2$.

I think the user is confused with the Euler class, and that does make sense as long as it's oriented

I'm fairly certain I can find bundles with the same Euler class which are no isomorphic, over a 4-manifold

Balarka Sen

@Danu I have a question on MSE to that effect.

Danu

@BalarkaSen really? :P

Balarka Sen

You're basically asking for nonisomorphic vector bundles with isomorphic compactly supported cohomology ring of the total spaces.

You can even find nonisomorphic vector bundles with diffeomorphic total spaces.

Not sure about the 4-manifold base thing.

Danu

@BalarkaSen Why am I asking that?

Balarka Sen

Well, uh, you're not but if you have such an example then Euler classes of those are the same.

Euler class = cup square of the PD of the zero section

Danu

12:00

JK you're 100% right

PD of the ZERO LOCUS of a generic sectino

but that's of course just the intersection of the zero section with itself

Balarka Sen

yeah exactly

Mike has given two answers here to that effect

Danu

its PD is then the cup square

I proved that in a seminar

That was one tedious proof

Balarka Sen

if you dig through them you're going to find examples

Leaky Nun

\documentclass[12pt,a4paper]{article}
\begin{document}
	\newcommand{\fib}[1]{
		\newcount \a
		\newcount \b
		\newcount \c
		\newcount \i
		\a = 0
		\b = 1
		\c = 1
		\i = 0
		\loop
		\advance \i by 1
		\a = \the\b
		\b = \the\c
		\advance \c by \the\a
		\ifnum \i < #1 \repeat
		\the\a
	}

	\fib{11}
	\fib{12}
	\fib{13}
	\fib{14}
\end{document}

89 144 233 377

Danu

@BalarkaSen how about this

Take the two non-isomorphic oriented rank three bundles over $S^2$

Now if I take the direct sum bundle over $S^2\times S^2$

Balarka Sen

12:12

yeah that works

Danu

That has zero Euler class

and... is not trivial (????)?

In general I should look to produce examples by adding trivial bundles

hmm?

Balarka Sen

Sorry, ignore that.

Aren't there zero Euler class but nontrivial rank k bundles on S^4?

Danu

Let's try to find out :D :P

Maybe it's easier to work with something with $H^2\neq 0$ so that we can use first Chern class

I mean what if you we just take a non-trivial line bundle $c_1\neq 0$ and then sum with $\Bbb C$

That has zero Euler class for sure but isn't trivial by the usual formula for Chern classes

That works on anything with nonzero $H^2$, right?

skullpatrol

Is a mod allowed to dismiss the flag of his message?

Danu

And the transition functions take values inside $SU(2)$, no?

No, this is weird haha

@skullpatrol I think so

Balarka Sen

12:23

Hey. Why not Milnor's S^3-bundle on S^4?

The exotic sphere example

Total space of that is homeomorphic to S^7, so it's automatically zero Euler class

I guess you want vector bundles though. Is this sphere bundle linear?

Oh yeah I think it is linear

I suppose that means you get a rank 4 vector bundle on S^4 with nonzero Euler class but which is not trivial (because the unit sphere bundle is Milnor's bundle)?

I am pretty sure fiddling with this will give you an example. I gotta go right now though, have to prepare for the one last exam

Danu

OK, "enjoy" ;-)

user193319

12:40

I know that if $\lambda$ is an eigenvalue of $A$, then $\lambda^k$ is an eigenvalue of $A^k$, but I am wondering whether the converse holds. That is, if $\lambda$ is an eigenvalue of $A^k$, then must $\lambda^{1/k}$ be an eigenvalue of $A$?

Danu

@user193319 No; take for instance $-\operatorname{id}$. It has eigenvalue $-1$ but its square has eigenvalue $1$.

Leaky Nun

@Danu well you can choose the value of $\lambda^{1/k}$ to make the converse hold...

user193319

@Danu I am not sure what you are getting at. $-I$ is not a power of $I$, if that's what you are aiming at.

Leaky Nun

@user193319 $(-I)^2 = I$

user193319

No. I said that it isn't a power of $I$; i.e, $-I \neq I^k$ for any $k \in \Bbb{N}$.

Leaky Nun

12:50

@user193319 $A = -I$

user193319

I think I have a proof of the theorem, but there is one step that bothers mean. Here is it: $\lambda \in \sigma(A)$ if and only if there exists $v \in V \setminus \{0\}$ such that $A(v) = \lambda v$ if and only if $A^{k-1}(A(v)) = \lambda A^{k-1}(v)$ or $A^k(v) = \lambda \lambda^{k-1} v = \lambda^k v$ if and only if $\lambda^{k} \in \sigma(A^k)$.

The only problem I see with the proof is, what if there is some $k$ such that $A^k(v) = 0$? This would imply $v =0$, which is a contradiction. I can't rule this out, nor can I find a example in which $A^k(v) = 0$ for some $k$.

Martin Sleziak

BTW there is a related post on main: If $\lambda^k$ is an eigenvalue of $A^k$, is $\lambda$ an eigenvalue of $A$?

user193319

Oh! I see. Thank you @MartinSleziak

Martin Sleziak

@user193319 I am not sure whether you claim that this is equivalence: $(\exists v\ne0) Av=\lambda v$ $\iff$ $(\exists v\ne0) A^{k-1}Av=\lambda A^{k-1}v$

There must be some problem with the above proof, as the linked counterexamples show.

Martin Sleziak

13:15

@user193319 This is, in my opinion, one of the problems in the attempted proof above. If you read it from right to left, then you know $\lambda^k\in \sigma(A^k)$, which gives you $A^kv=\lambda^kv$. But what do you use to replace the RHS by $\lambda A^{k-1}v$?

BTW I have reposted this in linear algebra chat room - maybe somebody interested will notice it there.

Danu

@BalarkaSen I just saw that the proof that $S^4$ is not even almost complex is really easy.

If it were almost complex, we would have $p_1(S^4)=c_1^2(S^4)-2c_2(S^4)=3\sigma(S^4)=0$. Since $H^2=0$, we find that $c_2=0$. But $c_2$ is the Euler class.

This is really exactly the same as what I said before, because you only need an almost complex structure to turn $TS^4$ into a complex vector bundle. Also, this is of course only easy modulo proving that the Pontryagin numbers or the signature are zero for oriented boundaries. But I remember showing you how to do it for the signature ;D Though now we're stuck using the signature theorem so you need to do some work in any case...

Secret

It has been 2 months and I still made that typo of "ordinary" to "ordinal"

Danu

@user193319 $I=(-I)^2$ but $\sqrt 1=1$ is what I was saying.

Of course, $(-1)^2=1$ as well so it's not such a great counterexample.

@MartinSleziak it's funny; the top answer on that post is the IMMEDIATE thought I had when I saw the question. Alas, I realised the $-\operatorname{id}$ example was even easier (to check explicitly), though not as "good".

Secret

13:40

in The h Bar, 49 secs ago, by Secret

Meanwhile, I forgot what is the complex number generalisation to a metric

sciencedirect.com/science/article/pii/S1110256X15000541

O, it is literally what is said on the tin

BAYMAX

Hi!

quick1

$\int_{-\infty}^{x} e^{\frac{a}{x}} dx$

how do I solve this integration?

Will Hunting

One does not solve an integration. One evaluates or computes an integral.

BAYMAX

yeah my bad!,good will hunting

:)

Leaky Nun

$\displaystyle \int_{-\infty}^{x} e^{a/x} \ \mathrm dx$

Not really sure if you can find its closed form

BAYMAX

yes , my latex looks different than yours!

also your profile pic changed! :)

actually this comes from a probability question

Leaky Nun

13:54

@BAYMAX show us the whole question

LegionMammal978

Hmm

BAYMAX

Let $X$ be a random variable having PDF $f(x) = \frac{1}{\lambda} e^{-\frac{\lambda}{x}} , x>0 , \lambda > 0$

find the $p$th Quantile?

LegionMammal978

What would an author mean if they say a value has an asymptotic upper bound of, say, $2^{n(1+o(1))}$?

BAYMAX

for which I had done $\int_{-\infty}^{x} f(t) dt = p$

now I have to find $x$

Leaky Nun

@LegionMammal978 like $2^{1.324342n}$

BAYMAX

13:58

$\int_{-\infty}^{x} \frac{1}{\lambda} e^{-\frac{\lambda}{t}} dt = p$

LegionMammal978

@LeakyNun I get that, but would this bound allow me to find a exact, "hard" upper bound for a given $n$?

Leaky Nun

@LegionMammal978 no idea

BAYMAX

$\int_{-\infty}^{x} e^{-\frac{\lambda}{t}}dt = p\lambda$

LegionMammal978

This is why I hate asymptotics so much :/

Leaky Nun

@LegionMammal978 more context?

LegionMammal978

14:01

@LeakyNun Was trying to get some upper bounds on the kissing number $k_n$ for $n=25..31$ (the standard tables only go up to $n=24$), and a loose upper bound is $k_n\leq2^{0.401n(1+o(1))}$

The problem is, I have no clue what bounds there are on the $o(1)$ term

Leaky Nun

@LegionMammal978 strange notation

LegionMammal978

@LeakyNun Why so?

Leaky Nun

I think, based on my guess, that $o(1)$ is between $0$ and $1$

@LegionMammal978 because you wouldn't use $o$ if you already have that term

LegionMammal978

Well, the paper the bound comes from is in Russian :P

Secret

wolframalpha.com/input/?i=integrate+e%5E(a%2Fx)

The integral is nonelementary

To get the wolfram alpha result, change of variable $\frac{a}{x}=u$ and then integrate by parts the $-\frac{a}{u}\frac{e^u}{u}$ formed

sorry what I said is wrong: The wolfram alpha result is a direct integration by parts of $1*e^{\frac{a}{x}}$

Hmm...

Leaky Nun

14:20

ya just use reciprocal substitution and then laurent series

Secret

14:37

777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777

nitsua60

@Secret You alright?

Secret

I guess so.

recently discovered that chat tags can be made this long

nitsua60

my-understanding-is-they-can-be-made-arbitrarily-long-subject-to-message-size-restrictions

Secret

well at least it does not strech the screen like those super long \$\$ messages

Will Hunting

@secret I just wanted you to know I don't mind your long messages and self conversations in this room. I quite enjoy them actually. =D

Leaky Nun

14:49

@WillHunting icon

Will Hunting

@LeakyNun Sorry, I don't understand what you mean by icon.

Leaky Nun

@WillHunting [your] icon [changed]

Will Hunting

@LeakyNun Oh yes, it is the latest pic I just took. It was taken to show my suffering.

@LeakyNun You can see all my pics on my instagram. =D

Secret

[Integral project] I am starting to wonder whether it is possible to capture all special functions into some kind of algebraic structure. Take the above integral as an example, it can be integrated simply because when $f(x)=\frac{a}{x}$, we call $\int xf'(x)e^{f(x)}dx$ the special function $a\text{Ei}\left(\frac{a}{x}\right)$

This means, of the $\mathfrak{c}$ many integrable functions, it just happens that finite many of them are interesting enough to earn a special name for them

This means in general, integrating an nonelementary integral is basically the same as asking the question of whether you can split it up into known integrals

Since infinite series are lesbegue integrals under the countable measure, it might explain why Waiting can often found the close form of some series. What is actually happening is we are just expressing one series in terms of integrals and other known series

So unless there really is some universal rationale on how people define special functions, the question of close forms is really the following:

Will Hunting

@Secret Well, that's because Waiting is the world's leading authority on series and integrals. =D

Secret

15:04

Definition: Given a Lesbegue integral with some measure $\mu$, a closed form is an element $f \in (\mu,\Bbb{F})$ such that it is a finite combination under addition, multiplication, exponentiation and composition of named elements $g \in (\mu,\Bbb{F})$

That is, suppose $h$ is the integrand. Then, the integral has a closed form if:

$$\exists f,g \in (\mu,\Bbb{F}), \int h d\mu = f = \prod_{n_1,...n_5 < \infty}(+^{n_1},-^{n_2},\times ^{n_3},\frac{-}{-}^{n_4},\circ^{n_5}) g_{n_1n_2n_3n_4n_5}$$

where the messy bracket is an ordered sequence of operations to be applied

E.g. $\sin (x^2) = \sin \circ (-)^2 \circ x$ ($\circ$ is right associative)

Therefore, an integral has an elementary antiderivative if $g$ are elementary functions

I am not sure, given the size of the space of all integrable functions, whether there can exists a function $k$ such that it can only be produced by an infinite composition (including operators like addition, multiplication, exponentiation etc.) of all known functions in the literature

DaneJoe

If I wanted to continuously deform e^x into ln(x), I would need a model for fractional iterations of e^x

I see nothing strange about this, so Can anyone tell me why people keep making up outlandish statements like "There is no entire (holomorphic everywhere) function f(z) with f(f(z))=z?" It's clearly easy to do something like ask mathematica to deform the exponential function into the half-exponential function which is shown extending over the entire domain.

and then from there to deform it into ln(x)

Secret

15:19

The above thing regarding the function $k$ can make a good MSE question, but probably way too broad to ask

DaneJoe

no questions more broad than this have already been asked on stack exchange

specifically regarding that topic

it was a stack exchange member who essentially said the existence of such a function would be a contradiction

and it was accepted as an answer

but according to topology...there's nothing wrong with it

we don't even have a taylor series for f(x), why would someone make the assumtion that the iteration wouldn't yield the proper range for e^x?

sorry, no, questions...*

Secret

@DaneJoe I would like to read the link to that, because that will basically imply every series and integral, as long they converge, will have a closed form if we actually named every single function in the space of all integrable functions

DaneJoe

oh wait never mind it wasn't accepted as the answer

but it had the most upvotes

also what's wrong with every integral having a closed form?

you know the functions already exist, but we simply haven't discovered them, right?

the numbers aren't random, they have to equation something

equal*

Secret

Well, one thing that I wonder about is whether there exists integrals that can only be computed numerically and has no closed form even in the most general sense

DaneJoe

the only time it wouldn't work in the general sense

is if for every partial

we needed to discover a new function

in other words that we will never know the generalization

but I don't know where that is the case

I guess it's possible

but with our current simple axioms of functional analysis

I see nothing wrong with the proposition that every convergent integral has some closed form in the form.

in the form of any combination of computational operators

There is already an answer for all the unanswered questions, it's just a matter of discovering them.

Secret

15:37

@DaneJoe Do you have the link to the question?

Dattier

Hi, $$f\in C^3(\mathbb R) \text{ with }f' \times f'''<0\\
\text{Is it true that : }\forall a,b\in \mathbb R^2, |f(a)-f(b)| \leq |f'(\frac{a+b}{2})|\times |a-b| $$

DaneJoe

I thought I posted the link

id I didn't my bad ill post it now

26

Q: Does the exponential function have a square root?

(asked by Nathaniel Hellerstein on the Q&A board at JMM) Is there a "half-exponential" function $h(x)$ such that $h(h(x))=e^x$? Is it unique? Is it analytic? Related question: Is there an invertible smooth function $E$ such that $E(x+1)=e^{E(x)}$? Is it unique? If so, then we can take $h(x)=E(E...

ca.classical-analysis-and-odes fractional-iteration

Secret

> so we are led to the absurd conclusion that $e^z$ takes the value zero infinitely often.

DaneJoe

well

that's not generally strange

because sin(x) does so

it just means it has a

hold on

I remember this term vaguely

let me find it

damnit

sorry

but they removed the term from wolfram's page for some reason

but anyway

shiva

math.stackexchange.com/questions/2400288/… please see this

DaneJoe

15:48

it's a point where a function changes from being monotonic to oscillating

it occurs in the cosine integral

but there is no reason to suspect that the half-exponential function does this that I can see

at least from a topological perspective

Mike Miller

@DaneJoe It's not generally strange, but it's also not true of the exponential, which never takes the value zero, since $|e^z|=|e^x e^{iy}|=|e^x|>0$

DaneJoe

@MikeMiller Right that's what I'm saying

the half-exponential function could take on 0 once

but only once

Shiva is english your first language?

@shiva *

I'm not being condescending I'm actually asking

Secret

Hmm, infinite number of roots does not sound very bad. It depends on whether it is countably many or uncountably many. Also since $e^x$ is already nonlinear wrt $x$, I don't see what's the problem that the half exponential will also be a nonlinear function hence it can very well map $0$ back to something nonzero when it is applied the second time?

DaneJoe

how could the zero of an exponential function be uncountable

zeros*

and yes the half exponential function has the property that ln(f(e^x))=f(x)

so it should be able to be mapped back

but for some reason it's excruciatingly difficult to explicitly define any form of the exponential function

I think I know where the infinite zeros comes from though

usually fraction iterations is defined about fixed points

e^x has no real fixed points

Secret

I am thinking about Let $a$ be in the kernel of the half exponential $f$. Then $f(a)=0$ and then $\exists b \in \Bbb{C}, f(f(a))=e^a=b\neq 0$

DaneJoe

15:54

yeah I don't think the half-exponential function would ever fail to be injective

I can see how it seems like a contradiction from a functional view

since the half-exponential function would span all real numbers

whereas e^x cannot

but topologically there is nothing wrong with the half-exponential function

but it should also be considered

that given how little we know about the function

we shouldn't be so quick to assume its iteration wouldn't yield e^x over at least some domain

It would seem a more generalized definition of a function is needed to reconcile these two conclusions

something that includes iterations of functions

Secret

If both $f$ and $e^x$ are linear functions, then yes we will have a problem since for linear function, we cannot escape from the kernel since for $l$ linear $l(0)=0$

But since both of these functions are nonlinear, then we can escape from the kernel as $f(0)$ can be multivalued. In that case, I think it is even ok for the half exponential to span all real numbers. Perhaps it can be defined in some continuous manner such that positive numbers maps to positive numbers, and negative numbers maps to some other negative numbers such that after two applications of $f$ it must map to positive num…

DaneJoe

The half-exponential function should span all real numbers

but not because it returns to 0 infinitely often

f(0) shouldn't be multivalued

the half-exponential function, in terms of homotopy, should be injective

which is to say that if you continuously deformed e^x into the half exponential function you would arrive at a function that has no strange properties, that is simply monotonically increasing

but grows a little more slowly than the exponential function

though eventually faster than any polynomial

Secret

You mean the half exponential itself is an injective function or the deformation map from it to $e^x$ is an injective map?

DaneJoe

the deformation map, at least so far

we can't currently prove the half-exp function itself is injective

shiva

@DaneJoe Not a first language

DaneJoe

16:08

at least not as far as I am aware

Okay, well then this brings up a good point

Shiva doesn't speak english, but they got downvoted for being unclear

that isn't right

this isn't the right room for that but it is the situation I expected

I would recommend using like google translator first if you can

typing it in your own native language as best you can, and then pasting the result

@shiva you shouldn't feel bad about it though, it's stack exchange's fault that they aren't accounting for the cultural diversity of its members

ys wong

Why do people always say that Complex Analysis is beautiful?

DaneJoe

because it lets you solve a lot of problems you shouldn't solve before

and there's a lot of unexpoected and yet simple results

unexpected*

like e^(-pi*i)=-1

er actually just pi*i

and also couldn't*

plus when you look at complex maps

ys wong

Is it diffcult?

DaneJoe

it takes a lot of training to get to it

then linear algebra

then differential equations

you need to finish all of calculus

then applied or functional analysis

is functional analysis itself difficult? it depends on the problem

ys wong

I currently finished Real Analysis

DaneJoe

16:17

sounds like complex analysis is a good way to go

complex analysis is heavily based on things like multivariable deriviatives

but also spans the other topics

ys wong

Does Complex Analysis share a lot of similar concepts with real?

The only difference being that in Complex u do things in the complex plane rather than the real line

DaneJoe

it has some similarities, it doesn't deal as much with discrete operators

but rather gets more into complex mapping and defining functions of a complex variable

so its based more on integrals

In my opinion it's harder because it's a bigger leap to go from only thinking about real variables to complex variables

but it should be equally as tedious as real analysis

polar coordinates will come up more too

Akiva Weinberger

@yswong Complex differentiable functions are much nicer than real differentiable functions.

For example, in complex analysis, if something has a derivative, then it is infinitely differentiable (its second derivative, third derivative, etc., all exist).

This is wildly not true in real analysis.

DaneJoe

@AkivaWeinberger would you agree that deriving them is harder? It's harder to work with descrete functions but deriving them is easier, whereas deriving complex functions is harder but using them is easier

ys wong

To me i think the fun starts from the integral portion onwards

DaneJoe

16:22

a lot of times all you're really trying to prove is that a function applies to complex variables in a similar way that it applies to real variables

ys wong

From cauchy integral formula onwards

DaneJoe

that e^x is defined not only for real x, but for any x+iy

Astyx

What's the set of derivatives of complex functions ? Is that something we know ?

Akiva Weinberger

The main conceptual difference is that, a function is differentiable if it's locally linear. And "linear" means something different for real functions then complex functions.

DaneJoe

Well it means something more general

Akiva Weinberger

16:24

@Astyx It's the same as the set of differentiable functions, at least on a simply connected domain, IIRC.

DaneJoe

just that it's mapped linear

Akiva Weinberger

The phrase I heard to explain linear maps in the complex setting is "amplitwist"

DaneJoe

complex analysis is like a combination of real analysis, linear algebra and calc 3

Akiva Weinberger

Basically, $f(z)=az$ stretches ("amplifies") the complex plane if $a$ is real, and rotates ("twists") it if it's imaginary

DaneJoe

complex numbers tend to act like ordinary numbers in some instances, like vectors in other instances, and like angles in others, so you should be comfortable with all those forms of math

Akiva Weinberger

16:25

and if $a$ has both a real and imaginary part, it does both

Note that $f(z)=\bar z$ and $f(z)={\rm Re}(z)$ are not linear or amplitwists

so something that locally looks like the conjugation function is not complex differentiable

(Amplitwists preserve angles and the orientations of those angles)

Astyx

So any differentiable function is the derivative of another function ? How do you prove that ?

Akiva Weinberger

(Unless the derivative is 0, because that means that it locally looks like a constant function.)

@Astyx No, every differentiable function is the derivative of another function, in complex analysis.

Remember that once-differentiable things are infinitely differentiable.

Astyx

Yeah, a mistake on my part that took an eternity to correct cause I'm on my phone

Akiva Weinberger

You show that integrals don't depend on the path of integration

and then you integrate it

Astyx

Cool

Akiva Weinberger

16:30

That's why I said simply connected domain, though. $\frac1z$, as defined on $\Bbb C\setminus\{0\}$, isn't the derivative of anything. (You can't define $\ln z$ for the entirety of $\Bbb C\setminus\{0\}$.)

(You get the weird multifunction thing where the value can drift in multiples of $2\pi i$ as you go around the origin)

Astyx

Right

Perturbative

17:07

Hey everyone!

Leaky Nun

@AkivaWeinberger can't I just integrate it?

Alessandro Codenotti

How do you integrate it? The integral is not path independent since the domain is not simply connected

Balarka Sen

@LeakyNun Locally, yes. Globally, no.

Alessandro Codenotti

not all paths are homotopic

Leaky Nun

@AlessandroCodenotti I see

Balarka Sen

17:13

If it was a derivative of something integrating it along any loop would give you zero.

But clearly, that does not hold.

ys wong

Whats the difference between Cauchy integral formula and residue integration?

How do u know when to use which?

Balarka Sen

Residue theorem is a generalization of Cauchy integral formula.

Alessandro Codenotti

The residue theorem is a generalization of Cauchy

Balarka Sen

SNIPED

Alessandro Codenotti

WTF

Balarka Sen

17:20

SNIPED AGAIN

Alessandro Codenotti

nobody answers for 5 minutes and then I get sniped as I type!

how is that possible

ys wong

Yup but which instance is it better to use Cauchy integral formula and which instance is it better to use residue theorem?

Perturbative

Humour me for a second, a vector field is conservative iff it is the gradient of some vector field right?

Alessandro Codenotti

In most exercise you'll want to use the residue theorem I think

Balarka Sen

@Perturbative Yeah. Well, gradient of some function, you mean.

Perturbative

17:26

@Balarka Yeah woops, meant scalar potential

Well in a test our lecturer set a question, where we were given a vector field $F = \nabla ln(xy)$, which is obviously conservative, we weren't asked to prove it was conservative, but we all got marked down for not showing that is was conservative

The question wanted us to draw some field lines and other boring stuff, but not to prove conservativeness

gxyd

17:48

Does anyone know of a manuscript that discusses "Weierstrass substitution"(also called tangent half angle substitution) in terms of the Risch algorithm?

There is one paper that discusses the Weierstrass substitution in symbolic integration but "not" specifically in terms of Risch algorithm, this one apmaths.uwo.ca/~djeffrey/Offprints/toms1994.ps

I posted the question now on site itself math.stackexchange.com/questions/2400454/…

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