Mathematics

(a) You can do it explicitly by a simple substitution. (b) Here's where you should think about symmetry. What do you notice about $f(x)$ and $f(-x)$ for the integrand?

Jake Rose

$f(-x)=-f(x)$?

Ted Shifrin

Right. So what does that tell you about area under the curve?

Jake Rose

=0

Ted Shifrin

Yup.

Jake Rose

Doesnt that imply that the entire integral =0?

Ted Shifrin

12:10 AM

Only when $x_0=0$.

Shift your coordinates and rethink.

GFauxPas

Ted, the last part was that the integral = 0, which is what the q asks

Ted Shifrin

Ah, OK, @GFauxPas.

GFauxPas

:) thank you

Balarka Sen

@Ted +1 on the answer. I still don't quite see what torsion being the Lie bracket really means though, I admit

Ted Shifrin

Well, all the algebra goes into torsion. 0 holonomy, 0 curvature.

Balarka Sen

12:14 AM

Well, maybe somewhat. The geodesics are the same, and somehow the twistyness all goes in not being able to complete a square with a flowlines of X and Y

Akiva Weinberger

You might have gotten a +1 on that answer, but I still have no idea whether geodesics in $\Bbb R^n$ with a strange connection are straight lines, or whether or not this is possible in $\Bbb R^2$.

Ted Shifrin

Now what does torsion actually mean? I mentioned spending weeks trying to understand that from Cartan. You have to think about holonomy for the affine connection, but instead of parallel translating vectors around little paths, we think about the base point and how it moves.

DogAteMy: You can make up a connection so that any curve you want will be a geodesic.

Akiva Weinberger

Compatible with the metric, though?

Balarka Sen

Ted's example is not compatible with the metric is it.

Ah no it is.

Ted Shifrin

I didn't think about that, but it probably is.

Jake Rose

12:17 AM

@TedShifrin Does the integral value change in some 'nice' way? As far as I can tell by shifting it all you do is make the integral either some real positive or negative value?

Ted Shifrin

Because everything is invariant.

Balarka Sen

'Cuz inner product of left invariant vector fields are constant

My metric is biinvariant

Ted Shifrin

@JakeRose: You can compute that value explicitly!! Let $x-x_0 = u$ and change variables.

GFauxPas

what's this notation $\int f(x) \, \mathrm d \lambda(x)$, same as $\int f d\lambda$?

Akiva Weinberger

Are $y_0$ and $z_0$ relevant?

Ted Shifrin

12:20 AM

Nope.

Yes, @GFauxPas.

GFauxPas

is there an advantage to writing it the left way?

Drew Brady

to put the argument of the function?

GFauxPas

good call

Akiva Weinberger

I guess if you want to consider $\int f(x^2)~{\rm d}\lambda(x)$

Ted Shifrin

Right. Just like you can write $\int_a^b f$ for the integral of $f$, but then if you want to integrate $x^2$, you really need $\int_a^b x^2\,dx$.

Akiva Weinberger

12:21 AM

or even $\int f(x)~{\rm d}\lambda(x^2)$ should make sense

Balarka Sen

@TedShifrin I suspect there's some nontrivial restriction on holonomy for a given connection to be compatible with some metric?

The holonomy needs to preserve a fiberwise quadratic form...

GFauxPas

you mean to say that $\int \phantom{x}^2 d \lambda$ isnt good notation? :P

Secret

$\int f(x)~{\rm d}\lambda(x^2) = \int f(x)2x\lambda ' (x^2)~{\rm d}x $

Jake Rose

@TedShifrin is it $x_0 \sqrt(pi)$

Ted Shifrin

DogAteMy: Are you forcing me to have the Euclidean metric or just some flat metric on the plane?

Yes, @JakeRose.

Secret

12:24 AM

How I parse $\int \phantom{x}^2 d \lambda$. It gives nonsense like:

$$\int ~{\rm d}\lambda^{~{\rm d}\lambda}$$

GFauxPas

\int \phantom{x}^2 d \lambda

you can write click tex expressions and vew the tex

Jake Rose

And then are the other two integrals (dy and dz) both equal to $\sqrt{pi}$?

show math as

$\pi$

Yup, @JakeRose.

12:26 AM

$\phantom{x}$

This opens a whole new world...

GFauxPas

$\displaystyle \int ~{\rm d}\lambda^{~{\rm d}\lambda}$ looks very mysterious

Secret

I am not even sure if titration of differential is well defined, need to check

Jake Rose

So then it evaluates to $x_0 \sqrt{\pi}^3$?

Ted Shifrin

OK, I'm outta here. I'll think more about your question, DogAteMy.

Secret

tetration, stupid Mac thinking too much about chemistry

Ted Shifrin

12:27 AM

@JakeRose, yes, but I'd write that as $\pi^{3/2}$.

Jake Rose

Yeah thats easier

Secret

10

Q: Exponential of the differential operator

I am not sure whether this question is even well-posed. But today I learnt that $e^Df(x) = f(x+1)$ where $D$ is differential operator and $$e^D \triangleq \sum_{i=0}^{\infty} \frac{D^i}{i!}.$$ (ref. Dan Piponi's answer) So I was curious as to whether the differential equation $$\frac{df(t)}{d...

differential-equations

there's exponential, but not really tetration

Jake Rose

Ive got the answer sheet and it says $x_0 (a \sqrt{\pi})^3$?

GFauxPas

what's 'a'?

Jake Rose

It was in the original expression

Secret

12:29 AM

3

Q: Logarithm of differential operator

Using the Taylor series expansion we have (for a sufficiently regular function $f$): $$ f(x+a)=\sum_{k=0}^n \frac{f^{(k)}(x)a^k}{k!} $$ So, defining the differential operator $D=\frac{d}{dx}$ and using the series expansion definition of the exponential function, we can write: $$ S_a f(x)=\exp(aD)...

reference-request logarithms operator-theory exponential-function differential-operators

Jake Rose

Why doesnt it like that notation?

GFauxPas

there you go

Secret

but to have $\ln D$, we need the function to be rapidly decreasing

after that, we should be able to define $e^{D \ln D} = D^D$

and so potentially, for a certain class of functions, we can have: $e^{~{\rm d} \ln ~{\rm d} \lambda} = ~{\rm d}^{~{\rm d}\lambda}$

Akiva Weinberger

I don't think $\ln D$ is defined, similarly to $1/D$

(as integrals are only unique up to constants)

But $\ln(1+D)$ should be fine

and thus $(D+1)^D$ I guess?

GFauxPas

well, we know what the log-derivative is, so $\ln D$ is simply the derivative-log

(don't treat that seriously)

Secret

12:33 AM

Yeah if you want the ln to be injective

so $e^{D \ln (D+I)} = (D+I)^{D}$ seemed valid

(I is the identity map)

Akiva Weinberger

What's $\ln(1+x)$ again? $x-x^2/2+x^3/3-\dotsb$, right?

Alternating?

So I guess $\ln(1+D)f=f-\frac12f''+\frac13f'''-\dotsb$

I don't know if there's a simpler form, like there is with $e^D$

Secret

ln of maps in general are more restrictive than their exp counterpart, usually only certain maps will have that series to converge

Drew Brady

Anyone around able to help me with part (d) of this: imgur.com/a/Q3c7o

GFauxPas

hint, please, anyone?: determine for which $\alpha > 0$ we have $\displaystyle \int_0^1 x^{-\alpha}\, \mathrm d \lambda(x) < \infty$

this one is worth two questions

in difficulty

Akiva Weinberger

Incidentally, there is a sense in which $e^{e^x}$ is not defined in the ring of formal power series but $ee^{e^x-1}$ does

Drew Brady

12:37 AM

My problem is regarding functional analysis

Akiva Weinberger

(though I think in practice we define the former to mean the latter)

I'mma go now for a bit

Secret

in particular, the map needs to be invertible, which I believe I+D is always invertible, I think (another trouble is that the details of D depends on which space we operate in, it behaves like a shift operator in polynomial and power series spaces)

but for general Banach spaces, I don't have the background yet to guess

$\int_0^1 x^{-\alpha}\, \mathrm d \lambda(x) = \int_0^1 x^{-\alpha}\, \lambda'(x) \mathrm dx$

so I guess when is $x^{-\alpha}\lambda'(x) < \infty$ in [0,1]

GFauxPas

I dont know what $\lambda'$ means, we didnt define that in class yet

Secret

Is $\lambda$ just some function on x, or something else?

GFauxPas

Lebesgue measure

Secret

12:42 AM

O...., then my above logic is all wrong

GFauxPas

:) it's okay, I didnt explain the notation in the exposition

Secret

I am not terribly good at lebesgue integration thus I cannot help without making tons of mistakes

GFauxPas

I forgot what $x^\ell$ even means when $\ell$ isnt rational

$e^{\ell \ln x}$?

Secret

Yes. In general, any exponent not rational is defined via exp

GFauxPas

$$\int_0^1 e^{-\alpha \ln x}\, \mathrm d \lambda(x) < \infty$$

hmm

user193319

12:48 AM

Problem: Suppose that for some $\epsilon \gt 0$, every $\epsilon$-ball in $X$ has compact closure. Show that $X$ is complete. Proof: Let $(x_n)_{n \in \Bbb{N}} \subseteq X$ be a Cauchy sequence. Then given that $\epsilon > 0$, there exists $N \in \Bbb{N}$ such that $d(x_n,x_m) < \epsilon$ or $x_n \in B(x_m, \epsilon) \subseteq \overline{B(x_m, \epsilon)}$ for every $n,m \ge N$. Choosing $m=N$, we see that $(x_n)_{n \ge N}$ is a sequence contained in a compact metric space, meaning it has a

... convergent subsequence. But that convergent subsequence is also a subsequence of $(x_n)_{n \in \Bbb{N}}$. Hence every Cauchy sequence has a convergent subsequence, which means $X$ is complete.

How does that sound?

Secret

(Completely unrelated observation) So completeness does not fully characterise what it means to be uncountable

Secret

1:36 AM

$$\int \sqrt{f(x)}dx$$

Let $f(x)=y^2 \implies x = f^{\leftarrow}(y^2) \implies dx = \frac{df^{\leftarrow}(y^2)}{d(y^2)}2ydy$

$$\int \sqrt{f(x)}dx = \int \frac{\pm 2y^2 df^{\leftarrow}(y^2)}{d(y^2)}dy$$

$$ = \pm 2 \left(\frac{y^3}{3}\frac{df^{\leftarrow}(y^2)}{d(y^2)} - \int \frac{y^3}{3} \frac{d^2f^{\leftarrow}(y^2)}{dyd(y^2)}dy\right)$$

majormaki

Hey, quick question: Does anyone know what $L^p(0,1;U)$ could possibly mean, where $U$ is a bounded (open) domain.

Secret

what's the context that it is being used?

Secret

$$\int \frac{y^3}{3} d\left(\frac{df^{\leftarrow}(y^2)}{d(y^2)}\right)$$

Let $z = \frac{df^{\leftarrow}(y^2)}{d(y^2)}$

$\int z d(y^2) = f^{\leftarrow}(y^2)$

majormaki

1:52 AM

The context doesn’t help.

I know that $L(0,1)$ is all $f\in L^p$ s.t. $f:[0,1]\to\mathbb{R}$.

Secret

yes

But it is not obvious to me why we need to specify an open domain U in such space

majormaki

You don’t

I suppose it’s not neccesary. It’s just that $U$ is ussually reserved for such a domain. That is just the notation.

Secret

Unless, it means those $f \in L^p(0,1)$ we are considering, are restricted to some subset $U$

i.e. $f|_U$ might be a possibility

majormaki

Perhaps, but $U\subset\mathbb{R}^n$. It is not an interval.

Maybe $f:U\to (0,1)$ and $f\in L^p$?

0celo7

@majormaki you'd write that the other way around

it could be vector-valued $L^2(0,1)$ functions

but having values in $U\subset \Bbb R^n$ isn't well-defined since $L^p$ functions are defined a.e.

But you could make sense of it, I suppose. Are you sure the context doesn't help?

majormaki

2:05 AM

Yes, and even if it did, I could not give it to you. I do not have it.

Thank you.

Secret

$L^p$ spaces are defined as $(\int |f|^{p} d\mu)^{\frac{1}{p}} < \infty$, I don't see why we cannot have $f$ being finite supported (i.e. a sequence such that all entries after the nth are zero)

then $f$ could be taken from a vector space and thus it becomes a subset of functionals that live in $L^p$ space

Secret

3:53 AM

$\int z d(y^2) = f^{\leftarrow}(y^2)$

$\int z2y dy = f^{\leftarrow}(y^2)$

Seth Taddiken

Why is it that when you have $v = \langle{x,y,z}\rangle$, $v^n = r^n \langle {sin(n\theta)cos(n\phi),sin(n\theta)sin(n\phi),cos(n\theta)} \rangle$?

Secret

looks like spherical coordinates stuff

but is v a vector?

Seth Taddiken

Sorry I just realized how out of context that was

Mandelbulb

The Mandelbulb is a three-dimensional fractal, constructed by Daniel White and Paul Nylander using spherical coordinates in 2009. A canonical 3-dimensional Mandelbrot set does not exist, since there is no 3-dimensional analogue of the 2-dimensional space of complex numbers. It is possible to construct Mandelbrot sets in 4 dimensions using quaternions and bicomplex numbers. White and Nylander's formula for the "nth power" of the vector v = ⟨ x , y , z ⟩ ...

I'm trying to understand the math for generating a Mandelbulb

I'm not sure why the 3D shape is defined the way that it is and I'm wondering what I need to learn in order to understand this

Secret

@XanderHenderson That's your speciality, you might want to help this guy out

Well, I am not even sure if <> is a vector there

Seth Taddiken

4:16 AM

I'd love some help, I think this stuff is so cool. I have a grasp on spherical coordinates but I don't know why the $n$ becomes a coefficient of $\theta$ and $\phi$... and how taking a vector to a power even works since dot and cross-products produce scalars, which can't be repeated...

anon

5:45 AM

@SethTaddiken the mandelbrot set uses complex numbers, the mandelbulb set uses quaternions. so you'll want to understand quaternions

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$Mathematics$ Mathematics