Mathematics - 2017-08-30 (page 6 of 8)

@Phase just ask

Phase

$\int_{0}^{ln(2)} \frac{1}{(1+e^{-x}}$ using the substituion $y = e^{-x}$, apparently transforms to $\int{0.5}{1} \frac{1}{y(1+y)}$, but I end up with $\int{0.5}{1} \frac{-1}{y(1+y)}$

huh. i seem to have forgotten formatting too lmao.

Why is the font so unreadable when its being edited

3:39 PM

@Phase maybe you forgot to swap the limits or something

Phase

Ah crud

god damn it I knew it was simple

ty

i swapped the limits subconsciously and forgot to change the sign RIP

good morning

@LeakyNun it works on 2 elements =p

You changed ur picture back leaky

@KasmirKhaan nice

3:42 PM

e e
a e

:D

@Faust7 Ja

@LeakyNun Is that enough ? i mean like i put that answer on homework ? =p

Hes nto working on the same problem is he?

@KasmirKhaan ae=e?

@Faust7 it's a different problem

a*e =a

3:43 PM

cause ill def tell him the answer

lol k

@Faust7 I dont wish to meet the other 6 fausts out there if all fausts are this way -.-

There is only one Faust

I killed the other 6

@Faust7 if you did that , you would not name urself faust 7

Faust is the protagonist of a classic German legend, based on the historical Johann Georg Faust (c. 1480–1540). Faust is a scholar who is highly successful yet dissatisfied with his life, which leads him to make a pact with the Devil, exchanging his soul for unlimited knowledge and worldly pleasures. The Faust legend has been the basis for many literary, artistic, cinematic, and musical works that have reinterpreted it through the ages. "Faust" and the adjective "Faustian" imply a situation in which an ambitious person surrenders moral integrity in order to achieve power and success for a delimited...

hmm nice

3:46 PM

@KasmirKhaan (aa)a = ea = e; a(aa) = ae = a

unlimited knowledge and worldly pleasures seems like a good deal for one little soul

Semiclassical

@Faust7 eh. unlimited in scope, but not unlimited in duration

@LeakyNun damn it :(

@Semiclassical Unlimited knowledge i feel like immortality shouldnt be that hard to figure out

Semiclassical

depends. that unlimited knowledge might just be enough to tell you that it's impossible

3:48 PM

Plus an Arch-Angel gets his soul back

@LeakyNun can you send me the solution on the 3by 3 case?

@KasmirKhaan but it is possible with 2 elements

@LeakyNun its cant be on 2 element

@Semiclassical true but id like to think not

@KasmirKhaan it can

3:49 PM

e
a e

those are fixed

if i make e*a= a then we have a group

why are those fixed?

As i dont particularly intend to die if i can help it

xe=e for all x in G

wait

Morning @EricSilva i had a question for you but i have forgotten it

so first colum is e
a

e
a

a*a =e ( because zx=e for all x in G)

3:52 PM

@KasmirKhaan why must a*a=e?

because zx=e for all x in G , it means that all elements have left inverse

@KasmirKhaan and then?

@LeakyNun each row must have an e

a*e=a

@KasmirKhaan why?

and if a*a=a then we dont have that critera fullfilled

I told u =p

3:53 PM

why must each row have an e?

because for each element it must have a left inverse

so?

zx=e for all x in G

so?

for every x in G there exist an element z in G such that zx=e

that must be fullfilled

3:55 PM

yes

how can that be fullfilled if second row is ( a,a)

ae=a and aa=a

if second row is a,a then ae=a and aa=a

and then?

a has no left inverse

that is not what we want ._.

@KasmirKhaan think carefully

have we exhausted all products of the form xa where x in G?

in order to know that a has no left inverse?

hmm let me think

hmm

e e
a a

4:01 PM

good

each column needs to have an e

:D

Hmm I think ill do the 3 element case another time

been here for 5 hours now :D

alright

well we did homomorphism and isomorphism today i need to do those as well

but after a break =p

Thanks alot again @LeakyNun

:D

np

surprised u did homomorphism and isomorphism so early

my first AA class we didnt even do homomorphism O.o

4:08 PM

@Faust7 we just introduced them, not theorems about them =p

we didnt even introduce a homomorphism

we do stuff different here =p

though technically isomorphism is a non-shitty homomorphism

very technical i can tell:D

=) its what i do best

4:10 PM

faust number 7 , am going to have a break for an hour or so and ill come back later-.-

I hope you by then tell me a good reason why u are number 7

@El'endiaStarman Maybe a simpler question will fetch an answer: what is $$\frac{\partial{}}{\partial{\theta}} \dot{\theta}$$ and why? Where $\dot{\theta} = \frac{d\theta}{dt}$.

@KasmirKhaan what r u talking about?

That was harder to do then than? i thought

@WillHunting

Just stumbled upon a problem: Can we regularize the following divergent series to some finite number?$$1-1+2-2+3-3+4-4+\dots$$

nfi

@SimplyBeautifulArt ...

4:22 PM

can u pair each term then sum all the zeros?

@Faust come on

what?

oh no you cant

@Faust No... that's like saying Grandi's series is zero =P

what are the partial sums?

1,0,2,0,...

i member someone telling me i couldn't do that already can someone expalin why again?

4:24 PM

@LeakyNun yup, $0$ and $(n+1)/2$

@Faust because the partial sums do not converge

@SimplyBeautifulArt and the average?

it oscillates between 0 and 1/2 right

so it isn't cesaro-summable

@LeakyNun The Cesaro sum goes to $+\infty$

The Abel sum goes to $+\infty$

@SimplyBeautifulArt how?

The Euler sum goes to $+\infty$

1/1,0,2/3,0,3/5,0,...

4:25 PM

so when i look at the partial sums i have to gorup the terms in a specfic way what way is that?

the partial averages fluctuate between 0 and 1/2

@Faust the usual way

@LeakyNun That's not how you Cesaro sum?

from left to right

@SimplyBeautifulArt what?

> The Cesàro sum of a sequence is defined as the limit of the arithmetic means of the partial sums of that sequence.

@LeakyNun well i have never done that before and the usual way appears not to be the logical way

Hrmf

4:26 PM

@Faust it is the logical way: it's how you sum numbers

Cesaro shouldn't work here

$$\lim_{n\to\infty}\frac{1+0+2+0+\dots+\frac{n+1}2}n$$

@SimplyBeautifulArt oops, sorry

It'll alternate between 0 and $\frac{n}{2n+1}$

right, its cesaro sum is infty

4:26 PM

I think

or maybe $n+1$ in the numerator

@LeakyNun perhaps if your brain works normally ima goggle it.

I'll be back later

But honestly don't know how to regularize it in any nice way

try Ramanujan summation

@SimplyBeautifulArt try euler?

Binomial transform

In combinatorics, the binomial transform is a sequence transformation (i.e., a transform of a sequence) that computes its forward differences. It is closely related to the Euler transform, which is the result of applying the binomial transform to the sequence associated with its ordinary generating function. == Definition == The binomial transform, T, of a sequence, {an}, is the sequence {sn} defined by s n = ∑ k = 0 ...

@LeakyNun A partial sum is the sum of the first n terms! you could of just said that

4:29 PM

@Faust well, lol

then yes the partialsums arent all 0

right

Im not stupid i just don't understand what your saying =)

anyone got any article son this Cesaro method?

Cesàro summation

In mathematical analysis, Cesàro summation assigns values to some sequences without a sum in the usual sense. The Cesàro sum of a sequence is defined as the limit of the arithmetic means of the partial sums of that sequence. Cesàro summation is named for the Italian analyst Ernesto Cesàro (1859–1906). The term summation can be misleading, as some statements and proofs regarding Cesàro summation can be said to implicate the Eilenberg–Mazur swindle. For example, it is commonly applied to Grandi's series with the conclusion that its sum is 1/2. == Definitions == A sequence {ai} is called Ces...

Sankyuu

4:33 PM

@LeakyNun $+\infty $

@SimplyBeautifulArt :'(

@SteamyRoot didn't try

@LeakyNun @SimplyBeautifulArt Thanks both of you thats very interesting!

Cesaro is nice in the sense that it's still somewhat intuitive

I just recently derived intuitively the Euler sm

4:35 PM

@SteamyRoot it reminds me of those intergrals

$$ \lim_{n \to \infty} \int_{0}^{n} x^2 $$

wierd

double subscript

still broken

change _ to ^

oh

Those integrals

they have a name

irrational? raddical?

definite?

they are alot easier to do with contour integration

Let $S_n$ be the nth partial sum. Let $$S(n,k)=\begin {align}S_n, &k=0\\\frac {S(n,k-1)+ S (n+1,k-1)}2,&k>0\end {align} $$

Euler sum of $S$ is $\lim S(1,k)$

4:41 PM

@SimplyBeautifulArt demonstration?

@LeakyNun at school :(

But it's intuitive this way I think

Theres alot of these

@Faust of what?

methods of summation didnt realise there was so many

@Faust principal value

@Faust lol, there are quite a few

4:46 PM

dam touch pad likes to click while i am typing.

Haha, Yeah, I used to have one

I disabled it

it works better than most and i like to use that feature to click occasionally so i leave it on

Where do you learn all these summation methods?

ive taken a reasonable amount of calculus and i didnt even know what a partial sum was...

@Faust xD

I enjoy adding up things you shouldn't add up :P

well some of these ideas seem to apply to integrals too

@El'endiaStarman I think they treat $\theta$ and $\dot\theta$ as two different variables

which is absurd, but

idk

4:50 PM

they must of run out of $\alpha$'s

@LeakyNun That much is clear. I'm trying to work out the justification behind that.

@Faust use all the Greek letters!

$\alpha \beta \ceti ? \gamma \omega \rao $

@El'endiaStarman does physics need justification?

ah i missed delta! facepalm

4:52 PM

Are you doing Lagrangian mechanics or something like that? It's normal to treat $\theta$ and $\dot\theta$ as different variables in that case @El'endiaStarman

@AlessandroCodenotti Yes, Lagrangian mechanics.

@AlessandroCodenotti :O

you should ask @Semiclassical, he's probably the one who knows the most about the topic here

@LeakyNun Ugh, not you too. Look, the math worked out in the end, so there has to be a reason it's valid (to some extent).

@El'endiaStarman /s

4:54 PM

@El'endiaStarman lol i miss physics for that reason right there

Little Rookie

Hello, i was comparing the definition of functional limits and continuity (real analysis), I noticed the definition of functional limits includes Limit Points and does not require the real variable $x$ to be in Domain of $f$.

Whereelse the definition of continuity does not include Limit Points and require $x$ to be in Domain of $f$.

Why is it so? Any help please.

@AlessandroCodenotti Thanks. :) [settles down to wait for Semiclassical]

Hes usually here hope hes ok...

14 hours ago, by Daminark

The connection is fine af too. Affine connection, you may say

The idea is that to do Lagrangian mechanics you work on the tangent bundle of the configuration space (and sometimes on the tangent bundle of that), this doubles the dimension of the manifold so instead of just having free coordinates for the position you also have them for the velocities and you effectively treat them as separate coordinates, you write $L(r_1,\cdots,r_n,\dot{r_1},\cdots,\dot{r_n})$ for the Lagrangian etc.

But I forgot the details, let's hope Semi arrives soon :P

5:08 PM

Welp, that went over my head quite quickly.

Should I change my username back to my real name?

@AlessandroCodenotti Wait, that might have actually cleared it up for me. Lemme see if I understand. When doing Lagrangian mechanics, to make the math easier, you treat the position and velocity of each particle as independent? So then, what ties them together is the.....Lagrangian itself?

@AlessandroCodenotti actually what is your profile picture?

@LeakyNun seven people looking into a well, seen from the bottom of the well

@AlessandroCodenotti :O

5:15 PM

its abuncha faces

nvm

But my words like silent raindrops fell and echoed in the wells of silence...

Thats oddly cryptic

@El'endiaStarman Hmm, it's more of a problem with finding a proper mathematical framework to talk about mechanics

@Faust Alessandro saying the word "well" made me recall that line from the song Sound of Silence.

Imago

hei

5:19 PM

I've seen this things in a mathematical physics course that was very heavy on the maths, I'm not sure about the physical interpretation of why this is correct

Imago

Given be the area of a 2-dimensional elephant is $A$. Shrinking the elephant now by its width, will also half its area. (at least if one considers the elephant area to be lebesque or riemann integrable)
However how does this change for 2-dimensional animals, one may consider "weird shaped"? Is there a quick example to see, when linearity is not preserved, but measuring area still makes sense?