serious-discussion

[twin flicker]

i cheated on my diet slightly for the past 2 days so i have to lock back in

[gusty bough]

lock in

[sturdy obsidian]

hI

[burnt ledge]

zased what kind

[polar temple]

is this a new form of "based"

do u have a heatwave as well

we have a 32° heatwave in England

[vital niche]

im in year 10 so im just doin newton's first and second laws relative to terminal velocity yh

[vital niche]

its so bad vro but rn its better

[burnt ledge]

Oh nice

[vital niche]

yh wbu??

[burnt ledge]

most of my degree was physics

[gusty bough]

Yep

[gusty bough]

if you like that, i'm pretty sure you'll enjoy lagrangian mechanics

basically newtonian mechanics but better

[vital niche]

thats sickk thankyou

[gusty bough]

It requires to know a tiny bit of Newtonian physics and ordinary differential equations (which once you learn derivatives or integrals at school, I encourage you to do) and you always have the same equation give a better coordinate system

So if you have a pendulum, it would be better to analyse the angle instead of (x, y) coordinate; you would go from 2 variables to a single 1! Then apply your formula and boom you got the equation of the movement as an ODE

That’s the ultimate basics of Lagrangian mechanics

[polar temple]

u in the uk too?

its good to see year 10s studying

i didnt study until y11 started

summer holidays too

do u do aqa triple?

[barren cairn]

Hai!

[zealous garden]

not very new

[polar temple]

now i dont know if i was right or if "zased" means "not very new"

probbaly tje first

[gusty bough]

Oh I thought zased was based with a typo

[zealous garden]

I'm saying zased is a form of based but it's not new

[torpid bay]

,w zased

[fathom swallow]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by yeatte137

[jade belfry]

hi

[sharp mulch]

This doxes your last name

[jade belfry]

how?

[ancient garden]

wjat the helly are you talking about?

[outer crag]

vro who is this

unc copying my pfp

[ancient garden]

its on google....

[outer crag]

most people's pfp's are

[latent edge]

,av 974792201143996416

[fathom swallow]

yeetsaltaccount's Avatar

Click here to view the image.

[latent edge]

Copying your profile as we speak

[outer crag]

💔

[steel vessel]

what happened to ur name

[dusky locust]

ANOTHER @outer crag??!!!

YOU'RE MULTIPLYING

[jade owl]

hi fiscussy 2

[dusky locust]

....what..

[soft wadi]

hey guyd

guys

[rain vale]

Huy

[rain osprey]

Hi

[eternal crow]

YEAST

huy ka rin

[uncut crest]

I have to download the new version of OneNote. Because the old version declares its end of service

I do not like changes

The old one is still working. I cannot find any reason to adapt a new one

Fk Microsoft

@microsoft

Now I have to adapt the new version

I'm so fking sad

fk

why

[untold cedar]

https://youtu.be/UIKGV2cTgqA?si=C8eGt9Q_feNIC-sB

now I don't feel bad i forgot how to do subtraction

my parents are laughing at me
they say you are a freaking math student and u can't count

no I can't

[median knot]

ufff 😭

[mighty knoll]

This was awesome thank you

[polar temple]

youre weird bro

😭 sorry

[zealous garden]

?

I am weird but nothing about that interaction supports this claim

[polar temple]

cuz why would you willingly say "zased" instead of "based"

its the first time ive seen it

[zealous garden]

I didn't say it homie

[polar temple]

even "based" is strange according to my hypocrisy

[zealous garden]

I told you what it means

[polar temple]

oh

mb

[zealous garden]

It was asked and I answered

[polar temple]

you're good bro

you probably are still weird

youre in a maths server

but thats the only thing making u weird afaik

[zealous garden]

https://tenor.com/view/koto-omg-hi-travis-touchdown-no-more-heroes-gif-25067651

<@&268886789983436800>

[blazing lintel]

30+37=67

67

[vivid quest]

what is this video about

i've seen people react so i'm curious now

nvm

it's explained in the first 3 second

i be like that

acting

then thinking

[vernal token]

Ah Tom Lehrer

[burnt ledge]

Cz its funny and i like it

[gray dove]

hello guys

[zealous garden]

Hey

[gray dove]

damn this channel pretty dead

[willow crystal]

It's more of a slower paced version of #discussion

rather than being dead

[burnt ledge]

Hiii

[vast wraith]

it means people have time to think

[vivid quest]

I don't use that extra time

[edgy tiger]

https://youtu.be/MMmBiYJyXog

THIS SHIT IS FIYA !!!!!

[jade belfry]

The sky is not blue. It is gray—painted over to hide the cracks. The sun is not yellow. It is a projection, hung like a stage light over an empty stage. And the stars? Only the lies we plant so no one notices the dark. There are still dinosaurs—but only one remains. We killed the rest and called it a “necessity.” Someone decided they should die by asteroid, hiding the real truth. We say volcanoes finished them, only because we can’t bear the truth staring back at us. And the stars? They’re just projections—images we plant to pretend everything is fine. But every time we say it’s fine, we strip away another piece of the world.

“The saddest thing isn’t the world is burning.
It’s that you have a cup of water, and you don’t dump it.”

[tough stratus]

yeah okay

[peak zephyr]

that guy sounds dumb

[fiery stump]

bro does not know how to put out fires

you need way more water than a cup, even for a campfire

[jade belfry]

you don't understand this is about apathy.

[outer crag]

Jeremy.

[peak zephyr]

Why be apathetic when you can lose your job instead

[silent mauve]

Jeremy? Jermy? Jerma? JERMA?

[mighty knoll]

Why not both?

[solid yarrow]

Big cup of water

[faint light]

i thought this was about global warming and i was like how tf would wasting water help

[kindred otter]

Have anyone done math projects how did you start?

[gritty heath]

asking your profs is a good start, but also if there is anything that interests you, you could work in that direction
what kind of math do you like, or what hobbies do you have

[kindred otter]

I mostly love calculus

[gritty heath]

can you code

im thinking if you can code, you can try writing an automatic differentiator
you can read up on this
https://en.wikipedia.org/wiki/Dual_number
and write some program
you'd probably want a DualNumbers class, a Function class with various hard coded functions, and the hope is you could try to implement a program that reads your input and gives you a closed form expression of your derivative, or a numerically evaluated value

idk how feasible is it tho

[zealous garden]

How terrible

[terse comet]

water makes things colder therefor dumping a cup of water on the ground cools the earth

hope this helps

[soft jungle]

Gurl...

[zealous garden]

I saw we build a fridge for the solar system

[faint light]

naruhodo…

[compact surge]

:3c

[vocal vessel]

Global warming solved!

[terse comet]

I hate how discord embeds these

Bit of a meme high card run, made it to ante 11

porple stake

@faint light i think plays this right

[mint canopy]

Love when stone joker works out

Those red holo stones do things to my brain chemistry

[fresh comet]

@soft jungle welcome to the mathcord!

[faint light]

😭 jesus

[soft jungle]

@fresh comet tyy

[terse comet]

the red seal holo was really helpful since high card is so low base mult

though I did get it to like lvl 20

[naive tundra]

did anyone here study in cpge or go to x/ens (pin me plz if you see this message)

[junior veldt]

pog

[serene heron]

I had a nice jokerless run a while ago

flush builds

[left summit]

Hey I’m new at uni
Can someone explain what CCS, MS, and SESA are ?

[raven tiger]

Hello, I’m new to the server I came here to ask a question and discuss about it and I have no idea where to post it so I joined this server to ask it and maybe have some answers or advices of where to ask it. The question is about a new "problem" to study that I am wondering wether or not I should try to study it or not, sorry it’s difficult to explain myself but basically I studied a problem during a research internship (which is more in theorical cs) and I found a generalization of this problem that I would like to study more mathematically and I don’t really know in which category it belongs to but the question would be about is it worth spending time studying the problem but my current question is where can I ask this question and add more detail and context ???

[agile fiber]

ask in one of the appropriate subject channels (e.g. if you think it's still theoretical cs then #theoretical-cs )

[raven tiger]

That’s the thing I don’t really think it’s theorical cs anymore the question goes to mathematicians I would say it maybe relates to advanced algebra but I’m not sure also, maybe I can ask it there and people will tell me if it’s not in the right channel ?

Thank you for your answer still

[agile fiber]

just make your best guess at which subject fits best

[mighty knoll]

Can you describe your project?

[raven tiger]

What do you mean exactly by "my project" ? I just posted the question in #advanced-algebra

I’m anxious now that the question is weird or incomprehensible or it does not belong here but it’s ok

[mighty knoll]

You’ll be fine, it sounds like pretty cool research

[steep nimbus]

Hello

Is anyone else here?

…

Alr

Im leaving

[fresh comet]

[tired carbon]

Hey guys, I finished calc bc last year, and for my senior year, I'm starting Calc 3/Multivariable - any tips? Did you guys enjoy this class?

[latent edge]

For resources you can check Khan Academy

I did not enjoy Calc 3

[spark scroll]

i didnt enjoy it either 🥀

mostly how it was taught tho

[dry steppe]

hey guys does anyone know if matricies are in calc ab?

[torpid bay]

it wasn't in mine

ab should just be like limits and derivatives, that's all (maybe some trig ig)

[fresh comet]

@dry steppe welcome to the mathcord

[dry steppe]

so how iportant would you say pre calc is to actual calc ab

[torpid bay]

the class I took was trig/calc, where precalc was the other option, so I can't really remember what it was sepcifcially. but in general, trig is pretty important for calc

i would have to look over the direct content again

gimme a sec

https://www.khanacademy.org/math/precalculus basing it off of this, Unit 2, unit 9, and unit 10 are the most important for calc. but in general you'll need to learn the other ones along the way at some point

trig, series, and limit+continuity

unit 4: rational functions are something that I see barely, and probability and combinatorics you'll end up taking an entire class about later to fully flesh out. While unit3: conic sections are the simple 2D equations that you'll need to be able to visualize, usually just visuallizing them is enough to get you on the right path if you ever see a problem with the equations from those. tho generally I just end up seeing a circle 95% of the time, and 4% is hyperbolas

vectors and matrices are great, and will be important later on, like in calc 3 or linear algebra, but It's fine to get a handle on them here

[zealous garden]

matrices are great
Yuck

Every theorem gets better when you remove the matrices sorry not sorry

"Given a symmetric matrix B and Invertible S, and matrix D=S^*BS
D has the same number of positive, negative, and null eigenvalues as B."
vs
"Two metric vectorspaces V,W are isometric iff V_+≈W_+, V_-≈W_-, and V_0≈W_0"

[torpid bay]

ah lel

$V \cong W \iff V_+ \approx W_+ \wedge V_- \approx W_- \wedge V_0 \approx V_0$?

[fathom swallow]

Yeatte

[zealous garden]

Yes

[torpid bay]

I'm not sure whats meant by approx?

[zealous garden]

Isometric, linearly isomorphic, same thing really

[torpid bay]

ah

[fiery stump]

omg i've proven two things isometric b4 in differential geometry

[zealous garden]

It's easy enough to prove the general equivalence on (in)definite metric vectorspaces, V+ etc. So all that remains is showing that every metric vectorspace admits a, not necessarily unique, decomposition into these (in)definite spaces

[zealous garden]

I think very different

That is for manifolds yes?

Pseudoriemannian manifolds?

This is for the local case of that I guess?

[fiery stump]

riemannian manifolds under particular charts

and induced riemannian metrics

[torpid bay]

my intro to diff geo was a lot of fun

the formulae are pretty nice looking too

[fiery stump]

i barely understood any of mine but i had fun too

[zealous garden]

Two metric vectorspaces $(V,B)$,$(W,D)$ are isometric iff there exists a linear map $S:V\to W$ s.t. $S$ is an isomorphism, and $$B(u,v)=D(Su,Sv)$$

[fathom swallow]

wraithlord_kojima

[zealous garden]

Very simple

[zealous garden]

Under particular charts?

[torpid bay]

B,D being metrics?

[zealous garden]

You mean more than just a smooth structure and metric tensor?

[zealous garden]

Symmetric bilinear forms yes

[fiery stump]

man i dont remember

[torpid bay]

pretty simple requirenents

[zealous garden]

We make no ask of nondegeneracy for many definitions

[torpid bay]

cones ❤️

[zealous garden]

You may notice that through linear representation of bilinear forms, we have that B(u,v)=u^TBv=D(Su,Sv)=u^TS^TDSv => B=S^TDS

[warm gull]

man i really really need a good geometry book

[zealous garden]

So it is pretty clear how our desired statement of isometries, the classification of metric vectorspaces, is equivalent to the typical Sylvester law of inertia

[torpid bay]

,w sylvester law of inertia

[zealous garden]

HAHAHAHA

Garbage wolfram as always

[torpid bay]

huh..... certainly reminds me of the jacobian and metric tensor

this is a nother form of it, right?

[zealous garden]

What?

,w jacobian

[fathom swallow]

Wolfram Alpha doesn't understand your query!
Perhaps try rephrasing your question?
Click here to refine your query online

Requested by wraithlord_koto

[zealous garden]

HAHAHA

[torpid bay]

jacobian is the change of basis matrix

with partial derivatives

[zealous garden]

Jacobian is too many things

You mean the differential?

The map between tangent spaces?

[torpid bay]

i think yeah ( also a thing that refers to too many things)

$J = \frac{\partial x_i}{\partial x_j}$ or smthn

[fathom swallow]

Yeatte

[torpid bay]

i cant quite remember the indices but yeah osmeting like this

maybe they were both on top?

[zealous garden]

The partial derivative in the ith coordinate direction, grab the jth component with respect to the basis. No?

[torpid bay]

did you see my diff geo notes that i posted a week ago?

yeah they're both up

$J = \frac{\partial x^i}{\partial y^j}$

[fathom swallow]

Yeatte

[warm gull]

can you please help me

it’s fairly simple

[worn ice]

https://hips.hearstapps.com/hmg-prod/images/riemann-zeta-function-1569526951.png?crop=1xw:1xh;center,top&resize=980:*

guys

[fiery stump]

im not clicking links to domains i dont know

[swift bronze]

2nded

[willow crystal]

!help

[quasi jetty]

To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.

[lost quartz]

@adham33

@wise siren

[ocean harbor]

u can tell its an image bc of png? not really a dangerous link but rather its embed fail

[dreamy ore]

we have notation like $a_n = 2\cdot 0.5^n$ and then $a_5 = 1/16$, why cant we have $f_x = x^2 + 3$ and then $f_{2.5} = 37/4$ and $f'_{2.5} = 5$

[fathom swallow]

shiru

[torpid bay]

rip i went to sleep and iddnt see this

[warm gull]

lol it’s ok

[zealous garden]

Idt so

[neat lintel]

Suppose that $TT^$ is invertible, show that $P=T^(TT^*)^{-1}T$ is a projection onto $(\ker(T))^\perp$

[fathom swallow]

Nicole

[neat lintel]

comp of kerT = imT^*

P^2=P

idk

[latent edge]

Is it safe to assume T is a bounded linear operator between Hilbert spaces?

[neat lintel]

finite dimensional vector space

[scarlet mango]

hi @mint canopy . I'm very very sorry to tag you but I couldn't help noticing a small error in your bio. The spoiler part would grammatically be right if it were written in one of the three ways: 1. I asked: Why are you here?
2. I asked why you are here
3. I asked "why are you here?"
If you don't blacklist me, thank you

Also, sorry for stalking

[unborn meteor]

LOL

[scarlet mango]

hello! Could you explain why this equation is true? I prolly won't understand but it wouldn't hurt to try

I'm a language nerd

[burnt ledge]

That is a definition

[scarlet mango]

welp

what is the intuition behind it?

[burnt ledge]

its like generalising the derivative

or partial derivatives if yoire familiar with them

[burnt ledge]

this tells you how much the output of a function changes when you change its input

[torpid bay]

oh i took shower am bajj

[scarlet mango]

that sounds like crazy work

[torpid bay]

i'ts required for quite a lot of things

or rather the concept of a derivative

u know derivatives?

[scarlet mango]

yep

can you show me the simplest application this thing may have?

[torpid bay]

it's probably to use it on vectors then yeah

[scarlet mango]

wait isnt that basically the normal derivative? Im a lil stupid

[torpid bay]

it's a little different because when you have 1 dimension input, and 1 dimension output, all you need is $\frac{dy}{dx}$

[fathom swallow]

Yeatte

[gritty heath]

If your function has multiple inputs and outputs you have to track all of thrm, so the jacobian assembles all the derivatives together

[torpid bay]

but when you have 2 inputs, then representing how much the output changes requires you to ask how much each input changes.

[scarlet mango]

How does it achieve that?

[scarlet mango]

That sounds true enough

[gritty heath]

You take the derivative of each output with respect to each input and throw the value in an array

That's it

[burnt ledge]

partial derivatives go a little bit further and they tell you how a function of several variables change when you change each of its variables

[gritty heath]

There are reasons why the array is shaped in some way but its fairly geometric and I dont geometry

[scarlet mango]

when you say value, you mean it could be a function or a scalar or 0, right? And by array, what does it look like?

[burnt ledge]

jacobian tells you how each of the outputs of a function with several outputs and several inputs changes when you change any of its inputs

[gritty heath]

You can find derivative at one point then its just a number

[scarlet mango]

yup thats true

[haughty lion]

A matrix where the (i,j)th entry is the rate of change of the ith output variable w.r.t. the jth input variable

[burnt ledge]

Yea sory thats what i meant, a derivative describes a function R -> R, partial derivatives encode the effects of each of the several inputs of a function R^n -> R, jacobian tells you each of the effects of all the inputs on all the outputs of a function R^n -> R^m

[torpid bay]

$\lbracket \frac{\partial z}{\partial x} \frac{\partial z}{\partial y} \rbracket= \frac{\partial z}{\partial u^i} = J$ for an example of 2 inputs, x,y and 1 output z

wait hold up

does it need to be same n for R^n?

[scarlet mango]

So, isnt it just a system of functions, each one being the derivative of our multivariable function wrt each variable?

[scarlet mango]

Yeah, I was asking whether we are talking general derivatives or at certain points

[torpid bay]

general derivatives

you can put in specific x,y values after

[scarlet mango]

so its a matrix? or is it just some kind of isomorphism?

[torpid bay]

to find a specific number for

it's a matrix yeah, it can be used to also change between coordinate systems

[gritty heath]

Yeah matrix

[scarlet mango]

In the left-most side, why is that product not in normal parentheses?

[fathom swallow]

Yeatte
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[torpid bay]

i dunno the [ ] latex

but its not a product

[burnt ledge]

yea a matrix

[torpid bay]

its a mtrix

https://en.wikipedia.org/wiki/Jacobian_matrix_and_determinant

[scarlet mango]

yep thats what I've known jacobian to be useful for

[scarlet mango]

how is u^i defined exactly?

[torpid bay]

i is just an index

$u^1 = x, u^2 = y$

[scarlet mango]

What is it?

[fathom swallow]

Yeatte

[torpid bay]

in this case

[scarlet mango]

ahh I see

[torpid bay]

imagine there's actual spaces in my latex

[burnt ledge]

In fact u can think of it like a linear transformation too, so say we have a function f: R^n -> R^m and we want to do something like a taylor expansion f(v+dv) = f(v) + [some matrix] dv + [higher order terms]

[scarlet mango]

Could we really use this definition in double integrals to change coordinates?

[torpid bay]

yes

[scarlet mango]

that seems demented

[scarlet mango]

Is there a solution process you can find that uses this matrix representation?

[burnt ledge]

if we want to find the best linear transformation to do this (similar to how f(x+dx) = f(x) + f'(x) dx + ... would do it for one variable) you find it is the jacobian

[torpid bay]

the only way i know how is to manually calculate it, by just takign the derivatie

[scarlet mango]

I never learned that f(x+dx)= f'(x)dx lol

[torpid bay]

$x = r\cos(\theta), \frac{\partial x}{\partial r} = \cos(\theta), \frac{\partial x}{\partial \theta} = -r\sin(\theta)$

[fathom swallow]

Yeatte

[scarlet mango]

Thats really weird lookin

[latent edge]

What discussy?

[burnt ledge]

taylor expansions are very important I think they can help alot in understanding

[torpid bay]

^

[scarlet mango]

so that equation is a taylor expansion?

[burnt ledge]

Yes

[latent edge]

No

[burnt ledge]

well the one that i wrote is

[scarlet mango]

lmfao

[burnt ledge]

youve slightly misquoted me there

[torpid bay]

[scarlet mango]

which one is that?

[latent edge]

This yes

[burnt ledge]

f(x+dx) = f(x) + f'(x) dx + ...

[scarlet mango]

but you mentioned taylor exp. right after writing f(x+dx)= f'(x)dx am I crazy?

[scarlet mango]

wait shit

[gritty heath]

Actually
3 blue 1 brown has a calculus series

I'm sure the jacobian is explained there

[scarlet mango]

I never learnt that thats a taylor expansion

[gritty heath]

Much better than a text medium like discord can ever explain

[molten raptor]

konnichiwa

[torpid bay]

it might be in a different form than usual

[burnt ledge]

What do you call that?

[scarlet mango]

I never understand his vids tho, its like for actually intelligent people

[scarlet mango]

I doubt that

[scarlet mango]

Thats interesting

[burnt ledge]

I keep clicking back to mathscussy when i want to come back to ts conversation and tehn i remember this is happening in discussy 2

[scarlet mango]

yeah sorry it started as a simple question

[torpid bay]

$f(x) = \sum_{n=0}^{\infty}{\frac{(x-x_0)^n f^{(n)}(x_0)}{n!}$ might be the form you know of

[burnt ledge]

Its no problem haha im just very forgetful

[torpid bay]

in this case it's just a little substitution to get to the x + dx form

[scarlet mango]

yupyup

[fathom swallow]

Yeatte
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[scarlet mango]

And we can sub dx-x_0 to be some constant? or am I crazy for this?

[torpid bay]

let x = y + dy, let x_0 = y or smthn

[latent edge]

The community is getting healthy

[torpid bay]

yeah

and then we get $f(y + dy) = \sum_{n=0}^{\infty}{\frac{(dy)^nf^{(n)}(y)}{n!}$

[fathom swallow]

Yeatte
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[torpid bay]

the first few terms being $f(y+dy) = f(y) + f'(y)dy + f''(y)dy^2 / 2...$

[fathom swallow]

Yeatte

[scarlet mango]

how did we arrive to that lmao

[torpid bay]

let x = y + dy, and let x_0 = y

[scarlet mango]

but how can we know that both of these will always hold true?

[burnt ledge]

we want the "x" in the formula (the input to x for the function we're expanding) to be y+dy here, and we want "x0" in the formula (the value where we know all the derivatives) to be y

[scarlet mango]

it would have to be that x_0= x-dy

[burnt ledge]

the taylor expansion formula tells you "how csn you find the value of a function at x when you know all its derivatives at x0" right?

[torpid bay]

it's just writing 2 new variables to be in terms of 2 old ones

[burnt ledge]

we are just spinning it slightly to figure out the value of the function at y+dy, when we know all the derovatives at y

[scarlet mango]

yes

[burnt ledge]

Which is effectively the same problem but with slightly different symbols

[torpid bay]

$y = x_0$ and $ dy = x-x_0$ if we want to define it in terms of the new ones being isolated

[fathom swallow]

Yeatte

[scarlet mango]

I see, and they just work things out themselves like usual?

[torpid bay]

yep

its juts a change of variables

2 new ones, 2 old ones

just like $a = x+1$ and $b = y+1$, it's guaranteed to be fine

[fathom swallow]

Yeatte

[torpid bay]

for an example of changing 2 old to 2 new

[scarlet mango]

when we write "y", btw, we mean an arbitrary variable, right? Cuz its usual to write y=f(x)

[torpid bay]

yeah

y just a vairalbe

[burnt ledge]

Yep

[torpid bay]

there's a nice little way to make the taylor series more compact which is really cool but prob later

[scarlet mango]

yes that makes sense

damn this went down a big ass rabbit hole

[torpid bay]

it goes furthur

[burnt ledge]

Hahaha yea

[torpid bay]

[scarlet mango]

why did we use the taylor expansion in the form we just mentioned again?

[burnt ledge]

It exposes the role of the derivative as the best linear approximation to a function

[torpid bay]

^

specifically the first derivative's role

being te one in J

[scarlet mango]

hmm I see

oh yeah btw

random

but

how do we use taylor series to approximate a multi-variable function?

[torpid bay]

it's just a lot more terms that's all

[burnt ledge]

Basically you just taylor expand with respect to each of the variables

[scarlet mango]

if we have f(x,y), can we somehow separate the x terms from the y terms and apply taylor series in f_1(x) and f_2(y) ?

[scarlet mango]

oh, thats way lazier than I thought

[gritty heath]

You can expand with respect to x first, then expand with respect to y for each term you got from expanding the first time

Sheer brute force works

[scarlet mango]

So that would give two different sums?

[gritty heath]

Its one giant sum

[scarlet mango]

or all the terms are in one sum?

okeoke got it

[torpid bay]

$f(x,y) = \sum_{n=0}^{\infty}{\sum_{m=0}^{\infty}{\frac{(x-x_0)^n(y-y_0)^m \frac{\partial^{n+m} f}{\partial x^n \partial y^m} (x_0,y_0)}{n!m!}}$ iirc

there's a few conditions, like interchangability of the order of derivatives

[scarlet mango]

can't we just use n for both sums?

[torpid bay]

nah, x and y are different variables, so like take z =x^2 y^2

$\frac{\partial^2 z}{\partial x \partial y} = 4xy$

[scarlet mango]

also, do you perhaps mean del(y^m) ?

[fathom swallow]

Yeatte

Yeatte
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[torpid bay]

ye

[scarlet mango]

that doesnt explain the reason they cant both have n lol

[torpid bay]

yeah I forgot the reasoning

[scarlet mango]

valid reason

yeah my mind expanded a little more from all this

[torpid bay]

i think it's because they are two seperate variables and so you just multiply the things together

[burnt ledge]

yea we just get $$f(x+\dd x, y+\dd y)=f(x,y) + \pdv{f}{x} (x,y)\dd x + \pdv{f}{y} (x,y)\dd y + \frac12\pdv[2]{f}{x} (x,y)(\dd x)^2 + \dots $$

[burnt ledge]

wtf why did all my parentheses disappear

[scarlet mango]

wait is that the same as the double sum we saw earlier?

[burnt ledge]

yea

[torpid bay]

$\pdv{a}{b}$

[fathom swallow]

Yeatte

[burnt ledge]

ive just written four of the terms instead of all infonitely many of them

[torpid bay]

huh, i so couldnve used that eralier

[scarlet mango]

nono, its just that im not accustomed to double sums very well, so I thought we would have a bunch of products

[burnt ledge]

I tend not to worry myself too much about the higher order terms because the whole point of the taylor expansion is to not care about them

[scarlet mango]

Thank you to everyone who contributed in this yapping sesh

it was really useful to learn some of those things which I'll see in like 2 years

[burnt ledge]

Gladly

calculus is pretty cool

[torpid bay]

you can losely think of d/dx and d/dy a multiplicative in a way

and so if I first took the taylor series for x, and then took the taylor series for y, the terms would essentially be all multiplied together

this is actually a situation where the compact taylor series makes it easier to see

[scarlet mango]

does that have to do with the roots of the original function?

[scarlet mango]

I didn't learn abt that

[torpid bay]

yeah, neither did i, I stumbled upon some approximations for f(x+h) like a few weeks back in a notebook and turns out, it wasn't just a good approximation, it was exactly the taylor series

I made it for help with coding simulation but it ended up being really cool

let's take $ \d {y}{x}$

texit?

u ded?

$\frac{dy}{dx}$

[fathom swallow]

Yeatte

[torpid bay]

and let's add the second derivative to it

$\frac{dy}{dx} + \frac {d^2y}{dx^2}$

[fathom swallow]

Yeatte

[torpid bay]

and see that both terms are derivatives acting on a function y

[scarlet mango]

how does that help with your coding simulation thing?

[burnt ledge]

\dv{y}{x} btw

[scarlet mango]

thats very true

[torpid bay]

$\big(\dv{}{x} + \diff[2]{}{x})y\big) = \frac{dy}{dx} + \frac {d^2y}{dx^2}$

oop

[scarlet mango]

when using powers, you should use \big( and \big)

[torpid bay]

$\big( x \big)$

[fathom swallow]

Yeatte

[scarlet mango]

or just left( and right)

lol its a bit big

,tex
\left( meow \right)

[fathom swallow]

fijokazż
Compile Error! Click the reaction for more information.
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Yeatte
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[scarlet mango]

is that the correct form? looks a lil wicked

[torpid bay]

$\big(\dv{}{x} + \dv[2]{}{x}\big)y = \frac{dy}{dx} + \frac {d^2y}{dx^2}$

ill just use partials for now

but just imagine they're both regular dervatives rn

[scarlet mango]

that isnt multiplication tho, its closer to matrix multiplication

[torpid bay]

$\dv[1]{y}{x}$

[fathom swallow]

Yeatte

[torpid bay]

ok good got it

yeah this is not multiplication

rather its operation

[scarlet mango]

does that mean that you apply the derivative operator on y?

[rocky shuttle]

composition

[scarlet mango]

or smth like that

[torpid bay]

yep

[scarlet mango]

yes

I see

[torpid bay]

we can apply this to the taylor series from before

[scarlet mango]

Hmm

can you show me the first few terms written like that? I cant see it

[scarlet mango]

I mean, I can't imagine it

[torpid bay]

$f(x+dx) = f(x) + dx\dv{f}{x}{(x)}+\frac{dx^2}{2!}\dv[2]{f}{x}{(x)}+\frac{dx^3}{3!}\dv[3]{f}{x} {(x)} ...$

where mah (x) go

[scarlet mango]

Isn't that how we wrote it earlier?

[torpid bay]

yep, just expanded out

[scarlet mango]

where?

[rocky shuttle]

\dv and \pdv shenanigans from the physics latex package

[rocky shuttle]

(mostly the reason why i don't use them)

[torpid bay]

there we go

[scarlet mango]

Yea but we didn't use the equation you mentioned

[fathom swallow]

Yeatte

[torpid bay]

rn its just the taylor series

[scarlet mango]

ohhh theres more

[torpid bay]

and now we can seperate the f(x) from the rest of the terms

[scarlet mango]

OHHHH

I seeeee

[torpid bay]

$f(x+dx) = \big(1+ dx\dv{}{x}+\frac{dx^2}{2!}\dv[2]{}{x}+\frac{dx^3}{3!}\dv[3]{}{x} ...\big) f(x)$

[scarlet mango]

yep

thats nice

is that the thing that helped you in your coding thingy?

[fathom swallow]

Yeatte

[scarlet mango]

or am I tripping? I think you were talking compact t. series

[rocky shuttle]

oh, the exp(d/dx) thingy

[torpid bay]

this is

let's have dx be h, to reduce the clutter a bit

[scarlet mango]

the what? e^(d/dx)?

[torpid bay]

$f(x+h) = f(x) + h\dv{f}{x}{(x)}+\frac{h^2}{2!}\dv[2]{f}{x}{(x)}+\frac{h^3}{3!}\dv[3]{f}{x} {(x)} ...$

[rocky shuttle]

f(x+h) = exp(h d/dx) f(x)

[torpid bay]

$f(x+h) = \big(1+ h\dv{}{x}+\frac{h^2}{2!}\dv[2]{}{x}+\frac{h^3}{3!}\dv[3]{}{x} ...\big) f(x)$

[fathom swallow]

Yeatte

[torpid bay]

now let's combine the h and d/dx a bit

[scarlet mango]

doesn't that give a different result tho?

Oh

OH

its the taylor series for e^x

[torpid bay]

yep

$\big(1+ h\dv{}{x}+\frac{h^2}{2!}\dv[2]{}{x}+\frac{h^3}{3!}\dv[3]{}{x} ...\big)=\big(1+ (h\dv{}{x})+\frac{(h\dv{}{x})^2}{2!}+\frac{(h\dv{}{x})^3}{3!} ...\big)= $

$\big( 1+ h\dv{}{x}+\frac{h^2}{2!}\dv[2]{}{x}+\frac{h^3}{3!}\dv[3]{}{x} ...\big) = \big( 1+ (h\dv{}{x})+\frac{(h\dv{}{x})^2}{2!}+\frac{(h\dv{}{x})^3}{3!} ...\big) = $

rip texit

[scarlet mango]

I think you didnt leave a space

after \big(

and the other

$\big( 1+ h\dv{}{x}+\frac{h^2}{2!}\dv[2]{}{x} +\frac{h^3}{3!}\dv[3]{}{x} ...\big) =\big( 1+ (h\dv{}{x})+\frac{(h\dv{}{x})^2}{2!} +\frac{(h\dv{}{x})^3}{3!} ...\big) = $

nope its just ded

[torpid bay]

anyway

yes

$\big(1+ (h\dv{}{x})+\frac{(h\dv{}{x})^2}{2!}+\frac{(h\dv{}{x})^3}{3!} ...\big)= e^{h\dv{}{x}}$

[fathom swallow]

Yeatte

[torpid bay]

putting this back into our original

we get

[rocky shuttle]

imagine you'd just leave h as dx, then you'd have e^d

[torpid bay]

$f(x+h) = e^{h\dv{}{x}}f(x) = \big( \sum_{n=0}^{\infty}{\frac{\big(h\dv{}{x}\big)^n}{n!}})f(x)$

[scarlet mango]

well i think this is the case where dx cant be multiplied by 1/dx. Or am I crazy?

[torpid bay]

which is y i wanted to switch to h lel

[rocky shuttle]

yeah, I don't advise using dx to denote an increment here

[scarlet mango]

i think you forgot to mult. e by f(x) but yeah thats true

[fathom swallow]

Yeatte

[torpid bay]

in this case $dx(\dv{}{x}) = 1$ actually so 👀

[scarlet mango]

this is a nice way to write this thing

[fathom swallow]

Yeatte

[scarlet mango]

wait really?

[torpid bay]

dx as a differential operator

this gets into diffetnerial geometry tho

where again, dx(d/dx) isnt multiplication but operation

anyway, the e^(h d/dx) can be generalized as well

$e^{h \dv{x}{g} \dv{}{x}} f(x) = f(g^{(-1)}(g(x)+h))$

[fathom swallow]

Yeatte

[torpid bay]

which is something for wayyyyy later

I recommend not using the compact taylor series tho, and just sticking with the ituition of the regular one

[scarlet mango]

thats insane to me, that it touches this area

[scarlet mango]

to me, it seems like the dx term is an increment, but d/dx is an operator, or is that crazy work?

[torpid bay]

for now you can think of dx as an increment, prob better for now

tho later, you will think of them as 'dual' to each other

[scarlet mango]

wait wtf is this lmao

[scarlet mango]

ohh I see lol

[scarlet mango]

I didnt understand the intuition of the regular one to begin with

[scarlet mango]

Doesn't duality mean action and meaning?

[torpid bay]

$\dv{y}{x}$ is the coefficient of the best linear function for a function y of x

[burnt ledge]

Yea all of this is nonsense you neednt worry about

[fathom swallow]

Yeatte

[scarlet mango]

I speak in nonsense

[torpid bay]

take $y = e^x$

texit hello?

baby steps texit smh

[burnt ledge]

try editing out the space after the $

[fathom swallow]

Yeatte

[torpid bay]

[scarlet mango]

that doesnt make much sense

[torpid bay]

let's say I wanted to approximate y = e^x at x = 1

for the first term, I just take y(0)

so let's write the n'th taylor polunomial as $y_n$ for now

[fathom swallow]

Yeatte

[torpid bay]

so $y_0 = y(0) = 1$

[fathom swallow]

Yeatte

[torpid bay]

which is good enough, it's close to (and precisely equal to) the e^x near x=0, which is what we want

but pretty far off for other values of x

so we take the next taylor polynomial

[scarlet mango]

didnt we want to find it for x=0?

[scarlet mango]

x=1**

omg

[torpid bay]

yeah, but I want to find it for other values near x

as well

[rocky shuttle]

the coefficient such that dy = (dy/dx) dx

[torpid bay]

let's take he next taylor polynomial

[scarlet mango]

I don't get the connection between the two statements

[torpid bay]

$y_1 = y(0) + \dv{y}{x}{(x)}|^{x=0} x$

[fathom swallow]

Yeatte

[torpid bay]

in thise case, taking the derivative of e^x at 0, is just 1

so the best approximation for a function of the type a + bx is 1 + 1x for e^x

[lone granite]

Calculus?

[torpid bay]

,w plot e^x and 1+1x

[fathom swallow]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by yeatte137

[torpid bay]

its exactly equal at x=0, but also much closer for values near x =0, than the first taylor polynomial was

[scarlet mango]

I see

[lone granite]

Affirmative.

[torpid bay]

and we can extend this intuition for all the rest of the taylor polynomials

[scarlet mango]

we can do the same thing here too?

[torpid bay]

$y_2(x) = y(0) + \dv{y}{x}{(x)}|^{x=0} x + \dv[2]{y}{x}{(x)}|^{x=0} \frac{x^2}{2!}$

and since the second derivative of e^x at x=0 is also 1 (wow)

it ends up being this

$y_2(x) = 1 + 1x + \frac{1x^2}{2!}$

[fathom swallow]

Yeatte

[scarlet mango]

is that 2 supposed to be there at the end?

[lone granite]

How much time did you spend learning these commands?

[fathom swallow]

Yeatte

[torpid bay]

,w plot e^x and 1+1x+1x^2 /2!

[fathom swallow]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by yeatte137

[torpid bay]

^ even better

[scarlet mango]

it takes 4 hours at most if you wanna learn sums, limits, integrals and derivatives imo

[scarlet mango]

very true

[lone granite]

Noted

[torpid bay]

it also takes 4 hours to forget how to do it and do just bootleg maatrices like ive done before

[lone granite]

[scarlet mango]

I mean I can see how the taylor series is perfect for these functions, but I could never understand the reasoning behind the inner machinations of it ya know?

[lone granite]

Name the Last Number before Absolute Infinity.

[torpid bay]

oh

none

iirc

[scarlet mango]

infinity is undefined i think lol

[rocky shuttle]

you just did

[scarlet mango]

so the number before infinity is also undefined

[lone granite]

Mission Passed.

It is possible to represent infinity minus one as a mathematical expression, but it does not actually equal anything or have any real mathematical value.

[scarlet mango]

I dont think humans will be able to comprehend what it means to be infinite in our lifetimes

[torpid bay]

infinity = non finite

[scarlet mango]

yeah maths is always useful even when it doesnt really make a lot of sense

[lone granite]

How does an Infinite Hotel run out of rooms?

[lone granite]

In maths, you are the maker in such concepts.

[scarlet mango]

yeah ive seen the vid bud

[rocky shuttle]

when a bigger cardinality arrives

[lone granite]

Video? Science.

[scarlet mango]

oh shit

[rocky shuttle]

Science? Math.

[scarlet mango]

I think infinity can only be interpreted but will nonetheless remain undefined forever

[lone granite]

Yes.

Math.

[scarlet mango]

Life.

[lone granite]

Cube.

[rocky shuttle]

Well, it's an umbrella term for a lot of (different) concepts in math.

[scarlet mango]

yeeee

[lone granite]

Try defining the difference between Infinity and Absolute Infinity.

[scarlet mango]

valid

interpretting maths is a weird isomorphism to interpretting life or some shit

[rocky shuttle]

https://tenor.com/view/jujutsu-kaisen-gojo-satoru-vs-gif-21280734

[scarlet mango]

idek what that means

[zealous garden]

Not undefined, doesn't exist

[torpid bay]

oh

yeah

you still there fijo

[scarlet mango]

yeeee

[scarlet mango]

I choose to believe it exists

[torpid bay]

remember $e^{h \dv{}{x}} f(x)$

[fathom swallow]

Yeatte

[scarlet mango]

yup

[torpid bay]

and how that makes a taylor series for x?

well

[scarlet mango]

you mean wrt to x?

[torpid bay]

yes

[scarlet mango]

yes

[lone granite]

Maybe it exists, beyond space and time. We just can't comprehend it.

Must be a dimension to see infinity

[torpid bay]

let's taek $e^{j \dv{}{y}}$

[fathom swallow]

Yeatte

[scarlet mango]

fr. I like to piss people off by choosing to believe contradicting things, but thats another way to put it

[torpid bay]

$e^{j\dv{}{y}}e^{h\dv{}{x}} f(x,y) = e^{j\dv{}{y}} f(x+h,y) = f(x+h,y+j)$

[fathom swallow]

Yeatte

[torpid bay]

$e^ae^b = e^{a+b}$ (usually)

[fathom swallow]

Yeatte

[torpid bay]

so, taking that

[lone granite]

Our mindsets are the same, it's just that you transfer it differently.

[zealous garden]

I wish

[torpid bay]

$e^{j\dv{}{y}+h\dv{}{x}}f(x,y) = f(x+h,y+j)$ would also be the taylor series of f with respect to x and y

[fathom swallow]

Yeatte

[torpid bay]

in this case we'll assume f is nice enough to where partial derivative commute

[rocky shuttle]

otherwise, use the Baker-Campbell-Hausdorff formula

[scarlet mango]

thats true

[scarlet mango]

yessss thats really beautiful

[scarlet mango]

what does that mean? that we dont have to use the del symbol?

[torpid bay]

ah

yeah

[scarlet mango]

whattttt?????

[scarlet mango]

Three people for one formula is insane

[torpid bay]

so $e^{j\pdv{}{y}+h\pdv{}{x}}f(x,y) = e^{j\pdv{}{y}}\big( \sum_{n=0}^{\infty}{\frac{\big( h\pdv{}{x}\big)^n}{n!}}f(x,y) \big)= \sum_{m=0}^{\infty}\frac{1}{m!}\big(j\dv{}{x}\big)^m{\sum_{n=0}^{\infty}{\frac{1}{n!}\big(h\pdv{}{x})^n f(x,y)}}$

[fathom swallow]

Yeatte

[torpid bay]

lowkey I'm never gonna use this

[scarlet mango]

its cool tho

btw

[torpid bay]

but just as a concept for the derivatives being seperate from each other

[scarlet mango]

shouldnt there be parentheses around the two sums? to showcase that its the operators applying to f?

[torpid bay]

yeah

assume they're there

[light thorn]

So, I got an offer about a co-seminar for my calculus 2 class, it says it will help with the more difficult subjects, but I'm not sure. Should I take it if it's even still available? It was supposed to start this morning

[torpid bay]

calc 2 is fun

[rocky shuttle]

what's the dual of a seminar?

[torpid bay]

quasinar

[scarlet mango]

Im no expert in the matter, but if you have time to spare, a little extra knowledge wouldnt hurt

[torpid bay]

or a pseudonar

[scarlet mango]

what does dual mean? Is it not connected to the yoneda conjecture or whats it called from cat theory?

[rocky shuttle]

like, for example, connected is the dual of nnected

[torpid bay]

dual just means that 2 things are connected in certain way to each other that's a little strong

like take dx and d/dx

[scarlet mango]

that doesnt clear it up

[rocky shuttle]

sorry, I'll stop with the bad puns

[torpid bay]

$dx = \dv{y}{x} dv$ but $\dv{}{x}=\dv{y}{x} \dv{}{y}$

[fathom swallow]

Yeatte

[scarlet mango]

ohh lol I thought it was fr sorry

[torpid bay]

in the first case its dy/dx, in the second case its the exact opposite

so If i used a change of coordinates, these two objects would change in the exact oppsoite way of each other, relating to the coordinate systems

[agile fiber]

duals are things that are in some way opposites or mirrors of each other, usually following a rule that dual(A) = B iff dual(B) = A and dual(dual(A)) = A

[scarlet mango]

what the heck

[rocky shuttle]

emphasis on usually

[zealous garden]

Many such cases

Sometimes they're anti-involutions instead of involutions

And sometimes it's just words on a page written by a monkey

[scarlet mango]

how is dx even equal to dy/dx dv? the second one is just chain rule i get that one

[torpid bay]

this is for the dx and dy in integrals

$\int dx = \int \dv{x}{y} dy$

[zealous garden]

In category theory you flip the arrows, in linear algebra you flip the arrows go to linear functionals on the space, in exterior algebra you take orthogonal complements, in geometric algebra you....

[scarlet mango]

wtf

[zealous garden]

Well, you do a thing

[scarlet mango]

how are these equal wth

[rocky shuttle]

Do the thing

[torpid bay]

oh

[fathom swallow]

Yeatte

[torpid bay]

hehe

[scarlet mango]

lol

but thats the good ol chain rule

Thats the second thing you showed

[torpid bay]

reverse chain rule technically but ye

[scarlet mango]

I was talking about the first

[zealous garden]

Never ask a Geometric Algebraist what "Inner product" means

[scarlet mango]

What does it mean in these terms i wanna know now

[zealous garden]

[rocky shuttle]

The answer would be "which inner product?"

[torpid bay]

of the 7 vectors product its one of the inner ones

[scarlet mango]

whichever you feel comfortable explaining? idk

[torpid bay]

not to be confused with the interior product

[zealous garden]

In GA an inner product is none of (positive definite, non degenerate, scalar valued, symmetric)

[rocky shuttle]

I don't feel comfortable explaining GA

[scarlet mango]

none of?

[zealous garden]

Correct, all of those are usually meant by one saying "Inner product" in other fields

[scarlet mango]

Lmao its like it gave y'all ptsd or something

[zealous garden]

No, I did

[scarlet mango]

yeah but what do you mean by none of (...)? like, i dont understand the linguistic structure of the statement lol

[zealous garden]

The inner product between two blades is the lowest generally nonzero term of the GP

[scarlet mango]

I see, cant wait to learn abt it

[zealous garden]

Extend to multivectors through bilinearity

[rocky shuttle]

Is that why you talk about metric vector spaces rather than just inner product spaces?

[zealous garden]

Somewhat, I also redefine Inner Product Space to merely mean any nondegenerate space

But yes

[scarlet mango]

whats a GP?

[rocky shuttle]

geometric product

[zealous garden]

Because that's the minimum requirement I need to make to access Geometric Algebras, and many theorems we care about don't really care if the form is positive definite I find

[scarlet mango]

wth are blades btw lol

[zealous garden]

You have a long way to go before you can wield a blade young padawan

Unironically though, a blade is a k-vector which is the wedge of 1 vectors

Geometrically, think of it as an oriented parallelogram

[rocky shuttle]

Imagine introducing sabers to GA

[zealous garden]

Or higher

[scarlet mango]

is it bad to ask what a k-vector is?

[scarlet mango]

again, i mean

[torpid bay]

$\hat{x} \wedge \hat{y} $ is a 2 blade, where $\hat{x}$ and $\hat{y}$ are 1 -blades

[fathom swallow]

Yeatte

[zealous garden]

Do you understand how awesome the double entendres are? A k-form evaluates a k-blade and gives a scalar, a form is a technique in martial arts

Like swordfighting specifically

[torpid bay]

i forget if its wedge or gprod

[zealous garden]

The techniques you use are called forms

Blades and forms

SPECIFICALLY swordfighting

It's meant to be

[torpid bay]

form 3, soresu was used by obi wan kenobi, one of the best users of form 3

[scarlet mango]

okay. first of all, what is a wedge? second of all, what is a 1-blade?

[zealous garden]

So a k-vector is a sum of k-blades

[torpid bay]

think of a 1-blade as a vector

[rocky shuttle]

And you use a dagger to reverse

[zealous garden]

A k-blade is easy to visualize, k-vectors can get complicated

[zealous garden]

OH MY GOD

[scarlet mango]

if a blade is defined by a vector, how tf is a vector defined by a blade lmao

[zealous garden]

They both stem from the definition of 1-vector as in a vector in V the vectorspace

Then we build up from these

[scarlet mango]

okay, and why is it called a blade?

[torpid bay]

mathematicians got bored

its just a name

[zealous garden]

So we have G(V), which has a space isomorphic to V which we will just call V. Vectors of V are 1-vectors. Blades are wedge products of 1-vectors, k-blades have k factors, k-vectors are sums of k-blades

[scarlet mango]

wait, 1-vector just means a vector? like, in general?

[zealous garden]

Cus it's a flat subspace

[zealous garden]

Kinda yeah

[torpid bay]

^ kinda yeah

[zealous garden]

Things get complicated MERELY because of how vector has historically been used

[scarlet mango]

what is a wedge? and a wedge product?

[zealous garden]

And linear algebra isomorphisms abused

[zealous garden]

Geometrically, think of it as building parallelograms and boxes out of your vectors

u^v is the parallelogram made by u,v,-u,-v laid end to end

[scarlet mango]

hmm I see

[zealous garden]

It's equivalent to any other parallelogram in the same plane, with the same area, and same orientation

And you just repeat to higher dimensional analogues

[scarlet mango]

what does it mean that k-vectors are sums of k-blades? I think i understand the above

[zealous garden]

Technically if you really wanna do GA there's a whole completely contradictory way to view things that is arguably better, but this is a historically fruitful starting point

Anyways I have a meeting, bye

[scarlet mango]

okeoke bye

[zealous garden]

X=u^v+w^z, but maybe you can't factor X into vectors at all!

[scarlet mango]

Thank you for teaching me your ways

[torpid bay]

a k blade is a special k-vector that can be represented as a wedge product of 1-blades

[scarlet mango]

how do we even add wedge products tho lmao

unless we define a plane where parallelograms are now single vectors

[torpid bay]

I think of it as like putting 2 different objects into a bag together

[scarlet mango]

that doesnt clear it up. It in fact makes it weirder

is there a visual way to see it? or is it some weird math shit?

[rocky shuttle]

it's just a superposition of blades

[torpid bay]

$1 + \hat{x}\wedge\hat{y}$ would be something where I have a scalar and a parallelogram, they' don't really add together into a single thing. but rather you can just manitupulate both of them at the same time and they don't interfere with each other

[fathom swallow]

Yeatte

[torpid bay]

so like

$(1+\hat{x}\wedge\hat{y})\wedge(\hat{z}) = \hat{z}+\hat{x}\wedge\hat{y}\wedge\hat{z}$

[fathom swallow]

Yeatte

[scarlet mango]

What does that mean? Im not very familiar with that term?

[torpid bay]

in the same bag, but no interferring

[scarlet mango]

what about 2^z?

[rocky shuttle]

a linear combination

[scarlet mango]

is it just 2z?

[rocky shuttle]

like "3 apples and 2 oranges" is such a combination

[torpid bay]

ye

[scarlet mango]

i mean yeah but is there no way to see it in a graph somehow?

[scarlet mango]

ohhhhh

I see

so we just show that we have these two things

[torpid bay]

i think

[scarlet mango]

its like when we do a+bi

[rocky shuttle]

you can draw multiple overlapping blades / parallelograms ig

[scarlet mango]

interesting

[torpid bay]

as examples of 0blade, 1-blade etc

[scarlet mango]

I cannot even fathom how these could be used in actually important fields of maths but its nevertheless interesting

[torpid bay]

well

actually

do you know quaternions?

[scarlet mango]

heard of em

is it 4 dimensional cube or something?

[torpid bay]

they are the vectors useed in 3D for rotations

[scarlet mango]

oh damn im far out

[torpid bay]

quaternions are like this

$q = a+bi+cj+dk$

[fathom swallow]

Yeatte

[scarlet mango]

uhm

what does this mean exactly?

[torpid bay]

it's a complex number, but more

[scarlet mango]

i,j,k are vectors first of all right?

ohhh

[torpid bay]

yeah

[scarlet mango]

oh okeoke

[torpid bay]

technically

[scarlet mango]

How is GA connected to this tho?

[torpid bay]

$\hat{i},\hat{j},\hat{k}$ are in vector form, while $i,j,k$ are in algebra form

[fathom swallow]

Yeatte

[torpid bay]

i,j,k, are actually bivectors

[scarlet mango]

huhhhh

[torpid bay]

2-blades, not 1 b-blades

^ i had stutter here

[scarlet mango]

so they each are represented in 2d planes?

[torpid bay]

yeah

[scarlet mango]

so we are talking 6 dimensions?

am I crazy for that?

[torpid bay]

it depends on how you count

[scarlet mango]

I suppose they are linearly independent

[torpid bay]

1 scalar, 3 vectors, 3 bivectors, 1 trivector

[scarlet mango]

the set containing them, i mean

[torpid bay]

yes

[scarlet mango]

oh wait. then i+j+k is just a trivector?

[torpid bay]

i+j+k is a bivector, i^j^k is a scalar i tihnk

lemme write them in k-blade form

[scarlet mango]

this is tantamount to smoking LSD

[zealous garden]

Good luck

[scarlet mango]

how is it a scalar?

[zealous garden]

0 kek

Meeting canceled

[scarlet mango]

is it because the vectors lie in different planes?

[torpid bay]

$1\ \hat{x} , \hat{y}, \hat{z},\ \hat{y}\wedge}\hat{z}, \hat{z}\wedge\hat{x},\hat{x}\wedge{\hat{y} \ \hat{x}\wedge\hat{y}\wedge\hat{z}$

[scarlet mango]

good for me I guess

[torpid bay]

for the various k blades

i,j,k are the 3 different 2 blades

[scarlet mango]

what am I even seeing here?

[zealous garden]

Hats are terrible

Why did you do this yeatte

[torpid bay]

remember that list of 1 scalar, 3 vectors, 3 bivector, 1, trivector? this is them listed out

[zealous garden]

Let's start here

[fathom swallow]

Yeatte

$1\\ \hat{x} , \hat{y}, \hat{z},\\ \hat{y}\wedge}\hat{z}, \hat{z}\wedge\hat{x},\hat{x}\wedge{\hat{y} \\ \hat{x}\wedge\hat{y}\wedge\hat{z}$
```Compilation error:```! Extra }, or forgotten $.
l.49 ...at{x} , \hat{y}, \hat{z},\\ \hat{y}\wedge}
                                                  \hat{z}, \hat{z}\wedge\hat...
I've deleted a group-closing symbol because it seems to be
spurious, as in `$x}$'. But perhaps the } is legitimate and
you forgot something else, as in `\hbox{$x}'. In such cases
the way to recover is to insert both the forgotten and the
deleted material, e.g., by typing `I$}'.```

[scarlet mango]

But, i dont understand. How is a+bi+cj+dk the same as the list you provided?

[torpid bay]

a is the scalar, bi + cj +dk are the bivectors

quaternions don't describe the whole k-blade space

you need biquaternions for that in this case

[scarlet mango]

jeez

this is dank

[zealous garden]

Complex Quaternions

Why ever say biquaternions

Just say complex quaternions, it leaves no questions

Biquaternions is like "what? Huh?"

[scarlet mango]

yeah because idek what to ask

[zealous garden]

Complex quaternions gives you the recipe

Quaternions with complex coefficients

[scarlet mango]

that makes it clearer

[torpid bay]

$i = y\wedge z \ j = z\wedge x \ k = x\wedge y$

[fathom swallow]

Yeatte

[scarlet mango]

hmm

its a lil hard to visualize

xyz are vectors tho right?

[torpid bay]

yes

and i would rotate the plane yz

in 3D you can think of them as rotate about the x axis

but it only works for 3D

[scarlet mango]

what do you mean by that? like, why would you rotate that?

[torpid bay]

hm, lets go to 2D for now

with a simple xy plane

[scarlet mango]

i'm listening

[torpid bay]

let's take the point (1,1) which we can represent as 1x + 1y

[scarlet mango]

yup

[torpid bay]

if I rotate te plane counter clockwise, im rotating through the plany $x\wedge y$

[fathom swallow]

Yeatte

[torpid bay]

or clockwise, but by convention counter lcokwise

[scarlet mango]

Hmm

what do you mean when you say "rotating through the plane x^y"?

[torpid bay]

if I take the point (1,1) on the plane xy and rotate it, i will always end up at nother point on the plane xy

[scarlet mango]

yes

[torpid bay]

so if we trace out it's path through the space, it will make a circle that lives entirely within that xy plane

and never leave it when rotating by $x\wedge y$

[fathom swallow]

Yeatte

[zealous garden]

no i, the imaginary unit

[torpid bay]

let's take the usual complex number for now

[scarlet mango]

lmfao true, didnt think it was a unit

[torpid bay]

if i take the point (1,1) it is then notated at 1 +1i

[zealous garden]

Ew

[torpid bay]

if I rotate it by 90 degs counter lcokwsie, I just multiply by i

[zealous garden]

Don't do this please

[torpid bay]

[scarlet mango]

-1+1i

[torpid bay]

yep, so notice how when I did that, it stayed i nthe complex plane

[scarlet mango]

do you mean "notated as", or "rotated at"? its like a typo fork

[zealous garden]

Why introduce all of this just to throw it away

You're wasting brain space yeatte

[torpid bay]

wait notated

[zealous garden]

I beg of you stop this

[torpid bay]

it is too late

[zealous garden]

Just show how to reflect through a hyperplane

[torpid bay]

I shall corrupt him

[zealous garden]

It's all so much simpler

[scarlet mango]

where else would it go?

[zealous garden]

I'm at work so I can't spend all my energy on guiding young fijo

[scarlet mango]

im not even sure what the purpose of this is anymore lmao

[torpid bay]

we can think of this 'i' rotate thru the complex plane as being like rotating the 1x +1y through the xy plane when we apply xy to the vector 1x +1y

[torpid bay]

i tried, I couldn't

[zealous garden]

Wtf do you mean you couldn't lmfao

[rocky shuttle]

how did we get here again lmao

[torpid bay]

we took a bunch of potions in mc

[zealous garden]

Use that P_v(u) = B(u,v) v/B(v,v)

[scarlet mango]

how do we apply xy to that vector tho? what does it mean to apply xy lol

[torpid bay]

$(x + y) \wedge (x\wedge y)$

[fathom swallow]

Yeatte

[zealous garden]

See yeatte