help-39

[glossy root]

But idk idk whatever eles

[light helm]

then it's p^(x-1) and p being prime
wdym

[glossy root]

Uh basically if x has 3 factors

Then it's p^3-1

Which is p²

So like

2

2² = 4 anf 4 = 1x4 and 2x2

Having only 3 factors

And etc

[toxic lichen]

so you want to count the numbers which have an odd number of divisors?

the english translation seems to be a bit wonky.

[glossy root]

Yup

[glossy root]

Yea lmao

[toxic lichen]

not all numbers with exactly x factors are of the form p^(x-1).

[glossy root]

._. Oh

[toxic lichen]

like

[glossy root]

I js asked chat gpt for that xd

[toxic lichen]

!nogpt

[pearl pond]

Please do not trust ChatGPT or similar AI tools for mathematical tasks, as they often generate output which "sounds correct" but has numerous factual or logical errors. Use of these AI tools to answer other people's help questions is strictly against server rules (see #rules).

[glossy root]

._.

[toxic lichen]

the number 2^4 * 3^7 has 40 factors but it is certainly not the 39th power of any prime

anyway

thinking about prime factorization won't help much here

[glossy root]

Anyways could u help with this question

@toxic lichen

[light helm]

try list/count the factors for the numbers from 1→10
and see if you can deduce what's going on

[toxic lichen]

think about it this way: for a natural number n, its factors (incl. 1 and n themselves) can be organized into pairs {d, n/d}

for what numbers does this work cleanly?

or maybe an easier question is, when could this mess up and how?

(this will eventually lead you to find a characterization of those numbers which have an odd number of factors, but patience)

[glossy root]

Uhhhhh

._.

[glossy root]

No like is there a method to immediately calculate it

Cus I only get like 2 minutes per question

[light helm]

yes, there is a trick that'll give you the answer almost immediately

consider the above info, to identify the trick/property to use

[toxic lichen]

if you do not know it beforehand then you are cooked

[light helm]

its worth it to try and realise it for yourself (and you'll also be more likely to remember)
rather than us telling you

[glossy root]

Ik 💀

[glossy root]

._.

Bro it's cus I been trying to figure this out for so long that's why I'm asking here 😭

[toxic lichen]

we are not an instant answer giving machine

[glossy root]

Yea no shit 💀

[toxic lichen]

think about it this way: for a natural number n, its factors (incl. 1 and n themselves) can be organized into pairs {d, n/d}
when can this go wrong?

try doing it for some number like 24

write out all the factors and pair them up in the way i described -- and see if you notice anything

[glossy root]

Wait what's d

[toxic lichen]

a divisor

to put it another way, the divisors of any natural number n can be arranged into pairs where the product of each pair equals n

[glossy root]

So js saying if we want like 6 then it's 6/3 + 6/2 + 6/1 + 6/6 ?

💀 wait no it's supposed to be 4

[toxic lichen]

no you are overthinking it

ok

how about this

can you follow a simple instruction that im about to give you? @glossy root

[glossy root]

Ye

(Probably)

[toxic lichen]

do exactly what i ask, no more, and no less:

write down a list of all the divisors of the natural number 24.
separate the numbers with commas and write them in order from lowest to highest.

[glossy root]

1,2,3,4,6,8,12,24

[toxic lichen]

ok

now try to arrange these divisors into pairs in such a way that the product of each pair is 24.

[glossy root]

1 x 24
2 x 12
3 x 8
4 x 6

[toxic lichen]

ok right

so you see that you can always* do this sort of pairing, yes?

[glossy root]

Yea

[toxic lichen]

ok

now do the same thing but for the number 36 instead.

list out all of its factors and try to pair them up in the same way.

[glossy root]

1 x 36
2 x 18
3 x 12
4 x 9
6 x 6

[toxic lichen]

do you notice anything different

[glossy root]

Uh there's a 2 increase in total factors?

Or the first 4 pairs first digits are the same

1,2,3,4

[toxic lichen]

well, the number's bigger. it's no surprise it has more factors.

and also 24 and 36 are both divisible by 2, 3 and 4 but this is not relevant to my point

[toxic lichen]

do you notice that the last pair in this is not like the others?

[glossy root]

You mean cus it's the same number?

[toxic lichen]

yes

6 cannot pair up with anybody except itself

now answer me this: why did this happen?

what makes 36 different from 24 that causes this lonely divisor to happen?

[glossy root]

Cus 36 = 6²

[toxic lichen]

or perhaps even more generally: precisely when do such lonely divisors happen?

[glossy root]

And square root of 24 doesn't have an integer solution

[glossy root]

Oh wait does this mean we count the square integer which is less then 9999 but more then 1000?

[toxic lichen]

you jumped ahead a bit.

what i wanted you to say is:

"This shit happens when the number we start with is a perfect square, because its square root will be the one factor that pairs with itself."

and yes, we need to count the perfect squares between 1000 and 9999.

[glossy root]

So so uh 32 - 99

99 - 32 + 1 = 68

[glossy root]

😭 had to add "this shit"

Anyways tq for ur help @toxic lichen

.close

[pearl pond]

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[toxic lichen]

do not call me "bro" please.

.reopen

[pearl pond]

✅

[toxic lichen]

could you edit that word out?

[glossy root]

💀💀

Why

Anyways sure

[toxic lichen]

cause i don't like being called "bro"

simple as, lol

[glossy root]

💀 are u a girl

[toxic lichen]

i am.

[glossy root]

Oh alright

Cya

.close

[pearl pond]

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[pearl pond]

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[untold pebble]

Hello, i have to prove whether these groups are isomorphic or not. Can somebody please help me? :
a) (Q, +) and (Q without 0, *)
b) (Reals > 0, *) and (R, +),
c) (R/Z, +) and (S, *)

[untold pebble]

for a) i've written that (Q, +) has only one element of finite order (0) and (Q without 0, *) has two elements of finite order (1 and -1), so they cannot be isomorphic

for b) i have no idea

and for c) i don't know what S is... I know Sn the symmetric group but i've never seen S

[toxic lichen]

if you have studied precalculus, or otherwise been through high school, then you know a homomorphism between the groups in b

though unfortunately if you dont know what is meant by S in (c) we can't help you either

[untold pebble]

the logarithm? log(x*y) = log(x) +log(y) it is an homeomorphism and it is also biijective so its an isomorphism

[toxic lichen]

homomorphism

no e

homeomorphism is a topology thing

but yes the log.

[untold pebble]

mh okay thanks!

"Verify that (R/Z , +) is isomorphic to (S, *)"

if somebody knows what S is please tell me

[main timber]

are you sure it's not in your previous lectures or something

[untold pebble]

yes im sure i've checked

[main timber]

I just had algebraic structures last semester and I don't really remember consistently using S for something, so it's likely its some notation your prof has been using a bunch 🙃

[untold pebble]

maybe it was a typo ?

[main timber]

my best guess is that it's the unit circle in the complex plane

[untold pebble]

oh yeah! might be thank you

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[pearl ridge]

Rona has a collection of pins. She knows she has fewer than 100 pins. She found that when she put them into piles of 2, there was one left over. When she puts them into piles of 3, again, there was one left over. The same thing happened, when she put them into piles of 4, However, when she put them into piles of 5, there was none left over. How many pins did Rona have?

Emily, Frank, Gary, and Julie have cats that are black, brown, white, or gray. The cats’ names are Buffy, Duffy, Fluffy, and Puffy. Buffy is brown. Emily’s cat, Duffy, is not gray. Frank’s cat is white, but is not named Puffy. The gray cat belongs to Gary. Which cat belongs to each student, and what is its color?

sorry i absolutely dont get it. i just need a simple solution cause this the kind of questions showing up on my 1hr exam. please helpp

[pearl ridge]

i quite get the 2nd question but not the 1st one

[fair creek]

first one is

classic

chinese remainder theorem

for n < 100

[pearl ridge]

what the prof taught us is getting the LCM. and i also do not know chinese remainder theorem T-T. i'm not a stem person sorry

[drowsy adder]

assume a number it should be such that if you take one off it should be divisible by 2,3,4 and the number itself should be divisible by 5

[pearl ridge]

i know the LCM of 2 3 4 is 12, so whats next? should i find the next lcm that is divisble by 5?

sooooo is the answer 25? cause the next lcm is 24, and if i add 1 it is 25. is it right?

[fair creek]

i'm not sure how it works withoit CRT but are you sure you can verify that no other numbers display the aforementioned property?

[pearl ridge]

im not sure but this is how we're taught. what is your answer on this?

[fair creek]

there's a solution via modular arithmetic

which gives a class of solutions

i haven't taken the time to calculate though

[pearl ridge]

owww that's sounds advanced, i'm not a stem student so the prof says she goes easy on our solutions and problems. i'll probably stick with what i did.

[fair creek]

okay sure that works

[pearl ridge]

Emily, Frank, Gary, and Julie have cats that are black, brown, white, or gray. The cats’ names are Buffy, Duffy, Fluffy, and Puffy. Buffy is brown. Emily’s cat, Duffy, is not gray. Frank’s cat is white, but is not named Puffy. The gray cat belongs to Gary. Which cat belongs to each student, and what is its color?
Buffy brown - Julie
Duffy black - Emily
Fluffy white - Frank
Puffy gray - Gary

is this correct?

[fair creek]

yes

[pearl ridge]

thank u so much

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[deft orchid]

<@&268886789983436800>

[pearl pond]

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[daring bay]

so when the vectors are in 1 plane then theyre collinear?

[vestal tapir]

they are coplanar they aren't always collinear

[daring bay]

hm ok

and when u have 3 vectors

and u gotta check if theyre linear dependent

then you do a overdefined linear system right

to check if theyre linear dependent

[rough forge]

ja bruder

du guckst ob nur die triviale lösung oder mehrere lösungen existieren

[daring bay]

omg

morgen kurzarbeit

der channel ist jz offen bis 23 uhr

was bedeutet aber komplanar nochmal?

[rough forge]

wenn die punkte/vektoren in einer gemeinsamen ebene liegen

[daring bay]

und kollinear ist doch dasselbe oder

[rough forge]

nein

kollinear heißen zwei vektoren wenn sie vielfache voneinander sind

aber nur weil vektoren koplanar sind, also in einer ebene liegen, sind sie lange nicht kollinear

[daring bay]

ja

ok

komme gleich mit der nächsten frage

also

eine geschlossene vektorkette

kannst du die erklären

[rough forge]

google es am besten

[pearl pond]

@daring bay Has your question been resolved?

[daring bay]

ok

ich versteh nicht ganz

wie man vektoren zeichnen muss

also hier zb

man zeichnet die punkte erstmal ein oder wie

[rough forge]

ja, und dann verbindest du sie

AB heißt ja von A nach B

[daring bay]

hä ok

also zeichnerisch ist billig

[rough forge]

💀

[pearl pond]

@daring bay Has your question been resolved?

[daring bay]

@rough forge

weist du wie ich hier vorgehen muss

alsoicz

ich weiß das bei einem parallelogram

die seiten 2 mal vorkommen

aber woher soll ich wissen welche vektoren AB oder AD sind

[rough forge]

mach dir halt ungefähr eine skizze

[pearl pond]

@daring bay Has your question been resolved?

[daring bay]

@rough forge

bro

,rotate

[daring bay]

18 = 2r-6s-4t
39 = 3r-9s
-6=2r+6s+2t

weißt du was ich falsch gemacht habe

does anyone know what i did wrong

<@&286206848099549185> this is an emergency

[pine temple]

What’s the emergency?

[daring bay]

do you know what i did wrong solving the system

[pine temple]

I don’t understand German, so I had to translate it

[daring bay]

no no no

thats a different

question

wait

im doing the b) here rn

i did something wrong solving the linear system

idk what tho

[pine temple]

Hmm…

[daring bay]

yeah

[pine temple]

Given the vectors a, b, c and D

a.) Check whether vectors a and b are collinear.
b.) Check whether vector d can be represented as a linear combination of a, b, and c. State this linear combination, if possible.

I’m not sure how to solve this exactly

What have you done so far?

[daring bay]

a system

[daring bay]

like here

[pine temple]

@daring bay I think you need to arrange them by component

So the numbers on Component 1 (The top) are in an equation

Oh wait

You did

Apologies

[daring bay]

np

i did something wrong tho

[pine temple]

96 =-36s-6t

[daring bay]

huh

[pine temple]

So you tried to do II times 2 - III times 3, right?

2(39 = 3r-9s)
-3(-6=2r+6s+2t)

[daring bay]

yes

OHHH

omg

okay

ok dont leave me

[pine temple]

I won’t

But I do think you’ll know what to do next

[daring bay]

yes ok wait

[pine temple]

?

[daring bay]

let me do

the thing

write it down

[daring bay]

so now we got 120=-48s

but

that seems wrong to

too*

cuz the answer for s is -3

[pine temple]

(78 = 6r-18s)
-(-18=6r+18s+6t)

This is what they multiply into

[daring bay]

wait what

yeah

[pine temple]

78 -(-18) is?

The keyboard filled it in automatically

[daring bay]

96

[pine temple]

=

6r - 6r

[daring bay]

0

[pine temple]

-18s - 18s

[daring bay]

-36s

[pine temple]

And -6t

What does that make?

[daring bay]

96 = -36s -6t

[pine temple]

Yes

[daring bay]

wair

wait

yeah ur right

[pine temple]

You also have 24 = -12s - 6t, which is correct

[daring bay]

why

[pine temple]

Why is it correct?

[daring bay]

no

omg its correct

i got the right answer

[pine temple]

Good job!

[daring bay]

we end up having

-72 = 24s

so -3 = s

now we can put that into II and solve for r

[pine temple]

Good!

[daring bay]

yeah so

s = -3

r = 4

and t = 2

so how can i put this into vector form

like d = ??

how do i write this

d = 4a-3s+2t

[daring bay]

thanks

ok im gonna do the next

exercise

[pine temple]

Ok

[daring bay]

yo so

did i type this in wrong

its the skalarproduct

rule

[rough forge]

sieht falsch aus

[daring bay]

warum aber

[daring bay]

ach

[rough forge]

du hsat die definition falsch angewendet

du multiplizierst beim skalarprodukt komponentenweise und addierst die sachen auf

[daring bay]

also x1 mal x2

und y1 mal y2

ich check

[rough forge]

ja

[daring bay]

ok ok ok

[daring bay]

prüfen sie rechnerisch ob b und c senkrecht auf einander stehen das heißt skalarprodukt = 0 dann senkrecht wenn nicht dann nicht

??

[rough forge]

ja

Hier noch mal ein merkblatt

[daring bay]

mit norm meinst du die länge

[rough forge]

Ja

[daring bay]

OK ok

kurzarbeit morgen

locker locker

[rough forge]

mal sehen

[daring bay]

nurnoch paar sachen dann fertig

[daring bay]

weißt du wie ich huer die g) mache

ich find keine yt videos dazu

[rough forge]

Das ist genau das, was ich dir letztes Mal gemalt habe

[daring bay]

also

[rough forge]

das mit dem R-Q

und R+Q

und was daraus folgte

[daring bay]

also..

[rough forge]

Also berechnen sollte straightfoward sein

[daring bay]

ja einfach 1/2 mal c usw

aber graphisch

[rough forge]

versuchs mal

[daring bay]

wie fange ich da aber an

[rough forge]

naja du gehst mal entlang c dann entlang a und irgendwann b entgegengesetzt

[daring bay]

also die hälfte mal c

gehe ich entlang?

[rough forge]

ja

versuchs mal und wir gucken

[daring bay]

okay

[rough forge]

so hätte ichs ungefähr gemacht

[daring bay]

hää

also wie gehst du da vor

[rough forge]

Ich habe die Vektoren a, b und c gemalt

[daring bay]

ich hab jetzt einfach irgend so ein scheiß gemacht rbudeee

ich checke garnichts

[rough forge]

von c gehe ich die Hälft

von a gehe ich die ganze Strecke

[daring bay]

abee

fängst du beim ursprung an

[rough forge]

von b gehe ich nur einen Drittel entgegengesetzt

[rough forge]

habe ich nicht festgelegt, sollte auch keine Rolle spielen

die Vektoren im Raum können überall liegen

Der Ursprung ist halt konventionell ein Bezugspunkt

[daring bay]

aber

ok ja

also wenn ich jetzt beim ursprung c male

dann muss ich + a addieren

[rough forge]

Also wenn man von dir erwartet das 1 : 1 nachzumalen wie in geogebra fände ich das fragwürdig

[daring bay]

gehe ich dann direkt von der spitze aus 2 nach unten 3 nach rechts und 2 nach oben

oder wie

[rough forge]

vor allem wirds nur noch unkenntlicher das in 3D 1 zu 1 zu versuchen zu malen

[daring bay]

so sieht die lösung aus

[rough forge]

also das ist ja heftig

R^3 Vektoren in einer Ebene

vielleicht sieht das nur so aus skull

[daring bay]

hä also

[rough forge]

Ok anscheinend musst du es echt richtig zeichnen

[daring bay]

er hat jz einfach

[rough forge]

bruder

stell dir vor es wären punkte

[daring bay]

ja

[rough forge]

die du da einfach rein zeichnest

[daring bay]

ja

[rough forge]

mit dem zusatz dass du da pfeile kritzelst

fertig

[daring bay]

laber nicht

als ob es so einfach ist

[rough forge]

das ist kein meisterwerk, aber es spiegt die Idee wieder

[daring bay]

warte bruder

guck mal

wenn ich

[rough forge]

du kannst halt sagen ein kasten entspricht eine einheit

[daring bay]

1/2c eingezeichnet habe

[daring bay]

dann mache ich + a

also gehe ich 2 in x richtung

[rough forge]

dann malst du (-2,0,1)

[daring bay]

hä

[rough forge]

statt (-4,0,2)

[daring bay]

ja

[rough forge]

ist ja die hälft

[daring bay]

genau

dann gehe ich von der spitze aus

von c jz

[rough forge]

ja

ja

[daring bay]

2 nach unten in x richtung

3 nach rechts

und 2 nach oben

[rough forge]

ja

[daring bay]

warum ist der hier aber nach oben gegangen

achso

nein

OK warte

WARTE BRUDE

ich habs gleich

[rough forge]

wie gesagt ist halt schwierig

aufm blatt

brauchst da schon kunst eine 2+ mindestens

[daring bay]

aber hier steht -1/3b

• und - wird zu plus oder

also 2 3 2 schonwieder für b

?

[daring bay]

minus und minus *

[rough forge]

Nochmal

du fängst mit (1/2)c an

[daring bay]

ja

dann a yeah

[rough forge]

von dort aus (Spitze) gehst du entlang a

[daring bay]

aber dann

[rough forge]

von der pfeilspitze jetzt

gehst du -(1/3)b entlang

[daring bay]

ja also

wird -6 zu 2

oder

für x

[rough forge]

also (2,3,-2)

[daring bay]

JA

aber

woher weiß man dann was P ist

[rough forge]

endgegner bist du

P ist die direkte Verbindung

[daring bay]

ach

das ist dann so eine vektorkette

??

[rough forge]

Von (1/2)OC bis (-1/3)OB

[rough forge]

jaa

[daring bay]

geschlossene vektorkette

[rough forge]

ja

[daring bay]

Fuck

ok

ey geh mal nicht schlafen bis 1 uhr

😭 😭

[rough forge]

ist das jetzt ein law von dir

[daring bay]

ok jz ist komplanar dean

das ist aber billig

[rough forge]

ja die würden in einer eigenen ebene liegen

bzw.

ja

[daring bay]

ok ok

ich geh auf toilette

geh nicht schlafen

[rough forge]

sonst krieg ich noch nen strafzettel für unterlassene hilfeleistung

[daring bay]

hö also

komplanar hat nichts mit lineare abhängigkeit oder basis in r3 zutun?

[rough forge]

doch schon

wenn 3 Vektoren koplanar sind dann sind die ja linear abhängig

die liegen ja in einer ebene

also in einem zweiidimensionalen teilraum von R^3

[daring bay]

also wenn t, r und s im lgs nicht 0 ist

dann sind sie linear abhängig?

[rough forge]

sind das die skalare?

,, t\mathbf{x}+r\mathbf{y}+s\mathbf{z} = \mathbf{0}

[rough forge]

wenn (t,r,s) = (0,0,0) dann ist die Gleichung immer erfüllt

so

[daring bay]

ja

[rough forge]

wenn aber das nur die einzige Lösung ist, dann sind x,y und z lin. unabhängig

das ist alles was du wissen musst

[daring bay]

also

wenn die skalare = 0 sind

[rough forge]

wenn im LGS auch nur die Lösung (0,0,0) existiert, dann impliziert das lineare Unabhängigkeit

wenn du eine weitere/andere Lösung(en) aber im LGS bekommst, dann heißt das sofort lineare abhängigkeit

[daring bay]

hä

ok warte

ich rechne gerade

[rough forge]

zockst du eig

so als 12. klässler

[daring bay]

ach digga

ich hab alles falsch gemacht

[daring bay]

ja

[rough forge]

oh

das überrascht mich um die uhrzeit

[daring bay]

ja

ich hab = 0

also

bei komplanar

[rough forge]

welche vektoren?

[daring bay]

aber das macht man nur bei r3

warte

ich machs nochmal

[rough forge]

,w rref{{2,4,-2},{0,3,1},{-1,0,5}}

[jolly parrot]

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Requested by adoniszgks

[rough forge]

also hast du als lösung nur (0,0,0)

[daring bay]

hä

[daring bay]

ja aber das ist falsch

[rough forge]

nein

[rough forge]

das zeigt dass nur die triviale lösung existiert, somit sind die nicht komplanar, weil sie nicht in einer Ebene liegen (sie spannen den gesamten R^3)

[daring bay]

ok warte

[rough forge]

was zockst du

[daring bay]

marvel rivals

zurzeit

[daring bay]

also

ich hab grad

ja also es ist nicht komplanar

das heißt sie sind linear abhängig?

[rough forge]

gegenteil

linear unabhängig

[rough forge]

.

[daring bay]

✅ ✅ ✅

jawoll

also guckmal

ich habe

ein überbestimmtes system gehabt

dann gab es ein widerspruch

das heißt nicht komplanar

digga wie merkt man sich das

hö

also komplanar = lineare abhängigkeit

[rough forge]

ja im R^3

[daring bay]

ja ok ok

ok nächste aufgabe

wir schaffen das digga locker

[rough forge]

das würde ich nicht so unterschreiben

im R^2 liegen die Vektoren alle in einer Ebene

trotzdem können manche linear unabhängig sein

z.B. (1,0) und (0,1) liegen in einer Ebene (schon weil die eine Ebene aufspannen) und sind linear unabhängig

koplanar heißt einfach nur, die Vektoren liegen in einer Ebene

im R^3 sind die zwangsläufig abhängig

[daring bay]

hä

und eine basis in r3

digga ich kann mir das garnicht bildlich vorstellen

[rough forge]

dein Zimmer ist ein R^3

die Kanten an deiner Wand sind die Vektoren

wenn du in eine Ecke guckst

die besteht aus 3 Kanten sozusagen

[daring bay]

ja

[rough forge]

deine 3 linear unabhängigen Vektoren

[daring bay]

ja

[rough forge]

da hast du deine Vorstellung

schwierig wirds ab R^4

bzw. unmöglich

[daring bay]

ja chill

[rough forge]

ja um die uhrzeit

[daring bay]

philosohpie jetzt oder was 😭 😭

also dings

ok ok

[rough forge]

hahahah

[daring bay]

bei der b )

muss ich wieder zeichnen und eif gucken

ob es eine verktorkette ist

[rough forge]

was hast du bei der a) gerade gefolgert

[daring bay]

linear unabhängig

[rough forge]

richtig

das heißt in der b)

[daring bay]

keine vektorkette

sowieso

[rough forge]

ES muss eine linear kombination geben für 3

[daring bay]

ja

hä

digga

[rough forge]

das ist doch gut zu wissen

[daring bay]

warte

[rough forge]

weil dann erwartest du im LGS eine eindeutige Lösung

[daring bay]

wenn sie linear unabhängig sind

dann zeigen die irgendwo hin

nh

[rough forge]

nicht ganz, wenn du die Linearkombination gefunden hast

das heißt

wie a, b und c skaliert und addiert werden müssen, dann werden die so liegen, dass der Weg d entsteht

Vektor kette

[daring bay]

ja

aber

das ist bei der b)

wenn ich die vektoren aus der a zeichnen würde

dann würden sie irgendwo hinzeigen oder

[rough forge]

ok ich male was

nein spaß ich gehe pennen

[daring bay]

waere

bitte

bitte mach es

[rough forge]

hahahahhahaah

also guck ma

so könnten deine vektoren iwo im raum liegen

wie wir wissen linear unabhängig

jetzt

exstiert ein vektor d im Raum

[daring bay]

ja

[rough forge]

jetzt linearkombination heißt nur wie muss ich a,b und c wählen um den gleichen weg zu konstruieren wo mich d auch hin bringt

wenn ich die zsm addiere sieht das so aus

a+b+c

[daring bay]

ja

[rough forge]

aber wir müssen bisschen fein tuning

z.b.

c sollte entgegengsetzt zeigen

-c

[daring bay]

ja

[rough forge]

und a und b vielleicht bisschen kleiner

So siehts deine Linearkombination am Ende aus

wir haben a,b und c bisschen runterskaliert

und c entgegengesetzt

[daring bay]

ahh

[rough forge]

die genauen Zahlen, das überlässt du der Mathematik

[daring bay]

das kann man dann auch umformen oder

wenn man zb

also wenn die abhängig sind

und man hat d = …

und dann nach c stellen

und so

[rough forge]

meinst du

wenn ich die Vektoren a,b und d habe dass ich dann eine Linearkombination für c finde?

[daring bay]

ja

[rough forge]

ja klar

[daring bay]

aber dings

[rough forge]

mathematisch heißt das, die gleichung nach c umformen

[daring bay]

warum hat der hier

0 = gemacht

für die kette

[daring bay]

ja

[rough forge]

wo, erkenne das nicht

[rough forge]

der hat d nach rechts gebracht

schaue dir mal die definition von vektorkette an

[rough forge]

im endeffekt heißt das geometrisch

[rough forge]

angenommen du gehst entlang c,a und b das waren 3 km

jetzt gehst du d rückwärts also -3km

dann erwartest du ja dass da 0 raus kommt

[daring bay]

aller

ja

gegen vektor

?

[rough forge]

ja -d meine ich

weil bro guck mal doch

wenn d = a+b+c ist dann ist 0 = -d + (a+b+c) equivalent

[daring bay]

ja

normal

[rough forge]

jetzt kannst du auch für a+b+c d einsetzten

0 = -d + (a+b+c) equivalent zu 0 = -d + d

das gleichts sich aus

[daring bay]

-d + d?

[rough forge]

ja

a+b+c = d war ja die beispiels linearkombination

[daring bay]

ja

[rough forge]

.

[daring bay]

ja

doch ja

[rough forge]

um wie viel uhr ist der test morgen

[daring bay]

genau

[daring bay]

9 45

kurzarbeit

vorletzte frage

[rough forge]

ok geil stell mir nen wecker um 9:45 nur um dann weiter zu pennen

[daring bay]

kann ich mir hier immer merken

AB = DC?

[rough forge]

Was wenn der Lehrer die gleichen Punkte nimmt sie aber anderes bennent

zum Beispiel er vertauscht B mit D

[daring bay]

ja deswegen

genau

deswegen

dann

AD

[rough forge]

Bro

mach folgendes

berechne AB, BC und AC

das geht in 20sekunden

weill dann gibts zwei situationen

du hast schon zwei parallele seiten

mit einer diagonalen

or gegenüber

[daring bay]

aber

ja

[rough forge]

ich würde einfach A,B und C im Raum zeichnen

[daring bay]

Ist A immer der pubkt unten links

ich kann mir das abee auch bisschien vorstellen

[rough forge]

so hätte ich es gemalt

dann hätte ich eine ahnung, okay AC sollte der Vektor diagonal im parallelogram sein

und AD sollte parallel zu BC sein

[daring bay]

ja

und AB BC

[rough forge]

das sind einfach die seiten

kanten

[daring bay]

ich musd auch noch mischungsprobleme können

hab ich 0 gemacht

[rough forge]

so kannst du es dann herleiten

so aber gleich ists soweit

[daring bay]

warum nicht DC?

[rough forge]

das spielt keine rolle

[daring bay]

dann wäre D = B-A-C

das würde gehen?

[rough forge]

beides parallelogramme

[daring bay]

danke bro

und

wenn ich überprüfen soll ob sie orthogonal sind

dann wieder skalarprodukt nh

[rough forge]

ja

[daring bay]

ok letzte frage

Bei A = a * h

woher weiß ich was Vektor a ist

ist es immer im parallelogramm AB?

[rough forge]

was ist jetzt a * h

was soll groß A sein?

[daring bay]

die fläche

[rough forge]

achso

norm vom kreuzprodukt

[daring bay]

ja

also ding

[rough forge]

Ja dann mache kreuzprodukt von AB und AD z.B

nimmst ein Punkt und die Nachabrn

[daring bay]

ja aber wenn ich zb die höhe brauche

dann muss ich A durch a machen

[rough forge]

du willst die fläche des parallelo

das ist die länge des normalenvektors

[daring bay]

ja

aber wenn da jz zb die höhe gefragt ist

dann tuh ich eif nach A/a = h umstellen oder

[rough forge]

mach pythagoras

bzw easier

Parallelo formel

A = h * a ja das meintest

h = A/a

[daring bay]

wäre a dann AB ?

[rough forge]

a ist die Grundseite

[daring bay]

ah

also

entweder

ja ok

je nachdem

ich könnte aber auch die obere seite nehmen nh

[rough forge]

ja

[rough forge]

Also länge von BC oder DA

[daring bay]

danke bro

ich hoffe es kommen keine mischungsprobleme dran

[rough forge]

versuch das wirklich zu skizzieren um wenigstens eine idee zu haben morgen

[daring bay]

ja

[rough forge]

noch nie gemacht denk eich

[daring bay]

mischungsprobleme?

die sind halt einfach

so anwendungsaufgaben mit LGS

[pearl pond]

@daring bay Has your question been resolved?

[pearl pond]

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[plucky carbon]

Can someone check this i feel like this is wrong

[light helm]

final value is fine,
though in the final step, make the cancellation of the root clearer

(make the cancellation line cut the radical bar as well, not just the 5 on the outside)

[plucky carbon]

got it thank you

.close

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[edgy wind]

so i’m doing implicit differentiation, the question is find d^2y/dx^2 of xe^y=4x^2

i got the first dy/dx as 8x-e^y / xe^y
so for second derivative i’m doing quotient rule but i’m lost at this point

[merry carbon]

,rcw

[edgy wind]

oh thanks for fixing the image, sorry about that

[merry carbon]

Don't worry about it, it happens sometimes

A bit of a bitch they gave that to you to do it implicitly

If anything though, in your quotient rule, make use of the fact you know what dy/dx is

[mental vine]

xe^y=4x^2?

let me do a quick check

[edgy wind]

yeah, it's tough

[mental vine]

e^y+xe^yy'=8x

dy/dx=8e^(-y)-1/x

[edgy wind]

so did i do the first derivative wrong?

[mental vine]

no i just simplified it

[edgy wind]

ahh

[mental vine]

y''=1/x^2-8e^(-y)y'

heck it looks ugly

[edgy wind]

wait

did you find it using quotient rule on the first y'

[mental vine]

y''=1/x^2-64e^(-2y)+8e^(-y)/x

wdym?

why do you quotient rule?

[edgy wind]

oh wait

nvm

[merry carbon]

(I also reiterate the previous point, use the fact you know what y' is)

[mental vine]

wait a minute! I think I see something

y''=1/x^2-64e^(-2y)+8e^(-y)/x

[edgy wind]

did you write this down or something haha, im a little lost how you got that answer

[mental vine]

y''=(1-64x^2/e^2y+8x/e^y)/x^2

you sub in y' to get this

[edgy wind]

ohhh

[mental vine]

^ this

now technically

[edgy wind]

i forgot to do that , i see

[mental vine]

64x^2/e^2y=64x^4/(xe^y)^2

and xe^y=4x^2

therefore

64x^2/e^2y=4

same reason

[edgy wind]

yeah i understand

[mental vine]

8x/e^y=2

so the whole thing=1-4+2/x^2=-x^-2

there you got it

d^2y/dx^2=-x^-2

[edgy wind]

yeah yeah, im just writing it down now

[mental vine]

ok :>

[edgy wind]

thanks

[mental vine]

np

[edgy wind]

.close

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[earnest yoke]

this is a puzzle/cypher that i have in my hw can anyone find the rule?

i cant see it

[pearl pond]

@earnest yoke Has your question been resolved?

[marble sigil]

it has to do with the encrypted message getting mixed up to the secret message, according to the numbers

[placid geyser]

I've seen a little bit of pattern but this doesn't apply to many letters

In the first one,
u is in the 2nd place and the number in 2nd place is 23
In the result, u is placed in the 23rd place

Similarly with the next a

But I think it might be a coincidence as it doesn't apply for many digits

Not even considering the missing letters

[pearl pond]

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[hoary nacelle]

Help

[hoary nacelle]

Is my solution correct

It’s really nice math cures sleeping disorders

[pearl pond]

@hoary nacelle Has your question been resolved?

[cursive wraith]

I'm not sure what is happening here, what are you trying to show?

[hoary nacelle]

It is because the LHS is a subset of RHS so I thought it would be handy to expand it a bit

[cursive wraith]

also writing \, intersections and unions in a line like that doesn't make a lot of sense. is {x} taken with the intersection of A?

[hoary nacelle]

It is more explicit, maybe redundant though

Yes I should make parentheses

Omg I am so dumb

I always make those stupid mistakes 🥲

[cursive wraith]

all you want is to show $(V_\epsilon (x) \backslash {x}) \cap \bar A \ne \emptyset$ right?

[jolly parrot]

rafilou is not not born in 2003

[cursive wraith]

so I'm guessing what you wanted to do is show $(V_\epsilon (x) \backslash {x}) \cap A \subseteq (V_\epsilon (x) \backslash {x}) \cap \bar A$?

[jolly parrot]

rafilou is not not born in 2003

[hoary nacelle]

Yes

I shouldn’t have expanded the RHS though it made it unclear actually

[cursive wraith]

yeah

[cursive wraith]

whenever you want to show a statement such as this one with "subset" chains

[hoary nacelle]

I just genuinely feel the explicitness can make things better maybe due to lack of formal math education, usually it made things only worse

[cursive wraith]

your "subset" chain should start with "$(V_\epsilon (x) \backslash {x}) \cap A \subseteq ...$" and should end with "$...\subseteq (V_\epsilon (x) \backslash {x}) \cap \bar A$"

[jolly parrot]

rafilou is not not born in 2003

[hoary nacelle]

Yes

That’s true actually more clear

Formalization has been so hard

[cursive wraith]

sometimes laying the grounds for your proof before going into details can help visualize exactly how to formalize it

[hoary nacelle]

And another one… I just have sleeping issues I feel I must do abit more curing it 🥲🥲

[cursive wraith]

here, without going into details, you saw that this statement was basically the whole idea

[hoary nacelle]

Yes sometimes I felt that way too… because I wanted to make thing clear usually turns to be making thing more complicated

Convoluted or involved

Yes exactly

[hoary nacelle]

It’s very hard to find a critical balance to make it even between explicitness and redundancy led confusion… maybe I can only solve this by actually attending math school… and I have always been trying to fix the issue like this

[cursive wraith]

I'm gonna finish talking about this before moving on to the other screenshots

[hoary nacelle]

And my notations are really problematic I joke about it often but truth is I almost can’t fix the issues with my notations another by product of not formal being math student

[cursive wraith]

why is the RHS non-empty?

[hoary nacelle]

Because X in A means it is already a limit point

Maybe I should just write this instead of that

Yes I should just write that sentence

[cursive wraith]

it's in A and it's a limit point of the closure, but why is it a limit point of A in that case?

Ik it might seem trivial but it doesn't directly come from the definition of a limit point

[hoary nacelle]

Since it is the limit point of the closure, then if it is in A then it is not in L, it is a limit point of itself

Or maybe I should say there exist. Y in V(x) cap clA since

[cursive wraith]

why? A and L might not be disjoint

in fact, you're trying to prove that x is in L

[hoary nacelle]

Yes that’s correct

I should do it this way correct l, bcs I assumed that if X in A the neighborhood must intersect point that is not x but it leads to gap

[cursive wraith]

I'm just gonna say

[hoary nacelle]

So I will say there is y such that y in A and V(x) it follows that y in V(x) cap clA with y=/x

This means y in A immediately follows that x is a limited point

Yes this makes sense

[cursive wraith]

I'm not quite sure about that

Let me just recap what we know and what we want to prove

we know: $x$ is a limit point of $\bar A$. So for all $\epsilon > 0$, $V_\epsilon(x)\setminus {x} \cap \bar A \neq \emptyset$

[cursive wraith]

what we want: for all $\epsilon > 0$, $(V_\epsilon(x)\setminus {x}) \cap A \neq \emptyset$

[cursive wraith]

ok but why is y in A?

it could just so happen that you picked y in L

[hoary nacelle]

This is by definition of limit point since we discussed the first case that y is not in L

[cursive wraith]

sorry I find it hard to follow you here

[hoary nacelle]

I will write it down 🥰🥰🥰

[cursive wraith]

your first case was that x is in L, and then it was trivial by definition

now your second case is that x is in A

(if you suppose x is not in L then you should arrive at a contradiction, because we want to prove x is in L)

[jolly parrot]

rafilou is not not born in 2003

rafilou is not not born in 2003

[cursive wraith]

so for now, we have existence of y in that set

but right now we don't know if y is in A or L

[hoary nacelle]

Yes that’s the case, my method for showing that y in A is actually wrong

[cursive wraith]

so here I believe you mean intersection with clA and not A

apart from that it's alright

also don't put "epsilon > 0" in the index

[hoary nacelle]

I should not just rely on intuition that x is in A then automatically a limit point.. the logic is flowed…

And I always make those flaw sadly 😭😭😭😭

[cursive wraith]

if you want to mention that epsilon > 0, make a statement for it

like "with epsilon > 0" after that

[hoary nacelle]

Okay I will note that… 😭😭😭 I really hope I can improve my notation though btw

[cursive wraith]

or "Let epsilon > 0" at the beginning

ok, I think it's much clearer now

but to be completely formal

you need to find an explicit value of "delta small enough"

so that you have what you claim

[hoary nacelle]

Sometimes I always forget that, I only remember it when I use exists e_0 >0 such that V_epsilon_0 subsets U, I will only remember in such case which is very very annoying though.. all small details and minor logic flaws

Can I say choose sufficiently small delta?

[cursive wraith]

the problem is right now, we don't know if taking delta smaller and smaller would even work, because it's not that trivial

I'm sure visualizing it can help

but finding an adequate value of delta would help the "credibility" of the proof

[hoary nacelle]

Can you teach me how to find explicit value for this case

[cursive wraith]

sure

[hoary nacelle]

Because the problem is abstract in formulation

[cursive wraith]

I'll do a drawing

brb

[hoary nacelle]

🥰🥰🥰

[cursive wraith]

ok so

the drawing is not the best but

as you can see, if we name d the distance between x and y

in the case that y is "close to the edge" of V_epsilon(x)

we want delta to be small enough so that V_delta(y) is completely inside V_epsilon(x)

to find a nice value, we notice that in the extreme case, d and delta would sum up to epsilon

so we need to pick delta smaller or equal to...?

[hoary nacelle]

Delta = epsilon -d

Delta < epsilon -d then it is always within the neighborhood

That’s brilliant

The drawing

[hoary nacelle]

Omg the drawing is so brilliant 🥰🥰🥰

I thought this one couldn’t pick explicit value

[cursive wraith]

yeah this value works when y is "close to the edge"

but in any case

[hoary nacelle]

I must properly define d right?

[cursive wraith]

won't be too hard

when $\delta \leq \epsilon - d$ we have that $V_\delta(y)\subseteq V_\epsilon(x)$

[jolly parrot]

rafilou is not not born in 2003

[cursive wraith]

but that's not exactly what we want yet

[hoary nacelle]

We need to show that there is a z

Such that z is not y or x

That lies within the delta neighborhood subsets the epsilon neighborhood

That means x intersects A closure at point in A that is not x itself

[cursive wraith]

z must not be x indeed

[hoary nacelle]

Then by definition X must be the accumulation point then the only if implication is finished

[cursive wraith]

the problem with our current delta

is that, if y is too close to x in the epsilon ball

[hoary nacelle]

It might be x

[cursive wraith]

then delta is so big

that V_delta(y) may contain x

[hoary nacelle]

So I should make a range for delta

[cursive wraith]

still using d = distance between x and y

what values of delta > 0 make sure that V_delta(y) doesn't contain x

[hoary nacelle]

Like delta in (0, epsilon-d)

This way it is not 0, and it is also not epsilon-d

[cursive wraith]

not quite

removing epsilon - d only doesn't solve the problem

[hoary nacelle]

Delta = max {delta , epsilon-d}

[cursive wraith]

delta can't be defined using itself

and I wouldn't advise for the "max"

[hoary nacelle]

Delta =min something

I feel like

Let me think

[cursive wraith]

here's a quick question to reply to that may set you on the correct path

[hoary nacelle]

1/2(epsilon-d) to epsilon-d

[cursive wraith]

what does it mean for x to be in V_delta(y)

the definition

[hoary nacelle]

It is a ball or a neighborhood with radius delta

Around y

[cursive wraith]

alright

and mathematically?

[hoary nacelle]

It is an open set containing interval (y-delta, y+delta)

So I need to make sure

That (x-epsilon, x+epsilon) intersect V(y) notin x

[cursive wraith]

say V strictly means the ball

so V_delta(y) in R would be (y-delta,y+delta) and only that