help-43

[carmine garden]

Is this better

[azure vault]

this is better

[carmine garden]

Cool! Thanks!

Now to prove it's of order $N$

[boreal girder]

What a wonderful world !

[carmine garden]

I think I need a break

sorry

thanks a lot again

.close

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[placid jasper]

B wins, winning stratergy move down

[sage sphinx]

can you explain why

[placid jasper]

area of a triangle is width x height / 2

B and C cant move

so width is always 10

current area is 45 because (9 x 10) / 2

= 45

moves left doesnt change the area of the triangle since the height doesnt change

the only way to change the area is to move down

the goal of the game is to get to the y axis coord 5

because ( 5 x 10 ) / 2 = 25

if A moves down on both goes, so does B and B wins

if A moves left and B moves left until reaches the y axis, it is then As go

once reaching y axis

[placid jasper]

so same thing applies since A must go down

so nevermind the stratergy is to follow what A does

if you follow As movement you will eventually either reach, y = 5, or the y axis

which is winnable in both cases

get it ?

[compact pewter]

@sage sphinx Has your question been resolved?

[sage sphinx]

sorry why are we reaching the axis

[sage sphinx]

kind of, i thnk its a bit confusing still 😭

[placid jasper]

because coords can only be positive

so once you reach axis you can only go down

[sage sphinx]

oh so by reaching y=5 you're saying b reaches y = -5?

[placid jasper]

what do you mean

basically in order to win you must reach y =5

you get that part?

[sage sphinx]

yeah

i was just thinking

doesnt C and B start at the same place basicaly

so wouldn't they both reach y = 5 at the same time

[placid jasper]

C and B are the two bottom dots

red on left and blue on right

[sage sphinx]

yeah

if they start at the same place

[placid jasper]

A is the top middle dot

and they both control A, one turn each alternating right?

oh i see what you mean, they have the same A that they are controlling alternatingly, they dont have different ones

otherwise yh theyd both reach y = 5 same time

but since its altrenating goes then only one can reach y = 5

[sage sphinx]

oh

[sage sphinx]

ohhhhh

[sage sphinx]

okay makes sense, thanks

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[agile moth]

is this wrong? how did it get to 280?

[stark marsh]

(361-81)/361 i don't see where is it wrong

[agile moth]

9^2 is 280?

[stark marsh]

$\frac{361}{361}-\frac{81}{361}=\frac{280}{361}$

[boreal girder]

yoboiqimmah

[agile moth]

why you minusing 81?

[stark marsh]

9^2

,w 1-(9/19)^2

[boreal girder]

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[agile moth]

i dont know what you're doing

its ok nevermind

[stark marsh]

$\frac{a}{b} \pm \frac{c}{d}=\frac{da\pm bc}{bd}$

[boreal girder]

yoboiqimmah

[agile moth]

dude...its ok. please stop. i dont like this latex stuff. rather just use words

[stark marsh]

ok 👍

[sturdy cedar]

[please help in my channel

[compact pewter]

@agile moth Has your question been resolved?

[agile moth]

help me and i'll help you

[winged lion]

[winged lion]

arithmetic rules, you can pull the exponent inside the fraction like
(9/19)² = 9²/19² and calaculating this gives you 81/361

now to subtract it from 1, the 1 needs to have the same denominator, so naturally you write 1 = 361/361

[agile moth]

@winged lion yea i figured it out thanks....

some people like to flex their latex skills a little too hard...

[winged lion]

i can flex mine too ;')

[stark marsh]

my guy i'm trying to help

[agile moth]

i apologize...sometimes it's just hard to follow along with the latex...sometimes its just easier to explain using words

[stark marsh]

i get your point but imo when it's math it's better explained with math imo

[agile moth]

no worries....appreciate your efforts nonetheless 🫶

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[fading grotto]

https://garticphone.com/en/?c=01410e3d48

[native shard]

lmao

[compact pewter]

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[flint wing]

u joined it?

[compact pewter]

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[floral kelp]

Guys i dont understand why we have to times 14!

[woeful schooner]

Put the 5 cars in a giant truck, now you have this truck and thirteen cars besides it, how many different ways can you arrange this truck and 13 cars?

[floral kelp]

OHHH

[floral kelp]

14!

[floral kelp]

Thank you so much!

[woeful schooner]

You're welcome

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[hard quartz]

can someone help me ques f

[hard quartz]

phi from-pi/2 to pi/2 right

my friend told me its a cone and phi from 0 to 2pi

can someone help me

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[potent berry]

Does this question make sense? It asks for at least one solution, but the answer given is for det of coeff matrix to be zero.

Even if the det of the coeff matrix is non-zero, it can have a unique solution, so I expect infinite values of \lambda to be possible.

[kind viper]

i think it means "which of the following lambdas give ≥1 sol"

well that is how i would've read it

but the given sol does not match up...

[potent berry]

but don't all the options give at least one solution?

Like if the det of the coeff matrix is non-zero, there is a unique solution, So there's $\geq 1$ sol for all $\lambda \neq -5$.

[boreal girder]

Percy

[kind viper]

let me guess, you're legally required to follow this book like gospel and any and all deviations are severely punished?

[kind viper]

yes that is correct

[potent berry]

no this is just a practice set, I'm just making sure I'm not confused as hell.

[potent berry]

ah okay so im assuming the question maybe wanted to say infinite solutions?

[hushed magnet]

det = 0 means infinitely many solutions or zero solutions

I'm not seeing where they show that its not zero

[kind viper]

,w rank {{3,-1,4,3},{1,2,-3,-2},{6,5,-5,-3}}

[boreal girder]

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[kind viper]

well it does have infinitely many solutions for that one value of lambda.

but yeah it is not shown.

[potent berry]

eh yeah it does and isnt shown lol

[hushed magnet]

I hate it when people who write solutions dont get whats going on

[kind viper]

the blind leading the blind

[potent berry]

hmm okay so ill just assume they meant infinite solutions not at least one lol

[hushed magnet]

there is no reason to use the det. you have to rref at some point anyway. just do it in terms of lambda

but no, blind people see det and use it

[potent berry]

yeah i dont know what you mean there

[hushed magnet]

you have to row reduce anyway to show that for lambda=-5 it has a solution

so just row reduce from the start

with lambda as a variable

[potent berry]

ah well i just find |A| and set it to 0

[hushed magnet]

thats just doing more work than you have to

I thought this was about being fast

[potent berry]

maybe i should learn that

[forest token]

what is row reduce?

[potent berry]

blame the indian education system

[hushed magnet]

gaussian elimination

[potent berry]

gaussian elim

[hushed magnet]

however you wanna call it

[potent berry]

anyhoo

.close

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[kind viper]

jee toppers use det

[hushed magnet]

that being said, if they dont even know that what they did is not enough...

[forest token]

oh we've never learnt to do it like that

[hushed magnet]

then I suppose it doesnt hurt if you also dont do it properly

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[potent berry]

i swear we werent taught the rref lol

[potent berry]

oh no

.close lol

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[hushed magnet]

why the fuck are you not taught that

its the single most important algorithm

dare I say in math

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[silver canopy]

\renewcommand\s{\op{sigmoid}}
can I get some help computing $\grad f$ for the function [
\6f{x,y,z} = \6\s{x-y+z} +\6\s{2x+3y-4z}
]

[silver canopy]

not sure how to calculate the derivatives again

[compact pewter]

@silver canopy Has your question been resolved?

[warped jasper]

that's just the chain rule
find sigmoid', then compose with the derivatives along each coordinate, and add the two sigmoid's gradients

also long time no see, lexa

[unkempt sail]

Cuz we're not really taught to understand we're just taught to apply formulas as fast as you can 🤡

[hushed magnet]

even in that framework, row reducing is a good mindnumbing thing to just teach to students without having them need to understand anything

how are you fucking not taught to solve systems

[unkempt sail]

I always convert it into a 2 variable system by cancelling out either variable

Makes it a bit easy

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[wild lion]

How can I formally prove b

I mean the way I can explain is whenever a is true b is true and simultaneous whenever b is true a is true. This condition will make sure the truth table of biimplication of a and b will always yield true.

But that is probably non rigorous right?

Also in 5 best I can think of is demonstrating two cases each corresponding to different truth tables

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[wild lion]

.reopen

.reopen

.reopen

<@&286206848099549185>

[knotty basin]

.reopen

what

.close

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[knotty basin]

.reopen

[compact pewter]

✅

[wild lion]

[knotty basin]

there that's resolved

[flint wing]

yw

[knotty basin]

as for the question i'll avoid misguiding you and i'll wait for someone more proficient to help you

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[lunar thistle]

I want to help but I don't know what this symbol means, I mean, I surfed the net and I didn't saw anything about it.

[compact pewter]

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[sharp grail]

I want someone to explain the gradient matrix and the equivalent gradient matrix line by line for me A

[compact pewter]

Please don't occupy multiple help channels.

[bleak dock]

.close

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[vapid yacht]

When I have

$f(x)$ transformed to $f(2x-1)$

[boreal girder]

Klein Bottle

[vapid yacht]

The order of transformations is

--> horizontal stretch by scale factor 1/2

--> horizontal shift right by 1/2

Right?

[kind viper]

yeah that works

[vapid yacht]

Are you certain that it's not the other way around?

[kind viper]

i am

[vapid yacht]

Gotcha

[kind viper]

if you wanted to do a translation first it would need to be by 1 not 1/2

[vapid yacht]

Got it. That makes sense

Thanks so much

Have a great day

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[wise apex]

Hey, I’m on the final stretch with 1 week left before the Matematik 1b final (gymnasiet, Sweden), and I need an A.
Does anyone have any solid tips for revision or what to focus on in this last week?
Thanks in advance I really appreciate any advice that could help me maximise my chances!

// im new to the server so please forgive me if i am not doing this right lol

[wise apex]

i basically just need some tips, thats all...

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[heavy basalt]

How to do c and d for 45

[strange ermine]

The same way you would do it for any functions

I hope you already did some "finding min and max" exercises

[heavy basalt]

Yeah ik

Local max is 2

Local min didn’t exist or something?

Cuz 0 is not the critical point

[strange ermine]

Yep, the local min doesn't exist

And the local max is at x = 2

You got it correct 💪

[heavy basalt]

What about concave up or down

[strange ermine]

It is concave down up until x = 3. There it changes concavity and it stays concave up from 3 to inf

[heavy basalt]

How u get 3

[strange ermine]

By studying the sign of the second derivative

[heavy basalt]

So we need to find f’’ too?

[strange ermine]

Yep, second derivative is needed to check where a function is concave up and concave down

The same way as the first derivative makes you study where the function is increasing or decreasing

[heavy basalt]

Ok got it 💀💀💀💀💀💀

Thx

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[unkempt grail]

Angle A is 30degrees, AC is sqrt(32+16sqrt3)
How to prove that perimeter is 16 ?

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[unkempt grail]

thx

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[slim crystal]

so i had to answer this:

[slim crystal]

and already did part A

does everything look correct here

[kind crane]

!15m

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[slim crystal]

<@&286206848099549185>

[tight badger]

Your reasoning is correct. But your values are not. Review this line of work on the LHS and RHS.

[slim crystal]

i thought i messed up here

im following a vid where my teacher is doing a similar problem and hers is 2cosθ/2

so how would mine work

would it be -cosθ? @tight badger

[tight badger]

Not quite. We are looking to isolate cos(theta) to find the x-values which correspond with the solution. How would you get rid of -2 on the RHS?

[slim crystal]

ohhh wait dividing by -2?

but will the LHS stay the same?

-squareroot2/2 will remain right

[tight badger]

Please refer to the Golden Rule of Algebra. If we divide the RHS by -2, we must do the same to the LHS. So what is (-sqrt(2))/(-2))?

[slim crystal]

ohh okay it will be positive

so squareroot/2

[tight badger]

Yes. (sqrt(2))/2.

[slim crystal]

so does that mean i have to change my explanation after

oh wait i dont i think

since its the saem on unit circle

right

[tight badger]

Your explanation is good. All you need to do is find the correct x-values on the unit circle, as (3pi)/4 and (5pi)/4 do not have a cos value of (sqrt(2))/2.

[slim crystal]

okay i get it so it would be like pi/4 and 7pi/4

[slim crystal]

do you think you can also look at Part B, im still working on it but if you can tell me if everythings good so far

[tight badger]

Of course.

[slim crystal]

thank you so much seriously

[slim crystal]

im still doing this i just fixed my mistakes

[slim crystal]

what do you think is it fine

[tight badger]

Yep, looks good so far. Take note that you set y = 0, not θ.

[slim crystal]

ill change that

[tight badger]

More importantly, do not forget what interval they are asking you to find the solutions in. There are more than 2 x-values where y = 0. But this step can be done either before or after dividing by 2.

(It may also help to graph this function, but I suppose it's really not necessary unless you'd like a visual to help you understand.)

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[strange pendant]

a = -3

3. Let ( \mathbb{H}_1 = {x \in \mathbb{R}^4 \mid x_1 - 3x_3 + 2x_4 = 0} ) and ( \mathbb{H}_2 = {x \in \mathbb{R}^4 \mid x_1 - x_2 + x_4 = 0} ). Find, if possible, a subspace ( W \subset \mathbb{H}_2 ) and all values of ( a \in \mathbb{R} ) such that

[W + \langle (a, 0, 1, -3) \rangle = \mathbb{H}_1.]

[boreal girder]

938c2cc0dcc05f2b68c4287040cfcf71

[strange pendant]

-3 -4 = 0

<(-3,0,1,-3)> is not in H2

,w nullspace {{1,0,-3,2}}

[boreal girder]

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[strange pendant]

H1 = <(3,0,1,0),(-2,0,0,1),(0,1,0,0)>

[vocal glade]

guys}

help me pls

[strange pendant]

W + <(-3,0,1,-3)> = <(3,0,1,0),(-2,0,0,1),(0,1,0,0)>

[vocal glade]

i need this for tomorrow

[strange pendant]

dim(A+B) = dim(A)+dim(B)-dim(AnB)

[vocal glade]

waaa

😔

[strange pendant]

dim(W + <(-3,0,1,-3)>) = dim(W) + dim(R) - dim(Wn<(-3,0,1,-3)>)

,w nullspace {{1,0,-3,2},{1,-1,0,1}}

[boreal girder]

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[strange pendant]

W = <(3,3,1,0),(-2,-1,0,1)>

<(3,3,1,0),(-2,-1,0,1)> + <(-3,0,1,-3)> = <(3,0,1,0),(-2,0,0,1),(0,1,0,0)>

,w rank {{3,3,1,0},{-2,-1,0,1},{-3,0,1,-3}}

[boreal girder]

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[strange pendant]

W has to be dim 3 mate

otherwise the sum would be direct

,w nullspace {{1,-1,0,1}}

[boreal girder]

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[strange pendant]

H2 = <(1,1,0,0),(-1,0,0,1),(0,0,1,0)>

W has 2 of this

H1 = <(3,0,1,0),(-2,0,0,1),(0,1,0,0)>

W = <(1,1,0,0),(-1,0,0,1),(3,0,1,0)>

,w rank {{-3,0,1,-3},{1,1,0,0},{-1,0,0,1},{3,0,1,0}}

[boreal girder]

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[strange pendant]

nah I have a better idea mate

(-3,0,1,-3) = a(3,0,1,0) + b(-2,0,0,1) + c(0,1,0,0)

-3 = 3a - 2b
0 = c
1 = a
-3 = b

oh im so fucking stupid

a-3+6 = 0

a+3 = 0

a = -3

<(-3,0,1,-3)> is not in H2

,w nullspace {{1,0,-3,2}}

[boreal girder]

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[strange pendant]

(3y-2z,x,y,z) = y(3,0,1,0) + z(-2,0,0,1) + x(0,1,0,0)

H1 = <(3,0,1,0),(-2,0,0,1),(0,1,0,0)>

a -3 -6 = 0

a -9 = 0

a = 9

<(9,0,1,-3)> not in H2

(9,0,1,-3) = a(3,0,1,0) + b(-2,0,0,1) + c(0,1,0,0)

9 = 3a -2b

0 = c

1 = a

-3 = b

(a,b,c) = (1,-3,0)

(9,0,1,-3) = (3,0,1,0) -3(-2,0,0,1)

(9,0,1,-3) = (3,0,1,0) + (6,0,0,-3)

(9,0,1,-3) = (9,0,1,-3)

W = <(3,0,1,0),(-2,0,0,1), w3>

,w nullspace {{1,0,-3,2},{1,-1,0,1}}

[boreal girder]

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[strange pendant]

H1 n H2 = <(3,3,1,0),(-2,-1,0,1)>

W = <(3,3,1,0),(3,0,1,0),(-2,0,0,1)>

,w rank {{3,3,1,0},{3,0,1,0},{-2,0,0,1}}

[boreal girder]

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[strange pendant]

H2 = { x1 - x2 + x4 = 0}
H1 = {x1 - 3x3 + 2x4 = 0}
W = <(3,3,1,0),(3,0,1,0),(-2,0,0,1)>
W + <(9,0,1,-3)> = H1

(x1,x2,x3,x4) = a(3,3,1,0) + b(3,0,1,0) + c(-2,0,0,1)

a(3,3,1,0) + b(3,0,1,0) + c(-2,0,0,1) = (3a+3b-2c,3a,a+b,c)

(x1,x2,x3,x4) = (3a+3b-2c,3a,a+b,c)

(3a+3b-2c)-3(a+b)+2c=0

3a+3b-2c-3a-3b+2c = 0
a(3-3) + b(3-3) + c(-2+2) = 0

so W is entirely contained in H1

we can also check if it has an intersection with H2

a(3,3,1,0) + b(3,0,1,0) + c(-2,0,0,1) = (3a+3b-2c,3a,a+b,c)

(3a+3b-2c) -3a + c = 0

3a+3b-2c-3a+c = 0

a(3-3) + b(3) + c(-2+1) = 0

3b -c = 0

3b = c

so it has an intersection with H2

W intersects non trivially with H2, but is not entirely contained in it

,w nullspace {{1,0,-3,2},{1,-1,0,1}}

[boreal girder]

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[strange pendant]

x(3,3,1,0) + y(-2,-1,0,1)

W = <(3,3,1,0),(-2,-1,0,1)>

<(3,3,1,0),(-2,-1,0,1)> + <(9,0,1,-3)> = H1

,w rank {{3,3,1,0},{-2,-1,0,1},{9,0,1,-3}}

[boreal girder]

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[strange pendant]

very interesting

whatever

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[wild musk]

Solve all triangles of ABC such that ∠C = 10deg, c = 3, and a = 5

I'm comfortable with
Case 1:

B = 180 - C - A, 
b = (3*sin(B)/sin(C)```

However, there's apparently a second possible case. This one I do not understand. Would love any n all explanation 🩷

[torn stratus]

yea

Its

SSA

Its a lil tricky

Use law of sines

[slate sand]

Does that make sense why there’s 2 solutions

[kind crane]

!nogpt

[compact pewter]

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).

[slate sand]

I mean it’s mostly correct other than the second last point

3 is definitely not greater than 5 and yet the ambiguous case is happening

[compact pewter]

@wild musk Has your question been resolved?

[wild musk]

ohhhh so between the two cases you flip of angle values of A and B?

then b = (3sin(16.8deg))/(sin(10deg)) instead of (3sin(153.2deg))/(sin(10deg)), but a and c are the same
ahhh yeah that all makes sense now actually

[wild musk]

thank you !!!

[wild musk]

thank you also!

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[split pelican]

Is this right? :3

[old mica]

looks good to me

[split pelican]

Tyyyy hi again

I have 2 more HAHAH but then I’m done that’s all for this unit I think

[old mica]

these also look correct

well done

[split pelican]

Gosh you’re THE BEST!! So helpful. Thank you so much 🥰🥰

YAYY thanks!!

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[warped cypress]

Sorry could somebody please teach me how to understand it

I am actually struggling to understand even with the help of some internet

[slate sand]

What don’t you understand

[warped cypress]

Number 6 please

Like A union B but not in B

[slate sand]

A U B’

That one?

[warped cypress]

Yes

[slate sand]

So take A

[warped cypress]

(Like why is it not like this)

[slate sand]

This is A by itself

[warped cypress]

Okay!

[kind viper]

because the overlapping bit belongs to A and thus belongs to A union (anything else)

[slate sand]

This is not B

Add them together and you get the picture they drew

[warped cypress]

Ohhh

[kind viper]

i also copied the illustrations for A and for B' from OP's picture

[warped cypress]

Ohhh

[slate sand]

U means you add them all up together

∩ means you only take what they both have in common

[warped cypress]

Let me try processss

Ohhh

[warped cypress]

How would u word this one btw

Nvm nvm

Do u mind if I ask few more please

[slate sand]

Draw each side by themselves

[warped cypress]

A union A and B , but not in both . Dosent that mean it’s just outside

[slate sand]

What does A’ look like? What does B’ look like?

[warped cypress]

Okay let me draw it now

[slate sand]

Are we adding them together or taking what’s in common?

[kind viper]

i dont think you should be trying to pull the word "not" far from the set(s) it attaches to

also btw

your picture has references for individual sets and their complements

look at those

[warped cypress]

Ohhhh

[kind viper]

like you should be keeping the notation

[slate sand]

Oh yeah, A’ U B’ means (not A) union (not B)

[kind viper]

^

[slate sand]

There are brackets there sort of to say “hey do this first”

[kind viper]

look at A', look at B', reason about their union

[warped cypress]

Oh

Ohhh

[kind viper]

tbf that is literally brackets' entire job lmfao

[warped cypress]

Let me draw the thing rq

[slate sand]

Even sets have their version of pendas or bedmas or whatever you call it

[warped cypress]

OHHHH

SO I union them together

OHHHH

thank you so much!!!!!!!!

[slate sand]

Cool drawing surface

[warped cypress]

Thanks it is very reliable

[kind viper]

yes

$\cup$nion vs. i$\cap$tersection

[boreal girder]

Ann

[warped cypress]

Let me see if I have any more questions rq rq

Thank so much once again!!

n mean have in common?

[kind viper]

yes

[warped cypress]

Ohhhh

[kind viper]

that's what your picture says

[warped cypress]

I get this one nowwww

[kind viper]

at the very top

[warped cypress]

Haha

Okay

I think I understand how set notation works now!

Thank you guys so much for ur help

cant believe I actually understand it so quickly I was sooo sooo confused before

Thank you!!!

Have an amazing day or night ann_dec & frosst !!

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[jagged ibex]

Is part I) because Z2 is a rotation of Z1 by 60 degrees? Therefore the angles formed by the other points are 60 degrees?

And how do you do part ii)?

[bleak dock]

that's correct so far for (i)

you've shown that angle AOB is 60 degrees (the rotation is around the origin)

but you haven't mentioned another fact

that would imply that AOB is equilateral

[jagged ibex]

Wat that???

[bleak dock]

this triangle has a 60 degree angle

[jagged ibex]

What is C is this case?

[jagged ibex]

And what property is this?

[bleak dock]

c would correspond to side BA on the diagram

[bleak dock]

follows from the cosine rule

well I used the picture for the shape of the triangle

[bleak dock]

what extra condition do you need for such a triangle to be equilateral?

[jagged ibex]

All sides are equal?

[bleak dock]

well yes but

[bleak dock]

you only need two of the sides to be equal AND AOB = 60 deg

[jagged ibex]

What is it then?

[bleak dock]

triangle AOB is isosceles

isosceles and 60 deg implies equilateral

[jagged ibex]

O true????

[bleak dock]

more specifically, you need to also show that OA has the same length as OB

how can you do that then given $z_2 = e^{i \pi/3} z_1$

[boreal girder]

south

[jagged ibex]

Could i divide z1 over? does that get me anywhere???

[bleak dock]

you'd probably be getting further away then

[bleak dock]

okay, how do you express "length" using symbols

length of a complex number

[jagged ibex]

Modulus

[bleak dock]

yep

so you need to show that $|z_2| = |z_1|$

[boreal girder]

south

[jagged ibex]

How do i get to that with this???

[bleak dock]

well you know you can express any complex number $z = re^{i \theta}$ right

[boreal girder]

south

[bleak dock]

what's r here for z2?

[jagged ibex]

yer

[bleak dock]

well you don't know that sorry

okay fair enough, if you do $\frac{z_2}{z_1} = e^{i \pi/3}$ and take the modulus of both sides

[boreal girder]

south

[jagged ibex]

ohhh modulus of rhs is 1

[bleak dock]

$|\frac{z_2}{z_1}| = |e^{i \pi/3}|$

[boreal girder]

south

[bleak dock]

yep!

[jagged ibex]

ok ok

[bleak dock]

so |z2/z1| = 1 implies |z2| = |z1|

[jagged ibex]

so that completely proves part i

How would i do part ii?

[bleak dock]

can you stop spamming this in every channel

[jagged ibex]

Whadahel

[bleak dock]

if you want to help just type something in this channel

[jagged ibex]

💀

Ok yea how parrt iii

ii sorry

[bleak dock]

okay that one's easier

you can try working backwards (remember to write $\iff$ for this)

[boreal girder]

south

[bleak dock]

sub in $z_2 = e^{i \pi/3} z_1$ into the equation and cancel

[boreal girder]

south

[jagged ibex]

Watchu mean by that

start from rhs then prove to lhs?

[bleak dock]

for a proper proof, you can't start off with z1^2 + z2^2 = z1z2

but if you reverse all your steps, it now works

[bleak dock]

that's why we need the double arrow

you write your proof backwards, so that the thing you get at the end (from assuming part (ii)) is the start of your proof

[jagged ibex]

Where do i start then? if im working backwards

[bleak dock]

you start from z1^2 + z2^2 = z1z2

[jagged ibex]

Yea im lowkey confused

[bleak dock]

and then do this

[jagged ibex]

If u can could u write it up for me so i can clearly see what u mean???

[bleak dock]

$(z_1)^2 + (e^{i \pi/3} z_1)^2 - z_1 (e^{i \pi/3} z_1)$

[boreal girder]

south

[bleak dock]

oh wait sorry for all the confusion

[bleak dock]

you can just proceed from here

you want to show this equals 0

[jagged ibex]

Ah ok, i tried it out i got 0

Tank you very muc 🔥 👍

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[knotty basin]

we had a question in the exam where it was basically "find the different ways you could arrange the letters of the word AMALGAMATE" given there must always be 3 letters bw the 2 M's. now i did this by making cases with A's i.e.
M no A M rest
M 1A 2 other M rest
M 2A 1 other M rest
M 3A M rest

and then internal permutations and all

however in retrospect cant i just fix the M and M, then find the internal permutations of the rest with 8!/4! and then * 6 for potential spaces for the M block?

[knotty basin]

i believe the answer was the same

im just curious as to why it works here, and when it wouldnt have

[kind viper]

yes you can do it the second way

i don't see a situation in which it wouldn't work except maybe if there's 3 Ms?

[knotty basin]

oooh wait im stupid the qs im thinking of forced you to have like the parallel to this qs would be "if there must be at least 2 As between the M's"

yeah here i can just do this

oh well

thank you

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[quasi wind]

solving this gives x={3,-2/3,0} , my answer key says answer is 0 and 3 ,i get why they excluded -2/3 but why did they INCLUDE 0?

[unkempt sail]

I also think it should be undefined at 0

I mean limit tending to 0 exists but not defined at 0

[quasi wind]

yes yes

[tacit robin]

They didn't include -2/3 because the LHS would be negative and the RHS positive

[quasi wind]

thank you but that was not my question lmao

[tacit robin]

I don't know about the zero part

[quasi wind]

ah ok no worries

wait its defined at x =0???

all my maths fundamentals are being killed infront of my eyes

how can a denominator be zero

[unkempt sail]

[quasi wind]

x =0 is a valid solution but why 😭

[compact pewter]

@quasi wind Has your question been resolved?

[compact pewter]

@quasi wind Has your question been resolved?

[quasi wind]

yea this makes sense

atleast i think it does

i dunno if my brain does

anyway thank yall 👍

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[austere solstice]

Hmm English pls?

[kind viper]

Let S and T be subspaces of R^4 given by <SEE OP>.

a) Find dim(S) and dim(T).
b) Find S ∩ T.
c) Find a basis of R^4 which contains a basis of S and a basis of T.
d) Find a vector v ∈ S+T which belongs neither to S nor to T individually.

[kind viper]

which letter are you doing rn?

[strange pendant]

2 = b

[kind viper]

they're numbered with letters in your original lmao

anyway ok

[strange pendant]

b)

[kind viper]

there are two ways to do b)

both equally valid

[strange pendant]

(a,a-b,b,a) ∈ S

[strange pendant]

you mean find cartesian equations of the plane for S and then solve the simultaneous system

[kind viper]

that would be the 2nd method

[strange pendant]

what would it be? intersection of like how many hyperplanes?

[kind viper]

to make this more precise, (a,a-b,b,a) is the expression for a generic vector in S.

plug this into each of the equations that define T, and get a system of 3 equations in two unknowns (a, b). solve that. this gives you a necessary and sufficient condition for then a(1,1,0,1) + b(0,-1,1,0) ∈ T.

this would be my first choice for method.

[kind viper]

idk what "it" is.

[strange pendant]

forget it

x1 -x2-x4 = 0
x1-3x2-x3=0
2x1 + x3 -3x4 = 0
(x1,x2,x3,x4)=(a,a-b,b,a) ∈ S

[kind viper]

x1 -x2-x4 = 0
x1-3x2-x3=0
2x1 + x3 -3x4 = 0
(x1,x2,x3,x4)=(a,a-b,b,a) ∈ S

yes

now rewrite all those eqs in terms of a and b

do it carefully and simplify with care but don't overthink.

[kind viper]

ok

so in other words, a(1,1,0,1) + b(0,-1,1,0) ∈ T if and only if a = b.

that means the generic vector of S ∩ T is a(1,1,0,1) + a(0,-1,1,0)

[strange pendant]

ye

[kind viper]

that means S ∩ T = {a(1,0,1,1) | a ∈ R}

= <(1,0,1,1)>

what you have done here is found a basis (of one element) for S ∩ T

[strange pendant]

ye , is a line that passes through the origin

[strange pendant]

what about c

[strange pendant]

dim T = 3?

[kind viper]

ok hold on

im looking at the right half of your work here

[strange pendant]

i wanted to write x4 yes

my bad

[strange pendant]

yep

[kind viper]

strange though, you're getting that any vector in T must satisfy -x4 = 0?

oh you messed up arithmetic

[strange pendant]

x4 = 0

[strange pendant]

where

[kind viper]

0 - (-1) = 1 not 0

[strange pendant]

my bad

[kind viper]

this would be the row-reduced matrix

you get dim(T) = 2

[strange pendant]

ok

c, what about it

[kind viper]

• start with v1 := (1,0,1,1), the only vector in our basis of S ∩ T.
• add one more vector v2 ∈ S so that {v1, v2} is a basis of S.
• add one more vector v3 ∈ T so that {v1, v3} a basis of T.
• add one more vector v4 ∈ R^4 so that {v1, v2, v3, v4} is a basis of R^4.

[strange pendant]

ok great

dim(S+T)=3

[compact pewter]

@strange pendant Has your question been resolved?

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[quasi python]

how to find the error on pade approximant for e^x
https://en.m.wikipedia.org/wiki/Padé_approximant

[bleak dock]

there's some theory that says the (m, n)th Pade approximant matches all the first (m + n) derivatives of the Taylor series

[quasi python]

for example this is order (8,8)

[quasi python]

what does that mean?

[bleak dock]

so you can use the Lagrange error bound here

[quasi python]

what is that

[bleak dock]

Google it

[quasi python]

this?

[bleak dock]

yep

[quasi python]

let's say I'm trying to find error on pade order (8,8)

[quasi python]

a is 4?

n is 8?

[quasi python]

what is f^(n+1)(z)

[bleak dock]

the (n + 1)th derivative of the function evaluated at z

[bleak dock]

this approximant should be centred at x = 0

[quasi python]

so I need to find the derivative of this?

[bleak dock]

no

[bleak dock]

you're missing some really essential concepts about Taylor series

[quasi python]

this one yeah?

[bleak dock]

yes

I'm mentioning things that you should already be knowing if you want to learn about Pade approximants

you're clearly not on the right level to do work with Pade approximants

[quasi python]

I am

I'm just using it tbh

[bleak dock]

is this for some kind of internal assessment / coursework?

[quasi python]

no, I'm making a program now, and it need pade

I need to know the error to make a better code

[bleak dock]

so m = 8 and n = 8 here

[quasi python]

yes

[bleak dock]

then substitute n = 8 + 8 = 16 here, a = 0

the derivative of e^x is always e^x

[quasi python]

I see...

[bleak dock]

what's your domain for the maximum error

like are you taking x to be in [-1, 1] or something

[quasi python]

x is [0, 0.01]

[bleak dock]

okay then, so f^(n + 1) (z) would be e^0.01

so the nice thing is that given any x in [0, 0.01]

you can substitute that x to find the maximum error for that particular x-value

[quasi python]

wait let me try calculate it

[quasi python]

aight so for order (8,8) it should be this yeah?

ty ❤️

[bleak dock]

yep!!

no worries

[compact pewter]

@quasi python Has your question been resolved?

[bleak dock]

.close

[compact pewter]

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[zinc jetty]

so i hav to find the bounds for the triple integrals of the following function

z^2+x^2+y^2=9

bounded by the cone z=root(x^2+y^2)

[zinc jetty]

in this question i am more interested in finding the bounds rather than actually computing a value

because that is the part i have trouble with

<@&286206848099549185>

[eternal pulsar]

Have you considered solving the equation?

[eternal pulsar]

!15m

[compact pewter]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[zinc jetty]

yeah i already did it

interms of z

z=root(9-x^2-y^2)

[eternal pulsar]

Hmm

I wouldn't necessarily do it that way

[zinc jetty]

so whats next

[zinc jetty]

ok then how

[eternal pulsar]

z^2=x^2+y^2 from the equation of the cone

So just subsitute the x^2+y^2 for z^2 in the first equation

[zinc jetty]

ok

so x^2+y^2=4,5

[eternal pulsar]

Yes

[zinc jetty]

ok good

now what

[eternal pulsar]

Find the z-plane of intersection as well

[full drift]

hey

[eternal pulsar]

so we know that x^2+y^2=4.5, so we have a circle of radius sqrt(4.5) that is parallel to the xy-plane

[zinc jetty]

ok so the cone is the lower bound

and z=root(9-x^2-y^2) is the upperbound

right?

?

[eternal pulsar]

yeah

[zinc jetty]

ok so

like this ?

sorry for the bad handwriting

@eternal pulsar

[eternal pulsar]

Looks good

[compact pewter]

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[carmine garden]

Prove that the centre of a group, is a normal subgroup
\
We begin by verifying the centre of a group is a subgroup
\begin{enumerate}
\item 1 by definition commutes with all the elements, it thus belongs to the centre of a group
\item Let $z$ belong to the centre of the group. Let $x$ be an arbitrary element in the group. Then $zx=xz$. Pre and post multiplying both sides by $z^{-1}$, we get $xz^{-1}=z^{-1}x$. Thus the centre is closed under inverses.
\item We now verify closure. Let $x,y$ belong to the centre of the group, we then wish to show $xy$ and $yx$ do too. Let $z$ belong to the group we then have $(xy)z=x(yz)=(xz)y=z(xy)$. Thus the centre is closed under multiplication.
\end{enumerate}
The centre is thus a subgroup.\
We now prove the centre is a normal group.
\
Let the group be represented by $G$, the centre by $C$, let $q\in G, t\in C$. We now wish to prove $qtq^{-1} \in C$. Let $y\in G$. $(qtq^{-1})y= qt(q^{-1}y)$\textbf{It's here I'm stuck}

[boreal girder]

What a wonderful world !

[barren horizon]

try to work with the $(qtq^{-1})$ first

[boreal girder]

Sepdron

[kind viper]

you're screwing yourself over by not using any names for the group or its center

call the group G and its center Z(G)

[carmine garden]

I realised that towards the end of my proof, will correct that when I update my proof

[carmine garden]

hmm>

[carmine garden]

for that I'd want to move qtq^{-1} to the end won't I

[barren horizon]

yeah, you do
but there's something very nice that you can do with the qtq^{-1} here

[carmine garden]

yea,I can rewite it as qq^{-1} t = t

so ty=yt

[barren horizon]

yup, you're close now

[carmine garden]

I now premultiply the LHs by qq^{-1} and post multiply the RHs by the ame

*same

that would give $qq^{-1}ty=ytqq^{-1}$, which can be simplified to give the desired result

[boreal girder]

What a wonderful world !

[carmine garden]

$qtq^{-1}y=yqtq^{-1}$

[boreal girder]

What a wonderful world !

[wheat pasture]

You could make your life a bit easier btw, from the fact that t is in the centre, getting to qtq^{-1} being in the centre is quite “easy” from noticing one thing…

[carmine garden]

that's what I've done, right

[wheat pasture]

Not in the easiest way, I’ll say

If t is in the centre of your group, and q is a general element of the group, then how do t and q relate to each other?

[carmine garden]

they commute

[wheat pasture]

Can you write that more explicitly?

[carmine garden]

tq=qt

[azure vault]

yes

and now in a way that makes "what we want" appear?

[carmine garden]

this proves the centre is a normal group?

[azure vault]

what do we want to show

we chose some arbitrary t in the centre

[carmine garden]

yes

[azure vault]

what's our end goal here?

for that t

or some related element

if we want to show the centre is a normal group

[carmine garden]

we want to show that for every element in the group qtq^{-1} is in the group

[azure vault]

for every q in the group

qtq^-1 is in the?

[carmine garden]

conjugate

[azure vault]

uh

qtq^-1 is a conjugate of t yes by definition

but what do we want to know about that conjugate

(sorry chartbit for taking your spot xdd)

[wheat pasture]

(There are two groups under consideration: which one do you want qtq^{-1} to be in?)

[carmine garden]

the centre

[azure vault]

yes

[wheat pasture]

You’re fine I like lurking anyway

[azure vault]

to show the centre is normal, we want every conjugate of the centre's elements to remain in the centre

so

for an arbitrary t in the centre

and an arbitrary q in the big group

we want qtq^-1 to be in the centre

[azure vault]

well now

coming back to this

is there really nothing that gets us what we want?

[carmine garden]

we could post mtiply by q^{-1}

and then use the fact that t is in teh centre

*the centre

[azure vault]

so what does post multiplying by q^(-1) give us

[carmine garden]

tqq^{-1}=qtq^{-1}

now get t=qtq^{-1}?

[azure vault]

yeees

so any conjugate of t

is still t

so naturally, in which group does it lie

[carmine garden]

the centre

ooh

okay

this was fun

thanks

[azure vault]

I need your cat emojis chartbit

[wheat pasture]

Similarly here you were done too, btw, you showed the same thing at this point last time, so didn’t need anything more to do

[wheat pasture]

It’s always amusing to use the various cat thumbs up for how good something is

I’m sure everyone knows me well enough by now “wait why’s the cat sad now ”

isn’t that the point of help channels

[azure vault]

the point of help channels? mmmh

good question

[carmine garden]

I'll close the channel for now?

thanks!

.close

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[compact pewter]

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[warped oar]

,hi

[warped oar]

Sorry.

.close

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[drowsy tusk]

A rotation of k degrees clockwise about the axis (1, 1, -1) is equivalent to a rotation of k degress anticlockwise about the axis (-1, -1, 1) yeah

[dusky nymph]

<@&268886789983436800>

[compact pewter]

@drowsy tusk Has your question been resolved?

[drowsy tusk]

ok i think it's just yes\

.close

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[uncut grove]

I've written an "investigation" on trigonometric functions and I need someone to kind of proof read and see if it makes sense and is followable?

[uncut grove]

it looks something like this

i was wondering if anyone here with experience in mathematical writing could go over the thing with me for around half an hour and fix the small details

[compact pewter]

@uncut grove Has your question been resolved?

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[zenith flower]

Let K\inR^m a compact set and f:K->R a continuos function. Show that graf(f) is a set with measure 0 in R^m+1.

[kind crane]

what's graf

[zenith flower]

(x,f(x))

The set of those elementos with x in K

[kind crane]

oh yes

can you prove it for m=1

say K=[0,1] and f(x) = x

or any simple f

[zenith flower]

Previously, I already did an exercise that says that if f:A→R is integrable, then the graph of
f has measure zero.

With A a box/rectangle

[zenith flower]

So this has measure 0

The graph

[kind crane]

right. integrals of integrable functions are continuous

[zenith flower]

My idea to prove that the graph of f has measure zero is to take a box A that covers K, and define a function g: ℝᵐ → ℝ by setting g(x) = f(x) if x ∈ K, and g(x) = 0 if x ∉ K.

i wanted to prove that g is integrable, but i can't

If g is integrable then (bcs graf(f)\ingraf(g) ) then graf(f) has measure 0

[kind crane]

the proof should be identitical with this except with hypercubes in R^m+1 instead of rectangles in R^2

[zenith flower]

I proved this for a rectangle in R^m.

[kind crane]

have you learned this yet

[zenith flower]

Thats fubini right?

[kind crane]

no

[zenith flower]

No, i didnt

[kind crane]

what about continuous functions are integrable

[kind crane]

then it's just this

[zenith flower]

Yes, but only for a box/rectangle in R^m (f:A->R)

[kind crane]

R^m is just a big rectangle

[zenith flower]

Xd

[zenith flower]

What do you mean? f si defined on K, but we dont know if K is a box/rectangle in R^m

K is compact.

[kind crane]

if K is compact can it can be written as a finite union of rectangles

[zenith flower]

It can be covered by a finite union but not always can be writren

as a finite union of rectangles

[kind crane]

covered with an arbitrarily small eps difference in volume

[zenith flower]

But K doesnt need to be a measured set

If that was the case then dK has measure 0 and because D(g)\indK the g is integrable and graf(g) has measure 0

[kind crane]

what is a measured set

[zenith flower]

Sorry, i mean "measurable"

Especifically Jordan measurable set.

[kind crane]

why do you need K is measurable?

[zenith flower]

I dont know if K is measurable or not

In general K is compact

[zenith flower]

I was thinking that you were thinking that K is measurable because you mentioned volume.

[kind crane]

i see

i don't know if i unconsciously made that assumption

if i did, then you might need to prove that every compact subset of R^m is measurable

https://math.stackexchange.com/questions/2536508/proving-every-compact-set-on-mathbbr-is-measurable

[zenith flower]

Wow

I didnt know that every compact subset of R^m is measurable

[zenith flower]

So with that information this is true

graf(f) \in graf(g) then graf(f) has measure 0

[zenith flower]

Im going to think this before seeing the proof, for now the exercise is finished.

Thanks riemann for the help

.close

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[serene blade]

where would i even start?

[dusky nymph]

determine whether the expression inside each abs val is positive or negative when x is near 6

[serene blade]

how would i determine

[dusky nymph]

well if x is near 6, what does x^2 - 10x equal, approximately?

[serene blade]

36-60

-24

i think

then i dont want either to be 0 ?

like numerator or denominator?

[dusky nymph]

well we're doing the first part

rewriting without abs vals

so now you know the expression inside the abs val in the numerator is negative when x is near 6

if we call that expression A, you know that |A| = -A

(for x near 6)

now do the same for the denominator

[serene blade]

-64

[dusky nymph]

yea

so that one is negative too

[serene blade]

but that =0

[dusky nymph]

-64 - 64 is not 0

[serene blade]

but absolute value

[dusky nymph]

ah right sry

yea but you're not going to write 64 - 64

you're going to write 100 - x^2 - 64

or a simplified version of that

i.e. you're not plugging in x=6

you will first simplify the entire fraction (that's step 2)

[serene blade]

oh

so if i plug in 6 and it is negative i multiple whatever is in the absolute value by -1?

[dusky nymph]

yep that is correct

[serene blade]

ok

[dusky nymph]

that's true because if A < 0 then |A| = -A

[serene blade]

this but simplify?