help-39

[prime bramble]

then find the minimal polynomial of A

and you should be good to go

[warm elbow]

We know $Av = \lambdav \implies A^5v = \lambda^5v \implies \lambda^5 = 1$ so we define $p(\lambda) = \lambda^5 - 1$. Thus $p(A)$) = 0. p(A) is not necessarily the minimal polynomial or characteristic polynomial, but if we have some mimial polynomial $m(\lambda)$, then $m(\lambda) | p(\lambda)$?

[jolly parrot]

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

[prime bramble]

absolutely!

[warm elbow]

oki

hmm

next uh

[prime bramble]

oh wait uh

hold on

[warm elbow]

does $p(\lambda) = \lambda^5 - 1$ necessarily split into linear factors?

[jolly parrot]

toast

[prime bramble]

over C, yes

every polynomial splits over C

[warm elbow]

ok thx, and

[prime bramble]

ah wait, you said linear factors

[warm elbow]

yeah

[prime bramble]

the answer is still yes

the roots to p are the 5th roots of unity

[warm elbow]

idk

oops

$p(\lambda) = (1-\lambda_1)(1-\lambda_2)(1-\lambda_3)(1-\lambda_4)(1-\lambda_5)$ so $m(z)$ is some product subset of $p(\lambda)$ for example $(1-\lambda_3)(1-\lambda_4)$, thus $m(z)$ has distinct linear factors thus it is diagonalizable?

[jolly parrot]

toast

[prime bramble]

everything there is right except for the claim that m(z) is some strict subset

m(z) has to have the same roots as the characteristic polynomial of A

[warm elbow]

hm

[prime bramble]

a couple of the details are a bit fuzzy to me right now

[warm elbow]

hahah thats okayy

[prime bramble]

do we know the size of A as a matrix?

[warm elbow]

just nxn

wait

no we dont

[prime bramble]

is n >= anything?

[warm elbow]

but it just satisfies A^5 = I

so i guess it would imply nxn

[prime bramble]

hmm, you know what?

I think you're right

[prime bramble]

wait sorry, I meant to react to this

[warm elbow]

hm

[prime bramble]

here's what I think

[warm elbow]

here is the theroem

A matrix is diagonalizable if and only if its minimal polynomial has no repeated roots

[prime bramble]

if n >= 5, then the char poly of A, c(z), needs to have at least five roots and m(z) needs to divide it while having the same roots, right

[prime bramble]

yes!

[warm elbow]

so i dont think it has to match the charactieristc polynomial then right

[prime bramble]

then because m(z) divides our p(z), and this p(z) is a product of linear factors, p(z) must be equal to m(z)

[warm elbow]

ye

[prime bramble]

so m(z) splits into linear factors and we're done

[warm elbow]

do we need to consider the size of n?

[prime bramble]

if n < 5, then m(z) must have less than 5 roots while still dividing our p(z)

[warm elbow]

ahh

[prime bramble]

in that case, m(z) is a strict subset as you claimed earlier

[warm elbow]

so if n >= 5, m(z) = p(z), n < 5, m(z) is a product subset of p(z) bc m(z) | p(z)

[prime bramble]

and it's in particular still split into linear factors

gg

[prime bramble]

mhm!

[warm elbow]

and the reason it splits into linear factors is simply bc of a standard result with complex numbers?

[prime bramble]

the roots of z^n - 1 are the nth roots of unity, and they're all distinct so yeah

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

[warm elbow]

tysm

this helped alot lol

back to suffering doing this fuckass review sheet 💔

[prime bramble]

sorry it took me a while to get the details down

it's been a hot minute since I've had to think about min polys

good luck with your review!!

[warm elbow]

yeah no worries hahah

i have one more question i couldnt figure out actually 😭

Let $\sigma$ be the largest singular value of a square matrix $A$. Prove that if $\lambda \in \sigma(A),$ then $|\lambda | \leq \sigma$

[jolly parrot]

toast

[prime bramble]

what's \sigma(A)?

[warm elbow]

uhh

[prime bramble]

the spectrum of A?

[warm elbow]

yes

[prime bramble]

what's a singular value again?

ah, sqrts of eigenvalues of A*A?

[warm elbow]

ye

[prime bramble]

iirc, the largest singular value is the operator norm of A, yeah?

[warm elbow]

yep

im not sure 100% what is a operator norm though

[prime bramble]

in this context, it's just a norm for the matrix A

the precise norm doesn't matter idt, though the common might be the Euclidean norm

that's the one where you compute sqrt(sum of |a_ij|^2)

in any case, the basic inequality involving matrix norms is $\norm{Ax} \leq \norm{A} \norm{x}$ for $x \neq 0$

[jolly parrot]

higher!

[prime bramble]

if you have access to this inequality, the problem should be very quick

[warm elbow]

oh hm

[prime bramble]

if not, then I unfortunately don't know how to do the problem as of yet

[prime bramble]

suggestion: fix an eigenvalue lambda and leq v neq 0 be an eigenvector. then ||use the defn of an eigenvector, take norms, and use the above inequality. recall that ||A|| = sigma and conclude||

[warm elbow]

i see give me on sec

okie

so

let $\lambda \in \sigma(A)$, then by definition, $Ax = \lambda x$ and $|Ax| = |\lambda x| = |\lambda | |x|$

[jolly parrot]

toast

[warm elbow]

we also know that $|Ax| \leq |A||x|$

[jolly parrot]

toast

[warm elbow]

so then $|\lambda | |x| \leq |A| |x| \implies $|\lambda | \leq |A|$

[jolly parrot]

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

[warm elbow]

but is $|A| = \sigma?$

[jolly parrot]

toast

[prime bramble]

yes, by definition

er, not by definition actually

it's a fact that I've forgotten how to prove

[warm elbow]

😭 hahaahaha

[prime bramble]

it seems like you're following Axler

I have no idea if he ever talks about the operator norm

if he doesn't, then this is an invalid proof

[warm elbow]

my teacher makes is own set of notes

loll

[prime bramble]

does your teacher talk about norms of matricies/lin maps?

[warm elbow]

he talks about the modulus of a matrix

[prime bramble]

hm, that seems to be a slightly different concept

[prime bramble]

mhm, that's not quite the same

[warm elbow]

well hm

htis question is in chapter 25 of his notes

lecture 25*

[prime bramble]

there's probably an intended proof

that doesn't rely on op norms

let me think

[warm elbow]

but idrk i never seem him define operator norms everywhere

[prime bramble]

mhm, it's probably not here then

[warm elbow]

maybe we can perhaps just use SVD?

[prime bramble]

I'm genuinely not sure how to do this problem then, I apologize

[prime bramble]

maybe? I'm not sure where you'd go with that approach, but I guess it can't hurt to try smth

[warm elbow]

no worries ahahaha

i wish he put solutions to thes eproblems

but i think he makes his finals directly out of these questions

with maybe slight modifications

[prime bramble]

icic

let me know if you find a solution to this problem

I'm curious to know how it's done now, but too tired to continue thinking about it

[warm elbow]

ahahaha no worries

[pearl pond]

@warm elbow Has your question been resolved?

[warm elbow]

actually @prime bramble

could we just do soething liek this

$|Ax| = |\lambda| |x| \leq \sigma |x|$

[jolly parrot]

toast

[warm elbow]

bc like

$\lambda \in \sigma(A)$ and $\sigma \in \sigma(A)$ and $|\lambda| \leq \sigma$

[jolly parrot]

toast

[warm elbow]

wait hmmm

we cant bc of

[prime bramble]

aren't we trying to prove that |lambda| <= sigma?

you can't just use it in the proof

[warm elbow]

yeah

u cant

nvm

[waxen agate]

Over C this statement is true for any n

[waxen agate]

Ah, it’s a different problem. I see.

[violet marlin]

hello

[midnight haven]

Hi

[warm elbow]

hi

[worthy quarry]

hiiii

[pearl pond]

@warm elbow Has your question been resolved?

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[thin eagle]

So for some reason the pattern I get is 0,1,2,3,5

[midnight haven]

Damn hate this question frr

Don't all the dots need to come down tho?

[naive zinc]

I assume it can be viewed as finding maximal overlap. Like triangle A and its reverse B

If you manage to move A and B such that their intersection C has the most many dots, then |A|-|C| will be the answer?

[midnight haven]

you mean B and C?

[naive zinc]

And I also guess there might be two sets of answers, one for odd, one for even

[midnight haven]

Then yea, that could be one way to go about it

[midnight haven]

Ye, thats what I was thinking

[thin eagle]

So you're suggesting I go further?

[naive zinc]

Yeah. My hypothesis is the patten might be recursive, P(n) has something to do with P(n-2)

[thin eagle]

I see, I'll try that.

[naive zinc]

Let’s explore few more terms together. And verify them first

Mine: p_6=||7||, p_7=||9||

[crystal dew]

don't troll in help channels please.

go to #chill for that kind of stuff.

[thin eagle]

Yea, got the same

[naive zinc]

My hypothesis is p_2n=n(n-1)+1, and I am coming up with a hypothesis for odd too

[toxic lichen]

<@&268886789983436800> troll

[naive zinc]

Or maybe the answer is mod 3 recursive. Maybe it’s time to consider whether there is any recurrence

Like I also believe p_(3n-2)=(3/2)n(n-1). so not sure I should proceed mod 2 or mod 3

[pearl pond]

@thin eagle Has your question been resolved?

[thin eagle]

I see, unfortunately I'll get busy with something else so I'll close the channel for now.

[pearl pond]

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[pearl pond]

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[naive zinc]

To construct pull back of f:A->C and g:B->C Consider how you can construct morphisms A prod B->C

[void grail]

🙌

[naive zinc]

What does you reply mean

Anyway, there are two morphisms A prod B -> A, A prod B ->B, use them, what are the two morphisms A prod B->C enter your mind?

[void grail]

I remember this diagram

looks like pullback

[naive zinc]

You are overthinking

[naive zinc]

^

[void grail]

the two projections π1 π2

[naive zinc]

To C

[void grail]

oh those are from c

[naive zinc]

Those are from A prod B to A and to B respectively, i need two from A prod B to C

[naive zinc]

You have f and g here

Really? Can’t see any?

[void grail]

A×B---pₐ-->A
|          |
pᵦ         f
|          |
v          v
B-----g--->C

f and g

<f,g>

[naive zinc]

No

[void grail]

Oh you mean to C--->(A×B) from A and B?

[naive zinc]

No

[void grail]

that's like pulling back along the projections

P--->C--->(A×B)

[naive zinc]

I can’t continue, f•p_a and g•p_b. Now prove by definition the equalizer of these two is a pull back of f and g

I don’t understand why you shift your attention and focus on something else, so I had to tell you what those two morphisms are

( any p: D->A, q: D->B such that fp=gq, <p,q> factors uniquely through this equalizer)

Have to go running

[void grail]

Thanks for your help @naive zinc

[void grail]

f:A→C, g:B→C

p_A,p_B: A×B→A,B

eq:
E→A×B
equalizes f∘p_A, g∘p_B

∀p:D→A, q:D→B,
f∘p = g∘q
→ ⟨p,q⟩:D→A×B
factors uniquely through eq

∴ E is pullback of f,g

[naive zinc]

This is what I said, which is a sketch not the whole proof

You need to prove it, in full

I have a 15 min walk to bus station

[pearl pond]

@void grail Has your question been resolved?

[naive zinc]

Arrived. e:E->A prod B being the equalizer, <p,q>=er for some r: D->E, as you desired. Showing uniqueness I left to your own
This has a more general result:
small (co)products + (co)equalizers -> small (co)limits. Left to your own, similar idea

[pearl pond]

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[teal rose]

Do i just apply mod term formula here

[teal rose]

Mid term

[eager jewel]

whats mid term formula

[tulip ore]

add a new name to the diamond list

[eager jewel]

?

[tulip ore]

is the midterm formula the one with the X?

[teal rose]

Like breaking the middle term

Not a formula

Like

[eager jewel]

yes\

[teal rose]

Turn 9x into 5x x 4x

[tulip ore]

5x + 4x instead

[teal rose]

Alr

[eager jewel]

yes

[tulip ore]

later on youll learn a method that goes by a lot of names

that method has you draw a big X

you put 9 at the top and 20 on the bottom

then on the left and right is where youd think and then figure out 5 and 4

[teal rose]

Wb thus

[tulip ore]

for this one, remember that $a^2b^2=(ab)^2$

[jolly parrot]

mtt

[eager jewel]

ab=x

[tulip ore]

yoda is then saying you can replace ab with x to make it easier to process

since all youre seeing is just ab

see if you can recognize or factor x^2 - 10x + 25

[teal rose]

Bro

I dont get itlile how to factorise it

[tulip ore]

heres something you can look for

you know the (x + y)^2 = x^2 + 2xy + y^2 thing?

[teal rose]

Yes

[tulip ore]

that took a while but now that its in your head,

Im gonna replace y with the the letter c

and you read it as standing in place for a number

(x + c)^2 = x^2 + 2cx + c^2

and you have x^2 + -10x + 25

[teal rose]

Oh

[tulip ore]

lol

[teal rose]

Then js use midterm

[tulip ore]

oh right that works too

mb

[tulip ore]

this is a special formula if the terms you split apart into are the same

lot more symmetry so its easier to recognize the pattern instead of noticing you have to split it in half

this is useful for example for x^2 + x + 1/4

thats (x + 1/2)^2 which is hard to see

[worthy urchin]

hey im a high school student very interested in maths and new to this server

[teal rose]

.

[compact ridge]

hi there, go chat in #discussion

[worthy urchin]

can i tell smthng i have completed grade 12th maths and also learned calculus

thats why im here

[verbal whale]

!occupied

[pearl pond]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[worthy urchin]

im in grade 9 and imbored so imhere to help others

where should i go to help others wiuth hard problems

can someone pls tell

[verbal whale]

CAN YOU READ THIS, PLEASE? @worthy urchin

[worthy urchin]

sorry

[grizzled ermine]

wlcm then

[tulip ore]

ok this one took me a while

idk how theyre showing you this if you dont even know the X method (otherwise known as the diamond method) yet

for this one, go look at xy - 3x - 3y

do you think you can factor xy - 3x - 3y or nah

[compact ridge]

if you write it as (x^2 + xy) - (3y + 3x) you should be able to continue factoring this

[tulip ore]

yea thats easier than what I was gonna go with

[pearl pond]

@teal rose Has your question been resolved?

[pearl pond]

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[fathom sleet]

I Wrote my answer to this Question as 3,8

Am I correct?

Its making an estimate

Can someone tell me if I got this correct

[compact ridge]

you need to be more precise than that

say, the biggest root looks to be around halfway between 9 and 10

[fathom sleet]

Would I get the marks for 3,8?

[compact ridge]

no

[fathom sleet]

Its 2 marks

[compact ridge]

you'd get 0/2

[pearl pond]

@fathom sleet Has your question been resolved?

[pearl pond]

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[bold lynx]

does anyone know how tho solve this? I dont know know any formula that can help

[autumn fossil]

maybe this could help

[toxic lichen]

i was about to suggest that you mark the centers of each circle and also connect them all

[bold lynx]

how can that help ?

[spiral coyote]

Area of shaded = area of triangle - area of 3 sectors

[bold lynx]

area of sector is about 32π/3 I think. the area of triangle is just √3/4*a squared right?

[light helm]

area of sector is about 32π/3 I think
area of each sector would be exactly that

[autumn fossil]

are you using calculator mid-problem again?

[bold lynx]

nah

thats why i got pie

[autumn fossil]

yes, and a is what?

[bold lynx]

lenght but i am kinda stuck tere currently watching a youtube totorial on how to solve

[autumn fossil]

think about how the length could relate to radius

radius is distance from center to any point on the circle

*circle, not triangle, sorry

[bold lynx]

yeah but the height is still gona be longer tho

[autumn fossil]

what height? You dont need height, you only need side of the triangle

try splitting the sides in the middle

[bold lynx]

wait isnt it heigh 8 base?

[autumn fossil]

height * base / 2? Yes, thats one formula for triangles area

[autumn fossil]

area of triangle is just √3/4*a squared
This formula works fine for equilateral triangle tho

so why not use it

[bold lynx]

i was doing that but got no iea what to do with the root

i never did that type until now

[light helm]

sqrt(3) = sqrt(3) is just a number
leave it as such

work with it same way as you would pi

keep values exact

[bold lynx]

the angle is just 360 divide 60

wait

root 3 is just 1.3

81.73

*1.73

[light helm]

are you using calculator mid-problem again?

[autumn fossil]

root3 is root3

[light helm]

leave sqrt(3) as sqrt(3)
sqrt(3) is irrational and can't be represented with a finite number of digits

[autumn fossil]

u dont need to plug decimal for everything right away

[bold lynx]

ok

[light helm]

and the question implies that you want your end result to have a radical and pi

[bold lynx]

so just leave the area at root 3 * 16?

[light helm]

that's not the correct area,
be careful of the value you use for a

[bold lynx]

value of a=8

[light helm]

no

8 is the radius of the circles
however the side length of the triangle isn't just the radius

as indicated by the pic above

[bold lynx]

wait so its lenght is 16?

ok so root 3 * 64 correct ?\

[bold lynx]

I am at the last part wheree its just sqrt 3 * 64-32=a sqrt b-c. I am stuckk how to remove the roots

[pearl pond]

@bold lynx Has your question been resolved?

[bold lynx]

do you know how to do that part

[spiral coyote]

32pi?

[spiral coyote]

Uve to compare

[bold lynx]

nah i removed pi from both sides

[spiral coyote]

U cant do thtTT

[bold lynx]

you cant?

[spiral coyote]

64 rt3 - 32 pi = a rtb - c pi

[spiral coyote]

How would u do tht?

[bold lynx]

idk divide both sides by pi

[spiral coyote]

Here we compare. LHS the coefficient of pi is 32 and RHS it is c so c=32

[spiral coyote]

Tht would give 64rt3/pi -32 = a rtb/pi -c

[bold lynx]

ohh i see

[spiral coyote]

Now whay can u say for a and b?

[pearl pond]

@bold lynx Has your question been resolved?

[pearl pond]

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[teal rose]

Can someone check and lmk if this is correct

[viscid shale]

,rccw

[viscid shale]

you missed half the problem in i)

You only calculated g(1)

oh, i just saw the right part, i guess thats a -3

[teal rose]

Oh yeah i juat realised

[viscid shale]

in ii) you plugged f(x) wrong, its 5x + 7, not 5x - 7

[teal rose]

Idk how to solve the last part

[viscid shale]

for iii) you can simplify it further too

and iv) is actually just a theory question

you dont have to do anything fancy

[lilac ocean]

it is not really a simplification

[viscid shale]

Do you recall the definition of an inverse function?

[teal rose]

Could you solve it? Im stuck

[lilac ocean]

what is the inverse of a function?

conceptually speaking

[teal rose]

Not sure

Like function moved to the other side

[viscid shale]

Thats how you algebraically find them

i want to believe that at some point you saw the definition of a function as something like this:

$f: A \to B$

[viscid shale]

Where A is the domain, and B is the codomain

And from here, you also saw what injective and surjective functions are

and why bijection is important

[stark sable]

\log_2(x - 1) + \log_2(x + 3) = \log_2(4x)

Can someone help me please

@viscid shale can u

[viscid shale]

!occupied

[pearl pond]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[viscid shale]

@teal rose

[lilac ocean]

actually, 0580 is igcse and they don't cover that

it's okay thoguh

[teal rose]

Yes

[lilac ocean]

right
you rearranged to get it in terms of this other value

so what would happen if you rearranged it for the other value again

[teal rose]

Can someone solve it and send me the working if possible?

[lilac ocean]

it's ok lets go through this

[viscid shale]

!nosols

[pearl pond]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[teal rose]

Ik how to do f inverse (x)

[viscid shale]

actually wrong factoid

[lilac ocean]

forget finding the inverse for now

[lilac ocean]

focus on this

[teal rose]

Uh i dont really get what u mean tbh

[lilac ocean]

okay, let me reword it

to find the inverse of y = f(x), you essentially set x = f(y) and rearranged back for y, right

what if you wanted to find the inverse of the inverse

what would happen

[teal rose]

Inverse of the inverse?

[lilac ocean]

yes

[teal rose]

Huh

[lilac ocean]

what would happen

[teal rose]

Is there a board where u can show me or smth

I rem this server had some white board u can plug fractions into

[lilac ocean]

hmmm

[viscid shale]

Ill give a really basic problem related to this idea.

Try to find the inverse of f(x) = 2x + 1

[teal rose]

Y : 2x+1
Y-1 divided by 2 = x

[viscid shale]

So whats the resulting function in terms of x?

[teal rose]

Y-1/2

[viscid shale]

1. parenthesis are important
2. thats in terms of y.

remember you "switch around" x and y at the end

[teal rose]

Whats a parenthesis-

[lilac ocean]

brackets

[viscid shale]

( )

[teal rose]

Yeah

Oh

Idk bro im stuck

X-1/2

So f(2) -1/2

[lilac ocean]

(x-1)/2 is good

[viscid shale]

We will call this new function g(x) now.

now, well do the following

[viscid shale]

Try to find both fg(x) and gf(x)

if you get the correct result, both should look the same, and also show a really important fact about inverse functions.

[teal rose]

Alr

Lemme do

[teal rose]

Can u check if this is accurate

[viscid shale]

youre missing the point in this, you dont even have to do any math to solve the problem in your paper.

[teal rose]

Is what i dis there accurate though?

[viscid shale]

Yes

[teal rose]

Also do i need to solve it further

F(2)

In which i add the values and further solve it?

[viscid shale]

Youre doing work just for nothing, but yeah, ig.

[pearl pond]

@teal rose Has your question been resolved?

[pearl pond]

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[sharp fractal]

seeing by your previous channel, have you been attempting your questions before posting? if not, can you please try to do so?

[static fable]

idk how to attempt this

i did with the other

[sharp fractal]

if you did, please go ahead and say what you tried and/or why you got stuck

[sharp fractal]

right, so what do you know about theta/2 ?

[static fable]

not alot

oh do you mean that it equals - 1/root3

[sharp fractal]

yeah, what about it? particularly its sign?

[static fable]

its negative

[sharp fractal]

and since its tan, what does that tell you about its quadrant placement?
also, isn't 1/root 3 familiar? do you have something you can reference it to?

[static fable]

its in the 2nd or 4th quadrant

idk what the 1/root3 means

[warped violet]

Just a sidenote, theta/2 does not equal -1/sqrt(3)

tan of theta/2 does.

[static fable]

tantheta

Ya

[sharp fractal]

yeah my bad, I meant to add tan but it skipped my mind lol

[warped violet]

No need to apologize. We all make mistakes 🙂

[static fable]

So

i dont know what the 1/root3 means

but i know that its in the 2nd or 4th quadrant

[sharp fractal]

do you know how to find a reference angle, especially now since you know what quadrant(s) youre working with?

[static fable]

im familiar with it

but i dont think im confident

in this case

and am i not meant to take hte absolute value menaing that it would be in the 3rd or 1st quadrant

or is that something else

[sharp fractal]

oh, I see where you're confused, just to remind you then, you should know that tan(30) or tan(pi/6) = 1/sqrt of 3. this is what I was hinting at earlier, but you might've forgot. can you move forward knowing this?

[static fable]

i wouldnt have known that

[sharp fractal]

it was part of foundational trig if you aren't familiar with that, if you'd like I can explain it to you pretty quick, although they are just values to remember/memorise

[static fable]

its ok

[sharp fractal]

^ I know you are concerned about the 1st and 3rd quadrant, but lets focus on the 2nd and 4th now, because it has already defined to you that tan is negative and where its two solutions lie

[static fable]

ok

[sharp fractal]

wanna go ahead and try and tell me where you get?

feel free to post pictures of your work, it's easier for me that way too

[static fable]

ok

im porety sure im not doing this right

[sharp fractal]

so you're on the right path, but heres the first thing - your question was given is radians, not degrees. and so was your range. I get why you used arctan here, but its application is redunant, as we have already defined what tan(1/sqrt3) is (pi/6). wanna give it another go with radians?

[static fable]

i donty really get what u mean by tan(1/squrt3) is pi/6

how does that help

[sharp fractal]

unfortunately the error from your previous work compounded here, giving you an incorrect final answer, but you are headed in the right path at the end (angle doubling, quadrants, etc)
you're really close, that arctan bit and turning it into degrees is what got these errors, so just fix 'em and we'll be back on track

(and PS, 360-18 isn't 362, keep an eye out for that)

[static fable]

lol bro i thought i did 380-362

tiredness getting to me

[sharp fractal]

are you familiar with this chart? (knowing that 30 degrees is equal to pi/6)

[static fable]

or whatever

Idek

[static fable]

yes

[sharp fractal]

is this a time sensitive problem? if not, you are free to go rest and come back another time. math is hard to grasp when you're tired

[static fable]

not off the top of my head but we have a formula table for it

[static fable]

na i just need to get study done exam tmrw

its only 8pm im good

[static fable]

i dont understand how this helps

[sharp fractal]

its the key to understanding the problem, its the reference angle we were talking about at first

[static fable]

oh so

theta/2 = 30degrees

[sharp fractal]

its the reference angle, not the solution, and also keep in mind we're solving in radians please

[static fable]

can i not convert it at the end

into radians

[sharp fractal]

its better to do all your work in radians especially if you're doing it in reference to the unit circle as not to cause confusion or calculation errors

do you have any issue with radians? if you'd like, I can clear them up for you

(if anything, doing it all in radians makes this way easier to solve, btw)

[static fable]

i prefer to do it at the end

ill try work in radians tho

ok so 30 degrees is the reference angle

[sharp fractal]

you know what, sure, lets solve it in degrees, but lets be extra careful on that convertion at the end

yeah perfect

[static fable]

ok

and since its in 2nd quadrant

i minus 30 from 180

and 360

[sharp fractal]

yes exactly 😭 you did it

give me those two numbers and we can move on to the last 2 steps

[static fable]

150 and 330

ok

5pi /3 +4pi
11pi /3 +4pi

theta = that

[sharp fractal]

why'd you add the 4pi?

[static fable]

because u add 360degrees n

for the general soluti

no

?

or because its not asking for the general solution i just leave it as 5pi/3 and 11pi/3

[sharp fractal]

the question said "within the range" you dont have to add it

yeah exactly

its just that

[static fable]

ok thx

[sharp fractal]

I hope you got everything, I'm sorry if I confused you at all. if you need any part cleared up better for you just ask

[static fable]

no i understand it

its ok

[sharp fractal]

and if you're done go ahead and close the channel :)

[static fable]

i think i prob need help

with another one

[sharp fractal]

go ahead, I might need to go soon though

[static fable]

i think its something to do with area of a sector formula

[sharp fractal]

yes, that'd be correct

show me anything youve attempted if youd like

[static fable]

all i have is 21 = 1/2 7^2 theta

idk what to do from here

[sharp fractal]

21cm is not the area of the full sector though, hence why this doesnt work

[static fable]

oh

idek then

[sharp fractal]

and by the way, we know theta, as it is (I think?) 1.2rad

[static fable]

oh

ok

idk what to do

[sharp fractal]

its okay, since you know the formula, we can just do this together,
area = 1/2 r^2 theta

so we just plug it in

1/2 x (7)^2 x 1.2

can you give me the value?

[static fable]

29.4

[sharp fractal]

yes great, so now we know the area of the entire sector AOB, we also know the shaded area as 21, can you find the area of the unshaded triangle?

[static fable]

minus

29.4-21

8.4cmsquared

[sharp fractal]

okay yay! now we have a triangle, wanna have a go at trying to find |OC|?

[static fable]

uh

do i use 1/2 ah

[sharp fractal]

yes

[static fable]

idk what a or h is tho

[sharp fractal]

what formula are you trying to use?

oh

my bad, lol

[static fable]

half the base times the height

[sharp fractal]

we cant use this, because this is not a right angle triangle

theres also no way to figure out the h anyway

[static fable]

1/2 ab sinc

i htink

[sharp fractal]

yes!

where |OA| is a, which we know is 7,

|OC| is what we're getting at,

and C is the angle AOC or 1.2 rad

(if you're using a calculator, make sure its in radian)

[static fable]

ye

|OC| i got this to be 5.137

ans 1.9

cm

hopefully

wiat

noi

idek

[sharp fractal]

can you show me your work? I wanna see what went wrong

if you can just show me what you plugged into the calculator

[static fable]

(1/2)(7)(sin(1.2))

i got 3.2621

then i minused that from 8.4

and got 5.1

[sharp fractal]

ah I see where the problem is, this is correct, but
minusing it from 8.4 is the issue
you need to divide them instead, not subtract

[static fable]

ohhhhhhhhhhhhh

8.4 divided by 3.2621

i got 2.5758

minus from 7

4.4

[sharp fractal]

yeah great job :) thats all there is to it

dont forget ur unit lol

cm

I gtg now I'm sorry, hope I was of help to you, have a good rest of the night

[static fable]

Thank you Bubsy u too

[pearl pond]

@static fable Has your question been resolved?

[pearl pond]

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[eager jewel]

i dont understand what they gave the function for

because they are not using it in the limit asked

also how to find the limit? i thought write arctan(1/a) as 1/a

[lone mesa]

tried expanding tan^(-1) = 1/a ?

[eager jewel]

i thought of that but i forgot arc tan expansion 😭

also i dont think it will help

[tulip ore]

theres a typo in this question, wheres the f in the limit?

[eager jewel]

thats what i said

but is it a typo?

[tulip ore]

that limit -> infinity

[eager jewel]

i also searched it up and got the same question

[tulip ore]

they must have scraped the question

that must be why these all look identical

make it digital without looking twice

so a typo in the question -> typo in the digital version

[eager jewel]

oh ok then

ill just leave it?

[tulip ore]

ngl what would happen if you replace the arctan with an f though

nvm that doesnt work

[eager jewel]

i still wouldnt know how to solve it 💀

[tulip ore]

you need an arctan to get the pi/whatever

and even then the limit would still -> infinity

[eager jewel]

yea

[tulip ore]

lol that was quick, now you can

[eager jewel]

alr then i wont waste more time

[tulip ore]

you know it -> infinity now

[eager jewel]

wait what

[tulip ore]

look at what you said

this means you agree that the limit would be -> infinity if we replace the arctan with f

[eager jewel]

oh i was saying yea to this

[tulip ore]

I dont see a way to fix this question

youll notice they add a lot

then only subtract 2 ln a

with the + a^2 there there isnt really a way to get this to -> a finite number

either way its like degree 2 vs degree 0

[eager jewel]

oh ok

yea then ill leave this q then

thank u

.close

[tulip ore]

np

[pearl pond]

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[pearl pond]

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[sinful tusk]

I have the answer key to this but I watched videos and everything i still dont understnad.

if 7 = 40, then 1 + 7 = 90deg. if 9 is 100. then its 190 and thats not 180 so im not sure.

I originally thought it was 50 for 7 instead of 40. and 2, I dont know how to find or how they found that.

[warped violet]

why are we labeling angles with numbers 🙁

countless other options

[sinful tusk]

im not sure its the district they kinda suck

[glass meadow]

Is PRST given to be a rectangle?

[sinful tusk]

yes

[glass meadow]

Then you're right, angle 7 is 50º

[nimble lily]

9 cannot be 100

just look at it

[glass meadow]

(if those are degrees)

[sinful tusk]

so then 9 should be 90 deg?

[glass meadow]

No

[sinful tusk]

how would i get 9?

[glass meadow]

180º - angle 1 - angle 7

[sinful tusk]

80º

[glass meadow]

Yes

I'm pretty sure they just mislabeled everything

[warped violet]

That would imply PRST is a square

[glass meadow]

If your answer key is saying angle 7 is 40º and angle 2 is 50º, they might be swapped in the picture

[warped violet]

If I were you, I'd first re-label the 9 angles labeled 1 through 9 to nine of the 24 Greek letters.

And work from there

In a subject where clarity and understanding are supposed to be profound, this seems rather counterproductive.

[sinful tusk]

Yea, thats what i was kinda thinking too since it mixes up sometimes as well. it seems like it did swap it. i got angle 7 as 50, angle 9 as 50, and angle 2 as 40. the picture is pretty confusing to me

[warped violet]

Yes exactly

[warped violet]

And then stuff like 7 = 40 follows

I know it's 'angle' 7, but nobody is going to be writing the symbol 500 times here

And then we get 1 + 7 = 90°

Which not only is already confusing right now, think about a month from now when you get back to it while studying

[sinful tusk]

yea, id rather label them with letters

[warped violet]

Simply defining $\alpha = \angle 1$ and $\eta= \angle 7$ makes reading $$\alpha + \eta = 90^{\circ}$$ much more sensible

[jolly parrot]

USS-Enterprise

[warped violet]

So, yeah I suggest doing that

Redrawing the diagram

And then trying again

[pearl pond]

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[pine jacinth]

A small archipelago is located at the junction of two moving tectonic plates.

The tectonic plate located to the southwest of the archipelago and represented by vector u
moves at a speed of 2.17 cm per year in the direction shown.
The tectonic plate located to the southeast of the archipelago and represented by v
moves at a speed of 2.23 cm per year.

What are the magnitude and direction of the annual displacement of the archipelago if it
corresponds to the sum of the displacements caused by the tectonic plates?

[pine jacinth]

how do i find the orientation of the vector u and v?

how do i do that without an orientation

orientation is counter clock-wise no?

alrigt

we only want the horizontal components?

alright

wait v doesnt have an angle

39

not sure how i can use it

if it said like it was perfectly between 2 vectors not one being more obtuse then i would do 180-80

100/2

but it says nothing about it

ohh

that doesnt help though

40

how?

oh

what about the other side

with the blue vector

the 41?

like this

not sure if i understand but thats not the angle for the blue or am i missing something

oh wait

you moved blue vector

?

oh ok

yup

(1,67;1,46)

yeah

found it

the norm is 4,38 and the orientation is 40,09

yeah

its that

i double checked

nope

why

oh cause its going to the left

but when i did the thing to find the x component

it gave me a positive number

gotcha

so its 0

wait

let me redo

yeah same

(0,0034;2,82)

what how

wait how can it be -2,23 tho

a vector cant be negatif

ive never done a number like this

like this problem

its very different

i see

i got (3,36;-0,1)

alright

norm is 3,36

what how

how can the x be at the y spot

oh yeah

because vertical is x

this number is filled with traps

i meant horizontal mb

the orientation os

-33,6

but wait

u have to do +90

i got 1,54 after adding 90

what idd u get

why subtract by 180

but u subtract when its in the 3rd quadrant by 180

2nd quadrant by 90

or add depends o nthe context

yeah

i see now

alright tysm for helping

have a good day

.close

[pearl pond]

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[signal sparrow]

use generating functions if possible

i just dk how to deal with the modulus and complex values i can solve if its real numbers

[somber adder]

!show

[pearl pond]

Show your work, and if possible, explain where you are stuck.

@signal sparrow Has your question been resolved?

[waxen agate]

Make the unit-modulus terms point almost in the same direction so their sum is large, while the smaller-modulus terms die out.

Actually, if number theory approach is valid, Kronecker’s theorem makes this rather trivial.

[pearl pond]

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[digital summit]

Hey, I'm trying to graph a line of best fit and I get an equation around y=1/3x + 58 but it doesn't work when you plug in the slope in c.
Am I messing up somewhere?
When I try to check with AI, not the best solution, but when I do, it says things like y=7x+40, but that's not even near the line but it checks out for c. I don't get it, doesn't the line have to have most of the points equal on each side?

[digital summit]

Also, another question I posted a little earlier;

[shut elm]

When you plugged in for part c did you make sure you plugged the value in for y and not x?

[untold otter]

pretty sure its b

[digital summit]

yea

90=1/3x+58
32=1/3x
96=x
it wont take 96 hours to get 90%....

[digital summit]

I meant for the equation, and anyway if you did graph that it would just go up and away from the points

[shut elm]

Yeah looking at it 1/3 appears to be a little low for the slope but let me run it with my calculator and see what I get for the line of best fit

I get approximately y=6.3253x+57.37308 as my line of best fit

[digital summit]

how? like 6/1x?

with the decimals

[digital summit]

i think i get it

[shut elm]

Kind of similar to the AI now that I look but with a closer intercept to yours

[digital summit]

i need a calculator lol

should you not use fractions when finding lines of best fit>

[shut elm]

Yeah but using a calculator I can only put in decimals

It's approximately 70/11

Although that number doesn't mean much

[shut elm]

Using the decimal tells us for every extra hour of study the grade increases by about 6, like you wrote here

[digital summit]

alright

whats the math behind the calculation? did you just pick two points and do y2-y1/x2-x1?

[shut elm]

I used the linear regression tool in my calculator but that would be how to do it by hand (without the tedious calculations)

Draw a line yourself that looks like it fits the data, and then choose two points on it, and do y2-y1/x2-x1

[digital summit]

what is linear regression xd

okay

[shut elm]

Fancy word for line of best fit

[digital summit]

ohhhh

okay thank you! :)

[digital summit]

Also do you have any suggestions for learning Geometry?

[shut elm]

No problem! I'll just add in an example with my line so you can see

[digital summit]

alright

[shut elm]

It kind of looks like it passes through these two points

So the slope is about 25/4 which is 6.25

Very close to the actual answer

Which is how I'd eyeball it on paper

[digital summit]

cool

thats helpful! :)

[shut elm]

Mmmm personally not sure since I never took it 😅 I would have just used Khan Academy since they do technically have written guides for each topic

And then anything I didn't understand based on that I'd look more into

That's what I do with my SAT students

Have them read it, do the practice, and then bring any questions to me

Since I consider application the most important

[digital summit]

that could work

are the guides like extensive or explanatory

[shut elm]

If you really want to learn about the meaning though a textbook is a good way to go, although I can't direct you to one

[shut elm]

Explanatory yes (on how to do the operations), not necessarily extensive but the videos should help with that

[digital summit]

yeah

If I do want to get into a higher class next year, I would have to test out of Geometry which would probably require a good understanding or better

can't hurt to always learn mroe

more* lol

[shut elm]

Very true 😆 well except for time

[digital summit]

true

I only got like 6 months if I want to do this lol

[shut elm]

I'm sure someone on here knows of a good way, hopefully they'll find your question

[digital summit]

Well see! :D

[digital summit]

Thank you for your help by the way!

[shut elm]

No problem! Good luck!

!done

[pearl pond]

If you are done with this channel, please mark your problem as solved by typing .close

[digital summit]

.close

[pearl pond]

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[sharp smelt]

Question (c)

[sharp smelt]

Now here's my issue

if this were of the form (1-t) I'm done

which can be obtained by a simple substution, I'm aware.

But is there a way around that

[naive zinc]

Consider Σ(-x)^k for k>=0. It will be a well defined sum since it will have finitely many terms

[sharp smelt]

ah

got it

thanks

Can I close this now

[naive zinc]

Sure

[sharp smelt]

tq

.close

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[hot bramble]

Let $(A,\tau_A)$ and $(B,\tau_B)$ be the subspace topologies, then we get the subspace topology of $C=A\cup B$ is $\tau_C={(A\cup B)\cap O=(A\cap O)\cup (B\cap O)|O\in \tau}=\tau_A\cup\tau_B$. So, $\tau_A,\tau_B\subseteq\tau_C$, so open sets in $\tau_A$ and $\tau_B$ are also open in $\tau_C$. To show that $\tau_C$ is connected, I suppose $U\in\tau_C$ is a non-empty proper clopen set of $A\cup B$ such that $(A\cup B)\backslash U=(A\backslash U)\cup(B\backslash U)$ is open. From here I don't really know what to do. I attempted to let $U=O_A\cup O_B$ for open sets $O_A\in\tau_A$ and $O_B\in\tau_B$, but expanding $(A\cup B)\backslash(O_A\cup O_B)$ became unwieldy very quickly.

[jolly parrot]

King of Ponies

[naive zinc]

$A \cup B=U \coprod V$ then consider $A=(A \cap U) \coprod (A \cap V)$ and $B=(B \cap U) \coprod (B \cap V)$

[jolly parrot]

Cogwheels of the mind

[naive zinc]

U, V open subsets of A cup B and \coprod means disjoint union

(Among four cases only two will be possible)

[hot bramble]

I don't think I quite understand, I haven't worked with the disjoint union before.

[naive zinc]

A space X having a clopen subset Y is equivalent to X being disjoint union of two open subsets Y and X-Y

(And X1 subset of X2 subset of X3, the induced topology on X1 from X3 is the same as (the induced topology on X1 from (the induced topology on X2 from X3)))

[naive zinc]

Like for example, A in U and B in V is impossible, A,B both in U gives you A \cup B in U

[hot bramble]

Now I am really confused. So if I understand, a clopen set Y of a topological space X defines an equivalence relation by X disjoint union X-Y = {(x,0) u (y,1) | x in X and y in X-Y}?

[naive zinc]

I said nothing like that

I said literally what I said

Y being a subset of a space X is equivalent to X being disjoint union of Y and X-Y
Then Y being clopen is equivalent to Y, X-Y being open
Then Y being proper is equivalent to Y, X-Y being proper

I said nothing about equivalence relation

[hot bramble]

Sorry, I over inferred what you were saying.

[hot bramble]

I think I slowly understanding, so you chose two opens sets that cover the union AuB and then defined A and B in terms of how they are covered by these two sets. But what are the cases you are talking about? Do you mean if either one of U or V are completely contained in either A or B, or if they overlap at some point in A or B? But then why is the disjoint union necessary here?

[naive zinc]

$A=(A \cap U) \coprod (A \cap V)$, then since A is connected, $A= A \cap U$ or $A=A \cap V$

[jolly parrot]

Cogwheels of the mind

[naive zinc]

(M=M cap N is equivalent to M being contained in N)

[pearl pond]

@hot bramble Has your question been resolved?

[hot bramble]

Oh, I think I see now. Since we choose $U\cup V=A\cup B$, then intersect this with $A$ gives $(A\cap U)\cup(A\cap V)=A$. Then since $A$ is connected, then $U$ and $V$ must overlap inside else $A\cap U$ and $A\cap V$ are clopen in $A$.

[jolly parrot]

King of Ponies

[naive zinc]

U and V being disjoint is given in the beginning

The conclusion is not what you wrote, it is A cap U=A or A cap V=A

[naive zinc]

Literally what I said

I don’t have more time on this, have to go running

[hot bramble]

Ah, thank you for trying, and apologies for not getting it.

[naive zinc]

Your issue is you shift attention from what I literally said to something unrelated

[hot bramble]

I think a part of the reason is why is that I am not fully understanding the things you stated. Like you defined A as the disjoint union of intersections, and I am still trying to figure out how this disjoint unions works to define A compared to just using the regular set operations.

So I am trying to rephrase what you stated in terms of the regular operations I understand, and I think my rephrasings are then either incorrect or besides what you initially stated.

[naive zinc]

I didn’t define anything. a space X being connected if and only if for any open U, V such that X is the disjoint union of U,V, we have X=U or X=V. I was proving X satisfies this, when X is your A cup B

Anyway, really got to go. Said all you need, everything at hand, just go through them again

[hot bramble]

Okay, I'll do my best.

[pearl pond]

Channel closed

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Use .reopen if this was a mistake.

[pearl pond]

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[humble root]

,w iqr of {1, 4, 5, 1, 2, 4, 5, 5, 11}

,w iqr of {1, 2, 1, 3, 2, 3, 4, 1, 8}

How did you get the IQR of B?

What was your Q1 and Q3?

[dense jasper]

orz u

[humble root]

Hmm wolfram lied to me

It said the IQR was 2.25

But it is actuallt 2.5

Must have been using a different method, but anyway

Anyways, I don't think interquartile range is the best measure of consistency, though it is one measure.

Try looking at the standard deviations.

Try looking at the standard deviations.

Looking at the standard deviations though, ||yes D is the correct answer.||

Hmm what did you get for the standard deviation?

o

No problem 👍

[pearl pond]

Channel closed

Closed by @flint wyvern

Use .reopen if this was a mistake.

[pearl pond]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[versed mica]

if its a practice exam you should just submit it

and itll tell you

👍

don't know how anyone is falling for this shit

😭

its a practice exam though bro

shouldnt count

and?

yea you want a good grade on your final

we get it

[cinder flower]

[versed mica]

practice exams arent graded

dawg

nah im just not a moron homie

[cinder flower]

knief you gained a lot of hubris during my year off

[versed mica]

opens a channel for every question "aM i RiGhT wITh A HerRE"

[versed mica]

i was calling out cheaters just as much last year?

i used to mod ping like crazy over it

no

im just laughing at you

[pearl pond]

Channel closed

Closed by @flint wyvern

Use .reopen if this was a mistake.

[pearl pond]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[royal harness]

hello is there a french speaker who can explain to me how to do question 5)c please. I struggled with it so much that I gave up 😔

[warped violet]

Can you perhaps try to give an English translation?

Would open the question up to many more helpers here

[shut elm]

5c says use the mean value theorem to demonstrate that...

[royal harness]

yeah sure just give me a minute to write it down in English

[royal harness]

I tried and I can't figure it out

[stray canyon]

je suis la

[royal harness]

oh salut

tu peux m'aider stp j'arrive pas à trouver la solution

[stray canyon]

je vais chercher

[royal harness]

azy merci ❤️‍🩹

[stray canyon]

je pense que deja tu dois voir un accroissement avec le $\frac {f(x) - 1} {x} = \frac {f(x) - f(0)} {x-0}$

[jolly parrot]

robins

[royal harness]

oui mais qu'est ce qu'on doit faire après

[stray canyon]

je sais pas c'est chaud en vrai

ok je crosi que je l'ai

enft non

mais tu dois surement utiliser l'hypothese de stricte décroissance de h'

comme ça tu peux minorer un h'(c_x) par h'(0) ou majorer par h'(x)

[pearl pond]

@royal harness Has your question been resolved?

[royal harness]

ouais personne ne l'a résolu

on avait 1h30 pour passer cet examen

le prof est taré

[stray canyon]

t'es en quoi ?

[royal harness]

remplacer l'expression de f(x)-1/x par h(x)/... ne pourrait pas marcher ?

[royal harness]

terminale

[stray canyon]

pas en france si ?

[royal harness]

non au maroc

[stray canyon]

ah ouais c'est cque je me disais

[royal harness]

mais quand même notre prof il est pas normal 😭😭

[stray canyon]

deja il fait des trucs comme ca et il fait pas un beau sujet en latex il abuse

[royal harness]

j'ai essayé de résoudre l'exercice sans Taf en posant une fonction auxiliaire mais quand je dérive pour voir ça monotonie ça devient hyyyyper long

bon je pense que je vais demander à mon prof de soutien

merci comme même @stray canyon !