help-49

[latent wadi]

We can all just agree this is a bs problem

The graph just sucks

[cedar coral]

0,2 is the intersection of the fucking horizontal asymptote with the y axis -.-

we're arguing it being on the graph, it isnt

cant believe i took so fucking long to identify that -.-"

[latent wadi]

Another thing I'm confused about, why would they specify the graph isn't defined on x=-1, but didn't mention the other vertical asymptotes?

[ripe charm]

I think because the bottom part crosses -1

[latent wadi]

I see

[cedar coral]

i dont think we have enough information...?

[ripe charm]

Alr alr

[cedar coral]

what does the rest of the thing look like?

[ripe charm]

?

[cedar coral]

cuz i see stuff written on the right

[ripe charm]

It's separate questions

The other 2 I did

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[leaden seal]

wouldnt b = 0.2?

[signal ibex]

-0.8 i think

[cerulean oyster]

From the starting point, you know that X is lower than the mean

Therefore, it has to be negative

So 0.2 is already wrong by default.

[leaden seal]

i mean by checking the value in a table

[cerulean oyster]

Im looking at this

[cerulean oyster]

@leaden seal

If you got your z-score wrong, you have basically nothing to work with in your accumulated probability.

[leaden seal]

so this would be -0.8

[cerulean oyster]

yes

[leaden seal]

and then find 0.8 in the table

[cerulean oyster]

yes, but, if youre trying to use that table in there, consider youre using the wrong one

at least for your current purposes

you can technically still find your desired value, but youll be doing extra work just cause

[midnight plank]

@leaden seal Has your question been resolved?

[leaden seal]

wdym?

by using a calculator?

[cerulean oyster]

By doing some basic arithmetics and use some reasoning

The table you have there tells you the accumulated prob from the center to a positive z

And you want this

[leaden seal]

so jus use the postive 0.8 becus its symmetric

[cerulean oyster]

Your table gets you this

[leaden seal]

i see what you mean the table is fom the centre

[cerulean oyster]

So you have to do
1 - (P(0 - Z) + .5)

[cerulean oyster]

You dont want 0-to though

you want the up-to

[leaden seal]

12.3%

[cerulean oyster]

By table you get the blue area

You then add 50%

and substract that from 100%

That technically gives you the red-right tail

But since Normal Dist is symetric, thats equal to your left-tail you wanted

[cerulean oyster]

wrong z

[cerulean oyster]

there you go

Still, id advice to, 1, get the common z-table

2. practice with it

[leaden seal]

which table?

[cerulean oyster]

This one, or the positive z-scores

Both just tell you the probability to the left.

0-to makes some calculations harder.

[midnight plank]

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[blazing nacelle]

$$G_{4}(\tau) = a, quad G_{6}(\tau) = b$$```

[blazing nacelle]

Let $a, b \in \mathbb{C}$ such that $a^{3} - b^{2} \neq 0$. Show $\exists \tau \in \mathbb{H}$ such that

$$G_{4}(\tau) = a, \qquad G_{6}(\tau) = b$$

I'd quite like help with literally just a first step

I have the discriminant function in terms of G4 and G6, but it does not fit nicely into the a^3 - 27b^2

[fallow scarab]

what's G_4 and G_6 and \mathbb{B}

[grand pond]

Varixiuqlhfbgraijbzjnqghppxnqmvw

[blazing nacelle]

mathbb{B} is a typo, and the G4 and G6 are the unnormalised eisenstein series of degree 4 and 6

H is the upper half plane

I have this for free

In fact here is just a screenshot of my question

[fallow scarab]

yes screenshots or pictures save everyone a lot of time

you also left out the 27

[blazing nacelle]

Apologies, I'm tired

Think i'll do contrapositive

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[grand pond]

Renato

[lavish venture]

do you have a screenshot of your work

[tidal turret]

yes, thank you for asking, but i got stuck mid way

[lavish venture]

so your last line says (n + 2)! + 2^n >= 2^(n + 1)

and this is where you got stuck?

try subtracting 2^n from both sides

[tidal turret]

this?

[lavish venture]

yea but (n + 1)! times (n + 2) is just (n + 2)!

[tidal turret]

good catch

[lavish venture]

mhm

and this is easy to show

[tidal turret]

we need more induction?

[lavish venture]

yea

[tidal turret]

hahaha nice!

[lavish venture]

its really straightforward though

[tidal turret]

let me try

[lavish venture]

nice

[tidal turret]

do you want to use induction to prove that 2^n is monotonically increasing function?

[lavish venture]

🤔

no

why?

[tidal turret]

i think we can take that as granted, i just don't want to be handwavy

[lavish venture]

where are you using that?

[tidal turret]

last slide

also, how are this p(n), p(1), p(n+1) called?

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[soft sky]

I'd like some help

[wispy mortar]

lets try to simplify the equation first

can u make any terms 0 as they approach infinity?

[soft sky]

I tried, the problem that I have that k can’t be zero

[plucky lark]

K is greater than 1 tho

[wispy mortar]

we don't care about k

try by taking n common from both denominator and numerator

[soft sky]

do i need to use the squeeze theorem?

ok i got it

i answered it

thank y'all ❤️

i got the lim equals 1

[mortal falcon]

how did you get that

[soft sky]

squeeze theorem

[mortal falcon]

what did you compare to?

[soft sky]

LHS = (sum from k=1 to n) (sqrt(n^2 - n))/(n^2 + n +n)
RHS = (sum from k=1 to n) (sqrt(n^2 ))/(n^2 + n )

LHS

RHS

and got that both limits get to 1

[midnight plank]

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[sudden bronze]

how to find the perimeter of this side

[mortal falcon]

find the radius of each of the arcs

that gives you horizontal and vertical lengths for each

[sudden bronze]

i got

22 on the right arc

and 12.6 on the left arc

[mortal falcon]

how did you get those?

one arc is radius 14, thats given

[midnight plank]

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[open fiber]

I guess this is a conceptual question

Because no specific matrix is provided

For a projection, I can imagine the eigenvalues are 0 (for vectors that are perpendicular to the original space) and 1 (for vectors already on it)

For reflection, I can imagine the eigenvalues are 0 (for any vector) and 1 (for any vector originally in the space)

I'm not sure how to produce the eigenvectors though

and I'm contradicted by this PDF

https://math.mit.edu/~gs/linearalgebra/ila5/linearalgebra5_6-1.pdf

[wraith dirge]

you use the properties of projection/reflection matricies to find the eigenvalues

for reflection, you know that R^2 = I

[open fiber]

Could you clarify?

I don't think I learned that principle

[wraith dirge]

when you reflect something twice, you get the original thing you reflected

[open fiber]

oh ok it seems obvious when you put it this way lol

[wraith dirge]

yeah it is kind of conceptual but the eigenvalues dont come out of nowhere

you use that concept to find them

for projection, you use the fact that P^2 = P for a projection matrix P. intuitively, this is because a projection applied twice is the same as a projection applied once

[open fiber]

so both of your assertions make sense but I'm not sure how to move forward from here

[wraith dirge]

i'll do projection as an example, you are trying to solve Px = Lx (L instead of lambda)

[open fiber]

okay

[wraith dirge]

use P^2 = P to get L^2 x = Lx

then you get L = 0, L = 1 as you already figured out

[open fiber]

oh shit, that's clever

[wraith dirge]

Same idea with reflection, if you can realize R^2 = I, then you get L^2 = 1, thus L = -1, L = 1

[open fiber]

so with R^2, we know that

RRx = Ix = x

so lambda^2x=1 so lambda= plus or minus 1

[wraith dirge]

yeah

the hard part is recognizing the properties of projection and reflection

[open fiber]

ok very helpful thanks

wait so for eigenvectors

the way I was taught to find eigenvectors was to find the basis of N(A-lambda I)

that's not applicable here given that we don't get the actual matrix

[solid iris]

for projections the L=1 case is Px=x. what x satisfies it?

[open fiber]

something originally on the subspace? the main trouble here is representing it

can I just write C(P)? Is that valid?

[solid iris]

what subspace?

[open fiber]

the one defined by the columns of P

is that right?

[wraith dirge]

i guess you could write it in words. like for L = 1, you could say "all vectors in the subspace you project onto"

the notation is not important

[solid iris]

talking about a projection assumes you already know the subspace it projects onto

you should just say c2b7's answer instead of making up notation

[open fiber]

okay

[solid iris]

now the L=0 case is Px=0. what x satisfies it?

[open fiber]

any vector, right? since 0 is a scalar multiple of any vector

[solid iris]

no

[open fiber]

wait stop

[open fiber]

sorry, mb, any vector perpendicular to the subspace

[solid iris]

yes

we fully answered projections so lets do reflections

[open fiber]

so for negative one the vector would be something that is perfectly reflected across the subspace, i.e. any vector perp to the subspace

for one it would just be any vector originally lying on the subspace

[solid iris]

yep

[open fiber]

got it

ty guys

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[solid iris]

np!

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[open fiber]

so I'm conceptually stuck

[open fiber]

I just have no idea how to go from here

answer isn't much help either

[solid iris]

tp'=Lp is an ODE

[open fiber]

ODE?

[solid iris]

differential equation

[open fiber]

wait thanks for the help but I just found an online source with the same question

https://askfilo.com/user-question-answers-algebra-2/in-each-of-the-following-cases-find-the-eigenvalues-and-35393333333132

for posterity

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[wraith notch]

$$ y' = (2x+3)(y^{2}-4)\ , \ y(0) = -1 $$

[wraith notch]

This was a problem on the test I have and I couldn't find the answer to it in the book, wait I messed up

[grand pond]

Alphurion

[wraith notch]

Yeah that looks right... I thi nk

[lusty python]

What's $y$ specifically

[grand pond]

1 divided by 0 equals Infinity

[lusty python]

And what are you trying to find?

[wraith notch]

it's an "initial-value" problem, ummm

It just said solve it

No other context

"solve this" and gave me those two equations

[fallow scarab]

Use separation of variables

[wraith notch]

so like $$ \frac{y'}{y^{2} - 4} = (2x + 3) $$

[grand pond]

Alphurion

[wraith notch]

But, I have to like, integrate it to find y right

[cedar pawn]

Mmhmm

[sharp coral]

the next step would be to integrate, yes

[cedar pawn]

Partial fractions time

[wraith notch]

$$\int \frac{y'}{y^{2} - 4} dy = - \frac{1}{4} ln | y + 2 | - \frac{1}{4} ln | y - 2 | + C}$$

please be righttt

NOOO

[grand pond]

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

[wraith notch]

yay

well I mean idk if its right

surely

so alldat............ equalsssssssss (2x+3) in theory

[sharp coral]

well that notation for the integral is a bit off

it should be [ \int \frac{y'}{y^2-4} \dd x = \int \frac{1}{y^2-4} \dd y ]

[grand pond]

cloud ☁

[wraith notch]

Oh

so if I integrate one side with dx, I cant just integrate the other side with dy

they both have to be in terms of one

variable

[lethal path]

yes, you would need $\int f(x) \ dx = \int g(y) \ dy$

[grand pond]

south

[sharp coral]

the common thing is to write $y' = \odv yx$ and then ``multiply the $dx$ to the other side'', i.e.
[ \frac 1{y^2 - 4} \dd y = (2x + 3) \dd x ]

[grand pond]

cloud ☁

[cedar pawn]

Then integrate both sides

[wraith notch]

Oh the other side

Its like $x^{2} + 3x + C$

[grand pond]

Alphurion

[wraith notch]

yayyy

[lethal path]

and you only need to write one +C, even though there are two integrations

[wraith notch]

so if I hypothetically insert -1 into the left side it should equal the right side

[lethal path]

cause if you have .... + c = .... + C'

that's just ..... = .... + (C' - c)

C' - c is a constant

[wraith notch]

Right

Would you guys be... upset with me... if I couldnt evaluate ln -1.....

[lethal path]

well ln(-1) is not a real number

[wraith notch]

Actually its ln 1 since its + 2

[lethal path]

you'd have to figure it out from e^(i pi) = -1

oh okay

[wraith notch]

the other one would be ln | -3 |

[lethal path]

yeah that's why ln |x| is very important

[wraith notch]

But do I know how to evaluate it? nop

failed alg 2 :3

so kawaii

Oh theyre like really long decimals anyways... but 1/4 of that...

[cedar pawn]

You don't need to evaluate the logs

Just simplify them as much as possible and then you can leave them as logs

[wraith notch]

sooo is ln 1 + ln 3 = ln 4 or is that like completely insane

[fallow scarab]

Wrong

Also you can use a calculator to answer your own question

[cedar pawn]

Why ln(1)?

Oh the condition

Okay

[wraith notch]

because ln | -1 + 2 | = 1

so ln 1

[fallow scarab]

Wrong again

[wraith notch]

o ok

[cedar pawn]

Why?

[fallow scarab]

.

[cedar pawn]

The ln(1) part is fine

Oh his notation is wacky

[cedar pawn]

You can't just add logs like that

You need to use log laws

[wraith notch]

I don't know any of them though so I can't really evaluate it without a calculator

[cedar pawn]

You don't need to evaluate it, just simplify it

And you should know log laws

[fallow scarab]

,tex .log rules

[grand pond]

riemann

[wraith notch]

so ln 1 + ln 3 is just ln 3

[cedar pawn]

Is that surprising to you?

[wraith notch]

It's something I didn't know

[cedar pawn]

I'll take that as a yes

Without a calculator, what's ln(1)?

[wraith notch]

I don't know it

[cedar pawn]

Okay so ln(x) is the inverse of e^x

In other words, ln(x) is the number that you need to raise e to in order to get x

[wraith notch]

0

[fallow scarab]

,calc log(1) + log(3)

[grand pond]

Result:

1.0986122886681

[fallow scarab]

,calc log(4)

[grand pond]

Result:

1.3862943611199

[fallow scarab]

Not the same

[cedar pawn]

How did you get this?

[wraith notch]

A number to the power of 0 is 1, therefore to get e to the power of something is equal to 1, the power must be 0

[cedar pawn]

Yes!

So ln(1) + ln(3) = 0 + ln(3) = ln(3)

[wraith notch]

Nice

[cedar pawn]

Which matches with the log law as well

ln(1) + ln(3) = ln((1)(3)) = ln(3)

ln(3) is just a number. You can treat it like a square root in the sense that you just leave it as it is

[wraith notch]

$-\frac{1}{2} ln 3 = x^{2} + 3x + C$

[cedar pawn]

Wasn't it -1/4?

[wraith notch]

well there was two hold on

[near aspen]

Pls help me solve this: Four million 500-kilowatt windmills in the states listed in the table could generate 91% of the total electrical use. If this were done, how many windmills would be in North Dakota? (Hint: It is not 11% of 4 million.)

[cedar pawn]

!occupied

[midnight plank]

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).

[cedar pawn]

?

How did you get 1/16?

[wraith notch]

Oh are they added aw man

[cedar pawn]

Noooo

[wraith notch]

wait what am I doing

shhhh

bro youre making me nervous

• 1/4 + -1/4 ...

-1/2...

yeah

[grand pond]

Alphurion

[wraith notch]

boom

[cedar pawn]

What are you doing? x.x

Still no

[wraith notch]

The problem

Why not

[cedar pawn]

5x + 5x =/= 10(2x)

You can't just add the terms in each product separately

[wraith notch]

Either the fraction multiplies or adds

I mean its a fraction but you get what I mean

[cedar pawn]

I don't

[wraith notch]

There are two -1/4, either we are using the properties of logs to say that they multiply and are 1/16, or we are saying because they are both negative they add up to -2/4 which = -1/2

unless you mean it simply it -1/4 still

but I 'm pretty sure you said no to that

[fallow scarab]

You should not be lazy and skip terms. Write out fully what you mean here

[wraith notch]

$-\frac{1}{4} ln 1 - \frac{1}{4} ln 3 = x^{2} + 3x + C$

[grand pond]

Alphurion

[wraith notch]

So it has been told to me that from the properties of logs that ln 3 and ln 1 become ln 3 in isolation

[cedar pawn]

Only if you have ln(1) + ln(3)

[wraith notch]

I was then informed the fraction was incorrect

So then what about for this problem, how would I go about this step?

[cedar pawn]

If you have 5x + 5y, how else can you write that?

[wraith notch]

10xy

5xy

5x5y

WHAT

well actually yeah thats right

uhhh

shoot idk

[fallow scarab]

Pretend x=2 and y=3

[wraith notch]

its just addition

right

[cedar pawn]

It looks like you could do with revising some algebra

[fallow scarab]

What's 10 + 15

[wraith notch]

25 prolly

[fallow scarab]

Prolly?

[wraith notch]

5(x + y) ??

[fallow scarab]

Finally

[wraith notch]

so you want me to....

[fallow scarab]

Apply that to left side here

[wraith notch]

so its -1/4 ln 3 then

[cedar pawn]

Yes

What does the other side equal?

[wraith notch]

-1

[cedar pawn]

?

How do you get -1?

[wraith notch]

Actually, I'm not sure

I don't even think I was supposed to plug in -1 actually

I shouldve plugged in 0

which would have led to -1/4 ln 2

[cedar pawn]

Be careful with what you're doing

[wraith notch]

which would make the other side equal to -1 because -1/4 ln 2 would be y(0)

[cedar pawn]

Rather than just doing random stuff

What does y(0) = -1 mean?

[wraith notch]

the function y when you input 0 is = -1

[cedar pawn]

input 0 into what?

[wraith notch]

so if this equation I plugged in with 0 is y(0) = -1 = x^2 + 3x + C

[cedar pawn]

?

Does y = x^2 + 3x + C?

[wraith notch]

No

[cedar pawn]

So why did you just write that? >.<

[wraith notch]

Because I don't know the solution, so I am going to try something rather than nothing

[cedar pawn]

It would help to think it through rather than just spamming

[cedar pawn]

^^^

[wraith notch]

You didn't speak for an entire minute, I'm not going to sit there and do nothing, I thought about it

I'm sorry if it wasn't right, but I clearly am not proficient enough to know it was an incorrect decision

[cedar pawn]

I need to go to sleep

[wraith notch]

Alright

.close

[midnight plank]

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[shut canyon]

a¹≅a, because a^1 is every arrow from 1 to a, in other words Hom(1,a)
I believe ∃!f : 1 ⟶ a, ∀ a in 𝒞

which is "select a"

[paper mantle]

i was gonna say some stuff but i just realized i know nothing about proofs lmao

[shut canyon]

and it is iso, because there is ! : a ⟶ 1

[shut canyon]

go on,
categories and logic are good ways to learn something about proofs, for example I found it cool to learn that proof can be seen as a path, and "showing" or "proving" we construct a path:

from the context of premises Γ to the conclusion φ

showing there exists: ∃
a path f(which is the name of the path, or arrow ⟶)
from the premises Γ
to the conclusion: φ

∃ f : Γ ⟶ φ

[paper mantle]

well im actually clueless on the proofs side

i only know a bit about induction

and from looking at that u can use induction to prove the 1^a = 1

[shut canyon]

provided the axioms/assumptions of categories (set of premises Γ), we are asking is can we go from a¹ to a and back, and from a to a¹ and back

we are asking: exists a path a¹ ⟶ a, and a ⟶ a¹?

[shut canyon]

how?

[shut canyon]

do you know implication? P ⟹ Q?

[paper mantle]

1^n = x

1^n+1 = x.1

1^n+1 = x

1^n+1 = 1^n

i think thats how u do it

cant recall

[shut canyon]

proof afik is you have premises and derive conclusions using logic,
it's like you ask a question and find answer

[paper mantle]

i mean i did just prove that 1^n = 1^n+1 no?

isnt that what induction means?

[shut canyon]

I need to learn induction

[shut canyon]

1. show it holds for 1,
2. then assume it holds for n,
3. see for n+1

I think

the way you define ℕ via Peano axioms is the blueprint also

[paper mantle]

tbh idek what ur saying rn lmao

no clue on proofs terminology haha

[midnight plank]

@shut canyon Has your question been resolved?

[shut canyon]

one sec, I don't yet udnerstand the example you gave

[paper mantle]

i mean its just induction

[shut canyon]

1ⁿ = 1 for any n ≥ 0
you can write it like this look: ∀ n, 1ⁿ = 1,
which is to say: for all n greater than or equal to 0 the nth power of 1 is 1

1ⁿ⁺¹ = 1 * 1ⁿ = 1 * 1 = 1

[paper mantle]

yeah thats what we did

[shut canyon]

maybe try 1 + 2 + … + n = n·(n+1)/2 with induction

[paper mantle]

hm?

but thats not on ur question tho

[shut canyon]

here is a template for induction, I hope it's correct,

1️. base: P(0)
2️. inductive hypothesis: assume P(k)
3️. step: show P(k) → P(k+1)
4️. conclude: ∀ n, P(n) holds

• P : proposition "…is P", or …has the property P
• P(k) : k has the property P
• P(k) → P(k+1) : if k has the property P, k+1 has the property P
• ∀ n, P(n) : every n has the property P

[paper mantle]

in induction u just prove that f(n) = f(n+1)

[shut canyon]

when you use induction, I heard it is good practice to say something like * "we will use induction"*,
and highlight the main steps, like "base case", inductive hypothesis and inductive step, is good form

[paper mantle]

i mean yeah thats usually what u do with any proof no?

or atleast i tought so

[shut canyon]

I guess so, and also some say proofs are like art, you can be creative with making proofs, I read this by Paul Lockhart in Measurement

[paper mantle]

imo u dont need to care much about the paperwork if u get what im saying

[shut canyon]

and it has an aspect of explaining and shining light and communicating so it's good to make accessible maybe

[paper mantle]

just demonstrate ur math and i think ur teachers will consider it

[shut canyon]

this is the natural numbers pattern in Lean4, I think it's analogous to Peano axioms and inductive proof:

inductive Nat
| zero : Nat
| succ : Nat → Nat

[paper mantle]

u see idk what that means lmao

[shut canyon]

not YET

[fathom onyx]

See, pragmatically, I'm not sure Kytsu's learning things in an order that makes sense here

[paper mantle]

yes

[fathom onyx]

The level of material you're reading, one should expect induction to already be a familiar tool to you

[paper mantle]

thats what i was thinking

[shut canyon]

understandable

[fathom onyx]

@paper mantle dw, "Peano axioms" is a set theory-based term, i.e. well into uni maths

It's to do with number construction at a fundamental level, and little to do with proofs in and of themselves

[paper mantle]

nice

[fathom onyx]

You may re-explain induction with them, though it is a little overkill

[paper mantle]

yeah i dont study proofs cause the olympiads i attend usually dont ask much about proofs in a strict way

[fathom onyx]

Olympiads aren't gonna need something like Peano axioms

[paper mantle]

noice

@shut canyon

so

induction

is prooving that a specific statement

is true

for n

and n+1

for example

[wise flicker]

What was the og q?

[paper mantle]

.

[shut canyon]

Peano axioms seem to say the same thing as induction, like isn't it it the same pattern?

[paper mantle]

my creativity to give an example at 6am is not good, if someone could id appreciate it lmao

[fathom onyx]

Peano axioms define the set of natural numbers via induction, yes

They aren't in and of themselves induction

[paper mantle]

oh thats very advanced shit

[fathom onyx]

I did say

[paper mantle]

but yeah now that i searched about it he should know a bit ab induction to know what a peano axiom is no?

[fathom onyx]

A perusal on MSE appears to give me some traction btw, regarding the original question:
https://math.stackexchange.com/questions/3627695/how-to-show-that-a1-simeq-a-for-all-c-objects-a-in-a-cartesian-closed-ca

[shut canyon]

thanks @fathom onyx

That's a cool way to visualise diagrams with the grid, I think

[shut canyon]

look there is the same question also via that link
https://math.stackexchange.com/questions/3463019/prove-that-a1-cong-a

I'll try to answer with my solution, I will probably be cast out

they will be like "ugh this person… do you even know Peano axioms?" 🐙

.close

[midnight plank]

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[soft sky]

I need help with this limit

[lyric charm]

!status

[midnight plank]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[soft sky]

I'm trying to use squeeze theorem

[lyric charm]

first off, the stuff inside the limit is quite clearly $\frac1n \sum_{k=1}^n \frac{1}{k}$. this might help

[grand pond]

Ann

[soft sky]

umm, how will it help?

[lyric charm]

well at the very least that's less shit to write every time you wanna refer to the function

[soft sky]

lol yea

[lyric charm]

but ok yeah so what are we squeezing with

[soft sky]

on LHS, I got that this whole expression is bigger than 2/1+n

I used AM-GM inequality to prove it

[lyric charm]

you mean 2/(1+n)

okay alright. so that's a lower bound.

what do you know about the sum $\sum_{k=1}^n \frac1k$?

[grand pond]

Ann

[soft sky]

i can say that this whole sum is less than $\sum_{k=1}^n \frac1n$

[grand pond]

KevinSnow

[lyric charm]

no it's not

sum[k=1,n] 1/n is just n/n aka 1

[soft sky]

oh yea right

[lyric charm]

ok lets see... do you know integration at all / do you have access to it

[soft sky]

no

im still at the beginning of calc

[midnight plank]

@soft sky Has your question been resolved?

[soft sky]

<@&286206848099549185>

[lyric charm]

rip

[soft sky]

bro this question is KILLING ME

can i use arithmetics? say that each of this terms tends to 0, so the whole expression tends to 0 aswell?

[slow thorn]

infinite times

0 times infinity is bad

[soft sky]

yea true

[midnight plank]

@soft sky Has your question been resolved?

[midnight plank]

@soft sky Has your question been resolved?

[fallen sparrow]

<@&268886789983436800>

[midnight plank]

@soft sky Has your question been resolved?

[small jasper]

Cosider grouping the terms by powers of $2$:
$$\sum^{n}_{k=1} \frac{1}{k}=1+\frac{1}{2}+\left(\frac{1}{3}+\frac{1}{4} \right)+\left(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8} \right)+\dots$$

[grand pond]

Civil Service Pigeon

[midnight plank]

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[zealous schooner]

I'm trying to apply the Central limit theorem to these two problems

[zealous schooner]

For the first one, we can take $U_n=#H-#T$ after $n$ flips, so that $U_n=X_1+\dots+X_n$ with $X_1,\dots,X_n$ i.i.d. as $\operatorname{Uniform}({-1,1})$

[grand pond]

kheer257

[zealous schooner]

Then we want the probability that U_n = 0

But I don't really know how CLT works for discrete random variables

[midnight plank]

@zealous schooner Has your question been resolved?

[fallow scarab]

Should just be CLT approximation to binomial distribution

https://en.wikipedia.org/wiki/De_Moivre–Laplace_theorem

[bold peak]

Isn't tossing a coin multiple times already binomial distribution

Do you really need CLT for that

[midnight plank]

@zealous schooner Has your question been resolved?

[midnight plank]

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[sturdy kindle]

for instance i have 5-x^2

and i have to put 1 in x,

is it 5-(1)^2 or 5(-1)^2?

[slender walrus]

the first one

[sturdy kindle]

why

[slender walrus]

order of operations

in the initial expression the power of 2 only applies to the x

[wraith dirge]

PEMDAS

1. Parentheses
2. Exponents
3. Multiplication & Division
4. Addition & Subtraction

[slender walrus]

and that x^2 is being subtracted from 5

in other words
5 minus (the square of the value of x)

for the second, you're treating it as though you had
5 * (-x)^2
which doesn't follow the established order of operations

sub or not
p - q
isn't the product of
p and
-q

[midnight plank]

@sturdy kindle Has your question been resolved?

[sturdy kindle]

ohhhh

[midnight plank]

@sturdy kindle Has your question been resolved?

[midnight plank]

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[dark cloud]

can someone explain to me what lateral surface area is?

[dark cloud]

if it's just the surface areas of two parallel surfaces how does that apply to a cone etc

[lyric charm]

it's the area of whichever part of a solid isn't its "top" or "bottom"

one speaks of it usually only for solids which have a clearly-defined top or bottom, namely:

• cones and pyramids
• prisms (incl boxes) and cylinders
• frusta

[dark cloud]

oh sweet

yeah that makes sense

thanks!

.close

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[midnight plank]

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[viral dagger]

shitt so sorry i fell asleepppp

[civic gazelle]

lmao

so we're currently on proving A, M, F, C concylic

we shall prove angle AMC = AFC

[viral dagger]

yeah

lemme think again

[civic gazelle]

any ideas on that?

[viral dagger]

AFC=180-AFE=180-EAC

then uhh

[civic gazelle]

keep going

[viral dagger]

tgat should equal DBC i think?

yeah

ohh wait

AST on that

DBC=BHC=AMC

done

[civic gazelle]

alr great

now we should be able to prove that E, M, F, B are concyclic

(im trying to figure out how to get D in there tho)

[civic gazelle]

ye it isnt too difficult either

[viral dagger]

BC seems tangent to (EMFBD)

[civic gazelle]

oh huh

i didnt notice that

try to prove BEF = BMF first

[viral dagger]

yeah ok thats what i thought the pair of angles was

BEF=BED-FED

BED=180-EAC

using the same angle chasing BED=AMC

so BEF=AMC-FED

hmm ok this is abit wierd

[civic gazelle]

hmm looks like you're overcomplicating stuff but anything works i think

you're forgetting EB//AC it seems

[viral dagger]

i mean i used it here

oh wait

BEF=ECA

[civic gazelle]

nice

and we just proved AMFC is a cyclic quad before so...

[viral dagger]

oh.

yeah fuckk i just realized that

[civic gazelle]

now we just need to get D in there

[viral dagger]

hold on brb gimme a sec

back

mm so probably something cyclic quad with D on the other 3 of the 4 points, then combine to get DMFB concyclic

[civic gazelle]

fuck this is harder than i thought lmao

[tawdry laurel]

[civic gazelle]

fake orz

[tawdry laurel]

[viral dagger]

lowkey DMBE seems promising?

[viral dagger]

huh didnt you say it wasnt too difficult

[viral dagger]

dis

[civic gazelle]

i made a blunder

was doing everything in my head lol

[civic gazelle]

this is the part im missing

[viral dagger]

samee

i also got a sol that depended on that

hold on lemme reconstruct the sol

DBM=BCM by alt seg on (BCM)
MDB=MBC by alt seg on (MDB)

[civic gazelle]

oh i found it

this is amazing

[viral dagger]

so DMB=BMC <=> DMA=AMC

AMC=AFC=180-CAE=AEB

so DMA=AEB

DMA=180-DMB=AEB=DEB (DEBM cyclic)

[viral dagger]

yeah this bit is the part that uses BC tangent

[viral dagger]

howwb😭

[civic gazelle]

*realizes i made another mistake *

[viral dagger]

wait hold on

BEA=180-CAD=DBC

alternate segment theorem is an iff right?

so BC is tangent

so were done no??

[civic gazelle]

it only means that BC is tangent to (BED)

doesnt mean D lies on that circle we want to prove

[viral dagger]

well unless i did something wrong, this should prove DEBM is cyclic

[civic gazelle]

we already did boil down the problem quite a lot

[viral dagger]

if DEBM and MEBF are cyclic then DEBFM is cyclic

then the radical axis theorem, then were done right?

[civic gazelle]

one interesting thing i discovered but havent been able to prove is that EM goes through the intersection of AC and (BMC)

[civic gazelle]

yes, but D is being a bitch and is stopping us

[viral dagger]

huh??

[civic gazelle]

we need to prove that D lies on the (MEBF) somehow

[viral dagger]

like i got BC tangent to (BED), then used that to prove DEBM is cyclic, doesent that incorporate the D?

[civic gazelle]

then you have to prove BC is tangent to (BEM) as well

[viral dagger]

shit

[civic gazelle]

i dont know if i should go for this

or i can pull out the phantom point trick

[midnight plank]

@viral dagger Has your question been resolved?

[midnight plank]

@viral dagger Has your question been resolved?

[civic gazelle]

yeah i think you must know about symmedians to get the property of DM

Once you know that you can solve the problem

@viral dagger yep, figured it out

[viral dagger]

shit symmedians

uhh i briefly studied them but i dont remember them at all

[civic gazelle]

yea from symmedian CD you have MDA = MAC = MBE
-> DMBE is a cyclic quad
-> problem solved

[viral dagger]

lemme pull up my notes

ok so this is my note, AD is the symmedian of ABC

so here in our diagram the symmedian is CD?

[civic gazelle]

yes

[viral dagger]

yeah D in our diagram is K here

[civic gazelle]

if you're unconvinced you can try to do this

(AOD) intersects (ABC) at E

AE will be the symmedian

and you'll get BED = CBA as well

it's just awful angle chasing

[viral dagger]

aaa im lost

wtff

[viral dagger]

i dont get how you got MDA = MAC = MBE

[civic gazelle]

you dont get why is MDA = MAC?

[viral dagger]

yeah

thats not in my notes of symmedian properties :<

[civic gazelle]

alr let's do this lemma first

given ∆ABC, O is the circumcenter, D is the midpoint of BC. (AOD) intersects (ABC) at E. Prove AE is a symmedian

[viral dagger]

hmm

the definition of symmedian is that the tangents of B and C to (ABC) intersect at X, and AX intersect (ABC) at E

[civic gazelle]

it's very looong angle chasing

[viral dagger]

so we want to prove AEX is colinear

[civic gazelle]

nah you dont need that

you only need to prove BAD = CAE

that's it

that's enough to make AE a symmedian

[viral dagger]

ok lowkey im hella stuck i have no idea how to prove this

[civic gazelle]

alr so BAD = BAE + EAD = BCE + EOD

EOD = BOD - BOE = BAC - 2BCE = 180 - BEC - 2 BCE

[viral dagger]

shouldnt EOD be BOD-BOE instead

[civic gazelle]

oops sorry

alr fixed

so BAD = BCE + 180 - BEC - 2BCE = 180 - BEC - BCE = EBC = EAC

and since this works both ways, EA will be a symmedian of triangle EBC as well

[viral dagger]

wait sorry is there a different definition of a symmedian??

[civic gazelle]

what do you mean by "definition" here?

[viral dagger]

the definition of a symmedian that i know is this

how does proving BAD=EAC lead to AE being a symmedian

[civic gazelle]

well the definition to me is you take a median then reflect it across an angle bisector

[viral dagger]

ohh wait

AM and the symmedian is isogonal?

[civic gazelle]

yes

[viral dagger]

is that what you were leading torwards?

ohhhhhhg

omg it makes sense now

[viral dagger]

ok so on applying the lemma

[civic gazelle]

yeah there's really only 1 specific point on arc BC that satisfies BAD = CAE

and it just so happens that that point also applies to your definition

[viral dagger]

mm i see

[civic gazelle]

do you have any questions on that?

[viral dagger]

so we want to prove COMD is concyclic

to apply the lemma right

[civic gazelle]

no

CD is a symmedian is enough

[viral dagger]

oh was the lemma just to show that you can prove symmedian like that?

[civic gazelle]

because the lemma goes both ways so if you have AE being a symmedian then AODE is a cyclic quad

[viral dagger]

no like i still dont get how you know CD is the symmedian of ABC

[civic gazelle]

oh yeah

[civic gazelle]

DBA = DCA
DBA = BCM from AST

so DCA = BCM

[viral dagger]

ohh

man my brain is so fucked after staring at this image for hours

[viral dagger]

up

[civic gazelle]

yes?

[viral dagger]

mmm on MDA=MAC

🤔

[civic gazelle]

oh yeah i havent explained the second part of the lemma

[civic gazelle]

going back to this image

so to not make you suffer through more angle chasing pain...

if i switch the roles of A and E

[viral dagger]

ohh wait

EA is the symmedian of BEC?

[civic gazelle]

yep

then it is smooth sailing from here

[viral dagger]

ok that part is in my notes

hmm

[civic gazelle]

you get CEA = BED

and then CEA = CBA so CBA = BED

this symmedian part prob took half the convo lol

[viral dagger]

ohh wait

MDA=CDB cause symmedian is isogonal conjugate of median

and CDB=CAB

[viral dagger]

so we get this

MAC=ABE cause parallel

MDA=ABE

ohh and that proves (MDEB) concyclic

which gets (MDEBF) concyclic

then the theorem

qed??

omgggg tysmmm

❤️

.solved tysmsmnmmmmm

[midnight plank]

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[civic gazelle]

. sorry i was away lmao

[viral dagger]

. but its correct right?

[civic gazelle]

. It is

. dumb reminder that we havent utilized all the givens somehow

[viral dagger]

. really? which part?

[civic gazelle]

. the symmedian part which took ages for me to figure out

[viral dagger]

. huh wdym?

[civic gazelle]

. we were so dead-on trying to prove BC is tangent to that circle

. anyways let's stop this so new helpees wont be confused

[twilit field]

I was thinking $Z_{120}, S_5, D_{60}$

[grand pond]

wai

[lyric charm]

ok, you have 3 non-isos to explain

[twilit field]

$Z_{120}$ is not isomorphic to either as its cyclic and neither of the other two are

[grand pond]

wai

[lyric charm]

ok and why are S_5 and D_60 not iso

[runic hamlet]

well there are certainly easier examples

about which you should think afterwards

[twilit field]

$D_{60}$ has an element of order 60, $S_5$ has none

[grand pond]

wai

[spiral rock]

There are abelian examples

[lyric charm]

why does S_5 have no element of order 60

[twilit field]

The order of the product of n- disjoint cycles is the lcm of their individual lengths

[lyric charm]

ok and why can't the lcm of a bunch of numbers adding up to ≤5 be 60?

[spiral rock]

A bit of an overkill solution: If there was an element of order 60 then the subgroup of S5 it generates is normal and hence its equal to A5, which is not cyclic

[twilit field]

wait, why is it normal

[spiral rock]

Subgroup of index 2

Its overkill because you use a really strong result regarding S5

[twilit field]

A decomposition of 120 would be 3x2x2x5x2, but when combination is chosen such that their sum is 5, their product , an upper bound on the lcm is less than 60

Is there any better justificatioon

[runic hamlet]

get your hands dirty and just go through all combinations

[twilit field]

okay, so the possibilities are elements of order 2, of order 3, or one of order 3 and another of order 2, their orders will be 2,3,10, none of which are 60

[lyric charm]

at least one cycle must have length a multiple of 5, but in S_5 this can only ever give order exactly 5 with a long cycle

[midnight plank]

@twilit field Has your question been resolved?

[twilit field]

Oops, forgot to close this

[quiet hinge]

Coming to the simpler examples that people were talking about, can you produce 3 abelian examples? ||Think of direct products||

[twilit field]

Will look into this after dinner

tq

.close

[midnight plank]

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[dapper nacelle]

hola!

i wrote th eq as a+ax+ay=60 and then wrote it as 1+x+y=60/a and then x+y=60/a -1

tried to use stars and bars

got 179 as answer

actually answer is 144

[visual tiger]

Can you explain your reasoning more in detail for 179?

[dapper nacelle]

basically we need to find no of solutions of the this modified equation

so by stars and bars you get
(60/a -1 +1)!/(60/a - 1)!(1)!

you get 60/a as the answer

the possible values of a are 1,2,3,4,5,6,10,12,15,30,60

sum them all uo

oh wait 1 is there

180 is what i get

[visual tiger]

Are you using stars and bars for x,y positive or non-negative?

60/a - 2 is the actual answer

[dapper nacelle]

ahhh

i took them as non negative

so eq would become x'+y' =60/a - 3?

from there it seems to work

thank you so much

.close

[midnight plank]

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[near geyser]

I didn't understand the part I we have to move from minus infinity

[lyric charm]

basically you need to imagine the line 9x+4y=c sweeping out the plane as c goes up

that's why this person is saying "from -∞"

[near geyser]

But why we are moving?

Can we not do it by +infinity to -infinity

Or it is restricted by condition<=1

[lyric charm]

that'll be downward

and then you'll instead want to keep moving until you FIRST hit the constraint

[near geyser]

For minima i will choose it

I guess

[wise flicker]

You preparing for upsc?

[near geyser]

Is this correct?

[near geyser]

Yes

[midnight plank]

@near geyser Has your question been resolved?

[small jasper]

3 is correct but a comment about the bound being attainable would be nice

[midnight plank]

@near geyser Has your question been resolved?

[near geyser]

Answer is correct but i am asking about the method

[small jasper]

3 is correct but a comment about the bound being attainable would be nice

The solution arrives at a correct bound

But in general, when using inequalities, you need to show that your bound is attainable

Ex. Let’s say that you have x,y positive and z negative

And you want to minimize |x+y+z|-(|x|+|y|+|z|)

The triangle inequality gives you that this is __<__0

So you may be tempted to say that the minimum is 0

But this can’t be attained since equality is only attained when x,y,z have the same sign

Which is impossible

In this case, the supremum is 0, but there isn’t a maximum

[midnight plank]

@near geyser Has your question been resolved?

[midnight plank]

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[verbal kindle]

how do you find the surface area of this shape? I thought of using the formula for a one by integrating 2pi x int from 0 to 3 of the line -4/3 x + 4

but thats not the right approach because one side is flat

so im not sure noww

[cedar mason]

im not even sure you can

Ellipses are annoying

[verbal kindle]

apparently you can

but im not sure how

[cedar coral]

you definitely can, and it's extremely annoying

[verbal kindle]

can you tell me how to approach it?

[cedar coral]

no, that's way beyond my skill in integral calculus

[midnight plank]

@verbal kindle Has your question been resolved?

[midnight plank]

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[last slate]

wouldnt the answer to this just be 96

[last slate]

2 · 2 · 4!

???????

[versed quiver]

depends on whether you are treating the groomsmen as two different people, or the same

[cerulean oyster]

given the wording, i dont think thats the case

[cerulean oyster]

can you explain how you come to that number?

[plucky lark]

I think they are distinct as ppl are distinct

wait a minute

[cerulean oyster]

@last slate

[versed quiver]

in that case, 5! x 2!?

[plucky lark]

they can be nondistinct mb

[cerulean oyster]

this is also wrong

Also, dont give answers plainly without any explanation behind

[versed quiver]

hmmmm, im pretty sure thats correct

[cerulean oyster]

Supposing that were right, explain how would you come to that solution

[versed quiver]

well i wasnt sure if i was right either

didnt wanna explain if my solution was incorrect

but i cant see what else it would be

what are you thinkin

[last slate]

okay so they want the bride and groom next to one another

thats S = { bride , groom } and S2 = { groom, bride }

then we have a remaining of random combinations for the 2 bridesmen and the 2 groomsmen

[cerulean oyster]

Yes, that limits quite a lot the possibilities for arrangements, if i have to give some advice, think of both as a "block" and then simply multiply by 2 whatever answers you come to.

[last slate]

i get what youre doing here it isnt wrong to think of it like this but why am i wrong

like

that would be if you only considered one position for the bride and groom

[cerulean oyster]

Youre missing a few combinations coming from the fact that the groom and bride can be placed anywhere

[last slate]

but i wanna do it this way

[last slate]

oooooh

oh shoot

[cerulean oyster]

my guess is that you assumed that they have to be in the center, right?

[gusty falcon]

why?

[cerulean oyster]

look at helpers lounge

[last slate]

well i assumed they are at the right but my mistake still applies to if you assume they are anywhere but as long as theyre next to one another in a fixed spot

so right left center

but they only stay there

what do i do then

[cerulean oyster]

the only condition is for the groom and bride to be next to eachother

[last slate]

Yes but they can be next to eachother anywhere like u said

[cerulean oyster]

How many positions would that leave for the block?

[last slate]

4

[cerulean oyster]

mb, you were right, i dont know where i got 5! from

[cerulean oyster]

Try to think of the edges too

Its basically how many 2 consecutive pairs from 6 elements you can get.

[last slate]

6 spots ib total

bride n groom take two

four remain

yes 4 spots

[versed quiver]

5 spots for the groomsmen and bridesmaids (black lines), think of how many spots that creates for the bride-groom block (red arrows)

[cerulean oyster]

We are talking about how many ways you can place the "block"

[versed quiver]

there should be another arrow on the far right lol mb

[last slate]

wdym by block

[cerulean oyster]

The "block" is essentially the pair of groom and bride, we dont really care in what order they come

How many ways you can fit two consecutive spaces?

[last slate]

5 ways

[cerulean oyster]

Well, now, for every placement of this "block" (which is the pair groom-bride, or bride-groom)

We have 4 free elements which we can arrange however we like

[last slate]

yes

[cerulean oyster]

what would that be?

[last slate]

4 · 3 · 2 · 1

[cerulean oyster]

so, 4!

[last slate]

Yes

[cerulean oyster]

Now, remember the "block" can be arranged as Bride-Groom or Groom-Bride

so we get an extra
*2

Finally, we just multiply all together
5 * 4! * 2
which is
5! * 2

[last slate]

you lost me here

[cerulean oyster]

lemme write an example

[last slate]

im thinking of the cartesian product

im trying to see how i can do it here

[cerulean oyster]

Imagine we place our "block" there

Which remember, is just the pairing of bride and groom

For this particular placement, the bridesmaid and groomsmen can arrange in 4! ways

But beyond the 4! ways to place the maids and men, you get this two for Bride-Groom and Groom-Bride

They are "similar" situations, but where the placement of the groom and bride get switched, but they still are beside eachother

[last slate]

can i think of all this via cartesian product

[cerulean oyster]

i dont really think so.

Not without having to do a huge one

the concept of permutations usually is easier

[midnight plank]

@last slate Has your question been resolved?

[last slate]

Ohhh I think i got it

You can arrange the bride and groom in 2 different ways as they take 2 spots so thats

| GroomBrideSet x BrideGroomSet | = 2

And those 2 arrangements can be arranged in 5 different ways so 2 * 5 = 10

Also we know that for the remaining ppl can be arranged in 4! Ways so ∴ there exists 10 4! Ways to rearrange the other guests and that simultaneously takes into account the amount of times the bride and groom can be rearranged next to one another

correct me if im wrong @cerulean oyster

i think the way im thinking of multiplying 10 and 4! Is a bit goofy

like there are 4! Ways to rearrange the groomsmen and the bridesmaids

and 10 ways to rearrange the bride and groom respectively

[midnight plank]

@last slate Has your question been resolved?

[midnight plank]

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[zealous schooner]

Can we have a random variable $X$ such that $\mathbb E\left[e^{tX}\right]=+\infty$ for all $t\ne 0$?

[grand pond]

kheer257

[zealous schooner]

I don't think X can be integer valued since [ \mathbb E\left[e^{tX}\right] = \sum_{n\ge 0} e^{nt}\mathbb P(X=n) \le \sum_{n\ge 0}\mathbb P(X=n) = 1 ] for all $t<0$

[grand pond]

kheer257

[zealous schooner]

Does this just mean X can't be discrete at all

[neon bone]

I think Cauchy Distribution is great example for that

[zealous schooner]

If $X$ is continuous with pdf $f_X$ then we want [ \int_{-\infty}^\infty e^{tx} f_X(x) \dd{x} ] to diverge to $+\infty$ for all $t\ne 0$

[grand pond]

kheer257

[zealous schooner]

What's that?