help-49

[odd lagoon]

idk if that is right

[fathom onyx]

(assuming relabelling i and j) can you describe what has happened to the vectors?

i.e. geometrically?

[odd lagoon]

no i dont understand what i have sketched anymore after you corrected me

[fathom onyx]

i (usually pointing right) has been moved to where j usually points (up)

and j now points left

[odd lagoon]

but i thought j did point vertically

and since i moved there

i thought i pointed vertically

[fathom onyx]

yh I've given the correction for you

[odd lagoon]

not in general but for this

[odd lagoon]

but isnt that

what i wrote?

[fathom onyx]

yh that's what I'm saying

[fathom onyx]

I'm giving you a follow-up question here

[odd lagoon]

but how is it a correction

if i did it right

[fathom onyx]

It's just your labels "i" and "j"

[odd lagoon]

is this right yes or no

[fathom onyx]

if you swap "i" and "j" (the letters there) then yes

[odd lagoon]

no i dont understand that still

because you want me to point i horisontally?

am i correct

in saying that

[fathom onyx]

Okay for the sake of clarity - let's say the original default vectors are i (pointing right) and j (pointing up)
and that the new vectors are T(i) (where i gets moved) and T(j) (where j gets moved)

[odd lagoon]

those are the default

basis vectors

[fathom onyx]

yh

[odd lagoon]

the identity matrix

and nothing is wrong?

[fathom onyx]

yh

Now, if I call T(i) and T(j) what those i and j vectors become if I multiply them by this matrix, where do I draw them?

[shadow schooner]

the identity matrix is associated to a function of R^2 to R^2 that does nothing to i and j

it packages nicely where they end up

[odd lagoon]

how do i draw i-hat vertically

on the j-axis

wouldn't that now be the i-axis

since i is now the vertical part

and i have a different horisontal part

[fathom onyx]

You're drawing what i becomes i.e. T(i)

Which in this case also equals j

[odd lagoon]

i think why im so flabberghastingly confused is because i didnt even realise it was becoming anything

[fathom onyx]

The axes do not change

[odd lagoon]

i dont even know what you mean by that

[fathom onyx]

I mean, the clue's in the name "linear __transform__ation"

[odd lagoon]

well i dont see it yet

i want to see

whatever it is you guys are seeing

[shadow schooner]

another way to visualize it

[odd lagoon]

but it is not making any sense to me

[odd lagoon]

isnt this

what i did?

[odd lagoon]

left part of the picture

is the basis vectors

[fathom onyx]

It is, I just needed you to update your labels

[odd lagoon]

or identity matrix

[odd lagoon]

wait but

where do the label stay the same

[fathom onyx]

But since you'd thought that meant you were wrong, I tried to re-approach this question

[odd lagoon]

does*

but i dont understand why i need to change the labels

so i am wrong

i dont think i am wrong i know i am

[shadow schooner]

fundamentally, any matrix represent a function that takes a vector (a,b) and sends it to another vector in the right side of my drawing

hence the T pointing from keft to right

[fathom onyx]

(je vais m'unvivre (/jk) et revenir, je te laisserai l'aider)

[shadow schooner]

comment t’as guess que je parle français, c’est inatendu

[fathom onyx]

bah j'ai pensé "touché" comme "blessé"

[odd lagoon]

im sorry

.close

[midnight plank]

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[shadow schooner]

no!

[fathom onyx]

.!etosih

.reopen

[shadow schooner]

ok i mean we tryed our best

[fathom onyx]

yeah ig

curse of switching to french for a bit

[shadow schooner]

nah c’est vraiment juste une sorte d’écureuil volant un polatouche

[fathom onyx]

ooohhhh :P

[shadow schooner]

clairement il s’est senti exclu, but i mean je me rapelle cope en algèbre lin aussi, ça orend du recul et plusieurs répétition pour bien apprécier cette merde imo

[fathom onyx]

faut qu'j'dorme quand même donc à la prochaine

[shadow schooner]

cya du québec

moi je vais bouffer

[midnight plank]

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[abstract birch]

can anyone direct me, what r the general steps?

[abstract birch]

first i need to prove its bijective right

only then its invertible?

[sharp coral]

"invertible" is equivalent to "bijective", yes

[abstract birch]

alright cool

so i just need to show bijectivity (is that a word)

and then find the inverse

then im done

[sharp coral]

yes

[abstract birch]

thank

.close

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[bitter tartan]

I need help

[bitter tartan]

am i using yu1 yu2 yu3 and finding dot products and norm

to start

[midnight plank]

@bitter tartan Has your question been resolved?

[midnight plank]

@bitter tartan Has your question been resolved?

[sharp coral]

do you know how to project a vector onto a subspace?

[midnight plank]

@bitter tartan Has your question been resolved?

[bitter tartan]

uhh

maybe

[sharp coral]

well to start with, do you know how to project a vector onto another vector?

[bitter tartan]

yes

[sharp coral]

if you have an orthogonal basis for a subspace, the projection onto that subspace is the sum of projections onto each basis vector

[midnight plank]

@bitter tartan Has your question been resolved?

[midnight plank]

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[sick cosmos]

.

#help-28

[sick cosmos]

Initially, only one elf is in a good mood (day 0). Every day, a cheerful elf has a chance to make a grumpy friend smile (and thus become cheerful) with probability p (not equal to 0). The elves form a complete graph, meaning every elf is friends with every other elf.

A grumpy elf becomes cheerful if at least one cheerful friend smiles at them. Importantly, an elf in a good mood can smile at multiple grumpy elves in the same day.

We define τ as the total number of days required until all n elves are cheerful. Compute E(τ).

I found some weird sum at the end by considering the variable giving the nb of days to go from k-1 cheerful elves to k cheerful elves, but Python simulations give me another result.
For p=1/2 and n=10, simulation gives 2.07 and formula gives 9 and decimals...

[last slate]

You can do this with binomial distribution or summation

Think with summation it should be smth like this

first day one elf is cheerful, they can make one of the other n-1 elves cheerful

with probability 1/2 or p

the expected number of days to go from 1 cheerful elf to 2 is 1/p

then you apply the same logic with going from 2 to 3 cheerful elves

and you generalize with summation

<@&286206848099549185> someone correct me if im wrong im not too sure lol

[runic hamlet]

an elf doesnt just smile at one other elf. so most of the time you will not go from k-1 happy elves to k happy elves

[sick cosmos]

how did you find this

[last slate]

my barely functional noggin

[runic hamlet]

in the python example, on day 1 approximately 5 elves will be happy. and then its not a surprise that on day 2 most elves will be happy. every one of the last 5 elves will be smiled at by 5 happy elves

[sick cosmos]

found that too

[sick cosmos]

i did that but it doesnt work

found this at the end $$E(\tau)=\sum_{k=2}^{n}\frac{1}{1-(1-p)^{(n-k+1)(k-1)}}$$

[grand pond]

Francis J. Underwood

[sick cosmos]

yh

[last slate]

actually.

it makes sense that it'd be 2.something

[sick cosmos]

uh

[last slate]

my summation went wrong

[last slate]

probability is 1/2

[sick cosmos]

no p belongs to (0,1]

[last slate]

wait so your probability isn't pre-determined?

[sick cosmos]

no

you dont know it

its a general formula for p

[sick cosmos]

the problem with this sum is that p=1 gives expected time as n-1

it should be 1

[last slate]

i think the python simulation shows it this way because of simultaneous smiles

in this model

[sick cosmos]

yes

but I dont understand why our method doesnt count that

bcs if there isi a simultaneous smile then just time to go from k-1 to k is 0

and we dont care about that

because the proobability of getting 0 cancels out in the expected time

[last slate]

i want to point out, i think setting the probability at 0.5 just makes it easier to simulate

so it's ok to pick a probability here since you can't rly find it i think.

[sick cosmos]

yh

[last slate]

i think the summation is wrong because it's

summing over the expected time to go from k-1 to k elves

it assumes one new elf per day

which is why it's closed to your sample size

being 10

it's over-simplified

[sick cosmos]

no bcs it can take 0 as value

but the problem is summing it gives a minimum time of n-1

[last slate]

ah

i see what you mean

yeah honestly im as confused as you are now

😔

[sick cosmos]

yh me too lol

[last slate]

@sick cosmos have you tried doing this iwth

markov chains

it might work

do you know how they work?

if it's not within your curriculum's scope that might just be unlucky tho

:(

[sick cosmos]

no idk

but you can explain

[last slate]

I unfortunately gotta dip atm I'm rly sorry

But i suggest youtubing it

[sick cosmos]

ok alr

but how could i apply it

[last slate]

i think you'll need to do it as a discrete-time markov chain where each state represents the number of cheerful elves within a day

you'll have your states like

S = {1,2,3,...,n}

and state n should be called "absorbing," meaning once all elves are cheerful, they stay cheerful

and the transitions from state k, you gotta compute the probability of k+j, for all possible j ∈ [0, n - k]

[sick cosmos]

alright

can I dm you for questions?

[last slate]

then X_t ∈ {1,2,....,n} should be the number of cheerful elves at time t

you'll need to calculate the transitional probabilities from there

then build the transition matrix and extract the submatrix Q

then you compute the fundamental matrix

then you finally get the expected time to absorption

[last slate]

.

this

[sick cosmos]

alr

[last slate]

but btw

this process is tedious as hell

don't do it for large sample sizes

you can use smth called a branching process or epidemic spread model too

they SHOULD be easier

[last slate]

i mean we got this channel lol

ok i gotta go now le wife shoutin at me cya

[sick cosmos]

if ever it closes

[sick cosmos]

alr

lol

[last slate]

@sick cosmos i'll solve it for ya in the car and get back to you in 30 mins

ttyl

[sick cosmos]

ok thx

[last slate]

@sick cosmos ok i managed to get it

[sick cosmos]

alr

can you send it pls

[last slate]

stochastic epidemic spread model but here's my answer

formula for estimating how many new elves become cheerful every day should be this

delta k = (n-k) * [1 - (1-p)^k]

k is the number of cheerful elves today
and delta k is number of new cheerful elves tomorrow
n: total elves:
p: probability a cheerful elf makes a grumpy elf smile

you set k = 1 t = 0

and while k < n

calculate how many new elves become cheerful

delta k = (n-k) * [1-(1-p)^k]

and you update the number of cheerful elves

k = k + delta k

and increase time

t = t + 1

then you stop when k >= n

the value of t is your estimated number of days until all elves are cheerful and shi and that'll be your E(T)

so like let's say day 0 yea

cheerful elves (k) = 1

new cheerful ones delta k

9*(1-0.5) = 4.5

then your total cheerful ones are 1 + 4.5 = 5.5

you do that for another and it's 9.9

aaaaaaaand that should be 2 days

approximately

cause you're at 9.9 elves at 2 days

so you just gotta round the elves up at that point

and call it a day

hope it helps?

[sick cosmos]

oh alr

but I dont understand how can I find E(T)

[last slate]

i mean

you just simulated it

took 2 days for the elves to be at

9.9

so

E(T) is just

approximately 2 days

[sick cosmos]

but it should depend of n and p?

[last slate]

yes this is assuming that

n = 10

and p = 1/2

[sick cosmos]

oh but then can I have a general formula

[last slate]

ok i can't do latex let me write it down on paint lol

one sec

[sick cosmos]

np

[last slate]

that should be E[T]

[sick cosmos]

alr

and so thats found how?

[last slate]

(n-k) is the number of grumpy elves that still gotta be cheered up

[sick cosmos]

yh but why did you inverse the delta k

is it a geometric law?

[last slate]

well you want to know how long it takes for all elves to become cheerful starting from 1 elf

and each day

elves spread their cheer to the others

you're calculating the numebr of new cheerful elves that join in each day

[sick cosmos]

yh

[last slate]

delta k represents the number of new cheerful elves that get infected with that cheer each new day

and its formula is what we got the inverse of

now like

the important point is that

delta k tells u how many cheerful elves will appear in one day

but you want to know how long it takes to go from k-1 cheerful elves to k elves

the inverse of k gives you the expected number of days to get those new cheerful elves

bc if for example let's say delta k is 5, that means 5 new elves will join today

so to go from k-1 elves to k elves, it would take 1 day if delta k is 5

but if delta k is 10 then it would take half a day!

[sick cosmos]

huh

[last slate]

so the time it takes to get to the next k-th elf is inverse proportional to how many new cheerful elves you expect to get on average

in one day

[sick cosmos]

ohhh

I think I understood

[last slate]

nice!

basically

you invert it becasue it's a rate of growth

it turns the rate into the time it takes to reach the next step

[sick cosmos]

yh alr

its like speed basically

[last slate]

exactly yeah pretty much

the speed of how quickly cheerful elves spread in this context specifically

and the inverse just gives you the time

just like you'd calculate travel time from speed and distance

ok i go do my groceries now you have a good day

🫡

[sick cosmos]

alr thanks

[sick cosmos]

wait but shouldnt delta k be equal to the number of grumpy elves divided by Tk where Tk is the time to get the delta k elves cheerful

so at the end we should end with smth like that $$E(T)=\frac{1}{1-(1-p)^{k}}$$

[grand pond]

Francis J. Underwood

[sick cosmos]

and the time to get to the next k-th elf will still be proportional to the average of delta k

[sick cosmos]

also shouldnt delta k be $(n-k)(1-(1-p)^{k(n-k)})$

[grand pond]

Francis J. Underwood

[sick cosmos]

because there is k cheerful elves that can make happy n-k grumpy elves

[last slate]

o yeah

same formula though lol

[last slate]

no

[sick cosmos]

oh yh wait no my bad

[last slate]

you got it?

[sick cosmos]

yh

[last slate]

i can explain why

[sick cosmos]

so this one is good?

[last slate]

no

now the model assumes that it's waiting for just one specific grumpy elf to become cheerful

you do not want your model to care which grumpy elf becomes cheerful first.

any grumpy elf becoming cheerful moves the model forward

they are independent

and each have a chance to become cheerful on any given day

[sick cosmos]

i dont understand

[last slate]

what do you not understand

[sick cosmos]

^

because then delta k is equal to n-k/Tk

and the n-k cancel

[last slate]

your logic assumes

all n-k grumpy elves become cheerful after

time T_k

yes?

that's not what we're doing here though

we only need one more cheerful elf to move from k to k+1

[sick cosmos]

no

wait yhh youre right then

[last slate]

hopefully that makes sense yeah

[sick cosmos]

why isnt the n-k at the exponent

evry elf independantly makes other elves cheerful

@last slate

[last slate]

cause the exponents count how many independent attempts

are made on a single grumpy elf

there are k cheerful elves

each trying to cheer up the same grumpy mf

[sick cosmos]

oh ok

alr

i think ive got it

.close

[midnight plank]

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[last slate]

🫡

[sick cosmos]

wait there is a problem

.reopen

[midnight plank]

✅

[sick cosmos]

for p=1 we dont have 1 as expected time

but all the elves would make grumpy elves cheerful right?

[last slate]

at p = 1, you don't get all elves cheerful in one day

you get exponential spread

smth like a chain reaction basically

[sick cosmos]

why?

but they are all friends

[last slate]

the formula gives approximately log(n)

[sick cosmos]

so if the first elf is happy

then the next day all the elves become happy

@last slate

[last slate]

no one day 0, there's only 1 cheerful elf

and on day 1, he can make some grumpy elves cheerful but not all, because the others are still grumpy

spreads quickly but not instantaneous

[sick cosmos]

he can

why not

p=1

so he can make anyone cheerful

[last slate]

because it takes time for it to spread

[sick cosmos]

huh

i dont understand tbh

like its a complete graph

[last slate]

it's a perfect success rate yeah

but you're not doubling it all at once lol

[sick cosmos]

huh

[last slate]

grows exponentially on a graph

but it's still sequential

[sick cosmos]

but p=1 gives perfect success rate

every grumpy elf can become happy

the probability that a cheerful elf doesnt make a grumpy elf cheerful is null

[last slate]

the model in of itself is sequential

[sick cosmos]

why

[last slate]

even with

p = 1

you get exponential growth

[sick cosmos]

huh

but why

[last slate]

because in the model

cheerful elves can't all act the same time.

cheerful elf #2 from day 0

can only start acting on day 1

[sick cosmos]

oh ight yh

but he can mzke multiple elves hzppy

[last slate]

did i misunderstand the question or something

let me read

@sick cosmos i gotta say btw, you ran a python simulation but is this from a uni book or smth?

you gotta check the answer

cause the method i used ran the same logic as the simulation im pretty sure

but im starting to get confused now

[sick cosmos]

from a book but no solution

[midnight plank]

@sick cosmos Has your question been resolved?

[sick cosmos]

@last slate

[last slate]

im out of ideas bro sorry lol

<@&286206848099549185>

if someone wants to revise

the chat

it's a lot though

[midnight plank]

@sick cosmos Has your question been resolved?

[wary trail]

is this the question?

[midnight plank]

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[sick cosmos]

yes

[midnight plank]

@sick cosmos Has your question been resolved?

[midnight plank]

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[sick cosmos]

.reopen

[midnight plank]

✅

[sick cosmos]

Initially, only one elf is in a good mood (day 0). Every day, a cheerful elf has a chance to make a grumpy friend smile (and thus become cheerful) with probability p (not equal to 0). The elves form a complete graph, meaning every elf is friends with every other elf.

A grumpy elf becomes cheerful if at least one cheerful friend smiles at them. Importantly, an elf in a good mood can smile at multiple grumpy elves in the same day.

We define τ as the total number of days required until all n elves are cheerful. Compute E(τ).

I found some weird sum at the end by considering the variable giving the nb of days to go from k-1 cheerful elves to k cheerful elves, but Python simulations give me another result.
For p=1/2 and n=10, simulation gives 2.07 and formula gives 9 and decimals...

Scroll up for precedent reasonings

[wary trail]

So for example, m cheerful elves gives 1-q^m probability for each elf to become cheerful?

[midnight plank]

@sick cosmos Has your question been resolved?

[midnight plank]

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[river rapids]

Which rule do I use

-10+2x or 10-2x

[thin tree]

,rccw

[lyric charm]

when x goes to 3 from above, what region are you in? x<5 or x>5?

[thin tree]

3.001 is >5 or <5

[river rapids]

X<5

X is less than 5

Right?

[thin tree]

I think yes

[lyric charm]

⚠️ ⚠️

[lyric charm]

lowercase x but yes

[thin tree]

What ann

Why

[lyric charm]

so you use the rule that applies when x<5

[lyric charm]

what if the breaking point were 3.0005

and not 5.0

[thin tree]

Like just an example

[river rapids]

Whyd my teacher take the rule when x is bigger than 5

He used 10-2x

[thin tree]

Try sending the full question

[river rapids]

,rcww

[river rapids]

It's tedious

[thin tree]

Hmm you didnt open mod

[river rapids]

I didn't what

[thin tree]

10-2x if x>3 and -(10-2x) is x>5?

[river rapids]

10-2x if x>5 and -(10-2x) is x>5?

Like that

[thin tree]

no

[river rapids]

Ẁĥÿ

[thin tree]

So

It's not 10-2x if x>5

3<x<=5 it's 10-2x

[river rapids]

Yes

Ohhhh

I see my mistake

[thin tree]

You were trying for 3+

[river rapids]

I didn't take the negative into acount

[thin tree]

Ok

[river rapids]

When redifinfg absolute value for 10-2x

I see my mistake

Tha ks

.close

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[sharp wave]

how do i start? im clueless

[lethal path]

you can rule out some options pretty quickly through $(1 + 40x)^{1/60} \approx (1 + 40x/60)$ and $(1 + 18x)^{1/54} \approx (1 + 18x/54)$

[grand pond]

south

[sharp wave]

This is only valid for n<<<1 right? n being 18 and 40 here?

[lethal path]

yeah so you would need to realise that the linear approximations would be an upper bound

[sharp wave]

how would it be an upper bound? 😭

it would be greater than the value near 0 right?

[sick cosmos]

no
It is 1-q^{m(n-m)}

[wary trail]

why (m(n-m))?

[midnight plank]

@sharp wave Has your question been resolved?

[midnight plank]

@sharp wave Has your question been resolved?

[shadow schooner]

finding upper bounds seems to be a good idea

you can exploit the monotonicity of the function at play and of the definite integral maybe

[sick cosmos]

because each of the m cheerful elves have n-m grumpy friends

[wary trail]

oh my question was for each elf which has m cheerful friends, there is (...) chance that it becomes cheerful next iteration

[shadow schooner]

using this i'm pretty confident A)and D) are true

By extension the upper bound I found also shows B) is false

for C) you would need a lower bound imo

[midnight plank]

@sharp wave Has your question been resolved?

[sharp wave]

sorry for the late response, could you please elaborate for A and D?

[frank iris]

Could one just solve the integral?

It would be a simple u-substitution.

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[autumn folio]

find the area between g(x), h(x), x = e and x = e^2, this is what I've done so far, got stuck here...

[autumn folio]

my draft seems to be correct

<@&286206848099549185>

summary page

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[silent torrent]

can someone help me ques 3.3c

[silent torrent]

a and b dont relevant to c

i have only fomula P(X<x)=1-e^(lambda * x) in my mind to solve this problem

[midnight plank]

@silent torrent Has your question been resolved?

[silent torrent]

<@&286206848099549185>

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@silent torrent Has your question been resolved?

[placid spoke]

the exponential distribution is memoryless, so him crossing the road immediately after a car passes doesn't matter

and you need him to take at least "t" seconds to cross the road

and you want that probability to be less than 0.2

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[sick cosmos]

yes then ur right

[sick cosmos]

Initially, only one elf is in a good mood (day 0). Every day, a cheerful elf has a chance to make a grumpy friend smile (and thus become cheerful) with probability p (not equal to 0). The elves form a complete graph, meaning every elf is friends with every other elf.

A grumpy elf becomes cheerful if at least one cheerful friend smiles at them. Importantly, an elf in a good mood can smile at multiple grumpy elves in the same day.

We define τ as the total number of days required until all n elves are cheerful. Compute E(τ).

I found some weird sum at the end by considering the variable giving the nb of days to go from k-1 cheerful elves to k cheerful elves, but Python simulations give me another result.
For p=1/2 and n=10, simulation gives 2.07 and formula gives 9 and decimals...

#help-49 message precedent reasonings

[midnight plank]

@sick cosmos Has your question been resolved?

[sick cosmos]

<@&286206848099549185>

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@sick cosmos Has your question been resolved?

[carmine sigil]

We can make a Markov chain and then diagonalize to find a formula for the probability that all elves are happy by the nth day, then take the first differences, and sum over them all to find the expected value. But that might wind up being tremendously complicated

And I don't think that it will generalize nicely

[midnight plank]

@sick cosmos Has your question been resolved?

[carmine sigil]

There's probably an argument from the linearity of expectations to be made here.

If n elves, k happy elves, p probability of converting, then on any day there is an expected value of (n-k) (1-(1-p)^k) elves that convert from grumpy to happy.

Thus we can define a recursive formula from this, which won't give us any exact values, but should be significantly easier to handle.

@sick cosmos ^

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[inland patio]

Suppose two linear operators $f,g$ are simultaneously diagonalizable (meaning there exists a basis such that the matrix representation of these operators is diagonal), then they are supposed to commute. Obviously diagonal matrices commute, but how do I show $fg=gf$?

[grand pond]

psie

[inland patio]

I guess I need to show they agree on the basis vectors that make their matrix representation diagonal?

This may seem basic, but I struggle with how to formalize this.

[runic hamlet]

well what is f(g(b1)) where b1 is one of those basis vectors

[inland patio]

hmm, g(b1)=a1b1+a2b2+...+anbn, and (a1,a2,...,an) is the first column of the matrix representation of g. Since it is diagonal, we actually have (a1,0,...,0) and so g(b1)=a1b1. Thus f(g(b1))=f(a1b1)=a1c1b1 by the same reasoning, where c1 is the 1,1 entry of the diagonal matrix representation of f. Likewise we obtain g(f(b1))=a1c1b1, so they agree. Thank you very much.

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[earnest forge]

I am required to solve this using L'Hopital's rule. It's going very long and ugly. Should I give up?

[earnest forge]

I have already solved it using a different method and got the answer by the way.

[polar star]

what was your answer?

[pearl hull]

Show your work

[polar star]

for lhopital rule, multiply by the conjugate on numerator and denom

but im guessing u used binomial expansion

[pearl hull]

There should be an overlap between using 🏥 and your method

[earnest forge]

That's what I did

Multiplied by conjugate and then divided both numerator and denominator by x

[earnest forge]

1

,w graph sqrt(x^2+x+1)-sqrt(x^2-x)

[grand pond]

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[earnest forge]

That's correct. y=1 is the horizontal asymptote.

[polar star]

yes use lhopital now

after simplifying

[earnest forge]

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[molten socket]

\lim_{x \to -\infty} \sqrt{\frac{x}{4x-3} }

[molten socket]

how does latex work

[keen herald]

$\lim_{x \to -\infty} \sqrt{\frac{x}{4x-3} }$

[molten socket]

limit x to negative infinity

[grand pond]

al-jebruh

[molten socket]

cool thanks

[keen herald]

Put it in $$

[molten socket]

so im gonna divide numerator and denominator by sqrt x

[brittle crater]

highest power strategy?

[molten socket]

but if sqrt x approaches a negative number, shouldnt sqrt x be imaginary so i cant actually do so

whats highest power strategy

[brittle crater]

u use the highest power of x from the denominator and divide the top n bottom by that so

$\lim_{x \to -\infty} \sqrt{\frac{x/x}{4x/x-3/x} }$

[grand pond]

a

[brittle crater]

srry idk how to use latex

[molten socket]

ok but in this process am i not dividing both numerator and denominator by sqrt x

where x appraoches negative infinity

so sqrt x is imaginary

or does limit not work like that

[brittle crater]

why are u dividing it by sqrt x

[brittle crater]

$\lim_{x \to -\infty} \sqrt{\frac{1}{4-3/x} }$

[grand pond]

a

[polar star]

just factor out x from numerator and denom so you have

sqrt(1/(4-3/x))

[molten socket]

oh ok

right

[brittle crater]

ye then 3/x lim goes to neg infinity it’ll b 0

[brittle crater]

$\lim_{x \to -\infty} \sqrt{\frac{1}{4} }$

[grand pond]

a

[molten socket]

alr ty

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[spark nebula]

Hello,
I need to find some notes or papers about hermite interpolation, i cant find any,
Do u guys have a reference or something which i can use to learn it’s mathematical formulation

[runic hamlet]

wikipedia?

you cant come in here with "I cant find anything" when there are very obvious things to find. presumably you excluded wikipedia? if so, you need to state why

[spark nebula]

So complexe, to be more specific i have to do a presentation where i explain this interpretation but what is found in wikipedia is just complex and comprehensive not like the notes you find in books for example

[runic hamlet]

the wikipedia article isnt that complex

a lot of books will be more complex if I had to guess

but I dont know any specific resources I can recommend

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[silent torrent]

can someone help me ques 1.29

[hollow helm]

What condition should be satisfied by the three segments for them to make a triangle?

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[umbral tusk]

subject: Decision 1 fm
topic: route inspection/chinese postman

ive found the 4 odd nodes and found that the arc combo with the least weight was 20.5 which was EG and BC. so i put that those arcs are the ones that need to be traversed twice but it wasnt the answer in the mark scheme, could someone explain to me the arcs that need to be traversed twice??

[midnight plank]

@umbral tusk Has your question been resolved?

[tulip moon]

what does the mark scheme say

I took decision 1 a few years ago and honestly the mark schemes are terrible

like they're wrong a good 20% of the time

[umbral tusk]

really? so far I've never encountered a d1 mark scheme that's wrong, but the textbooks are horrible especially for core pure 1

[tulip moon]

discrete 2 was terrible

but discrete 1 definitely had some mistakes as well iirc

[umbral tusk]

oooh u took both decision modules? what's the difference in content between them

[tulip moon]

just completely different topics

discrete 2 is a lot more painful

[umbral tusk]

oh 😭

[tulip moon]

none of it is particularly difficult it's just remembering a bunch of algorithms

core pure 2 imo is easier than core pure 1 as well

as long as you get good at calculus

oh wait no i'm stupid it's saying BA AC cause that's BC

you're correct you've just misread the mark scheme

[umbral tusk]

Ohhh 😭😭 i was so confused where the BA and AC came from but now it makes sense they've just split it

tysmmm!

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[restive pivot]

Is an eulerian cycle the same thing as a circuit?

[summer geode]

no

[restive pivot]

These 5 definitions are accurate right? @summer geode

[summer geode]

im just looking at google rn but yeah looks right

[restive pivot]

In that case isn't this definition of a circuit

The same as that of an Eulerian cycle

[summer geode]

a circiut doesnt have he extra condition that each edge is traversed

eulerian cycle does

[restive pivot]

Ah I see

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[wary trail]

Can i have a check of this please

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@wary trail Has your question been resolved?

[wary trail]

,w solve y"+y'+y=0

[grand pond]

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[sick cosmos]

found this, is it correct?

[sick cosmos]

for $p \ne 1$

[grand pond]

Francis J. Underwood

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[twilit sentinel]

How come does concatenation of two languages in P also belong to P?

[twilit sentinel]

Won't we have to check for all possible splits

Spiralling into NP?

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[remote spade]

Proof 3rd equation of motion(v^2 = u^2 - as^2)

[tawdry laurel]

What have you tried

@remote spade

[remote spade]

Elimating t from equation of motion 1 and 2

Eliminating

But i get wrong answer

[sudden yacht]

By the way, are you sure it isn't
v² = u² - 2a•s?

Otherwise the dimensions do not agree

[remote spade]

I just want help in simplyifing this

Step 2 . Substitute (t) in the second equation of motion (s=u(\frac{v-u}{a})+\frac{1}{2}a(\frac{v-u}{a})^{2}) (s=\frac{uv-u^{2}}{a}+\frac{1}{2}a\frac{(v-u)^{2}}{a^{2}}) (s=\frac{uv-u^{2}}{a}+\frac{(v-u)^{2}}{2a}) (s=\frac{2uv-2u^{2}+v^{2}-2uv+u^{2}}{2a}) (s=\frac{v^{2}-u^{2}}{2a})

[grand pond]

Prime

[remote spade]

Opps

Yeah

This

[tawdry laurel]

Ah

[remote spade]

You see those last 2 equations

[tawdry laurel]

Indeed

[remote spade]

I cant understand that

[remote spade]

2uv cut by 2uv
then there is left
(2u^2 + v^2 + u^2)/2a

Then how it turned into v^2 - u^2/2a?

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[proven cradle]

hi so i have been able to find a

[proven cradle]

but i dont get how to find b

i get this

[potent veldt]

https://mathleaks.com/study/kb/rule/tangent_and_intersected_chord_theorem

[proven cradle]

got it ty

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[dark path]

Only gcse question, but i get stuck after a few angles in and ended up with the wrong answer

[dark path]

I get that ABCD is a kite

and ABOD is quadrilateral

so BOD is 180 - y cus the tangents meet at 90 degrees to the radius

and then i went wrong some point after that

[tough shale]

If BOD is 180 - y, what would be BCD?

[dark path]

oh oops

half

90 - y/2

so would obc be thtat too

so 90-y/2 + 2x = 90

guys am i stupid i just solved that and got 180 - 2x = y

i screwed up wait

[tribal temple]

(Are you sure this is a kite, though?)

[dark path]

oh no

maube

[thin tree]

OBC is not congruent to ODC

[dark path]

OOPS

thanks guys

i keep messing up im failing my gcse

why do i keep getting -y/2 = -2x

[tribal temple]

How are you going about it?

(Also as another way you could start, is that within the circle, you have two isosceles triangles, though they aren’t the same - that allows you to “chase angles”)

[dark path]

90 - y/2 + 2x = 90
-y/2 + 2x = 0
-y/2 + -2x

=**

at end sorry

[dark path]

thats how i tried last time but i defimitelt messed up again

[tribal temple]

Where did you get the first line from?

[dark path]

uhmmmmm

because OH NO

I DID IT WRONG

okay give me a second

[dark path]

45 - y/2 + 2x = 90

[tribal temple]

(The coloured angles you should be able to get in terms of x quite easily )

[dark path]

purple is x right

and doc is 180 - 2x

180 - 2x + 180 - y = BOC

[tribal temple]

What about the green and blue angles?

[dark path]

green is uhm 90 - 2x

and blue is same

[tribal temple]

So you can also get BOC another way too

[dark path]

i got -4x

dod i ,ess up

omg

[tribal temple]

Should be +4x you get

[dark path]

OOSP

yes i see sorry

okay so what do i do next]

[tribal temple]

Well now we have angle BOC as 4x, and also angle DOC as 180 - 2x

What would angle BOD be?

[dark path]

180 + 2x

[tribal temple]

Are you sure?

[dark path]

wait how do i know whicgh way the angle is

i thought i added them together to get that like big one on outside of kite ABOD

[tribal temple]

Fair point I’m referring to the smaller option of the two

[dark path]

oohhhh

180 - 2x

[tribal temple]

Yep, then what do you get the angle that’s supposed to be y?

[dark path]

I SEE

tjank you so much

[tribal temple]

It’s been a pleasure you’ve gotten it in the end

[dark path]

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[rain wasp]

k = 0.71 per year
does this mean k = 0.71/t?

[potent veldt]

k = 0.71. The units are 1/year because k is multiplied by y(which is in mass) and (1 -y/m) (which is unitless), and dy/dt's units have to be mass per unit of time

[rain wasp]

ah i see

weird result

oh i subbed it wrong

should be good right

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[inland patio]

I'm just trying to comprehend a basic example of a Markov chain in my book. It helps when I think about this in terms of rolling a die. So the situation is that you either are in the suburbs or the city of a certain metropolitan area that remains constant over time; these are the two "states". If you live in the city, you roll die 1 with outcomes 'stay in the city' with probability 0.90 or 'move to suburbs' with probability 0.10. On the other hand, if you live in the suburbs, you roll die 2, with the same outcomes, but probabilities 0.98 of 'staying in the suburbs' and 0.02 to 'move to city'. Suppose we would like to find the probability of a certain person moving from city to city to suburb. Is this an application then of the law of total probability?

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[inland patio]

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[chilly adder]

Solve for $x$ and $y$:

$$\left(\sin^2x+\frac{1}{\sin^2x}\right)^2+\left(\cos^2x+\frac{1}{\cos^2x}\right)^2=12+\frac{1}{2}\sin y$$

[grand pond]

@chilly adder

[midnight plank]

@chilly adder Has your question been resolved?

[chilly adder]

<@&286206848099549185>

[crisp valley]

i'd say treat the two parts of each section being squared as a and b

then expand it using a^2+2ab+b^2, sub back in for both

then try writing cos in terms of sin

mix it back

badabing badaboom

[chilly adder]

i tried that and i got 8th degree expressions on the left

I got on the left:

$$\frac{\sin^8x+2\sin^4x+1}{\sin^4x}+\frac{\cos^8x+2\cos^4x+1}{\cos^4x}$$

writing $\cos^2x=1-\sin^2x$ seems tedious...

[grand pond]

@chilly adder

[chilly adder]

i thought of checking the boundedness on the left;

so i diffed the LHS and set it to 0, i got

$$\frac{1}{(1-u)^3}-\frac{1}{u^3}-1-2u=0$$

[grand pond]

@chilly adder

[chilly adder]

didnt seem like im going anywhere lol

[exotic stratus]

Have you tried using, $\left(a + \frac 1 a\right)^2 = a^2 + \frac 1 {a^2} + 2$?

[grand pond]

HitenTandon

HitenTandon

[midnight plank]

@chilly adder Has your question been resolved?

[chilly adder]

i did that too but what to do after that

[brisk iris]

What do you get after using this ?

[midnight plank]

@chilly adder Has your question been resolved?

[midnight plank]

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[dreamy lichen]

How'd I prove that DE is parallel to AB?

[dreamy lichen]

A is not necessarily 90°, E is midpoint of BC, AD bisects <A

an equivalent condition to prove would be hat EDP are collinear here (P is midpoint of AC, D lies on thales circle above AC)

[chrome vessel]

Extend CD to intersect AB at F

try and see if you can prove that P is the midpoint of AC with this

[dreamy lichen]

Wow

i just figured out exactly the same thing at exactly the same time lol

reflect C across AD

But yeah, that works

[last slate]

Damn telepathy

[dreamy lichen]

you figured out quite quickly though, me and ZED were stuck on this for half an hour already

if not more

[chrome vessel]

well you have angle bisector and a right triangle

its begging to be extended

[dreamy lichen]

once we extend it, we get this nice little triangle BC'C

and ED just connects the midpoints

and thus must be parallel to BC' and so BA

[chrome vessel]

yeah so parallel comes through

[dreamy lichen]

this was the og question btw

this was a damn good question, stealing it to my collection

[dreamy lichen]

thanks btw!

[last slate]

This zambak book has tones of them if you want you can check them out

[dreamy lichen]

Yeah, ill definitely take a look at them

thanks

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[chrome vessel]

np!

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[shut canyon]

S is a finite set of points in the plane,
△ is a non collinear triangle in S
∴ ∀S ∃△ ⊆ S [∀p ∈ S, p ∉ int(△)]

to prove, suppose the contrary:
¬(∀S ∃△ ⊆ S )[∀p ∈ S, p ∉ int(△)]) ≡
(∃S ∀△ ⊆ S )[¬(∀p ∈ S s.t. p ∉ int(△)])) ≡
(∃S ∀△ ⊆ S )[∃p ∈ S s.t. p ∈ int(△)]))

Then every triangle in S has a point in its interior, and then S is infinite. Which is a contradiction, therefore the original statement is true.

[shut canyon]

Is this good reasoning?

Oh S is in the plane.

[chrome vessel]

justify the last statement: if every triangle in S has a point in its interior, then S is infijite.

[shut canyon]

I am trying to rephrase the conclusion. It is , in other words there is a set in the plane, in which for every triangle there is a point in it's intererior. So any triangle we select, there is a point inside it, which means there is another triangle that can be made in that triangle, and then there is another one inside that, and so on and so forth.

[chrome vessel]

yeah but the justification should be a little bit more formal. You have to use the non-collinearity condition, and use your argument right here to construct distinct points in S (and prove that they are distinct)

[midnight plank]

@shut canyon Has your question been resolved?

[shut canyon]

I'm thinking about it, thank you for your help.

I think adding \△ shows p is distinct from the vertices of the selected triangle ∀p ∈ S\△, p ∉ int(△)

Thank you @chrome vessel

[chrome vessel]

I don't quite understand what you mean here

[shut canyon]

¬(∀S ∃△ ⊆ S )[∀p ∈ S\△, p ∉ int(△)]) ≡
(∃S ∀△ ⊆ S )[¬(∀p ∈ S\△ s.t. p ∉ int(△)])) ≡
(∃S ∀△ ⊆ S )[∃p ∈ S\△ s.t. p ∈ int(△)])) (1)

Then there is a triangle (a,b,p), a,b ∈ △, a ≠ b, and by (1) there is a point p' ∈ S\(a,b,p) s.t. p' ∈ int(a,b,p). This is an infinite process, that must stop if S is finite. Contradiction!

[shut canyon]

I mean p is in S minus the vertices of the triangle

[chrome vessel]

Ok you do the same process on (a,b,p') to get p'', and so on and so forth, but you have to show that p'' is not the same as p. And when the process continues, you have to show that p''' is not the same as p' and p, etc

[shut canyon]

∃p' ≠ p ∈ S\(a,b,p)

[chrome vessel]

but im talking about $p''\neq p'\in S\setminus (a,b,p')$

[grand pond]

qwertytrewq

[chrome vessel]

just from this, it does not say that $p''\neq p$

[grand pond]

qwertytrewq

[shut canyon]

Okay

[chrome vessel]

so a little bit of justification is needed

[shut canyon]

I don't see it yet, but I sort of see what you mean, perhaps it can be shown with mathematical induction.

[chrome vessel]

that is, maybe the process of choosing these points eventually repeats

[chrome vessel]

hint: $int((a,b,p))$ does not contain $a,b,p$, and $int((a,b,p'))\subseteq int((a,b,p))$

[grand pond]

qwertytrewq

[chrome vessel]

and yes, induction should work

[shut canyon]

Okay, thank you so much @chrome vessel . This is still a bit like brain contortion for me.

[midnight plank]

@shut canyon Has your question been resolved?

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[night gyro]

what kind of answer is this how do i make sure i get this type of questions correct since there so random?

[sharp coral]

you know that usually there are a lot of children at the pool on a hot day, so to explain the outlier you just have to give a plausible reason why there are not very many children at the pool on a particular hot day

[night gyro]

i said the temperature could have been hotter

the answer tht they gave i would hv never got tht

[sharp coral]

well the temperature is one of the hottest on the chart, so that doesn't really make sense as an answer

it doesn't need to be a math exam specifically, just some reason why children would choose not to go to the pool on a hot day

[night gyro]

most of the children had a illness that day

how abt tht

[midnight plank]

@night gyro Has your question been resolved?

[sharp coral]

sure, that works

[tribal temple]

(usually a lot of leeway is given for these, and any "reasonable" answer would be accepted for them, you'll see the markschemes will allow for quite a range of responses)

[midnight plank]

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[last slate]

shortcut for problems of this form? nth derivative of a fraction

[last slate]

like if you have $\frac{\partial ^{n}}{\partial z^{m}}\frac{f(z)}}{g(z)}$

$\frac{\partial ^{n}}{\partial z^{n}}\frac{f(z)}{g(z)^{m}}$

[grand pond]

Tibbs.

[prisma axle]

I'm sure you could write down some sort of recurrence relation for this, but generally you just crank the algebra wheel. Without knowing the specific functions you don't get a whole lot of utility out of it

[last slate]

i saw somewhere that something does exist, but i dont know where. something like princeton guide to advanced physics, or a web page. saw it in passing so i cant help much more than that

[prisma axle]

Yeah it most definitely does, I've seen something like it when dealing with Bernoulli polynomials. But again it's just an exercise in wheel-cranking

[last slate]

https://www.physicsforums.com/threads/finding-the-nth-derivative-of-a-fraction.305175/ there's this. maybe it's a different way of writing what i've already seen

[last slate]

all good. thank you.
i shall crank those wheels then lol

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[midnight plank]

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[last slate]

do we derive or integrate

[golden comet]

well think about it

if you were to draw a velocity-time graph, how do you get the acceleration-time graph?

[last slate]

delta v / t

[golden comet]

do you solve for the area under the graph or the slope of the graph?

[last slate]

slope

[golden comet]

therefore do you derive or integrate for slope?

[last slate]

derive

i lovw u Mr madmod

[golden comet]

exactly

[last slate]

thank u

[golden comet]

np

[last slate]

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[last slate]

integrate!?

wrong way

you want to go to acceleration. which is the time differential of the velocity

[midnight plank]

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[radiant roost]

there's a hint that says assume a >= b >= c and show a | bc
then bound lcm(a,b,c) and show lcm(a,b,c)=bc

i see that lcm(a,b,c) | bc so <= bc but how do i show >= bc?

[midnight plank]

@radiant roost Has your question been resolved?

[empty cove]

since ab >= bc and ac >= bc we have lcm(a,b,c) >= 3/4bc

[radiant roost]

yeah

[empty cove]

so now we know it divides bc and it's >= 3/4bc, right?

[radiant roost]

oh

you're right

thank you 🙏

wow

[empty cove]

np :)

[radiant roost]

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[radiant roost]

.reopen

[midnight plank]

✅

[radiant roost]

the last hint says this

but this doesn't seem true

3 | (2+1) but 3 doesn't divide 2*1

i guess they mean it works in the context of the problem?

i don't see why

oh it may work for primes other than three

since 3bc = a(b+c)

[radiant roost]

but what does that do?

[radiant roost]

it gives us p | b,c and p^2 | bc

[midnight plank]

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[midnight plank]

@radiant roost Has your question been resolved?

[radiant roost]

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[midnight plank]

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[chilly tiger]

I know that (sin(x))^2+(cos(x))^2=1, but for the others am I supposed to use this identity?

This looks like way too much unnecessary algebra but I don't know 🤷

[woeful turret]

try using 1-2sin^2x = cos 2x

[chilly tiger]

Oh, so $(sin(2x+5))^2 = (sin(x-5))^2 \implies sin(2x+5) = sin(x-5), sin(2x+5) = sin(5-x)$ right

[grand pond]

TeslaRoad

[chilly tiger]

Oh yea, tysm for the help

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[molten bay]

How to find normal to the ellipse if we are given a line

[shrewd tusk]

well the tangent should have the same slope as the line so put dy/dx=m, where m is slope of line, then you'll get (x,y) on the ellipse which lies on the normal

then yk the slope of normal must be -1/m, so you can use 1-point form and get the eqn of normal

[midnight plank]

@molten bay Has your question been resolved?

[shrewd tusk]

uh which part is unclear?

[midnight plank]

@molten bay Has your question been resolved?

[midnight plank]

@molten bay Has your question been resolved?

[slow thorn]

what line

[midnight plank]

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[astral talon]

could you translate this please?

[tidal turret]

is sending

[astral talon]

can you just confirm one thing - the question is asking average SPEED and not VELOCITY?

i believe it's asking for velocity (the vector quantity) because the answer for speed isn't one of the options

can you define average velocity for me?

[tidal turret]

velocity is a vector

speed is the magnitude of the vector

[tidal turret]

in spanish is called velocidad media

maybe it means speed

[fresh sparrow]

average velocity?

[tidal turret]

yes bevause speed can't be negative

[tidal turret]

I think so

[astral talon]

.

[tidal turret]

displacement over time it took

no?

[astral talon]

more specifically, the total displacement divided by the total time

[tidal turret]

question would be to define displacement now

[astral talon]

i'll assume you know what displacement means

my question is can you find the displacement here

[tidal turret]

lowkey I dont

[astral talon]

😭

[tidal turret]

tf - ti

[astral talon]

okay what do you understand by displacement

[tidal turret]

xf- xi

[astral talon]

in words?

[tidal turret]

is like,

is like the

subtraction of the final position with the initial position

[tidal turret]

idk

[tidal turret]

change of position of an object

[astral talon]

from a more conceptual standpoint, the displacement is the measure of the shortest distance covered.

so let's say i start from an origin and move 30m left, and then 40m right, i have a DISPLACEMENT of only 10m right