help-19

[main cape]

I didnt understand it well

Sure ill show my work

[modest nacelle]

Tips, manipulate the fraction so that the numerator no longer contains x

[main cape]

x?

[manic sleet]

you have to prove it via epsilon-delta?

[main cape]

Yep

But it doesnt click

The method

Its unorthodox to what ive learnt in highschool

[manic sleet]

so you must show that (\forall\epsilon>0,\exists N\in\mathbb{N}:n>N):
[\left|\frac{2n-3}{n+2}-2\right|<\epsilon]

apparently I suck at LaTeX

[clever fjord]

PajamaMamaLlama

[main cape]

Whats that 😂

[sand horizon]

So n is ?

[manic sleet]

"coding" language for math

[main cape]

So i have to get n alone?

[sand horizon]

Yup

That'll be the rank for a given epsilon

Beware that n is a natural number and not a real number

[main cape]

Yeah

[sand horizon]

You just shown that for any epsilon > 0 there exist n in N such that |u_n - 2| < epsilon which is the definition of u_n tends to 2

[cinder plover]

anubhav bro kat jaoge mit jaoge

[stark flame]

bumboclat

[main cape]

Its done?

[sand horizon]

Pretty much yeah

[main cape]

Oh

But i felt that i didnt do anything

😂😂

[sand horizon]

Not meant to be hard

[main cape]

Thank u for ur time

❤️

[odd edge]

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[odd edge]

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[vocal willow]

can anyone help me with this?

[wooden python]

do you know

properties and laws of log

[vocal willow]

not that much

[silk stream]

Its basic

[vocal willow]

wait

i think i get it

what exponent do i give to the base? to match the number of the yk

[gusty fiber]

@vocal willow

you still there?

[vocal willow]

yep

[gusty fiber]

let’s start with a

what do you raise to 2 in order for it to become 1?

[vocal willow]

0?

[gusty fiber]

sorry mistype

yeah that’s the answer

what do you raise to 5 in order for it to become (5)^2

[vocal willow]

2

[gusty fiber]

correct

[vocal willow]

and number 3 is 0?

[gusty fiber]

unfortunately not

what do you raise to 3 so that it results to 3

[vocal willow]

exponent of 1

[gusty fiber]

yes

so every time you see a logarithm

phrase this question in your head

“what do I raise to (small number in the bottom) in order for it to become (big number on the right most)?”

[vocal willow]

thank broo

[gusty fiber]

yeah do you get it?

[vocal willow]

also i just wanna ask a clarifying questuob

[vocal willow]

yes

[gusty fiber]

whats your question

[vocal willow]

not related to this

if its addition its just putting them next to each other

and if - is always using the deision

[gusty fiber]

wait

I understand 2

but number 1

[vocal willow]

idk i think its a typo or something

thanks for the help dude

im okay now

.close

[odd edge]

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[exotic oracle]

how can we prove this using the defintion

[sand horizon]

C'est quoi la définition de la limite que tu as ?

[exotic oracle]

pour tou epsilon posi. il existe alpha posi. tel que 0<|x-a|<alpha implique |fx-l|< epsilon

[sand horizon]

Tu connais la notion de voisinage ?

[exotic oracle]

non

mmm je connais "au voisinage" de quelque chose cad de l'approcher

comment on peut dire well en frc?

[sand horizon]

Un voisinage d'un point c'est un intervalle autour de ce point oui

[sand horizon]

Bien ?

[exotic oracle]

ah oui

[exotic oracle]

donc je la connais

masi ca sert a quoi?

[sand horizon]

A rendre les choses un peu plus simple mais comme il demande la définition de la limite on va faire avec la definition

[exotic oracle]

je ne vois pas ou on faut commencais

[sand horizon]

L'écrire pour (f ○ g)(x) -> l

[exotic oracle]

j'ecrivais les defintions

les trois quon a

[sand horizon]

f est continue ?

[exotic oracle]

on avais pas entrer dans la contuinité

donc je ne peut pas l'utiliser

[sand horizon]

Ok

[exotic oracle]

c'est trop dur 😭

[sand horizon]

Oui ducoup

[sand horizon]

Non, faut juste pas se perdre dans toutes les écritures mais sinon c'est comme des lego les définitions vont s'embriquer les unes dans les autres

[exotic oracle]

je pense que j'aura labandonner

[exotic oracle]

oui moi suassi je la trouver facile on thoery mais je suis perdu dans tous les ecritures comme vous avez dire

[odd edge]

@exotic oracle Has your question been resolved?

[gleaming turtle]

This is false as written (assuming I understood what it's asking). You need an extra condition to make it work. For example, if f(x) = 0 for all x, and g(x) = 7 when x=0 and g(x)=13 when x=/=0, then lim_(x -> 0) f(x) = 0 and lim_(y -> 0) g(y) = 13, but lim_(x -> 0) g(f(x)) = lim_(x->0) g(0) = 7.

(One such condition would be that f(x) =/= b in some neighborhood of a.)

[exotic oracle]

no its not?

its pretty intuitive i think

[gleaming turtle]

It's usually true

But not always

Because of how limits don't say anything about the value of the function at the actual point

[exotic oracle]

there is some fluff at top

u understand french?

[gleaming turtle]

No, but I can try to understand anyways

[exotic oracle]

i think it should be understandable

[gleaming turtle]

Yeah so the fluff isn't enough to solve the issue

[exotic oracle]

how so

[gleaming turtle]

Did you see this example I gave^

[exotic oracle]

yeah

[gleaming turtle]

It'll come up in the proof when you need to show that 0 < |f(x) - b| < delta but you'll only know that |f(x) - b| < delta

[exotic oracle]

no? we have to prove that if 0<|x-a|<alpha then f(x)-b|< epsilon

or am i missing something

[gleaming turtle]

You'll see when you write the proof

[exotic oracle]

ive been trying for hours man

i swear

[gleaming turtle]

You already wrote the definition of a limit, so this is what it says for f and g (choosing different letters for each to hopefully make things less confusing when we combine them):

• For every epsilon' > 0, there exists alpha > 0 such that 0 < |x-a| < alpha implies |f(x)-b| < epsilon'
• For every epsilon > 0, there exists beta > 0 such that 0 < |y-b| < beta implies |g(y)-l| < epsilon

Does that make sense?

[exotic oracle]

i dont think we need different epsilons?

[exotic oracle]

yeah i got there

[gleaming turtle]

It doesn't need to be different, but you're allowed to choose a different letter and it can help if we want to set one of them to a specific value

[exotic oracle]

wait

epsilon is a value that very close to zero right?

do i remeber that correctly?

[gleaming turtle]

Epsilon isn't a specific number, but the idea is that if it's true for every epsilon > 0 then it's true for epsilon as small as you want.

[exotic oracle]

ok i understand

[gleaming turtle]

What we want to prove is that

• For every epsilon > 0, there exists alpha > 0 such that 0 < |x-a| < alpha implies |g(f(x))-l| < epsilon

[exotic oracle]

yeah

[gleaming turtle]

So when we need to prove a "for all"/"for every" we use "let". So the first line of the proof is going to be "Let epsilon > 0."

We need to eventually get to |g(f(x)) - l| < epsilon, yeah?

[gleaming turtle]

So it makes sense to use the second bullet point here, since it says the same thing if you pick y=f(x)

[exotic oracle]

yeah sorry for the late reply

we have to prove fx - b< beta?

wait thats wrong

[gleaming turtle]

Almost

0 < |f(x)-b| < beta

[exotic oracle]

and we have |f(x)-b|< epslon

[gleaming turtle]

Well what we know is

For every epsilon' > 0, there exists alpha > 0 such that 0 < |x-a| < alpha implies |f(x)-b| < epsilon'
We can choose epsilon' to be whatever we want, it doesn't have to be epsilon

(this is why I chose to use a different name here)

What should we choose epsilon' to be, do you think?

[exotic oracle]

beta?

[gleaming turtle]

yep!

[exotic oracle]

but thats way too obvious no?

[gleaming turtle]

it is the obvious choice, yes, but that's what the other person meant by saying it's like lego

[exotic oracle]

so im just stupid?

i think thatll be enough to write a proof

[gleaming turtle]

The difficulty in this proof is getting used to working with these sorts of definitions and proofs, not the content itself

quantifiers, especially

[exotic oracle]

thanks @gleaming turtle and @sand horizon

[gleaming turtle]

And one last thing

[exotic oracle]

yeah ill practice more

[gleaming turtle]

You see where the issue is with the statement?

You'll get |f(x) - b| < beta

[exotic oracle]

yeah

[gleaming turtle]

But you need 0 < |f(x) - b| < beta

[exotic oracle]

i get it now

[exotic oracle]

cant we work around that?

[gleaming turtle]

You'll need an extra condition like f(x) =/= b for all x in I or that g is continuous. Otherwise the f and g I gave you earlier are a counterexample

[exotic oracle]

question; why would it being continous change anything?

[gleaming turtle]

If g is continuous then l=g(b) so you'll get |g(f(x)) - l| = 0 when f(x)=b

So you can deal with that case separately

[odd edge]

@exotic oracle Has your question been resolved?

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[stiff finch]

I need help with this two-part answer. I tried to put this answer I got in different ways, and it keeps saying it is wrong

[stiff finch]

😢

<@&286206848099549185>

[stoic cloud]

you gotta wait a tincy bit longer before ping helpers my friend 🙃 (4 min)

[stiff finch]

Oh sorry

Thought it was the 15

[stoic cloud]

Nah it’s good.

[outer hinge]

im having a massive shit rn
anyways bottom area is (M-4x)(M-2x)

[stiff finch]

Hm

How about above

[outer hinge]

ts an open box so you just take the volume and minus the roof

[stiff finch]

So how do i find my part b, im so confused

Because for c i know the graph is the first one is correct

[vapid bolt]

i think the diagram is a bit confusing the box dimensions should be:
h: x w: 32-4x l: 32-2x

[outer hinge]

ok so if you got V(x) you reduce it to like

a quadratic or something
and find its minmax

[vapid bolt]

cuz of the locking tabs

[sly yoke]

Use his formula for the bottom area and you'll see where it goes to zero

[stiff finch]

Okay there's another part for this i'm about to take the picture

for the X

[outer hinge]

find x in v(x) = max

[sly yoke]

Ok, do you know derivatives? Or are you supposed to lol?

[stiff finch]

I don't know tbh I take online course of math in college and this what i'm assigned

The second picture and the first picture are the same picture just broken apart

Wait

Is x=4?

i had 5.3 before

[stiff finch]

This part right here i can't seem to get the right answer

is B [0,16] instead of (0,16) or am i wrong?

[sly yoke]

ok, you could derivate the function obtained by multiplying the terms of V(x). Since you don't know derivatives I think you can just look at the graph

sorry for the delay btw

try to see for what X the curve has the largest result

[stiff finch]

can you reply to what image your telling me about so i don't get lost

[sly yoke]

For b consider the formula for the volume (32-4x)(32-2x)x. The upper limit is not 16, because there is a lower x that anything higher than it would output a negative volume

[stiff finch]

Oh okay that’s interesting

[sly yoke]

btw I'm trying to not give straight up answers cause I figured you might prefer that

[stiff finch]

I’m deadass so confused and lost im trying to get it done before it closes at 11;59, like your giving me the step by step process but this is imo really hard to understand

[sly yoke]

I think all of the graphs have the peak at about the same x. What course is this for?

[stiff finch]

MATH 175 pre-calculus

[stiff finch]

The correct graph is the top left one

I got that part correct

I just needed the A) and B) part because I’m confused and keep getting it incorrect

[sly yoke]

Ok, I'm sorry for this. In that new formula, because of the (32-4x), anything for x higher than 8 will output a negative volume, which is wrong. So the limit is (0,8), they don't include 0 and 8 because with those numbers the volume would be zero

[sly yoke]

gotcha

[stiff finch]

I’m a chemistry major sorry if I sound dumb for my major, math is really hard for me

[stiff finch]

Okay see now that makes sense

[sly yoke]

Did you get the new formula for the volume? It was (32 - 2x)(32 - 4x)x

[stiff finch]

Yeah I got that correct eneded up figuring that out

I didn’t figure the (0,8) on my own though

[sly yoke]

No problem at all. I'm here to help! Sorry if I sound unfriendly, english is not my main language and I'm trying to be effective in my communication here

[sly yoke]

great!

[stiff finch]

No you sound perfectly fine I don’t get offended easily, unless you were to just say “you’re dumb.” Your being nice and calm and understand I really do appreciate it

[stiff finch]

Thank you so much @sly yoke and @outer hinge

[sly yoke]

You're welcome! Did you get everything you needed?

[stiff finch]

Yes great experience and got my work done

I truly thank you guys

.close

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[sinful cave]

hm dumb question but did I mess up here?

[sinful cave]

I got y right

so I tried clearing for x, and Im stuck at the blue part

I know that the answer is the red solution

how can I get there from that fraction?

honestly I mustve messed up somewhere, just cant tell where

[odd edge]

@sinful cave Has your question been resolved?

[sinful cave]

just noticed I messed up the sign

specifically, I multiplied y by r, and forgot the sign when moving it to the other side

.close

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[remote cloud]

Why does this require the map to be invertable? is surjectivity not enough?

[forest sky]

the point is that by proving injectivity we can then prove surjectivity (because both are equivalent to invertibility when dimensions are the same)

[odd edge]

@remote cloud Has your question been resolved?

Channel closed

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[remote cloud]

.reopenm

.reopen

[odd edge]

✅

[remote cloud]

@forest sky I get that but i dont understand how having the map be invertable proves this

[forest sky]

because if you have a linear map T from V to W with dim(V) = dim(W) then T injective iff T invertible iff T surjective

so we prove injectivity and that implies surjectivity

[remote cloud]

No, like

I dont understand how the exsistance of such a polynomial is proven with the map being invertable

[forest sky]

"for all q, there exists a polynomial p such that Tp = q" is just stating the defn of surjectivity for that map

[remote cloud]

Ok, so if we only proved surjectivity would this work?

why does it also need to be injective?

are we using injevtivity to prove surjetivity

to show that a solution exsists

[forest sky]

yes, that is what we are doing

[remote cloud]

ok so the main condition we need is surjectivity, not invertability

but because they have the same dimention, we can use injectivity to imply surjectivity which is what we need

[forest sky]

it's reasoning in the form
"T is injective, T is surjective iff it is injective, therefore T is surjective"

[remote cloud]

the fact that its invertable isnt really needed it for this

ok, this came right after the definition about invertablility so i was curious

So, suppose we proved that its surjective

would that be enough?

[forest sky]

yes so there's a bunch of thing equivalent to a linear map being invertible (including injective and surjective). this is an example of where one of those equivalencies can be usedul

[forest sky]

sure, but that's what we're doing here. just in a more roundabout way because injectivity is easier

[remote cloud]

ah ok that makes snese

sence

sasjlf

so what is important is, if u cant prove surjectivity right away

and the dimentions of the domain and co domain line up

you can prove injectivity?

[forest sky]

yes, and more generally if you want to prove one of the statements equivalent to invertibility but another one is easier to prove, just prove whichever is easiest and then the thing you want to prove follows

here is a list of two dozen such equivalencies, which you build up over the course of a linear algebra class:
https://mathworld.wolfram.com/InvertibleMatrixTheorem.html

[remote cloud]

Ok cool, so for same dimention its jsut a chain

Like

Is invertable $\iff$ is injective $\iff$ is surjective

[clever fjord]

BOSS

[remote cloud]

and then we use either one to prove the other two in cases its needed.

[forest sky]

yes

alternatively if you prove any of them is false then all the others are also false

[remote cloud]

so invertable maps are only between two codomains of the same size

[forest sky]

a map can only possibly be invertible if the domain and codomain have the same dimension, yes

[remote cloud]

Also, is an isomorphism just an invertable function which is a bijective function

like are they not all the same definition?

[forest sky]

an isomorphism is a bit more specific because it also has to "preserve the structure" of the spaces in question

in linear algebra "preserves the structure" means being linear. so an isomorphism is a bijective linear map

[remote cloud]

Also, is an isomorphism just an invertable linear map which is a bijective linear map ***

Ok yeah, and the last two are also the same

why do we sometiomes say isomophism and other times say invertable linear map

its just he same thing right/

[lofty spade]

they mean the same thing in the context of vector spacs

[forest sky]

in linear algebra there are a bunch of terms which are basically equivalent

so pick your favorite

[lofty spade]

sometimes you want to emphasize different things

like that the two vector spaces are isomorphic

[remote cloud]

w ok

[remote cloud]

Ok, i just wanted to make sure im not going crazy in thinking all these definitions mean the same thing

[lofty spade]

there are other kinds of isomorphisms, like group isomorphisms and ring isomorphisms, but they don’t come up in linear algebra

[remote cloud]

in group theory i remember there was more to an isomphism

yeah

I did those before this class and i dont remember those things being the exact same

alr tyty

.close

[odd edge]

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[remote cloud]

.reopen

wait

[odd edge]

✅

[remote cloud]

so there is only one vector space per each dim n?

or like has the same structure?

[forest sky]

well there is only one vector space of each dimension up to isomorphism

which might leave out a bunch of information that isn't part of the vector space structure

[forest sky]

but as far as the vector space structure goes (taking linear combinations etc) they are the same

[remote cloud]

hmm ok so its just like groups

but more contained on the size of a generator

[forest sky]

this is actually very useful because it means every problem about finite dimensional vector spaces and linear transformations can be reduced to a problem about F^n and matrices

[remote cloud]

Wild okm

thats really cool

so again like cyclic groups or the mod groups in group theory

[lofty spade]

(because you can always choose a basis)

but some vector spaces don’t have a “natural” basis

[remote cloud]

so if its basis 2

i can just say its

[forest sky]

yeah it's all completely basis dependent

[remote cloud]

$F^2$

[clever fjord]

BOSS

[remote cloud]

so that means like

[lofty spade]

like the subspace ${(x, y, z) \in F^3: x+y+z=0}$

[clever fjord]

soup_norm

[remote cloud]

$\mathcal{P}_n(F) \cong F^{(n+1)}$

[clever fjord]

BOSS

[forest sky]

indeed

[remote cloud]

ok cool its starting to make mroe sense

they are not exactly the same tho

they just have the same structure

[forest sky]

they have the same structure as vector spaces

[remote cloud]

so i would have to define a map

and then i can do all my work in F^n

but at the end i would have to bring it back to the origninal vector space

[lofty spade]

yes, that’s what you can do if you only care about the vector space structure

[lofty spade]

yes

[remote cloud]

okok

im gussing there are some important pre-defined maps?

[forest sky]

well the canonical map will be the map where you map each basis vector to the corresponding basis vector of the other space

this of course requires you to pick bases of each space in the first place

[remote cloud]

ok wait going back to a previous lemma

There is one that states there is always a map from the basis of some vector to the same number of elements in another vector

Can i choose these vectors?

[lofty spade]

(this is only canonical once you’ve got a basis)

[remote cloud]

or is it just given by the lemma

[lofty spade]

same number of elements? what do you mean?

[remote cloud]

let me find it

[forest sky]

you can choose them, yes

[remote cloud]

ok cool just wanted to make sure

so if we choose the basis

then its surjective

[forest sky]

you can choose to map the basis vectors of the domain to any vectors in the codomain. once you do that then the output for every other vector in the domain is fixed though

[lofty spade]

which vectors do you want to choose?

[remote cloud]

Basis of V to basis of W

if they are the same dim

[lofty spade]

remember it can’t be surjective if V has a strictly smaller dimension than W

[remote cloud]

then its an isomorphism right?

[forest sky]

if we choose the list of w's to be a basis then we get an isomorphism, yes

[remote cloud]

^

ok cool

i was just going to use that to prove the theorem lol

Like if we have two maps, of the same dim, i can just map them togetehr

and thats an isomorphism

[remote cloud]

for this

[coarse shale]

hello

[forest sky]

yes so that proves the backward direction

[remote cloud]

ill do the forward one in as second

but yes ok good to know im on the right track

[odd edge]

@remote cloud Has your question been resolved?

[odd edge]

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[wet ridge]

hmm

can you assume the new speed?

as 3/4v?

hello>

[wicked olive]

I think you could apply distance constant approach

[torn tulip]

yeah

[wet ridge]

try useing 3/4(V)(t+35/60) = vt ?

[torn tulip]

d-50/3S/4

[wet ridge]

no

the distance is constant

[torn tulip]

yeah but i am not getting the answer

[torn tulip]

but

d-50,d-74

[wet ridge]

no I mean the distance to its destination

you can equate to equations

[torn tulip]

i did

[wet ridge]

how?

did you equate them

[torn tulip]

$$
\frac{D-50}{3S/4} = t + 35
$$
and
$$
\frac{D-74}{3\frac{S}{4}} = t +
15

are these correct?

[clever fjord]

Rabbit
Compile Error! Click the reaction for more information.
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[wet ridge]

hmm yes

[torn tulip]

then i substitute

but i got s=96

[wet ridge]

I guess its a calculation error

[torn tulip]

i changed 35 into hour too

[wet ridge]

is ti 24?

so what you want to do is take the difference in time b/w the two cases which is 1/3rd hour

(\mathbf{V}): Normal speed (\Delta t=(35-15)\text{\ min}=20\text{\ min}=\frac{1}{3}\text{\ h})
$$
\frac{24}{\frac{3}{4}\mathbf{V}}-\frac{24}{\mathbf{V}}=\frac{1}{3}
$$
$$
\frac{32}{\mathbf{V}}-\frac{24}{\mathbf{V}}=\frac{1}{3}
$$
$$
\frac{8}{\mathbf{V}}=\frac{1}{3}
$$
$$
\mathbf{V}=8\times 3
$$
$$
\mathbf{V}=24
$$

[clever fjord]

qfalcon_

[torn tulip]

where didyou get 24?

[wet ridge]

24 is the distacne extra

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[short terrace]

,rccw

[short terrace]

Worthless piece of shit bot

[pale bison]

When using BIDMAS does the indices outside the bracket apply before the 5 on the outside

[short terrace]

Yes indices first

[pale bison]

Ok

[short terrace]

The 5 on the outside is a multiplication

[pale bison]

.close

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[pale bison]

Help with question 12 d

[pale bison]

How do I find the time at which max height is reached?

[austere jolt]

are you allowed to use calculus?

[pale bison]

Idk

Probably

[austere jolt]

if yes, derivatives kill this problem in one shot
if not, find the vertex of the function h(t)

[pale bison]

Turning point ?

Omg wait I’m stupid

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[left quarry]

Find every shape that's have a special name and prove it why

[short terrace]

…what

[left quarry]

.

Do it

[short terrace]

No thanks

[left quarry]

10 robux

[short terrace]

Best wishes though

[pale bison]

What is this boss

Good luck with whatever that is

[short terrace]

I’m gonna do one thing

!xy

[odd edge]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[short terrace]

In case that makes it slightly better

[austere jolt]

this looks like an architect's fever dream, and I agree that the original question would be interesting to see

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[pale bison]

Question 19

I’m confused

[mystic nova]

Sorry but this channel is closed

[pale bison]

Oh

[mystic nova]

You'll have to claim another channel

[pale bison]

.reopen

Ok

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[sand matrix]

Is my h'(y) = 0?

Also this is solving an ODE that started off as not exact but using a method to make it so and to solve using that.

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[cursive sigil]

Does anybody have any tricks on how to find domain and range of equations ? I’m very bad at this and always feel like I’m just guessing if it’s not just X cannot equal somthing or if it’s a root 3 equation

[safe vapor]

There isn't a universal trick for domain & range. Domain can be fairly easy by just remembering the domains of common discontinous functions (ie. sqrtx, lnx, rationals, etc.)

[odd edge]

@cursive sigil Has your question been resolved?

[cursive sigil]

For the root of three is the domain and range always ( negative infinity , infinity ) ?

[safe vapor]

root of three?

you mean cube root?

$\sqrt[3]{x}$?

[clever fjord]

Stitches

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[left quarry]

Nope

[thorny shore]

Just write down the function in terms of both the dependent variable and independent variable: y = ... and x = ...

Then, see what shouldn't fit, like checking for division by zero or negative inside root, etc.

This should suffice for most.

[mystic nest]

ghi tiếng việt đi e rồi để người ta dịch ra t.a

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[empty dune]

Hello guys 👋
I am trying to learn dsa and am kinda stuck on the maths part
I want to take some time and try to learn some maths so it would become easier
Any good books/resources on number theory and combinatorics would be grateful 😄
Thanks in advance 🙏

[empty dune]

.close

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[proven harbor]

I recommend khan academy

Tho chatgpt is also really good \s

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[empty dune]

Thank you ver much😄

[empty dune]

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[somber remnant]

Does anyone have any ideas on how to do number 5a?

[wooden python]

have you done limits of rational functions before

[somber remnant]

yes

[wooden python]

ok, what do you usually do with these?

[somber remnant]

if we r talking about finding the limits of rational function i will just substitute x into the function and see if its 0/0 or inf/inf

or any other intermediate form

if it is then i will do L'hopital next.

[somber remnant]

If it isnt then either i have the answer or it does not exist.

[somber remnant]

its case by case tho if it can cancel each other out i will cancel it out first

[wooden python]

i mean if all you know is l'hôpital then i guess you need to l'hôpital.

there are better ways for rational functions tho

[somber remnant]

i dont have a shed of experience on how to use mean thingy tho

[wooden python]

mean thingy?

[somber remnant]

the mean value theorem

i just know its definition

[wooden python]

the question doesnt say to use MVT...

[somber remnant]

but never used it before

[wooden python]

and it isnt applicable here either

[somber remnant]

we r talking about 5a right?

@wooden python

mmm ima just to do this tmrw ig

its fairly late for me

appreciate ur help tho

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[wooden python]

hlgjsADK:G

fuck

sorry

i was looking at the wrong thing the entire time im so sorry.

i had my eyes on 6a the entire time

sorry

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[lavish mesa]

what is the motivation/visualisation behind these

[quasi sparrow]

https://math.libretexts.org/Bookshelves/Analysis/Introduction_to_Mathematical_Analysis_I_(Lafferriere_Lafferriere_and_Nguyen)/04%3A_Differentiation/4.06%3A_Section_6-

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[vernal ibex]

1. What makes a differential equation linear
2. What makes a differential equation homogeneous
3. Why do we care about whether a function has constant coefficients
4. Why are questions 1-3 important for solving a differential equation?

[odd edge]

Please don't occupy multiple help channels.

[amber veldt]

what have you tried, what do you know, etc etc

[sand horizon]

1. you can superpose solutions and apply linear algebra things
2. dy/dx = F(x,y) where F(tx,ty) = t^nF(x,y) (F is homogenous of degree n so)
3. in DE ? Not the same methods when its not
4. to solve it tbh

[quasi sparrow]

https://tutorial.math.lamar.edu/Classes/DE/SecondOrderConcepts.aspx

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[clever fjord]

brandon

[sinful jewel]

If you’re in a city with a university or college, then you could just go to the library. They should have some resources. This should also be free

(Are you from Germany? Where did you find this problem?)

Ah ok. Bc this was a problem from the BWM (Bundeswettbewerb Mathematik) and I tried (and failed) it

[sinful jewel]

Maybe also Archive?

Yeah… next year maybe?

Or you could try the books from springer. They have a lot

A that’s unfortunate. Yes it is

Aber bist du denn in einer Uni Stadt. Weil dort gibt es das fr alles. (Habe mal in einer normalen Bibliothek geschaut, da war die Ausbeute sehr mau…)

Yes there actually is a video from DorFuchs about it. It’s quite understandable. After that it seemed pretty easy

Nah, das natürlich blöd. Manchmal kann man bei denen einen Account machen und dann online ausleihen

[sinful jewel]

Also one of my friends has a solution if you just want a latex pdf

[sinful jewel]

Jaa, das sollte auch lange reichen. Ich hatte das halt mal probiert und bin nach 30 Seiten abgestorben. Man muss halt echt alles noch mal selbst durchdenken und die Aufgaben sind auch schwer. Davor hält das Buch dann aber auch lange…

Aber ich bin auch nicht so alt wie du (außer du hast Klassen übersprungen) also kann es bei dir ja auch ganz anders sein

Ja das war zweite Runde. Also die anderen habe ich gemacht (hoffentlich richtig, dann AIMO). Die gingen eigentlich. Ich saß nur leider sehr lange an 1a), weil mir der richtige Ansatz fehlte. Man musste das alles mit Latex schreiben, also wenn du was haben willst kann ich es gerne schicken

Ja, 3 war am einfachsten tatsächlich

Wait a sec

Ups, habe mich gedoxt...

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[mystic saffron]

Can anyone help me with simultaneous equations, ive soent a week trying to understand them and its so hard too

[quasi sparrow]

do you have a specific problem/question

[mystic saffron]

I mean not really but let me find a example

[quasi sparrow]

do one problem correctly first

[mystic saffron]

What do you mean

[quasi sparrow]

i mean do one problem before doing a lot of them

[mystic saffron]

Alright

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[torn sonnet]

why cant this be simplified to 13/29

[amber veldt]

$$\frac{1 + 3}{2+3} \ne \frac 3 3$$

[clever fjord]

gfauxpas

[torn sonnet]

wouldnt u do $$\frac{1+3}{2+3} = \frac{1}{2}$$

[clever fjord]

vfx

[amber veldt]

that's not true either

[torn sonnet]

oh

right

mv

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[split olive]

Okay so this all makes sense, but how did width go from 20->16 cm??? Like I get it cannot be 20cm cause 28-20(2) is negative and volume cannot have a negative dimension but still

[split olive]

Why the number 16

Like length and height both were

-2(4) and then the equation was gotten for them

But why those ones and not height? Is it because height is only reflected of x in the equation while they were

-2(x)

[cold sage]

its just 28-12 no?

since x is 6

the width being 20 is the x=4 case

the notation on the left makes me sad though, should be 0 rather than V(x)

and the V(4)

now I wont be able to sleep in peace

[odd edge]

@split olive Has your question been resolved?

[split olive]

I have no idea

I don't think my teacher did that

[split olive]

What so we plug in x because each is equal to l - 2x?

Or w -2x?

[cold sage]

they did two cases
x=4 and x=6 since x=20 wasnt acceptable

then they just used w=28-2x and l=32-2x

for each one

[split olive]

Okay but why would the factor be used

Like for two different dimensions

Is it cause it gives us the value of x? The value of x could be 6 though no?

I think I don't understand the context of this honestly

I get it's volume and the different dimensions

but why did we use x -4? It was the factor that led to the rest of the values?

[cold sage]

x here is the height

they then wrote the length and width in terms of x
and then wrote V in terms of x by V= lwh

then they set V=1920 and solved for x

[split olive]

Okay but we got two valid factors

(x-4)(x-6)

Why did we use teh former

[cold sage]

it was always going to factor into 4(x-4)(x-6)(x-20)=0

though how they got 4 simply is a mystery, you could try the rational root theorem or guessing, whichever floats your boat

[split olive]

Oh I did

But I'm saying why did we then do 32 -2x and make x = 4

For this

Why didn't we plug in x = 6

Or another one

I think I'm tripping

[cold sage]

x=4 and x=6 are both valid heights, since they give us the volume we wanted

thats why we were solving V=1920 in terms of x

we only excluded x=20 because it wouldnt make physical sense

mathematically, it would also be a solution

thats why at the bottom they have two sets of h, w and l

one for x=4 and the other for x=6

[split olive]

Ohhh

So the reason why we used 4 is because the overall height is less

[cold sage]

lost me a bit there

they used 4 AND 6

[split olive]

So to equal the volume it has to be subtracted by 4 because the factor of the height is less

[cold sage]

confusion

[split olive]

OKay so it's 1920 = (x-4)(32-2x)(28-2x)

[cold sage]

no

[split olive]

If we had used x-6 we'd be doing

32-2(6)(28-2(6)

Not by 4

[cold sage]

1920=x(32-2x)(28-2x)

[cold sage]

we arent just choosing x-6 and x-4 arbitrarily
they were things we got by solving this

[split olive]

Oh

But x = 4 or 6 right? So the x we put into that equation will also change the value for the other two factors

[cold sage]

yeah

[split olive]

I see

Can we choose either or no?

[cold sage]

theyre both possible solutions
so you have to state both

[split olive]

LIke can I just choose 6 instead and use that dimensions

Why did my teacher only do one?

[cold sage]

they state the x=4 solutions

then the x=6 solutions

[split olive]

Maybe she just wanted to?

[cold sage]

they did both

[split olive]

Oh

OHHHH

Wait

Oh

Dude I thought she used the left side

and the final was the right

I didn't know how everything changed

[cold sage]

aha, no, different cases

[split olive]

I see

So we have to actually show both

Thank you so much man

[cold sage]

always state em all

[split olive]

You just saved me a mark on my test prob

or 2

[cold sage]

no worries

[split olive]

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[cloud galleon]

why is the third one's dericative not 3?

[cloud galleon]

so far i know option 1 and 4 are correct, but am confused about the third

f(2)=-1
f'(2)=3

how does one find the derv of the third?

[dim bramble]

Maybe u forgot the chain rule for sin(πx)?

[cloud galleon]

Oh?

So 3pi cos (pix)?

[golden sonnet]

yea

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[near swan]

in CRT, what if the left-hand side does not have an x but something like, say, 3x (and different values for each expression)? how do we solve the system, then?

[haughty fulcrum]

if 3 does not divide the modulus, you can multiply by 3's inverse

if 3 does divide the modulus, you divide by 3

cases like 3x=2(mod 6) would be considered no solutions

[near swan]

but 3x ≡ 12 (mod 6) works?

x ≡ 4 (mod 2)

[haughty fulcrum]

yes

[near swan]

all right, thank you!

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[near swan]

[repost because i have forgottten how we did this] how do we calculate these again?

[near swan]

oh

so we break 2537 into 43 and 59, and solve it?

but is still complicated

[wicked olive]

Yes

But ig using euler’s is easy

[near swan]

we find phi(43) and then do 1819 ^ (phi(43)) ≡ 1 (mod 43)?

[wicked olive]

Yeah the only problem with euler is that the power is too small

So ig crt is the only way here

[near swan]

okay, thank you

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[pearl prawn]

I'm out of idea after writing out
f(x) = 3x for x>0
f(x) = x for x<0

[wooden python]

so far so good

do you know how to graph the lines y=3x and y=x

[dim bramble]

Maybe you forgot abt the f(x+3) = f(x) part?

[pearl prawn]

ya, because I don't really know how to deal with it

[proven harbor]

if f(x + T) = f(x)
the T is called the period of f(x)

[pearl prawn]

period? like function of sin cos?

[wooden python]

sin and cos are famous for being periodic yes

anyway

graph y = x for -1 ≤ x ≤ 0 and then y = 3x for 0 < x < 2

show us the result

[dim bramble]

.

[odd edge]

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

[proven harbor]

this means that the graph starts to repeat after that

so what you draw between -1 to 2

just copy paste that everywhere

[proven harbor]

yes like that

[pearl prawn]

ok, so I repeat until it reach [-4, 5]
and I just calculate the y value and find the range

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[toxic rose]

if 3 naturals a,b,c are in AP and a+b+c=21, find the number of ordered triplets of a,b,c?

[toxic rose]

so i reduced it solving $x+y=6$ for $x,y \geq 0$

[clever fjord]

rak³en

[wooden python]

how did 6 happen

can you show your steps for the reduction

[toxic rose]

ok

first set $a=x,b=x+y,c=x+2y$ to reduce it to $x+y=7$ such that $x \geq 1, y \geq 0$ and then replace $x$ by $x'+1$ so that $x' \geq 0$

[clever fjord]

rak³en

[wooden python]

you dont know that y > 0

[toxic rose]

uh?

OH

so for eg.

[wooden python]

decreasing APs are still APs

[toxic rose]

okay cant come up with one but yeah

[toxic rose]

yes

[wooden python]

its much easier to think in terms of the fact that b=7

8, 7, 6

[toxic rose]

set a,b,c = 7-d,7,7+d?

so i have 15 choices for d ic ic

wait wtf the answer key says 13?

oh okay so d cant be 7 or -7

right so 13

thx

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[south shore]

Hi, can I ask a question here?

[wooden python]

yes. you've already claimed this channel just by posting in it

post your question

[south shore]

Oh, thank you

[wooden python]

in the future you can (& should) simply post your question from the get go.

[south shore]

oh wait nvm I'm so stupid. I understand it

thank you Ann

how do I close this?

[short terrace]

You can still post the question here lol

[south shore]

it's ok

[short terrace]

But if you're done, you close it by typing .close

[wooden python]

.close to close the channel

[south shore]

.close

[odd edge]

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[south shore]

thank you

[short terrace]

If you ask questions in the future, just post them directly to the channel!

[south shore]

ok

[odd edge]

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[dim bramble]

Question.

Let d_n := { 2n, if n is prime; n, otherwise } (and take d_1 = 1). For n∈ℕ define
I_n := ∫0^1 x^{x/d_n},dx,
where x^{x/d_n} at x=0 is taken by continuous extension (value 1). Evaluate
L := lim{n→∞} n(1 − I_n).
Give a complete, rigorous proof.

[dim bramble]

can anyone help?

[noble forge]

maybe

[dim bramble]

Can anyone help?...

[dim bramble]

Will u?

[low crow]

I'm out of my league here but if no one else comes along in 15 minutes you can ping Helpers as the Bot said

[dim bramble]

Alr thanks

[wooden python]

looking at $J_n := \int_0^1 x^{x/n} \dd{x}$ seems like a good idea

[clever fjord]

Ann

[dim bramble]

Nvm

i figured it out

The Limit L does not exist

Still thank you tho

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[mellow cape]

can anyone teach me how to find out square and cube roots quickly plz.

[lofty spade]

in what context?

[mellow cape]

wdym ?

[lofty spade]

for fun? in an exam? with or without a calculator?

of any positive number or just square numbers?

[mellow cape]

without a calculator

any positive number

[lofty spade]

and just for fun, right?

[stoic cloud]

Well you could memorize the cubes and squares — probably start with those so you could recognize certain roots.

[lofty spade]

there’s a numerical square root extraction method somewhere on internet, maybe wikipedia, let me try to find it (though it only gives approximate solutions)

[mellow cape]

no like if i need to solve a question with square or cube roots

[mellow cape]

school wont give such easy questions

[lofty spade]

generally they won’t ask you to evaluate the square root without a calculator, right?

[stoic cloud]

If you’re dealing with roots that will result in irrational answers, there are probably strategies littered all over YouTube and wiki.

[mellow cape]

man im indian so unfortunately, no calc for me

[mellow cape]

yea but like there are too many so i like wanted to ask someone which is the best

[stoic cloud]

Well do you know exactly what’s going to be on your assignment? Ex: radical simplification, rearrangement, blah blah blah

[lofty spade]

can you send an example of a question where you need to calculate the square root of a given positive number?

usually there are other methods to get the answer that aren’t as tedious

[mellow cape]

man im in 9th this kinda shi has'nt been tought to us

[stoic cloud]

🤔

[mellow cape]

well tbh i didnt have any in mind i wanted to know a trick or smth to help me in the future

im sry is this was kinda an irrelevant question

[lofty spade]

you can try some of these methods https://en.wikipedia.org/wiki/Square_root_algorithms but usually they shouldn’t be necessary

[lofty spade]

it’s a reasonable question, don’t worry

[stoic cloud]

I feel like it would at this point.

[lofty spade]

like usually you can do other (and more exact) things to get the answer

[stoic cloud]

Rad. Simplification is pretty much taught right after you learn radicals no?

[mellow cape]

nah only like the radical sign thingy

[stoic cloud]

Interesting.

[stoic cloud]

Well do you know exactly what’s going to be on your assignment?

Could you still possibly answer this question?

[lofty spade]

if you’re doing these algorithms, you could memorize the squares of the natural numbers up to 30 or 32 (since sqrt(1000) is 31.6…)

that might help

[mellow cape]

well ok then, thanks @lofty spade , @stoic cloud

[odd edge]

@mellow cape Has your question been resolved?

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[toxic rose]

A is a set of cardinality $n$. What is the number of ways of choosing two subsets of A such that they have exactly 2 common elements?

[clever fjord]

rak³en

[woeful briar]

What have you tried

[wooden python]

are the subsets allowed to be any size

[toxic rose]

well heres what i tried

let P and Q be the two subsets

with cardinalities p and q

[wooden python]

ok and what then

i will say i have a much more elegant way in mind but continue yours

[toxic rose]

then the number of ways of choosing p and q $= \binom{n}{2} \sum_{0 \leq a,b \leq n-2} \binom{n-2}{a} \binom{n-2}{b}$

[clever fjord]

rak³en

[toxic rose]

does that give nc2 * 3^(n-2) as the answer

[wooden python]

it does

and you can arrive at that much easier

first pick which 2 of A's n elements go in P \cap Q

then for each of the other n-2 elements independently of the others

you have 3 choices: go in P only, go in Q only, or go in neither

[toxic rose]

yes i found that on the web

[toxic rose]

question is this correct?

[wooden python]

it uhhh doesnt look that way

but it's also a very ☠️ way of doing it even if it is correct

[toxic rose]

what do u think am missing

[toxic rose]

indeed

@wooden python ?

[wooden python]

what.

sorry i cant rly indulge you on this extremely complicated and needlessly difficult solution strategy

[toxic rose]

😐

[tepid pelican]

I agree with Ann that this is needlessly complicated, but once u choose the 2 common elements, that sum is supposed to count the number of ways you can form 2 disjoint subsets out of n-2 elements, right?

If you wanna suffer less, try finding a different way to count that 2nd part (counting n.o. ways in which u can form 2 disjoint subsets of n-2 elements)

[toxic rose]

already done lol i am trying to suffer more for no reason

[toxic rose]

yes, i hope

[tepid pelican]

oh, well

enjoy your suffering

it actually doesnt even look correct, shouldnt it be (n-2)Ca * (n-2-a)Cb

[toxic rose]

i wrote that initially but thought it was wrong

oh wait

yes

[tepid pelican]

and it still overcounts

[toxic rose]

but why..?

[tepid pelican]

you can choose the same subset with a = 7 and b = 3 and with b = 7 and a = 3 e.g.

it will be counted in both

[toxic rose]

wait those are the same?

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[manic grail]

why du/dυ= 0?

[manic grail]

t-u = υ <=> u= t- υ => du/dυ = -1?

[wooden python]

i think you mean v and not ypsilon there btw

[manic grail]

there is no v

this is ypsilon

[wooden python]

nope i am afraid that is v

$v$

[clever fjord]

Ann

[manic grail]

how is that v

[wooden python]

$v \upsilon$

[clever fjord]

Ann

[wooden python]

ypsilon is wider

[manic grail]

v is v wdym

[wooden python]

but also i dont remember ever seeing lowercase ypsilon as a variable name anywhere in math

[manic grail]

i havent seen in any text v be written like that

[wooden python]

$\mathrm{v} v \mathbf{v}$

[clever fjord]

Ann

[wooden python]

idk what to tell you, this is what v looks like in the LaTeX font.

[manic grail]

yea the first is a v

ok v wtv

[manic grail]

why is the derivative 0 and not -1 like i showed here

[wooden python]

anyway uhhhh

yeah ngl i got nothing either

it is mighty confusing wtf they're trying to achieve here

[manic grail]

wdym

is my calculation wrong

[wooden python]

i looked over your pic 5 times yet cannot figure out why ∂u/∂v is supposed to be 0

[manic grail]

this is the integral

and we set t= u + v

the jacobian should be 2 from my calculations

over the region Ruv

oh wait actually maybe ik

@wooden python i see why its 0 we cant do the u= t-v because in the jacobian we transform each variable and u stays the same its not dependant on v

its u=u , t= u-v

the transformation

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[frigid canopy]

To start $Y(\omega) = e^{X(\omega)} \forall x \in \Omega$, where $\Omega$ is the sample spae and $\omega \in \Omega$

[clever fjord]

wai

[frigid canopy]

$P(Y(\omega)) = P(e^{X(\omega)})$

[clever fjord]

wai

[frigid canopy]

the issue is working with the right hand side now

[low locust]

probabilities of single numbers for continuous distributions are always zero

so thats not the correct thing to look at

[frigid canopy]

so a subset of the sample sapce instead

[low locust]

well I see two options

either you have the transformation formula for PDF's

then you just plug in

a different option would be to find the CDF of Y first and differentiate that

and there are surely also other options

[frigid canopy]

so basically this

$f_Y(y) = P(Y≤y) = P(e^{X}≤y)$

[clever fjord]

wai

[leaden karma]

log-normal distribution 🥰

[frigid canopy]

so I need to integrate over $\frac{1}{√{2π}} e^{-x^2/2}≤y$

[low locust]

yes but you dont need to actually compute the integral

[clever fjord]

wai

[low locust]

nvm

[frigid canopy]

hmm?

[low locust]

you should never ever have the expression for the density in such an inequality

you want to rewrite P(e^X <= y) to something of the form P(X <= something) so that you can write this as integral_-infty^something (density of X)

[frigid canopy]

<@&268886789983436800>

[frigid canopy]

so I want X≤ln(y)

[leaden karma]

So what does the CDF look like

[frigid canopy]

Let $A={x \mid \frac{1}{\sqrt{2\pi}e^{-x^2/2}}}$.
\ $\int_{A} N(0,1)$

[clever fjord]

wai

[frigid canopy]

this is a bit of a mess rn

just a min

[low locust]

forget this. thats needed for more difficult situations. not here

[leaden karma]

What

[frigid canopy]

so I use the other method you described?

[low locust]

this

[low locust]

its the same as this. but at a slightly lower level of formality

this set A_y only seems to confuse you

[frigid canopy]

X≤ln(y)

so $\int_{-\infty}^{\ln(y)} \frac{1}{\sqrt{2π}} e^{-x^2/2}dx$

[clever fjord]

wai

[frigid canopy]

and now I differentiate this

[low locust]

yes

[frigid canopy]

that would give me $\frac{1}{y} \frac{1}{\sqrt{2π}} e^{-(\ln(y))^2}/2}$

[clever fjord]

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

[leaden karma]

Yes

[frigid canopy]

cool. Thanks!

now to think about why this works

I think I get it