help-0

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[naive grove]

Hello! 👋
What is in this example of the tripel alogrithm the d_vj, d_vi and the d_ij?

The triangle before a number comes when d_vj + d_iv is smaller then d_ij like in the formula on picture 3.

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[spare loom]

Guys, how do i find factors of number fast, im slow at calculating

[spare loom]

Im doing it by dividing and disibility n stuff

I dont rven know how disibility works

[red nacelle]

gcf?

use euclid's algorithm

you set up the long division thing

"how many of the divisor fits in the dividend"

multiply the divisor by that number

bring it down

then subtract

if you get a number greater than zero you redo it

this time you divide the divisor by the remainder

if you finally get a remainder of 0 then the quotient is the gcf

also note that every even number (2, 4, 6, 8, 10, ...) are all divisible by 2

so if you have a fraction 2/4

you divide by 2 on the numerator and denominator to get 1/2

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[verbal lagoon]

Hi. Do these seem fairly okay to you?
Do I at least get half marks?

[median oar]

Your full name is in the pictures btw @verbal lagoon

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[verbal lagoon]

Do they seem correct?

[median oar]

It also does say that maybe you shouldn’t discuss the content of this paper with anyone else

[verbal lagoon]

Oh

But these are past papers

So it's fine

Yes??

They are past papers from my college

It's fine ??

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[halcyon ore]

im stuck idk what to do

[wispy osprey]

Me niether

[halcyon ore]

what the bot doin bruh

@lone heart lock in twin

.close

[halcyon ore]

.reopen

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[halcyon ore]

.reopen

[lone heart]

✅ Original question: #help-0 message

[halcyon ore]

yeah this please help me 😭

[vale wigeon]

yeah your first line is wrong

[halcyon ore]

omg

[vale wigeon]

i would say that your best course of action is to do 2^ both sides

that gets you $\frac{x-a}{x+a} = 2^c$

[ocean seal]

Ann

[halcyon ore]

how did u get there so quick whaaat

[vale wigeon]

||note i said 2^ both sides, and not both sides ^2 -- huge difference||

[vale wigeon]

simplify $2^{\log_2(x-a)-\log_2(x+a)}$ tbh

[ocean seal]

Ann

[halcyon ore]

😭

let me think about this deeply

[vale wigeon]

it ain't THAT deep

it's like, knee deep as far as log shit goes imo

[plush steppe]

that's once again not how you do logs at all

[vale wigeon]

maybe upper leg deep

like mid thighs

waist deep at MOST

[halcyon ore]

I understand it now

I blame it on doing too much studying today

[halcyon ore]

ok u got me this time I thought i was cooking

[plush steppe]

you should take a break

[halcyon ore]

how did u go from

2^log_2((x-a)/(x+a))

[plush steppe]

combine the left and convert to exponential form

[halcyon ore]

to the thing

[halcyon ore]

soon I'll just sleep its like 5:30 rn

[halcyon ore]

how from this tho

[plush steppe]

$2^{\log_2(x)} = x$

[halcyon ore]

oh thats just a log rule

I dont remember this one but bet

[halcyon ore]

I get it till here now

now let me think about next steps

$\frac{x-a}{x+a} = 2^c$

[ocean seal]

Conquestor

[deep dune]

Uhhh

[halcyon ore]

x-a = (2^c)(x+a)

[deep dune]

Gotta eliminate the x from the denominator somehow

[deep dune]

Yep

[plush steppe]

it isnt really taught often but i think it's good to recognise the form

[halcyon ore]

ic

[deep dune]

But there's x on the RHS

[halcyon ore]

ill remember it

[halcyon ore]

ye I'm thinking about ts

[deep dune]

...

[plush steppe]

@deep dune can you let this guy think about the problem

[deep dune]

Alr

[halcyon ore]

x-a = (2^c)(x+a)

x - a = (2^c)x + (2^c)a

[deep dune]

Go on

[halcyon ore]

• (2^c)a - a = (2^c)x - x

idk why it became bullet point

hol up

-(2^c)a - a = (2^c)x - x

[deep dune]

🤯

[halcyon ore]

hmm

I dont know how to get rid of 2^c from right side

cuz i cant divide it off

[plush steppe]

try to think about it

[halcyon ore]

wait can I use logs again or sum

[plush steppe]

no

[halcyon ore]

hmm

WAIT

OMG

-(2^c)a - a = x((2^c)-1)

[plush steppe]

👍

[deep dune]

Note that a, b, c are constants btw

So you don't need to care about them rlly much

[halcyon ore]

o

now let me try simplify a bit before dividing

[deep dune]

Good choice

[halcyon ore]

-(2^c)a - a = x((2^c)-1)

(-a((2^c)+1))/((2^c)-1) = x

I dont think this was the method

[deep dune]

Omg

[deep dune]

That has to be the method

[halcyon ore]

wait let me write it

down

[plush steppe]

simplify it

[halcyon ore]

ye i was jus writing it on the ipad

so i can read it better

than allat brackets

[plush steppe]

you should learn latex bro

hop on the latex train

[halcyon ore]

after my exams i promise

i have so much months after it

ok do i jus

[plush steppe]

what year are you in?

[halcyon ore]

multiply by -1

[plush steppe]

yh

[halcyon ore]

year 11 but doing year 12 work

my school goes up to year 13

[plush steppe]

idk why i asked that when i have no clue what age year 11 is

[halcyon ore]

2009

15 turning 16

[plush steppe]

oh cool

[halcyon ore]

so when i multiply by negative 1

i just proved it right

[plush steppe]

yep

[halcyon ore]

why does the a go outside the thing though im kinda curious

like outside the fraction

[plush steppe]

it's the same as multiplication

[halcyon ore]

this also might be real simple

ic

well

[plush steppe]

$a\frac{b}{c} = \frac{ab}{c}$

[halcyon ore]

OH yeah

cuz its basically

a/1

[plush steppe]

(only unless youre not doing mixed fractions)

[halcyon ore]

and top times top bottom times bottom

makes sense

so they couldve left a in the fraction

they just chose not to

[plush steppe]

yep

[halcyon ore]

or is putting a outside considered a simpler form

oh okay

great

[plush steppe]

no

[halcyon ore]

btw this is kinda a side thing but

whats the best place ot start learning latex

laTeX

[plush steppe]

my personal preference is to have it inside the fraction

[halcyon ore]

yea i would prefer that too probably makes it easier to read for me

[plush steppe]

https://www.overleaf.com/learn/latex/Learn_LaTeX_in_30_minutes

i used this

[halcyon ore]

oh wow

how long did it take u

truly 30 minutes?

[plush steppe]

there are some more resources in the pinned message of #latex-help

[plush steppe]

not that long

[halcyon ore]

oh beautiful

oh

i might just learn it when i wake up later then

[deep dune]

I use overleaf all the time

[halcyon ore]

tysm both of you

[halcyon ore]

oo ic

oh i forgot ann helped me earlier too haha

thank you all THREE of you

very very helpful

and ill remember that new log rule now

byebyeee

have a great day/night

.close

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[dusty grove]

How do I put x is not equal to 1 on desmos Google is useless

[modern sedge]

in what graph?

what exactly are u trying to do

[dusty grove]

I wanna graph a multiple rule function

[modern sedge]

oh and u want desmos to do that empty circle thing to indicate that the point isnt within the domain?

[dusty grove]

I wanna grapg this

[modern sedge]

,rccw

[vale wigeon]

what are these sixes

[dusty grove]

Thats a comam

Comma

Arabic ones ig

[vale wigeon]

oh i was about to say it looks upside-down lmfao

[modern sedge]

idt desmos can do this

[vale wigeon]

yeah i dont think it can graph this neatly

uhh hmm

hold on let me try

[dusty grove]

Ok

[fickle heath]

Why not just plot x³+2x and the point (1, 3*1) ?

[dusty grove]

What

[vale wigeon]

btw both expressions give the same value at x=1 anyway don't they

like f(1) is gonna be 3 even if you went and plugged 1 into x^3 + 2x anyway

[fickle heath]

The point (1, f(1))

[dusty grove]

Ye

[vale wigeon]

but also yeah you can plot individual pts

that is kinda your only option

[modern sedge]

u can pick the hollow circle in options then

[dusty grove]

So f prime of 1 is 5

[vale wigeon]

well ok like

{cond: expr, cond: expr, ...}

[dusty grove]

I can just write f(x) as x³+2x right?

[vale wigeon]

this is the desmos format for piecewise functions

you can put an expression without a condition at the very end, and that's gonna get graphed for all x's that DON'T satisfy any of the previous conditions

[vale wigeon]

indeed you can unless your teacher forbids it

[dusty grove]

Whyd my teacher forbid it

Not like he can change math as he wills

Its either right or wrong

[vale wigeon]

idk sometimes teachers can just arbitrarily say "YOU ARE NOT ALLOWED TO DO XYZ THING AND YOU CAN'T DO SHIT ABOUT IT BECAUSE ARBITRARY AUTHORITY AND FUCK YOU"

in less exaggerated terms, but something like this

or maybe you have a teacher who's strict on applying her and only her methods

[sour blade]

you can probably js graph this on desmos by using the function $f(x)=\frac{(x^3+2x)(x-1)}{x-1}$ and then use the point (1,3)

[ocean seal]

yoda on ketamine

[modern sedge]

u dont even need to do the x-1 thing

[dusty grove]

Na I asked him he said your right you could that

One time

[fickle heath]

You cannot plot f(x) = x³+2x without the point x=1 since there are points arbitrarily close to it

Any of these points could "look like" the point at x=1

[dusty grove]

Its okay I got my prob solved

[vale wigeon]

ok good

thats kinda what i was hoping

the "ultra-strict teacher" thing is an ANTI-preference for me

[dusty grove]

He's doesnt enforce anything he likes to hear what we have to say

[vale wigeon]

that's good

[dusty grove]

Like better ways to solve problems

[vale wigeon]

that's how teachers should be

[dusty grove]

N stuff

Ye he's goos

Good

But he keeps making mistakes in tests

I be telling him this function should be different but he won't listen🫠

Whats the function that makes a right angle with the tangent line cakled

Called

[vale wigeon]

normal

if you're talking about the straight line that is at right angles to the tangent.

[vale wigeon]

yeah that is called the normal

[dusty grove]

So the slope of the normal is -1/slope of the tangent

?

[fickle heath]

In general, yeah

Of course the slope can't be zero to do that

[dusty grove]

It could

But the tangent would be x=C

[fickle heath]

tangent_slope * normal_slope = -1

[dusty grove]

Ik

[vale wigeon]

unless they are a pair of horizontal and vertical

[dusty grove]

Y= C and X=C

Well thanks ann and Nel

And there was someone else

Yoda

.close

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[tired bane]

can someone check this idk why I got everything except the -16.. they had 2 rows of zero as the final rref

[gray isle]

careful with your signs

eg for your
R3 - 3R1

you messed up
-11 - 3(-3)

[ember depot]

u made mistake in 3rd step

[ember depot]

yup

so it becomes -11+9 = -2

what u did was -11 - 9 = -20

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[vocal belfry]

i need help with the third one and fourth

[lavish cave]

always helps to let there be an element, x

[junior vigil]

what does it mean for x to be in A? for x to be in B? for x to be in A u B?

first you are showing A is a subset of B, then A u B =B, then you are showing if A u B =B, then A is a subset of B

that is what 3 is asking

[lavish cave]

yeah, I'm not sure if you can use something like $A \cup B = A \cup (B - A)$

[ocean seal]

south

[junior vigil]

well if x in (A u B) then x is in A or x is in B (this is inclusive or, so x could be in both)

[lavish cave]

mhm, that's simpler

[vocal belfry]

i know

[vocal belfry]

i tried it but didnt work

[junior vigil]

which direction are you having trouble proving, first part or second part?

two sets are equal if each is a subset of the other

[vocal belfry]

second

if x∈A, then x∈A∪B. Since we have the hypothesis A∪B=B, this implies that x∈B. Therefore, A⊆B.

[junior vigil]

that looks good

you showed that every element of A is an element of B, which is precisely what you do to show one set is a subset of another

[vocal belfry]

got it

[junior vigil]

is that all you had questions about? :3

you said you had questions on 4 too right

[vocal belfry]

i solved it

it was easier than i expected

i used contradiction

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[fading galleon]

How to write formal proofs for geometry

[vale wigeon]

uh

tf they mean approximated

oh chord AB

[vale wigeon]

is this your very first time writing a geometry proof

you might wanna try some easier proof questions i think

[vapid steppe]

isnt rtheta is just the arc length

which is like almost the same as the chord length

[winter pewter]

you could try joining A and B then perform some sine law

[winter pewter]

if only that was acceptable

[fading galleon]

It is but how to prove it

[lone heart]

@fading galleon Has your question been resolved?

[austere fossil]

for very small values of theta (given in the question) it is a valid approximation
tanx=sinx=x

so yeah

[fading galleon]

Is the arc length formula just an approximation ?

I thought it was exact

[austere fossil]

im sorry i read find the length of AC

if you need to find AB then yeah its precisely Rtheta

[fading galleon]

How do I write the proof

In formal language

[austere fossil]

not my cup of tea, sorry.

[plush steppe]

idk if proving that will help solve the question, but you can do so by considering that the arc length is the proportion of the angle theta to the sum of angles at a point multiplied by the circumference of the circle

[austere fossil]

okay my bad i misjudged, it asks you to find the length of chord AB

its just a trigonometry question then

[plush steppe]

wait a sec, the area of triangle ABO is 1/2 r^2 sin theta.

[austere fossil]

notice that triangle OAB is isosceles and then try to find the side length AB using trigonometry, then finally use the approximation "tanx=sinx=x" for a small angle x and you will get your result.

[plush steppe]

exactly what i was thinking

[austere fossil]

it involves a construction and youre done

[fading galleon]

Do we really need to use all that can't we just go from the circumference formula to that

[plush steppe]

i mean we dont know how strict the marking is

they might not accept using arc length to approximate chord length

[fading galleon]

It's just A level

[plush steppe]

going from showing exact expression for chord AB to showing that it can be approximated would definitely be accepted

[fading galleon]

But isn't the chord exactly the same as the arc length

[plush steppe]

arc length is the length on the circumference from A to B

chord length is the length of line joining the two points straight from one another

[fading galleon]

Whats a chord length

[plush steppe]

the length of the chord

[fading galleon]

Is it a line connecting A and B

[plush steppe]

yes

[fading galleon]

Oh

[winter pewter]

as i said before you could join A and B and use sine law in triangle AOB

then use small angle aproximations on the result to prove it

[fading galleon]

But the angle seems big

Almost 45°

Ok nvm

It says small

.close

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[vocal belfry]

.close

[lime spruce]

I want help

[lone heart]

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[old widget]

There are n lines which has 분산 of 8 in length and average of 8 in length, when creating a triangle by cutting each line to 3 lines,what is the average area of triangle made by these lines?

[grave matrix]

[bit busy, can't rly help, but to assist others:
분산 means "variance" here]

[lone heart]

@old widget Has your question been resolved?

[lone heart]

@old widget Has your question been resolved?

[lone heart]

@old widget Has your question been resolved?

[old widget]

<@&268886789983436800>

[molten crow]

The "3 lines" are to be equal in size or do they have random lengths between each other?

[old widget]

They divide into equal sizes

[molten crow]

I wonder if the average of the area is influenced by the act of squaring.

Whatever, my best idea:
Knowing that the line is broken down into three equal parts, this must mean that they form an equilateral triangle.

From there you could calculate the area of it using the average length
Im not precisely sure if this step creates a difference in the resulting average though

[lone heart]

@old widget Has your question been resolved?

[steep escarp]

이분 어제 저녁에도 있으셨는데 아직까자 resolved 안되셨나 보네

[lone heart]

@old widget Has your question been resolved?

[crisp wigeon]

So an equilateral triangle?

[old widget]

Yes

[crisp wigeon]

You know the fact that variance = mean of squared quantities - mean ^2

[old widget]

Isn't it in wrong order?

Because you have to sum the pow(abs(value - average),2) first then divide by count of elements

[crisp wigeon]

Well I gave a proof

[crisp wigeon]

Area of equilateral triangle is √3/4 a^2 so mean of area is just √3/4 *(variance+square of mean )

Giving the final ans as 18√3 unit area

[old widget]

What the heck are these

[crisp wigeon]

I am not sure you don't get the symbols or smth?

Could you show how your books write it?

[old widget]

I remember it was not written like that

[delicate glacier]

Usually theres little hooks on the end of the lines

But thats also a valid way to write it

[old widget]

What the heck is even that?

[late dew]

Summation

[crisp wigeon]

It mean like summing over given something

So like (x1+ x2+...xi)/n is commonly written as Sigma (that symbol) xi and you write i goes from where to like 1 to n here all down or sometimes the upper limit is written above while lower is written down

If you show me how your book writes the mean i could try writing it like that

[delicate glacier]

Its sigma, just like me

[deep dune]

i'd just do something like a greek epsilon

[lone heart]

@old widget Has your question been resolved?

[old widget]

So it is not an operation but a loop

[wraith ferry]

Um yeah? But technically it is just using the operation of "addition".

Like if you have a bunch of numbers you want to add up, you can simplify the working out by adding that symbol.

[topaz ibex]

help channels really are just as bad as people say they were...

[topaz ibex]

whyyyyyyyyyyyyyyyyyyyyyyyyyyyy

[old widget]

?

[topaz ibex]

thats greens theorem in your name and your name and pfp are green wow

[woeful pulsar]

from the previous help channel, we got to area of equilateral triangle but we are kinda stuck at changing one formula for the variance to the other formula for the variance

@old widget if you can say which textbook you have, I can try to find it in there

I would think many textbooks have the derivation of one formula from the other

[old widget]

I cannot

[thick ledge]

nice to see you here

[mortal anchor]

Anyone online

[steep escarp]

I think I have seen this (아라스킨충님) session for approximately 3days

[lone heart]

@old widget Has your question been resolved?

[old widget]

There were no progression made after getting formula of area of triangle

근데 저런 처음보는 기호는 어쩌다가 나온거지

[steep escarp]

?

무슨 기호여?

[old widget]

이거도 있고

[old widget]

저 줄그은것도

[crisp turtle]

$=x$

[ocean seal]

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

[crisp turtle]

는 x의 평균이란 말이에요

$$=x = \frac{\sum_{i = 1}^n x_i}{n}$$

[ocean seal]

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

[crisp turtle]

$$\sum_{x = a}^b f(x)$$

[ocean seal]

shiru

[crisp turtle]

는 대충

var sum = 0
for (int x = a; x <= b; x += 1){
  sum += f(x);
}
return sum

이고요

[fathom birch]

n개의 선분의 평균과 분산이 모두 8일 때, 각 선분으로 만든 n개의 정삼각형의 넓이의 평균은?

이건가요?

이게 맞다면

선분의 길이가 l일 때, 정삼각형의 넓이는 l²×sqrt(3)/36이다.
따라서 구하고자 하는 평균값을 M이라고 하면 M = E(X²)×sqrt(3)/36이다.

V(X) = E(X²) - (E(X))²
=> 8 = E(X²) - 8²
=> E(X²) = 72이므로
M = 2sqrt(3)이다.

이렇게 구해집니다.

'확률과 통계'의 범위로 서술하였습니다.
이보다 더 낮은 수준의 풀이는 모르겠네요

[old widget]

E() 는 또 뭐지

[fathom birch]

평균이요

[crisp turtle]

기대값

[old widget]

그렇게 두개인가요?

[ripe kraken]

awesome

[steep escarp]

와 천재 한국인들 개간지네

[old widget]

X는 어디서 나왔나요?

[fathom birch]

X를 선분 l의 길이에 대한 확률변수로 정의한 채로 서술했습니다

[old widget]

확률변수요?

[rustic brook]

이분은 왜 맨날 이서버에..

[ember bane]

이 대화는 벌써 여러 날 동안 열려왔습니다...

[old widget]

생각해보니까 실제로 평균과 분산 모두 8인 배열을 직접 찾으면 직접 넓이 계산할 수 있는건가

[fathom birch]

되긴 해요

선분길이
8-2sqrt2
8+2sqrt2

정삼각형 넓이
(72-32sqrt2)(sqrt3)/36 (72+32sqrt2)(sqrt3)/36

넓이 평균 = 2sqrt3

[onyx acorn]

<@&268886789983436800> not sure what's going on here

[hushed locust]

don't spam others' help channels.

[deep swan]

<@&268886789983436800>

[onyx acorn]

did you just ping mods on a mod? after flooding a help channel to boot?

[grave matrix]

"Hello, police? I'm robbing someone"

[old widget]

? 이게 진짜 되네요

[steep escarp]

Guys don’t accuse them they r having a “serrrrrrrrious” conversation on this channel

[magic onyx]

hello

gyus

hey hey hey

listen up plz

[winter pewter]

ts still up

[keen obsidian]

is this a competitive programming problem

?

you could ask for help in erichtos swervcer

[icy coral]

,help

[ocean seal]

A brief description and guide on how to use me was sent to your DMs!
Please use ,list to see a list of all my commands, and ,help cmd to get detailed help on a command!

[fluid hull]

this is not how u ask a question whatsoever

[woeful pulsar]

you can go to #❓how-to-get-help to learn how to get your own channel

[keen obsidian]

bro ts still up

[old widget]

I think I got it

[steep escarp]

Finally…!

[open spruce]

bro why this channel still occupied

[rustic mist]

This channel has been occupied since 1st november?

[thick ledge]

fr how

[keen obsidian]

is this a competetive programming problem?

[old widget]

I got the answer

[steep escarp]

Yay!

[onyx acorn]

!done, if you are.

[lone heart]

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

[tacit arch]

.close

[lone heart]

Channel closed

Closed by @tacit arch

Use .reopen if this was a mistake.

[lone heart]

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[vast harbor]

$\theta \equiv 2\theta \equiv 3 \theta \pmod{\pi}$

[vast harbor]

Does this mean:
$$ \theta \equiv 0 \pmod{\pi}$$

[vast harbor]

And how to prove

[mortal trellis]

how would you solve x=2x

[vast harbor]

But it's like +k \pi

k in Z

[mortal trellis]

the great thing about modulo is: doesnt matter

theta=2theta+kpi

[vast harbor]

I see thx

.close

[lone heart]

Channel closed

Closed by @vast harbor

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[lone heart]

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[upbeat wedge]

What's a dummy variable

[upbeat wedge]

I'll sleep soon, so i shall check if anyone answers later, pls ping me

[inner belfry]

It’s a variable that is used for convenience in the expression as a placeholder. In other words, it doesn’t have a meaning by itself. Think about, for example, how the variable in an integral can be changed without changing the result.

[upbeat wedge]

Hmmm

[inner belfry]

[\rb{
\int\map fx,\dd x=\int\map ft,\dd t
}]

[ocean seal]

Cooly

[inner belfry]

Here, x and t are dummy variables.

[upbeat wedge]

Ye

Well

Here's another example that confuses me

Say f(5x^2)=3x

How to find f(x)

It's related to dummy variables btw,
Idk how many ways i can solve this but one way is supposing U=5x^2

Then f(u) =3√(u/5)

Then f(x) is 3√(x/5)

I don't get how we can go from u to x in the last step, didn't we let u=5x^2 at the beginning

[inner belfry]

When we write f(u)=…, what we really mean is “the rule that tells us what f does to its input”. It doesn’t matter whether the input is x, u, t, or, . Once we’ve found the rule, we can rename the dummy variable however we like.

[upbeat wedge]

Im starting to feel better about it, can u explain that again if u don't mind??

Maybe more clearer?

[inner belfry]

I’ll try

[upbeat wedge]

Ty

[inner belfry]

When we write f(u) = 3sqrt(u/5), we’re describing the rule or the formula that defines the function f. The letter inside the brackets (be it u, x, t or anything else) is only a dummy variable, meaning it stands for the input of the function. Once we figured out what the function does to an input, we can replace the dummy variable with any symbols we like. So after finding f(u) = 3sqrt(u/5), we can write f(x) = 3sqrt(x/5); this doesn’t change the function, it just renames the input variable.

Is that clearer?

[upbeat wedge]

Ye i think that's good

Do u mind if i say what i understood and u correct me?

[inner belfry]

Sure

[upbeat wedge]

Ty

So
When we have f(5x^2)=....
We have the rule that gives us what the function does to the whole 5x^2, not just x

So our goal is to find what it does to a single input like t,u or whatever, like whatever the function does to x, it will do so to any other variable

Whatever it does to g(x) ( so it's f(g(x)) ), it will does to g(t) as long as g is the same

Am i right? I just got this thought of g and i wanted to see if it's correct

[inner belfry]

Yes, I think you got the idea. I’ll just rephrase it slightly here.

When we have something like f(5x^2) = …, that tells us how the function f acts on the whole expression 5x^2, not just on x. Our goal is to find a general rule for what f does to any input; we often call that input t, u, or x.

In other words, whatever f does to 5x^2, it will do the same to any expression of the same form, like 5t^2. So if we can find the rule connecting the input to the output, we can replace 5x^2 with a single variable u (or t), find f(u), and then rename the variable to get f(x).

Generally, when we write something like f(g(x)), it means we’re applying f to whatever the inner function g gives us. The rule that defines f doesn’t depend on the letter used; it depend only on the input value. So whatever f does to g(x), it will also do to g(t) or g(u), as long as the function g itself is the same. In other words, f(g(x)) and f(g(t)) follow the same rule, so the variable name is just a label for the input.

[upbeat wedge]

Ye that's what i meant

Ty buddy, u helped me a lot

Btw, how can i learn more and deeper about stuff like that, functions and calculus

Which course? Real analysis?

[inner belfry]

Yes, real analysis definitely

[inner belfry]

No problem

[upbeat wedge]

Any others?

[inner belfry]

Well, Calculus too

Mostly in this two

For the first two years of university

[upbeat wedge]

Any recommended textbooks?

[inner belfry]

Hmmm. You might get a better response in #book-recommendations but I’d personally say Calculus Volume 1 by OpenStax (which is a free online textbook), Elementary Calculus by Michael Corral, and Introduction to Calculus: A Computational Approach. Then maybe you can look into Real Analysis in Understanding Analysis by Stephen Abbott.

[upbeat wedge]

Okay, thanks very much bro, i appreciate it alot

Gl

[inner belfry]

Take care

Type .close if you’re finished here

[upbeat wedge]

Alr ty

.close

[lone heart]

Channel closed

Closed by @upbeat wedge

Use .reopen if this was a mistake.

[lone heart]

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[median oar]

looking at a problem like this

[median oar]

surely it's true that each part by themselves are both open and closed

like (-10, -1) is both open and closed

since clearly [0, 3) is clearly open

and so is {6}

so the complement of (-10, -1) would be open

which would also mean that if i have X = (-1, 0) U (0, 1)

then (0, 0.5] is closed

[lone heart]

@median oar Has your question been resolved?

[median oar]

.close

[lone heart]

Channel closed

Closed by @median oar

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[lone heart]

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

[lapis umbra]

anyone know how to do this?

[lapis umbra]

I can't decipher what the top part means

just looking at it, it makes sense that it's just 6k

it seems like 6* 1, 6 * 2, 6*3

[tacit arch]

right

so for the first 3 terms you can factor 6 * 6 * 6 out

[lapis umbra]

wait what why is that

if it's just 6k

then you can just use that as the top

[tacit arch]

i mean first 3 terms of the product

6^3 * 3!

try to find a general pattern for the kth term

[lapis umbra]

how do u know it's this

[tacit arch]

i used this

[tacit arch]

then did this

and used 6^3 = 6 * 6 * 6

[lapis umbra]

but why are you factoring out 6?

[lapis umbra]

if that is true, isn't it just 6k?

[tacit arch]

to do this

[lapis umbra]

because k goes from 1 to infinity

so it's just 6 * k forever

and k changes as the index goes up

so if u have 6k, then doesnt it give you 6 + 12 + 18, which is what you want?

[tacit arch]

i don't know why you're trying to jump ahead so much

[lapis umbra]

wdym

[tacit arch]

you haven't even gotten the general kth term yet

[lapis umbra]

isnt that the general k term? 6k

[tacit arch]

this is the 3rd term

k = 3

well with 3!

[tacit arch]

meant to quote this one

[lapis umbra]

huh

[tacit arch]

does 6k have the form 6^3 * 3! for k = 3?

[lapis umbra]

no, but wait how do you know that second general form is the right one?

[tacit arch]

.

just for the numerator

[tacit arch]

you meant top as in numerator right?

[lapis umbra]

but why are you doing this? isn't it already just telling you what the general term is? it says 6k in the top

yes

[tacit arch]

no

[lapis umbra]

so 6 x 12 x 18

[tacit arch]

yes that's the numerator for k =3

[lapis umbra]

and then it has the part that says 6k

[tacit arch]

when k = 3, 6k = ?

[lapis umbra]

18

[tacit arch]

right so you stop at 18

6 * 12 * 18 is the numerator

[lapis umbra]

but how do you know the index is three there? because there are only three terms?

[tacit arch]

k is the index

that's implied when the sigma doesn't have anything underneath

[lapis umbra]

oh, so then here, what does the .... mean

because i think when they do the +...+ then put something like 6k that shows the formula is 6k

based on what u said they probably should have put 6 x 12 x 18 x .... x (6k)

[tacit arch]

$\cdot$ is multplication

[ocean seal]

riemann

[tacit arch]

$+$ is addition

[ocean seal]

riemann

[lapis umbra]

ok

so the .... means a bunch of multiplications?

but i thought the dots had to hover to mean that, not be on the floor like a period

[tacit arch]

both are used

[lapis umbra]

oh, well my professor sometimes does the same thing but with pluses

she writes +...+

does that mean plus and a bunch of multiplications?

so like if there was something like

6+12+18+...+6k

[tacit arch]

... is generic for repeating the pattern

1/9 = 0.1111.....

[lapis umbra]

oh

[tacit arch]

e = 1 + 1 + 1/2 + 1/3! + 1/4! + ...

[lapis umbra]

how do u know that

[tacit arch]

https://en.wikipedia.org/wiki/E_(mathematical_constant)

[lapis umbra]

ok

[lapis umbra]

um ok

so this here basically means

that the pattern is repeating forever?

then what does that 6k mean

so like

6x1x6x2x6x3x6x4x6x5x6x7x6x8x6x9x6x10?

[tacit arch]

no

the pattern repeats until (6k)

[lapis umbra]

oh

but isnt 6k forever?

[tacit arch]

first term is only 6 because k=1

[lapis umbra]

it goes to infinity?

[tacit arch]

no k is fixed in the fraction and that's finite

[lapis umbra]

but is this right?

[tacit arch]

for k = 10 sure

[lapis umbra]

oh i see what you mean

so you can bring out 10 6ses and you're left with a factorial

is that right?

[tacit arch]

yes

[lapis umbra]

okay so it's

[tacit arch]

you can also use a calculator to check your own work

[lapis umbra]

6^k * k!?

[tacit arch]

,calc 6^10 * 10!

[ocean seal]

Result:

2.194196594688e+14

[lapis umbra]

ok

well this notation was confusing, they should have put dots before the .... to show the pattern continued

[lapis umbra]

is this right?

[tacit arch]

yes

[lapis umbra]

wait i still need to solve it

[tacit arch]

yes

[lapis umbra]

ok

if ur taking a limit and you get something like (k+4)^1/k

isnt that just 1, because 1/k goes to 0, and then anything to 0 is 1

ok i did root test on the problem and i got

6 * k root of k factorial

k root of 6k!

i'm not sure how u take the limit of this

[tacit arch]

ratio test is probably better

[lapis umbra]

ok

ok

i got 6(k+1)

over a lot of ks

so that means it's 0, and so that's less than 1, so it converges?

is that the right answer?

well, it has to be, diverges isnt the right answer and it's never none of the above

[lone heart]

@lapis umbra Has your question been resolved?

[lone heart]

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[alpine saddle]

help

[zinc haven]

what exactly do you need help with? do you not understand the transformations or...?

[gray isle]

also there are inappropriate use of = signs there

[alpine saddle]

solving efractoins when there is = between them

[gray isle]

is there a specific question you're stuck on?

[alpine saddle]

cross multiplying ths

can I chat later? because I have to go somewher r

sorry

[gray isle]

close with .close and come back later when you have the time

[lone heart]

@alpine saddle Has your question been resolved?

[lone heart]

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[half seal]

Translation:
Expression (given expression) for all values a,b exist in R that are defined, is identital to: and you can see the possible answers, with the last one meaning "I dont know"

[tranquil radish]

Well what have u tried so far?

[half seal]

Im not good at math so im sorry if im stupid but, I tried multiplying (a+b) with up and bottom until I got (a+b)^3, (a+b) then divided which led to (a+b)^2. then I kept doing simmilar things but it led to it being wrong

[tranquil radish]

Does the question say show ur work?

[half seal]

No

[tranquil radish]

Uhm maybe u can try subing in a value for a and for b and see which expression is the same

[half seal]

Okok bet, thank youu

i found the solution and figured out how to do this 🙏

thank uu

.close

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[kind pasture]

anyone here that uses geogebra that gets why i get a different result than the solution? mine is the first pic

[fluid hull]

second one is correct

[kind pasture]

yeah, ik, but why do i get the wrong answer on the first one

[fluid hull]

probably glitched out lmao

i dont see anything different in any of the 2 eqs

[lone heart]

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[night mountain]

Can anyone show me how to draw the FBD with an artificial cut here.

[hushed locust]

start with choosing where you want the cut and which side you want to show

[lone heart]

@night mountain Has your question been resolved?

[night mountain]

I was thinking I would cut at B, or to the right of B which would give me a reaction force at A, TA to the left and then T0 to the right. The right hand piece would have TB going to the left (I think) and then no other forces, so TB would just be zero. Is this accurate?

But if that were the case, do I even need to make a cut? If no, does that mean I only need to make a cut if there are multiple torques acting at once? And in this case, since there is only one torque, a cut would not be necessary?

[hushed locust]

well generally i would not cut exactly at the point of application of a concentrated force or torque

generally you need to make as many cuts as there are distinct areas of the shaft (it's a distinct area if there is a force or torque applied on one side and not the other)

[lone heart]

@night mountain Has your question been resolved?

[night mountain]

Yeah that makes sense. How would you go about dividing this shaft?

And the division of the shaft would only be to create the torque diagram, correct?

I do understand what you said about the egeneral rule where you need to make as many cuts as there are unique areas of the shaft, but considering there is only one force in this shaft, the artificial cut would just reduce the length of the shaft where the left side would contain all of the meaningful information and the right side would the have, for instance TB = 0

[hushed locust]

well in this case, the main divisions would be between A and B and between B and C. so make one cut in between each, at some variable distance x (you may or may not use this distance, depending on the problem). then chose one side of the shaft to use for each cut (doesn't matter which really)

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[digital crypt]

can someone helpe me with this pls

[limpid turret]

!status

[lone heart]

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

[outer island]

6 7

[digital crypt]

i don t know where to begin f is surjective so f x have at leste a solution bt idk 1

[median oar]

f(E) = F is an equality between sets right?

[digital crypt]

yes

[median oar]

How do you prove equality between sets

[digital crypt]

by doing two inclusions

[median oar]

So try that

[digital crypt]

ok thx ill try

sry to disturb u again i found the first iclutio bt idk how to prove the second one

.solve

.close

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[vale slate]

I know the solution is correct but wanted to ask what I could have done better with my solution:

Problem: Give an example of a vector space $V$, a basis $v_1,\cdots,v_n$ of $V$ and a sub space $U$ of $V$, so that no vector $v_1, \cdots ,vn$ lies within a basis of $U$

Solution:

Let $V=\mathbb{F}_{2}^{2}$.Then $B_V={\begin{pmatrix} 1 \ 0 \end{pmatrix},\begin{pmatrix} 0 \ 1 \end{pmatrix}}$ is a basis of $V$. The vectors $v_1,\cdots,v_n$ are linearly independent because the equality $\sum{i=1}^n a_i v_i = 0$ only holds for coefficients $a_1=a_2=\cdots=a_n=0$. We choose the subspace $\langle \begin{pmatrix} 0 \ 0 \end{pmatrix}, \begin{pmatrix} 1 \ 1\end{pmatrix} \rangle$ with the Basis $B_U = {u} = \begin{pmatrix} 1 \ 1 \end{pmatrix}$. The condition is satisfied since $u \notin B_V$ and $u \neq \begin{pmatrix} 1 \ 0 \end{pmatrix}$ and $u \neq \begin{pmatrix} 0 \ 1 \end{pmatrix}$

I feel like reading this is cumbersome, any tips on what I could improve?

[north rover]

Seems like you copied text according to discord...

[vale slate]

Yup 😄

[north rover]

Its fine just put some _ and ^ in and you should be fine

[vale slate]

I fixed it 🙂

[north rover]

Uh not quite

V=F2

[vale slate]

True

There we go

[vale crag]

the sum is still f-ed up

[ocean seal]

Baker

[vale slate]

Ill go fix it and come back

.close

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[vale slate]

.reopen

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✅ Original question: #help-0 message

[vale slate]

Problem: Give an example of a vector space $V$, a basis $v_1,\cdots,v_n$ of $V$ and a sub space $U$ of $V$, so that no vector $v_1, \cdots ,v_n$ lies within a basis of $U$

Solution:

Let $V=\mathbb{F}{2}^{2}$.Then $B_V={\begin{pmatrix} 1 \ 0 \end{pmatrix},\begin{pmatrix} 0 \ 1 \end{pmatrix}}$ is a basis of $V$. The vectors $v_1 ,\cdots,v_n$ are linearly independent because the equality $\sum{i=1}^{n} a_i v_i = 0$ only holds for coefficients $a_1=a_2=\cdots=a_n=0$. We choose the subspace $\langle \begin{pmatrix} 0 \ 0 \end{pmatrix}, \begin{pmatrix} 1 \ 1\end{pmatrix} \rangle$ with the Basis $B_U = {u} = \begin{pmatrix} 1 \ 1 \end{pmatrix}$. The condition is satisfied since $u \notin B_V$ and $u \neq \begin{pmatrix} 1 \ 0 \end{pmatrix}$ and $u \neq \begin{pmatrix} 0 \ 1 \end{pmatrix}$

I forgot to add my question ... I wanted to know what I can improve upon with respect to my solution text as I feel it is cumbersome to read.

[ocean seal]

Baker

[tiny fable]

I would honestly delete this one line, I think it's very clear that that's a legitimate basis and it just makes it more fluffy to include it

Also, I'm assuming <> means "span of"? in which case the span of (0,0) and (1,1) is the same as the span of (1,1), so (0,0) is redundant

and technically a basis is a set, so you shouldn't say a vector is a basis on its own

[vale slate]

Yes <> means span, the notation was introduced by our professor. So the whole line can go ok. And with the vector point you where I say B_U = {u} = ( 1 \ 1) ?

[tiny fable]

yeah I would just say B_U = {(1,1)}

[vale slate]

Oh ok

[tiny fable]

the rest i think is good

[vale slate]

Perfect thank you very much! 🙂

That makes it much more comprehensible. What would you say I can do to improve my proof writing ? Just read a lot of proofs or are there books I could read?

[tiny fable]

I think just doing proofs yourself is the best way to go tbh. a lot of people (myself included) start off by just writing down wayyyy too much because they feel like they have to. I would recommend asking yourself at every step "would anyone in my class understand this without me explaining why", and use that to judge for yourself

[vale slate]

That sounds good, although - unfortunately - I am at a point where I am a Postgraduate who hasn't had much exposure to math during Undergrad studies (I studied Business Information Systems), I find it hard to sometimes answer these questions. I cannot discern, yet, between necessary and unnecessary information in a proof.

[tiny fable]

yeah i know how you feel, i had a friend in the same boat during my graduate studies. if you're friends with anyone in your program, it would probably be best to ask them for their opinions too

practice makes perfect tho ultimately

[vale slate]

Yeah let's see if that will happen I study in germany via a long distance university ... no campus unfortunately but study groups are a thing I will check up on that.

Thank you very much, I will probably be back asking more questions 😄

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[tribal yarrow]

I have some sort of a stupid question, does anyone have tips and trick how to do mental math calcs from hex to binary?

So if you see something like 0xAD

Youll instantly know its
1010_1101

[regal condor]

Going from a base k to base k^n or vice versa is easy

For hex to binary in particular, think of it this way

A is 10 in decimal which is 1010 in binary

D is 13 in decimal which is 1101 in binary

Essentially when you're going from k^n to k, you can take it one digit at a time

[tribal yarrow]

Is there a way to skip converting to decimal altogether

Apparently my work is reading hex dump and i need to know which bits are flip by a glance

[regal condor]

I mean not really, but you can just memorize it

[tribal yarrow]

So im trying to think to remember like when we kids a mult table

[regal condor]

It will take barely a second once you get used to jt

[tribal yarrow]

I see

[regal condor]

The idea is, you can deal with single digits rather than the whole number

Hullo cloud

[undone rampart]

There is a way, no?

[tribal yarrow]

My work is something like

A1 35 9F 37 13
23 19 83 65 10
61 42 85 16 14

Its pretty exhausting if i have to do calculations when i see like 1000 lines of these

[hushed locust]

every digit in hex corresponds exactly to 4 digits in binary. so if you just learn the conversions for each digit then you replace each hex digit with the corresponding binary digits

[regal condor]

I mean it's memorization

[tribal yarrow]

Ig i need to practice converting 1 to F in to binary like a table then

[regal condor]

What exactly are you supposed to do with these

[regal condor]

Yeah ig

[tribal yarrow]

Some bits in the data represent some errors that i need to match with the specs table

[undone rampart]

Meant something like this

[tribal yarrow]

And i need to quickly look up which bits are flip to find the errors

[undone rampart]

The values yea
But process

[tribal yarrow]

Also something annoying there are some flags that take 5 bits instead of 4 bits

[regal condor]

Yeah you don't need to take the whole thing to decimal

[tribal yarrow]

So you kind of concatenate the other hex num

[undone rampart]

Js the A to F

[regal condor]

But you still are either memorizing what each hex corresponds to, or going to decimal and then to binary

[tribal yarrow]

Yeah i get how you change the valu, maybe ill need to remember to forego the calculations

[regal condor]

Oh yea 0-9 is the same

[tribal yarrow]

Okay then, table it is

[regal condor]

If it's 1000 lines surely you're allowed to automate it somehow?

[tribal yarrow]

Yesn't

[regal condor]

That's weird

[tribal yarrow]

Was told to get used to manually reading it

Some tools apparently is unable to export the hex dump

So you have to read it directly

If i can have access to the dump ofc im automating it

[regal condor]

I think that's the point where I'd decide "okay fuck this, not for me"

Best wishes to you though

[tribal yarrow]

Unless if we wanna go crazy and put a camera on the output screen

And do ocr

Lmao

[tribal yarrow]

Thanks for the suggestion tho

[regal condor]

Lmao

Good luck

[tribal yarrow]

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[lone heart]

@split swift Has your question been resolved?

[inner belfry]

What have you tried?

Unconditionally, X and Y are independent, so Cov(X,Y)=0. But oncne we condition on the triangular region X+Y>1, the variables become negatively coorelated: if X is large, then only small values of Y keep X+Y>1, and vice-versa.

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@split swift Has your question been resolved?

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[faint monolith]

IDK WHAT TO DO

idk what to do

so 5t=2npi +- 4t

[cinder tundra]

Remember that 4 cos³θ – 3 cosθ = cos 3θ.

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[faint monolith]

THANKS

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[warped topaz]

The last pic is the solution

[regal condor]

What do you not understand

[warped topaz]

I understand we take the derivative and set it equal to 0 to find the extrema, but then why are they saying f(-1) = -1

Why are they doing that

[regal condor]

The extrema either occur at points where the derivative is zero, or at the endpoints

[warped topaz]

Ahhh

[regal condor]

If you're looking at a closed interval to be specific

[warped topaz]

So the endpoint is lower than the extrema, so its the absolute min, right?

[regal condor]

Yes

[warped topaz]

And same for the endpoint being the absolute max

Ok got it

Thanks!

❤️

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[humble hare]

How do I prove it

[regal condor]

,rccw

[regal condor]

Well, there's 4 things. Calculate the mean and median

[humble hare]

Ok lemme just write that down

Ok so the mean is 77k and the median is 74k

Where do I go from here

<@&286206848099549185> sorry for ping

[woven oyster]

I will make the calculation myself and tell you

I'm getting mean as 8.4852k

I'm getting median as 8.4852k

Hence they are same

[humble hare]

could you tell me the method too

[woven oyster]

For mean just add all the values and divide by count (here 4)

For the median we have two values sqrt(50) sqrt(98) . So adding then and dividing by 2 will give the answer

[humble hare]

i did and i got 77k in a square root and 74k in a square root

[woven oyster]

First take square root and then go for calculation

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[unborn burrow]

im not so confident about this proof

[unborn burrow]

Let me see if I can reason it

So we took some arbitrary epsilon > 0

but since g is continous at L we know that there exists some distance delta that we can make y close enough to L such that g(y) is close enough to g(L)

since f range belongs in the domain of g and f converges to L

we know there exists some distance we can make x close to a scuh that f(x) - L is less than delta

but if f(x) - L is less than delta then that must imply that |g(f(x)) - g(L)| < epsilon

just feels like theres possibly something missing tho

like I didnt really use the |x - a| < delta_2 anywhere

[sour verge]

The delta_2 is what ensures that |f(x) - L| < delta

[unborn burrow]

hm yeah thats true

[sour verge]

You pick delta by continuity of g so that whenever |f(x) - L| < delta, you have |g(f(x)) - g(L)| < epsilon

Then you pick delta_2 such that whenever |x-a| < delta_2, you have |f(x) - L| < delta.

So all in all, if you have |x-a| < delta_2, then you have |f(x) - L| < delta, so by continuity |g(f(x) -g(L)| < epsilon.

[unborn burrow]

ok thanks

.solved

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[coarse crag]

helloo

why is it not 2?

does cerrtain mean 100%?

[fluid hull]

yes

[coarse crag]

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[rapid ocean]

How do you get that blurb on the side that says math

@strong orbit

Ki baz shooshtarizadeh

Alexis

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[sturdy echo]

Please explain how to solve 13th and 14th sum

[eager rampart]

,rccw

[nimble fern]

what have you tried so far?

[sturdy echo]

I wrote the function in inverse and tried to find out range using the given domain for inverse as a function's range is its inverse's domain

But I got wrong answer

[nimble fern]

i see

i think you can first check the where the local minimum is located at and it's corresponding value

[nimble fern]

e.g. for Q13 you can use method of completing square

[lone heart]

@sturdy echo Has your question been resolved?

[nimble fern]

Have you tried the method mentioned above?

[lone heart]

@sturdy echo Has your question been resolved?

[rocky wren]

does that say x is an element of the Range?

[sturdy echo]

Wait I check

[nimble fern]

gtg

[sturdy echo]

Alr I will tell u if it worked or not

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[covert abyss]

have this exercise if someone needs it

[lunar saddle]

[lavish cave]

can you send this as an image instead?

[lunar saddle]

!nopdf

mb

[lavish cave]

!nopdf

[lone heart]

Please post images (such as PNGs or JPGs) of the question rather than other filetypes such as PDFs which have to be downloaded. Non-image downloads can potentially contain viruses or other security risks.

[lunar saddle]

smh

[covert abyss]

gimme a sec then @lavish cave

[lunar saddle]

ty

[lavish cave]

oh wait this isn't even your channel

.reopen

[lone heart]

✅ Original question: #help-0 message

[covert abyss]

dun @lavish cave

[gray isle]

This channel is set to close.
Please make a new one.

[alpine sable]

Hi guys

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@midnight hemlock Has your question been resolved?

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[fresh flame]

I’m not sure what technique to start

By parts?

[wraith ferry]

Hi! Notice how the bounds are from -pi/4 to pi/4? I wonder if we can exploit one of the properties of the inner function.

[fresh flame]

ye im guessing its just 0 because 0 because its an odd function

but if I were asked to just find the antiderivative what would I do?

not the definite integral

[wraith ferry]

Hmm

I don't see any immediate trick to be honest. You could expand sin(5x) and somehow get something in terms of sin(x), then you could rewrite tan^2(x) into 1 + sec^2(x) and try and cancel out some terms. It looks long and tedious, but that might be the approach.

,w integral tan^2(x)sin^3(5x)

[ocean seal]

Results provided by WolframAlpha

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Requested by redfernstation

[wraith ferry]

Yeah good luck with that

[fresh flame]

lol ok thx

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