help-39

[past perch]

.reopen

ok wtv i'll explain

now let's say for example in a different scenario, you start at the origin again and this time move right with a positive velocity

youre still moving away from the origin

HOWEVER

lets go back to this

[pearl pond]

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[past perch]

you moved left first

[past perch]

then after you moved left for some time, now you decide to move right with a positive velocity

now because youre on the left of the origin and now moving right, youre moving TOWARDS the origin

YOU -------> origin

EVENTUALLY you will reach the origin again

if you CONTINUE at your positive velocity

you will CROSS the origin and now start moving away from it

origin -------> you

therefore now even though your velocity is positive you're moving AWAY relative to the origin

so for part d, it's asking when does the particle start from the origin, move left, stop, then move right and reach the origin again. it's asking for when is the NEGATIVE DISPLACEMENT EQUAL TO THE POSITIVE DISPLACEMENT

we know the displacement is the area under the curve

we can see by congruency that the negative right triangle is the same as the positive right triangle

you moved left 9 metres till time = 3

then you stopped

then you moved right

for 3 more seconds

and find the area

youll see you moved right 9 metres again

so basically at the end of the line

you reached the start again

left 9
right 9
overall 0

applying this same logic can you now solve f and G

[terse tartan]

So what was the answer for d?

[past perch]

t=6

.close

[pearl pond]

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[terse tartan]

VERY longwinded way of saying what I said.

.reopen

[past perch]

why do people complain about in depth explanations that focus on building concept rather than giving a solution

[terse tartan]

I'm not complaining.

Very detailed. I like it.

[past perch]

oh nvm sorry 😭 thank you

@heavy hill

[pearl pond]

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[brittle quail]

Stuck with this

[keen wolf]

Try x+5 =t instead

@brittle quail

[brittle quail]

I can get it with that sub but the exercise asks for that substitution

@keen wolf

[keen wolf]

Dt = dx (1/2root x+5)

X=t**2 -5

@brittle quail

[keen wolf]

Yep now split and solvd

[desert timber]

are you asking me to solve

[keen wolf]

Lol i thought it was wd _nesumo

[brittle quail]

I'm solving rn

Like this right?

[desert timber]

yeah

[sonic wigeon]

how do i join a live

[brittle quail]

Shouldn't then this be the result?

[compact ridge]

A what

[sonic wigeon]

a group

[compact ridge]

This entire server already is one

[pearl pond]

@brittle quail Has your question been resolved?

[pearl pond]

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[swift spindle]

how do i prove that cos 70 + sin 40 = cos10

[rustic tendon]

,w cos(70 degrees) + sin(40 degrees) - cos(10 degrees)

[jolly parrot]

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[rustic tendon]

write cos 70 as sin 20 and use the sum to product formula

[swift spindle]

how do yk that cos 70 is sin 20?

[rustic tendon]

cos x = sin (90 - x)

[swift spindle]

ohh

okok thx

.close

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[unborn quarry]

Can somebody explain how to get angle 31 ?

[pearl pond]

@unborn quarry Has your question been resolved?

[pearl pond]

@unborn quarry Has your question been resolved?

[dense jasper]

Use the dot product

,w arccos(10/(sqrt(68*2)) in degrees

[jolly parrot]

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[dense jasper]

https://www.wolframalpha.com/input?i2d=true&i=arccosDivide[10%2CSqrt[136]]

[pearl pond]

@unborn quarry Has your question been resolved?

[pearl pond]

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[calm wing]

this is my task

[calm wing]

this is the suggested solution

this is my solution

wait nvm i see my mistake

i can't just ignore 1/(x^2+y^2), can i?

i guess it's because i don't take into account 1/(x^2+y^2) when i dot product with the derivative of the parametrization

but i don't entirely understand why my method doesn't work, given that x^2+y^2=a^2 for the entire integral

oh i see a typo

fuck

it should be (-a sin theta, a cos theta) injstead of (a cos theta, a sin theta)

yeah it works out like this

awesome

thanks everyone (the voices)

.close

[pearl pond]

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[pearl pond]

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[inland laurel]

number of ordered pairs (a,b) such that gcd(a,b) = 6 and lcm(a,b) = 24 ?

[inland laurel]

a. 0
b. 1
c. 3
d. 2

how to approach this problem?

<@&286206848099549185>

anyone here

[placid geyser]

do you know this thing:

if we take a = dx, b = dy

where gcd(x, y) = 1

gcd(a, b) = d and lcm(x, y) = dxy

??

[inland laurel]

no idont know

please explain the concept @placid geyser

[placid geyser]

see, like, if x and y are co-primes

then d is their highest common divisor

[inland laurel]

gcd(x,y) = 1

[placid geyser]

yes

[inland laurel]

so d = 1?

[placid geyser]

no

[placid geyser]

sorry, i meant: d is (a and b's highest common divisor)

hence, gcd(a, d) = d

[inland laurel]

what u mean by d is a

[placid geyser]

d is the greatest common divisor of a and b

[inland laurel]

and a and b are co primes?

[placid geyser]

no, a and b are any 2 numbers given in the question

x and y are the co primes we considered

a = dx and b = dy

[inland laurel]

didn't understand

[placid geyser]

let's start from scratch

so, take 2 co-primes x and y

such that a = dx and b = dy

[inland laurel]

why

[placid geyser]

for example, if a = 16 and b = 12

a = 4×4
b = 4×3

here, x and y would be 4 and 3

[placid geyser]

just wait and see

[inland laurel]

alr

[inland laurel]

yes

[placid geyser]

and d would be 4, right?

[inland laurel]

yes

[placid geyser]

the common multiple

[inland laurel]

yes

[placid geyser]

it's also the greatest common multiple

[inland laurel]

yes

[placid geyser]

so, gcd(a, b) = d

makes sense?

[inland laurel]

greatest common multiple is same as greatest common divisor?

[placid geyser]

oh, sorry

i meant divisor

multiple is a whole different thing

[inland laurel]

all good

yes makes sense

[placid geyser]

now, coming to lcm,

[inland laurel]

but

but

[placid geyser]

yes?

[inland laurel]

4 is multiplied there

why do we say is greatest common divisor

[placid geyser]

cause 4 is the greatest number that is a divisor of both 12 and 16

no other number greater than 4 can do the job

[inland laurel]

alr

[placid geyser]

what method did they use to teach you to find the lcm and hcf of 2 numbers in general?

do u know the one where you write 2 numbers beside each other

[inland laurel]

prime factorization

[placid geyser]

and u divide by prime numbers

for both

the short division method

[inland laurel]

and if hcf look out for common factors of lowest power

and if lcm look out for all factors of highest power

[placid geyser]

do u know this method?

[inland laurel]

yes

i know

but why do we take them together

[placid geyser]

common factors

2 is common factor

3 is a common factor (twice)

[inland laurel]

2x3x3 is lcm them

[placid geyser]

yes

[inland laurel]

common factors are lcm?

[placid geyser]

no

for lcm, u do 2x3x3x1x3 right?

multiplying the leftovers too

[inland laurel]

yes

[placid geyser]

so, 2x3x3 is the gcd?

[inland laurel]

im confuesd

😭

[placid geyser]

for gcd, u do 2x3x3 right?

generally

[inland laurel]

i do

[placid geyser]

and lcm is 2x3x3x1x3 which is the same as (2x3x3) x (1x3) = d x (1x3) = dxy

[placid geyser]

there, 1 and 3 are the remaining co-primes

[inland laurel]

18 = 2x3x3 = 2 x 3^2 54 = 2x2x3x3 = 2^2 x 3^2 lcm (18,54) = 3^2 x 2^2 = 9x4 = 36 hcf (18,54) = 2 x 3^2 = 2x9 = 18

i do this ^

very shit concept

idk why even that works?

[placid geyser]

18 = 2×3×3×1 = 2×3²
54 = 2×3×3×3 = 2×3³

[inland laurel]

yes

mb

im very bad today

😭

[placid geyser]

so, the common prime factors are 2, 3 and 3

[inland laurel]

yes

[placid geyser]

to find the greatest factor, we can multiply up those numbers right?

2×3×3

[inland laurel]

yes

18

[placid geyser]

that is the same as what is done here

[inland laurel]

hm

[placid geyser]

selecting the lowest power is the same as picking the 3s that are common and leaving the others

[inland laurel]

oh

[placid geyser]

that is how gcd works

[inland laurel]

got it

[placid geyser]

and as for lcm,

both of them should be able to divide it right?

[inland laurel]

both of what???

[placid geyser]

first, starting with the definition of a multiple

[placid geyser]

a and b

[placid geyser]

let's do this first

a multiple of a number "x" can be divided by "x" right?

[inland laurel]

yes

[placid geyser]

so if we want a common multiple, of a and b, it should be divisible by both a and b

[inland laurel]

yes

[placid geyser]

and the smallest one of them is called the lcm

[inland laurel]

oh

the least

one

[placid geyser]

yes

[placid geyser]

coming back to this,

remember how you take the higher power for lcm?

isn't it the same as first taking the lower power, then adding the 3s and other numbers you left out?

[inland laurel]

lcm(18,54) = 2x3^3

18

mb

2x3^3

= 54

[placid geyser]

Yes

[placid geyser]

??

[inland laurel]

u wanna say that\

take

2x3^2 + 3

21

Im taking rest i think my mind is not working now

ill ask u later this on dm ig

[placid geyser]

Okay

[pearl pond]

@inland laurel Has your question been resolved?

[inland laurel]

Yes

[placid geyser]

Click on the ✅

In the bot's reactions

[pearl pond]

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[stiff mauve]

anyone?

[pearl pond]

@stiff mauve Has your question been resolved?

[rustic tendon]

@stiff mauve what have you tried?

do you know the definition of conditional expectation?

[stiff mauve]

i genuinely have no idea how to start

[stiff mauve]

yeah

[rustic tendon]

honestly your notation is tripping me up a bit

[rustic tendon]

maybe start with that then and see what happens

is E(Y|x) = E(Y| X = x)?

[stiff mauve]

f(Y | X = x) = f(Y, X = x) / f( X = x) right

[rustic tendon]

yeah

[stiff mauve]

might be yeah

since X = x i guess

mistake

[rustic tendon]

right then expand with that

[stiff mauve]

uh

[rustic tendon]

It will be $x = \sum_y yP(Y = y | X = x)$

[jolly parrot]

dyxn

[rustic tendon]

but like i've never seen the notation Y|x ~ POI(x)

[stiff mauve]

isnt poission and expontenial, continuous distributions

[rustic tendon]

Poisson is discrete

[stiff mauve]

ow ok

Idk man @rustic tendon help

[rustic tendon]

yeah sorry man I just don't get the notation here, you should ask your teacher, wait for someone else, or maybe post it in #probability-statistics

[stiff mauve]

okay thank u for your time

[pearl pond]

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[dense iris]

So I did this question and I got it wrong

because I thought only the first line equation would use the negative reciprocal as the gradient

could someone explain to me why both of them use the neg reciprocal?

[merry carbon]

How do you mean here?

(could you maybe show how you did it?)

[dense iris]

ahh its rly messy handwriting

ill show u

[merry carbon]

Assumedly you're setting x = 4 here - how comes you have sqrt{8}?

[dense iris]

AHHHHH

careless

used the y value by mistake

[merry carbon]

It does happen

Other than that, I'm mostly happy with what you've done, a small comment here (though noting that you may not have intended to show anyone this!), be careful not to set y equal to the gradient here

[dense iris]

oh yeah

What i was originally wondering tho was

why have they used the neg reiciprocal for the curve

I thought it was only for the line

[merry carbon]

As in this bit here?

[dense iris]

ye

[merry carbon]

If so, they're noting the fact that your normal gradient is -1/(dy/dx), so you can get an equation that way as well

The more "natural" (imo) way would be to get the tangent gradient via the normal and go about it like that, but they wanna give you as many ways to do it, hence the oe (or equivalent)

[dense iris]

ok great thank you

.close

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[fossil jewel]

anyone have any ideas? i dont know where to start

[harsh gyro]

i found a solution but it's rather involved

im not sure if there's something simpler

the idea is to split into two cases

case 1: for every point in M, there exists an open ball around M containing no other points

case 2: for some point in M, there are other points in any open ball around M

[fringe robin]

use hausdorff?

[pearl pond]

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[royal aspen]

How do I solve this frq? I'm just genuinly confused on where to even begin. I know I can use integral from a to b of 1/2 r^2 dtheta but besides that I have no idea

[chrome plank]

a) you need to find for what values of theta does the curve cross back over itself. then you can integrate between these values

[royal aspen]

oh ok that's not that bad

how do you even begin to attempt b)

[chrome plank]

find the largest value of r

[royal aspen]

like the critical point?

[chrome plank]

It might help if you instead, plot a graph of r vs theta

(instead of plotting on the cartesian grid)

[royal aspen]

wouldn't the critical point be a maximum though?

and thus be the farthest point?

[chrome plank]

red is (x,y) coordinates, green is (theta, r) coordinates

[chrome plank]

what do you mean by "critical point"?

[royal aspen]

when dr/dtheta = 0 I think

[chrome plank]

yes

when dr/dt = 0, the graph will be flat, and r will be a local maxima or minima

(unless it's a saddle point, but there are none here)

[royal aspen]

so all I have to do is just solve for when dr/dtheta is 0? (is it dr/dtheta or dr/dt)

[chrome plank]

pretty much

you'll get two solutions between 0 and 3/2pi, only one of which will be a local maxima

(see the green plot)

[royal aspen]

ok

thanks

[pearl pond]

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[worn turret]

How do I find an equation of a reflected line given an equation y=mx+c reflection line y=2*x?

[pearl pond]

@worn turret Has your question been resolved?

[frozen night]

what have you done so far?

[worn turret]

not sure how I can get the equation of the reflected line

[frozen night]

you want a transformation that you apply to the thing you are reflecting

and you are given a transformation

remember, matrices are just systems of equations

[pearl pond]

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[pearl pond]

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[outer hare]

Are factor groups just the collection of left cosets of a normal subgroup (strictly the set, i understand the opperation is coset multiplication)?

[outer hare]

like for G\H, H being normal, say |H| = d, then G\H is just H,dH, 2dH, etc etc until d-1

[lilac jackal]

yes, we also call them quotient group

[outer hare]

ok so like

we know SL is normal in GL right

so the cosets would just be

det(1), det(2), etc etc

det(1) being id

[elfin stirrup]

If G is a Group and H is a subgroup then

G\H has cosets as elements gH for all g in G

If H is a normal subgroup then G\H is actually a group if you define the operation as

gH * hH = (gh) * H (yes this product is well defined one has to check that)

[outer hare]

got u

so quotent groups only work with normal subgroups

because thats needed for the opperation to work

[elfin stirrup]

yes otherwise you cant define the product this way

or the operation is not well defined

[outer hare]

ok thats what i thought

perfect tyty

.close

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[charred tinsel]

Can I get some help on this?

[charred tinsel]

Just wanna make sure I’m going in the right direction

I just got to this trig stuff, and I’m kinda stuck

[edgy wren]

going good

[charred tinsel]

Is this good?

[edgy wren]

looks correct got close enough answers when chekcing

[pearl pond]

@charred tinsel Has your question been resolved?

[charred tinsel]

I have one more

[edgy wren]

ok

[charred tinsel]

Please check my work so far. Did I do ok with my formula?

[edgy wren]

yeah going ok,
my brain was a little off forgot about the -2ab when I was calculating

[charred tinsel]

This specifically applies to Law of Cosines when solving for interior angles right?

[edgy wren]

yeah but you can just find 1 angle then use sine rule for the rest

I feel that approach is eaiser for me

[charred tinsel]

can u elaborate?

[edgy wren]

like you found 31 as one of the angles you can continue finding the other angles using the sine rule (becuase you have 2 known sides and an angle)

[charred tinsel]

Got it. I'll leave now. Thanks for the help.

How do you leave btw?

[compact ridge]

type .close

[charred tinsel]

.close

[pearl pond]

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[toxic lichen]

<@&268886789983436800> scam

[pearl pond]

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[brave portal]

I've got a geometry problem I just have no idea how to solve:
I'm given a triangle ABC with two cevians AF and BE, where point F divides line BC into segments BF:FC with the ratio of 3:2 and point E divides line AC into segments AE:EC with the ratio of 6:2.5. These cevians intersect at point K. (image of figure attached) The task asks for me to find the ratio of AK:KF.
Due to my country's (frustatingly) limited curriculum, I likely can't use mass points as a method of solving the problem as we weren't taught it. Ceva's theorem is also off-limits for the same reason.
The effort of trying to prove that the cevians are perpendicular wasn't fruitful because I'm pretty sure I lack info. I tried to draw a line parallel to cevian AF so that it touches point B, but for whatever reason I can't come to the conclusion that the line segment BK would be perpendicular to both of these parallel lines (figure 2). I didn't find any successful application of the right angles despite failing to recognize that fact in my previous tries (law of sines, Pythagoras's theorem and trigonometric ratios all don't work due to insufficient info or useless results).
Drawing other lines to create new shapes, such as a parallelogram with BE and EC as its sides, also hasn't harnessed any useful info. Trying to prove that triangle ABC is a right triangle also failed because the perpendicular bisectors of the triangle meet outside the triangle (figure 3, tried my best).

I do know that, according to my textbook, the right answer should be 4:1.
I'm likely missing an obvious step or two, so any pointers would be appreciated.

[dapper kraken]

try this: since BCA can be anything/its not related to the ratios and the way the question was setup, the answer should be constant for whatever triangle ABC that fits the ratio criteria, assume BCA is an ideal angle such as 90 degrees and continuing it shouldnt be too hard

[pearl pond]

@brave portal Has your question been resolved?

[brave portal]

ill try this

.close

[pearl pond]

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[pearl pond]

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[craggy lark]

if i plug the formula on wolfram alpha it wont work, where is the mistake in my calculations... (day 3 of asking for this)

[rich jolt]

whats the question

[long flicker]

@craggy lark

Have you heard of “integration by parts”?

[rich jolt]

can u post that

[long flicker]

https://www.mathsisfun.com/calculus/integration-by-parts.html

[craggy lark]

bro i was looking for a new representation of the gamma function

if you ibp you get (x)!

[rose robin]

it should be xdu in after the first u sub
and then I have no idea about the rest

[pearl pond]

@craggy lark Has your question been resolved?

[craggy lark]

x!=gamma(x+1)=xgamma(x)

then simplified the x you are left with just gamma(x)

[rose robin]

ah mb, missed that you switched from x! to gamma(x)

[craggy lark]

dwdw

[pearl pond]

@craggy lark Has your question been resolved?

[pearl pond]

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[midnight haven]

what happened here

[pearl kite]

cot(x) = 1/tan(x)

[midnight haven]

no i mean

what does the question want

[pearl kite]

from what ive understood

u basically show it converges to alpha as n-> infinity

[midnight haven]

woah whattt

what does converge mean

[oak ivy]

Get closer to 0?

[midnight haven]

oh

what does 3x = cot(x/2) have to do with getting closer to 0

[fleet pollen]

Basically what x is at n = infinity

[fleet pollen]

If it converges, that means that when n is infinity, x is a defined number (meaning it isn’t undefined or also infinity)

[pearl kite]

is there something missing

like show me part a of the question

[midnight haven]

wait

[pearl kite]

because it doesnt define what alpha is

at all

[midnight haven]

here

[fleet pollen]

Since when n is infinity, x converges to a, that means that at really large values of n, the x’es are really close to each other

[midnight haven]

oh okay

kinda got it

[fleet pollen]

So we can set x _ n and x_n+1 as the same number

And after that, we can just solve what x_n is

And that equation gives us 3x_n = cot(x_n/2)

For when n approaches infinity

[midnight haven]

let me try to analyse give me a min

okay im lost

[pearl kite]

u know alpha is the point x such that cot(x/2)-3x = 0

u understand that right?

[midnight haven]

nope

hold on

[pearl kite]

its what u did in part a

[midnight haven]

let me understand what converges is again

[pearl kite]

you showed there exists some value for x

which we denoted as alpha

[midnight haven]

it is when the number is definded right?

[midnight haven]

oh yea

[pearl kite]

such that at x=alpha we have a root e.g = 0

[pearl kite]

converging is that as you go to infinity you reach a specific point/number

diverges means as you go to infinity you go to infinity

[midnight haven]

oh so we rearranged to equation

[pearl kite]

now lets suppose that the iteration formula converges

[midnight haven]

and to be converged it must equal the alpha thingy

[pearl kite]

that means x_n+1 and x_n is gonna be the same (as n->infinity) lets call that point L

when you do the algebra

[fleet pollen]

For example, in the geometry equation 1, 1/2, 1/4, 1/8… if it goes on infinitely we eventually reach 0, so we say it converges to 0

[pearl kite]

you'll get 3L = cot(L/2)

thus using your knowledge from part a

[midnight haven]

oh

[pearl kite]

you've shown

it has converged to alpha

[midnight haven]

wait let me try another question

give me a minute

[pearl kite]

the wording of ur question is ugly

ur examiners are ruthless

[midnight haven]

yea cambridge is a bitch

[pearl kite]

cambridge international a levels?

[midnight haven]

yea

[pearl kite]

good luck

edexcel is usually a lot nicer

even OCR

[midnight haven]

@pearl kite

[midnight haven]

ik their system is better as well

[pearl pond]

@midnight haven Has your question been resolved?

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[calm wing]

does my answer look good?

[urban coral]

how do you treat sups if (M) is unbounded

[jolly parrot]

peanutbutter

[urban coral]

would you say (\sup M = \infty)?

[jolly parrot]

peanutbutter

[urban coral]

and why does a smallest x exist with that property

that's way stronger than you need

you know that if (\alpha = \sup M), then (M\cap (\alpha - 1/n, \alpha]\neq \emptyset) since (\alpha) is the least upper bound

[jolly parrot]

peanutbutter

[urban coral]

or said another way, you know that (\alpha - 1/n) is not an upper bound of (M)

[jolly parrot]

peanutbutter

[urban coral]

so there is an (a_n\in M) with (\alpha - 1/n \le a_n \le \alpha)

[jolly parrot]

peanutbutter

[urban coral]

also, it looks like you are trying to define a_n to be a set

a_n = {x with some property}

and how do you know that it converges to the sup

i think you'd probably be expected to check

an example where an (x) like yours doesn't exist is when you have something like (M = (99/100,1)) and you take ``the smallest'' (x) such that (x\in (1/2,1]\cap M = (99/100,1))

[jolly parrot]

peanutbutter

[pearl pond]

@calm wing Has your question been resolved?

[calm wing]

ooo good point

hmmmmmm

[calm wing]

sorry notational quirk, ignore the { }

i want every a_n to be the smallest x such that x > sup M - 1/n

[calm wing]

that's sorta what i am going for tho

[urban coral]

(this doesn't necessarily exist)

[calm wing]

why not

[calm wing]

i can find you an x from that set given an epsilon nvm

[calm wing]

if we drop the "smallest" condition will my thing work then

[urban coral]

Yeah

[calm wing]

alright i was also uncertain about that part lol

[urban coral]

Although can you prove that the sequence converges to the sup

[calm wing]

yes

[calm wing]

when you proof the limit it basically becomes this

so by squeeze thm it would be the supremum

i'll treat the infinite case separately

[rustic tendon]

shouldn't the left inequality be strict?

[calm wing]

it should

[calm wing]

hmm

i am not sure about that one

$a_n = \inf (M \cap [n, +\infty))$

[jolly parrot]

artemetra

[calm wing]

idk if i can use inf tho

@urban coral sorry for the ping but is there a way to avoid using inf?

acutally i don't have to use inf

i can just pick any element

[rustic tendon]

can't you just use the Archimedean property

[urban coral]

it doesn't really matter

[calm wing]

M doesn't have to have integers

or wdym

[urban coral]

if (\sup M = \infty), then for every (n), there is an (a_n\in M) such that (a_n \ge n)

[jolly parrot]

peanutbutter

[pearl pond]

@calm wing Has your question been resolved?

[urban coral]

yes, ive resolved it.

[pearl pond]

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[pearl pond]

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

[eager jewel]

2

took e^iA = a
e^iC = c
e^iB = b

then got determinant
(abc)(1)(1)
(1) (abc) (1)
(1) (1) (abc)

not able to simplify

then i tried taking angles eg A = pi/2
B= pi/3
C = pi/6

didnt work

[urban coral]

oh lord!

can you expand the determinant?

[eager jewel]

u want me to expand that?

it will get too ugly wont it

[urban coral]

yeah. it could get ugly

but you know, yoda

i believe in you

[solid pier]

[ping dealt with]

[eager jewel]

hm?

[eager jewel]

;/

@rustic tendon :D

[urban coral]

there was a scoundrel while you were away

[eager jewel]

oh

[rustic tendon]

substitute pi/3, pi/3, pi/3

🔥

[eager jewel]

how will it form a triangle

😭

[rustic tendon]

there you go

[urban coral]

oh yeah it is multiple choice

you don't even need to care about what is going on

[eager jewel]

oh huh

[sterile imp]

I mean if A + B + C = 180 which = pi

So my gut says it’s -1

And call it a day

[rustic tendon]

also if you consider a straight line a triangle... A = B = 0, C = pi

[sterile imp]

PFFT

[eager jewel]

🙏

[eager jewel]

how can u consider st line a triangle? but let me try that

[sterile imp]

Trust me, I’m good at multiple choice PFFT

[rustic tendon]

abstraction?

the only property that defines a triangle in this universe is the sum of the angles

[eager jewel]

OH

[rustic tendon]

this universe = universe containing that mcq and you

does it work 💀

[eager jewel]

trying it one sec

no im getting the ans as 0

NO wait trying again

[rustic tendon]

,w det{{1, -1, 1}, {-1, 1, 1}, {1, 1, 1}}

[jolly parrot]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by dyxn06

[urban coral]

i got -4

hey

[
e^{2iA} e^{2iB} e^{2iC} - 3 + 2 e^{-iA} e^{-iB} e^{-iC}
]

[jolly parrot]

peanutbutter

[eager jewel]

yes same

oh

[sterile imp]

I was close enough 😔

[rustic tendon]

ah did you just brute force open it?

[eager jewel]

@rustic tendonlove u

[urban coral]

i just put it into chatgpt

[sterile imp]

-1 and -4 doesn’t look much different

[urban coral]

i wasn't calculating it

[sterile imp]

Work smarter not harder

[urban coral]

you know, i think you might be onto something

no, you're definitely not

[rustic tendon]

that's why sub 0, 0, pi

[sterile imp]

I think I’m on something actually

[urban coral]

i did no work at all

oh

[sterile imp]

JOKING ofc

[pearl pond]

@eager jewel Has your question been resolved?

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[royal galleon]

I want to show that the set of negative integers are unbounded below. So I know the set of positive integera are unbounded above. So I can multiply the set of positive by -1 which creates the set of negative integers. So by multiplying by negative one all I am doing is flipping the direction. So the amount of elements is the same but now it starts from -1 and goes down by -1. Which shows it is unbounded below. But I feel like this is not a strict proof

[rustic tendon]

are you working with supremums and infimums?

[royal galleon]

Yes

[royal galleon]

But how can you apply those concepts to this?

[rustic tendon]

Do you have access to the axiom of completeness?

[royal galleon]

One second let me check

[rustic tendon]

Oh yeah just use 1.27 then

and find a contradiction

[royal galleon]

Ye but how can you interaate theoguht the elements of the set of negative numbers like positive numbers

[rustic tendon]

What?

[royal galleon]

Like positive numbers are inductive set so I can say it is bounded then say there is n+1 so there is a contradiction

[rustic tendon]

just do n - 1 here

[royal galleon]

Ye but how do we know n-1 exists in the set of negative integers

The only reason we knew n+1 existed in positive integera was because it was an inductive set

[rustic tendon]

if n is an integer then n - 1 is also an integer

I assumed you already had some sort of construction for Z^-

what if you just did -(|n| + 1) then?

[royal galleon]

Wdym by that?

[rustic tendon]

you can iterate over |n| right?

[royal galleon]

Are those absoule value signs?

[rustic tendon]

yeah

[royal galleon]

So your saying it doesn't matter if it is an inductive set?

[rustic tendon]

i don't think so

[royal galleon]

Wait what says that n-1 exists in the set of integers?

[rustic tendon]

closure? 😭

[royal galleon]

Ok I'll look into that thx

.close

[pearl pond]

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[pearl pond]

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[karmic karma]

i know that the value of f(x) is a maximum at 6

[karmic karma]

but how do i find the actual value

[open rivet]

plug it in

[karmic karma]

into where

[open rivet]

into f(x)

[karmic karma]

i dont see the equation for f(x)

[open rivet]

ah, i see

there is none

you have to estimate

[karmic karma]

so i just start counting the blocks?

[versed wigeon]

You just find the highest point and compare to Oy because it says maximum value? Nvm it's f'(x) 😭mb

[open rivet]

there is an easier way

[karmic karma]

how so

[open rivet]

remember what f'(x) means

[karmic karma]

ok i get what f'x but not sure how im supposed to use it to evaluate

[open rivet]

f(x+1)=f(x)+f'(x)

this comes from f'(x)=f(x+1)-f(x)

[karmic karma]

is this like a formula

also im still confused because i dont have f(x)

[open rivet]

you have f(2)

and you have f'(2)

so you can find f(3)

[open rivet]

remember what f'(x) means

[karmic karma]

i got 110 as my approximation

[open rivet]

sounds about right

[karmic karma]

but i still dont get how you get f(x+1)=f(x)+f'(x)
the value of the function + the slope gets you the next y value?
y=x^2
f(1) + f'(1) = 3
f(2) = 4

[open rivet]

do you know the defintion of f'(x)?

[karmic karma]

the tangent line of a point right

oh

yea i get it

maybe

[open rivet]

$f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$

[jolly parrot]

Bonk

[open rivet]

generally, we let h go to 0

to get the proper value

but we can approximate it by choosing a small h

for example, h = 1

then you get...?

[karmic karma]

f(x+1)-f(x)

[open rivet]

tada

🙂

[karmic karma]

i just thought of it as the current value + its slope = next value

thanks 👍

[open rivet]

yeah, thats the same as this

but more informal

[open rivet]

here you take steps of size 1

but you can also take steps of size 2

or size 1/4

etc

[karmic karma]

how would i think of that using my informal way

[open rivet]

using h=1 here

the smaller your h gets

the better your approximation gets

[karmic karma]

so like (f(x+1/2) - f(x))*2

= f'

ok yea

.close

[pearl pond]

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[open rivet]

yup 🙂

[pearl pond]

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[orchid token]

why are they saying t_(r+1)th term? if for example I want to find the x^2 term of (2+x)^4, my n = 4, p = 2, q = x, and r = 2 to get the right ans of 24x^2

[orchid token]

or is it still techincally the same cuz there is a 0 term?

idk

how am i supposed to be interpreting this

would x^2 term be T_3? so r = 2? idk

[dense jasper]

I'm assuming that they're basing the terms off of increasing powers of q

So the first term would have q^0

the second term would have q^1

and so on

Ex. (p+q)^4 = p^4 q^0 + 4p^3 q+6p^2 q^2+4pq^3+q^4

Here, we're considering p^4 q^0 to be our first term, 4p^3 q to be our second term, etc.

@orchid token

[orchid token]

oh so the third term in the expansion has the x^2 power?

[dense jasper]

mhm

[orchid token]

oh kk

[light helm]

its not intuitive to number it this way

[orchid token]

right cuz (a+ b)^3 has 4 terms

[light helm]

either way you'd use r for the power you want

[orchid token]

oh kk

so i can js think of it like if i want to find the x^2 term in a sequence to just set r = 2

so r is the power i want

and n is the the power of the binomial

[light helm]

yeh

[orchid token]

kk thank you

.close

[pearl pond]

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[low wadi]

Im so lost, tried using ai and that confused me even more

[orchid pendant]

what is the goal

to simplify them?

[low wadi]

If this is simplifying. to eventually solve the code

[orchid pendant]

what have you done

[low wadi]

Ignore the circled part. Thats level 1 (this is all ai btw)

[placid geyser]

!nogpt

[pearl pond]

Please do not trust ChatGPT or similar AI tools for mathematical tasks, as they often generate output which "sounds correct" but has numerous factual or logical errors. Use of these AI tools to answer other people's help questions is strictly against server rules (see #rules).

[orchid pendant]

what is confusing you

[low wadi]

Everything 😂

[orchid pendant]

okay focus on the 10s, keep them separate

and then multiply out the decimals

or their respective property

ex, for R, (4 * 10^4)/(3.63 * 10^-4)

= 4/3.63 * (10^4)/(10^-4)

~1.1 * 10^8

[low wadi]

Im still lost

.close

[pearl pond]

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[pearl pond]

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[merry carbon]

Would you like us to just admire your handwriting? (it is really good )

(or do you have a question you wanted to ask us? )

[leaden osprey]

Omg thank u lol

Idk I’m just confused

I just tried something else

Does this work

I used 3pi/4 and the other one on the unit circle

[merry carbon]

The solutions 3pi/8 and 7pi/8? Those seem fine sure

[leaden osprey]

Am I allowed to do that tho

Just using the opposite sides

[merry carbon]

Well, I mean, considering that tan is pi-periodic, the "opposite sides" are pi apart, so as per the diagram, they're fine

(for sin and cos, you need to be careful, as those aren't pi-periodic, but 2pi-periodic)

[leaden osprey]

So only if its opposite sides

Can I do this with anything that opposite

[merry carbon]

As in, if u = 3pi/4, or u = [any of the other points you marked] were a solution to tan(u) = [what you wanted], then so would u = 7pi/4 [= 3pi/4 + pi], or any of the "opposite" points (which would correspond to adding or subtracting pi)

[leaden osprey]

Wait what

[merry carbon]

In your one, you circled 3pi/4, you're also allowed to circle 7pi/4, for tan

(of course, I'm being quite careful in my phrasing, to make sure it's as clear as possible: the fact you needed to divide means you need to be careful how you go about it!)

[leaden osprey]

Oh yea the 2x

But if it were another line on the circle

Like

Pi/4 and 5pi/4

Would it still work

[merry carbon]

Yes, it would, as would 5pi/6 and 11pi/6 etc etc

[pearl pond]

@leaden osprey Has your question been resolved?

[leaden osprey]

Thank you

[pearl pond]

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[pearl pond]

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[proper flare]

If a is prime to b and y, and b is prime to x, prove that ax + by is prime to ab.

[fringe goblet]

can anyone help me solve this : Find no. Of integral values of a for which (x - 3a)( x - a - 3) ≤ 0 for all x €[1,3]

[proper flare]

Hell naw

[proper flare]

!occupied

[pearl pond]

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

[proper flare]

!help

[pearl pond]

To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.

[subtle geyser]

u good?

[oak ivy]

Lol hes not

[proper flare]

?

[subtle geyser]

do u have a question @proper flare

[oak ivy]

.

[subtle geyser]

oh

[toxic lichen]

when you say "prime to" do you mean "coprime to"

[proper flare]

Yes.

[toxic lichen]

so
(a,b) = 1
(a,y) = 1
(b,x) = 1

is what we are given

yes

[proper flare]

Yes.

[subtle geyser]

bezout's identity?

[proper flare]

Huh?

[subtle geyser]

have u learned bezouts identity

[proper flare]

It's allowed.

[subtle geyser]

what do u mean integral values?

i dont get the q

[toxic lichen]

so we have:
\begin{itemize}
\item $m_1 a + n_1 b = 1$
\item $m_2 a + n_2 b = 1$
\item $m_3 b + n_3 x = 1$
\end{itemize}

[jolly parrot]

Ann

[toxic lichen]

don't encourage channel hogging btw

[subtle geyser]

oh my bad

[toxic lichen]

we have this, where m_i and n_i are some integers

and we want to find a linear combination of ax+by with ab that adds to 1

[proper flare]

Did you mean n_2 y in the second one?

[toxic lichen]

yes i did my bad

typo there

[vital crescent]

Another way to do this is by showing a Lemma: if m is coprime to n, and l is coprime to n, then show that ml is coprime to n.

[proper flare]

Got it

Let (ax + by, ab) = d
Let (a,d) = d_0
d_0 | d and d | (ax + by)
So d_0 | ax + by
So (ax + by)/(d_0) is an integer.
(a/d_0)x + by/d_0 = integer
by/d_0 = integer

Let (b, d_0) = d_1
Then d_1 | d_0 and d_1 | b
d_0 | a. So d_1 | a.
Since (a,b) = 1, d_1 = 1
So (b, d_0) = 1

Now let (y, d_0) = d_2
d_2 | d_0 | a. So d_2 | a.
d_2 also divides y.
Since (a, y) = 1, d_2 = 1

So (b, d_0) = 1 = (y, d_0)
But by/d_0 = integer
This is only possible when d_0 = 1, because if d_0 >= 2, then this would mean that d_0 has common factors with b and/or y.

So d_0 = 1 = (a,d)
In the same way, we can show that (b,d) = 1
But d | ab. We can say that this is only possible when d = 1. So d = 1 = (ax + by, ab).

Is this proof correct?

[vital crescent]

yeah just need to justify why d|ab and (b,d)=(a,d)=1 implies d=1

[proper flare]

Let ab/d = k
ab = d(k)
(dk, d) = 1
d(k,1) = 1
d(1) = 1
d = 1

Is this correct?

k is a positive integer, and because 1 | k, (k,1) = 1.

Because if a | b, then (a,b) = a.

<@&286206848099549185>

[pearl pond]

@proper flare Has your question been resolved?

[pearl pond]

@proper flare Has your question been resolved?

[vital crescent]

why is (dk,d)=1?

the third line

you have to essentially prove that (b,d)= (a,d)=1 implies (ab,d)=1

[pearl pond]

@proper flare Has your question been resolved?

[pearl pond]

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[pearl pond]

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[sharp smelt]

Trying to prove or disprove the existance of a sequences, such that for every $n$ in $N$ it's possible to find $n$ consecutive one's somehwere in the sequence

[jolly parrot]

What a wonderful world !

[sharp smelt]

I was thinking of $a_n=1$

[jolly parrot]

What a wonderful world !

[edgy stone]

that works

[placid geyser]

Does the sequence have to have a proper form? Can it be an infinite sequence?

By proper form, I mean "can be expressed in common mathematical functions"

[sharp smelt]

I don't follow

[placid geyser]

So it doesn't have to be?

[cursive wraith]

a_n = 1 works anyways

[placid geyser]

It does though

I was thinking of the digits of π

Anything is possible >x<

[cursive wraith]

then prove it does satisfy the statement

[placid geyser]

π has an infinite number of digits, hence any sequence number can be found in there

[tropic saddle]

thats not how that works

[placid geyser]

It's not?

[cursive wraith]

nope

[tropic saddle]

0.111111..... is infinite but doesnt contain the digit 2

[placid geyser]

Well...

[tropic saddle]

0.101001000100001... is irrational and doesnt contain the digit 2

[cursive wraith]

"any finite sequence can be found in its base x expansion" is the property of a universe number

[placid geyser]

Is there any proof that π does not?

[tropic saddle]

we dont know about pi

[cursive wraith]

is there any proof that pi does?

We THINK pi is a universe number

never had one proof

[placid geyser]

Random strings of numbers are found though

[cursive wraith]

but can you prove ALL of them are there?

[placid geyser]

That's rational

π is irrarional

[cursive wraith]

then consider this

[tropic saddle]

reasons why I gave the second example

[placid geyser]

Hmmm

[tropic saddle]

as if I could look into the future

[placid geyser]

There r people who say so

[cursive wraith]

Then I'm looking forward to meeting them

in the meantime

[placid geyser]

But is defined to follow a pattern

[cursive wraith]

we don't know if pi is a universe number or not

[tropic saddle]

well yes and?

[placid geyser]

π on the other hand, appears to be random for us so far,

[tropic saddle]

appears to be

[cursive wraith]

appears, so far

xdd

[placid geyser]

Yes

[cursive wraith]

"and that's the proof folks!"

[placid geyser]

Yes lol

[cursive wraith]

-_- I thought at least you would see the irony

Imagine if everything that 'oh it seems to be the case' was assumed true and people relied on those statements without ever checking on if there was a valid proof for it, in order to build more statements

What would math look like then, when we realize one of those "oh it seems to be the case" statements at the foundation of maths turned out, not to be the case?

(circa Hilbert problems and set theory stuff)

So anything neither proven nor disproven cannot be taken as true or false

and while "pi is a universe number" is some very interesting claim

it stops there

[placid geyser]

How r we even supposed to approach this problem of proving/disproving this statement?

That π has all strings of numbers

Just finding all the digits?

[cursive wraith]

that's a different question, I feel like we've strayed far off from the original question from the owner of this channel

you can open a new one to discuss it if you want

[tropic saddle]

there is a reason why we dont know

showing claims like this so far has only been done for very special examples

[cursive wraith]

like 0.01234567891011121314...

pi's digits are unpredictable, and we're still computing them

[vital crescent]

iirc there is a paper that shows that weird numbers like 0.1010010001... have measure 0

i forgot the specifics of the statement

[tropic saddle]

its somewhat easy to show that almost all numbers are normal

[cursive wraith]

non-universal numbers indeed have measure 0 on [0,1]

[tropic saddle]

which is an even stronger claim

[vital crescent]

oh yeah thats what i was looking for

[cursive wraith]

it's an easy statement to prove

take some string of digits that's not allowed, say its length is n

the probability that it doesn't show up on a real number between decimal 1 and decimal n is 1 - 10^(-n)

(someone correct me if the exponent is wrong)

the probability it doesn't show up between decimal n+1 and 2n is the same, 1- 10^(-n)

betwen 2n+1 and 3n, same probability

etc...

[placid geyser]

What does it mean?

[cursive wraith]

so the probability that string doesn't show up in a real number in [0,1] is smaller than (1-10^(-n))(1-10^(-n))(1-10^(-n))... = 0

since there is a countable amount of possible strings of digits of finite length

measure of countable union of events of measure 0

= 0

almost every real number is a universe number in base 10

and well, since there's also a countable amount of bases

almost every real number is a universe number in every base

[vital crescent]

oh didnt realize its that simple i never looked at the proof

[cursive wraith]

measure is an extension of the notion of "length"

at least the lebesgue measure over the real numbers

sorry for polluting your help channel @sharp smelt we can take a new one if you don't want us to continue 😭

[vital crescent]

the examples they give, where it some string of integer never occurs, is a "small" subset of real numbers

[sharp smelt]

nah, it's fine

I'm writing another answer rn in overleaf, so this helps keep the channel open

[cursive wraith]

xddd

alright so, to sum up

we know if we pick a random real number, it's almost certain that it will be a universe number

however, pi is not a random real number

in fact it's among the set of definable real numbers

which if I recall correctly is countable

so also of null measure

so, looking at the first digits of pi

we have a small idea as to if pi is indeed normal/universe or not

but we don't have the tools to prove it

you can easily show rational numbers are not normal/universe

because we know something about their base x representation

we don't know crap about pi's base x representation however other than the first trillion digits (in base 10)

which sounds like a lot but pales in comparison to infinity

[tropic saddle]

funnily enough there are formulas for computing the nth digit of pi in base 16

[cursive wraith]

yeah I do remember that

[proper flare]

yes

[pearl pond]

@sharp smelt Has your question been resolved?

[pearl pond]

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[sonic mist]

hello dear server, its me again :/ i feel genuinely embarassed

[sonic mist]

the questionmark is the only place i didnt understand

the 2/3s and stuff are just the teacher explaining a fact dont mind them

[hazy pilot]

(-1)*(-1) = +1

[sonic mist]

yeah

but

-1?

[hazy pilot]

1 + 1/cos(x) < 0

so because the absolute values

it will be -(1+1/cos(x))

[sonic mist]

we invert it yeaaahhh how did i miss that

thanks :>

.close

[pearl pond]

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[pearl pond]

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[rapid wagon]

Let I(x) be the antiderivative of $arctan^2(x)$ express using I(x) the antiderivative: $\int \frac{ln(1+x^2)}{1+x^2}dx$

[jolly parrot]

prograce

[rapid wagon]

How

[hazy pilot]

let ln(1 + x^2) = u

[toxic lichen]

some nontrivial integration by parts fuckery

[rapid wagon]

Ok so dx=du(1+x^2)/2x