help-36

[floral zenith]

Right

[fossil kiln]

-# Oh and area of triangle using determinant would be quicker i think

[final saddle]

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[old quarry]

.alralr next time ill use that

[novel vapor]

I want to prove
f(x)=x^2
f(x) -> 16 when x -> 4
Using definition

[final saddle]

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

[novel vapor]

2

[faint edge]

What definitions are you working with here?

[dull nest]

Hii

[faint edge]

f(4)=4^2=4(1+1+1+1)=4+4+4+4=8+8=16 is reasonable to me for ex. But if we are being super formal about things, it depends a little bit on what you take as primitive.

[worldly mesa]

epsilon-delta probably

[faint edge]

Ah, I forgot they were doing limits.

[novel vapor]

Well im stuck here
|x-4||x+4|<epsilon

[faint edge]

Do you know that lim f(x) g(x) = [lim f(x)][lim g(x)]?

You can bound x

[novel vapor]

yea i guess?

[novel vapor]

What can you explain more

[faint edge]

The quick general way is this. You have f(x) is a product of the identity function with itself.

[novel vapor]

what is f(x) and g(x) here

[faint edge]

This is a completely different approach

The idea is you can say x is within some fixed constant of 4. This allows you to derive a bound on x+4 that you can use to get the delta you want.

[faint edge]

Both would be the identity function in this case. f(x)=g(x)=x.

[novel vapor]

Its like calculating the limit of two points -4 and 4?

[faint edge]

Which trick?

I'm describing two different ways to do the limit.

[novel vapor]

Can we do that?

[faint edge]

Say abs(x-4)<4, then -4 < x-4 < 4, so 0 < x < 8, so 4 < x+4 < 12, then abs(x+4)< 12.

You can pick delta as small as you like so, if you pick delta smaller than 4, you always get a bound on abs(x+4) that you can use this way.

[novel vapor]

What
There are no restrictions on Delta?

[faint edge]

It has to be positive and it has to make your eps/delta proof go through. But other than that you can make it as small as you like.

[novel vapor]

Oh i thinks that’s right

[faint edge]

The limit definition starts as "for all eps>0, there exists a del>0 ...", so, if some del < 4 makes the proof go thru ur fine.

[novel vapor]

We have the statement
|x-c|<something
We can choose delta less than something

Not necessarily equal to it

[faint edge]

I'd think of it as, somebody else gives you an epsilon, you get to pick a delta as small as you like to make your proof go thru. So, in particular you can pick delta smaller than any fixed constant you want.

[faint edge]

Up here my fixed constant is 4

You could also pick abs(x-4) < 0.0000000000001 for ex

Or uhh

Well delta < that ugly decimal

[faint edge]

Actually yeah, I'm mixing up what ur saying here this is fine.

For any fixed constant k, you can choose to take delta < k.

Do you see how this helps?

[novel vapor]

Yea

I'm thinking about the solution now

Well we need to bound delta then

[faint edge]

From here using my work abs(x-4)abs(x+4) < ??? < epsilon

What is ???

[novel vapor]

Uhhh

[faint edge]

Use this

[novel vapor]

Maybe the upper bound of x+4

[faint edge]

Thst times abs(x-4) yah

[novel vapor]

Times |x-4|

yea

Now
9|x-4|<epsilon

|x-4|<epsilon/9

Now we didnt forget that we bound delta

[faint edge]

Did I do arithmetic wrong?

Why 9?

[novel vapor]

Oh

Not 9

12

Epsilon/12

Then the solution is the minimum between 4 and epsilon/12

Cuz delta must be less than 4

[faint edge]

Yep exactly

Can you pick other deltas?

Besides the min?

[novel vapor]

Chenging 4 you mean?

I can bound abs(x-4) by 20 for example

Any number will work

[faint edge]

An easier way to see is that you took del=min{4,eps/12} but you could have taken anything less than that min.

[novel vapor]

Epsilon/100000000000000000000000?

[faint edge]

No, what I'm saying is that if del>0 makes ur limit proof go thru, so will any smaller delta.

[novel vapor]

Or min(3,eps/12) work too

[faint edge]

Yah that's true.

[novel vapor]

Min(0000000000….,1,eps/12)

[faint edge]

I'd probably avoid setting delta = min and just take delta < min since it's easier to avoid some odd technicality.

it shouldn't matter tbh.

[novel vapor]

Why?

Its fine to let del=min..

[faint edge]

Yeah no that is fine

You just develop habits and feel weird not doing them lmao.

Do you see why the other trick works?

The lim f(x) g(x) one?

[novel vapor]

Why

I think its the easiest way to prove what i sent

Showing limf(x)=4

[faint edge]

16

[novel vapor]

And limg(x)=4

What

Not 4?

[faint edge]

Ah I see what you mean

Your f is f(x)=x^2 but the notation I used to describe the prop reused f

My bad

[novel vapor]

Let it h(x)

[faint edge]

The trick is just if h is the identity function f(x)=h(x)h(x) so apply the property.

[novel vapor]

Show
h(x) -> 4 when x -> 4

Del=eps

[faint edge]

Yep that works.

[novel vapor]

Obviously

[faint edge]

Then apply the product property for limits.

[novel vapor]

Yeah it work

[faint edge]

What if I ask you to compute lim x^3?

[novel vapor]

Better than min

[faint edge]

(At x=4)

[novel vapor]

Limf(x)=limh(x)*h(x)*h(x)

with h is the identity function

[faint edge]

Presumably you could use both tricks yah

[novel vapor]

Bruh is there some good notation to write limits?

[faint edge]

But one is much worse than the other lmao

[novel vapor]

Specifically when discussing

[faint edge]

$\lim_{x\to 4} f(x)?$

[soft zealot]

DootDooter

[novel vapor]

Im not writing all of that bruh

[faint edge]

Sometimes for shorthand ppl do lim x->4 f(x) or variations of that.

[novel vapor]

Okay thats look good

[novel vapor]

What is it

[faint edge]

You might have to clarify if ppl don't get it.

The identity trivk

*trick

You can derive continuity of polynomials in general with that one.

Both are good to know though.

Sometimes you have extra info in a problem that you can constrain to squeeze out a limit.

Whoever made this q probably wants you to use the bounding trick.

[novel vapor]

Dont tell me that’s there more other tricks

[faint edge]

Sometimes all of analysis feels like inequality and limit tricks

[shadow marlin]

to be fair, the "tricks" are usually very natural once you understand them

[final saddle]

@novel vapor Has your question been resolved?

[novel vapor]

Well here i can use the same trick as before
limf(x)g(x)
Limf(x)+g(x)

[faint edge]

Wdym?

[wintry sigil]

Guys is white monster safe to drink

[wintry sigil]

Is this the question?

[wintry sigil]

All of these questions are pretty simple

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[novel vapor]

Maybe for non overthinkers

[latent dragon]

Please don’t go into help channels saying “questions are pretty simple.”

[novel vapor]

.close

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[wintry sigil]

Sarcasm

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[smoky bramble]

Say, if f((x+7)/(2x-3))=3x-7
can't I just substitute x as x(2x-3)/(x+7) to solve for f(x)

[smoky bramble]

or will there be some edge case behavior that won't allow that

[honest grove]

cn you help me

[digital sable]

what do you mean exactly

[honest grove]

in mths

number pttern

[smoky bramble]

say uhhh

,, f(\frac {x+7}{2x-3})=3x-7

[soft zealot]

festive shitposter

[honest grove]

https://discordapp.com/channels/268882317391429632/1455103697225912474 in this question i need help

[smoky bramble]

therefore, \

$f(\frac {x+7}{2x-3} \times \frac{x(2x-3)}{x+7})=3(\frac {x(2x-3)}{x+7}) -7$

[soft zealot]

festive shitposter

[smoky bramble]

does this kind of logic work?

[tired walrus]

no

[tired walrus]

!occupied

[final saddle]

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).

[digital sable]

unfortunately not

[smoky bramble]

shucks

[tired walrus]

if you wanna do this sorta thing you need to find the inverse function of (x+7)/(2x-3) and then substitute x = that

[smoky bramble]

so the matrix inverse thing is necessary

damn

f • T = g
f = g • T^-1 right

.close

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[merry trench]

basically put it as t

[crystal pawn]

…

[proper dagger]

you opened a channel btw. do you have a question?

[merry trench]

.close

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[grave wharf]

can anyone check this

[grave wharf]

if its wrong could you try 12π/365 instead of 14π/365 and 28.32 instead of 28.37

[sturdy cypress]

apply a couple pounds of force

[final saddle]

@grave wharf Has your question been resolved?

[grave wharf]

what?

[final saddle]

@grave wharf Has your question been resolved?

[grave wharf]

. close

.close

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[oblique current]

i'm back 💔 please @ me when replying :))

[rugged merlin]

Are you doing a

[oblique current]

i've done a, i'm doing b now

[rugged merlin]

Okay

What have you done

[oblique current]

for it to not slip, Fr ≤ uR?

[rugged merlin]

Yea

Aka u need to find F_max = uR

[oblique current]

FrX + P = (19/2root3)W = 2W + P, P = ((19/2root3)-2)W

[rugged merlin]

You need to do some casework

Like

[oblique current]

Fr is the frictional force and what does uR mean again? 😭

[rugged merlin]

The ladder can flip towards the wall or away from the wall

[oblique current]

yes

[rugged merlin]

Normal force multiplied with friction coefficient

[oblique current]

is it like the max force needed for an object to start moving from rest?

[rugged merlin]

I think this is correct

[oblique current]

yh it is, it's the lower half of the range

[rugged merlin]

Thats when its slipping away

[oblique current]

and P is larger than that

[rugged merlin]

Yes

Min force

Needed

[oblique current]

oh yh

[rugged merlin]

Well

In reality the other case is the same

[oblique current]

I get how to algebraically manipulate it, i just need to understand these cases and Fr being ≤ or ≥

[rugged merlin]

Mhm

[oblique current]

so in this case, why would this be slipping away?

is it just cause it's -2?

[rugged merlin]

The ladder wants to slide out left, so friction acts in the right same direction as P, to help hold it

[oblique current]

yes

[rugged merlin]

So its the difference between

P + F_max = R_y
And
P - F_max = R_y

[oblique current]

ohh, and then the other case is when P + Ry = FrX?

[rugged merlin]

Well

Yeah

[oblique current]

ok hold on

[rugged merlin]

Yeah so do

The exact same thing again

Ill brb 5 minutes but for c and d just write out what you think and ill review it

[oblique current]

okayy

it's actually FrX that changes direction right? and not P, P stays going towards the wall

[rugged merlin]

Hey

I'm back

[rugged merlin]

Yes

[oblique current]

heyyy

[oblique current]

ok now i've got both -2 and +2, i still don't get how to tell them apart as to which is the lower end of the range and ≤ P

[rugged merlin]

Well i mean

Surely the -2 one is going to be the lower end

It is smaller, obviously

[oblique current]

oh yh 😭

that's so silly of me

[rugged merlin]

x - something is smaller than x + something

[oblique current]

forget i ever asked that

[rugged merlin]

😭

Issok it happens

[oblique current]

i've done part c and part i of part d

c is just that weight acts from the midpoint

di is the magnitude gets smaller

and now i'm stuck on dii

[oblique current]

way too often in my case 😭

[rugged merlin]

Lemme read it rq

[oblique current]

okayy

[rugged merlin]

Ok so

Obviously nothing changes if ur force is below F_max

Like we talked about earlier

[oblique current]

mhm

[rugged merlin]

Yeah si lile

You have R_y ± F_max as your range

[oblique current]

yh

[rugged merlin]

Ans you determined in i that R_y decreases

If R_y decreases what happens to R_y ± F_max

[oblique current]

gets smaller

[rugged merlin]

Ding ding ding

[oblique current]

:p

ok let me think it through once more

[rugged merlin]

Aighty

[oblique current]

yaya we've done the question

[rugged merlin]

Yeppieee

[oblique current]

i've done 4 questions over the past 3 days 😭

[rugged merlin]

😭

[oblique current]

god save me i hate applied

[rugged merlin]

If you are learning i guess thats better than nothing

[rugged merlin]

Whatcha doing?

[oblique current]

and my one note isn't synced either so i'll have to copy it all up when school begins 😭

[oblique current]

first year of a levels

applied is compulsory, i'd throw this part of maths out the window if i could

[oblique current]

but i guess that means i get to go over the process again 😭

[rugged merlin]

Oh it was this bad?

Darn

[oblique current]

not like i want to but

[oblique current]

it's just like physics and i've never been a fan of it, and i'm not that good at it either

[rugged merlin]

Surely you can figure out a way to sync it...?

[oblique current]

moments, forces and stuff like that

[rugged merlin]

Yeah i hated classical mechanics too 😭 I only dig the electrical stuff lmao

[oblique current]

i'm in china rn for my holiday, i can't install a vpn so i can't sign in to microsoft and sync it

[rugged merlin]

Oof

[oblique current]

yh. but i've got 2 questions left

[rugged merlin]

Thats so unfortunate

[oblique current]

in my pack of exercises 😭

[oblique current]

ikkk

[rugged merlin]

Two thirds done tho so u r close

[oblique current]

i had the choice to bring my home ipad where i can install whatever and have it all up to date but i chose my bigger, newer school ipad

they're are 12 questions in total, i did some at school before

but my friends helped so ig that's how i made it so far 😭

[rugged merlin]

Well good luck with the rest!!!

[oblique current]

yepp, trying to give them a go before coming to seek comfort in here 😭

i'll leave the last two for tomorrow 😭

the next question has tension in it

[rugged merlin]

Oh yeah its like

[rugged merlin]

Hella late in china

[rugged merlin]

The tension questions are easierrr

[oblique current]

it's only 10, i sleep at like 12 but i just don't know if i want to do anymore lol

[oblique current]

maybeee

[rugged merlin]

Maybe do something fun before eep

[oblique current]

maybe, if i can think of anything to do 😭

i don't really do much but the day goes by so fast

[rugged merlin]

Felt

Maybe u should game

[oblique current]

i love gaming

roblox is blocked here tho

and the person i play minecraft with is asleep rn

i play codm too but that's blocked too 😭

so i have nothing

[rugged merlin]

Misery

[oblique current]

ikr

can't wait to go bacl

gonna go grind ranked with him 😭

force him to play wiht me

.close

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[fossil kiln]

$ABCD$ is a square. Find the other 2 vertices if $A(3,5)$ and $C(5,-3)$.

[soft zealot]

ch3rry

[fossil kiln]

$r=\sqrt{17}$

[heady moon]

looks good

[fossil kiln]

The x,y eqn we have

Every point on BD will satisfy?

[tired walrus]

$\sqrt{17}$

[soft zealot]

Ann

[heady moon]

yes if AC remain diagonal of square

[soft zealot]

ch3rry

[fossil kiln]

-# u get the point

I dont understand

Is r supposed to be a constant value for aby x,y?

[jagged wave]

(this is a dupe of help-44?, which was just closed for this same question)

[fossil kiln]

Yh got too messy

[heady moon]

r will depend on midpoint

of diagonal

[fossil kiln]

So r changes for every x and y?

[heady moon]

what do x and y mean

[fossil kiln]

A point tht would satisfy BD

As in line BD

[heady moon]

right it will change

so will theta

[fossil kiln]

Theta cant possibly change thts the slope

[heady moon]

yh but for a different square it will

[fossil kiln]

im talking about line BD

[heady moon]

like here we are keeping lines AC and BD fixed

[fossil kiln]

No square wht

[heady moon]

right

[fossil kiln]

Well consider BD is fixed

Erm anyways

So r has different values for every point

[jagged wave]

(you mentioned earlier you were using the parametric equation for a line?)

[heady moon]

yes

[fossil kiln]

And the eqns here

They're wrt to (4,1)

[fossil kiln]

(That is what i used senora)

[heady moon]

yes

[fossil kiln]

So yeah makes sense

If i put (4,1) r is 0

Cool

I kinda get it now

Anything you'd like to add?

[heady moon]

nothing beside the fact forcing parametric is..

not suggested

[fossil kiln]

Wut

[heady moon]

like js get centre then you can draw line perpendicular to AC

[fossil kiln]

Oh ik

Im js learning to apply parametric form of a line thts why

.close

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[heady moon]

got u

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[tiny kraken]

I can't understand this solution

My doubts:

• Why are there R at the end and also the beginning ( I would assume they are the same one )
• Why are there 9 places to insert?
• Why partitioning 18 into 9 groups is binom(27,9)

[jagged wave]

question gets cut off but i assume the full final sentence is "in how many ways [can claudia make such a necklace?]"

[tiny kraken]

yeah

I also want to share my attempt which gave a different answer:

• I so did form a skeleton: R_R_R_R_R_R_R_R_
• Each gap contain at least 2 beads so I used stars and bars
• R_R_R_R_R_R_R_R_ can be simplify into x1+x2+x3+x4+x5+x6+x7+x8 = 32
  => x1+1 + x2+1 + x3+1 + x4+1 + x5+1 + x6+1 + x7+1+ x8+1 = 32-8=24
  So basically it's binom(23,7) here

[winter lava]

this is before accounting for rotation of the necklace?
and we're using all 40 beads?

[odd seal]

Why are there R at the end and also the beginning ( I would assume they are the same one )
The necklace doesnt have full beads, and i am gonna say the expected arrangement of beads is such that there is some string exposed, so the skeleton would be instead
_R_R_R_R_R_R_R_R_
Hence the 9 spaces.

[tiny kraken]

huh that makes sense

[tiny kraken]

that's the whole solution tbh, ig we avoid rotation by fix a red bead

[odd seal]

in this situation there would be no issue of rotation

[winter lava]

https://en.wikipedia.org/wiki/Necklace_(combinatorics)

[odd seal]

Why partitioning 18 into 9 groups is binom(27,9)
this is basically stars and bars

[tiny kraken]

Okay I can see how that makes sense, but why are we inserting 18?

[winter lava]

on this page they account for rotation

[tiny kraken]

Ohhh

[tiny kraken]

no wait

that doesn't make sense

[odd seal]

the 14 beads are already known to be between the red ones

[tiny kraken]

That mean on two end there can be 0 bead

[odd seal]

yes

[tiny kraken]

But that's invalid case

[odd seal]

this would be the necklace

are you a darkmoder? sorry the string disappeared

[tiny kraken]

Nah dw I can see it

If we enclose the necklace then between two end red beads there won't be any blue bead

[odd seal]

well yes, but that would be a combinatorial bracelet (what I mean is based on how I was taught, necklaces dont have rotational symmetry)

so like the green stuff is the clasps, and you dont consider that the red beads need any blue beads there

[tiny kraken]

[odd seal]

and this solution that you have seems to follow similar idea

[tiny kraken]

Okay that make sense ig

Huh wait

so if it's combinatorial bracelet

then how would I do it?

[barren pebble]

Do you know about Euler candy division problem?

[jagged wave]

to be clear what is the definition of necklace you're using

[tiny kraken]

That's just another name for Stars and Bars

[barren pebble]

The solution says there are 7 spaces between the red beads so i'm assuming the necklace here is just a string and not an enclosed loop

[odd seal]

if its a bracelet, you would have to consider symmetry as well

[tiny kraken]

I thought it's what they call combinatorial bracelet that I was told

[odd seal]

and ofc there would be only 8 spaces

[tiny kraken]

Okay so

Hm

Okay let's say it is combinatorial bracelet

[barren pebble]

Uhm sure, do you know the number of solutions to x_1+x_2+...+x_n=k with n,k,x_i ∈ ℕ?

[tiny kraken]

I know mate, read what I sent

And all the messages above

[barren pebble]

Oh yeah

[tiny kraken]

Then I could do this

[barren pebble]

Well the answer is basically that, but with a different definition of a necklace

[tiny kraken]

Yeah

Okay new question, how would I do this problem if it's a combinatorial bracelet

[winter lava]

doesn't stars & bars give C(27, 8)?

[tiny kraken]

This seems valid in the case

Can someone verify that

Then taking account for rotation

Okay most of them would have 40 different rotation right?

And some would have less

[winter lava]

yes

[odd seal]

wait, not even that, it should be 26C8 right?

[tiny kraken]

Lmao

Okay so let's number each position, and let's say a1,a2,a3,a4,a5,a6,a7,a8 are position of red beads

{a1,a2,a3,a4,a5,a6,a7,a8} = {a1+k,a2+k,a3+k,a4+k,a5+k,a6+k,a7+k,a8+k} mod 40 for some cases

we want to find those cases where after some k rotations it repeats itself

[winter lava]

the only possibilities for the # of rotations are 40, 20, 10, and 5

[tiny kraken]

huh

[tiny kraken]

what's #

[winter lava]

number of rotations

[tiny kraken]

ah

[winter lava]

if you divide the necklace into n segments, n must divide both 8 and 40

[tiny kraken]

hm

[winter lava]

or...

wait

[tiny kraken]

why not 15 and 25 and 30 and 35

40 doesn't seem valid

[winter lava]

a random neckace has 40 rotations as you said before

[tiny kraken]

huh hang on I am processing

Ain't 0 rotation and 40 rotations the same?

[winter lava]

i'm talking about the number of ordered 40-tuples that can be rotated to match a given 40-tuple

there is always at least one such tuple

so it's never 0

[tiny kraken]

Okay that means we take only the 40 rotations and ge rid of other ones

Yeah I was thinking keeping the original 0 rotation

A bit confusing

[winter lava]

i think we are talking about different things

[tiny kraken]

Uhh

I wad thinking of rotating the original one by k rotations, in such it repeat itself

[heady moon]

We on dis?

[tiny kraken]

Kinda

[winter lava]

ok then the only possible numbers are 40, 20, 10 and 5 i believe

a rotation by 15 is the same as a rotation by 5

same with 25

so we don't count those

[tiny kraken]

wydm by the same?

[winter lava]

if the necklace is the same after rotating by 15, then it is the same after rotating by 5

[heady moon]

What the soln suggests is this kinda necklace and not an enclosed one, ig

[winter lava]

because you can rotate by 15*3 = 45 = 5

[tiny kraken]

hm

[tiny kraken]

Okay I am confused what did this mean

[winter lava]

i think fionna just got interested in the case that it's an enclosed necklace

[tiny kraken]

Why are we multiplying

[heady moon]

Ic

[winter lava]

you can rotate it by 15 three times if you want, and it should stay the same

[tiny kraken]

Oh so that's an example okay

[winter lava]

yes it means any necklace that stays the same after rotating by 15, also stays the same after rotating by 5

[loud sundial]

orz

[tiny kraken]

no orz, I am still confused

[winter lava]

the number of identical segments has to divide both 8 and 40, so it's 1,2,4, or 8
if there's 1,2,4, or 8 identical segments, that means the rotations are 40, 20, 10, or 5 respectively

[tiny kraken]

Why 40?

[winter lava]

because there are 40 beads and each segment must have an equal number of beads

so the number of identical segments has to divide 40

[tiny kraken]

No?

I meant

Gimme a min

[winter lava]

hmm actually

it might be important that some necklaces stay the same after rotation by 15

if u plan to use burnside's lemma

but once you calculate the number of necklaces that stay the same after rotation by 5, you know that it's the same for 15 and 25, etc

[tiny kraken]

Still, isn't if it stay the same after 15 rotations then it stay the same after 25,20 or just 5k as well?

[winter lava]

(i think there might another way to do it, that avoids burnside's lemma)

[tiny kraken]

Sorry it's 12 am here Im not so awake

[tiny kraken]

I'm still having some doubts about this

[winter lava]

if it stays the same after 15 rotations, then it stays the same after 5k rotations for any integer k

[tiny kraken]

yeah

hm

so

How many cases it does that

[tiny kraken]

I have no idea what burnside lemma is

[tiny kraken]

Don't answer it

yet

let me think for a min

okay so in general it would be an arrangement with the first bead being number 1,2,3,4 or 5

then other beads are a1 + 5k

so it should be 5 cases only

[winter lava]

i think 5 is right, the segment can be BRRRR, RBRRR, RRBRR, RRRBR, or RRRRB

[tiny kraken]

Is this right?

[winter lava]

for a closed necklace?

[tiny kraken]

yes but yet account for the rotation

wait no

ohhh

huh

[tiny kraken]

It shouldn't be x8+1 and x1+1here

,w (binom(24,7)-5)/40 + 1

it's still fraction

Something went wrong

[tiny kraken]

nvm I confused myself

[tiny kraken]

Wait why do I feel like this standalone is a valid solution

[heady moon]

,w ((binom(24,7)-5)/40) + 1

[soft zealot]

Results provided by WolframAlpha

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

Requested by ihavebecumdeath

[tiny kraken]

It's not right

[heady moon]

Nope

[tiny kraken]

No

I mean that alone

Not subtracting any other cases

[heady moon]

Why+1?

[tiny kraken]

there's only 1 way to get RBBBBR in this

[tiny kraken]

this

in R_R_R_R_R_R_R_R_ which is placing 4 blue beads into each gap

[tiny kraken]

read the above

[heady moon]

Ok

[tiny kraken]

R_R_R_R_R_R_R_R_ this's the skeleton, so basically each gap has to have at least 2 beads

[winter lava]

but then you're only counting necklaces with R in the first position

[heady moon]

Right

[tiny kraken]

Hmm, we can let red bead be the first position in any case, no?. I mean we can pick one red bead be the first position

[winter lava]

i'm not so sure

[tiny kraken]

It's 1am rn damn

Imma sleep, I will be back tomorroe

I still need some help with question 2.12 and this one

And question 2.12 looks scary

[heady moon]

This works

Removes rotations

[tiny kraken]

Are you sure? like extremely sure? cuz in some problems I fix a position and it doesn't give me the right answer

So now I don't even know when we can do that

[heady moon]

What? Wouldn't fixing give you 23C7

[winter lava]

fionna wanted to count them with rotations being distinct

like RBB and BRB are distinct

so i think you can't fix the first bead

[heady moon]

Yeah so start from one red bead as reference point

[tiny kraken]

If you meant
R B
B B R B

[heady moon]

Linearly are not

[tiny kraken]

I would consider it the same

[heady moon]

Makes sense

[winter lava]

you said this

[tiny kraken]

No I meant

That solution alone

Seems to solve this

[tiny kraken]

this

[winter lava]

i can't follow your train of thought

[tiny kraken]

Okay so

[heady moon]

You say we take em distinct, hmm..

[tiny kraken]

I have feeling that this right here might be the solution for the new question I asked

[tiny kraken]

This question

[tiny kraken]

I know mate, but how can we make sure that's all the cases? and not repeating?

[winter lava]

what does the third bullet mean?

[tiny kraken]

R_R_R_R_R_R_R_R_ solve this using bars and stars

Basically

[winter lava]

you had an underscore at the beginning

[tiny kraken]

Yeah it's a typo and it also confused me myself, sorry

[winter lava]

no because you will overcount. you will count these as distinct:
R BBBBBBBBBB R BB R BB R BB R BBBBBBBBBB R BB R BB R BB
R BB R BBBBBBBBBB R BB R BB R BB R BBBBBBBBBB R BB R BB

[tiny kraken]

Hmmmmm

Yeah

okay so how would I proceed then?

I meant I know there're 5 case it repeats itself

[winter lava]

i know how to deal with the rotation but i'm having trouble with the condition that R's are separated by two B's

[tiny kraken]

How many ways can we arrange treating each rotation as distinct way

Yeah

That, idk how to count that

[winter lava]

imagine n(k) is the number of necklaces that stay the same after rotation by k
then you do (1/40) (n(0)+n(5)+n(10)+n(15)+n(20)+n(25)+n(30)+n(35))
which is the same as (1/40) (n(0)+4n(5)+2n(10)+n(20))

this is burnside's lemma

[tiny kraken]

,can a highschool student learn this/

Would it worth the effort

[winter lava]

umm i think it might turn up in like IMO type stuff

other than that, it's probably not useful to you

[tiny kraken]

Ah, I'll leave that for another day then, I am practicing combinatorics for an exam and it's at most as hard as some of the difficult problems in AIME

So ig IMO stuff won't be useful

[heady moon]

Hmm.. basically, cyclic shifts of solutions of x1,x2,...x8

[winter lava]

yes

[tiny kraken]

we could try to figure out how many repetition caused by that shifting

[tiny kraken]

I still have trouble with this bh

[winter lava]

yeah me too that's why i can't solve it

[tiny kraken]

Okay that's for tomorrow I'm dying rn

It's 1:32 am

Thanks you all

Ly

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[loud sundial]

Moderators good I give you guys Q

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[untold kiln]

In context of calculus, I'm learning to integration techniques. In one of the problems from the book I'm using, I can't tell how they got the part that is underlined yellow, in the following passage:

[untold kiln]

From attempting this, the following pink underlined part is as far as I got:

How did the author get the underlined equation?

[formal trail]

note that your equation is dx/du = u/x

but really it would be much easier if you differentiated implicitly

[untold kiln]

how is dx/du = u/x?

[formal trail]

the denominator of your equation is x according to the equation on the left

[untold kiln]

ah

so then x/u . 1/x = 1/u

I still don't know how to get to the yellow underlined equation.

[formal trail]

you can get the yellow by differentiating both sides of $u^2 = x^2 + 4$ implicitly

[soft zealot]

cloud ☁

[formal trail]

although in a more roundabout way you get it by rearranging dx/du = u/x

[untold kiln]

The book has implicit differentiation in an appendix. But, I'll try getting there via rearranging method.

Thanks

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[crystal pawn]

..

[rugged merlin]

<@&268886789983436800>

Oh

[crystal pawn]

bro js gon

[rugged merlin]

Proactive mod intervention?

[crystal pawn]

probably

[cyan kayak]

banned from another ping, mass delete got it

thanks though!

[crystal pawn]

u gotta.. close

[cyan kayak]

#help-36 message

[crystal pawn]

ohh mb

[cyan kayak]

;D

[rugged merlin]

<@&268886789983436800>

Round two

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[tame smelt]

I want to ask say I have a matrix/tensor with values between -1 and 1 real numbers. How can I scale that with another matrix/tensor with values between 0 and 5 in such a way that I can reverse the original first matrix without knowing the second one?

[vital crag]

can you even do that with a single number

[tame smelt]

No but maybe there is some clever way with a matrix

What if I don't have bounds

Just want to scale a matrix by another one

And then reverse it without knowing the second one

[vital crag]

define "scale a matrix by another matrix"

do it for 2x2 matrices

[tame smelt]

Just scale multiply each value by another one

Make it larger

[vital crag]

too vague

scale by another what?

[tame smelt]

Scale by tensors actually 2 tensors the same size

[vital crag]

give an explicit example with two 2x2 matrices

[tame smelt]

What do you mean?

[1, 3]

[6, 2]

[6. 6

Maybe I could add some "magic" number as a base so I can use that number to revert to the original [1,3]

[obtuse hedge]

Is this elementwise multiplication

[tame smelt]

yes no dot

[vital crag]

yea you just proved why you can't "go in reverse"

how do you know 6 "came from" 1 * 6 or 2 * 3

[vital crag]

these are called hadamard products. should read this to learn about them https://en.wikipedia.org/wiki/Hadamard_product_(matrices)

[tame smelt]

Ok how about is it possible to find such a number that:

[1, 3] * x = a

Where a / mean(a) = [1,3]

[tame smelt]

I will read this thanks

maybe this is what I need

But also what about my suggestion?

ok whatever it's dumb

Thanks

.close

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[tame smelt]

.reopen

[final saddle]

✅ Original question: #help-36 message

[tame smelt]

Actually I have another question

Can I scale a tensor by a single value such that the tensor divided by its mean equals another tensor

How can I find such value is it possible?

I'm mostly thinking if this is possible in the context I described earlier but after I do the elementwise multiplication to get the first tensor by dividing by its mean after I multiply each element by the same value as needed x

So basically it would be:

[a1,a2] * [b1,b2] * scalar-x = A where A / mean(A) == [a1,a2]

[glossy zephyr]

Just as advice, when you want to refer to matrices / tensors

use A not a

[tame smelt]

ok

[glossy zephyr]

it isnt an actual rule but more of a common use

[glossy zephyr]

Depends what kind of product you are intending to use here, but this would require the product of [a1,a2] * Bx to be colinear to [a1,a2] to start with

Since mean(A)^(-1) is a scalar too.

[obtuse hedge]

Hadamard product from before

[glossy zephyr]

Yeah, guessed so. I think for the identity to be possible it would be required for b1 = b2

[obtuse hedge]

Agreed

[glossy zephyr]

which basically transforms the hadamar product into a fancy scalar product

[obtuse hedge]

Doing the math out on paper to check though

[tame smelt]

Hmm

Isn't colinear geometry?

[glossy zephyr]

Yes, but its a term that applies all over linear algebra too

basically, two elements of any vector space are colinear if one is the result of a scalar product of the other

obviously, in their respective vector space

[tranquil pine]

isnt linear algebra just degree 1

i am uneducated

[glossy zephyr]

youre confusing linear equation with linear algebra

[tranquil pine]

whats linear algebra

[obtuse hedge]

Based on my calculations it looks like this identity holds whenever $\frac{a_1}{a_2} = \frac{b_2(b_1-1)}{b_1(b_2-1)}$

[soft zealot]

Coolempire2026

[obtuse hedge]

Kind of interesting

[glossy zephyr]

really broad topic where we study mathematical spaces and elements and the translations between them

Where they have scalar product and addition > and depending on the case, a lot more types of products

[tranquil pine]

do u start ts in uni?

[glossy zephyr]

You kinda see the basics in HS without knowing it.

[tranquil pine]

okayy

[glossy zephyr]

For example, polynomials and vectors are both part of the study of linear algebra

and those are given in HS level

You eventually learn of more advanced structures like matrices, tensors, etc...

[tame smelt]

I was thinking would it be easier instead of doing multiplication then maybe just make it:

[a1,a2] * [b1,b2] + scalar-x = A where A / mean(A) == [a1,a2]

[glossy zephyr]

+ ?

[tame smelt]

addition

[glossy zephyr]

you cant add a scalar to a matrix

[tame smelt]

Of a single scalar across all elements

[obtuse hedge]

Oh I should note I did this with the scalar being 1 (i.e. without a scalar)

[glossy zephyr]

there surely is some operation that does that.

But i think you can just define it on your own

[obtuse hedge]

Which should allow you to solve for b2 based on your choice of b1, and given a1 and a2

[glossy zephyr]

Well, adding the same number to all elements of matrix usually doesnt lead to a colinear matrix.

So this is prob a lot different

[obtuse hedge]

I agree

Unless it's 0 in which case it's just the situation I solved for

[glossy zephyr]

Yeah, ofc.

Ill write the problem as Also, for the hadamar product (A * B) + x =/= A * (B + x)

[tame smelt]

So it's solvable this way then?

[glossy zephyr]

I guess we are going with the first here

[obtuse hedge]

I assume so as well

[glossy zephyr]

btw, ill rewrite the problem to A,B,x,C so i can write less, lmao

where C / mean(C) = A

So : A*B + x = C where C / mean(C) = A

C and A have to be colinear still for that to be possible

So we have a system of equations of the form

knowing C/mean(C) = A --> C = Amean(C)

With m being the mean, a shared number across all equations

For all elements of the matrices*

[tame smelt]

So it's not possible then?

[glossy zephyr]

Its really restrictive

there must an m and x that satisfy all the equations at the same time

[tame smelt]

Ok makes sense

[glossy zephyr]

Quick question, are A and B row/column matrices or just some matrix?

anyways, i had an idea on this, but i think this will only be possible if (not conclusive tho) A has some sort of "inverse" st.

A x A^-1 = Identity Matrix

when A = n x m, A^-1 m x n and I will be n x n

I asked for row/column because then C x A^-1 will be a 1x1 "matrix"

aka a scalar for our purposes

only if C x A^-1 = mean(C) then you will have the equation come true

no proof, just an idea i had, and i have 0 clue how to prove this is (if its) true anyways, lmao

[tame smelt]

I guess I could make it bigger than the output I have if needed so like maybe only portion of the values are the original matrix

As long as I have them the way I want so like if it's:

[a1,a2] * [d1, d2, b1,b2, c1, c2] * scalar-x = A where A / mean(A) == [l1, l2, a1,a2, l3, l4]

Not sure if this could help

[glossy zephyr]

hadamard prod. wont work for diff. sized matrices tho

[tame smelt]

No what I mean is to extend the resulting matrix C as needed with values that I don't care about but in a given limit so that C / mean(C) == A

[glossy zephyr]

[a1,a2] * [d1,d2...c2] is the problem

[tame smelt]

Ye anyway it seems bogus

Thanks for the help anyway I'll close for now if I come up with something else I'll reopen

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[lapis wedge]

im having a bit trouble understanding why adding the equations in a system of equations give the same solution as the system

[lapis wedge]

So like if we have 3x + 4y = 6 and 5x - 2y = 1

then 8x + 2y = 7 has the same solution as the system above

what is it about adding these two equations that forms a new singular equation with a solution to the system

.solved

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[bitter fossil]

Hi, I have a question about when to use the unilateral vs the bilateral Laplace transform. For instance, x(t) = cos(at). The unilateral LT is s/{s^2+a^2} but The bilateral LT doesn't exist.

[worthy wren]

in what context are you learning about bilateral laplace transforms

bilateral laplace transforms mathematically differ just by the integration bounds

so the bilateral laplace transform may not exist because it consists of -inf to inf bounds (u'd need your integrand to converge over that entire domain)

[bitter fossil]

Sorry I forgot to mention, I'm studying Signal and System

I understand they differ in integration bounds, but I'm not sure their usecase

[worthy wren]

i havent done signal and systems but im guessing if u care about the context for what happens before t=0 for the system, u'd need to use bilateral

[bitter fossil]

Thank you, but the original problem doesn't involve system. Only finding LT and its ROC

[worthy wren]

i mean if the signal is defined only for t>0 then use the unilateral for ROC

since clearly L_bilateral{cos(at)} wont exist, it is meaningless to ask its ROC

[bitter fossil]

I guess the given problem is not really clear

it should have been cos(at).u(t)

[worthy wren]

then the ROC is the same as the unilateral one and so is the LT

cuz u(t) = 0 for t<0

[bitter fossil]

Yes, but it's not wrong to say there is no bilateral LT

[worthy wren]

indeed

there will be a bilateral LT

[worthy wren]

what i said here applied to the function cos(at), not cos(at)*u(t)

[bitter fossil]

Yeah, you only wrote cos(at)

[worthy wren]

anyways for finding ROC, u can always first start with bilateral LT

if u see it doesnt exist then u can try finding the unilateral LT

if ur function is piecewise continuous and of an exponential order, there will be some nonzero ROC at least

[bitter fossil]

alr, thank you

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[oblique current]

i'm back 💔

[onyx peak]

So what's the issue here?

[oblique current]

i just don't know what to do

[onyx peak]

a good way to start would be to model both the boundary curves as equations

can you find the equation of that parabola and that line?

[oblique current]

x^2 - 6x? and y = 2x + 60?

[onyx peak]

x^2 - 6x

almost

you probably did x(x-6)

but you can still multiply it by some constant without changing the roots

[oblique current]

yh i originally put an A at the front

[onyx peak]

yeah, that's good

and after that, you need to make sure it passes through (10, 80)

[oblique current]

ah yes

[onyx peak]

y = x^2 - 6x
doesn't pass through that

[oblique current]

A = 2

so 2x^2 - 12x

[onyx peak]

so y = 2x^2 - 12x and y = 2x + 60

now we need to actually describe the region

would u know how to describe e.g. this region using an inequality?

The one above the parabola

[oblique current]

y ≥ 2x^2 - 12x?

[onyx peak]

yep great

and the R region is the region which is
A) above the parabola and above the line
B) above the parabola and below the line
C) below the parabola and above the line
D) below the parabola and below the line

[oblique current]

B

[onyx peak]

and do u know how to describe

below the line
using an inequality

[oblique current]

y ≤ 2x + 60

[onyx peak]

y ≥ 2x^2 - 12x and y ≤ 2x + 60
together, they describe the region

both above the parabola and below the line

[oblique current]

oh, is that it?

[onyx peak]

yes

[oblique current]

😭

[onyx peak]

if you wanted to, you could write it slightly more compactly

2x^2 - 12x ≤ y ≤ 2x + 60

both both ways should be fine

[oblique current]

okok

i'm asking the stupidest questions god 😭

[onyx peak]

Nah, this kind of questions is pretty common. I wouldnt know how to do it either if nobody told me

[oblique current]

this is the second question i've over complicated lmao

but now i know it's not so hard

so thank you again lmao

[onyx peak]

np

[oblique current]

.close

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[knotty elk]

Is this rigorous enough?

[scarlet onyx]

what do you want to prove ?

[knotty elk]

x^2 - y^2 = (x-y)(x+y)

I got stuck because the associative law of addition is mentioned in the book for 3 terms and not 4. And I dont think I can remove the brackets just like that without a proper proof or justification

[scarlet onyx]

if you have the distributivity you can develop the first expression

[knotty elk]

Can I just write that
(x^2 - xy) + (yx-y^2) = x^2 -xy + yx -y^2 = x^2 - y^2

Because the associative law states that if (a+b) + c = a + (b+c)

There are 3 terms and we interchange the brackets between 2 terms a&b or b&c

[leaden moon]

cool

but splitting into variables is not really necessary

[knotty elk]

Wdym
Can u please elaborate?

[leaden moon]

$(x - y)(x + y) = x(x + y) - y(x + y) = x^2 + xy - yx - y^2 = x^2 - y^2$

[soft zealot]

1 divided by 0 equals Infinity

[leaden moon]

also your $\in$ looks a lot like $\epsilon$

[soft zealot]

1 divided by 0 equals Infinity

[scarlet onyx]

$(x^2 - xy) + (yx - y^2) = x^2 + (-xy + (yx - y^2))$

[soft zealot]

etnz the etnah

[scarlet onyx]

you can consider the second bracket a single variable and use the associativity on 3 terms

then you use the associativity again

[knotty elk]

Thanks guys ❤️

[leaden moon]

hm

similarly

an exercise for ya

prove that $a^3 \pm b^3 = (a \pm b)(a^2 \mp ab + b^2)$

[soft zealot]

1 divided by 0 equals Infinity

[leaden moon]

<@&268886789983436800> user id: 423444344037703680

[final saddle]

@knotty elk Has your question been resolved?

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[prisma crag]

what is other than regular prism?

[mint orbit]

you mean like, what other shapes?

[prisma crag]

regular...

word

[mint orbit]

Well you say what is other

what do you mean other than?

[prisma crag]

like sloping..

[mint orbit]

its really not clear what youre asking

[prisma crag]

regular vs sloped

[mint orbit]

what about it?

[tepid ridge]

So, you have to move the number to the right then minus it and x3 radical

[prisma crag]

someone said the definition of regular prism is The base is a regular polygon and the lateral edges are perpendicular to the bases.

what is regular polygon

[tepid ridge]

First all of all what’s the equation of the parabola and line

[mint orbit]

@tepid ridge can you stop responding to random people in here please

[mint orbit]

its exactly what you just defined it as

[prisma crag]

what is regular polygon???

[mint orbit]

oh, regular polygon just means all the sides and angles are the same

so triangle, square

pentagon

hexagon

[prisma crag]

but trapeze isn't...

[mint orbit]

it can or it can not be

if its a square its regular

[prisma crag]

so if its not regular polygon then its sloped

[mint orbit]

no

you could have it be all parallel

what makes it regular is that the polygon base is a regular polygon

like you can have this

no slopes anywhere, all the edges coming out of the trapezoid are parallel

but its not a regular prism

[prisma crag]

which is sloped again..

[mint orbit]

i assumed you meant like, twisted

[prisma crag]

can you explain this? lateral edges are perpendicular to the bases.

like show on the image

makr

[mint orbit]

thats what they are here

they extend out at 90 degree angle with everything

compare to like this thing

the base is a square or rectangle

but the lines extending dont make 90 degree angle

so even though the base might be a regular polygon, the shape isnt a regular prism

we could make the lateral edges extend perpendicularly by changing the base to be the side which is a parallelogram

but then, the base isnt a regular polygon

does that help?

[prisma crag]

ya

but what if the lateral edges were in parallel to the bases

[mint orbit]

if the lateral edges are parallel to the bases, and the bottom and top are regular polygons, then its a regular prism

[prisma crag]

no...

The base is a regular polygon and the lateral edges are perpendicular to the bases.

[mint orbit]

ah, sorry, got em mixed up

well the lateral edges cant be parallel to the bases

otherwise they wouldnt be lateral edges

and we wouldnt have a prism

at all

[prisma crag]

so the green one isn't regular?

[mint orbit]

its like so hard to tell from these drawings

is the base a pentagon?

[prisma crag]

nvm

but where would the lateral edges be parallel

in what prism

[mint orbit]

it doesnt make a prism

itd be 2 dimensional

[prisma crag]

like can you draw?

[mint orbit]

it wont make sense

the shape in entirely contained in the plane, theres no perspective here

[prisma crag]

what is this 💀

[mint orbit]

i told you it wont make sense lol

[final saddle]

@prisma crag Has your question been resolved?

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[minor sandal]

Q5 part c please

[barren hound]

,rcw

[minor sandal]

I was getting -1 and 1

In solution manual: infimum doesn't exist

[barren hound]

how did you get -1 and 1

[minor sandal]

x²<1

[barren hound]

remember when you multiply by a negative number you flip the < to a >

[minor sandal]

Thanks

.close

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[hexed forge]

You also from India

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[torn dome]

For part 1 I got x^2 < 1 and x^2 > -1

How do I find the range of x from here?

I mean i cant take the root in x^2 > -1

[onyx peak]

When is x^2 > -1?

for what x?

What even are the possible values of x^2?

[silver dew]

-1<-x^2<1

[torn dome]

all values ig

0,1,2,3..

[onyx peak]

and x^2 < 1?

[torn dome]

so can I say x > -1

[onyx peak]

No

x^2 > -1 is always satisfied

it works for all real numbers

[torn dome]

(0,1)

[onyx peak]

it's practically an irelevant condition

[onyx peak]

x can be real

it can be negative too

[torn dome]

(-1,1)?

So whhat does that prove

[onyx peak]

The conditions x^2 < 1 and x^2 > -1
are equivalent to
x in (-1, 1)

[silver dew]

It’s just that Σt^n converges iff -1<t<1, you just plug in different t for your questions

[onyx peak]

the conditions work only for x in (-1, 1) (and all x in (-1, 1) satisfy the conditions)