help-23

[safe radish]

@umbral folio Has your question been resolved?

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[lean otter]

How do i find the Rule in function notation for this one?

[lean otter]

What do you mean by Rule

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[pine wren]

does anyone know of a calculator for if a series diverges or converges with steps?

[safe radish]

@pine wren Has your question been resolved?

[soft matrix]

Just use the tests
Btw yeah u can use symbolab

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[ebon seal]

How to do this?

Er mentions something about finding normal vectors but i don’t get how that gets us the ans

And by line l, do they mean the line that always passes through both the planes?

[safe radish]

@ebon seal Has your question been resolved?

[quiet plume]

The idea is that the normal vector $\vec{n_1}$ of plane $\Pi_1$ is orthogonal to $\Pi_1$, and similarly the normal vector $\vec{n_2}$ of plane $\Pi_2$ is orthogonal to $\Pi_2$.\
If the line is contained within \emph{both} planes, then it has to be orthogonal to \emph{both} normal vectors. \
So if you find a vector orthogonal to $\vec{n_1}$ and $\vec{n_2}$, then you have the direction vector of the line.

[flat frigate]

Azyrashacorki

[ebon seal]

Man what is orthogonal

[quiet plume]

Perpendicular

[ebon seal]

I don’t understand ur 2nd line

Why does that line have to be perpendicular to both normal vectors?

And first of all

[ebon seal]

This is correct?

[quiet plume]

Any line in P_1 has to be perpendicular to the normal vector of P_1. Does that make sense to you?

[ebon seal]

@quiet plume is this correct?

[quiet plume]

The planes intersect at a line, I don't know what you mean by "always passes through both planes."

It does pass through both planes, since it's the intersection

[ebon seal]

I don’t understand what line they intersect at

I can’t visualize it

Do u know like a good 3d online tool so i can visualise it?

[ebon seal]

This is correct

[quiet plume]

There are three cases for the intersection of two planes.
A bit like how lines in 2D either cross at one point, are parallel, or are coincident(the same).

The planes might intersect at a line, be parallel and never touch, or be totally coincident.
See this Desmos graph with the three cases to get a visual representation of those cases.

[quiet plume]

If you agree with this, then any line in P_2 is perpendicular to the normal vector of P_2.

[ebon seal]

Ok wait

[ebon seal]

Obviously, yes

[quiet plume]

But if they intersect at a line, then this line must be part of both planes

So in particular, it has to be perpendicular to both normal vectors simultaneously

[ebon seal]

@quiet plume

Do u see the green line i made?

[quiet plume]

Yes

[ebon seal]

Is that the line l?

[quiet plume]

Yes

[ebon seal]

Ok

Now I understand

[quiet plume]

Those are arbitrary graphs though they don't have anything to do with the planes in your problem

[ebon seal]

Ye I understand

[quiet plume]

So are you good with finding the normal vectors for the plane and then finding a vector normal to both of those?

[ebon seal]

How do i find the normal vector for the planes tho?

I don’t have a point on the planes

[ebon seal]

Yes, correct

So dot product=0

[quiet plume]

Those are direction vectors for the planes

You can compute the cross product of the red vectors to get the normal vector.

[ebon seal]

But

Does the plane not have infinite direction vectors?

How does a plane have only 2 direction vectors?

I understand we can pick any 2 direction vectors

[quiet plume]

The plane is spanned by 2 linearly independent vectors.

[ebon seal]

And do cross product

[ebon seal]

What does this mean?

[quiet plume]

It means that yes, you can go in infinitely many directions on either plane, but the point is you can start from the point (2,4,6) and add linear combinations of the red vector to generate all the points on the plane P_1.

Just like you can go in infinitely many directions on the 2D plane, and yet you can write any point as a linear combination of the basis vectors (1,0) and (0,1).

[ebon seal]

What does ‘add linear combinations of the red vector’ mean?

[quiet plume]

Adding multiples of either

Like $a(2,0,3) + b(0,-4,5)$

[flat frigate]

Azyrashacorki

[ebon seal]

Ohk

So basically

What ur saying is

U can input any value into theta1 and phi1

Like infinitely many values

And u’ll have all the infinite possible direction vectors

Of the plane

Correct?

[quiet plume]

Yep

[ebon seal]

Got it

Ok lemme try now

[ebon seal]

Ok so i got the normal vectors of both planes now

[quiet plume]

Great! Now compute the cross product of those to get the direction vector of the line

[ebon seal]

Got it

Line l will have same position vector?

[quiet plume]

Well it will have to go through the point (2,4,6) since that point is in both planes

And then that direction vector you just found

[ebon seal]

Yes

[ebon seal]

Yes

Fuck im getting the direction vector wrong tho

What are you getting?

Lemme try mine again

Oh nvm i got my mistake

Bruv i forgot how to find determinant

How do i compute this again?

I mean I remember we eliminate row and column

But i forgot abt the alternating signs thing

[quiet plume]

First entry of the first row. You multiply that entry, i, by the determinant of the matrix you get by eliminating the row and column that go through i.

[ebon seal]

Yeah yeah i got it

Alr bruv i think i got this now

Sorry for wasting so much of ur time

And thank you very much

[quiet plume]

It's alright it's not like I was stuck to the screen, and even then it wasn't that long

Good luck in your endeavors

[ebon seal]

Thanks brudda

.close

[safe radish]

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[heavy stirrup]

hi i need help with an algebra equation ; x² + 2x +5 =0
and you have to use the quadratic formula for this. I've done it but i dont think its the right answer im supposed to get

[tardy mango]

!show

[safe radish]

Show your work, and if possible, explain where you are stuck.

[tardy mango]

,rotate

[tardy mango]

,rotate 180

[heavy stirrup]

ohh

-2 ± 4(?) sorry i didnt understand

[tardy mango]

4i

[lean otter]

the square root of -16 is not -4

if you squared -4 you would get 16

[heavy stirrup]

another question: how do you do the checking for the answer? to see if its correct and all

[glacial cairn]

You put it back into the expression (x² + 2x +5) and check that it equals what it should equal (0)

[safe radish]

@heavy stirrup Has your question been resolved?

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[umbral solar]

Guys I can't seem to get this right been trying for over a hour

Its finding determinant of that matrix

i get 279 but correct is 99

.close

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[ebon junco]

I am stuck on this question

[ebon junco]

It seems very pigeonholey, because of the 2^n + 1, but i couldnt make any meaningful progress

[arctic raven]

it sounds like you can draw a graph with colored edges, whose vertices are the N cities and whose edges are colored via one of n colors depending on which airlines offer service between the two cities on that edge

haven't seen anything to do with that yet

[ebon junco]

yep

[arctic raven]

I was thinking that maybe there has to be a 3-cycle all in one color as well, but the wording makes it sound like it could be ever as small as 5

[ebon junco]

wdym

as small as 5?

[arctic raven]

well like maybe the smallest odd cycle in a single color is of length 5 instead of 3

[ebon junco]

oh

i was thinking about proving that one of the airline's graphs cannot be bipartite

then it would have to have an odd cycle

[arctic raven]

that could be good

I was just thinking you could probably show one of them at least has a cycle of any kind

[ebon junco]

even if it's an even cycle?

[arctic raven]

it'd be a start, I guess

[ebon junco]

i dont know how to use the 2^n + 1 information

[arctic raven]

sorry I still haven't worked it out myself. do we know the number of cycles of odd length in K_{2^n + 1}?

[ebon junco]

what is K?

[arctic raven]

the complete graph of order 2^n + 1

so a graph with 2^n+1 vertices and all of its edges

[ebon junco]

oh

N C 3 + N C 5 + ... ?

[arctic raven]

I think it's more than that, because the order in which you choose 5 vertices to make a cycle matters

sometimes

[ebon junco]

true

[safe radish]

@ebon junco Has your question been resolved?

[ebon junco]

you multiply each N C K by (K-1)!/2?

[arctic raven]

yeah

I haven't found much joy from that yet

ok I'm starting to have vaguely sensible thoughts

if you do end up having no odd cycles in any of your airlines, then just as you suggested, you can make bipartite representations for your graph, with edges restricted to only one airline

so imagine airline A_i induces a bipartite representation for your graph H_i

we can find the (maximum) number of edges that each H_i can have, and then the complete airline graph will have at least as many edges as the union of all the edges of the H_i's

surely that number must be too small, or maybe too large lol

and by surely I really mean hopefully

[ebon junco]

if you color every city red or blue, for each airline

for the bipartite thing

then there will be 2^n possible states

so 2 cities must have same coloring on every graph

but that means 1 graph cannot be bipartite

since there are 2^n + 1 cities?

[arctic raven]

I'm worried because the colorings don't necessarily interact with one another here, right?

[ebon junco]

thats what im thinking about rn

doesnt it still work?

because if they are WLOG, both blue, then they cannot be connected by that specific airline

and if they are always the same color, they cannot be connected by any of the airlines

[arctic raven]

ok I've finally caught up and I believe you

the red-blue imagery messed with me a bit because A1's red is different from A2's red

but I see now that there's at least one city that matches another city's parity in every airline

[ebon junco]

yes, the same pair of cities is always the same color for every airline

[arctic raven]

yes

that's very good

[ebon junco]

my head still hurts from tihnking about it

[arctic raven]

so every airline A_i induces partite sets V_(1,i) and V_(2,i), so that every city's parity within all the airlines is encoded via some tuple over {1,2}^n

there are 2^n such tuples that each city can be given, but with 2^n+1 cities, there must be a pair of two cities whose parities are encoded via the same tuple by the pigeons

[ebon junco]

yes

thats a better way of putting it

[arctic raven]

these two cities do not share an edge because they lie in the same partite sets in all bipartite graphs induced by each of the airlines, but this contradicts the graph being complete

well done

[ebon junco]

thanks for the help!

[arctic raven]

you did the hard part, very nice job lol

[arctic raven]

by the way, in case you're curious, this was wrong and did not help at all lol

the total count of edges in each edge-induced subgraph was not generally less than the number of edges in the complete graph

[ebon junco]

did you do something like N^2/4 * n?

[arctic raven]

I stuck with N = 2^n+1 and it allowed at most 2^(n-1) * [2^(n-1) + 1] edges in each subgraph

and n times that is generally larger than the total number of edges, being 2^(n-1) * [2^n + 1]

[ebon junco]

i also tried that earlier

[arctic raven]

hey well thanks for sharing that problem and baring with me, even though I didn't know the answer or solution path

bearing with me? 🤷

bearing with me

[ebon junco]

dw, i had spent 3 hours getting nowhere before asking here

it was good to have someone to verify with

the problem was from "The art and craft of problem solving"

[arctic raven]

nice, thanks for sharing

[safe radish]

@ebon junco Has your question been resolved?

[safe radish]

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[west hedge]

Can someone tell me the where the vertex of the graph lands?

Then I can throw you something that you would never imagine

[river oriole]

i can see its crazy from the graph

why the hell isnt it on the y axis

[west hedge]

Just tell me the vertex based on the function given

[river oriole]

(0,-5)

[west hedge]

What

How

[river oriole]

what

2x^2-5

[west hedge]

I made a mistake shit

[river oriole]

lmao huh

[west hedge]

I thought if I differentiate it, then I will get 4x - 5

Then the x coordinates of the vertex will be 5/4

[river oriole]

i mean

you executed it wrong

5 is a constant

its just 0

[west hedge]

Yea, I’m fked up

[river oriole]

it happens dw

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[slim urchin]

Hi guys I really need help with this statistical inference question

[slim urchin]

my attempt is the first pic, the second pic is the question

I can't figure out whether the red circled formula (inclusion exlusion) is relevant (my attempt at a solution is the 2nd image).
I thought divisions 2 and 3 might have an overlapping bit with the 5 winning numbers, and same for divisions 5 and 6 for the 2 supplementary number bit, but does this mean I only subtract those relevant overlaps and treat everything else the same? or just follow the formula as written there?

<@&286206848099549185>

[safe radish]

@slim urchin Has your question been resolved?

[safe radish]

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[dull sequoia]

so i found this online

[dull sequoia]

https://ocw.mit.edu/courses/18-443-statistics-for-applications-fall-2006/03b407da8a94b3fe22d987453807ca46_lecture3.pdf

but my uni gives the example of

Let ${X_i} \sim \exp$ with mean $1/\theta$, then $\mathcal I(\theta) = n/\theta^2$

[flat frigate]

frosst

[dull sequoia]

then it says this

but shouldn't it be $\sqrt{n}(\hat\theta - \theta) \to N(0, \theta^2/n)$

[flat frigate]

frosst

[dull sequoia]

and then if you want the variance to become 1 you divide by the sqrt of the current variance

so it'll be $\frac{\sqrt{n}(\hat\theta - \theta)}{\frac{\theta}{\sqrt{n}}} = \frac{n(\hat\theta - \theta)}{\theta} \to N(0, 1)$

[flat frigate]

frosst

[fast badge]

I(theta) is a quantity of the distribution not of the number of samples. It shouldn't depend on n I believe

[safe radish]

@dull sequoia Has your question been resolved?

[safe radish]

@dull sequoia Has your question been resolved?

[safe radish]

@dull sequoia Has your question been resolved?

[safe radish]

@dull sequoia Has your question been resolved?

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[torpid abyss]

yes

$3/4 = 9 / 12$

[flat frigate]

Nguyễn Tuấn Minh

[torpid abyss]

9 > 5

so??

$1 - 5/13 = 8/13$

[flat frigate]

Nguyễn Tuấn Minh

[torpid abyss]

$1 - 3/4 = 1/4$

[flat frigate]

Nguyễn Tuấn Minh

[torpid abyss]

$1/4 = 8/32 < 8/13$

[flat frigate]

Nguyễn Tuấn Minh

[torpid abyss]

$\Rightarrow 1 - 3/4 < 1 - 5 / 13$

[flat frigate]

Nguyễn Tuấn Minh

[torpid abyss]

so??

$\Rightarrow -3/4 < -5/13 \Rightarrow 3/4 > 5/13$

[flat frigate]

Nguyễn Tuấn Minh

[torpid abyss]

ok??

@sullen crescent

[safe radish]

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[upper anchor]

Hi, could someone help with this text problem please? I have translated it so if there is something that doesnt make sense pls let me know 🙂

A linesegment runs through point (1,2) and has it's 2 endpoints on the coordinate axes. Determine shortest possible length for the line segment

[cunning pasture]

x+y = 3?

Maybe

[upper anchor]

How did you approach it?

[waxen inlet]

you can write a line as y = ax + b, since the line goes through (1, 2), you can apply that to the formula and get a + b = 2
using this piece of info, you can rewrite the line as y = ax + 2 - a,
its 2 endpoints are on the coordinate axis
so we know its 2 endpoints are (0, 2 - a) and (a-2/a, 0), use the distance formula to write a function that calculates the distance by different values of a, calculate its derivative and solve f' = 0

[upper anchor]

thank you, I will try!

[cunning pasture]

actualyl x+y is not 3

there's a shorter lenght

nvm this

[safe radish]

@upper anchor Has your question been resolved?

[cunning pasture]

this is the shortest lenght

[safe radish]

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[steep magnet]

cos(2t) = 2sin(t)cos(2t)

2sin(t) = 1 and cos(2t) = 0

sin(t) = 1/2 and cos(2t) = 0

t = arcsin(1/2) and 2t = arccos(0)

t = pi/6, 5pi/6 and 2t = pi/2, 3pi/2

t = pi/6, 5pi/6, pi/4, 3pi/4

Im not sure what to do next though

[bold sandal]

Why exactly would cos2t = 0?

Also when looking at the figure it seems to not be right for there to be 4 intersection points

Only 2sint = 1 looks to be the equation for t you are looking for which then just gives you 2 values for t

[steep magnet]

If you have (x+1)(x+2)=(x+1), your solutions are x+1 = 0 and x+2 = 1, right?

[bold sandal]

No

You divide both sides by x+1 in this case

Which leaves you with x+2 = 1

[steep magnet]

And the x+1 is also a solution, no?

[bold sandal]

What do you mean?

[steep magnet]

Because it could also be a solution for x... idk this is just what I learned

Am I thinking of something else?

[bold sandal]

x+1 = 0 would be a solution because x+2= 1 is just the same

[steep magnet]

Ok I picked a bad example then

[bold sandal]

Yes

But generally

When we have

$a \cdot b = a$ we divide both sides by a
$a \cdot b / a = a/a$

[flat frigate]

thijs2725

[bold sandal]

Which then gives us b = 1

Only notable important thing to remember here is we assume a does not equal 0 here

[maiden ledge]

solution is x

[bold sandal]

Because we cannot divide by 0

[maiden ledge]

= -1

[bold sandal]

That was a example xd

[bold sandal]

This was the question

[maiden ledge]

rhe?

[bold sandal]

@steep magnet are you still with us?

[steep magnet]

Yes im just reading

I am trying to understand

[maiden ledge]

i need help

[steep magnet]

#❓how-to-get-help

[maiden ledge]

with dirichlet's approximation

[bold sandal]

Open another help channel pls

[maiden ledge]

that would be tiresome

[steep magnet]

Ok... so we just have t = pi/6 and 5pi/6

[bold sandal]

Yes

[maiden ledge]

to learn to openchannel

[bold sandal]

Now just plug those values for t back into your equations for x and y

[steep magnet]

I got (1/2, 1/2)

for both

Is that a point on the graph

[bold sandal]

Could be

[steep magnet]

We dont know for certain?

[bold sandal]

Based on intuition it could be true

But you would need the answer sheet to know for certain right haha

But I would think it s correct

[steep magnet]

Your name is dutch haha do you speak dutch

Cause the answer sheet is in dutch 😅

[bold sandal]

Ja

[steep magnet]

I dont really understand it...

[bold sandal]

Weet je zeker dat dit dezelfde opgave is haha

[steep magnet]

Mijn nederlands is niet goed 😝

[bold sandal]

It doesnt seem like the same exercise haha

[steep magnet]

Oh oops!!!

[bold sandal]

This looks like something about triangles

[steep magnet]

That makes sense why I coudlnt understand it now LOL

So the only point is (1/2, 1/,2), which is our answer, right?

[bold sandal]

Yes

[steep magnet]

Oh wait

Cos(2t) = 0 is a solution

[bold sandal]

But the answer sheet also says cos2t = 0…

Oh nvm my bad sry sry sry

[steep magnet]

Yeye that makes sense cause of course y=x goes through (0,0)

[steep magnet]

Npnp haha

[bold sandal]

You were right in the first place

😅😅😅😅

[steep magnet]

I understand now tho

Thank you so much for your help!

❤️

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[steep magnet]

Fijne dag jonge 😝

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[cunning pasture]

ding dong

[cunning pasture]

=)

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[thorny dock]

WHat is meant by the initial profile of a PDE?>

[thorny dock]

ie

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@thorny dock Has your question been resolved?

[mossy lotus]

I think it just means the initial condition or the solution at t=0

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[cursive sandal]

I don't understand how it did partial fraction to it

[manic path]

,rotate

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[rapid summit]

can anyone tell me a way of doing this with permutation isntead of picking a number using combination and then arranging it using factorial?

[rapid summit]

if there isnt a way pls explain why u cant do it iwht permutation

[solar hazel]

isntead of picking a number using combination and then arranging it using factorial
that's... what a permutation is

(n choose k)*k! = n permute k

[rapid summit]

yes ive tried solving it with permutation but i kept getting the wrong answer

i tried all the posibilities here

idk if im doing it wrongly

i did (26p2 x 9p1 x 5p1) + (26p1 x 9p1 x....

[solar hazel]

that doesn't make sense to me

like why would that count everything

[rapid summit]

oh, could u pls explain why?

[solar hazel]

26p2 x 9p1 x 5p1 is like
ways to pick 2 capital letters for the first 2 characters (where order matters), then a digit for the 3rd character, then a symbol for the 4th character

it does not count passwords of form, e,g, symbol letter letter digit

it's much easier to deal with this by choosing from each character type first then permuting

is there any reason you don't want that?

[rapid summit]

alr i understand now tysm

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[stark yew]

Is this how u do it

[stark yew]

Question one

Domain I got x<3

[hearty egret]

how could you get x<3?

[stark yew]

Wait I messed up

Is it not this

[lean otter]

whyd u changr f(x) bruh?

[stark yew]

To its inverse ?

[lean otter]

no bro

read the question again

[stark yew]

Ok

I found the inverse of the function

[lean otter]

bro ur coeff is -1/2

[stark yew]

Should it be this

[lean otter]

ye now its good

uh

[stark yew]

??

[lean otter]

2 sides

so there should b more than 1 answer

[stark yew]

Fhe first equation here is inverse

I’m pretty sure it’s just one

So you don’t elimate the fraction right

Or do u because I did that

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[stark yew]

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[final schooner]

help

[final schooner]

Question is, why cant this be a basis for P^2

[peak estuary]

whats the dim of P^2

[final schooner]

should be 3 if i watch tho the combination

[peak estuary]

?

can you give some basis of P^2

[final schooner]

?

dont understand it

[peak estuary]

what is a basis

[final schooner]

a "combination" of vector dont know how we say it in english

[peak estuary]

no

its a linearly independent spanning set of vectors

do you know what these words mean

[final schooner]

yes

[peak estuary]

then can you give one example of this

[peak estuary]

can you give an easy example in the case of P^2

[final schooner]

two matrices instead of 4 than

[peak estuary]

no

bases always have the same number of elements

[final schooner]

yeah if its was matrices of 1*2 it should be good

[peak estuary]

ok sure if you also change the space, yes

so I asked about P^2. that space has no matrices

so a basis also cant contain matrices

[final schooner]

so whats the point for my question ?

[peak estuary]

well I am trying to build up some ground from which we can then approach the problem

[final schooner]

ok so whats next

or can you just explain me how they came to that basis

[peak estuary]

well its not a basis

and I am trying to lead you there

[final schooner]

yeah i understand that its not a basis cuz of not the sale dimension but they asked to found a basis if this one wasnt one

so my question is how do i find a basisi for p2

ow wait

[peak estuary]

it doesnt help that {x, 1-x^2} isnt a basis either

[final schooner]

bruh

you telling me my professor is wrong

translated question: Why can't the following set be a basis for P2? Is this collection generating P2? If not, find a basis and the dimension of the subspace produced by this set.

@peak estuary

<@&286206848099549185>

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[elder grotto]

Help

[cerulean crane]

What do you need help with?

[elder grotto]

I was working on this question and I wanted to know why I can't simplify it further. Let me work it out

[rough storm]

!original

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[elder grotto]

$\frac{3ab+4bc}{2ab}$

[flat frigate]

UndercoverWasabi

[elder grotto]

Then I worked it out

$\frac{b(3a+4x)}{2ab}$

[flat frigate]

UndercoverWasabi

[elder grotto]

$\frac{3a+4c}{2a}$

[flat frigate]

UndercoverWasabi

[elder grotto]

Why can’t I just remove (don’t know how to call it in English) the a’s here?

[median vigil]

because the 4c does not have a factor of a. so you would have to first split it off

[elder grotto]

Ah thanks

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[simple gazelle]

i am trying to prove the area of an ellipse is pi * ab, why is it negative??? has i ever???

[tardy mango]

https://math.stackexchange.com/questions/3639024/negative-sign-in-calculating-area-of-an-ellipse-using-parametric-equations

[simple gazelle]

oh my someone has the same exact issue

thank you

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[magic heath]

Hi. I'm struggling with a Lambda Calculus question so I'm hoping I can get some help.
I need to reduce (λz.λu.zu)u(zλw.ww) and I have done so and reached u(zλw.ww) as a normal form, however, I checked interpreters online https://lambdacalc.io/ and https://lambster.dev/ and they reduce to (zu), and I don't understand the steps required to reach that normal form. Any help would be so appreciated, thanks

[safe radish]

@magic heath Has your question been resolved?

[magic heath]

no <@&286206848099549185>

[wheat comet]

unfortunately non there

[polar pulsar]

Could you link the question

?

[magic heath]

Reduce (λz.λu.zu) u (zλw.ww) to normal form

[polar pulsar]

oh

Lambda calc is something I wasnt taught yet

sorry ig

[magic heath]

no worries 🙂 thanks anyways

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[hardy lion]

what have you tried

[snow robin]

put it in between $$

[hardy lion]

add dollar signs to the left and right

[flat frigate]

Lizzy

[hardy lion]

thats a good start

so you want to match coefficients

you have 0n+4 on the left side

and (a+b)n+2a on the right

yup

set a+b=0

you know a

as n varies, you want to make sure both sides are equal

since you have 0n on the left side, the only way for the right side to match is it also have a coefficient of 0 for n

the nth term will cancel out with the (n+2)th term

honestly i think its best to write it out

that way you dont accidentally miss something

if you have a/n and -a(n+c), then the nth term will cancel out with the (n+c)th term

or at least the half of it where they match

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[hardy lion]

youre welcome

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[gritty umbra]

i need help with absolute values in inequalities. I have three items and I have tried answering them. I wanted to check here if I did it correctly

[gritty umbra]

|x + 9| >= -6

= means is greater than/equal to idk how to write it properly

i came up with (-15,-3)

[rough storm]

are you sure its -6?

[gritty umbra]

yes

i mean thats what it said on the book

[rough storm]

!original

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[rough storm]

what is the definition of absolute value?

[gritty umbra]

here are the original problems, we were asked to answer them in interval notations

sorry for the bad quality

[rough storm]

alright, then this is wrong

[rough storm]

tell me this

[gritty umbra]

the positive value of something

[rough storm]

yes

so ANYTHING inside absolute value is made positive (well nonnegative in the case of 0)

so when is |x+9| greater than or equal to -6

[gritty umbra]

it became +6 after i put it in the format thing

oh nvm

[rough storm]

huh

[gritty umbra]

so what do i do with it?

[rough storm]

do you agree by our definition of absolute value that |x|>=0 is always true

[gritty umbra]

yes

[rough storm]

there's nothing special about x here. x is a variable, so it makes sense that |x+3|>=0 for all x, |2x+3|>=0 for all x, so on

and if something is greater than 0 is it greater than -6?

[gritty umbra]

no

yes

[rough storm]

yes!

so what x works for |x+9|>=-6

[gritty umbra]

-3

[rough storm]

are you familiar with how to write all real numbers in interval notation

[gritty umbra]

no

[rough storm]

wait, is it 6 or -6

[gritty umbra]

dont mind what I said there, i was talking about the |x| < k = -k < x < k thing so since -6 is my k, when i put it in the -k it became positive

[rough storm]

all real numbers can be written in interval notation as

$(-\infty, \infty)$

[flat frigate]

🫎 Moosey 🫎

[gritty umbra]

hmm

[sour vale]

Guys I need help

[gritty umbra]

sorry, but you need to occupy another channel 😆

[sour vale]

Sorry I’m new

[gritty umbra]

#help-35 is free

its alright

[gritty umbra]

what's next

[rough storm]

thats it

for 15

since |x+9| is always greater than 0, it must always be greater than -6, so it holds for all values of x

[gritty umbra]

uhh how do i write the final answer

[gritty umbra]

with this

[rough storm]

that's your final answer for 15

[gritty umbra]

infinity and infinity??

[rough storm]

(-3,3) is read as 'the interval from -3 to 3, not including -3 and 3

so (-infinity, infinity) is read as the interval going from negative infinity to infinity. This is exactly all real numbers

[gritty umbra]

so whatever real number you put on x it will always be >= -6?

[rough storm]

|x+9| will always be greater than -6

whatever number you put in for x

[gritty umbra]

is the solution just like this then?

click for full image

[rough storm]

okay this is the issue

[gritty umbra]

erm, is it okay for you to elaborate? i want to learn

[rough storm]

we need to think more clearly about the piecewise definition of absolute value of x

does the word piecewise mean anything to you

basically, |x| is defined as x, for x>=0, and -x for x<0

this means that |x+9| is x+9 for x+9>=0 and -x-9 for x+9<=0

[gritty umbra]

i dont have a strong grasp about this, sorry

[rough storm]

okay

let's think about it like this then

when does the function inside become zero? Like what x makes x+9=0

[gritty umbra]

-9

[rough storm]

so we know that below -9, our function will look like -x-9

and above -9, it will be x+9

[gritty umbra]

hmm

[rough storm]

https://math.libretexts.org/Bookshelves/Algebra/Advanced_Algebra/02%3A_Graphing_Functions_and_Inequalities/206%3A_Solving_Absolute_Value_Equations_and_Inequalities

[gritty umbra]

what now

[rough storm]

this might help more

[gritty umbra]

okay

[rough storm]

maybe it makes more sense if i rearrange it to be |x+9|+6>=0

you see that |x+9| is always greater than or equal to 0, and adding 6 to that won't change that, so it will always be true for any x

[gritty umbra]

i just watched the organic chem tutor's vid and understood it in less than 2 mins 😭😭

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[median shore]

I just need help with these 2 problems I js forgot how to do it

[torpid fable]

What u did so far is good all you need to do now is solve for x

[rancid sand]

Use trigo (lol)

[median shore]

Yeah I forgot how to do that 😭

Multiply x?

[torpid fable]

Yeah

Then divide sin48/tan20

[median shore]

Wait what

This is what I got

[median shore]

^

[torpid fable]

Is that 48 suppose to look like an exponent or did u forget and tried to squeeze it in

[median shore]

I tried to squeeze it in

[torpid fable]

Yeah so now just divide both sides by sin48

[median shore]

How?

[torpid fable]

Calculator for rhs, for lhs sin48/sin48=1

Rhs means right side

[median shore]

Okay thanks I got it

Idk how to end it but you can end it

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[west hedge]

I’m fked

[west hedge]

Guys

Help me

The solution draws a triangle

Why

Please

Someone elaborate

[gleaming quiver]

-12 is right

[west hedge]

Yes Steve

[gleaming quiver]

what do u need?

[west hedge]

I don’t know how to solve it

I’m so ashamed

[severe pond]

tragic

perhaps you should read their solution

[devout shale]

That triangle kinda remind me of Mickey

[severe pond]

word

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[jovial sentinel]

1 = cosx + sinx
what should I do from here?

[lean otter]

divide by sqrt2

[jovial sentinel]

im confused on why I would do that

[lean otter]

to make it smth like sin(x+a)

[jovial sentinel]

which trig identity is that

[lean otter]

addition

of angles

[jovial sentinel]

im not sure what you mean

oh

this?

[lean otter]

yes

[jovial sentinel]

still confused on how i get from sinx + cosx = 1 to sin(x+a)

[lean otter]

did you divide

[jovial sentinel]

well i dont know why we divide

.close

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[sudden lagoon]

Can this answer be simplified?

I think you can rationalize the denominator but idk if it makes sense to do that. And idk how i would do that

[empty gyro]

You could do $\sqrt{\frac{e^x}{\pi}}$ but that's about it

[flat frigate]

SWR

[eternal carbon]

what swr said is good

also your answer is very slightly wrong

review the steps you took to get your answer

[sudden lagoon]

i dont see how? I just substituted my u and a into the formula

[severe pond]

a should be a constant in that formula

[sudden lagoon]

so then sqrt((pi)(e^x)) cant be a?

[eternal carbon]

correct

[severe pond]

yea a is a constant

[eternal carbon]

perform a u-substitution first to make the integral easier to work with

[severe pond]

you can factor an e^x in the denominator to cancel a square root e^x

or just do u sub

because the e^x in the numerator will cancel

[sudden lagoon]

can you elaborate on this? idk what exactly youre recommending

[severe pond]

u=e^x

du=e^x dx

which is precisely the numerator

but don’t use your trig sub formula

it doesn’t work

a is supposed to be a constant

the denominator will be piu-u^2

then from there it should be relatively easy

[sudden lagoon]

im sorry, but i dont completely understand this

because piu isnt a constant either

i dont get how to complete the u sub

[severe pond]

well that is the u sub

@sudden lagoon

you have to integrate 1/sqrt(piu-u^2)

just complete the square in the denominator

inside the square root

[safe radish]

@sudden lagoon Has your question been resolved?

[sudden lagoon]

i got 2arcsin(sqrt(u)) / (sqrt(pi))

[lean otter]

@severe pond please help this guy

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@sudden lagoon Has your question been resolved?

[safe radish]

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[sinful cliff]

Hello, I worked out this summation to be the answer to x^n - y^n when x >= y. And I see that it also works for when y >= x. But is there a way to prove it?

[soft matrix]

Use binomial theorem

And it doesn't seem to be xⁿ - yⁿ

[sinful cliff]

It does seem to be in my tests tho

Wanna try?

[languid siren]

I think it is

You’re skipping the i = 0 term

[soft matrix]

? What do u think it is

[sinful cliff]

Yes, intentionally so

[languid siren]

x^n - y^n

[soft matrix]

Ahhh fck right

Kk it's indeed xⁿ - yⁿ

[sinful cliff]

How did you know?

[languid siren]

.

[soft matrix]

Separate the yⁿ term and then use binomial theorem

[sinful cliff]

Well yeah, then it becomes (y + (x - y))^n - x^n which is x^n - y^n

[soft matrix]

I'm wondering if we can prove this using strong induction

[sinful cliff]

I wondered if you used a different method

[soft matrix]

Try induction

[sinful cliff]

Well I derived it from using n-dimensional cubes

So it's kinda already proved. Twice, if we consider the binomial theorem too

[soft matrix]

Uh?

[sinful cliff]

I'm not interested in getting acquainted with strong induction at this time

[soft matrix]

ok
Well then you're done with it

[sinful cliff]

I can explain how I used n-dimensional cubes if you want

[soft matrix]

Go ahead

I'm listening

Or rather reading

[sinful cliff]

I first assumed x to be >y
I divided space inside the x^n n-cube on each dimension into 2 intervals: those belonging to the smaller cube and those outside of it. So in each dimensions I divided the big cube into [y] and [x-y].
Considering that each space has n dimensions, there are 2^n different places inside the bigger n-cube that we can sum up and get the bigger n-cube. But I decided to group them into those with the same coefficients. It's done by using (n choose i). Need I explain why? And then it's multiplied by the appropriate dimensions and then y^n is subtracted. Done

What do you think

[soft matrix]

I don't understand any of that shit considering I'm the dumbest person in geometry

[sinful cliff]

🥲

[soft matrix]

What's the smaller cube here?

[sinful cliff]

y^n

since x is assumed to be >y

[soft matrix]

Ok

Why are there 2ⁿ spaces inside the bigger cube (by the bigger cube u mean xⁿ ryt?)

[sinful cliff]

yes, x^n

Because on each dimension those spaces can occupy either the [y] zone or the [x-y] zone

So there are 2 of them

Hence 2^n variations

[soft matrix]

Ok so your proof is based on that these are all the different points inside the bigger cube but outside the smaller cube

[sinful cliff]

Not points, but rather subspaces but yes

[soft matrix]

Yea yea same thing

[sinful cliff]

And then I just sum them up

Without the subspace which is [y] on all dimensions

[soft matrix]

How exactly did u select the spaces outside yⁿ

[sinful cliff]

Wdym select

[soft matrix]

Like how did u come to the fact that those(the original question)are the spaces out side yⁿ

[sinful cliff]

Because x^n - y^n is just the big cube minus the small cube

So all the space of the big cube outside the small cube

[soft matrix]

ik that
Nvm

[soft matrix]

Why does the original exp show all the spaces out side the yⁿ cube

?

[sinful cliff]

I might be misunderstanding your question, but are you asking me how I got the expression itself?

[soft matrix]

Yea how did u arrive to the fact that the no of spaces outside the smaller cube are equal to the expression

[sinful cliff]

Yea, I'm writing that up

[soft matrix]

tysm

[sinful cliff]

Argh

PC crashed and the text is lost

Gonna rewrite

[soft matrix]

[soft matrix]

U didn't copy?

[sinful cliff]

Um no

I didn't expect my PC to shut down...

[soft matrix]

looks like il have to wait another half an hour but sure go ahead I'm curious regarding your argument

[sinful cliff]

Thanks 🙂

[soft matrix]

[sinful cliff]

All the subspaces can be represented as n-elements sequences. For the sake of simplicity let's assume n=4. Then we will have {y, y, y, y}, where [y] can be replaced by [x-y]. The elements of the sequence here represent not only the length in the given dimension but also the specific occupied space, hence if even 1 element is different between sequences then they occupy different space, don't have intersections and are safe to sum up. Here I start summing up sequences(shapes) with different counts of dimensions switched to [x-y], their number equal to i. E.g. i=2 means there are 2 [x-y] elements in the sequence. The dimensions of these shapes will be (x-y)^i * y^(n-i) because, again, the elements are both the specific occupied space and the length in the given dimension. The trick to find different combinations here is to think of the elements in the sequence as of numbers and then use combinatorics. So we have n numbers and the length of the sequence is i because that's how many [x-y] we will have. We get a sequence of numbers which represent the specific occupied spaces in the initial sequence. Then we just find the number of different ones. So essentially the number of combinations also known as (n choose i). Then sum up the variations with different i. That's it

👀

[soft matrix]

WAIT AHH I GET IT NOW
So u selected the I spaces outside the yⁿ box and multiply then with their dimensions to obtain the given expression so indeed the volume of the part xⁿ - yⁿ will represent the given expression
Hmm but one thing why would we sum up all value of i to get the volume

[sinful cliff]

Because i represents different count of [x-y] in the sequence. And we want to have them all

Because if they are different (and they are, at least because they have different count of [x-y] in them), then they are an unaccounted part of the bigger cube, hence we add it to what we have

[soft matrix]

Ok i understand
Well atleast from my side the proof seems correct
And this might be the 2nd most elegant proof I've ever seen I'm my life
This proof is absolutely beautiful
Il suggest to go to Combinatorics channel to see if anyone can find any mistake or something

from my side it's w

[sinful cliff]

Actually I proved the binomial theorem in the same fashion

At least for natural values of n

[soft matrix]

Damn come to the dms i would love to see it

[sinful cliff]

It's literally the same, just with a couple tweeks

[soft matrix]

Ahh so you must have taked a cube of side x+y in the nth dimension and did the same

[sinful cliff]

Yea

[soft matrix]

nice try to look on google if a similar proof exists

[sinful cliff]

If it doesn't, then what

[soft matrix]

If not then post your proof

[soft matrix]

Write a research paper

[sinful cliff]

Wow

I am a researcher now

[soft matrix]

Ur proof deserves that respect

[sinful cliff]

How do I know if anyone has done this before?

[sinful cliff]

I tried googling but didn't see anything similar. But I might just be bad at googling

[soft matrix]

hmm ask other senior ppl on the server
Don't show them the proof just state how you have done it

Like using what u have done it

[sinful cliff]

I'm touched

[soft matrix]

[sinful cliff]

But the proof is already on that sever, can't they just look up my messages?

[soft matrix]

I think someone has done it before long long before
https://www.math.brown.edu/tbanchof/Beyond3d/chapter2/section04.html

[sinful cliff]

They've simply counted it up to the 4th degree and haven't provided a way to generalize, no?

And also they haven't done it for x^n - y^n

[soft matrix]

I'm pretty sure that if they can do it for 4 someone must have did it for n
But il suggest still to post your proove on the web or some where

[sinful cliff]

I'll try to see if there are any similar papers

And also skim through textbooks

[soft matrix]

[safe radish]

@sinful cliff Has your question been resolved?

[sinful cliff]

Thanks for your interest, praise and help. Dm me if you want. I'm going

.close

[safe radish]

Channel closed

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Use .reopen if this was a mistake.

[soft matrix]

want what?

[soft matrix]

Wlc btw

[sinful cliff]

If you want to talk to me about something

[soft matrix]

Ohh ok
Cya

[safe radish]

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[shell ingot]

wait help how did they get 6 in this

[shell ingot]

gelp

[lean otter]

bcz

if x = 3 then the functions is 0

[mossy lotus]

Its range of f

you simplified the function to f(x) = x + 3. But the domain is restricted to R - {3}. So, value of f at x=3 should not be in the range

at x = 3, x + 3 = 6

[shell ingot]

ily mwa

.close

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[hollow jewel]

hello , i am getting two unmathed results for the same problem and i cant get what i am doing wrong in the second way .well i have to find the distance between the point K(-1,2) and the line 4x+3y+12=0 , i can do it by taking the formula |Ax+By+C|/root(A^2+B^2) the result base on that is 14/5, i also thought to solve this like this :i am taking another point of the line the E(0,-4) and finding KE=(1,-6) and then i calculate the projection of KE to the direction vector (3,-4), then using pythagorean theorema

[safe radish]

@hollow jewel Has your question been resolved?

[hollow jewel]

<@&286206848099549185>

also i am finding the same with the way i proved the first formula using the magnitude of the cross product

[safe radish]

@hollow jewel Has your question been resolved?

[hollow jewel]

<@&286206848099549185>

[desert swallow]

You can only ping for helpers once 15 minutes after you've posted your question

Anyway, KE is not (0,3)

[hollow jewel]

@desert swallow yeah you are right but still i am getting that the projection is (27/37,-162/37) with magnitude that doesnt match with the other results

[desert swallow]

You don't need to the projected vector, just its magnitude.

Seems like you divided by 37 instead of 25

Try calculating it again

[hollow jewel]

you are right mate i was doing wrong calculations thanks for your time and sorry i stacked and chatgpt was saying sth about i choosed the wrong point anyways have a good one

.close

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[storm ice]

If I have this and want to evaluate the value for k = sqrt(5/3)

i am getting lost along the way

so start off like this

simplifies to this

oh nvm

.close

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[earnest tree]

Hey guys, does anyone know how to do this? Can someone teach me?
Calculate the equation of a circle: endpoints of a diameter are at (1,5) and (4,3) Thank you!

[brave wolf]

so you need to find the center and radius

center is the midpoint of the diameter

radius is half of the length of the diameter

[earnest tree]

yesyes

[brave wolf]

so what would be the midpoint of line segment between (1,5) and (4,3) (which would be the diameter)

[earnest tree]

we only took the lesson yesterday so i dont know much sorry 😭

75/2?

[brave wolf]

How did you calculate that?

we need to find midpoint of the purple diameter

[earnest tree]

Ohhh wait sorry its 20

[brave wolf]

20 is what?

the midpoint? Or length?

[earnest tree]

midpointtt

[brave wolf]

midpoint is a point, it has 2 coordinates

it isnt just 1 number

we need to find the coordinates of midpoint

to find them, note that x coordinate of the midpoint should be the average of x coordinates of (1, 5) and (4, 3)

[earnest tree]

oooh okayy

[brave wolf]

same with y coordinate

[earnest tree]

would it be (3,4)? 😭

[brave wolf]

almost

3 is not the average of 1 and 4 though

(1 + 4) / 2 is the average

aka arithmetic mean

[earnest tree]

ooh okayy okay

it would be 2.5 then?

[brave wolf]

so what would be the midpoint?...

correct

[earnest tree]

(2.5,4)

[brave wolf]

so we now know the center of the circle

[earnest tree]

thank you so much 🙂

then it would be the radius leftt

[brave wolf]

now we just need to find the radius

yes

do you know how to find distance between 2 points?

[earnest tree]

distance formulaa?

[brave wolf]

yes, that works

[earnest tree]

WAITT yes i did this in the previous question

thank you so much

[brave wolf]

btw it will be probably easier to calculate the length of diameter and divide it by 2, rather than calculating distance of the points on circle from the center

if you do it with diameter, you wont have to deal with fractions that much

[earnest tree]

thank u broo btw sorry if it took so long regarding mine bc this topic is new to me haha

[brave wolf]

np, it didn't take long, you were quite quick

[earnest tree]

Im still a little confused abt the radius part thoo 😅

[brave wolf]

okay

[brave wolf]

let's calculate the length of the diameter first

ill just repost the pic

[earnest tree]

okiii

[brave wolf]

it's the distance between points (1, 5) and (4,3)

[earnest tree]

i will use this for the distance formula rather than the center numberr?

[brave wolf]

both would work

[earnest tree]

im only confused about that part

[earnest tree]

ohhh

[brave wolf]

you can use the center if you want to

but you will have to deal with numbers like 2.5

so doing the diameter and then dividing by 2 is probably a bit easier

[earnest tree]

yesyes it will be correct if i use (1,5) and (4,3) right?

[earnest tree]

oooh

[brave wolf]

yes, you just need to divide the diameter by 2 in the end

[earnest tree]

omg okayy thank you so muchh 🙂

[brave wolf]

Have you solved it btw?

if so, you can close this with .close

[earnest tree]

Slr, my final answer is (x-2.5)^2 + (y-4)^2 = 2

[brave wolf]

hmm the radius looks wrong

[earnest tree]

ohh mannn

[brave wolf]

$\sqrt{\left(1-4\right)^{2}+\left(5-3\right)^{2}}=\sqrt{9+4}=\sqrt{13}$

[earnest tree]

illt ry to find it out, thank you 🙂

[flat frigate]

MæthIsAlwaysRight

[earnest tree]

OH

TY

[brave wolf]

this should be the diameter

[earnest tree]

bro thank u so much 😭