help-23

[split kayak]

So this turns into $2\sqrt{2\cdot7}$

[trim slate]

ok wait wait

so when we have 7 x 2^3 we can go lower so we do correct?

[split kayak]

Yes, when we operate with square roots, we usually want to simplify as much as possible.

[trim slate]

im guessing you go lower because you dont want ^3 but rather ^2?

[split kayak]

yep.

[trim slate]

okok

[split kayak]

ill try to showcase it in a simple line $\sqrt8 = \sqrt{2^3} = \sqrt{2^2 \cdot 2} = \sqrt{2^2}\cdot \sqrt 2 = 2\sqrt2$

[trim slate]

so the problem would go from 7x2^3 x z^7 to now 7 x 2^2 x 2 x z^7

okok

[split kayak]

thats why we use prime factorization here, it makes everything much easier to simplify

[trim slate]

so does that mean now we have simplied 56 as much as possible ?

[trim slate]

yesyes

[split kayak]

Well, you can simplify a bit more here.

[trim slate]

but shouldn’t i wait to simplify that part after handling x^7?

[split kayak]

Since square roots distribute over multiplication, you can solve separately.

[trim slate]

so i just jeep the 2 and do 2x7?

[split kayak]

$\sqrt{56z^7} = \sqrt{56} \sqrt{z^7}$
And we just solved $\sqrt{56}$

[split kayak]

now, to be honest, sqrt(z^7) is purely a notation problem.

[trim slate]

how do i write sq root on keyboard?

is it just ^

wait no

just sqrt?

[split kayak]

Depends on your cellphone, some of them dont have it
people will understand you if you write sqrt(x)

[trim slate]

so 2sqrt(14) * sqrt(z^7)

is what i have rn correct?

[split kayak]

Yep

[trim slate]

okok

[split kayak]

we can "simplify" a bit sqrt(z^7) too

[trim slate]

what does this meN?

mean*

[split kayak]

that there is a way to write it in a bit more "fancy" way

[trim slate]

ohh okok

[split kayak]

when you have the square root of an odd exponent: $\sqrt{z^7}$

[split kayak]

You can rewrite it as: $z^{\frac 72}$

[split kayak]

So you have a fraction as exponent

[trim slate]

okok wait

OKOK I SEE

[split kayak]

If its an even number, you divide by two and simplify.

[trim slate]

did you get 2 because thats the lowest?

[trim slate]

OHHH okok

[trim slate]

but 7 isnt even?

[split kayak]

Nope, it has to do with this: $\sqrt x = x^{\frac 12}$

[split kayak]

x = x^1 right?

Well, the square root is multiplying with the "1" exponent.

[trim slate]

is that just a formula?

[split kayak]

for our case, 7 * 1/2 = 7/2

[split kayak]

more of an identity of exponents

$\left(x^a\right)^b = x^{a\cdot b}$

[split kayak]

And since the square root is like an exponent of 1/2

you divide by two the number

[trim slate]

so i should always know to re write it as sqrt(x) = x^1/2? if its smth like x^7?

[split kayak]

yep.

and then simplify to x^(7/2)

[trim slate]

okok

wait

do i keep sqrt (x) as x or am i makinf it z?

wait im a little lost

[split kayak]

x is just a symbol we use

it can be z, a, b, a greek letter

etc...

[trim slate]

okay so using z^7 how would it look like with it filled into sqrt(x)=x^1/2

[split kayak]

sqrt(z^7) = (z^7)^(1/2) = z^(7/2)

[split kayak]

again, simplifying the two exponents into their multiplication is thanks to this property

[trim slate]

so this?

[split kayak]

youd have to erase the sqrt.

[trim slate]

wait wait whay do i do with this?

[split kayak]

is the property of exponents that allows you to simplify your z exponent.

we already did the sqrt 56

ill go from there

[trim slate]

right

okok

[split kayak]

$2\sqrt{14}\sqrt{z^7} = 2\sqrt{14}\left(z^7\right)^{\frac 12} = 2\sqrt{14}z^{7\cdot \frac 12} = 2\sqrt{14}z^{\frac 72}$

[split kayak]

Always remember that the square root can be rewritten as a fractional exponent.

[trim slate]

okok wait im going through it to make sure i understand how you got everything

wait when u did 7 x 1/2 did you just get that bc 7 x1 =7 and / 2 stays the same?

[split kayak]

yep

its just rational arithmetics

[trim slate]

okok

so is that the full step or is there more?

[split kayak]

that should be it

At this point depends on who you ask / your professor.

But you could also write it as $2\sqrt{14z^7}$

[split kayak]

Refer to the first part of this to see how this is the case

[trim slate]

im doing khan academy for own personal learning so ill see rn

um

it says the answer is

2z^3 sqrt(14z)

[split kayak]

okay, yea, that is also another way

they did the same as we did with the 2

z^7 = z^6 * z right?

[trim slate]

wait

it says

wait ill send pic

[split kayak]

yep, they did the same with the z as we did with the 2.

With a bigger exponent, but the same idea

[trim slate]

ohhh i see

wait im confused then if its also how you can do it then how is it wrong for how we did it

how are we supposed to know which to do

[split kayak]

Depends a lot on what theyre asking for

But yeah, all 3 options are the same thing

[trim slate]

“remove all perfect squares from inside the square root”

[split kayak]

Yeah, they want us to have no numbers with exponent inside the sqrt

[trim slate]

meaning they want no numbets above the z?

[split kayak]

Thats what exponents are, yea

the z^7 becomes z^6 * z, once you take the sqrt it becomes z^3 * sqrt(z)

since 6 divided by 2 is 3.

[trim slate]

ohhh

[split kayak]

and well, 1 divided by 2 is 1/2 which is the normal sqrt.

[trim slate]

okok i see

so rhats my bad for not giving enough context

but it still uses the same odeas

[split kayak]

yep.

[trim slate]

.cllose

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[river needle]

need help with c

[river needle]

somebody help plz

[radiant ice]

!show

[safe radish]

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

[glacial meadow]

In case no one is able to help you, you could try the linear algebra channel as well

Since this seems like an affine geometry problem

(do wait a bit before doing so though)

[river needle]

i just need help with part c.)

[radiant ice]

what dont you understand about it

[river needle]

how it's solved bruh

[radiant ice]

okay so for l_1 what are the x y and z values

+lamda

[glacial meadow]

Well, if it's a hint you want: your points P and Q will be points in L1 that are a distance of 35 apart from A(why?)

[river needle]

shouldn't p and q be 70 apart?

right

u said from A

yea cuz AP forms the radius of the circle

[glacial meadow]

Yup

Well, since you know this and the form of a generic point in L1, you could try to get it from here

[river needle]

AP = lambda(2, -3, 6)

?

i get lambda = +-5

[glacial meadow]

Not really, you have two main ways to do this:
1: get a point in L1, which will be of the form (4+2λ,4-3λ,-5+6λ), compute the distance from this point to A(subtract the coordinates of A and take modulus) and equate it to 35

[glacial meadow]

I think that's right, actually

Since you know that the points A, P and Q are all in L1, all you need are the points "above A" and "below A" a distance of 35 from A, which you can get by getting a direction vector of L1 of modulus 35 and both adding and subtracting it to A

Since the vector you're given has modulus 7, λ=±5 added to point A is what does the trick

[river needle]

i just dont understand one thing

lambda = 5 does not get me to point A from the origin

lambda =2 does

or are we talking bout a different lambda here?

[iron estuary]

λ doesn’t measure distance from the origin

λ = 2 gives point A because that’s where the intersection lies on l1

[river needle]

when i plug in lambda = 2 for l1

yea

lambda = 5 will me lead to a different point on the line tho

[iron estuary]

for part (c) we measure distance from A along the direction of the line

[river needle]

so we're talking bout a different lambda here

?

[iron estuary]

since $\lVert(2,-3,6)\rVert = 7$, moving $35$ units from $A$ corresponds to $\lambda=\pm 5$ relative to $A$ i.e. $\lambda = 2 \pm 5$

[flat frigate]

anflo

[iron estuary]

no no no it’s the same λ just a different reference point

think of λ as a slider on the same line

A happens to be at slider position 2

[river needle]

so from A, lambda needs to be scaled more?

[iron estuary]

to go 35 units away from A you move the slider 5 steps forward or backward

no no scaling is happening

λ always measures how many direction vectors (2,−3,6) you move along the line

from A going 35 units means moving ±5 direction vectors because each one has length 7

[river needle]

the way im seeing it is

if u start from A then u scale lambda ur direction vector by 5 to get to the point Q on the circumference

if u start from the other point, u scale lambda by 2 to get to Q

is that right?

[iron estuary]

almost but remember the scaling is always of the direction vector, not of λ itself

[river needle]

oh yea mb

ty

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[light tiger]

Help

[light tiger]

What is the equation of quadratic equations?

[quasi bison]

"equation of quadratic equations"?

[light tiger]

Yup

[quasi bison]

...

[quasi timber]

I’m assuming they mean standard form? Like 𝑎𝑥² + 𝑏𝑥 + 𝑐..?

[quasi bison]

are you expecting ax^2 + bx + c = 0 or do you maybe mean the quadratic formula?

!xy

[safe radish]

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

[quasi timber]

@light tiger

[quasi bison]

⏳

inb4 he says "nevermind i got it"

[vital bison]

An equation of quadratic equations is a.s. a quadratic equation /s

[radiant ice]

#discussion

#❓how-to-get-help

also #calculus

[quasi bison]

hello and welcome to the server. this channel is currently occupied. please claim your own channel if you would like to get help

[safe radish]

@light tiger Has your question been resolved?

[safe radish]

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[north dragon]

so the thing is

[north dragon]

im having a LOT of problem with when to add / subtract pi in inverse trigno functions

can any1 reccomend me smth to do like watch a vid or a trick to get it right

not even able to understand solutions atp

nvm

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[hybrid heath]

I don't get the idea of R and dimensions

[vital bison]

R^4 because each vector has 4 components, each of which is a real number. 2D space because the span of the two vectors is of dimension 2

[quasi bison]

what does 9n vector mean?

[vital bison]

I think that says "In" and not "9n"

[hybrid heath]

oh

[quasi bison]

do you know how to imagine lines and planes in 3-dimensional space

[hybrid heath]

yes..it is I

[hybrid heath]

what is plane

[quasi bison]

infinite flat surface

(geometry) A flat surface extending infinitely in all directions (e.g. horizontal or vertical plane)

[hybrid heath]

i have heard about it

but don't know

[quasi bison]

difficult

and you're doing linear algebra now without having this preparation?

[hybrid heath]

I have just started..
i have done matrices only in L.A.

[quasi bison]

ok but you have not done vector stuff before

and since you just said you don't know what a plane is, you must not have a lot of intuition for 3D geometry

[hybrid heath]

isn't geometry a completely different topic?

[quasi bison]

maybe if you like walled-garden thinking, then yes it is

[hybrid heath]

so what are the prerequisites for learning Linear algebra

[quasi bison]

i'd say probably familiarity with vectors in 2D and 3D

[hybrid heath]

i mean i know the basics

wait let me show you

[quasi bison]

the concept of plane is part of the basics and you're missing it\

[hybrid heath]

plane is made in 3d

then please suggest me a video on it

@quasi bison ?

[quasi bison]

i cannot possibly suggest you one (1) video on an entire topic

look up vector stuff on khan academy idk

there is a lot

[hybrid heath]

I am following a udemy batch..and i thought that they have taught me enough..but you are saying that its not enough...with whom do i agree more

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[winter whale]

how did we get 20 + theta

[halcyon mango]

initially the angle between the pawl and the wheel is 20 degrees as stated in the figure. due to the forces initially it moves by theta so at eq, the angle should be 20 + theta right?

[winter whale]

hmm i kinda get what you're saying

[halcyon mango]

but..?

[winter whale]

Guess there's no but

Thanks

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[lost jewel]

Cut a regular square pyramid along its edges (which may involve multiple cuts) to create a singular flat shape. The mirror image or/and rotations of a shape is not considered. In how many distinct ways can this cutting be performed?

[lost jewel]

These are some of the shapes that can be obtained but I haven't seen the pattern yet

[fathom adder]

You can also have a square and one triangle attached to it. Then all the triangles are on that that triangle

[mossy lotus]

count how many of its edges u cut in those examples

[mossy lotus]

theres also a pattern of 1 triangle attached to square by itself, and the other 3 triangles opposite to it

[lost jewel]

Yes, I'm aware of that

[lost jewel]

oke

[lost jewel]

Oh they're all 8, hmm

[mossy lotus]

indeed

(you cut 4 edges apart)

[split fulcrum]

yakuu baktep

[mossy lotus]

and now you have to choose which 4 from the 8 edges that exist

[mossy lotus]

traitor beam, wheres your greeks

[split fulcrum]

i will become greek soon

[lost jewel]

Just that?, what if we cut it into 2 pieces?

Should we find all the cases for that

[mossy lotus]

well yes, you have to ignore those cases ofc

eg cutting all three sides from same tria face is not good

[lost jewel]

we could even cut the square out

[mossy lotus]

or cutting all 4 edges of base is also not good

[lost jewel]

yeh

[mossy lotus]

yea, but then it wont be one piece mad pirate noises

[lost jewel]

Hmm

what if we get a mirror image

[mossy lotus]

yea, honestly, since this is somewhat of a small problem, you can just list out all possible cuts and count that way

[lost jewel]

Hates it when I have to do that

Letmme try

Am I missing any cases? I can't tell

[lost jewel]

I just found out that method won't work well cuz some time it doesn't fully cut the pyramid into a flat shape

[nimble vine]

maybe

[lost jewel]

That's the 6th one I drew it wrong mb

[nimble vine]

ok

[lost jewel]

Still, is that all?

[nimble vine]

ig

[lost jewel]

Okay I would love if more people verify it

I don't trust myself

[nimble vine]

same

[mossy lotus]

can you show an example? coz I cant see a case where it doesnt make it flat while keeping it a single piece

[lost jewel]

beat me, ig a week of daily 4 hours sleep finally caught up on me

I couldn't find one either

[lost jewel]

Anyway can someone check if these are all the cases

[lost jewel]

Ig the only way to do that is to cut out a triangle, I was a bit dumb

[mossy lotus]

2 can have another variant right?

where the middle two triangles are attached

[lost jewel]

Oh yeah how did I miss that

oh wait

[mossy lotus]

oh nvm, im idiot, vertices dont attach

[lost jewel]

Yeah

[nimble vine]

maybe

[lost jewel]

Oh

Yes yes

So 8

[nimble vine]

maybe

why do you need this?

[lost jewel]

Uhh, it's my hw

[nimble vine]

oh

is q word to word

[lost jewel]

not sure what that mean, if you mean did I translated it correctly then yeah

[nimble vine]

hmm.. then original

[primal bone]

I think you might still be missing one more net

[nimble vine]

enlighten us

[primal bone]

in 2 min...

[lost jewel]

I can't show it here for some personal info, but I have been working on the same problem as you so I don't think there're any flaw in it, sorry

[nimble vine]

alr

[primal bone]

You've considered the case where three triangles are still attached to each other - but have you considered which of these triangles is still attached to the square? @lost jewel

OH WAIT nvm that's your 5th one there

[lost jewel]

I hate cases bash, I can't tell if it's all the cases or not

[primal bone]

I think you've got all of them tho

[nimble vine]

lock it its enough

[primal bone]

[1] being when all 4 triangles are attached to the square

[2] being 3 triangles attached

[3] through [6] for 2 attachments

And [7] being one attachment

[lost jewel]

and [8] here being also one attachment which I missed

cool cool

Alright love y'all

.close

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[lost jewel]

A verification from Wires is enough for me

And oppenheimer himself

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[lost jewel]

Given a tetrahedron $ABCD$. It's known that $\angle BAC=\angle CAD=\angle DAB=60$ and $AB+AC+AD=BC+
BD+CD=6$.Compute its volume

[flat frigate]

Fionna The Unemployed

[lost jewel]

Diagram

also status 1

mate I made a new ticket

[nimble vine]

k

[opaque fern]

I reckon try to represent all thr edges with the law of cosines for a starter

[lost jewel]

I am avoiding trigs btw

But okay

[opaque fern]

Aight I'm out then

This sounds horrible without trig

[lost jewel]

It's still horrible with trigs

[opaque fern]

Well its helpful here bcuz like

[lost jewel]

Eventho I don't like trigs ig I still give it a try

[opaque fern]

You can use the fact that for a triangle with sides x and y enclosing a 60 degrees angle, you can relate that with the third side, a, with

a > (x+y)/2

[lost jewel]

I have no idea where that comes from

Am-Gm?

[opaque fern]

I think this fails if the triangle is an equilateral though

[opaque fern]

law of cosines basically

[lost jewel]

Okay I can't even prove it

in 10 min

[opaque fern]

I can write a proof for you?

[lost jewel]

sure

[opaque fern]

But also I think my idea with that is that if you can manage to retrieve some relationship between the sides that reduces the sum to be in terms of one side then the thing is solved

[lost jewel]

I'm still waiting for the proof

[opaque fern]

You know that [
a^2 = x^2 + y^2 -xy
]
So if you chug that in the inequality you have [
x^2 + y^2 -xy \ge \8{\4{x+y}2}^2
]
So do some simplification you get
[
(x-y)^2 \ge 0
]

[opaque fern]

That inequality is true for all values

[lost jewel]

Okay now idk why that inequality is true

[opaque fern]

If you think about it if you like fix the sum of x + y to be something, the length of the third side will be minimum when it is an equilateral

so when x = y

If you vary the values the third side will get longer

[lost jewel]

I have to show why that true tho I can't just say that

[opaque fern]

I mean thats an algebraic proof

[lost jewel]

But I have to prove that inequality

[opaque fern]

That proves the inequality?

Like you mean a derivation?

[lost jewel]

yeah isn't that the same thing, derive the inequality you show above

the x^2+y^2-xy >= (x+y)^2/4

[opaque fern]

I mean deriving it isnt the same as proving it is true

[lost jewel]

ehh I can't tell the different tbh, but how would you do so

by that I meant deriving it

[opaque fern]

Ok well

,w optimise x^2 + y^2 - xy subject to x + y = 6

[flat frigate]

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[opaque fern]

Okay so @lost jewel

There's two ways to do this tuat i can think of

[lost jewel]

I'm listening

[opaque fern]

Either with lagrange multipliers or by some uh intuition

Are you familiar with the former?

[lost jewel]

No I'm a highschooler

[opaque fern]

Aight

So like

[lost jewel]

irrelevant but I feel like this is the kind of inequality that appears in a problem and then never be used again

[opaque fern]

Our constraint function is $x + y = S.$ In other words, we are trying to minimise
\e{equation}{
a^2 = x^2 + y^2 -xy \Implies a^2 = S^2 - 3xy
}
To minimise this you'd need to maximise $xy$, so substituting $y = S - x$ you get [
x(S-x) = Sx- x^2
]
Which is downward-facing quadratic with a vertex of [
x = \4S2
]
Since $y = S - x$ you have $y=\5S2$ as well. Putting this in (1) gets you
\e{align}{
a_{\t{min.}}^2= S^2 - \4{3S^2}4&\Implies a_{\t{min.}}^2 = \4{S^2}4 \[1ex] &\Implies a = \4S2
}

@lost jewel

[opaque fern]

This tells you the minimum value of a is at (x+y)/2

Hence a > (x+y)/2

[lost jewel]

You substituting x+y=S and using it in a function of x itself, now that's what I didn't know we could do

If I have x^2(x-1)+4x(x-1)+5

and let S=x-1

Can I do the same thing

It should have a minimum at x=-2

,w minimum x^2(x-1)+4x(x-1)+5

[flat frigate]

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[lost jewel]

hm

maybe if x>1

It works

,w minimum x^2(x-1)-4x(x-1)+5 where x>=1

[flat frigate]

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[lost jewel]

Okay it broke my mind I have no idea how that works

[opaque fern]

lets see

uh

,w minimise S^3 + 6S^2 +5S+5

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[opaque fern]

@lost jewel

[opaque fern]

this is the same as this

its just shifted by 1

cuz its S = x - 1

[lost jewel]

yeah I see but I can't use the substitute S=x-1 fully here

Like what you did

[opaque fern]

u can tho

[lost jewel]

Can you elaborate

[opaque fern]

give me a bit to latex

[lost jewel]

Sorry I'm literally having some food right now so I might reply a bit slowly

[opaque fern]

eat well

if you let $f(x) = x^2(x-1) + 4x(x-1) + 5$ if you sub in $S = x-1$ you get [
f(S) = (S+1)^2S + 4(S+1)(S) + 5
]
do algebra and you get [
f(S) = S^3 + 6S^2 +5S+ 5
]

[lost jewel]

No I mean using substitute S and also x in the same function

Like what you did

[opaque fern]

oh its bcuz S in our case is actually constrained

like we have x + y = 6 = S

[lost jewel]

eh

x+y isn't 6

[opaque fern]

i think it is tho

[lost jewel]

but it says here AB+AC+AD=6

[opaque fern]

im talking about this

i mean the whole preceding stuff was a hypothetical example

to give you a derivation

[lost jewel]

hm

That's still weird how it works

If restricted x+y=6, but if x+y !=6 ik it should be the same but if x+y=constant doesn't that mean we have a restriction for x? is f(x) in this case a quadratic function with that restriction?

[opaque fern]

yes

[lost jewel]

why?

[opaque fern]

i think all of this is shadowing what you do in multivariate calculus

with lagrange multipliers

[lost jewel]

Shoule I take it in highschool?

self taught

Nah I'm kidding

[opaque fern]

if you graph like

z = x^2 + y^2 -xy

it would look like a bowl or somthing called an elliptic paraboloid

[lost jewel]

Should it be a quadratic on a slant surface?

[opaque fern]

when you specify x + y = 6 you are specifying a plane that intersects with that bowl

it is a quadratic bcuz its a parabola

i can show you

one sec

this is kinda bad but u can see it ig

[lost jewel]

So, f(x) is a quadratic/parabola on a surface

Yeah I can see why

[opaque fern]

yea

[lost jewel]

It's okay 3D geometry are main focus in my syllabus along with combinatorics

[opaque fern]

oh thats fun

[lost jewel]

I would hit myself if I can't see it

[lost jewel]

Nah

[opaque fern]

[lost jewel]

Hang on

[lost jewel]

I can't say this's fun

[opaque fern]

yikes

ok anyways this is all so tangential to your actual question

circling back

uh

[opaque fern]

my idea was

[lost jewel]

Yeah

That's it tbh

[opaque fern]

oh

[lost jewel]

BC>= (AB+AC)/2
BD>= (AB+AD)/2
CD>= (AC+AD)/2

[opaque fern]

yeah

[lost jewel]

so BC+BD+CD>=AB+AC+AD

the equality holds when they are all equal

[opaque fern]

yep

exactly

si they are all 2

[lost jewel]

I could prove x^2+y^2-xy>(x+y)^2/4 true

with x,y>0

But your was much stronger case

[opaque fern]

anyways this was fun

i love optimisation questions

[lost jewel]

well you know multivariable calculus and I do not

If I saw these what I would do is using proven and allowed inequalities

[opaque fern]

u dont technically need it

as u have seen

[lost jewel]

Ig? but I can't see myself coming up with such way

Can we apply multivariable cal for the question in help-25?

[opaque fern]

one sec

lemme see

[opaque fern]

yeah u can

,w optimize (a+1)(b+1)(c+1) subject to 1/a + 1/b + 1/c = 1

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[lost jewel]

a,b,c>0 also

[opaque fern]

well u can definitely use lagrange multipliers

i just dont know how to use it with wa

[lost jewel]

Ikw

[opaque fern]

it actually becomes super rivial wih that

trivial

[lost jewel]

My mentor may has introduced us to lagrange multipliers

Hang on

Imma check my recorded lectures

I lost it

But I remember he drew something like this

I had no idea what he said tho

[opaque fern]

[lost jewel]

It was about optimization problems given systems of equations

It's also multivariable

But I zoned out and didn't understand what he said

[opaque fern]

thonk

[lost jewel]

Is this lagrange multipliers?

If not what it is, it looks important when dealing with optimization problems

I found the record

This's what he drew, he was talking about imagine pulling out with a force F smth smth

[nimble vine]

crazy we moving to lagrange

for stubid piramid

[lost jewel]

I have a different stubid piramid problem if you want to try

[nimble vine]

is that solved?

[lost jewel]

yes I solved it

[nimble vine]

alr

[lost jewel]

Is that a yes or no

[nimble vine]

ok

[lost jewel]

aight

Imma close this

I'm still not sure if that's lagrange multipliers or not tho

.close

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[safe radish]

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[delicate patio]

I am clearly doing something wrong, I suspect I am doing th wrong row operations

the answer is that x3 is free, and x1, x2 and x4 are pivot columns,

[safe radish]

@delicate patio Has your question been resolved?

[gray mulch]

You don't wanna use the regular Gaussian elimination when you can get a "unit" value with a certain linear combination

So here we want to get a unit value for the variable x2, so you just subtract R2 from R3, then use it to eliminate the x2 variable in R2

So the mistake was that your third row shouldn't contain the variable x3
(0 0 0 -11 | 11) instead of
(0 0 6 -11 | 11).
You had 2 -2 and 3 -3 respectively on those positions, so R3 - R2*3/2 would yield 0 0 on positions x2 and x3

[delicate patio]

Let me read it, one moment, see if I understand it

Ah I see what you did, still not fully sure why, what do you mean with shouldnt contain the variable x3?

[gray mulch]

It gets eliminated with the variable x2 on the same step

[delicate patio]

You mean when I do my R3 - (3/2) * R2?

-3 - (3/2) * -2?

[gray mulch]

Yes

[delicate patio]

Oh damn I just miscalculated

that does yield 0

and not 6 like I pretended

[gray mulch]

Ye, you generally don't wanna use that tedious fraction multiplication method

Use the one I showed in red

If possible that is

[delicate patio]

Yeah, or I could just do (1/2) * R2 first

[gray mulch]

Ye, that works too

[delicate patio]

mhm wow well lesson learned

i cannot do simple calculations so avoid having to 😂

thanks mate!

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[mighty mango]

the second derivate with respect to x is...
k^2
1/2 k^2
2/k^2
1/k^2

[mighty mango]

yeah I tried two different approaches and both resulted in a complicated first derivative so I didnt want to take the second derivative

[quasi bison]

is y a constant

[mighty mango]

no

its a function of x

[quasi bison]

uhh

[open wedge]

idk if multiplying with the conjugate works

[quasi bison]

oh hold on

[opaque fern]

implicit differentiate

[quasi bison]

this is an implicit thing

[mighty mango]

yes

[quasi bison]

can we find y explicitly in terms of x then

[mighty mango]

I did try that lex 😡

[open wedge]

lmao

[quasi bison]

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

[opaque fern]

there is a quick way to do implicit differentiation via partial derivatives

[quasi bison]

y + sqrt(y^2 - x^2) = 1/2 k^2

uhh god

[mighty mango]

yes I did that

then differentiated

[quasi bison]

oh did you not do more algebra to it

[mighty mango]

which cuased a horrendous first derivative. i didnt want to take the second derivative

[quasi bison]

$\sqrt{y^2-x^2} = \frac12 k^2 - y$

[mighty mango]

I did not.

[flat frigate]

Ann

[open wedge]

try moving one square root and then square it

[quasi bison]

$y^2 - x^2 = \frac14 k^4 - k^2 y + y^2$

[flat frigate]

Ann

[quasi bison]

$-k^2 y + \frac14 k^4 = -x^2$

[flat frigate]

Ann

[quasi bison]

turns out algebruh

was the solution all along

[open wedge]

$k - \sqrt{y - x} = \sqrt{y + x}$

$k^2 + y - x - 2k\sqrt{y - x} = y + x$

[flat frigate]

1 divided by 0 equals Infinity

[mighty mango]

wow

all i had to do was square twice 😡

[quasi bison]

no good

[open wedge]

learn more algebra

[quasi bison]

or maybe yes good actually

[open wedge]

i'd move the roots to one side and then square again

[quasi bison]

that cooks

[mighty mango]

yeah i see how both approaches work

ty ann and infinity

[open wedge]

np

[open wedge]

both of them are just squaring twice

[mighty mango]

🤨 amazing advice to the beginner calculus-1 student

[open wedge]

normal advice from someone who didn't learn calc in their curriculum

[mighty mango]

yeah lex is also like

a highschooler

[open wedge]

lol

[mighty mango]

.close

[safe radish]

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[mighty mango]

they wanna be a nurse when they grow up

[open wedge]

im a middle schooler

[opaque fern]

not a highschooler

[mighty mango]

yes you are.

[opaque fern]

aight

[open wedge]

what grade is lex in

[open wedge]

pronouns?

[opaque fern]

he

[open wedge]

i thought you're a she 😭

[mighty mango]

you were never him.

[open wedge]

im lucky i asked for the pronouns

😭

[open wedge]

ngl practice some algebra

not rlly painful

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[cosmic sky]

Can someone check my solution to this please?

[gilded vault]

this is more or less correct although the proof could be more rigorous

a good way to show that the function is surjective would be to actually define the inverse function

so this would be:

• Find the largest triangular number less than i: let this be $T_n$.
• Then $y=i-T_n$
• $x=n+2-y$

[flat frigate]

Arnavutköy

[gilded vault]

and then u show that if $f(x_1,y_1)=f(x_2,y_2)$, then $x_1=x_2$ and $y_1=y_2$

this is more or less based upon the same arguments which u have

[flat frigate]

Arnavutköy

[faint hornet]

you are not trying to show surjectivity though

[gilded vault]

"prove that the set of values of this function are all natural numbers"

[faint hornet]

well it’s a bit ambiguous I guess

[gilded vault]

"for any natural i=f(x,y) the numbers x and y are defined unambiguosly" is injectivity meanwhile

and then i guess u need to prove that its always a natural number to start with

but

yeah

[faint hornet]

I interpret it as proving that the only “outputs” are natural numbers and this is injective

so I suppose it’s up to interpretation

[gilded vault]

wait the word 'all' means it is surjective right

this btw is the canonical transformation from $\mathbb{N}^2\to\mathbb{N}$

[flat frigate]

Arnavutköy

[faint hornet]

well yes however it can also be interpreted as all outputs are natural rather than all outputs cover every natural number

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[delicate tulip]

HELPGE PLEASE
Im asked to prove the following:

[delicate tulip]

this is what I have so far from the little knowledge I got on this topic

how do I prove it tho? What even does it mean to prove, what am I asked to show, in what way

^ I replaced the variables b and c with their definition on top and factored out the a in case its not clear what I done

[verbal cloud]

it's the right idea

you could write it more formally I guess

[delicate tulip]

definetly

[verbal cloud]

if you're aiming for that, you only need to change the order of your arguments

[delicate tulip]

am I not missing the proof itself? Would what I have get full points in a exam?
if not please correct me @verbal cloud

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[verbal cloud]

I mean it's missing the sentences so probably not

[verbal cloud]

the correct order should be like
a|b so b = ak ; a|c so c = al
xb + yc = ... = a(xk + yl)
therefore a | xb + yc

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[alpine oxide]

draft answer, hence didnt bother to specifiy sums, but would this work

[alpine oxide]

(Lin Alg 2)

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[rain nova]

https://sites.google.com/view/ruixiang-zhang/home/teaching/math258_fall2021?authuser=0 hey does anyone know how to prove 1 or where i could go to figure out a specific solution, i hav a rough idea of the solution but dont kno how to write the proof tbh

[rain nova]

any help/advice appreciated 😄 😄

[uneven knoll]

i recommend asking in #real-complex-analysis or #advanced-analysis, there you would get a faster response

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[rain nova]

bet ty

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[real gulch]

How to prove that a field extension is purely inseparable iff it's an epimorphism? I'm stuck proving that in characteristic 0, the only epimorphisms are isomorphisms.

[feral linden]

What do you mean “iff it’s an epimorphism”? What is “it”?

Anyway, E/F is algebraic and purely inseparable and char(F)=0 then E=F

(Because char(F)=0, algebraic E/F is always separable. Thus both separable and purely inseparable, only happens when E=F) I haven’t thought about not algebraic extension cases though

[keen tulip]

in the category CRing, the followings are equivalent: 1. f is an epimorphism and 2. the multiplication map mu : L otimes_k L to L, x otimes y mapsto xy is an isomorphism

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[iron wolf]

I'm learning Asymptotic methods right now, can someone please explain what these solutions are doing after they substitute x=E^aX

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[iron wolf]

<@&286206848099549185>

[plucky elk]

Can also try #dynamical-systems or #advanced-analysis

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@iron wolf Has your question been resolved?

[iron wolf]

its fine i figured it out

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[tall yarrow]

https://cdn.discordapp.com/attachments/486841106298961930/1455092511017009337/File_20251229-122742.jpg?ex=69537800&is=69522680&hm=272aa5a5387108f1aa052f29a620a0fad67d399fdfa794dc32fd5b9fc97bdedf&

[tall yarrow]

Anyone please help me with this question

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[lament rivet]

@tall yarrow U can use byparts

[winged flare]

.close

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[tawdry summit]

For context I want to create a rectangular border in minecraft, but since that's not possible I have to draw fake borders(squares) around the player that overlap with the rectangles edges. The data I'm working with is player's x and z coordinates. the outer rectangles minX, minZ, maxX, maxZ coordinates.

What I tried was finding the wall that the player is closest to by calculating the distance to the west, east, south and north walls and then using the smallest value I got from that to draw the border around the player with that distance, but if the player is in a corner of the rectangular border it will display the border correctly for one side of the rectangle, but not the other. I'm wondering if there's a way to calculate the centerX, centerZ of the square border so that it perfectly overlaps with both sides of the rectangle. (Check the second image for a better explanation of the issue).

[magic junco]

Could you elaborate on how this actually works? I’m confused about the mechanism behind this

I understand that you are trying to create a rectangle border, but how does creating a border around players resolve this?

[tawdry summit]

So the big rectangular border is what I'm trying to achieve, but because of limitations in minecraft you can't just display a rectangular border so I'm trying to display square borders(which minecraft does support) to act as the border. Basically I'm trying to achieve this by drawing a border around the player that moves with them.

I'm not sure how to better explain this

[magic junco]

yes, so far so good

[tawdry summit]

I'm trying to figure out how to calculate the location of the border around the player

[magic junco]

My confusion is that do you want it to look like a rectangle “from a higher perspective” or “from the player themselves”

[tawdry summit]

The border is going to be millions of blocks in each direction so the perspective isn't important I think. I'm trying to create an illusion of a rectangular border using square borders.

[tawdry summit]

location as in the radius/diameter of the square and centerX and centerZ for the border iself

https://streamable.com/7co7h7

Maybe this helps

The square border around the player always needs to be within the confines of the rectangular border. The player can move closer and further from the border, but it should always overlap with the rectangle.

No idea if this helps, but here's the code I'm using to calculate how close the player is to the rectangular border.

[safe radish]

@tawdry summit Has your question been resolved?

[tawdry summit]

@meager igloo Any help is appreciated

oh wrong ping

<@&286206848099549185>

[low echo]

Hello there

[tawdry summit]

Hi

[fringe estuary]

so your trying to find how close the player is to the border you created around them?

[tawdry summit]

No I'm trying to find the size of the border, centerX and centerZ coordinate for the border to create it around the player

[fringe estuary]

ah is the border size not always static? What variables cause it to change size?

[tawdry summit]

Oh wait my bad. Its static yes

so only centerX and centerZ coordinates for the border around the player in a way that they never go past the rectangle

[fringe estuary]

Can the player not go above the border, there's like a ceiling?

[tawdry summit]

the player can go as low and as high as they want within the border so that's not important

[digital sparrow]

lets say the square has side length a and the rect has dimensions lxb
then if the x cooords is less than a then the square will overflow

so we can just set its x_coord to be "a" instead of that of the player

[tawdry summit]

Does that behave correctly for corners as well? It might end up drawing one side correct, but the other side is incorrect.

[digital sparrow]

you will have to do the same thing for the other side*

ie the z coord

total 4 if statements

box_x = player_x
box_z = player_z

if box_x < mixX + a:
  box_x = mixX + a
if box_z < minZ + a:
  box_z = minZ + a

if box_x > maxX - a:
  box_x = maxX - a
if box_z > maxZ - a:
  box_z = maxZ - a

something like this

idk c so i used py

[fringe estuary]

Your method is to find the closest wall to the player but is the distance to all walls not supposed to be static? Would finding where the player is at to the the walls and comparing it to how far away the walls/player are supposed to be not a better solution?

ah wait im just confused

[tawdry summit]

Aren't the boxes x and y coordinates always the same? It can only be a square.

[digital sparrow]

the box_x and z represent the centre of the box

[tawdry summit]

Ohhh

[digital sparrow]

the box is the fake border

[tawdry summit]

Yep I see what you're doing

I'll give that a go. I hope it works

[digital sparrow]

nice

[tawdry summit]

Thanks for the help!

.close

[safe radish]

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[digital sparrow]

npp

[safe radish]

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[limpid lodge]

Determine all values of $\alpha$ for which the point $(\alpha, \alpha ^ 2)$ lies inside the triangle formed by the lines:

x + 3y - 1 = 0

x + 2y - 3 = 0

5x - 6y - 1 = 0

[flat frigate]

ch3rry

[quasi bison]

why are the lines not in mathmode also

but ok,

!status

[safe radish]

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

[limpid lodge]

All ik is we're supposed to use the formula for position of a point wrt to a line

[vital bison]

Yeah I think it's just three inequalities, and you find the intersection of the ranges for alpha

[quasi bison]

wrt to

i mean you'd want to at least graph these lines and see where this triangle is relative to each of them right

it's gonna be above 2 of the lines and below 1, or vice versa.

[limpid lodge]

i looked at the soln

They first find the vertices of the triangle

[open wedge]

🔥

[limpid lodge]

Is there any way to do it without tht?

[open wedge]

they find the vertices of the triangle because they want to bound $x$ and $y$ right?

[flat frigate]

1 divided by 0 equals Infinity

[quasi bison]

ig you can rawdog to find the intersections of y=x^2 with each line and puzzle sth out from that

[vital bison]

Position of a point wrt the line is still probably more efficient

[quasi bison]

power of a point?

[limpid lodge]

Maybe smth with slopes, no?

[quasi bison]

where and how are you gonna apply power of a pt here

[nimble vine]

eh maybe determinant

[vital bison]

Assuming you know how to solve quadratic inequalities, it's probably the easiest

[vital bison]

I meant the position, sorry

[limpid lodge]

wdym

[digital sparrow]

finding the vertices would be the easiest

however another alternative is to do it wrt to (0, 0)

just plot the lines and you will see whether it should be >0 or <0

[vital bison]

So if you have a line ax + by + c = 0, and a random point (s, t), computing as + bt + c will tell you where that point is with respect to the line

If that quantity is greater than 0, its on one side, if its lesser than 0, its on the other side, and if its equal to 0, it obviously lies on the line

If you draw out your triangle you will notice that points inside the triangle are above two of the lines and below one of the other, or vice-versa

[limpid lodge]

Right but that's not the method i was asking for

[vital bison]

That will give you three total inequalities in alpha that you solve, and that will be the range

[vital bison]

Alternatively this

(a, a^2) lies on y = x^2, so you can find the potential intersections with the triangle and use that to find the values you need for alpha

[limpid lodge]

Ohok

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[safe radish]

@stark ether Has your question been resolved?

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[thin lion]

Hello is the is correct?

[thin lion]

,rotate

[safe radish]

@thin lion Has your question been resolved?

[obsidian oracle]

there are a few weird things with that

Mainly how do you justify this exactly

and also, what does $y_n\to \infty$ mean to you

[flat frigate]

Santafilou2003

[obsidian oracle]

because this is not the correct way to move to the conclusion

[thin lion]

Hm from definition of divergence

For all L in R, there exists an N in Naturals st for all n>=N we have |yn| >=L

[thin lion]

I’m tryna say that for sufficiently large n, the terms in the sequence (yn) are greater than the terms in the sequence (xn)

[obsidian oracle]

That's not divergence to infinity (+infinity)

Find the correct definition of divergence to +infinity first

[obsidian oracle]

Also react to this with a ❌ if you don't want the channel to close

[thin lion]

Is it for all L>0?

[obsidian oracle]

no the L is not the problem

you're asking for the absolute value of y_n to grow as large as you want

then y_n either grows into the positives

or in the negatives

or both

how does that force y_n to go to +infinity?

[thin lion]

Sorry, idk what im missing in definition of divergence, that’s the definition I was told in my lecture

[obsidian oracle]

I'm certain the real definition doesn't involve the absolute values

[thin lion]

Oh wait

😔

Yes ur right

Oh wait

Do I use max N here?

[obsidian oracle]

?

[thin lion]

Define N* = max(M,N) then for all n >=N* I have that yn >= xn >= L

[obsidian oracle]

What even is M

you already know that y_n >= x_n for all n natural

[thin lion]

That’s to say that after a certain threshold, yn >= xn

[obsidian oracle]

there is no such threshold... M = 1

It's part of your initial assumptions

[thin lion]

then im lost

[obsidian oracle]

Define N* = max(M,N) then for all n >=N I have that yn >= xn >= L

[thin lion]

Yh I think my absolute values are messed up

[obsidian oracle]

get rid of all your absolute values since they come from a faulty definition

[thin lion]

Alright

[obsidian oracle]

So there is no need for that anymore

[thin lion]

So that’s all I need to say after I define xn diverging to inf?

[obsidian oracle]

Using the definition for divergence of x_n to inf

For any L, you can find N such that .... x_n >= L

[thin lion]

Yep mb

[obsidian oracle]

so for that same L and N

.... y_n >= x_n >= L

you proved y_n -> inf

[thin lion]

Oh alright

Maybe I over complicated it

Tysm

.close

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[lament rivet]

I already got the answer using brute force If u can tell me a good method

[quasi bison]

B ∪ C consists of sets T ⊆ {1,2,3,4,5,6,7} such that at least one of the following is true:

• 1 ∉ T
• 2 ∈ T
• sum(T) is prime

a good way to count them seems to be:

• count the sets matching criterion 1
• count the sets which FAIL criterion 1, but match criterion 2
• count the sets which FAIL criteria 1 AND 2, but match criterion 3
• add up all these counts, knowing you have not overcounted anything

[lament rivet]

Ye we can compute n(B) using its complement

But I dont know what to do about C

[quasi bison]

you don't actually need n(B) here

you don't need n(C) either

just view B ∪ C as its own thing as i outlined

[lament rivet]

No I need to find n(Bcomplent intersection C(complement))

[lament rivet]

I alr did this

[quasi bison]

that was your brute force?

[lament rivet]

ye

[quasi bison]

exactly as i described?

[lament rivet]

Pretty much

[quasi bison]

i dont think you can improve on that then

[lament rivet]

I did better than that I think

I just made cases so that it was B complement intersection C complement

But I had to make 21 cases

[quasi bison]

21 cases???

[lament rivet]

Ye

[quasi bison]

that's a lot

[lament rivet]

Exactly

[half folio]

limx->0 (8^x-2^x)/x = ?

[quasi bison]

ok right so do you wanna walk through my method

[quasi bison]

!occupied

[safe radish]

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).

[lament rivet]

I think there will be more cases if I use ur method

[lament rivet]

But sure

[lament rivet]

exactly u have a method where I need to go over 3 criterian's In my method I need to go over 1

[quasi bison]

ok but yours requires 21 subcases

[lament rivet]

Ur's will have more when u calculate the 3rd case

[lament rivet]

They were 21 ways to get the sum to be composite

[quasi bison]

coprime?

do you mean prime

[lament rivet]

No sry composite

Idk why my brain just stopped working for a min there

[quasi bison]

guh

[lament rivet]

Well If u can come up with a better method pls do let me know

[quasi bison]

so formally i had seven cases for sum T, but none of these were that bad

[lament rivet]

I dont think the ans is correct

The ans is one more

107

[quasi bison]

did i miss one?

oh yes i did

{1,4,5,7}

good catch, thank you

[lament rivet]

Damn

[quasi bison]

don't call me "bro".

[lament rivet]

Sry mb

[quasi bison]

i have a bio for this server now

& i explicitly say not to call me that in there

so uh yknow

but anyway ok

[lament rivet]

I didnt see it sry

.close

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[safe radish]

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[bright osprey]

Exercise 9

[bright osprey]

Attempt ↓

How do i prove that it can be less than OR equal to x³e^x/6

Instead of it being strictly less than

[safe radish]

@bright osprey Has your question been resolved?

[bright osprey]

<@&286206848099549185>

[viral geyser]

hi

[bright osprey]

Hey

[viral geyser]

no offence but are you every gender and have all the pronouns ever discoverable

and since this is a help channel what do you need help with

[bright osprey]

no i actually like pressed them all when i joined the server

ill take them off if its against the server rules

[bright osprey]

@viral geyser i need help with this

[viral geyser]

no no it’s alr it’s not against the rules

[viral geyser]

Alr

[bright osprey]

alright

[viral geyser]

and you need to prove what is equal to x^3 e^x/6

[bright osprey]

i wanna prove that e^x - 1 - x - x^2/2 <= x^3e^x

but so far im only at e^x - 1 - x - x^2/2 < x^3e^x

should i just write "it is sufficient to take e^x - 1 - x - x^2/2 <= x^3e^x"

since according to taylor-lagrange formula e^x - 1 - x - x^2/2 = x^3e^c
with c belonging to (0,x)

[hardy panther]

why is everything so advanced😭😭😭😭😭

[safe radish]

@bright osprey Has your question been resolved?

[bright osprey]

It's real analysis

[loud wren]

you're not given taylor series yet, right?

[loud wren]

hmmph. I feel like this isn't enough, cause y'know, rigor, but if you can find an example with equality that should do it

[bright osprey]

its taylor's formula right?

the one used for approximating a function

we were given taylor-lagrange and taylor-young

although taylor-young has nothing to do with the question

pretty sure taylor-lagrange is just taylor's formula with lagrange's remainder added at the end

[bright osprey]

so i pretty much just plug in any value of x>0 and it should be correct?

[ripe raven]

now i might be wrong bc i took the class a while ago

but isnt showing that its < a stronger result

and u can just say <=

A<B => A<=B

[loud wren]

That's my best guess

i'm not too good with the acutal analysis part 😂but as mochimiso said, along with what I'm thinking right now, the bounding should be fine

[bright osprey]

i think thats correct

i guess im done here

thanks for the help

.close

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[maiden kernel]

What exactly does exp() mean? „Converts the sum inside into a multiplier“
Does anyone have an example to make it easier for me to imagine? exp(3+3+3)= 3*3?

[hardy vale]

exp() is used to denote general exponential functions

its commonly used in the case of like