help-49

[dawn dagger]

the premise is already false

[tidal turret]

yes, otherwise we would have a complex w or something

maybe I am tripping hard

[dawn dagger]

indeed

let's introduce quaternions as well

[civic lynx]

project the neighbourhood onto the Riemann sphere for best results

[tidal turret]

like, let u = -3<-2, then u^2 + w^2 <= 1, then 9 + w^2 <= 1 then w^2 <= -8

[dawn dagger]

how about you just finish the proof

you are doing silly stuff atp

[tidal turret]

how do I justify this upper bound works?

[dawn dagger]

you just need to explicitly state the neighborhood you consider around the origin, for the upper bound

write down why what you wrote is true and can be used

[tidal turret]

well we consider the neighborhood of (0-2,0+2), something like that?

[dawn dagger]

then you don't consider all paths

this would be just a line

like you neglect what happens below and above

a disk doesn't

[tidal turret]

we consider the disk of radius somrthing

[tidal turret]

I am getting a better idea now of what you mean

[tidal turret]

just consider the disk, x^2+y^2 of radius 2

[dawn dagger]

for example

[tidal turret]

all radius work?

[dawn dagger]

again no

radius 2 is indeed necessary

we may go lower but that's it

[dawn dagger]

think of the yellow strip where the inequality works

so we pick a nice neighborhood (disk) which contains it

or a subset of that

remember, w was independent, for the inequality to work

[tidal turret]

the radius can be at max 2

[dawn dagger]

yes

[tidal turret]

because otherwise u something something

[dawn dagger]

otherwise there are points where the inequality fails

[tidal turret]

yeah

[dawn dagger]

,, u\ge -2 , \lr , u+2 \ge 0 , \lr , e^{u+2} \ge 1 , \lr , w^2e^{u+2} \ge w^2 , \lr , u^2+w^2e^{u+2} \ge u^2+w^2

[tidal turret]

🔥

consider the disk u^2 + w^2 of radius 1, then, the following upper bounding holds

well we justified the upper bound, how to conclude the limit is zero

[dawn dagger]

well that's because u>=-1 implies u>=-2 but not the converse

[dawn dagger]

anyway this means

we can pick a neighborhood that satisfies u>=-2 or a subset of that

it can be a square, strip, or a nice disk

[tidal turret]

now what?

[dawn dagger]

idk

recall what we did and wanted

[dawn dagger]

first of all, what was the problem

[tidal turret]

we need another upper bound

[tidal turret]

prove that the limit approaches 0

[dawn dagger]

no i meant, the problem was there was no justification for your upper bound

that's why we went through this journey

now you should know

explain now why this upper bound works and is true

then showing the limit of the upper bound being 0 should be easy

[tidal turret]

the upper bound works if the radius of the disk is less than or equal to 2, the disk guarantee we cover all the neighborhood of between (0,0) within a radius of 2, which will cover all the possible paths?

idk if I am explaining myself correctly

[dawn dagger]

yea

that being said

[tidal turret]

you got to be joking me

[dawn dagger]

however it is still important to learn

not always find the best solution blindly

claiming something to be true without justification is always wrong

at least in math

[vague seal]

always

[dawn dagger]

and that at least is a lesson for future problems to keep in mind

that will save you

and make up the time you may feel have lost now

[tidal turret]

I didn't lost any time

I found it interesting the disk and neighborhood argument

is just, that this was my first weeks in university. I found it a little bit too rigorous, this simplification you did to the problem was better for my level of maths

[dawn dagger]

well it's for the better, what if you encounter a problem, that doesn't simplify nicely

you never know that from the getgo, but the good part is, you were trained for that

[tidal turret]

ok

how to continue from here?

[dawn dagger]

up to you, i am kinda done

i think said and explained everything what needed to be

you can now choose how to complete the proof, and then post it for verification

[brisk iris]

Sorry just a question, do you think this thing converges ?

[dawn dagger]

i checked with wolfram

[brisk iris]

Honestly just asking, it does ?

[tidal turret]

, w lim (x,y) to (2,3) of (ln (y-2) * (3x-6)^2)/(e^x*(y-3)^2 + (x-2)^2)

[grand pond]

Results provided by WolframAlpha

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

Requested by renatoxbz

[brisk iris]

Hmmm

[dawn dagger]

other than a factor 9 missing, it should go to 0

[brisk iris]

Thanks

[dawn dagger]

using this especially

it should be apparent

[tidal turret]

factor of 9 where?

[dawn dagger]

(3x-6)²=9(x-2)²

[tidal turret]

thanks

well, let me try to finish this, thing is tricky

[dawn dagger]

looks good, and i would be happy to see, if you could derive the implication

[tidal turret]

[dawn dagger]

what if i lied

[tidal turret]

e^(u+2) > 0 and w^2 >= 0

[dawn dagger]

yeah the premise is given

[tidal turret]

well

[dawn dagger]

well if you can't derive it, what is the justification for using it

[tidal turret]

idk how to prove it formally speaking

[dawn dagger]

e^(u+2)w^2 >= 0

there is somehow a u^2 term

[tidal turret]

you can just add u^2 both sides

[dawn dagger]

exactly

complete it

[tidal turret]

e^(u+2).w^2 >= 0
e^(u+2).w^2 + u^2 >= u^2

[dawn dagger]

so far what i wrote

[tidal turret]

[dawn dagger]

i gtg but in case you are actually hopeless: ||the rest is cross division (with no sign actually flipping because the terms are positive when dividing) ||

[midnight plank]

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[cedar pawn]

Have you drawn it?

[gilded current]

yes

[cedar pawn]

Something lik this?

[gilded current]

yess

I drew perpendiculars to the point b/w green and blue ylw and blue green adn yellow

like dis

[cedar pawn]

b/w?

[gilded current]

Don't mind the name because Imma delete it later

[cedar pawn]

I feel like this can just be brute forced

But I'm wondering if we can avoid that

I can't see an obvious way to do this otherwise

[gilded current]

how brute forcE?

[midnight plank]

@gilded current Has your question been resolved?

[midnight plank]

@gilded current Has your question been resolved?

[midnight plank]

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[snow dawn]

Here, your exponent n is \frac{1}{2}. Using the power rule \frac{d}{dx}(x^n) = nx^{n-1}: \frac{d}{dx}(x^{1/2}) = \frac{1}{2}x^{(1/2)-1} = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}} ​

[snow dawn]

Can someone please explain what this means? I don't get how the answer is 1/(2 sqroot x) instead of 3.x^3/2/(2)

[lyric charm]

you are supposed to lower the exponent by 1 not raise it

[snow dawn]

I lowered it by 1

1/2 - 1 is 3/2

[lyric charm]

1/2 - 1 is not 3/2

1/2 - 1 is -1/2

[lyric charm]

this should have raised a red flag: how did you SUBTRACT 1 from a number only for it to become BIGGER somehow?

[snow dawn]

Oh true that where be my spectacles

[lyric charm]

1/2 + 1 would be 3/2

[snow dawn]

That still doesn't solve the problem

Even after simplification

[lyric charm]

$\dv{x} (x^{1/2}) = \frac12 x^{-1/2}$

[grand pond]

Ann

[lyric charm]

does this make sense to you now yes or no

[snow dawn]

Here, your exponent n is $\frac{1}{2}$. Using the power rule $\frac{d}{dx}(x^n)$ = $nx^{n-1}:$ $\frac{d}{dx}(x^{1/2})$ = $\frac{1}{2}x^{(1/2)-1}$ = $\frac{1}{2}x^{-1/2}$ = $\frac{1}{2\sqrt{x}} $

[grand pond]

Normie

[lyric charm]

slight

but ok yes that's all correct what you're copying

where is your confusion?

[snow dawn]

Oh I see

We first multiply the power then subtract 1 from the power

[lyric charm]

... yes, that's what the power rule says.

[snow dawn]

I was like -1/2 multiplied by but ok

.close

[midnight plank]

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[noble briar]

could i have some help with volumes of revolutions?

[tawdry laurel]

post your question here and then we can help you

[noble briar]

im struggling with the under the x axis part

like understanding it

🥲🥲🥲

[topaz epoch]

you should convert into cartesian then use the formula for volumes of revolution

and it's revolved around the x-axis

[noble briar]

ye i did that

wait hold on ill sned what im up to

ignore the top part

the left was my attempt of trying to convert it to cartesian form

but it didn’t work out

so i found this formula

[topaz epoch]

you should convert the entire thing to cartesian form such that there aren't any t's remaining

[noble briar]

top right

[topaz epoch]

you can rewrite sin(t) in terms of cos(t)

[noble briar]

but it’s sin(2t)

oh wait what

[topaz epoch]

I'm talking about your y = 2x sin(t)

[noble briar]

ohh

oh yeye ok

i’ll try that

like that?

[topaz epoch]

$\sin^{2}\left(x\right)+\cos^{2}\left(x\right)=1 \implies \sin\left(x\right)=\pm \sqrt{1-\cos^{2}\left(x\right)}$

[grand pond]

Roy

[topaz epoch]

so we have y = 2x times what?

[noble briar]

its sin^2

[topaz epoch]

mhm

so you should end up with $y=\pm 2x\sqrt{1-x^{2}}$, $V = 2\pi\int_{0}^{1}\left(2x\sqrt{1-x^{2}}\right)^{2}dx$

[grand pond]

Roy

[noble briar]

where did the 2pi come from ?

is that the negative area part?

thats what im struggling to understand

the point that says

• the fact that there is now part of the shape under the x axis. how will that effect the volume

i cant visualise and undertsand the under the x axis aprt

hello?😭

um i asked ai

and they put a 2 at the front? 😭 im so lost

[mortal falcon]

0 to 1 only gives you half of the shape

[noble briar]

y?

[mortal falcon]

Look at the image

The object goes from -1 to 1

we just broke it in half at the origin

[noble briar]

oh yesyes i get that

but hows thtat linked to the part below the x axis

[mortal falcon]

we got the part below the x axis by revolving the curve around the x axis

the y = expression is the height above the x-axis right

if we revolve that around the x-axis, we get the entire volume “above” and “below”

[ruby depot]

Which way is under?

[noble briar]

oh?

its only above?

wait what

[mortal falcon]

Well okay the +/- gave you below too but you can think of it as revolving one or the other

bc it covers the entire rotation

[noble briar]

oh this

?

[mortal falcon]

I was going off of this

But yeah I assume that was used there

[noble briar]

ok then instead of going from 0 to 1 and doing -1 to 1

is what i did right?

cuz it’s +/-

so i split it up

but then that gives u

0

[mortal falcon]

Why did you make the second one negative

I don’t understand what you’re doing here actually

[noble briar]

cuz the equation before that was +/-

ok so

[mortal falcon]

if you’re integrating from x= -1 to 1, then you need a function describing y in terms of x from -1 to 1

What is that function

[noble briar]

2xsin(t)

and i want the volume around the x

so the formula

[mortal falcon]

if you actually have a proper function from -1 to 1 and you revolve it around the x axis then there is no second integral to compute

[noble briar]

integral of pi y^2

[mortal falcon]

The rotation is done around the x-axis, there is no additional “below the axis” to compute

[noble briar]

huh?

ok wait lemme redo everything

[mortal falcon]

your - integral doesn’t make sense

[noble briar]

oh wait

how do u get the cartesian equation from x=cos t and y=sin(2t

[mortal falcon]

the way I would do it is y = 2sin(t) cos(t) = 2sin(t)x = 2x sqrt(1 - x^2)

where the sqrt is +/- depending on t

[noble briar]

ok

[mortal falcon]

Luckily the sign doesn’t matter bc we’re squaring anyway

[noble briar]

huh

[noble briar]

ohhh this make sense

[mortal falcon]

This is clever but you should be careful with your bounds

you want t to span the curve exactly once along the x-axis

[noble briar]

right

this is with the cartesian form

[mortal falcon]

You put the +/- outside of the square again

It’s inside the square

[noble briar]

oh?

[mortal falcon]

sin(t) = either + or - sqrt(1 - cos^2 (t))

the sign depends on the quadrant t corresponds to

[noble briar]

right

ok

[mortal falcon]

But the point is that regardless, after squaring the sign is gone

[noble briar]

IH

OH

righttt i squared it omg

okok yes

?

did i do this correctly?

[mortal falcon]

I’ll assume you meant to change the +/-, the integral setup looks fine and from there it’s just power rule stuff yeah

[noble briar]

omg

ttytytyty

so like

for the 'under the x axis'

[mortal falcon]

the +/- corresponds to above and below the x axis, but bc we’re anyway rotating around the x axis it doesn’t matter what sign we take (and mathematically this is bc we square to get the area, we don’t care abt sign, just distance from the x axis)

[noble briar]

"the fact that there is now part of the shape under the x axis. how will that effect the volume. "

right ok

so it doesnt effect the volume?

[mortal falcon]

if we used the entire t range from 0 to 2pi then we would have double counted the volume bc the curve double counts

Bc of how we substituted we manually just picked the x range -1 to 1

so we only took 1 curve from -1 to 1 and didn’t double back ever

[noble briar]

ohh ok

[mortal falcon]

if you look at your parametric method, if you just set the bounds properly you would have gotten the same answer

The trick is that 0 to 2pi would make the answer different from the true volume

Bc each cross section at an x value only depends on the y coordinate once

[noble briar]

for the bounds in the parametric method

[mortal falcon]

but the parametric visits each x value twice

[noble briar]

do i have to solve sin(2x) for -1 and 1?

or is it cos(x)

[mortal falcon]

cos(x) would be easier

[noble briar]

right

[mortal falcon]

Like it’s clear that our period is 2pi bc of the cos

[noble briar]

so it wouldve been 0 and pi

[mortal falcon]

Since sin(2x) just halves the period

I think so

Equivalently pi to 2pi

And that method fully keeps the sign of our square root implicitly with the trig, it just again doesn’t rlly matter

[noble briar]

yes ok lemme try the parametric method

[noble briar]

right

[mortal falcon]

That seems not good

Why not decompose the sin(2t) into 2sin(t)cos(t) before doing the rest

[noble briar]

ok

[midnight plank]

@noble briar Has your question been resolved?

[grim forge]

I think after the 3rd step, you should have done the substitution u = cos t instead

[noble briar]

oh ok ill try

i kinda got the same hting?

[grim forge]

Can you show?

[grim forge]

sin^2 t = 1- cos^2 t

[noble briar]

omg ur right

omg yay

ok

@grim forge yay tyty

ok so um

i got that but the next part is the same but the volume revolved around the y axis

so its basically the same right

js finding the equation in terms of y

x=f(y)

[grim forge]

Ye around y- axis it should be int(pi x^2 dy) I think

[noble briar]

ye okok

i need help understanding this formula

this is for finding the volume revolved around the y axis

[midnight plank]

@noble briar Has your question been resolved?

[noble briar]

.close

[midnight plank]

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[steel cedar]

Why do i get a different answer for b. part

[steel cedar]

I honestly don't understand these arctan signs thingy

[midnight plank]

@steel cedar Has your question been resolved?

[steel cedar]

<@&286206848099549185>

[golden crescent]

h

[golden crescent]

signs like this right

[steel cedar]

I mean the arctan

The quadrant thing

[golden crescent]

oh mb

[steel cedar]

The only difference from the answer key is the phase

jeez

.close

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[calm junco]

for this, do you just have to expand the terms in the expansion and spot a pattern or something? I dont see how else to use 2.109 to show this result

[runic hamlet]

I would just multiply out both sides and hope everything fits

[calm junco]

as in everything on line 2.116 ?

[runic hamlet]

yes

[calm junco]

and sub in the expansions?

[runic hamlet]

yes

[calm junco]

ok

let me try

@runic hamlet

i only expanded up to power of 2

but im not sure what to do with this...

[runic hamlet]

hmm yeah ok that looks mildly painful

are you actually asked to show this

[calm junco]

nope, just trying to do it to justify it to myself lol

[runic hamlet]

yeah not sure

I would expect that this pops out if you spend some more time on it but I dont think thats a particularly good use of that time

(I'm talking on the order of hours. not minutes)

[calm junco]

alright, yh i will skip this for now then

book is quite dense so I dont want to spend too much time on small bits

[runic hamlet]

I didnt actually know that there was a version for non-commuting operators

[calm junco]

and then not be able to go over the more important bits

[calm junco]

it looks odd lol

thanks anyways

ima close it now then

.close

[midnight plank]

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[jade magnet]

20 = (n)(2a + d(n-1))/2
40 = (n)(2a+d(n-1))
40 = 2an + dn^2 - dn
0 = dn^2 + (2a-d)n - 40

idk what to do

2a isnt necessarily even if d and n is odd

oh wait

[lethal path]

pause right there, 40 = (n)(2a+d(n-1))

that implies that n is even and 2a + d(n - 1) is also even

since 20 is a multiple of 4

[jade magnet]

oh

oh

bruh

idk how i didnt notice that

ok nvm

ik how to do the question now

[lethal path]

well, that's one possible way to make a multiple of 4 (I know n can also be a multiple of 4)

no worries!

[jade magnet]

so either d is even or n-1 is even

[lethal path]

yeah so d can be even, exactly

[jade magnet]

but n-1 cant be even

so d has to be even

and n is even

so is it II and III only?

[lethal path]

but then you have this other case

[waxen willow]

n = 5 , a = 2 , d = 1?

[lethal path]

guessing numbers doesn't help

[jade magnet]

i mean either way

[jade magnet]

would this work

[lethal path]

but indeed, you should check what happens when n = 4 or n = 8

[lethal path]

you don't have the logic figured out yet

pause before you rush to conclusions

[waxen willow]

That is a counter example that n doesnt have to be even

[lethal path]

cool

then it'd be clearer if you had just stated that

[waxen willow]

2 + 3 + 4 + 5 + 6 = 20

[lethal path]

but appreciated

[waxen willow]

You cannot assume n is even bc LHS is even, especially if RHS has 2 factors

[lethal path]

yeah I realised, so we needed to start over to figure out case II

[waxen willow]

In fact for every question you should either prove it or give a counter example. It is easier to first try and find if there is a counter example and then if unable to , you should prove it.

[waxen willow]

This is also an example that III doesnt have to hold as common difference is 1 (not even)

[lethal path]

oh right, so yeah we really should have checked something simple like 9 + 11 = 20

so that means I is not true

[waxen willow]

Also i think 20 is more trivial counterexample

A sequence with n = 1

[lethal path]

then actually for II, you could just have the series {20} with 1 term

yeah

you realised too

damn yeah the quadratic is absolutely not the right approach

there are just too many free variables

[waxen willow]

I mean with so much freedom you can probably guess most of the things will be false

[lethal path]

exactly

wow

[lunar sleet]

Does this work?

[lethal path]

yes!

[midnight plank]

@jade magnet Has your question been resolved?

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[modern pendant]

Anyone else use McGraw hill connect before?

[modern pendant]

.close

[midnight plank]

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[grim forge]

Can someone tell me if this is a valid proof of De Morgan's Law? This is supposed to be a 11th grade level proof.

[lyric charm]

even if it's valid, all of those element listings are mega extraneous for sure

makes it hard to read, hard to verify, and easy to fuck up

[graceful drum]

what if the sets have an uncountable number of elements

[lyric charm]

yeah that too

i would say it is possible to salvage your idea by like... drawing a Venn diagram and acknowledging that U ends up partitioned into 4 (disjoint) sets & then expressing both sides of the equality in terms of those

[grim forge]

This is 11th grade. We're not even supposed to know about countably infinite ones

[graceful drum]

are venn diagrams allowed

[lyric charm]

you're not supposed to even know about the existence of sets like N or Z or Q?

naturals, integers, rationals respectively?

[grim forge]

Ok excpet those

[lyric charm]

that knowledge is considered illegal for you??

not even "the set of all even numbers"?

i think if you're in 11th grade and doing set theory then those things should be well within your comprehension

did your teacher punish someone for even speaking of such sets as a thing that might exist??

[grim forge]

Fine. We're not supposed to know about 'countable' and 'uncountable'. We're just supposed to know what infinity is

[graceful drum]

i mean you could prove it using venn diagrams as ann said or prove that any element of one set is an element of the other (X⊆Y & Y⊆X then X=Y)

[lyric charm]

still, your proof relies on finitude and that's completely unnecessary and just makes your life worse.

[grim forge]

I don't think they are. I asked the teacher and she said she preferred one without diagrams

[grim forge]

Ok

[lyric charm]

dispreference is not the same as forbidding.

[grim forge]

Ok fine. Forget about the teacher. Forget about uncountably infinite sets. Could you just explain how to do it without diagrams?

[lyric charm]

sure

do you know how to prove statements about set inclusion?

i.e. S ⊆ T

[grim forge]

Yeah

Uh actually, could you give an example of what you mean?

[lyric charm]

i mean such statements as $A \cap B \subseteq A \cup B$

[grim forge]

Just so you know, I was taught the basic operations, like union intersection, subtraction, complement. Thats it. And also subsets, I guess

[grand pond]

Ann

[grim forge]

Oh yeah

[lyric charm]

right then you should also know that to prove some two sets are equal you prove their subset-hood in both directions

[grim forge]

Yes

[lyric charm]

that's what cherryman was saying

[grim forge]

Ok. So pick an element in one set and use it to show that for all elements of that set, those elements are also present in the other set?

And do the same in the other set

[lyric charm]

that's a somewhat wonky way to write it...

[grim forge]

Im not very good at this proofs thing...

[lyric charm]

for your specific case, the proof for (A ∪ B)' ⊆ A' ∩ B' needs to look like this:

Let x ∈ (A ∪ B)'.
[some arguments later]
Therefore, x ∈ A' ∩ B'.

[grim forge]

Understood

Then I do the same thing for the other way round right? After that I can say
=> (A union B)' = A' intersection B'

Anyway, thanks @lyric charm and @graceful drum . We have a exam tmr, and nobody in my class knows how to prove it, so I decided to come here for clarification.

.close

[midnight plank]

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[midnight plank]

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[neat sun]

Here say d²f/dx1dx2 do we apply derivative wrt x1 first or x2

[crude pier]

x1 first

[neat sun]

U sure?

[dawn dagger]

oh wait no

[crude pier]

yeah looking into it as well and apparently not

[dawn dagger]

,, \f{\p^2f}{\p x_1\p x_2} = \f{\p}{\p x_1} \lp \f{\p f}{\p x_2}\rp = f_{x_2 x_1}

[crude pier]

cringe notation if I'm being honest

[neat sun]

Hmm

Oki

Thanks

.close

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[tidal turret]

a) find the domain of f
b) graph f using ggb
c) is it possible to extend f to the circle C such that it is continuous, if so how and why?

[ruby depot]

Which one are you asking?

[tidal turret]

everything if possible

[thin cypress]

The domain of f is where f exists, in other words what values of x and y make the function have a valid output?

[ruby depot]

To find try thinking backwards.

[tidal turret]

R^2 except when 1 - x^2 - y^2 = 0

[thin cypress]

yes!

[ruby depot]

What is your problem with b?

[thin cypress]

Do you need help with graphing using GeoGebra?

[lyric charm]

i mean the geogebra thing is like "open the 3D canvas -> type the function in" innit

[tidal turret]

did we solved A correctly?

[tidal turret]

yes :D

[ruby depot]

Yes

I would like to ask you what shape it has. The undefined region.

[tidal turret]

a circle

with radius 1 and center origin

[ruby depot]

Yes using this you can plot the shape of it with your hand too which you can try.

[tidal turret]

?

[ruby depot]

I mean like how should the graph behave.

The circle doedn't mean much here but it is useful what I want to say is.

[tidal turret]

idk how to graph f(x,y) by hand

[ruby depot]

Sorry let me rollback a second.

[ruby depot]

We can kinda do it. Want to try?

[tidal turret]

sure

[ruby depot]

Alrgiht, first as a starting point and a hint does x^2 + y^2 ring a bell?

[tidal turret]

wdym?

[ruby depot]

What comes to your mind when you see x^2 + y^2

[tidal turret]

you want to graph a circle?

[ruby depot]

Have you heard about the polar cords?

[tidal turret]

what?

[ruby depot]

Polar coordinates.

What you are used to is Cartesian coordinates which has two "length" basis x and y. Polar coordinates in the other uses theta and r, one length(r) and one angle(theta).

[tidal turret]

whats your point

[ruby depot]

The conversion between the two are as fallows:

$$x = r \cos{\theta}\ ,\ y = r \sin{\theta}$$

[grand pond]

Qwert

[ruby depot]

And the other way around is:

$$r = x^2 + y^2\ ,\ \theta=\tan^{-1}{y/x}$$

[grand pond]

Qwert

[ruby depot]

Sorry I meant r^2 = x^2 + y^2

If we the conversion between them the function becomes:

$$f(r, \theta) = \frac{20}{1-r^2}$$

[grand pond]

Qwert

[ruby depot]

Which also means the function does not depend on the theta/angle.

[tidal turret]

can you elaborate?

[ruby depot]

It really is bit of a complicated topic to fully go trough here it was not the question so I am going to keep it short.

Normally you would have a point that is defined by the distance from the origin in both directions (x,y). But in polar coordinates we the point have a distance from the orgin directly(the shortest path/line) and the angle it is at.

[ruby depot]

Yeah looks good.

[tidal turret]

now what?

[ruby depot]

As I stated the function does not depend on the angle.

So for now lets think a slice of the function theta = 0

And here r is not negative(not because of the square, the distance is always positive)

So only think about the right side of the graph. Positive r

[tidal turret]

can you do a drawing or something

[ruby depot]

I will hold you a minute.

[tidal turret]

ping me

[ruby depot]

Sorry about my hand writhing but here:

@tidal turret

[tidal turret]

what?

[ruby depot]

I am sorry. Let me try to explain it once more. Did you understood the polar coordinates first?

[ruby depot]

Zoom in maybe?

[tidal turret]

its not a cone

[ruby depot]

Yes I ploted g(x,y) = x^2 + y^2 not your function.

[tidal turret]

put it in ggb

[ruby depot]

We can see what we want it is true.

[tidal turret]

what?

[ruby depot]

Sorry the square root of it.

Not r^2 just r

I believe you are bit puzzled...

[tidal turret]

how so?

[ruby depot]

Probably I was not able to explain it.

Can you quickly go trough the question again? Writing it out may help.

[tidal turret]

wdym?

[ruby depot]

I mean like try solve it from the start maybe?

To help you understand how to do it.

I am sorry 🙁

[tidal turret]

sorry about what

[ruby depot]

Sorry that I wasn't able to explain it.

[tidal turret]

explain what

[dawn dagger]

Renato doesn't even know what's going on

[ruby depot]

😭

Ok, did you understood the polar corrdinates?

[dawn dagger]

tbh I would have just went with cartesian

[ruby depot]

I felt like using polar is more natural but maybe.

[dawn dagger]

yea i see

[tidal turret]

I dont think I follow

what should I do for c?

[ruby depot]

c will come with this too but you can take the limit or the function is simple enough to see it.

[tidal turret]

what?

[ruby depot]

Ok we are not using the limit either, that is calculus stuff. For this question it is not needed.

Idk at this point, do you know calculus?

Ok ok lets go back to basics and not care about drawing it ourself.

[tidal turret]

what limit?

[ruby depot]

Limit of the function as you aproach the circle.

[tidal turret]

like for example (1,0)

[ruby depot]

Yes.

[tidal turret]

,w lim (x,y) to (1,0) of 20/(1-x^2-y^2)

[grand pond]

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[tidal turret]

whats your point?

[ruby depot]

C is you can not.

That is the answer.

From one side it goes to -inf from one to inf.

[tidal turret]

there is no lateral limits here

[dawn dagger]

The limit in polar cordinates seems to be more useful, i believe

[tidal turret]

how so?

,, \lim_{r \to 1} \frac{20}{1-r^2}

[ruby depot]

Rather then for a point (x,y) -> (1,0) you do the whole circle r->1

[grand pond]

Renato

[tidal turret]

ok that is cool

[ruby depot]

Well then this is the end of this question. Learning more about polar coordinates is useful so try giving it some of your time. Hope I was able to help. Have a great evening.

[tidal turret]

polar cords is hard dude

[ruby depot]

But it is really is useful and makes questions easier if you know it.

[tidal turret]

.close

[midnight plank]

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[unkempt skiff]

how do i do the last question

[unkempt skiff]

i thought it's area under the curve

for

vt grpah

[small jasper]

Displacement is the signed area, distance is the total area regardless of sign (aka you add the area regardless of if you're above or below the t axis)

idk which one you did

[unkempt skiff]

this was what I did

wait

was I supposed to to 0-2 interval then do 2-5 interval

[small jasper]

,w 6t^2 - 42t + 60 = 0

[grand pond]

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[small jasper]

yes

you're supposed to split at t=2 and t=5 for the reason I said above

[unkempt skiff]

oooooohhhh

I see

Ty :D

[midnight plank]

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

[unkempt skiff]

so I need to do 3 intervals right

[unkempt skiff]

0-2, 2-5, 5-14?

[small jasper]

correct

[unkempt skiff]

Thx

omg

[unkempt skiff]

I got the erong answer

uhhh

Or is it supposed to be absolute?

[midnight plank]

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

[unkempt skiff]

yep i got it

it was total rea

total area

[midnight plank]

@unkempt skiff Has your question been resolved?

[midnight plank]

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[surreal moon]

!status

[midnight plank]

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

[median urchin]

uh

i got 60 and 120

<@&286206848099549185>

[gentle hornet]

What have you tried

[median urchin]

extending EF and DC to intesect

labeled it as M

FM and CM are both 10

[gentle hornet]

How did you get this?

[median urchin]

pythagorean theorem and angle bisector theorem

[median urchin]

idk how to use

[gentle hornet]

Damn this is a good question

[median urchin]

lol

[gentle hornet]

Well you could always just bash

You have an angle

[median urchin]

how?

[gentle hornet]

In DEM, it's a 3-4-5 triangle

[median urchin]

ye

[gentle hornet]

Where the angle opposite to the 3 side is always 37°

[median urchin]

yea

[gentle hornet]

So just find all angles and use slight trigo to get the other 2 sides

It should work imo

[median urchin]

yea that the problem

idk trig

[gentle hornet]

Oh you're supposed to solve with just geometry?

[median urchin]

yea

[gentle hornet]

Hmm

Yeah I see this working with trig but I'm not sure about geometry

OOH

WAIT

Triangles EBF and FCM are similar

By AA criteria as EB is parallel to CM

[median urchin]

ok

[median urchin]

so?

[gentle hornet]

Crap that doesn't give anything too useful

I'm stumped man sorry

Maybe ping helpers again

[median urchin]

ok

<@&286206848099549185>

[midnight plank]

@median urchin Has your question been resolved?

[median urchin]

<@&286206848099549185>

[median urchin]

<@&286206848099549185>

[vocal warren]

can you use trig or nah?

[waxen willow]

Oh its this problem again

[median urchin]

yea'

[median urchin]

no

[vocal warren]

what can you use

sine theorem?

[median urchin]

ye

as long as u walk me thru it

[vocal warren]

bro wdym

so then also trig

if I walk you through it

anyway I will give this a try

[median urchin]

don't makeit too hard

[median urchin]

kk

[brisk iris]

I think I found a way but it’s super complicated

[median urchin]

ok

[brisk iris]

Do you want me to guide you through it anyway ?

[vocal warren]

have you found the sides?

FC is 2sqrt10

[median urchin]

ye

[brisk iris]

@median urchin ?

So my strategy is splitting the área in three

Área 1 is obviously EFD. Easy to find

Área 2 is DFC

[median urchin]

oh

[humble wolf]

,rccw

[median urchin]

kk

[brisk iris]

Area 3 I’m still thinking

My god … I’m dumb

I found ND, it’s easy

Ok Forget about splitting áreas

[median urchin]

ok..

[brisk iris]

What you have to do is find ND. Then it’s over

[median urchin]

yea

[brisk iris]

Cause Área = ND * AB and we know AB

[median urchin]

how do u get it

like

[brisk iris]

Ok let’s take it step by step

[median urchin]

ye

[brisk iris]

Look at triangle EDF, can you find the Value of the angle EDF ? Using trig

[median urchin]

no

looks like 30

[brisk iris]

Of course you can

Use tan EDF = 6/12

Did you get it ?

[median urchin]

2/root(5)?

[brisk iris]

I Dunno in degrees it’s around 26

Is It what you found ?

[median urchin]

ig

[brisk iris]

?

[median urchin]

ye

[brisk iris]

Good

Now can you give me the Value of angle NDE ?

[median urchin]

128

[brisk iris]

How ?

[median urchin]

NDCadds to 90

oh

38 mb

90-52

[brisk iris]

Perfect it’s not exactly 26 but you get the idea

[median urchin]

yea

[brisk iris]

Now using triangle NDE, you can find ND using trig

You have DE = 12 and the Value of the angle

[median urchin]

ok

then?

[brisk iris]

You found ND, it’s all you need

Your área is ND*AB

[median urchin]

yea but how i find it

[brisk iris]

Using this

[median urchin]

hm

[brisk iris]

cos NDE = ND/12

[median urchin]

okay but wait

im not good at trig lol

[brisk iris]

You just found angle NDE Right ?

[median urchin]

ye

i got 50 answers that why

[brisk iris]

You didn’t answer this

[brisk iris]

Wdym ?

[median urchin]

cos is adjacent over hypotenuse right

[brisk iris]

Yes and adjacent is ND hypothenuse Is DE

[median urchin]

okay

[brisk iris]

Give me the Value of cos NDE

[median urchin]

ye im doing

15?

[brisk iris]

Ok let’s back up

You know the Value of NDE

[median urchin]

yea

[brisk iris]

Use your calculator to compute cos NDE

[median urchin]

ok

0.119

[brisk iris]

Good

[median urchin]

no

.95507364404

i put 58 not 38

[brisk iris]

It’s supposed to be 0.788

[vocal warren]

is the whole area 48?

[median urchin]

oh

ye'

[brisk iris]

Start by computing the cos what did you get ?

[median urchin]

.7880107536

multiply them?

[vocal warren]

do you know the answer?

[median urchin]

no

[brisk iris]

Good now multiply by DE = 12

[median urchin]

9.45612904328

[brisk iris]

Now That’s ND

[vocal warren]

@brisk iris did you calculate the area?

[median urchin]

so its 94?

[brisk iris]

And the área Is ND*AB

[median urchin]

ye 94

[brisk iris]

No I didn’t calculate

I just found the way to

[median urchin]

looks wrong

[vocal warren]

cause I get something different than you

[median urchin]

ye i dont think its 94

[vocal warren]

you dont know how much it is bro...

[brisk iris]

But the reasoning stands

Do you want me to do the math ?

[vocal warren]

no it's fine but I get 48 even number

I did it without calculating the sines and cosines

[brisk iris]

How Did you donit ?

[vocal warren]

what I did is:

1. find DF
2. find FC using cosine theorem
3. find angle DFC with sine theorem (it is 45 degrees)
4. I found angle DCF (pi - angles)
5. DAB is the same as DCF and ADC is pi - DCF so I have all the angles in triangle ADE, then use sine theorem to find AD, after AD I construct the height from D and use that right triangle to find the height

just check if you also get DFC as 45 degrees, everything else should be correct

the angles I got for the parallelogram are pi/4 + theta and 3/4 pi - theta

[brisk iris]

But you followed my reasoning, it’s fast and I don’t see any mistake

[vocal warren]

but I get 48 and you get 92

[brisk iris]

Ok let’s challenge this

Triangle DEF has an área of 12*3

Which Is 36

[vocal warren]

half of that

but you can imagine that the whole thing wont be more than 72 then

[brisk iris]

Yup

Let me check the math

[vocal warren]

alright, dont worry, I might be wrong too

[median urchin]

its not 48 its not 94

[vocal warren]

wasnt it 92

[vocal warren]

how much is it then lol

[median urchin]

idk\

[vocal warren]

how you know it's wrong then?

[median urchin]

i verified

[brisk iris]

Try 96

Please

You used too many approximations

96 should work

@median urchin

[median urchin]

uh

yay

5th try

ty lol

[median urchin]

lemme put solutuin

[brisk iris]

It was just too many approximations

[median urchin]

dw

ima read them both

[vocal warren]

lol

[brisk iris]

This is a fast way to get the answer

[vocal warren]

totally forgot about those theorems

[brisk iris]

And an advice to avoid approx errors. Leave the whole formula and compute last minute

[midnight plank]

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[meager ore]

Prove that if $\mathcal{A}$ is a basis for a topology on $X$, then the topology generated by $\mathcal{A}$ equals the intersection of all topologies on $X$ that contain $\mathcal{A}$.

[grand pond]

Halex

[meager ore]

T(A) is a topology and also contains A

[vocal laurel]

can anyone explain 1 of these to me. Im confused as to what the final answer is suppose to look like

[jaunty ivy]

!occupied

[midnight plank]

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).

[pearl hull]

Claim your own channel

[vocal laurel]

oof sorry

[midnight plank]

@meager ore Has your question been resolved?

[midnight plank]

@meager ore Has your question been resolved?

[midnight plank]

@meager ore Has your question been resolved?

[midnight plank]

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[inland patio]

Consider a nondecreasing, right continuous and say bounded function $F$ on $\mathbb{R}$. Let $(x_n)$ be a real sequence converging to $x_0$. Is it true that $$\liminf_n F(x_n)\leq F(x_0)?$$ My current approach to show this gets very cluttered. What I do is consider cases of how $(x_n)$ behaves. Is there a more straightforward way?

[grand pond]

psie

[subtle blaze]

Does F = 1_(x>=0) break this

Consider the sequence 0, 0, 0, …

Oh it doesn’t

Ah I see

[inland patio]

Yes, it holds I think.

[subtle blaze]

If you have a convergent sequence can’t you just take a monotonic subsequence that also converges

Would that make it easier to do cases?

[inland patio]

well, yes and no, we might have that the sequence is eventually <= x_0, so there is no monotonic subseqeunce that decreases to x_0 so that we can't use right continuity 😔

[subtle blaze]

Wait how

Between (-inf, x0) and [x0, inf) at least 1 of these interval must include infinitely many terms of the sequence

[inland patio]

yes, if it's only (-inf,x0), then we are in the case I was considering. The sequence might also oscillate around x0, and so have infinitely many terms in both intervals

[subtle blaze]

If a point is on the left it automatically holds by non decreasing

If it’s on the right you know it’s continuous so it’ll get arbitrarily close to F(x0)

[inland patio]

and if it's in both?

[subtle blaze]

You can just delete all the points on the left

Now they’re all on the right

[inland patio]

but there might be infinitely many on the left?

[subtle blaze]

Including points that are equal or less than F(x0) can’t make the liminf greater than F(x0)

Since F is non decreasing

[inland patio]

ok

[subtle blaze]

You can then take a sledgehammer of ||right continuity as a topology then show decreasing sequences converging to x0 would converge under the image of F since F is continuousy||

[inland patio]

I'm still a bit unsure how we can just delete the points on the left if there are infinitely many of them.

by the way, this claim about liminf and F I saw here about the quantile function, see Fact 4 where this claim is made 🙂

[subtle blaze]

Let x_n be a sequence, inf{x_n} <= liminf {x_n}

Do you agree

[inland patio]

wait, let me think

[inland patio]

inf{x_n} is an increasing sequence and so I would say yes, I agree

[subtle blaze]

Okay

So if I have a sequence x_n, and inside that a subsequence y_m

inf x_n <= inf y_n

Wait wrong way

[inland patio]

yes, makes sense

[subtle blaze]

Let’s call these sets X and Y, suppose that X \ Y >= Y

We have inf {X} >= inf {X \ Y}

Fuck I’m getting these signs going the wrong way all the time

If no one else says anything lemme figure out what exactly I want to say first gimme a moment

[inland patio]

no worries :), but which sets are X and Y now?

[subtle blaze]

Liminf X = min(liminf X \ Y, liminf Y) <= liminf X \ Y

Something like this I think

This is X and Y being the ordered sets of the sequence

Y is a subsequence of X

[inland patio]

Ok.

[subtle blaze]

Y are the points on the left of x0

X is all the points in the sequence

[inland patio]

but recall we are considering liminf of the sequence F(x_n), not of x_n

[subtle blaze]

Ya but then you use the sledge hammer of right continuity

[inland patio]

ah ok

[subtle blaze]

The image of convergent sequences under continuous maps are convergent

If you show a decreasing sequenc converges then the image (under a right continuous map) will as well

You can definitely go through the epsilons and deltas if you wish but I would rather not 💀

[inland patio]

how do we use this inequality then exactly?

[subtle blaze]

The liminf of the whole sequence X is less than or equal to liminf of the sequence missing the points less than x0 (the set Y)

[subtle blaze]

That’s what you asked here

We show liminf X \ Y is F(x0) by continuity

So then liminf X <= liminf X \ Y = F(x0)

[inland patio]

ah ok, so X \ Y are all the terms F(x_n) with x_n on the right of x_0, is that correct?

[subtle blaze]

No not yet

Without the F

[inland patio]

Hmm, I think when there are infinitely many terms in (-inf, x0) and [x0,inf), I still struggle understanding what we have to do exactly. Are you trying to say that we are taking a subsequence with only terms in [x0,inf)?

[subtle blaze]

Yeah

[inland patio]

ok, but there would be subsequence with terms eventually in (-inf,x0) too. We have to consider those too.

ah wait, we did perhaps already

[subtle blaze]

But those don’t matter because F is non decreasing

If at some point, none of your points in the subsequence is above x0 your limsup cant even be greater than x0 let alone your liminf

[inland patio]

Ok. So to summarize; either there are infinitely many terms in (-inf,x0) -- we use non-decreasingness of F. Or there are infinitely many terms in [x0,inf) -- we use right-continuity. Or there are infinitely many terms in either interval, and then any subsequence will eventually be one of the two options we have already considered. Is that correct?

[subtle blaze]

Hmm

We first start by using the fact that if a sequence converges, then every subsequence also converges

That’s how we can specifically analyse the cases

[inland patio]

indeed

[subtle blaze]

Now we have that liminf of X <= liminf X \ A for any A ⊂ X

We also have that X converges so liminf X = lim X

So liminf X \ A = lim X as well

So if we remove stuff from X (removing A) so that X\A is either of the cases you wrote down, then bla bla bla

Either X < x0, or X > x0, or both, at some point in the sequence

The first one is simple by nondecreasing F

[inland patio]

right

[subtle blaze]

If X >= x_0 then there is a subsequence that’s monotonically decreasing, whose liminf is equal to liminf X because X converges

If it’s both, you can delete the < x0 part of the sequence then you’re in the X >= x0 case

[subtle blaze]

I guess we still ended up doing cases in the end

Maybe we can try the full continuity way?

X converges to x0 implies liminf X = x0

Nah I guess it hinges on both right continuity and non decreasing

[inland patio]

yeah

[subtle blaze]

If X converges wrt std top. then X converges wrt right cont. top. and since F is continuous, F(X) converges wrt right cont. top. so liminf F(X) = lim F(X)

[subtle blaze]

The < is only for when sup X = x0 right?

Yeah I’m not 100% sure how to argue this properly

[inland patio]

Hmm, I think we simply have < when the limit F(x_n) doesn't exist, or?

[subtle blaze]

https://en.m.wikipedia.org/wiki/Càdlàg oh goodness me this is a whole thing

[subtle blaze]

Oh yeah we wouldn’t know what to do about increasing sequences

Yeah I’m not sure :/

[inland patio]

Well, thanks for the help nevertheless.

.close

[midnight plank]

Channel closed

Closed by @inland patio

Use .reopen if this was a mistake.

[midnight plank]

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[minor heath]

HELPPPPPP

THIS IS NO CALCULATOR BTW

[hasty herald]

cooked

[rare maple]

Do you need help with all the parts

[minor heath]

yes

pls

[rare maple]

For the first part did you try graphing the function

[minor heath]

no calc

[rare maple]

Without calculator

[minor heath]

exam question no calculator

like

[rare maple]

A rough sketch

[minor heath]

you only have pen and paper

ohhh

ok

intercept

is y = 1