help-49

[languid mica]

Like

[blissful pier]

(\sum _n 5a(n)=5\sum_n a(n))

[grand pond]

PajamaMamaLlama

[humble torrent]

Well write out the sum and see

[languid mica]

Sum of xn + x^n

[blissful pier]

like that?

[fallow scarab]

https://en.khanacademy.org/math/ap-calculus-ab/ab-integration-new/ab-6-3/v/sigma-notation-sum

[languid mica]

How would i write it on the right side

[fallow scarab]

https://math.libretexts.org/Bookshelves/Precalculus/Precalculus_(Stitz-Zeager)/09%3A_Sequences_and_the_Binomial_Theorem/9.02%3A_Summation_Notation

[languid mica]

(\sum _n a(n) + b(n) =\sum_n a(n) + \sum_n b(n))

Like this ?

(\sum _n a(n) + b(n) =\sum_n a(n) + \sum_n b(n))

Fire

[blissful pier]

that is also a property of finite summation, yes but read more from Riemann's links

[languid mica]

No like

[blissful pier]

\(a\)

[grand pond]

PajamaMamaLlama

[languid mica]

Is sum a(n) + b(n(

Aplying sum on b(n) aswell

Or i need to put parenthesis to show i wanna include b(n)

[fallow scarab]

read

[languid mica]

Thats basically what im asking

[languid mica]

Bro

Read the question

[blissful pier]

depends

[languid mica]

I dont wanna know identities

[blissful pier]

for that I usually include parentheses

[languid mica]

I wanna know what the notation means

[blissful pier]

to clear up any ambiguity

[languid mica]

Like example

For sin operator

We use sin(something)

And sin is applied to everything in parenthesis

[blissful pier]

you're asking about:
(\sum_n(a(n)+b(n))) vs. (\sum_n a(n)+b(n))?

[grand pond]

PajamaMamaLlama

[fallow scarab]

explained at the top

[languid mica]

No bro

Like sin(something)

[blissful pier]

this?

[languid mica]

What sin is applied on ends on the )

On sum where does it end?

[fallow scarab]

and below the identities

[languid mica]

Yes

Are they the same?

[autumn canopy]

Yes they are the same

[blissful pier]

it would depend on context

but most of the time yes

[languid mica]

Oh ok

So sum is just applied to anything thats on the right of the Σ

[autumn canopy]

If n is coming from the sum and you don't have it anywhere else, it's pretty clear

[blissful pier]

but if you're writing it, I prefer to use the former to clear up any ambiguity

[languid mica]

Whats that with parenthesis?

Like here are outside parenthesis needed to show that C isnt included in sum?

Or not

I put cause i thought sum doesn't end its just whatever is on the right of Σ

[autumn canopy]

Yes, this is how I would write it, in the second picture you might leave off the inner () because you have a_k in the second summand and it's clear the k could only come from the sum

But it's still better like this

[languid mica]

Ok thanks

[autumn canopy]

Some people might write stuff like $\dst \sum_{k = 1}^\infty \arctan\l(\frac{1}{k^2}\r) + 5$ though, without worrying about this

[grand pond]

Kepe

[languid mica]

In this expression

[autumn canopy]

(This would be written badly but from context it should become clear)

[languid mica]

Is 5 included to the sum

Or no

[blissful pier]

exactly the ambiguity good parentheses resolves

[autumn canopy]

There is no yes/no here, it depends on context entirely, this is written badly

[languid mica]

Oh ok

[autumn canopy]

You look at the line above this

[languid mica]

Yea got it thanks

Another thing

[steel crest]

I don't see any reason to think the 5 is part of the summation?

[languid mica]

been hearing about Γ and ζ function what are those in ? Like course

Idk if i should have learnt them already or they are later on

[autumn canopy]

Yeah, surely it wouldn't be included there if we are reasonable

[blissful pier]

it could be, depends on the context, even if it is ridiculous it's a demonstration

[steel crest]

analytic number theory

and other places

[blissful pier]

those are special functions but if you're only in Calc III from your previous questions you don't have to worry abt them rn

[autumn canopy]

Gamma(x) is the generalization of the factorial

zeta usually the Riemann zeta function

[blissful pier]

(\Gamma (n)=\int_0^\infty t^{n-1}e^{-t}dt=(n-1)!) and (\zeta(s)=\sum_{n=0}^{\infty}\frac{1}{n^s})

[grand pond]

PajamaMamaLlama

[languid mica]

I just finished calc 3 theory

[blissful pier]

(might've messed up the integral but those are the defintiions)

[languid mica]

Im solving challenging for me problems so i learn everything to the fullest

And ill then move on to a new

Idk which prob complex analysis

[blissful pier]

complex analyssi is usually a higher divison course

I might suggest LinAlg

and if you're big into theory then Axler

my university used Lay's Linear Algebra

[languid mica]

I already did

Idk what level i have completed

But i did most matrix stuff and transformations etc

[blissful pier]

subspaces, vector spans, orthgonality, etc too?

[languid mica]

Are those linear algebra 1 or 2?

I probably only did 1

[blissful pier]

I haven't seen linear algebra split into multiple classes?

[languid mica]

Oh

[blissful pier]

so I am unfamiliar with the terms and wouldn't be able to comment

[autumn canopy]

For me it was split up into two (Germany)

Those would be middle-late la1

[shadow schooner]

split in two for me too

[languid mica]

In Aristotelian university here in Greeceit is

I mean those i think are same semester with complex analysis

For this university

[autumn canopy]

la2 is multilinear algebra (properly defining the determinant, dual spaces, tensor stuff)

[languid mica]

Like la1 is second semester

La2 is 6th-7th

[autumn canopy]

huh

[languid mica]

Same with conplex analysis

[blissful pier]

Ahhh I see, alr yeah I haven't heard it called la2 before

[autumn canopy]

The canonical forms, singular value decomposition

diagonlization as a special case was la1

[languid mica]

Where would i study that

[autumn canopy]

Here la1 is first semester, la2 second

[languid mica]

You know any website first saying definitions and then theorems then afterwards giving problems to solve

[languid mica]

la2 might be different here idk

[autumn canopy]

Are you a math major?

[blissful pier]

idk of any websites, but you've kinda just described a textbook

[languid mica]

Im not in uni yet

[languid mica]

Yes online ?

I used math.lamar.edu or wtv

[blissful pier]

https://broman.dev/download/Linear Algebra and its Applications 5th Edition.pdf

[languid mica]

Had most def/theorems

[blissful pier]

Here's Lay's Linear Algebra

[autumn canopy]

Sheldon Axler's book is put up online

[languid mica]

Oh ok

[autumn canopy]

Though it's pretty big

[blissful pier]

Axler's Linear Algebra Done Right

[languid mica]

Wdym pretty big

[autumn canopy]

Like 400 or so pages

[languid mica]

I thought la was small course

Like la1

[autumn canopy]

la1 is one semester, same as la2

My lecture notes were ~100 pages respectively

[languid mica]

Just finished highschool so idk im kinda just hoping that the websites i use are valid

[autumn canopy]

So a total of around 200 pages

[languid mica]

Damn

I only did a lot of transformations and matrix stuff equations etc

[autumn canopy]

Yeah that's good, sounds like the beginning of la1

[languid mica]

Yea i thought the rest is much more advanced since wasnt in the book i was using

So i just thought id wait for later on

[autumn canopy]

Also it depends on if you do actual abstract linear algebra like a math major or just the computational version of it

I heard for engineering students etc. they have a computational version

[languid mica]

Nah i am going to be in engineering uni in octomber

But till them

I just wanna do pure maths

For fun

[autumn canopy]

cool!

[languid mica]

Im probably gonna keep doing math till i finish all courses in math uni even if its after engineering

[autumn canopy]

By the way, this is a classic: Prof. Strang has video recordings of his MIT lectures https://ocw.mit.edu/courses/18-06-linear-algebra-spring-2010/video_galleries/video-lectures/

You might complement this with Axler's book

I think Strang doesn't do any dual spaces or tensors for example

[languid mica]

Is this all la 1?

Like will i learn all of la1 if i follow all those vids

[autumn canopy]

I'd say this covers la1 and a little bit of la2

[languid mica]

Ok

Ill follow this course

But when i get home im away rn and its not good on phone prob

[autumn canopy]

Atleast that's how it should be roughly at my uni

[languid mica]

Rn i just do random stuff

[shadow schooner]

imo parts of LA empowers a lot of math both more applied things like stats/optimization/differential equations and more pure stuff like functional analysis, differential geometry, even abstract algebra. The list goes on.

[languid mica]

Im doing int a/x^n +b dx rn

[autumn canopy]

Yeah functional analysis is basically just linear algebra and then trying to apply a bunch of analysis to it

[languid mica]

Where n=2k kEN

[languid mica]

Mk

[languid mica]

Can you dm me this

Link

So i remember

.close

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[analog carbon]

i know its not hard i js forgot how to do this

[modern sapphire]

there is only a handful of circles related theorem in school maths. Which one do you think fits this situation?

You have two chords intersecting, and you know the lengths of three parts

[analog carbon]

hmmm

abbc = cddc

[modern sapphire]

if you want me to validate your guess, then it would be better if you specify what ab, bc etc. are

[analog carbon]

Ok

soooo a is 2 and b is x and c is 3 and d is 4

[modern sapphire]

umm, that seems so wrong

[analog carbon]

idk bro

[modern sapphire]

can you answer this?

[analog carbon]

\😭

[analog carbon]

wait maybne its like

a^2 = dc

[modern sapphire]

Do you know a theorem called power of a point?

[analog carbon]

no

[modern sapphire]

oh okay, how about similar triangle?

do you know triangle similarity?

[analog carbon]

all angles of a triangle add up to 180

degrees

[modern sapphire]

umm, thats true for all triangles. Similarity is like when two triangles are same shape but different sizes, like one is a zoomed out/scaled version of the other

[analog carbon]

oh like kinda like tranformations

[modern sapphire]

you seem all over the place

can you tell me what grade this is?

[analog carbon]

geometry

circles

[modern sapphire]

well duh

[analog carbon]

i think its geometry hnrs

[modern sapphire]

let me redraw the diagram

[analog carbon]

alr

[modern sapphire]

if you remove the third chord that has z, the remaining figure would look something like this

[analog carbon]

ohh so x=z

[modern sapphire]

no

i didnt say that

[analog carbon]

u dilated the circle

[modern sapphire]

i did no such thing

i am only ignoring the third chord for the time being since it has no use in finding x

[analog carbon]

ohh ok

then how'd u get 6

[modern sapphire]

the length 6 comes from the fact that 6 = 4+2

[analog carbon]

ye

ok

i understnad

[modern sapphire]

these two combined are 6 units long

[analog carbon]

ye

[modern sapphire]

so uhh, yea

if you join these points

[analog carbon]

then do i do b(a+b) = d(c+d)

[analog carbon]

ohh

ae times ed = be time ec

[modern sapphire]

can you tell me what the a,b,c,d you say are?

[modern sapphire]

oh ok

[modern sapphire]

yea exactlyl

[analog carbon]

ty broo

saved my life

[modern sapphire]

thats the power of a point theorem

[analog carbon]

eliminate

and simplify

i got it

many thanks

[modern sapphire]

just curious, since you dont know the name, what do you call this result, if not power of a point?

[analog carbon]

the x value

[modern sapphire]

[analog carbon]

DUDE

i dont knoiwww

[modern sapphire]

like how did you know this is a thing?

what made you say ae times ed = be time ec

[analog carbon]

i did these problems before so i remember

the equations

im js very washed

[modern sapphire]

oh ok

[analog carbon]

alr so x=4

ty

[modern sapphire]

yea

[analog carbon]

have a nice night or day

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[modern sapphire]

ty, you too

[midnight plank]

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[molten bay]

All options are wrong

[rose trout]

How come?

[modern sapphire]

well, the last option looks closest to the answer

[hoary pulsar]

Remember you can't just take the sqrt of 625 and 0.0025 to get the answer

Since they're combined into 1 number they'll interfere with one another

[molten bay]

How can you say?

I don't think it is good

.close

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[hushed mauve]

sorry to poke on a closed channel but the last one is the answer

[molten bay]

,w √625.0025

[grand pond]

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Requested by andy03738

[hushed mauve]

ah

well i'm guessing they want it rounded

but aight

[hushed mauve]

you probably should have went one or two digits further

[midnight plank]

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[neat silo]

Two congruent rectangles ABCD and CEFO are placed inside a circle with center O and radius 5. Find the CG/GO.

[neat silo]

Is BCF = 90 degrees?

[lyric charm]

"BCF = 90°" and "B, C, F collinear" contradict each other

[neat silo]

why

[lyric charm]

if three points are collinear they form a 180° angle not 90°...

[neat silo]

oh mb its ABF = 90 degree and B-C-F collinear

[lyric charm]

anyway triangle ABF is right and angle ABC is also right so yes BCF are collinear

[neat silo]

alr mb for the typo

ty

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[molten bay]

Option C can not be possible because they are asking of P(2)

And B,A are not linearly independence

[sudden yacht]

Exactly

So D is the correct answer

[molten bay]

If it was 4t^2 then option A could be correct too?

@sudden yacht

[sudden yacht]

Yes

Instead of 4y² you mean, right?

[molten bay]

Yeah

[sudden yacht]

Yep

[molten bay]

Thanks Alberto Z.

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[lapis pawn]

I feel pretty stupid asking for help for this ( lack of sleep ) but I completely forgot how to find the perimeter

[lyric charm]

1. is this all that they give you
2. are you currently in a test

[mystic condor]

Its the sum of all the sides so if you figure out the lengths youll get it

[lyric charm]

3. the perimeter of any shape is the sum of all of its side lengths

as oakley just said

[lapis pawn]

I know this but can’t find a way to get that last length

[lyric charm]

i think saying "completely forgot" is an unjustified exaggeration.

[mystic condor]

Treat that segment like a triangle

[pine thicket]

i think its safe to assume that all angles except the two on the left are right angles

[lyric charm]

is it really safe

maybe it's given outright and we don't even need to assume

[pine thicket]

oh true

but i figured you would need to otherwise the problem isnt solvable

[buoyant linden]

cuck question for not being explicit

[lapis pawn]

Nah not really I usually don’t get questions like these anymore so I forget what to do

[pine thicket]

do you have any ideas to start?

[lapis pawn]

Nope

I haven’t been given anything to work with so I’m stuck

[lyric charm]

is there anything in the question except the diagram

at all

[frozen finch]

6+2+2+3+a+8

a^2=(8-3-2)^2 + (6-2)^2

[lyric charm]

!nosols

[midnight plank]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[lyric charm]

and also we dont know that the right-looking angles really are right

[frozen finch]

my bad.

[frozen finch]

oh god.

well in that case

this has too many variables

[lapis pawn]

But you would have to presume it is one

To solve the question

[lyric charm]

i think the current verdict is that this question is underspecified and cannot be solved yes

[frozen finch]

yes

[lyric charm]

you should tell your teacher

that this question is missing info

[frozen finch]

if you arent told thise are 90°s, you basically have here a series of equations that give no solution

[lyric charm]

thats what i am saying.

[frozen finch]

and your teacher

if they didnt say "this is 90 degrees"

you have no answer to tell your teacher.

[lapis pawn]

Ok thx for the help

[frozen finch]

np

but yeah man, your teacher gave you no info, and thats a teacher mistake.

[midnight plank]

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[twilit field]

Let $G$ be an open subset of $\R$, then $G$ is the union of a collection of open interval such that
\begin{enumerate}
\item The pairwise intersection is empty
\item The collection is countable
\end{enumerate}

[grand pond]

wai

[twilit field]

I need help with part (1), rather writing a proper proof

[flat spire]

wai

[twilit field]

Let $x \in G$ be fixed , define $\Gamma = { V_{\varepsilon}(x) \mid V_{\varepsilon}(x) \subseteq G}$. Define $A = \bigcup {V{\varepsilon}(x) \in \Gamma}V_{\varepsilon}(x)$. Now consider $y \in G \land y \notin A$. We define $B$ similarly, if $A \subseteq B$, for some epsilon neighbourhood of $y$, we redefine $A$ to be the largest $\varepsilon-$neigbourhood containing both sests

[grand pond]

wai

[twilit field]

is this idea right y/n please

[runic hamlet]

idea yes, execution not yet

[twilit field]

tq

[twilit field]

As in I have to describe why it terminates?

[runic hamlet]

among other things, sure

[oak nymph]

What can you say about two open intervals in R that intersect

[twilit field]

their intersection is open

[oak nymph]

A statement about their union would be nicer

Other than it being open as well, we which know

[twilit field]

Their union is open, or something else?

[oak nymph]

Their union is another open interval

And do you know that every open set in R is a union of open intervals?

[twilit field]

yes, prove that a while ago

[oak nymph]

Well then a proof should strike you immediately

These facts are enough

[twilit field]

Hmm, but I'm trying to prove we can construct a disjoint union

[oak nymph]

Ofcourse, that is what they want us to prove

[twilit field]

I'd first like to proceed with my original idea if that's fine

[oak nymph]

Sure

[runic hamlet]

you literally need that fact already to talk about the set A

[twilit field]

what fact

Let $x \in G$ be fixed , define $\Gamma = { V_{\varepsilon}(x) \mid V_{\varepsilon}(x) \subseteq G}$. Define $A = \bigcup {V{\varepsilon}(x) \in \Gamma}V_{\varepsilon}(x)$. Now consider $y \in G \land y \notin A$. If there exists $\epsilon'$ such that $A \subseteq V_{\varepsilon'}(y)$, which is also an interval as the union of two open sets whose intesection is non-empty is an interval. Then redefine $A \text {to be } A \cup V_{\varepsilon'}(y)$ , else we define $B$ simialrly to the way we define $A$, but using $y$ in place of $x$.

[runic hamlet]

the fact that the union of the intervals is again an interval

otherwise the set A would be useless to you

[grand pond]

wai

[midnight plank]

@twilit field Has your question been resolved?

[twilit field]

<@&286206848099549185>

[oak nymph]

Do you want a hint to continue with this?

@twilit field

[twilit field]

yes please and (b), ist it right so far?

[runic hamlet]

you do know that you have to satisfy both conditions at the same time, yes?

[twilit field]

countability too?

[runic hamlet]

yes

[twilit field]

lemme think about that

[oak nymph]

Well i dont specifically know how to get from this to a solution, but id like to suggest start off by giving a proof sketch first, since this problem demands some formalism (atleast the way i did it)

[twilit field]

Are you asking me for a proof skecth?

[oak nymph]

Yes

[twilit field]

Okay, we start off by fixing an element, say,x, in G, and taking the union (A)of all its epsilon neighbourhoods of x. If there's no other elment in $G$, we're done, if not, we consider a point outside (A), which is in $G$. If any epsilon-neigbourhood contains $A$, we re-define $A$ to be the union of the original A and all epsilon neigbhourhoods of $y$, if no eps- neighbourhood contains $A$, we define the union of all epsilon neighbourhoods of $y$ to be a new set, and continue this process

[grand pond]

wai

[oak nymph]

Okay, but you havent addressed the important problems that may arise with this

One of which you pointed out yourself

[twilit field]

that it may not be countable ?

[oak nymph]

Not that, the first is the termination

Second, do you see a problem with "if any epsilon neighbourhood contains A, we re define it..."

[twilit field]

redefining is a problem, I suppose?

[oak nymph]

No, it leads to "unexpected behaviuour" so to speak

Lets say your open set was (1,2) U (3,4)

[twilit field]

oh, worded badly, if there exists an epsilon neigbourhood containing A

[oak nymph]

Still

Its a problem

[oak nymph]

Lets say we picked x = 1.5

Then what is A?

[twilit field]

(1,2)

oh

[oak nymph]

Now pick 3.5=y

[twilit field]

that's not in G

[oak nymph]

Typo

[oak nymph]

Now continue running your proof

There is an epsilon neighbourhood of 3.5 containing A

[twilit field]

then(3,4)

[twilit field]

right, but it also contains points not in G

[oak nymph]

Yes

[twilit field]

So I have to construct it more carefully

[oak nymph]

Yes

Here is a sketch i have, which you can see if you want:

||You know the open set is a union of open intervals, call that set of intervals S.||

||Say two intervals A and B in S are related if they are "connected" i.e there are intervals J1,J2,J3,...,Jn in S such that J1=A and Jn=B, and Jk intersects J(k+1) for k in {1,2,..,n-1}||

||this relation is an equivalence relation, and the union of all sets in an equivalence class is in turn another intervals||

||Equivalence classes are disjoint, giving you disjointness.||

||Now prove that there are countable many classes:
Every class is an interval, so we can find a rational in it then...||

[twilit field]

okay, thanks

[twilit field]

Issue, haven't done connectedness yet

or even compactness

[oak nymph]

I meant connected in a non formal way

Hence the quotes

Think of the intervals in S as vertices in a graph. And draw an edge between two vertices if they intersect, then there being a path between two vertices (i.e the two vertices being "connected") is the same as the equivalence relation i have written

[twilit field]

Could you explain this bit too please

[oak nymph]

I worded that badly. Strictly speaking the equivalence classes are not intervals: what i meant is ||the union of all intervals in a class is inturn an interval, and in each of these intervals you can find a rational. Since the intervals are disjoint, you have an injection into the rationals||

[twilit field]

ah

that's smart

thanks

[twilit field]

.close

[midnight plank]

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[midnight plank]

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[blazing moat]

how do I solve this? I don't understand what's going on 😭 ts sat problems are so weird 😭

[ebon mica]

what can you notice about the 2 equations

[blazing moat]

second one is *3 of the first one

[ebon mica]

mhm

so that means...

how would a point on the first graph appear be related to the second graph

[blazing moat]

uhhh idk 😭

[ebon mica]

well

the 2 graphs are the same, right?

because one is just the other times 3

[blazing moat]

yea

[ebon mica]

so if one point is on one of the graphs

then its automatically on the other one

because both of them are the same

would you agree with that?

[blazing moat]

yea

[ebon mica]

so now you just need to find any one point from either graph in terms of r

do you think you can do that

[blazing moat]

5x + 8y = 9 --> y = -5/8x + 9/8 it kinda looks similar to the answer but idk how to relate with r

😭

do i just do x = r

[ebon mica]

[blazing moat]

oh ok

thanks for helping :)

i get it

.close

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[ebon mica]

it doesnt necessarily have to be x = r though

it could maybe be y = r too

[blazing moat]

yea

like any number

r

[ebon mica]

try it out and ping me if you run into any problems

[midnight plank]

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[dusty portal]

What is $\sum_{n=1}^{\infty}\frac{H_{n+2}}{n(n+2)}$?

[grand pond]

;(

[frozen talon]

who is H?

[dusty portal]

$H_n=\sum_{k=1}^n\frac1k$

[grand pond]

;(

[frozen talon]

have you tried treating it like it's telescoping?

[dusty portal]

Yeah, but I don't know how exactly to go about that

[frozen talon]

maybe you can find some recursive relation on ({\left(\frac{H_{n+2}}{n(n+2)}\right)}_{n=1}^{\infty})

[grand pond]

Flip

[midnight plank]

@dusty portal Has your question been resolved?

[dusty portal]

Ugh

My internet went out, sorry

I’ll be back in a few

[steel crest]

my intuition says

to.split 1/(n(n+2)) first

partial fractions

[dusty portal]

But I don't know what to do after that

[steel crest]

what do you get for the pfd?

[dusty portal]

$\frac12\qty(\frac1n-\frac1{n+2})$

[grand pond]

;(

[steel crest]

hm, okay

and you have, lets see

idk.the answer btw

[dusty portal]

Me neither

Actually wait it's 1/2 apparently but no verification

[steel crest]

1/2(H[n+2]/n-H[n+2]/(n+2))

looks promising

what happens if you write H[n+2]=H[n]+remainder

[dusty portal]

remainder=1/(n+1)+1/(n+2)

[steel crest]

so we have

not writing 1/2 because im on my phone and parens are annoying

H[n]/n +(1/(n+1))/n + (1/(n+2))/n - H[n+2]/(n+2)

[dusty portal]

[steel crest]

now do the same

[dusty portal]

Ookie

So uh

[steel crest]

for H[n+2] on the right most term

maybe cancelations happen

[dusty portal]

$\qty(\sum_{n=1}^{\infty}\frac{H_n}n-\frac{H_{n+2}}{n+2})+\frac52$

[dusty portal]

You're right lol

[grand pond]

;(

[fallow scarab]

yes flip was right

[dusty portal]

Thank you for cooking 🧑‍🍳

[steel crest]

[dusty portal]

Idk how to do it though :((

[steel crest]

that will be $4

[dusty portal]

Now I know

[steel crest]

oh yes you do

[dusty portal]

Am just lazy sometimes ig 😔

[fallow scarab]

don't forget to show your tablet showing 18%, 20%, and 22% tip

[steel crest]

just write out the first few terms

[fallow scarab]

you haven't done 100 partial fraction decompositions like the rest of us

[dusty portal]

So this'll be 1+(1+1/2)/2+5/2=1+3/4+5/2=7/4+10/4=17/4

[dusty portal]

But ->:((((((((((((((

Ok danke everyone

[steel crest]

wait

the answer is 1/2 how

[dusty portal]

Idk

I never verified

[steel crest]

gl

[dusty portal]

💀

I'm satisfied for now

[fallow scarab]

hmmmmmmmmmmmm

[dusty portal]

,close

.close

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[dusty portal]

[steel crest]

id expect a pi term

because zeta(2)=pi²/6

.reopen

[midnight plank]

✅

[steel crest]

do you know that result

oh wait

WAIT

[dusty portal]

wut?

[steel crest]

i missed something obvious

sum H[n+2]/(n+2)

can be reindexed

instead of n+2 going from 1 to infinity

wait reindex the other one

[dusty portal]

No

[steel crest]

if you dont want negative indices

[steel crest]

uou don't need it

sum H[n]/n for n>=1

is the same as

[dusty portal]

I solved here, it telescopes

[steel crest]

not 7/4?

oh bc from before

5/2

so its 17/4 and 1/2 from before so its 17/8

okay nm

ty

[dusty portal]

.close

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[earnest mica]

.close

[chilly cobalt]

I have that $a_1 = \frac{1}{3}$ and $a_{n+1} = \frac{2a_n^2 + 1}{3}$. How do i show that this sequence approaches 1/2?

[grand pond]

Copter

[chilly cobalt]

this is part of a bigger problem but this is all i need right now

[lavish venture]

first show it’s convergent then take the limit of both sides

[chilly cobalt]

induction?

how do i show the limit part ;-;

[white gate]

once it is convergent then lim a_n = L

[lavish venture]

L = lim a_{n + 1} but also L = lim a_n

[white gate]

L = (2L^2 + 1) / 3

[lavish venture]

you have a quadratic

[chilly cobalt]

ohhh okay

this showed up in an olym and usually stuff like this is never used so😭

[white gate]

believe me, this will show up a lot

and could be even more

[chilly cobalt]

sadge

wait is my solution even correct

ill type the original problem one sec

Let $f: \mathbb{R}^+ \rightarrow \mathbb{R}^+$ be a function such that $\ f(3x) \geq f(f(2x)) + x \forall x \in \mathbb{R}^+ \$ find the largest constant $c$ such that $f(x) \geq cx$ for all x

[grand pond]

Copter

[chilly cobalt]

and then c = 1/2 yeah?

[white gate]

wait

so the initial problem you gave is related to this problem?

[chilly cobalt]

yep

[white gate]

oh i see

i see how it relates now

[chilly cobalt]

i let $f(x) \geq a_n x$ and that implies $a_{n+1} = ...$

[grand pond]

Copter

[chilly cobalt]

this should be right then

[white gate]

yeah

[chilly cobalt]

LETS GOOO

tysm man!!

.close

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[chilly cobalt]

absolute cinema

[white gate]

nice problem tho

[chilly cobalt]

imosl stuff

[midnight plank]

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[frozen arch]

how do u do this

[sudden yacht]

By recalling the definition of real and imaginary part...

There's not much to explain, honestly

[gaunt nimbus]

ye..

[sudden yacht]

If you write z = a + ib, this is the cartesian form.

a is the real part
b is the imaginary part

[gaunt nimbus]

If you write $z = a + ib$, this is the cartesian form.\
\
${a}$ = ${\Re(z)}$ is the real part \
$b$ = $\Im(z)$ is the imaginary part

[grand pond]

k

[grim vector]

And if there is nothing its 0

[sudden yacht]

Thanks 👍

[gaunt nimbus]

von Neumann..

the ordinals 😭

[frozen arch]

oh thanks lol

.close

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[soft dagger]

What's the intuition behind matrix multiplication?

[gaunt plume]

I guess the most useful / intuitive way of thinking about them is as linear functions on vector spaces

If you know how that works

[chilly cobalt]

vectors

[soft dagger]

What exactly does that mean? Why and how did they come up with it? Any video suggestions?

[unique juniper]

https://math.stackexchange.com/questions/271927/why-historically-do-we-multiply-matrices-as-we-do

this might help

[gaunt plume]

I actually don't know how to type vectors in latex but I'll try without. In the two dimensional case, the matrix (a b) (c d) sends (1,0) to (a,c), and (0,1) to (b, d)

Basically each of the columns says where the basis vectors go

And then it turns out that matrix multiplication is equivalent to composition

Then from that you can work out the actual rules of matrix multiplication

If that makes sense

[unique juniper]

just this bit might answer you... maybe?

[soft dagger]

Ty gang

[unique juniper]

linear algebra stuff i think

[twilit field]

Another way is to think of this as a way to compose functions

[lyric charm]

watch essence of linear algebra by 3b1b

unironically he does an excellent job explaining this intuition

[little anvil]

Mat(u)*Mat(v)= Mat(u•v) with u and v two linear applications

[royal glen]

man I fucking love math

[midnight plank]

@soft dagger Has your question been resolved?

[midnight plank]

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[jade magnet]

so it doesnt say that at either points P and Q that the tangents are the same right?

[lyric charm]

tangents?

to whom

[jade magnet]

oh wait mb

one tangent

because (a^2)x is a straight line

actually wtf am i talking about

[lyric charm]

yeah wtf are you talking about indeed.

[jade magnet]

so like we know (a^2)x = x(b-x)^2

[lyric charm]

yes

[jade magnet]

so a^2(x) = x(x^2-bx+b^2)

so (x^2-2bx+b^2) = a^2?

assuming x isnt 0

ofc

[lyric charm]

(b-x)^2 = x^2 - 2bx + b^2.

[jade magnet]

yeah sorry idk why i forgot to type that

wait so does a equal b-x when x is not equal to 0?

because if it does i can substitute that

wait actually no i cant

[jade magnet]

or i could write it as x^3-2bx^2+(b^2-a^2)x = 0

[lyric charm]

i think you're kinda overthinking it

[jade magnet]

yeah probably

[lyric charm]

massively at that

you've got it down to $(x-b)^2 = a^2$

[grand pond]

Ann

[lyric charm]

thus $x-b = \pm a$

[grand pond]

Ann

[jade magnet]

oh

so x = a + b or x = b - a

but since we want P it'll be b-a

the tangent of the curve is 3x^2 - 4bx + b^2 so we can just substitute b-a this time now

2b^2 - 4ab + 3a^2

is the gradient

acc no i made some sort of mistake

clearly

a(3a-2b) means i should have got 3a^2 - 2ba not 3a^2 - 4ba + 2b^2

wait i know why

[jade magnet]

i put -2bx instead of -4bx

okay NOW i should get the right answer

okay yeah that fixes the issue

okay i get it now

idk why it took me this long

.closed

huh

.closed

.close

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[twilit field]

I'd like to prove if a set is compact, then it's is closed and bounded

[frank wolf]

mhm

[lyric charm]

remind us what defn you are using for compactness?

[twilit field]

If a set is compact every open cover of it has a finite subcover.

[frank wolf]

So you've got a subset $X \subset \mathbb{R}^n$

[grand pond]

Pseudo (Cat theory #1 Fan)

[frank wolf]

which is compact in the subspace topology

[lyric charm]

hold on

this is not phrased correctly

[frank wolf]

and you want to show that, as a subset of $\mathbb{R}^n$, it is closed and bounded

[twilit field]

Let's say I'm working with R

[grand pond]

Pseudo (Cat theory #1 Fan)

[lyric charm]

do you mean "we call a set compact if ..."

is that your definition yes or no

[twilit field]

A set F is compact if and only if every open cover of F has a finite
subcover

and F \substeq R

[lyric charm]

is that your DEFINITION yes or no?

or is your definition something else?

[twilit field]

this is my defn

[lyric charm]

i could have sworn i saw ppl imply you're not using this as defn back in hlounge

[twilit field]

Well, this is my defn

[frank wolf]

i see

[lyric charm]

oh no i was wrong

[frank wolf]

so which of compact => closed, compact => bounded do you want to start with

[lyric charm]

contraposition can get at least half of the job done

[frank wolf]

hm true

[twilit field]

comapct -> bounded

[frank wolf]

i see

[lyric charm]

1. unbounded => not compact
2. not closed => not compact

#1 is easy if you understand what a set not being compact means

[frank wolf]

yes i would second Ann's suggestion

[twilit field]

yea thought of that too, was going to do that

just want it verified

[lyric charm]

first what letter do you want to use for your set

pick one and stick with that choice throughout

[twilit field]

F

[lyric charm]

ok

can you formulate what it means for F to not be compact

[twilit field]

A set F is not compact if there exists an open cover of F without a finite subcover

[lyric charm]

ok

so let's suppose now F is unbounded

construct an open cover of F which does not admit a finite subcover

[twilit field]

(n-1,n+1), n\in Z

[lyric charm]

well the union of these guys is R so it covering F is obvious

[frank wolf]

-# strictly should you not intersect these with F?

[lyric charm]

can you prove that no finite subcollection of these intervals can possibly cover F

[lyric charm]

mehhhhhhh

we're not considering F as a topological space in and of itself here

[frank wolf]

i guess it depends how you define open cover

i see

[twilit field]

yes, one minute

Let there be a finite subcover covering $F$. Then its union is bounded

[lyric charm]

what's G

[grand pond]

wai

[twilit field]

typo

[lyric charm]

why is the union of that finite subcover bounded

[twilit field]

As there are finitely many intervals, we can take min of all the n-1 elements, and max of all the n+1 elements, and these will exist

[frank wolf]

that sounds reasonable

I think you can also do this in a more direct way

[twilit field]

as in not by contrapositive?

[frank wolf]

if you take this as an open cover of F, and assume F is compact

then it has a finite subcover

which, by your argument above, is bounded

so F is also bounded, since it’s contained in a bounded set

[frank wolf]

mhm

it doesn’t really matter though, both arguments are good

[twilit field]

now to prove if F is not compact it need not be closed

[frank wolf]

wait hm

you want to prove compact => closed

contrapositive is not closed => not compact

[frank wolf]

I can’t quite parse what this is

[twilit field]

oops

right

[frank wolf]

also remember that not closed isn’t the same as open!!

[twilit field]

We first prove if F is not closed or not bounded, then it's not compact. If $F$ is not bounded, consider the open cover $(n-1,n+1)$. No finite collection of this contains $F$, thus $F$ is not compact. Let $F$ not be closed, there then exists a limit point of $F$ not in it. Let this limit point be $q$. Then let an open cover of $F$ be the interval of the smallest eps neigbourhood containing all of $F \cup ( \frac{1}{n}, \frac{n \cdot t}{n+1})$

[grand pond]

wai

We first prove if F is not closed or not  bounded, then it's not compact.  If $F$ is not bounded, consider the open cover $(n-1,n+1)$. No finite collection of this contains $F$, thus $F$ is not compact. Let $F$ not be closed, there then exists a limit point of $F$ not in it. Let this limit point be $q$. Then  let an open cover of $F$ be  the interval of the smallest eps neigbourhood  containing all of $F$ \cup ( \frac{1}{n}, \frac{n \cdot t}{n+1})$
```Compilation error:```! Missing $ inserted.
<inserted text> 
                $
l.1422 ...neigbourhood  containing all of $F$ \cup
                                                   ( \frac{1}{n}, \frac{n \c...
I've inserted a begin-math/end-math symbol since I think
you left one out. Proceed, with fingers crossed.

Preview: Tightpage -1310720 -1310720 1310720 1310720
[1{/usr/local/texlive/2023/texmf-var/fonts/map/pdftex/updmap/pdftex.map}]
(./796939110815629322.aux)```

[twilit field]

The second part is very badly worded

[frank wolf]

yeah I can’t quite tell what you’re trying to say

[twilit field]

We define $F$ to be the smallest $\varepsilon-$ neigbourhood containing all of $F$.Let this be $F'$. We then define an open subcover $G$ to be ${S \mid S = F'\cup ( \frac{1}{n}, \frac{n \cdot t}{n+1} )}$

[grand pond]

wai

[twilit field]

where n \in \Z \setminus {0}

[twilit field]

Is this better?

We define $F$ to be the smallest $\varepsilon-$ neigbourhood containing all of $F$.Let this be $F'$. We then define an open subcover $G$ to be ${S \mid S = F'\cup ( \frac{1}{n}, \frac{n \cdot t}{n+1} )}$

[grand pond]

wai

[frank wolf]

I don’t understand how you’re defining F again

Isn’t F a subset of R

[twilit field]

I messed up big time

sorry

[twilit field]

$F'$ is the smallest neigbourhood containing all of $F$

[grand pond]

wai

[twilit field]

and $t$ is a limit point of $F$ not in it

[grand pond]

wai

[twilit field]

so we define an open subcover of $F$, to be $G = {S \mid S = F' \cup (\frac{1}{n},\frac{n \cdot t}{n+1}); n \in \Z \setminus{0}$

[grand pond]

wai

[twilit field]

read this as smallest open set containing all of F

[twilit field]

Does it make more sense now?

[lyric charm]

does that exist

[twilit field]

I was doing some research into that, apparently not?

just did that

[lyric charm]

btw your proof is suffering from a lack of paragraphs and textual structure

writing everything in a single paragraph is a good way to lose your reader

[twilit field]

I'll re-write it properly

[lyric charm]

have some mercy on them

anyway

right now you're trying to prove that a non-closed F has an open cover w/o a finite subcover, yes?

[twilit field]

yes

[lyric charm]

ok

so you say that F has a limit point outside itself

and you named it... t, by the looks of it

do you want nudges towards the right construction or would you like me to just give it to you

[twilit field]

nudges please

[lyric charm]

you dont know how F is shaped around t or for that matter anywhere else. so the opens in your construction would want to be "big" somehow in order to be sure to cover F

i say big in a decidedly informal sense

[twilit field]

So I need to construct an open set containing ideally all but an epsilon neigbourhood of t

[lyric charm]

you're thinking in the right direction here yes

i will also advise against trying to write it in interval notation

[twilit field]

I was thinking of writing it as a set difference of R and a closed set

[lyric charm]

easiest to write it in setbuilder notation

also has the bonus that itll generalize to R^d

cause the cover that you gave for unbounded => not compact is also generalizable

[twilit field]

${x \mid \exists \varepsilon : V_{\varepsilon}(x) \var \cap {t} = \varnothing}$

[grand pond]

wai

[lyric charm]

uhhh

what is $V_{\ep}(x) \delta$?

[grand pond]

Ann

[lyric charm]

also what is that set...

like what's it meant to be

[twilit field]

wait, where did the delta come from

ooh

[lyric charm]

presumably when you typed \var.

$\var$

[grand pond]

Ann

[lyric charm]

hm.

idk what that is doing there though

[twilit field]

I meant to type varepslion earlier on, must have not removed that

${x \mid \exists \varepsilon : V_{\varepsilon}(x) \cap {t} = \varnothing}$

[grand pond]

wai

[lyric charm]

taken at face value, the set you've just described is R \ {t}.

which does not do us much good.

[twilit field]

lemme try a bit harder

[lyric charm]

remember you want to describe a COLLECTION OF SETS, not merely one set.

perhaps parameterized by something.

[twilit field]

Parameterised by t

[lyric charm]

npoe

nope*

t is fixed

it's a point that is a limit pt of F but doesn't lie in F

i say "trying hard" may lead you to breaking your own back

[twilit field]

so think simple

[lyric charm]

yes think simple

[twilit field]

A part of the open cover can be the sets of the form (t-n-1, t-n) and (t+n,t+n+1) , where n in in N

Then we have to find another open cover in [t-1,t+1]

[lyric charm]

mmmm

you're kinda introducing unnecessary complexity here

[twilit field]

hmm?

[lyric charm]

here is something to help you remove one of the layers of it

(a, +∞) is an open set

so is (-∞, a)

for any real a

[twilit field]

yes

[lyric charm]

think about your "all but an epsilon nbhd of t" again

[twilit field]

as t is a limit point, every eps neighbourhood intersects F at some point

[lyric charm]

every epsilon nbhd of t intersects F at some point, yes

that's gonna be important later

[twilit field]

in that case we can't construct an eps neigbourhood around t, can we?

[lyric charm]

when you say "can't" do you mean "unable" or "unhelpful"

[twilit field]

unable

well, especially if it's sandwiched b/w two parts of F

Could I have the construction please, or should I think about this a while longer?

[lyric charm]

eh let me spare you the misery

the collection of opens that i was specifically thinking of is something like: $$E_{\ep}(t) := { x \in \bR : |x-t| > \ep}$$ with $\ep > 0$

[grand pond]

Ann

[lyric charm]

E stands for exterior

but don't think too hard about that

if you want your collection to be countable you can take only E_{1/n}(t) into your cover

[twilit field]

Eh, that's fine

[lyric charm]

$\bigcup_{\ep>0} E_{\ep}(t) = \bR \setminus {t}$

[grand pond]

Ann

[lyric charm]

and try to think about what happens if you take the intersection of some finite subcollection of these "exterior-ball" sets

[twilit field]

We get a subset of the exterior of t

[twilit field]

this is so simple . 😔 Should have been able to think of this

[lyric charm]

RA is not a good place to learn to cook up such simple constructions

[twilit field]

wdym

Ok, yeah, this is really simple, I get that

[twilit field]

wait, why is this false

[twilit field]

tbh I've never had to do this before

atleast not that I can recall

[lyric charm]

you're taking the word "exterior" too seriously

[twilit field]

We get a subset of R\ {t}

[lyric charm]

i mean, no shit.

im gonna say you need to actually sit down and write out what happens if we take a finite subcover of my cover.

[twilit field]

okay, I'm lost

[lyric charm]

ok so consider the subcover consisting of $E_{\ep_1}, E_{\ep_2}, \dots, E_{\ep_n}$ (dropping the $(t)$ from each one cause i cbf to write it), and let's also say $\ep_1 > \ep_2 > \dots > \ep_n$ for convenience.

[grand pond]

Ann

[lyric charm]

what's the union of just these guys

[twilit field]

E_{\eps_n}

[lyric charm]

ok and what can you say about E_{ε_n}

keeping in mind your goal of showing that my open cover does not admit a finite subcover

[twilit field]

it consists of all numbers more than eps_n+t and less than -eps+t

[lyric charm]

technically correct but misses the point

[twilit field]

it doesn't cover all of the subcover?

[lyric charm]

words ouch

[twilit field]

every finite subcover misses out some point of the open cover

[lyric charm]

simpler tbh

$F$ has nonempty intersection with the $\ep_n$-neighborhood of $t$, so it has at least one point lying in $V_{\ep_n}(t)$ and thus outside $E_{\ep_n}(t)$. thus $F \nsubseteq E_{\ep_n}(t)$.

[grand pond]

Ann

[twilit field]

ah right

thanks

I'll close this now, thanks so mcuh!

*much

.close

[midnight plank]

Channel closed

Closed by @twilit field

Use .reopen if this was a mistake.

[midnight plank]

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[safe onyx]

I got halfway through and got undefined so imma restart getting help 😓

$F\left(x,y,z\right)=xz \vec i+x \vec j+y \vec k$

[grand pond]

smeagol

[safe onyx]

$S \to y=\sqrt{9-x^{2}-z^{2}}$

[grand pond]

smeagol

[safe onyx]

$a(r,\theta) = r \cos(\theta) \hat i + \sqrt{9-r^2} \hat j + r \sin(\theta) \hat k$

[grand pond]

smeagol

[safe onyx]

S parameterized

Then the normal would be

$a_r \cross a_{\theta}$

[grand pond]

smeagol

[safe onyx]

$a_r = \cos {\theta} \hat i - \frac{r}{\sqrt{9-r^2}} \hat j + \sin{\theta} \hat k$

$a_\theta = -rsin(\theta) \hat i + 0 \hat j + r \cos{\theta} \hat k$

[grand pond]

smeagol

[dawn dagger]

You forgot the square root

[grand pond]

smeagol

[safe onyx]

thank you!

so the cross product is

$n = \frac{r^2 cos \theta}{\sqrt{9-r^2}} \hat i - ( - r \sin^2 \theta - r cos^2 \theta) \hat j + \frac{r^2 \sin \theta}{\sqrt{9-r^2}} \hat k$

[grand pond]

smeagol

[safe onyx]

$n = \frac{r^2 cos \theta}{\sqrt{9-r^2}} \hat i + r \hat j + \frac{r^2 \sin \theta}{\sqrt{9-r^2}} \hat k$

[grand pond]

smeagol

[safe onyx]

does this look ok so far?

[dawn dagger]

,w (cos(theta), -r/sqrt(9-r^2), sin(theta)) x (-rsin(theta),0,rcos(theta))

[grand pond]

Results provided by WolframAlpha

Click here to refine your query online
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Requested by adoniszgks

[dawn dagger]

Should be right so far.

[safe onyx]

j should be positive here

since it says y is positive

[dawn dagger]

Yes, that's why I said it's right so far.

[safe onyx]

yeah

[dawn dagger]

The cross product a x b = - b x a is not commutative

What were your next steps?

[safe onyx]

next I need to make n a unit vector and divide it by it's magnitude?

$|n| = \sqrt{ \frac{r^4 \cos^2 \theta}{9-r^2} + r^2 + \frac{r^4 \sin^2 \theta}{9-r^2}}$

[dawn dagger]

If I recall correctly we didn't have to...

[grand pond]

smeagol

[safe onyx]

oh right

cause the magnitudes cancel out

so I just need n dot F

[dawn dagger]

Yeah exactly

[dawn dagger]

What are the bounds btw?

[safe onyx]

r goes from 0 to 3
theta goes from 0 to 2pi

[dawn dagger]

Ok nice!

[safe onyx]

$F(a(r,\theta)) = r^2 \cos(\theta) \sin(\theta) \hat i + r \cos\theta \hat j + \sqrt{9-r^2} \hat k$

[grand pond]

smeagol

[dawn dagger]

Looks good.

[safe onyx]

that dotted with n is

[dawn dagger]

,w (r^2cos(theta)sin(theta), rcos(theta), sqrt(9-r^2)) . (r^2cos(theta)/sqrt(9-r^2),r,r^2sin(theta)/sqrt(9-r^2))

wth

[safe onyx]

$\frac{ r^4 cos^2 \theta \sin \theta}{\sqrt{9-r^2}} + r^2 \cos \theta + r^2 \sin \theta$

[grand pond]

smeagol

[safe onyx]

looks the same

[dawn dagger]

yea

[safe onyx]

it just is bulky underneath

then I can integrate than rdrdtheta?

[dawn dagger]

in fact the bulky had nothing to do with the request

[safe onyx]

computer brain fat

[dawn dagger]

The left term integral seems painful.

[safe onyx]

$\int^{2\pi}_0 \int^3_0 ( \frac{r^5 \cos^2 \theta \sin \theta}{\sqrt{9-r^2}} + r^3 \cos\theta + r^3 \sin \theta ) drd\theta$

[dawn dagger]

\int_{0}^{2\pi}

[grand pond]

smeagol

[dawn dagger]

I think for the left term integral wrt r I would do || u = 9-r² ||

[safe onyx]

desmos says it's undefined

[dawn dagger]

The other terms r^3(cos(theta)+sin(theta)) should be 0

because you integrate them over [0,2\pi)

[safe onyx]

gotcha

[dawn dagger]

Probably because of r=3 division by 3, I suppose

[safe onyx]

ok now it's not so angry

I took care of the r^3costheta and r^3 sin theta

why solve the r side when all the cos sin are 0 heh

[dawn dagger]

Hmm yeah, given that r^5/sqrt(9-r^2) will return some constant, again the integral over [0,2pi) of cosine and sine will be 0.

[safe onyx]

yep

love all this work to get just 0