help-23

[copper wraith]

first thought z-z'=0

second---z/z'=1

z'/z=1

all assumption would be correct?

[burnt notch]

Well, z - z' = 0 surely works

The other unfortunately work only if the number is non 0

But apart from that they're all the same thing yeah

[copper wraith]

x^2+y^2=/0

[burnt notch]

?

[copper wraith]

sqaure of real number is 0 only when both are x=y=0

so exclude the origin my statments are correct?

[burnt notch]

I don't get what you want to achieve

[safe radish]

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[void cosmos]

Hello, I need some help with matrix algebra. Can anyone help me?

[void cosmos]

I completed a set of problems, but the thing is the textbook has no answer key, so typically I run it through GPT or something which is fairly accurate for most times, but I believe its making a few errors.

[normal moss]

Post away

[void cosmos]

So here are the problems I completed for Problem Set 2.3 in Gilbert Strang's MIT 5th edition Linear Algebra Textbook. Posting the images of the problems for the problem set. Please note that I did not complete the questions in a sequential manner and did not do all of the problems in the pages, just did some of them.

Also if you have any tips of general improvement for my work please let me know.

[safe radish]

@void cosmos Has your question been resolved?

[safe radish]

Please post images (such as PNGs or JPGs) of the question rather than other filetypes such as PDFs which have to be downloaded. Non-image downloads can potentially contain viruses or other security risks.

[bleak kite]

@void cosmos

which ones did you not do?

[void cosmos]

Hey Benze, I just got Chat to recheck it again and it is confident on most of then, so I will just then request for you and others to check on the remaining ones left that chat isn't fully sure about.

[bleak kite]

Sure, No problem👌

[void cosmos]

So if you could look at these questions specifically: (Chat isn't too sure about these ones)
3)
9(a), 9(b)
18)
19 except for the extra “find another non-diagonal matrix whose square is
𝐼
I” part, which I’m not fully sure you answered clearly
21 still looks incomplete to me unless I’m missing writing that is too faint to read

[void cosmos]

Ty.

[bleak kite]

Looking at the 18th one

it could either be orthogonal

or it could be idempotent

otherwise they would not ask such a high power

[void cosmos]

For question 3 I recognize I made a silly my dumbass accidentally made it into a Lower Triangle instead of upper lol.

[bleak kite]

(Just a second I'm doin these rn)

[void cosmos]

Thanks and then compare it to the work I have in my PDF please thanks.

[bleak kite]

(.....pdf i dont wanna open i dont wanna get virused or smth-)

[void cosmos]

I'll just post images then lol no virus I've been a longtime member of this server. Almost 6 years lol.

[bleak kite]

it's not about you its the fact i dont do it anyway

[void cosmos]

Here is my work for the questions, I haven't attempted Q21 yet, so no needa check that one as I'll do that later so if you can just contrast my work to the questions for 3, 9, 18 and 19 that'd be great.

[bleak kite]

3 matches mine

You're clear there

[void cosmos]

Hmmm, but because they said U doesn't that mean upper, so shouldn't I have it in the upper format?

[bleak kite]

Nah i think this is fine

[void cosmos]

They didn't explicitly mention it though in the question, so I didn't think too much of it,

[bleak kite]

I just plugged and chugged till i got it too didn't put any thought in it

[void cosmos]

Yah I think thats fine doesn't really matter honestly. Alrighty keep going

[void cosmos]

Yeah with these things they tend to get a bit monotonous, so its like just cranking away lol. I'd honestly probs double check my work, but its almost 4 am for me and I am so dead tired lol.

[bleak kite]

elimination matrices is not exactly my thing T-T

So I'll leave the ninth one to you too

I know how to do it

but I ain't fully sure about it

[void cosmos]

No worries, I appreciate the honesty.

[bleak kite]

18th one

F^100 part bothers u ryt

[void cosmos]

9 looks good to me double checked it and everything works out.

[bleak kite]

n=100 in there ashte

[void cosmos]

Yep I got that.

[bleak kite]

the rest of 18 is basic multiplication

[void cosmos]

Can you check EF and FE for me pls real quick.

[bleak kite]

sure

[void cosmos]

Hmmm, wait I just checked EF and FE looks good to me.

E^(2) looks good to me.

[bleak kite]

EF is
1 0 0
a 1 0
b c 1

[bleak kite]

okay

19 is a bit weird

it could be a element switching places type

one sec

[void cosmos]

Hmm I checked PQ and QP those look good.

For M^(2) = I, I just utilized a permutation matrix and so multiplying a row exchange matrix by itslef just undoes previous operation, so we get the identity matrix.

[bleak kite]

P removes Q and Q removes P is the main idea of this thing

So that means you can either go one step forward in the pattern or one step backward

tht should be it

[void cosmos]

Hmmmm was the way I did it fine would you be able to see?

[bleak kite]

Sure if you've come to the same conclusion dont overthink it's fine

i didnt bother i saw two matrices and decided one does things and the other reverses it

[void cosmos]

Cool TY so much! 🙂

I really appreciate you.

[bleak kite]

np

[bleak kite]

me too

[void cosmos]

Thank you. Can I add you in case I have the occassional question or something?

[bleak kite]

I'm in highschool

you prolly know more

[void cosmos]

I won't bombard or anything, just for the occassional question.

[void cosmos]

Oh lol, I'm in Uni.

[bleak kite]

well strang is a book they teach us to some extent

[void cosmos]

Ohhhh shoot my roots go back to India, but I have been raised in Canada for majority of my life.

I've heard about JEE.

Really tough exam and I have deep respect for anyone who attempts it.

[bleak kite]

we're going off topic

.close()

(we'll meet again chill)

(dont hv to add me for tht)

[void cosmos]

Nah, nah, I added you lol. Take my request kindly, if u can.

.close

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[midnight steeple]

#probability-statistics message

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@midnight steeple Has your question been resolved?

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[lean otter]

x^2 - 4 |x| + 3 < 0

My answer came x belongs to (- 3, -1) U (1, 3)
But the workbook answer says only (1, 3)
Is my process alright?

[lean otter]

do I also need to take x^2 = |x|^2 , I thought it was unnecessary

[thin bridge]

ur work is fine

was there anything else stated in the question?

[lean otter]

yeah the workbook had my answer as option D but the answer showed B -> (1, 3)

alright

[thin bridge]

,w solve x^2 - 4|x| + 3 < 0

[lean otter]

thanks for verifying

[flat frigate]

Results provided by WolframAlpha

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

Requested by ramonov

[lean otter]

what

oh yueah

that is same as (-3, -1) U (1, 3)

thanks for verifying again

.close

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[smoky sandal]

Tom plays a game, he is fighting the boss. Initially the boss has infinite hp, Tom has 100 hp. If Tom hits the boss once he gains 30 hp from the boss and if the boss hits him, he looses 10 hp. There is 50% chance of landing hit for both. If Tom's hp becomes 0 game over. Find probability that Tom looses

[fiery remnant]

under these conditions isnt the probability just 1? because there's no way for him to win...

the boss has infinite hp

[smoky sandal]

maybe

[split fulcrum]

sam beam

[fiery remnant]

waterrbeam

[smoky sandal]

beam

[safe radish]

@smoky sandal Has your question been resolved?

[smoky sandal]

@vagrant jolt

https://tenor.com/view/im-crine-lmfao-idk-funny-gif-17839070261635906856

<@&286206848099549185>

[pale dock]

!15m

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Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[pale dock]

That's first of all

Second of all

Someone asnwered your question and you said maybe

Then sent a meme gif for some reason

And that's very weird

If you think the question wasn't correct or you didn't understand the answer then that's fine but it feels very weird to ping for this

[smoky sandal]

its been 20 mns

[pale dock]

Mhm my bad

I just saw your last message

[smoky sandal]

that was because of

[pale dock]

???

[pale dock]

Is this NOT your question?

[smoky sandal]

yes that happens everytime i try to to ping helpers

[pale dock]

Can you clarify your question, please?

[smoky sandal]

what is not clear ?

[pale dock]

What your question is? The pinned question has already been answered by sam.

So what do you need help with?

[smoky sandal]

you can have infinite moves, so idk if its 1

[pale dock]

Okay well let's put it this way

They asked for the probability of tom losing, correct?

But since there are only 2 options, winning and losing, we can instead say that the chance for tom losing is 100% - the chance of winning

So if the boss has infinite HP, what are the odds that tom will eventually get the boss to reach zero health?

[smoky sandal]

ok

if the hp does not get to zero tom not loose or win

[smoky sandal]

but the number of moves is also infinite

[pale dock]

Well, infinity is not a number so you can't have something that goes on forever

What you could have is an indefinite repition where tom always gains HP faster than the boss can damage him

But you can't reach an infinite number of moves

[smoky sandal]

oh but that does not mean game over, the game can go on forever?

[austere nimbus]

The game can go forever

But it's probability is 0

Your question is interesting and requires a but of combinatorics

[smoky sandal]

hm.. why

[austere nimbus]

Do you know how to do basic combinatorics

[austere nimbus]

I'll try to bring you there

[smoky sandal]

a bit

[austere nimbus]

Okay good

Do you know about binomial coefficients

(Lemme chekc the english name)

[smoky sandal]

yes

[austere nimbus]

Ok

Good

So we're gonna first look at a simpler problem

If the boss hits and if they hit the boss, let's first assum that the hp gained and dealth are equal

[smoky sandal]

ok

[torpid osprey]

probability that tom loses is 1 minus the probability that he wins.
boss has infinite health so can never die, so tom can never ever win in the first place, so the probability he wins is 0
1 - 0 is 1. the probability that tom loses is 1.

[austere nimbus]

The problem is thatwe don't know the probability that neither wins

[torpid osprey]

0

[austere nimbus]

How

Thzt's whereim tryingto get

[torpid osprey]

what constitutes a draw? there's no draw state

[smoky sandal]

sorry i gtg brb

[austere nimbus]

Draw: the game co tinues indefinitely

[torpid osprey]

probably negligible

[austere nimbus]

It's pretty hard to prove

[torpid osprey]

is it even worth the effort

[austere nimbus]

Yes

I think

[torpid osprey]

then happy helping

[austere nimbus]

Im going to prepare some drawings for when they come back

[pale dock]

The way I'd explain it is just through coin flips

Like the chance of 2 heads in a row is 1/2^2

And 5 in a row is 1/2^5

And it gets smaller

Until it eventually reaches zero at infinity

Which means that it could never happen

[austere nimbus]

@smoky sandal here are some intermediate questions, might not be perfect so ping me if you don't understand, need help or have doubts.

Ig question 6) is to look at @torpid osprey's argument and adapt it

4. is tricky so maybe we'll have to talk about it

[quartz hamlet]

Hey hii everyone, honestly I am no maths freak but my girlfriend left me so I felt lonely so I just joined discord

[austere nimbus]

#discussion is the best channel to talk about casual stuff

Alternatively #serious-discussion or #chill

[safe radish]

@smoky sandal Has your question been resolved?

[safe radish]

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[lavish magnet]

For question b I used s=ut+1/2*at^2 but i got the wrong answer because I did -9.81 and the correct answer didn't use a negative. So my question is why is the gravitonal pull not negative here?

[covert rain]

maybe because they oriented the positive vertical axis downwards

[safe radish]

@lavish magnet Has your question been resolved?

[dreamy moss]

<@&268886789983436800>

[lavish magnet]

.close

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[blazing swallow]

@amber glade @wraith ember

[amber glade]

That's odd. What happened?

[wraith ember]

yeah

[blazing swallow]

in the future you can type .reopen to reopen the channel to avoid that happening (you closed it beforehand)

[wraith ember]

fahhhh

[amber glade]

Ohhh, my bad.

I'll make sure I do that next time.

[blazing swallow]

all g! you can click on the forwarded message to bring up whatever you said last

[amber glade]

Sounds good.

[wraith ember]

@blazing swallow if possible can you forward other txts too

or @amber glade go to help 25 and type .reopen to chk if it works

ok W

[blazing swallow]

.close

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[cold trellis]

.repon

.reopen

$baby you light up my world like nobody else$

[flat frigate]

maso7

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[flat frigate]

Renato

[spiral saddle]

do I need to apply the arquimidean property or the axiom of the supremum?

I am not sure how to proceed with 11a)

[proud tree]

consider |l-r| and the definition of the limit here

<@&268886789983436800>

[quiet plume]

In particular, the idea is if the sequence converges to l, then eventually r should lie outside the epsilon-interval you set around l.
|r-l| in this case is the distance between l and r. Try to use this as Green suggested.

[spiral saddle]

I dont get it

[spiral saddle]

an -> l <=> n >= n0 => |an - l| < epsilon

what does l and r has to do with this

[quiet plume]

All I'll say is that r-l > 0 (and this should ring a bell from the definition)

[spiral saddle]

what?

[spiral saddle]

epsilon > 0 and r - l > 0

[quiet plume]

Indeed.

[spiral saddle]

i dont get it

we cant use transitivity

[quiet plume]

The definition tells you something about any choice of epsilon > 0.

[spiral saddle]

what?

i dont understand

[severe pond]

the sequence is eventually contained in any epsilon neighborhood of l

r is some distance from l

you’re on a?

[spiral saddle]

ye

[severe pond]

in the definition of the limit this holds for every epsilon neighborhood

the distance from r to l would be a smart choice here

what is that distance?

[spiral saddle]

epsi

[severe pond]

no what is the distance from r to l

recall that r > l

[spiral saddle]

0

[severe pond]

that would be if r = l

[spiral saddle]

i mean

> 0

[severe pond]

ok

but

what is the distance

if you had to write an expression for it

like distance = …

[spiral saddle]

who knows at this point

[severe pond]

what’s the distance between 3 and 4?

[spiral saddle]

r - l

[severe pond]

yes

[spiral saddle]

|3-4

[severe pond]

so you know that the sequence is eventually contained in an epsilon interval with radius r - l around l

can you picture this on a number line?

place l anywhere

then r somewhere to the right of it

draw an arrow above the number line from where l is to where r is then draw an arrow in the opposite direction of the same length, starting at l

that will create your interval

then just imagine a bunch of points from the sequence being trapped inside there

[wild cape]

has this been satisfied?

[severe pond]

?

how the hell are we supposed to know if he understands every textbook example and prior exercise

no one works a full time job helping renato

we just pop in on our own time

[wild cape]

this is more a question to renato

[severe pond]

"TO OTHER HELPERS:"

[wild cape]

@spiral saddle remember i asked you to not open a channel without having at least one concrete idea of solving the problem?

[spiral saddle]

can u make a drowing

[severe pond]

what wasn’t clear here?

[spiral saddle]

my idea was using the definition of convergence of a sequence aswell as the arquimidean property or the axiom of the supremum

[severe pond]

how would the archimedean property be used here?

[wild cape]

those are all very vague ideas

the closest idea is use definition of convergence but thats still very vague

[severe pond]

and the limit of a sequence doesn’t have to be the supremum of the sequence

this sequence isn’t necessarily monotone increasing

[wild cape]

you should first ask for ideas from the professor or ppl who already passed the class

[spiral saddle]

either use the archimidean property to find a lower bound for the epsilon

[severe pond]

🤔

0?

[spiral saddle]

if I would know a concrete idea on how to solve everything then I wouldnt be needing help, just a will to execute it

[wild cape]

it takes several concrete ideas to solve this. i simply need you to have at least one coming in

and i dont believe you asked the professor or senior classmates for ideas before coming here

[severe pond]

why does it matter who he asks

so long as they know what they’re talking about

[wild cape]

you probably dont have the context of a very long convo i had with ren the other day

[severe pond]

i don’t

[spiral saddle]

can I get some help or not?

[severe pond]

this is what i had in mind

[spiral saddle]

2l - r?

[severe pond]

yes that’s just the point at a distance r - l from l on the left

it doesn’t really matter here

you don’t have to label it

subtract r - l from l and you get 2l - r

which translates to going a distance r - l to the left of l

i didn’t draw points in the sequence outside of this interval but there can be

it’s just that there would only be finitely may of them

[spiral saddle]

l - (r - l) = l - r + l = 2l - r
I see

[severe pond]

because the sequence has to eventually be contained in this interval

[spiral saddle]

why are you showing me this

[severe pond]

a visual aid to get a better intuition for the problem

it should be clear from this image why a is true

do you see how a_n < r for all of the n pictured here

[spiral saddle]

yes

[severe pond]

now you just need to write it with symbols

how would we describe all of the sequence members being in this interval

what can we say?

[severe pond]

(eventually)

before we go further, you understand what i said about it being possible that finitely many of the sequence members aren’t in that interval right

[spiral saddle]

no

[severe pond]

ok then you should’ve asked

[spiral saddle]

|an - l| > |r - l|

[severe pond]

🤔

what

[spiral saddle]

i was trying to describe all the sequence numbers in the interval

[severe pond]

no that’s not quite right but we will get there

[severe pond]

this is more important for now

[spiral saddle]

but maybe its

2l - r < x < r

[spiral saddle]

how

[severe pond]

in the sequential limit definition there is something about "there exists N such that for all n > N, …"

the finitely many indices before N are irrelevant then

[spiral saddle]

why are you saying finite

[severe pond]

we don’t care if a_1, a_2, …, a_N don’t fit in that epsilon neighborhood, so long as we can find such an N so that for all of the indices afterwards, the sequence members are in that neighborhood

[severe pond]

these first N makes up a finite list

of sequence members

[spiral saddle]

ok, yeah because n >= N

[severe pond]

like imagine i had the sequence:
1, 0, 0, 0, 0, 0, …

clearly this converges to 0

we don’t care if the first sequence member doesn’t fit in every epsilon neighborhood around 0

and it doesn’t have to be the first

we can have something like

[spiral saddle]

ok

[severe pond]

1, 0, 0, 0, 0.5, 0, 0, 0 etc

and now the 0.5 might be a problem for some epsilon intervals

but we can just take the points after it

which is an infinite list

[spiral saddle]

yeah

[severe pond]

so here we can take the last point that isn’t in this interval and just throw that first part of the sequence out

which i didn’t draw

but imagine there are some other points in the sequence not in the interval

there can only be finitely many of them

because we have to cluster around l

so the sequence essentially shrinks around l

we can’t stay far away infinitely often

[spiral saddle]

how do we find a suitable N?

[severe pond]

we don’t need to make it explicit

we just use the definition

[spiral saddle]

i mean I am still stuck on how you grasped that a) is true

[severe pond]

recall that for any epsilon > 0, we can find such an N so that blah blah blah

so if we come up with some epsilon > 0 then by definition we are guaranteed the existence of some N, whatever it is

we said r - l > 0

so by definition there is some N so that…

you tell me

[severe pond]

hmm

not sure what you’re asking exactly

like how do i have the intuition that this is true?

i’ve done tons and tons of these problems and familiarized myself with it

a lot of people just see the definitions and don’t concretely understand what it’s saying

if you understand it conceptually it should be no issue

[severe pond]

i know that a_n has to get "closer" to l than any other number which means it’ll eventually have to be below r since l is below r

[spiral saddle]

the definition of convergence of a sequence used that forall n in N there exists some n0 in N such that n >= n0

[severe pond]

the sequence is clustering around l

[severe pond]

no?

that’s just an obviously true statement, take
n_0 = n

[severe pond]

there is some N so that for all n > N,
|a_n - l| < epsilon

ever seen this

[severe pond]

yes

[spiral saddle]

yes but

how to connect it with r < l

[severe pond]

we already said which epsilon we are interested in

[severe pond]

^

[spiral saddle]

r - l is our epsilon??

[severe pond]

yes

|a_n - l| < r - l

manipulate this to show that a_n < r

[spiral saddle]

ok

l - r < an - l < r - l

2l-r < an < r

[severe pond]

no

yes

funny

aren’t those the endpoints of the interval i drew earlier

[spiral saddle]

you have a solid grasp of the def

[severe pond]

👍🏻

[spiral saddle]

is it that easy for u?

[severe pond]

this?

yes

you’ll get comfortable with it eventually

it’s like tying your shoes

[spiral saddle]

say i have a sequence bn that converges to l when n -> inf, then here the left side endpoint will be what? and right side endpoint?

[severe pond]

well in general it’s l - epsilon and l + epsilon

if you want an interval around l with radius epsilon

here i had a specific choice of epsilon in mind

because i wanted a_n < r

which translates to r being the right endpoint of one of these epsilon intervals

[spiral saddle]

you make it seem easy

[severe pond]

you should get comfortable with epsilon intervals

this is the best way to think of convergence

[severe pond]

i’ve spent more time on it than you have

my fault

💀

[spiral saddle]

bae?

[severe pond]

have

b is next to h

[spiral saddle]

alr I appreciate the help, though I cant say I am confident about this stuff

[severe pond]

try the next one

[spiral saddle]

this is my first exercise using the definition of convergence of a sequence

[severe pond]

use almost the exact same idea for part b

see where you get

[severe pond]

^read some of these again

where i explained my thought process

[spiral saddle]

lim an = l <=> |an - l| < l - r

[severe pond]

there are some unstated conditions there

[spiral saddle]

jajaja

[severe pond]

but yes l - r is a good choice of epsilon

[spiral saddle]

sorry, how can I be more formal about it?

do you mind helping me with the wording

[severe pond]

how about i word the first one for you

[spiral saddle]

yes

please

[severe pond]

since a_n —> l and r > l <—> r - l > 0, there exists n_0 in N such that
|a_n - l| < r - l <—> l - r < a_n - l < r - l <—> 2l - r < a_n < r whenever n > n_0

ok maybe latex would be better

but i think you can read this well enough

keep it simple

if you don’t like the whenever n > n_0 afterwards you can just replace it with for all n > n_0 before the inequality equivalence’s

[spiral saddle]

sinice an -> l and l > r <=> l - r > 0, then there exists some n0 in N such that
|an - l| < l - r <=> r - l < an - l < l - r <=> r < an < 2l - r

[severe pond]

perfect

[spiral saddle]

what

[severe pond]

but just say

whenever n > n_0

at the end

or say for all n > n_0 after the such that

[spiral saddle]

n >= n0?

[severe pond]

it doesn’t matter

if your class uses >= then use that

both work

i use >

[spiral saddle]

but this proof is complete?

[severe pond]

yes why wouldn’t it be

we’ve shown that there is some n_0 such that for all n >= n_0 the inequality holds

[spiral saddle]

we never found one such n0

[severe pond]

we did

[severe pond]

we guaranteed the existence of it here

we don’t need a formula for it

just need to show such an n_0 exists

we know it exists because a_n —> l and r - l > 0

or in the second case l - r > 0

and for any positive distance such an n_0 exists

[severe pond]

which comes directly from the definition of convergence

[spiral saddle]

oh right

I see

what about c)?

[severe pond]

thanks

well let’s think about this for a second, if r >= l then r > l or r = l

we already handled the case when r > l yes?

[spiral saddle]

what

[severe pond]

what was confusing

[severe pond]

there are two cases here

[spiral saddle]

no because now we have an <= r

[severe pond]

r > l or r = l

[spiral saddle]

in the other exercise we had

an < r

[severe pond]

ok but if a_n < r then a_n <= r

[spiral saddle]

oh right

ok fair enough

[severe pond]

so the r > l case has been resolved

now what if r = l

is is true that if r = l then there is some n_0 such that a_n <= r for all n >= n_0

i.e. is a_n eventually bounded above by l

[severe pond]

this just says a_n <= l for all n >= n_0

[spiral saddle]

as long as it is bounded above by L then an <= r forall n >= n0

[severe pond]

is this true if a_n —> l

[severe pond]

sure because l = r

[severe pond]

^

[spiral saddle]

how do you know an is bounded above by l

[spiral saddle]

yes

[severe pond]

well a_n <= l means a_n is bounded above by l

[severe pond]

and we had r = l here

[spiral saddle]

similarly an -> r

[severe pond]

are you sure?

remember my picture

[spiral saddle]

dude, lim an = L

[severe pond]

yes

[spiral saddle]

that by itself means that an -> l

[severe pond]

yes

go on

[spiral saddle]

an -> l <=> |an - r| < epsilon

[severe pond]

but epsilon needs to be > 0

[spiral saddle]

you said an is bounded above by L

[severe pond]

if l = r then you can’t make l- r or r - l epsilon

[spiral saddle]

yes mb

I deleted that

[severe pond]

no, i said that’s what the question is asking

is it true that a_n is bounded above by l if a_n —> l

eventually

[spiral saddle]

we need to use n >= n0 somehow and find a suitable n0 maybe?

[severe pond]

well try thinking of it visually

[spiral saddle]

uff

[severe pond]

if i have a sequence of points converging to l, do they have to eventually cluster from below l

or can they reside above l as well as they shrink around l

[spiral saddle]

they can reside above l aswell as they shrink around l

[severe pond]

yes

so then this isn’t true

consider a_n = 1/n

[spiral saddle]

because L is not an upper bound for every number in the sequence

[severe pond]

this converges to 0

but 0 is not an upper bound for the sequence

we are entirely converging from above here

[spiral saddle]

1/1 = 1

[severe pond]

not sure what you mean

that’s the first member of the sequence

sure

1, 1/2, 1/3, 1/4, …

[spiral saddle]

what I mean is that the sequence goes

1. 0.5, 0.3, 0.25

[severe pond]

yes

and it converges to 0

[spiral saddle]

is 1 an upper bound?

[severe pond]

yes

0 is the limit of the sequence though

which is why i mentioned it

[spiral saddle]

I shee

can you help me write a formal argument for c)?

[severe pond]

start it off by saying if r >= l then r > l or r = l

and mention how the r > l case has already been handled

in part a

[severe pond]

maybe note this

something like

if r >= l then r > l or r = l. if r > l then by part a it follows that there exists n_0 such that a_n < r whenever n >= n_0 so a_n <= r

and then can you try the r = l case

[spiral saddle]

what am I supposed to say about d)?

[severe pond]

can you show me your solution for the r = l case first

[spiral saddle]

r != l

[spiral saddle]

oops, ok

[severe pond]

you good?

[spiral saddle]

no

[severe pond]

where are you stuck

[spiral saddle]

r = l

[severe pond]

so we established it’s not necessarily true if r = l

yes?

[spiral saddle]

oh, you mean I need a counterexample

[severe pond]

that works

[spiral saddle]

right

i was trying to show it was true

[severe pond]

oh

[spiral saddle]

an < r => an <= r?

because < is more strict

and because a <= b is by definition a < b or a = b

and in an OR we just need one of them

[severe pond]

i already wrote out the r > l case

[severe pond]

yes

[spiral saddle]

ok

how did you know the equality case was false

how did u figured?

[severe pond]

i translated the symbols into something concrete which i knew wasn’t true

[spiral saddle]

for r = l is it possible to assume its true and get to an absurdity

[severe pond]

sure

[spiral saddle]

h0w

[severe pond]

so the statement becomes if r = l then there exists n_0 such that a_n <= r = l for all n >= n_0

where a_n —> l

a subtlety here is that a_n is an arbitrary sequence with limit l

[severe pond]

so it’s really saying for all sequences a_n that converge to l, this holds

to contradict it youd need to find a single sequence with limit l so that for every n_0 there is some n >= n_0 such that a_n > l

take a_n = l + 1/n

the contradiction is it doesn’t hold for every sequence which is the first condition we had

[severe pond]

this has limit l but a_n > l for every n in fact

do you follow?

[spiral saddle]

i see

[spiral saddle]

for every n there exists some n0

[severe pond]

i negated the statement

we’re trying to show it’s false

so show its negation is true

[spiral saddle]

p -> q = not(q) -> p

[severe pond]

you mean not p

[spiral saddle]

contrapositive?

p -> q = not(q) -> not(p)

[severe pond]

i didn’t mention that but sure

[spiral saddle]

you are not using it?

[severe pond]

i mean contradiction and contrapositive are much of the same

just a difference in wording usually

[spiral saddle]

why you negated

[severe pond]

^

proving the negation is true is the same as proving the original is false

[spiral saddle]

right thats contrapositive

[severe pond]

whatever youd like to call it

doesn’t matter

thanks bro

[spiral saddle]

goddamn

[severe pond]

just withdrew 2500

🙏🏻

[spiral saddle]

mrbeast

[severe pond]

fuck that guy

[spiral saddle]

🤔

[severe pond]

maybe it’s easier to see with symbols

[spiral saddle]

forall n in N there exists some n0 in N negated is forall n0 in N there exists some n in N?

[severe pond]

$\lnot(\forall (a_n){n \in \mathbb{N}} : a_n \to \ell, \exists n_0 \in \mathbb{N} ,, \forall n \geq n_0, ,, a_n \leq \ell) \iff \exists (a_n){n \in \mathbb{N}} : a_n \to \ell ,, \forall n_0 \in \mathbb{N} ,, \exists n \geq n_0 : a_n > \ell$

[flat frigate]

knief

[severe pond]

a bit sloppy

but you get the point

[spiral saddle]

i shee

[severe pond]

i gave such a sequence

a_n = l + 1/n

[spiral saddle]

seems hard

[severe pond]

for every n_0 just take n = n_0

then n >= n_0 and a_n > l

[severe pond]

it’s not so bad, just find a sequence that converges to l from above

[spiral saddle]

if n = 1

a1 = l + 1

[spiral saddle]

this converges to l but is not upperly bounded by l

[spiral saddle]

wdym from above

[severe pond]

like the sequence members are coming from above l

they shrink towards l but they are on the right

so not bounded above by l

[spiral saddle]

so l is a lower bound

[severe pond]

sure

it’s easiest that way

[spiral saddle]

what

[severe pond]

like the easiest thing to do is make l a lower bound

we didn’t have to

[spiral saddle]

but

[severe pond]

yes i did make it a lower bound

i’m just saying this wasn’t our only option

it could shrink around l from both sides

something like

a_n = l + (-1)^n/n

l still isn’t an upper bound here

but it’s also not a lower bound

now the sequence is approaching l from both sides

but what’s important is that it isn’t only coming from below

this is what the question was asking

[spiral saddle]

aghh dude

I dont follow

is there a simpler route

[severe pond]

the original question was asking if a_n —> l, does l have to be an eventual upper bound for a_n

we found a sequence that showed this wasn’t the case

namely a sequence that converges to l without l being an eventual upper bound

you can do this either by having l be a lower bound or by l being neither a lower bound or an upper bound

[spiral saddle]

I see

[severe pond]

if l is a lower bound then all of my sequence members are coming from the right of l

if l is neither a lower bound or an upper bound then i have sequence members getting closer to l on both sides

i gave examples of both

in both cases l isn’t an upper bound

[spiral saddle]

but l = r

[severe pond]

yea

[spiral saddle]

oh right an <= r and so an <= l

ok I get it now

[severe pond]

you can use l and r interchangeably since we are working under the assumption that l = r

that was the case we have been focusing on

[spiral saddle]

an <= l for all n in N means l is upper bound of an

[severe pond]

it just mentioned eventually being an upper bound

so

for all n >= n_0

[spiral saddle]

we know an -> l we just need to find a sequence that has limit l but that l is not an upper bound

[severe pond]

is there some n_0 so that for all n afterwards, l is an upper bound

[severe pond]

yes

[spiral saddle]

but take for example an = l + 1/n

take n0 = 2

then forall n in N such that n >= 2

it is true that l is an upper bound and that an converges to l

@severe pond sussy

[severe pond]

how is it an upper bound

[spiral saddle]

my bad

my bad

[severe pond]

you’re claiming that l >= a_n for all n >= 2

so l >= l + 1/n for all n >= 2

[spiral saddle]

my bad

[severe pond]

this is equivalent to 0 >= 1/n for all n >= 2

not true

[spiral saddle]

yeah srry

math is tough

[severe pond]

yep

[spiral saddle]

can you help with d

[severe pond]

i can

send it again

did you get a chance to word part c though

[spiral saddle]

no

I mean i got it tho

[severe pond]

alright

[spiral saddle]

we spent like 30 min or more talking about it, until I finally understood c)

[severe pond]

that’s ok

[severe pond]

ok so if r is an eventual strict upper bound for a_n what can we say about l

in relation to r

there are three possibilities to consider

can you guess what those might be

[spiral saddle]

if l < r then this is true

r != l

is required i think

[severe pond]

why?

[spiral saddle]

I am trying to verbalize it

suppose r = l, then an -> r and we know that an -> r and r being a upper bound is not valid for all sequences

[severe pond]

this is true by part a

[spiral saddle]

yes

[severe pond]

hmm

it doesn’t have to necessarily be true

is it ever true

we don’t care if it’s always true, just if it’s ever true

is it possible that l = r

ok the better way to say it is, is it possible that l is an eventual upper bound for a_n

i’m using l and r interchangeably here assuming l = r

[spiral saddle]

my intuition from the drawing tells me no, because we can have sequences that get outside of the radius

but, idk

[severe pond]

i think you’re misunderstanding what the question is asking

[severe pond]

it’s asking for the set of possible l such that that condition holds for some sequence converging to l

we know that every l < r works because of part a

regardless of the sequence a_n

if l = r there are in fact sequences a_n for which this holds

[spiral saddle]

yes for example the constant sequence an = c, c is an upper bound aswell as a lower bound of an

[severe pond]

well we want a strict upper bound

we want a_n < l

so like

take a_n = l - 1/n

l is a strict upper bound

and the limit of a_n

so this works

[spiral saddle]

right

[severe pond]

so it’s possible that r is the limit of a_n here

all we know about r is that it’s an eventual strict upper bound

we are asking if l can equal r

i.e. can r be the limit of a_n

we just answered it

it can

it’s possible

so can l = r

and l can definitely be less than r

so l <= r

[spiral saddle]

what is the difference between c) and d)

[severe pond]

can l be greater than r?

[severe pond]

c is asking if l <= r implies r is an eventual upper bound for a_n. d is asking for a condition on l if r is an eventual strict upper bound for a_n

[spiral saddle]

dude

if l > r and r is an strict upper bound

then l is another strict upper bound

[severe pond]

true

what else

[spiral saddle]

you are asking if its possible that an -> l aswell as l is an strict upper bound

yes it is possible

take an = l - 1/n

this converges to l and l is an strict upper bound

[severe pond]

hmm

but you’re missing something

r is a strict upper bound that is less than l

if r is below l while also being a strict upper bound, how can a_n get close enough to l to converge to l

[spiral saddle]

don't get it

[severe pond]

imagine a number line

r is some point to the left of l

and every sequence member is below r

can a_n converge to l?

or is the sequence too far away

[spiral saddle]

no, it converges to r

[severe pond]

not necessarily

let’s say a_n = 1/n, r = 2, l = 3

clearly this won’t work

my points need to cluster around l

since r < l there is some fixed distance from my sequence to l

i have a_n < r < l

i can never get in between r and l

so i can never get close enough to l to converge to l

[spiral saddle]

if l > r and an -> l then r is never an upper bound

[severe pond]

well it might be an upper bound for finitely many of the terms in the beginning

it’s just not an eventual strict upper bound

it can never be an upper bound for the tail

[spiral saddle]

how to formalize

[severe pond]

so we already established that l <= r

now we are showing it’s impossible for l > r

ok

so assume l > r

we are trying to contradict l being the limit of the sequence while r being an eventual strict upper bound

since r is an eventual strict upper bound there is some n_0 such that a_n < r for all n >= n_0. but a_n —> l so there exists N >= n_0 such that
|a_n - l| < l - r whenever n >= N. hence
r - l < a_n - l < l - r <—> r < a_n < 2l - r

contradiction

since a_n < r

[severe pond]

btw i said there exists N >= n_0 here because we know first that there is some N so that for all n >= N, |a_n - l| < l - r from the limit definition but i also want N >= n_0 to contradict the eventual upper bound so i can just take the max of N and n_0

if N was less than or equal n_0 then choose n_0 as N

[spiral saddle]

nice!!

[severe pond]

else, keep N

[severe pond]

thank you thank you

math is fun

[spiral saddle]

is it?

[severe pond]

if you’re autistic enough, sure

💀

[spiral saddle]

what?

[severe pond]

well in the usual limit definition you just say

there exists N in N

[spiral saddle]

n0

[severe pond]

yes but we already used n_0 for a label

[severe pond]

i specified that i wanted N to be >= n_0 where n_0 comes from r being an eventual upper bound

it’s subtle but important

[spiral saddle]

ok i see

[severe pond]

i did that because i wanted both conditions to hold simultaneously

i wanted both r to be an upper bound for those n and for a_n to be within that distance from l

but clearly they couldn’t both be true

since i got contradicting inequalities

so our assumption that l > r was wrong

it can never be true

hence

l <= r

[spiral saddle]

why do you need an n0 for the strict upper bound, there exists some n0? like if r is an strict upper bound its that an < r forall n in N

[severe pond]

it doesn’t say it’s a strict upper bound for all n

it says

for all n >= n_0

[spiral saddle]

i shee

[severe pond]

so i wanted to choose my N >= n_0 as well

[severe pond]

so that i get this

[spiral saddle]

i see

hardest shit I have done in my life

[severe pond]

it only gets worse from here

💀

[spiral saddle]

why do you need N >= n0

?

[severe pond]

so that a_n is both bounded above by r and within a distance l - r from l

would it be easier to add an extra step where you make K = max(n_0, N)

then take n >= K

it’s the same thing

[severe pond]

and have no condition on N other than N in N

[spiral saddle]

i shee

jajaj

math really is nasty

[severe pond]

so why study it

🤔

[spiral saddle]

well not math, this exercise

really got me sweating my ass on the chair

[severe pond]

😭

it’s over now

[spiral saddle]

im sorry

[severe pond]

what for

[spiral saddle]

is great when you understand it but painful when you not, ykwim

[severe pond]

yep

much more enjoyable from this side

[spiral saddle]

jajaja

you probably also have your ups and downs

[severe pond]

yep

plenty of times smashing your head in a wall trying to figure something out

[spiral saddle]

I appreciate the help, sorry for making u waste so much time

[severe pond]

it’s alright

4 hours

[spiral saddle]

yes

[severe pond]

hopefully you got something out of it

[spiral saddle]

i will re do it after a break

this exercise

i think it gets better with time

[spiral saddle]

i did learnt a lot today, though this shit is tough

I was having an easy time with the exercises 1 to 10

[severe pond]

it takes a lot of focus and effort

you need to invest a lot of time into it

[spiral saddle]

exercises 10 and 11 stuff started to get complicated real fast

[spiral saddle]

i see

[severe pond]

alright well i’m going to get off discord now

good luck sir

[spiral saddle]

this exercises made it look like my linear algebra questions were easy

shit definitely started to get real very fast

[severe pond]

not surprising

there are a lot of moving parts you need to keep track of in analysis

[spiral saddle]

cya, tyvm for your help sir

[severe pond]

very precise

[severe pond]

you’re welcome

[spiral saddle]

.solved

[safe radish]

Channel closed

Closed by @spiral saddle

Use .reopen if this was a mistake.

[safe radish]

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[safe radish]

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[normal moss]

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Available help channel!

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[safe radish]

Please don't occupy multiple help channels.

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[primal bone]

https://klipy.com/gifs/focalors-hydro-archon-6

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Please don't occupy multiple help channels.

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[twilit cargo]

hi, want to double check my working with someone, all the notations and stuff too please

thanks!

[tardy mango]

saying that a sum is equal to the limit of its terms is wrong

you should say you're using the test for divergence and then show you computing the limit

and as always, writing out arithmetic with infinity is abusive as well

the rest is fine

[twilit cargo]

i see, so absuive as in not recommended or illegal in maths?

good now?

[warm warren]

Not good practice. Some profs may flare up at the use of it.

[twilit cargo]

i see, so should i just skip that step and just right 3 + 0 + 0 ?

like in the next one?

gotcha

[twilit cargo]

^

[warm warren]

Well i cant see the bottom part

its only the first half

[twilit cargo]

sory

[warm warren]

you may want to include $\lim_{x\to \inf} 1/x$ instead of $1/\inf$

[flat frigate]

Annie Maqionde

[twilit cargo]

ah so like that rt

[warm warren]

yup all good

[twilit cargo]

thank you

.close

[safe radish]

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[desert pasture]

How is it that $x^n = \overbar{a(x)b(x)}$

[flat frigate]

Wai
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[desert pasture]

isn''t 1 in P too

[gaunt elk]

here ?

[desert pasture]

yes

[gaunt elk]

no 1 isn't in P else it would be the whole ring by multiplicativity

[desert pasture]

one more thing

[gaunt elk]

Ok by hypothesis all the coef of f are in P except the first one

[desert pasture]

I don't get the contradiction

a_0 in P^2 fine

[uneven knoll]

Well no it's not fine

Because you are assuming that a_0 is not in P^2

[gaunt elk]

let a_0 and b_0 be constant terms of a and b

[desert pasture]

oops, right

and as 1 not in P that is possible

makes sense

tysm again

[uneven knoll]

Yeah 1 is not in P because a prime ideal is by definition an ideal ≠(1) such that ...

[desert pasture]

.close

[safe radish]

Channel closed

Closed by @desert pasture

Use .reopen if this was a mistake.

[safe radish]

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[simple galleon]

I heard in this YouTube video that all parabolas are similar but I didn't really understand the proof.
I want to start by creating a definition of similarity that uses the coordinate plane.

[simple galleon]

,tex the two functions $f$ and $g$ are similar iff $\\forall x,\frac{f(x)}{x}=\frac{g(x)}{x}$

[flat frigate]

bored amogi

[simple galleon]

Does this work? Why or why not?

[west hinge]

can you not express f and g in terms of ax2+bx+c and dx2+ex+f?

[upbeat swan]

Well

[simple galleon]

yea

[west hinge]

and then divide by x and show that they're equals up to a constnat

[upbeat swan]

First of allyour definition doesnt work for x=0

[simple galleon]

oh thats true

[west hinge]

how does that work though??

what is the definition of similar??

[simple galleon]

i got no clue

[west hinge]

it was up to a constant now it's equals

[upbeat swan]

The definition should just be

[west hinge]

and then even if its up to a constant not all parabolas are 'similar'

[humble egret]

isnt this supposed to be fx=kgx

[upbeat swan]

f(x) = k g(x), for all x!=0

[west hinge]

because you can ensure a=kd but then what about b=ke?

[simple galleon]

idk tbh

[upbeat swan]

Dividing by x makes no sense

[simple galleon]

he was saying something about the 'aspect ratio'

which i imagine would be y/x

[west hinge]

I think what you're looking for is:

suppose f and g are 2 parabolas

Tf=g for all f,g belonging to the set P2(F)

(set of degree 2 polynomials over a field)

and T is a linear transformation

?

[simple galleon]

whats a 'linear transformation'?

oh a transformation that preserves linearity right

uh maybe? im not sure

[west hinge]

lmao idek what your question is

[simple galleon]

he did say that all circles are similar, maybe we can start with that instead?

[humble egret]

the question looks simple enough what is this linearity thing lmao

if f(x)=4x^2+6x equals to g(x) if and only if f(x)= Kg(x) then k is a whole number and it should look something like 2g(x)=2(2x^2+3x) is it not?

[simple galleon]

why not x=0?

[simple galleon]

um, f(x)=4x^2+6x=g(x) iff f(x)=kg(x) is false?

unless k=1

[upbeat swan]

"There exists a real number k such that f(x) = k g(x) for all x in the domain"

[simple galleon]

that wouldnt work--translation is an isometry and translation isnt covered under that definition

[humble egret]

if k is one it is false and shouldn't it be different and special for every function?

[simple galleon]

could you prove how thats false?

oh sorry i forgot its f(x)=4x^2+6x=g(x)

or am i misunderstanding you?

[upbeat swan]

Yeah i think you re right

Let me see

[humble egret]

because f(x) is not equal to g(x) for the specific f(x) we are dealing with for k=1, k can have values specific for g(x) u are dealing with when multiplied altogether u should be getting fx . hey i think what u are looking for is something more rigorous and maybe i would not know that, what are u working on ?