help-38

[acoustic snow]

no?

oh i forgot for all n > n_0

[lean prawn]

Note that this is slightly nontrivial but it is true

but I think we need a concrete upper bound to solve the original problem

[acoustic snow]

right

so isnt this what you were saying?

[lean prawn]

so here's a picture of what I'm trying to do

So maybe there's some sequence that converges to c'

now note, that that sequence can bounce around above and below c' for all of infinity if it chooses

the only requirement is that it gets closer and closer to c' in the epsilon-delta fashion

We're looking for some upper bound for that sequence

We can't choose c' because that's not necessarily an upper bound

[acoustic snow]

ahh

right

[lean prawn]

So a clever hack is to choose c=(c'+1)/2

And that is actually gonna be an upper bound, not necessarily for the beginning part of the sequence, but after some N

[acoustic snow]

yeah

[lean prawn]

And the reasoning is that we can choose epsilon=(c-c')/2, and within that range everything is gonna be below c for sure

So the definition of the limit says that there exists some N such that after N, every term is within (c-c')/2 distance away from c', and such terms are gonna be bounded above by c

[acoustic snow]

and what do we need epsilon for?

[lean prawn]

and also c<1

[acoustic snow]

right

[lean prawn]

which gets us back to this, after a certain point all the terms of the sequence are bounded above by some quantity c, which is strictly less than 1

[acoustic snow]

right

[lean prawn]

So we find that every time you jump ahead 2024 terms in the original sequence, we multiply the term by c<1

but of course we can keep doing that indefinitely, getting us arbitrarily close to 0, no matter what positive term we started on

[acoustic snow]

wdym

[lean prawn]

This tells us that

$a_{n+2024}<c\cdot a_n$

$\forall n\in\mathbb{N}$

[solid kiln]

Stipendi

[lean prawn]

So for instance a_{2025}<c*a_1

[acoustic snow]

ah

[lean prawn]

And a_{2025+2024}<c*c*a_1

etc.

[lean prawn]

Yeah this was wrong, every time you jump ahead 2024 terms in the sequence, multiplying by c<1 forms an upper bound for that term, is what I should've said. But in any case we find an arbitrarily small upper bound for the sequence, starting at some potentially very large index

[acoustic snow]

yeah

[lean prawn]

which is what it means for the sequence to converge to 0

[acoustic snow]

right

but how do i formalize this

[lean prawn]

haha that's gonna be a good exercise for you

[acoustic snow]

the tldr is that starting from a certain place a_n+2024 < c*a_n

wait

doesnt that just tell us that the subsequence $a_{n+2024k}$ converges to 0?

[solid kiln]

sibber

[acoustic snow]

for any natural k

why does that prove that a_n converges to 0?

[lean prawn]

You mean

$a_k=a_{1+{2024(k-1)}}$

?

[solid kiln]

Stipendi

[acoustic snow]

oh yeah that

take every 2024th element form a_n

[lean prawn]

Well if you let epsilon>0 be fixed, then just choose n such that c^n < epsilon, which exists because c<1, and now N = 1+2024*n has the required property for the statement you're trying to prove

[acoustic snow]

what is N?

[lean prawn]

A quantity such that if n>=N (and note this is a different n, sorry about that) then |a_n-0|<epsilon

[mild yew]

+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++p

]

[lean prawn]

Did you fall asleep on your keyboard?

[acoustic snow]

where did c^n come from?

and what is this n were choosing

lol sorry im confused

[lean prawn]

no worries

I really regret that I used n twice for two different purposes so I will restate the same thing with a different variable name

For any epsilon>0, there exists some positive integer t such that c^t < epsilon

So maybe c=0.98 (we know it's less than 1) and maybe epsilon=0.01

If you make t big enough then eventually 0.98^t < epsilon

[acoustic snow]

right

[lean prawn]

Now, by what I said here, if you jump ahead in the sequence 1+2024t terms, then that term will be bounded above by c^t

because for every 2024 jump you get another c factor

[lean prawn]

this thing

So, if we choose N=1+2024t, then for any term a_n such that n>=N, we get that it's gonna be bounded above by c^t

but c^t itself was bounded above by epsilon

so therefore the terms are bounded above by our epsilon

and since epsilon>0 was arbitrary, that shows that the sequence converges to 0

[acoustic snow]

but this is only starting from a certain n

or lets call it n_0

[lean prawn]

ok you're right I'm missing some details

[acoustic snow]

so what we have is this

[lean prawn]

I guess you have two options, one would be to appeal to the fact that you have a subsequence that converges, the other option is maybe to say that if the upper bound works for some $a_n$ then it also works for $a_{n+2024}$, and remember that my argument for why $a_{1+2024t}$ is bounded above by epsilon is perefctly sound, but what about $a_{2+2024t}, a_{3+2024t}, \dots, a_{2023+2024t}$

[solid kiln]

Stipendi

[acoustic snow]

we can say

that every subsequence like this converges to 0

and that all of those subsequences together = a_n (starting from n_0)

like intuitively that makes sense

but how to formalize lol

[acoustic snow]

right?

[lean prawn]

And I guess you could argue for example that if one of those is not bounded above by epsilon, then add 2024 until it is, and keep repeating that until all of them are bounded above by epsilon, which will terminate in finite steps because you only have 2023 terms to go through and each of them will terminate in finite steps, and keep track of how many times you have to jump ahead, and add that to your previous N and that's your new N

[acoustic snow]

ohh yeah i forgot n isnt a real number

[lean prawn]

Of course I can't make any promises because I don't know who's grading your homework but the submission probably needs to be less formal than you think as long as it contains the right ideas

I would think that everything I've said, written in your own words, is probably good enough

[acoustic snow]

i do have an idea of how formal they want it to be

[acoustic snow]

anyway now im trying to formalize this part

[lean prawn]

Basically if you can find $t_1$ such that $a_{1+2024t_1} < \epsilon$, then surely you can perform the exact same argument to find $t_2$ such that $a_{2+2024t_2}<\epsilon, \dots, t_{2023}$ such that $a_{2023+2024t_{2023}}<\epsilon$, and now $N=1+\max{t_1, \dots, t_{2023}}\cdot 2024$ will work

[solid kiln]

Stipendi

[acoustic snow]

right

[lean prawn]

basically what I said here written more formally

[acoustic snow]

yeah

hmm i actually dont think i need to exlpain that

i mean that

i can just directly say that $\forall n > n_0$ $a_{n+2024k} < c^k a_n$

[solid kiln]

sibber

[acoustic snow]

no?

[lean prawn]

yess very good

[acoustic snow]

so now how do i say that it converges to 0 lol

[lean prawn]

I need to be super careful with every message I send because if I confuse myself I will definitely confuse you

[acoustic snow]

hahaha

hmm i can say that $c^k a_n \rightarrow 0$

[solid kiln]

sibber

[acoustic snow]

without proof

and a_n+2024k < that

so all of those converge to 0

[lean prawn]

I would think about this inequality for $n=n_0+1, n_0+2,\dots, n_0+2023, n_0+2024$

[solid kiln]

Stipendi

[lean prawn]

For each of those values of n, we "clearly see" (or at least I see) that there exists some k such that we get an upper bound that is arbitrarily close to zero

and in particular can be made smaller than epsilon

[acoustic snow]

just to make sure, your epsilon is my c right?

one sec let me show you how i defined it

[lean prawn]

And since there is a consecutive list of 2023 terms that are all bounded above by epsilon starting at index (compute the index here), it means that after that index all the terms will be bounded above by epsilon

[acoustic snow]

i think its enough to say the union of all those sequences is equal to a_n after that n_0

or thats almost the same thing

[lean prawn]

Sorry this is not quite right, the terms are not consecutive, but they can be made consecutive by choosing the largest of the found indexes and doing some black magic calculations

[lean prawn]

Yeah kinda the problem with this is that you're almost taking on more than you can chew

I mean technically there is no problem and you absolutely can do it that way

but it's a bit painful

[acoustic snow]

if i just add a small explenation for this part i think the proof would be enough

[lean prawn]

because here I'm saying like let me find some t2 such that 2+2024t_2 is a nice index, and then I'm actually handwaving the fact that you first need to pass what you call n0 and then you can give the c*c*...*c argument

[acoustic snow]

wdym handwaving

[lean prawn]

whereas you're being maybe a bit more honest about how the indices are computed, at the cost of getting indices where you're not quite sure what the n in a_{n+2024k} actually is

[acoustic snow]

ah

yeah

but this way i can just union all of those n's and get the origina sequence

(ignoring the first n_0 - 1 elements)

so i wouldnt need to explain that for every 2023 it repeats etc

[lean prawn]

maybe nitpicking but your first inequality is basically saying

$\frac{a_{n+2024}}{a_n}-L<\epsilon$

[solid kiln]

Stipendi

[lean prawn]

while the definition of the limit would have absolute values around the LHS

[acoustic snow]

oh yeah no this is a

how do you say

feature?

something thats derived from limsup

that we proved in class

[lean prawn]

ohh ok I actually don't know a lot about limits so I'll just take your word for it lol

yeah ok the sup part is probably saying that the sequence doesn't do the "bouncing around" that I warned you about

so I think we overengineered the problem and you were originally right that the limit itself would've sufficed as an upper bound XD

sorry about that

[acoustic snow]

that $\limsup a_n = L$ iff $\forall \epsilon > 0$ $\exists N \in \mathbb{N}$ s.t. $\forall n >= N$ $a_n < L + \epsilon$ and something else

[lean prawn]

but if it excites you at all then you can feel proud that this argument would generalize to a regular limit as well

[solid kiln]

sibber

[lean prawn]

well idk actually

in any case I'll continue reading your proof

[acoustic snow]

it does say that it can bounce around L (< L + epsilon) but also that the limit itself (not limsup) will be around that same L (< L + epsilon) (where L is the limsup)

wait no

lmao

its essetially saying that if L is a limsup for a_n then starting from a certain index, a_n < L + epsilon

(but it doesnt say that a_n > L - epsilon)

(which is why thats only true if L is a limsup)

[lean prawn]

The second line is pretty sus because if L=1/2 then c=3/2 which contradicts what you say later

[acoustic snow]

oh right

$\epsilon$ should be $\frac{L + 1}{2} - L$

[solid kiln]

sibber

[acoustic snow]

thanks for catching that!

[lean prawn]

The line

$\lim_{n\rightarrow \infty} c^k a_n =0$

[solid kiln]

Stipendi

[lean prawn]

is pretty sus because if that's true for some k then you can just take c^k out of the limit and you have what you wanted to prove

[acoustic snow]

thats because c < 1

[lean prawn]

but obviously it can't be that easy

[acoustic snow]

no c^k is a sequence from k to inf

[lean prawn]

should the limit have k\rightarrow\inf ?

[acoustic snow]

actually yeah that does need some explenation

[acoustic snow]

oh

oops

brainfart

i confused myself with that

yeah that should be tha limit as k approaches inf not n

[lean prawn]

then it makes sense but what you have after doesn't really follow convincingly

[acoustic snow]

that should also be the limit on k

let me send the fixed version

i should also emphathize that c > 0

[lean prawn]

Isn't

$\bigcup_{n\ge n_0} {a_{n+2024k}}_{k=0}^\infty$

the same thing as

$\bigcup_{n\ge n_0} {a_{n}}

or do I not understand what this notation means

[acoustic snow]

i dont know what the bottom thing means

oh

[solid kiln]

Stipendi
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[acoustic snow]

this defines the series to be on k

so if it were a function it would be a(k) = n + 2024k (where n is a fixed n >= n_0)

[lean prawn]

okay I see what you're saying

[acoustic snow]

the union means merge all of sequences for all n's >= n_0

[lean prawn]

yeah I think it's good

well done

[acoustic snow]

:)

thanks!

now the second harder part

[lean prawn]

np ^^

oh no

[acoustic snow]

hahaha

[lean prawn]

I was like "lemme just quickly check if I can get some instant dopamine on the math server and then I'll start working on my thesis"

[acoustic snow]

hahahaha

yeah unfortunately these exams are hard

so youre grad not undergrad like you role says :)

[lean prawn]

it's my bachelor thesis and it's mostly done but I haven't submitted it yet so I think that means I'm still an undergrad

I don't think I want to take the postgrad role even when I do graduate because I don't want additional pressure on me

[acoustic snow]

you do a thesis for a bachelors?

[lean prawn]

yeah we do that in Finland

[acoustic snow]

oh

[lean prawn]

it's much smaller than a master's thesis

[acoustic snow]

cool

[lean prawn]

but honestly I probably could've passed my thesis off as a master's thesis

but it's too late now, I overengineered it and I have to live with my decisions 😂

[acoustic snow]

hahah :)

thats cool tho it shows passion

[lean prawn]

haha thanks

Sometimes I wish I could enjoy normal things like video games

makes it super awkward the one time I agree to play video games with my friends just to please them

XD

[acoustic snow]

enjoying less normal things makes you more unique and interesting

[acoustic snow]

ah yeah lol

anyway thanks a lot for the help!

.close

[trim jolt]

Channel closed

Closed by @acoustic snow

Use .reopen if this was a mistake.

[lean prawn]

no worries!

[acoustic snow]

maybe normal is the wrong word here

conventional

[lean prawn]

I'm quite unconvential in many ways

[acoustic snow]

which is cool!

[lean prawn]

aww

[trim jolt]

Available help channel!

Send your question here to claim the channel.

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• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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[vast light]

Hey when solving for a right triangle missing angles

[vast light]

How do I know which trig function to use

SOH CAH TOA

I have my b given the opposite so I know it would be sin or tan

[prisma stone]

Can you send the question

[vast light]

Yea

Problem is 43

That’s the angle given

[prisma stone]

What is the question asking for

[vast light]

It wants me to solve the right triangle

Finding the missing angles

C and A

[prisma stone]

sin(x) = opposite / hipotenus
sin(83.7) = 3.21 / c

You can use tan or cot to find a

[vast light]

Oh ok why wouldn’t it be tangent?

Instead of sine

I’m just confused on which trig function to use first

[prisma stone]

If you want to find hipotenus you should be looking to use sine and cosine

[vast light]

Right so if I have for instance B and C I need to find I should always look for C first?

[prisma stone]

I mean there is no "first" or "last" really

[vast light]

Ok thank you that’s all I needed!

[trim jolt]

@vast light Has your question been resolved?

[trim jolt]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[trim jolt]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[marsh forum]

Prove that if the limit of $f$ at $a$ and that of $g$ at a, both exist, $\lim_{x\to a} (f(x)+g(x))= \lim_{x\to a} f(x)+ \lim_{x\to a} g(x)$

[solid kiln]

Veni, vidi, perii

[marsh forum]

so I have $|f(x)-L_1|< \varepsilon_1; |g(x)-L_2|< \varepsilon_2$

[solid kiln]

Veni, vidi, perii

[marsh forum]

adding them and using the triange inequality, I get

$|(f(x)+g(x)-(L_1+L_2)|< \epsilon_1+\epsilon_2$

[solid kiln]

Veni, vidi, perii

[lime sphinx]

you can set \epsilon_1 and \epsilon_2 to any values you want > 0 like epsilon over 2 for instance to complete the proof

[marsh forum]

Now $\epsilon_{1,2} \leq 1$

[solid kiln]

Veni, vidi, perii

[marsh forum]

so $|(f(x)+g(x)-(L_1+L_2)|< 2$

[solid kiln]

Veni, vidi, perii

[marsh forum]

But at the same time I know |x-a|< \delta$

hmm

[lime sphinx]

oh you seem to have a bit of trouble with the epsilon delta proof mechanism.

here is a setup

[marsh forum]

Hm

I see

thanks

so $|x-a| < \delta \implies |f(x)-L| < \epsilon_1$

[solid kiln]

Veni, vidi, perii

[lime sphinx]

yes for any a

[marsh forum]

Don't we usually find a delta given an epsilon though

[willow urchin]

since we're talking about limits we need 0<|x-..|

[lime sphinx]

depends on the definition given

[willow urchin]

definition of limit excludes x=a

you might be thinking of continuity

[lime sphinx]

not in france i guess

[lime sphinx]

oh maybe analysis was a long time ago

ah yeah would make sense sorry

[marsh forum]

this is a calc 1 course here

I mean we have done some calc at school, but at uni this is my 101 course

analysis in year 2

anywaym I digress

[marsh forum]

what about this

[willow urchin]

this is a more abstract limit so we need to think different

[marsh forum]

Then why not write the implication backwards

[willow urchin]

pola's work is right (except there should be 0<|x-..|)

[lime sphinx]

yeah you still need a delta that will make y or a close enough to x so both hypotheses can be used

[willow urchin]

look at delta1,2 to find a good delta

[marsh forum]

Not sure How I'd do that tbh

like $|x-a|<1$

say

[solid kiln]

Veni, vidi, perii

[marsh forum]

so $x \in (a-1,a+1)$

[solid kiln]

Veni, vidi, perii

[marsh forum]

I mean as the limit exists for both

there exists some delta , such that delta_1 \leq \epsilon_1

and $\delta_1 \leq \epsilon_2$

[solid kiln]

Veni, vidi, perii

[willow urchin]

lets look back to pola's post

https://i.gyazo.com/0ce53b3fe5783010deba14418836743a.png
https://gyazo.com/c5ad9d5617ccd4e6a0db142732b28ca5

call these implications 1) and 2)

youre writing the final inequality |f+g-L1-L2|<.. where 0<|x-a|<delta with delta TBD

give a sufficient condition on delta so we can use 1)

[marsh forum]

$\delta<\epsilon_1$

[solid kiln]

Veni, vidi, perii

[willow urchin]

that doesnt make sense

[marsh forum]

Yea, just realised

idk

tbh

[willow urchin]

we want to use 1) to say |f-L1|<e1 but first we need 0<|x-a|<delta1

[marsh forum]

yes

[willow urchin]

we're at 0<|x-a|<delta with delta TBD, give a sufficient condition on delta so that 0<|x-a|<delta1

[marsh forum]

$\delta< \delta_1$

[solid kiln]

Veni, vidi, perii

[willow urchin]

that works, actually we can be looser with delta<=delta1

0<|x-a|<delta is already strict

[marsh forum]

we also have $\delta < \delta_2$

[solid kiln]

Veni, vidi, perii

[willow urchin]

take d<=d2

now define d so that d<=d1 and d<=d2

[marsh forum]

so $\delta < \frac{\delta_1+ \delta_2}{2}$

[solid kiln]

Veni, vidi, perii

[willow urchin]

you sure?

is the average smaller than both?

[marsh forum]

no

[willow urchin]

think of a function which is <= both inputs

[marsh forum]

Hmm, I also know $a<\sqrt{ab}<\frac{a+b}{2}<b$

[solid kiln]

Veni, vidi, perii

[marsh forum]

maybe $\delta_1<\delta_2$

[solid kiln]

Veni, vidi, perii

[willow urchin]

thats not guaranteed

at this point we dont need more restrictions, just think up the right function

[marsh forum]

Is it a simple function

[willow urchin]

kinda

[marsh forum]

min(something)?

[willow urchin]

yes

[lime sphinx]

yes

[willow urchin]

$\min(a,b)\le a,b$

[solid kiln]

RokettoJanpu

[marsh forum]

oh right

$\leq$

[solid kiln]

Veni, vidi, perii

[marsh forum]

I was thinking of strict inequalities

so $min(\delta_1,\delta_2)=\delta$

[willow urchin]

no set delta AS that

[solid kiln]

Veni, vidi, perii

[willow urchin]

now everything is in place

[marsh forum]

Hmm, so $|x-a| \leq \delta_1 \implies \forall \varepsilon_1 |f(x)-L|< \varepsilon_1$

wait

that's wrong

it's for all epsillon

[solid kiln]

Veni, vidi, perii

[willow urchin]

lets just build off pola's post, ill repost it

https://cdn.discordapp.com/attachments/908078008609431674/1278540372439597066/IMG_2671.png?ex=66d12d02&is=66cfdb82&hm=8265a796c45b007b99cc2f2cfbc2f82a5bc9cd8f77bf06aff4e0455ca0faaf1a&

[marsh forum]

yes

okay

[willow urchin]

dont forget 0<|x-a|

er wait

[marsh forum]

so $0<|x-a|< \delta_1 \to |f(x)-L_1| < \epsilon_1$

[willow urchin]

@lime sphinx the foralls should be on x

[solid kiln]

Veni, vidi, perii

[marsh forum]

and $0<|x-a|<\delta_2 \to |g(x)-L_2|< \epsilon_2$

[solid kiln]

Veni, vidi, perii

[marsh forum]

so I add them up now

so $|f(x)-L_1|+|g(x)-L_2|<\epsilon_1+ \epsilon_2$

[solid kiln]

Veni, vidi, perii

[willow urchin]

if youre finishing the proof you need to write it in the right order

whats delta

[marsh forum]

right

$0<|x-a|< min(\delta_1,\delta_2) \implies |f(x)-L_1|+|g(x)-L_2|<\epsilon_1+ \epsilon_2$ $

[solid kiln]

Veni, vidi, perii
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[willow urchin]

lets write it like this

in this exact order

set $\delta=\min(\delta_1,\delta_2)$

[solid kiln]

RokettoJanpu

[willow urchin]

if $0<|x-a|<\delta$ then $0<|x-a|<\delta_1$ and $0<|x-a|<\delta_2$, so applying 1) and 2) we get
$$|f(x)+g(x)-L_1-L_2|\le|f(x)-L_1|+|g(x)-L_2|<\ep_1+\ep_2=\ep$$

[solid kiln]

RokettoJanpu

[marsh forum]

Hmm

okay

wait, we get $|x-a| < min(\delta_1, \delta_2)$

rigth

[solid kiln]

Veni, vidi, perii

[willow urchin]

ofc its just from how we defined delta

[marsh forum]

so I have $|f(x)+g(x)-(L_1+L_2)| \leq \varepsilon_1+ \varepsilon_2$

[solid kiln]

Veni, vidi, perii

[marsh forum]

or $|h(x)-L_3|\leq \epsilon$

[solid kiln]

Veni, vidi, perii

[marsh forum]

This just proves the limit exists though, doesn't prove that they're equal

Right

[lime sphinx]

maybe concrete numbers will make more sense. give me any epsilon say 0.01. we can guarantee with our hypotheses that there exist delta_1 >0 such that f(x) is less than 0.005 (epsilon/2) away from L_1 IF x is less than delta_1 away from y. same thing with g(x). there exist delta_2 >0 such that g(x) is at most 0.005 away from L_2 IF x is less than delta_2 away from y. If we set delta = min(delta_1,delta_2) then delta is smaller than both delta_1 and delta_2. Furthermore, if x is less than delta away from y then

[marsh forum]

No, I get that

[lime sphinx]

oh and sorry about my loosely written up setup, it caused more confusion than good

[lime sphinx]

I think $L_1 = \lim_{x\to a}f(x)$

[solid kiln]

pola_touche

[lime sphinx]

by definition same thing for L_2

so it proves it exist and they are equals

[trim jolt]

@marsh forum Has your question been resolved?

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[open arrow]

Hello

[open arrow]

I will send the question

Y = one side of red

Y also = 1m

X = I side of black

The distance between black and red is 2m

What is the perimeter

Of black

[olive verge]

What is P?

[open arrow]

P = perimeter

I tried solving it yesterday

With people on this server

Ignor wit

It is false

[olive verge]

I see that wait

[open arrow]

Yellow and green = black or big red

[olive verge]

Where is this 2 m distance?

[open arrow]

Cyan line

[olive verge]

from centre of the line or from the vertex?

vertex of red to vertex of black?

ok i see

[open arrow]

Nethers vortex

[olive verge]

nearest?

[open arrow]

Sorry hold up

I may have forgotten

What’s vortex

[olive verge]

Vortex mean the point

[open arrow]

Edge of red

[olive verge]

ok i see it now

[open arrow]

To black

Is this soemthing that normal to learn

In year 8

[olive verge]

is it 24?

[open arrow]

How did you get that?

[olive verge]

Wait

[open arrow]

Ok

[trim jolt]

@open arrow Has your question been resolved?

[open arrow]

I’ve Ben waiting for a whole

[tranquil grove]

what grade is this?

because I don't think it is possible to get an exact answer without using trigonometry

also it says y = 1m and y = one side of the black octagon

[trim jolt]

@open arrow Has your question been resolved?

[open arrow]

8

I meant to say

X = 1 side of black

Can you teach me triganomatry

[tranquil grove]

it'll take too long

[open arrow]

Simple form

[tranquil grove]

but theres a good way to approximate it through just algebra

[open arrow]

How’s

How??

Is triganomatey

What grade math

[tranquil grove]

just imagine the octagons as circles. we know the formula for circles so it will be easier

trigonometry is like 11th grade

[open arrow]

Some year 11

Said soemthing about

Turn it into a circle

And divide it

By the side

The times 2

I forgot

[tranquil grove]

idk

because the red octagon has a perimeter of 16, we can approximate it's radius through 2πr = 16

[open arrow]

Is there like lecture in this server

I don’t think it’s 16

It should be 8m

[tranquil grove]

right

2πr = 8

so the radius is about 1.27

[open arrow]

Yea

[tranquil grove]

wait r u sure the perimeter of the red octagon is 8

on the paper it says P = 16

[open arrow]

That was my anwser

At first

[tranquil grove]

oh

[open arrow]

But realise how wrong I am

[tranquil grove]

anyways because the radius is 1.27 add 2 to get the radius of the black octagon

[open arrow]

Ohhh

[tranquil grove]

so the radius of the black octagon would be roughly 3.27

[open arrow]

Cuz that’s the distance ?

[tranquil grove]

yes

[open arrow]

And if that’s the radius

[tranquil grove]

now just use 3.27 as your radius in 2πr to get an approximate answer

for the perimeter

[open arrow]

Ok

7.98

Approximately

Are there any lecture in this server

That teach trig

[tranquil grove]

where did you get that number from

[open arrow]

2x pi

X radius?

[tranquil grove]

you shouldve gotten like 20

[open arrow]

Wait

Wait a

I did 1.27

Instead of 3.27

20.54

[tranquil grove]

yes

[tranquil grove]

there are

in the pre-university tab there should be a chat called geometry and trigonometry

[open arrow]

Ok

Well the thing is

I don’t think I am in pre uni

Nvm

Found it

If u have spare time

Can u teach me trig

[trim jolt]

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[still plinth]

if I get infinity as the answer for a limit do I put infinity or do I just say "does not exist"

[left oriole]

depends on the context

do you have a specific problem statement?

[still plinth]

yeah

number 3

[trim jolt]

@still plinth Has your question been resolved?

[left oriole]

the limit is -infty, you would either write that or you would write DNE
both are correct, but the first gives more information about why it doesn't exist

if this is for a class then you would do whatever your instructor requires in this case

[still plinth]

oh ok

yeah you're right it is -inf

so I guess it doesn't really matter then?

[left oriole]

not unless you have an instructor that has a preference

mathematically either answer is correct

[still plinth]

ok then, I was just helping out a friend so I'm not sure about instructor preference

thanks for the info though

.close

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[shell basalt]

what did i do wrong

[main sigil]

We dont know what you did

and i dont even see the question

[shell basalt]

graph it

i was screenshotting lol

i thought it starts at the median

also

the solution has it starting in line with the y axis

WAT ABOUT THE TRANSLATION ???

[main sigil]

does the solution look like this?

[shell basalt]

NO..

[main sigil]

oh wait, it's translated by pi/4

i see

[shell basalt]

I in fact do not see 😂

[main sigil]

You need to complete the graph

this is just one tiny segment of the graph

the graph extends indefinitely in both directions

[shell basalt]

so like what u drew

[main sigil]

Yeah, but extend it such that it ends at (0, 0) instead

[shell basalt]

oh

how do u knoiw that

f(0) ☠️

[main sigil]

You can plug x = 0 here

[shell basalt]

yuh

i should start with the endpoints shouldnt i

[main sigil]

sqrt(2) * -sqrt(2) / 2 + 1 = -1 + 1 = 0

[main sigil]

I think that doing the translation first is better

then you can finish the graph and check where to put the endpoints

[shell basalt]

how would u go about it

[main sigil]

Same way as you did it here, but then extend it

[shell basalt]

ooh right

ok

thank u

.close

[trim jolt]

Channel closed

Closed by @shell basalt

Use .reopen if this was a mistake.

[main sigil]

Hmm this still seems somewhat off

[shell basalt]

oh

.reopenn

.reopne

.reopen

[trim jolt]

✅

[shell basalt]

jesus

median is y=1 right

[main sigil]

yeah, it is

but at x = 6pi/4, it should be 0

[shell basalt]

and sin graphs usually start at the median

thats a 5

my bad xD

[main sigil]

oh, I see

you could also mark the point of intersection with x axis then

that would be 3pi/2

[shell basalt]

yeah i should have

do sin graphs usually start at the median

on the y axis

[main sigil]

if they are not translated

[shell basalt]

yeah

ok thank u

[main sigil]

then yes

[shell basalt]

i appretiate the help

.close

[trim jolt]

Channel closed

Closed by @shell basalt

Use .reopen if this was a mistake.

[shell basalt]

.reopen

[trim jolt]

✅

[shell basalt]

nvm one more question

when im finding x interceopt

what do i do from here

or rather

how do i find the intercepts within my domain

if they arent at a quarter cycle since its translated

its -pi,2pi

@main sigil

[main sigil]

this is the domain?

[shell basalt]

yeah

im struggling to envision it

[main sigil]

2(x+pi/6) = -pi/6 + 2kpi
2(x+pi/6) = 7pi/6 + 2kpi

[shell basalt]

wait what

[main sigil]

is it sin(2(x+pi/6))?

[shell basalt]

this is what it gave 💀

this is it

sorry

[main sigil]

same thing

[main sigil]

so this is a problem of type sin(u) = -1/2

[main sigil]

when is sin(something) = -1/2?

[main sigil]

when something = 7pi/6 + 2kpi or something = -pi/6 + 2kpi

[shell basalt]

k?

why the k

[main sigil]

k is just some integer in Z

[shell basalt]

for this u can have -pi/6, 7pi/6?

[main sigil]

and 2pi is the period

[main sigil]

sure

but -pi/6 + 2pi also works

as does -pi/6 + 4pi

and 7pi/6 - 2pi

that's why I added the + 2kpi

[shell basalt]

ohhh

thats why it does the weird shit on my calcualtor

but if its within the domain what are the only solutions

and how the fuck do i get all 6 of them for the graph....

[main sigil]

you need to simplify this now

2(x+pi/6) = -pi/6 + 2kpi
2(x+pi/6) = 7pi/6 + 2kpi

I think it should end up looking like this

[shell basalt]

oh dear im done for

[main sigil]

It's not that bad

[shell basalt]

i cant tell if im meant to know this for my test tommorrow

[shell wraith]

hello im new

[shell basalt]

because the practise questions are all over the place

[main sigil]

2(x+pi/6) = -pi/6 + 2kpi
2(x+pi/6) = 7pi/6 + 2kpi

Divide both sides by 2
x+pi/6 = -pi/12 + kpi
x+pi/6 = 7pi/12 + kpi

Subtract pi/6 from both sides
x = -3pi/12 + kpi
x = 5pi/12 + kpi

[main sigil]

Good for you, you have enough material to practise

[shell basalt]

but idk if we need it or not

im so lost

[main sigil]

well, practising it is definitely not gonna do any damage

[shell basalt]

im going to try and understand what u just sent one sec

[shell basalt]

ok ok

im following

[main sigil]

Good

x = -3pi/12 + kpi
x = 5pi/12 + kpi

So now that you have this, you can just plug in some values of k and see which work out

(by work out I mean that the resulting x will be within the domain)

[shell basalt]

when do i stop

plgging in

is this all tech free

[main sigil]

when you are no longer getting new solutions

[main sigil]

yep

[shell basalt]

fucking hell

what like

1,2

3,4,5

[main sigil]

and negative as well

[shell basalt]

ah man

[main sigil]

we can start at k = -3

[shell basalt]

how would u do it

is that just a guess

[main sigil]

sort of

-3pi/12 - 3pi is definitely way smaller than -pi

as is 5pi/12 - 3pi

so we got no sols here

let's try -2 now

-3pi/12 - 2pi is also smaller than -pi

as is 5pi/12 - 2pi

if we try k = -1 now

-3pi/12 - pi is smaller than -pi, but not by a lot

5pi/12 - pi is slightly greater than -pi

so this gives us our first solution

[shell basalt]

are u doing all of that in ur head...

[main sigil]

yeah, is there anything that was too unclear?

[shell basalt]

no i just cant compute that fastr

[main sigil]

I'm not even computing it to be honest

e.g. -3pi/12 - pi is smaller than -pi

because the left sides differs by -3pi/12

[shell basalt]

i know that

but like

[main sigil]

so I dont need to compute anything to compare it

[shell basalt]

the 5pi/12 - pi

[main sigil]

this one is greater than -pi by 5pi/12

so I again dont need to compute much

[shell basalt]

wait how do u know that

im so sorry

[main sigil]

5pi/12 - pi = -pi + 5pi/12

it's just -pi + 5pi/12

greater than -pi by 5pi/12

this is something you will be able to do fairly quickly after some practising

can you try k = 0 now?

[shell basalt]

ok

well

what am i comparing there is no + pi

[main sigil]

nothing much

[shell basalt]

theyu work

[main sigil]

just the -3pi/12 and 5pi/12

and yes, both work

[shell basalt]

wait

yea

[main sigil]

so now we have 3 sols

try k = 1

[shell basalt]

both work

[main sigil]

that's 5 sols now

try k = 2 now

[shell basalt]

first not second

the -3pi/12 works

[main sigil]

right

so that's 6 sols

[shell basalt]

oh

[main sigil]

you could try k = 3 and see that neither one works

and then obviously no larger k is going to work either

[shell basalt]

so u try whole numbers

how did u know to start at -3

[main sigil]

yes, k can only be a whole number

[main sigil]

when we start at -3, there is gonna be the -3pi term

so its sufficiently small to be smaller than -pi

if you want to speed up the process a bit btw, you can do it by just marking -3pi/12 and 5pi/12 on the x-axis and then keep adding pi to them and marking it until you are out of domain

[shell basalt]

true

ohh okj

i see

i may have become significantly more ready

for my test

[main sigil]

That would look like this

[shell basalt]

that was quite the journey

i didnt really learn that in class

but was going through the textobook and there was questions for it

and i was so lost

[main sigil]

interesting

maybe the test will actually be significantly easier than this

[shell basalt]

i fucking hope so

ive been rear ended and blind sighted

simultaneously

because my teacher does not communicate

and we skipped half the chapters

so idfk what we doing

but wish me luck

thanks for ur helpo

[main sigil]

gl :)

[shell basalt]

.close

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[wind cloak]

help! how should I approach this problem?

[distant geode]

You can rewrite X and Y then try to denote it

[strange ice]

ig expand everything 💀 (first impression)

[wind cloak]

tried it

[naive otter]

a²cd + b²da + c²ab + d²bc = 6abcd
a²db + b²ac + c²db + d²ac = 8abcd

[naive otter]

^

[wind cloak]

how does this help though?

[naive otter]

It doesn't 🥲

One moment b

[strange ice]

lol

this is the furthest i can go
a/b + b²da + c/d + d²bc = 6abcd
a²db + b²ac + c²db + d²ac = 8abcd
a/b + b²da + c/d + d²bc - a²db + b²ac + c²db + d²ac = 2abcd
a/b+c/d = 2abcd - 8 - b²da - d²bc

[distant geode]

So you have 4 numbers right? when you add up it all you get 2
A.6
B.8
So you now need to find the value just two of these fractions added together.
Let’s guess that the sum of those two fractions is 2. If our guess is correct, then the rest of the fractions in the first result should add up to make 6.
If a/b + c/d = 2 then the remaining fractions should add up to 6-2=4 so overall it is 2

[meager bloom]

You can't just guess sum as 2

What if there were no options

Its given in option so you can guess all 4 tries

[naive otter]

Okk , so a/b +c/d = x
Then ,
x + b/c +d/a = 6
Again , a/c+b/d+c/a+d/b = 8
Or a/b×b/c + b/c×c/d + c/d×d/a + d/a×a/b = 8

= b/c × (x) + d/a ×(x) = 8

Or x(b/c +d/a) = 8

Now b/c +d/a = y

Therefore xy = 8

And x+y = 6

@wind cloak heyy

[strange ice]

oooo

[naive otter]

,w solve xy = 8 and x+y = 6

[strange ice]

but which is which

[solid kiln]

Results provided by WolframAlpha

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

Requested by xor_11

[wind cloak]

oh my

[wind cloak]

that's insan

e

[strange ice]

so 2 or 4

[naive otter]

💀 I've only slept 4 hrs yesterday

That makes me crazy

[naive otter]

Anyone

[wind cloak]

bro's a 🐐

[naive otter]

Nahh. Bro

[wind cloak]

fr

[naive otter]

I. Only love algebra, you r more goated than me

[wind cloak]

I am just a normal guy

[naive otter]

Everyday common guy me too 🙂

[wind cloak]

thank's to other helpers too❤️

[strange ice]

average joe

but wait the question is still yet to solve

[naive otter]

Huh?!

[wind cloak]

erm

why

[strange ice]

whats the answer then

[naive otter]

Both are

[strange ice]

oh

i see

[naive otter]

Eqns are symmetric so , multiple answers possible

[strange ice]

its a multiple choice with multiple answers

[wind cloak]

probably

yeah

[naive otter]

Bye have a god day y'all !

[wind cloak]

.close

[trim jolt]

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[wind cloak]

fr i was typing !done

then i realized it is my channel

[strange ice]

🇵🇰

[wind cloak]

that's out of the context

[strange ice]

dw it's from somewhere else

its an emotion pretty much

[wind cloak]

what

[strange ice]

ull know if u know my avatar

and in that disc server

[wind cloak]

idk what's that

[wraith hinge]

how could math literally relate to pakistan

[trim jolt]

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[livid thunder]

dont even know where to start

[shrewd ridge]

it's 1

[livid thunder]

?

[wraith hinge]

1 will fail for p=2

[wind cloak]

2+1=3

[wind cloak]

yeah

[shrewd ridge]

then i go with 3

[livid thunder]

2+3=5

[split chasm]

same fail

[livid thunder]

i dont think this problem is meant to be solved by guessing

[split chasm]

consider the odd/even cases

[shrewd ridge]

i've never seen a problem that isn't, but it could be my first

[trim jolt]

@livid thunder Has your question been resolved?

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[marsh forum]

I don't get part (b)

[marsh forum]

nvm

I think I got it

I break the matrix into multiple components, don't I?

Wait

I don't get it

I mean y is a 1 by m matrix, right

OK. part b makes no sense to me

[marble wharf]

what about it makes no sense to you

[marsh forum]

Let A be an arbitrary m by n matrix

$\begin{bmatrix}
a_{11} &a_{12} &\hdots&a_{1n}
\
\vdots
\ a_{m1}&a_{m2}& \dots&a_{mn}\end{bmatrix}$

[solid kiln]

Veni, vidi, perii

[marsh forum]

and y is a 1 by m matrix

so $\begin{bmatrix} y_1&y_2 & \dots&y_{m} \end{bmatrix}$

[solid kiln]

Veni, vidi, perii

[marble wharf]

yes. and?

[marsh forum]

$\mathbf{y}A= \begin{bmatrix} \sum_{i=1}^n a_{2i}x_{ji}\\sum_{i=1}^n a_{1i} x_{j2} \
\vdots \end{bmatrix}
$

and so on

[solid kiln]

Veni, vidi, perii

[trim jolt]

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[marsh forum]

what's the best way to express 11

[marsh forum]

I was thinking $(B \cap C) \setminus (A \cap B \cap C)$

[solid kiln]

Veni, vidi, perii

[vast spoke]

There's no best way, but there's many standard ways:

,w c and b and not a

[marsh forum]

I know, but what other ways exist

[vast spoke]

There's infinite other ways, 7 standard ways are shown above

[marsh forum]

Hmm, okay

thanks

Also for c) $(A \cup B \cup C) \setminus (A \cap B \cap C)$ works?

[solid kiln]

Veni, vidi, perii

[vast spoke]

It's correct and probably the cleanest way to write it

[marsh forum]

Thanks

.close

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[marsh forum]

I was thinking this would be given by the matrix $\begin{bmatrix} 0&1\1&0\end{bmatrix} \times \begin{bmatrix} 0&-1\1&0\end{bmatrix}$

[solid kiln]

Veni, vidi, perii

[trim jolt]

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[graceful nova]

hi i dont understand how we go from this step to the next

[steady tulip]

Laws of exponents

[graceful nova]

i figured as much but how

the law is (am)^n = a^n * m^n

[steady tulip]

(a^x)(a^y) = a^(x+y)

[steady tulip]

This one won't apply here

[graceful nova]

2+(k*2)^(2k+1) right

isnt this whats going on here

[steady tulip]

Yes

Wait no

The last 2 terms are being multiplied

[graceful nova]

OHHHH

Yea i just noticed

thanks

honestly bad notation for their end tbh

[steady tulip]

Kinda

[graceful nova]

super long induction question as well, i was so confused

thanks!

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[worthy sinew]

this is a question from boolean algebra

where we have to minimize the literals

did i do it wrong if yes then where?

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[worthy sinew]

?

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[finite tusk]

how can i solve this integral?

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[limber pike]

did u try by sibustion

i doubt it will work

half angle sibisution wouldnt work either

ohh

@finite tusk uhh i think u can do it by trigromneitc sibsition

[finite tusk]

i tried that

it didn’t work

[limber pike]

how come?

can u send me

ur attempt?

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[finite tusk]

.reopen

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✅

[finite tusk]

i couldn’t do it by trig sub because it’s not a perfect square and turning it into a perfect square is gonna take to much time so i don’t think it is the right solution

[limber pike]

i mean u oculd turn it into a perferct

square

oh im dumb

i forgot

soryr sorry sirry

forgot everything i siad

hmmmmmmmmmmmmmmmm

[finite tusk]

it’s an exam example and the way that you’re suggesting is gon take alotta time so i don’t think it will be the right way to actually solve it

[limber pike]

yeah yeah yeah i get it

hmmmm i took calc 2 a year ago so i forgot a lot of stuff sorry

[finite tusk]

it’s okay dw

[limber pike]

did u try doing it by parts

or did that not work

i think it will be too long

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[uneven oyster]

I need help with a proof, lets say i've got the sequence of functions arctan(nx)

[uneven oyster]

i need to investigate the uniform convergence of arctan(nx)

[grim sparrow]

what have you done so far?

[uneven oyster]

I've found the limit function

Which is kinda easy

But im having trouble proving that the sup{|fn(x)-f(x)|} is not zero

The limit function f is pi over 2 for positive reals, -pi/2 nor negatives and 0 when x=0

[whole coral]

Are the sequence functions continuous? What about the limit function?

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[uneven oyster]

The sequence is continuous, the limit function is not

Is there a theorem i am missing? Like if the sequence is continuous and the limit function is not, then the sequence does not uniformly converge to its limit function?

[whole coral]

yep

Contrapositive of that: if the f_n's are continuous and f is not, then f_n cannot converge to f uniformly

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[radiant rapids]

Anyone know how to solve the last two questions

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[radiant rapids]

<@&286206848099549185>

[odd heath]

Wait. Is that possible?

[carmine mauve]

cos is defined on the entire real number line. So you can indeed compute cos( - 430769/73 pi )

It's arccos (the inverse of cos) which is only defined on the domain [-1, +1].

But cos of anything produces a number on the domain [-1, +1].

take a look 🙂
https://www.desmos.com/calculator/2qouqvuoh6

In other words, arccos( cos( x ) ) = x

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[glossy jay]

hi

[glossy jay]

14xy(81x²-121y²)
/
7xy(9x+11y)

is the answer 2xy(9x+11y)?

.close

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