help-36

[soft zealot]

MCAntDJC

[whole halo]

which is?

oh its the $T'=-2T^2-\pi^2/2$ isnt it

[soft zealot]

mtt

[low idol]

Yes

[whole halo]

thats not enough information: other than the differential equation, were you given any other information?

[low idol]

Just recognizing that $T(x)=\frac\pi2\cot \pi x$

[soft zealot]

MCAntDJC

[whole halo]

recognizing is the wrong word if youre given the result

what is the original form for T(x) that you were given

[whole halo]

was it only the differential equation, or was it the differential equation and also the series representation?

[low idol]

I was able to derive the differential equation from the series by using PFD (and I started with the series)

[whole halo]

...derive...

can you talk in voice chat?

[low idol]

On a phone?

[whole halo]

thats what phones are for, arent they

so far the responses Ive been getting back dont seem to understand the problem I am asking, regardless of how long you take to craft them

usually when we talk this out, we can answer this question much faster

no need to type latex on a phone, that takes forever

yes or no, are you open to talk?

[low idol]

Not really, because people around me are sleeping 😔

[whole halo]

in that case this is going to take forever

is there a reason why theres always a few minutes between responses?

[low idol]

I'm always anxious or nervous, then usually overthink my responses

[whole halo]

yea thats what I guessed

theres something I usually think of when in a situation like that
you have to realize youre not like dealing with confidential information or in front of the UN

youre just answering questions, and wrong questions can be independent of bad communication

wrong answers, then, can also be independent of bad communication

I want you to quickly answer this question: where did you get this problem from?

[low idol]

I made it up, was playing around with a series ($S(x)$) and thought to compare its square to its derivative; then I got an equation for it, rewrote in terms of $T$, and solved for $T$.

[soft zealot]

MCAntDJC

[whole halo]

there you go, that wasnt hard was it

the basic premise was $S(x)$

[soft zealot]

mtt

[whole halo]

then from there, everything else was derived, found, rediscovered, etc.

Ill take a shot in the dark and say you care about what couldve been better in us getting to this point?

[low idol]

I got the right result with $S$, but in trying the same with $T$ I got an error...

[soft zealot]

MCAntDJC

[low idol]

I think it comes from

[whole halo]

so do you see any problems in us taking 3 whole hours just to ask where you got the problem from, or?

maybe we can figure out what went wrong and try to improve on it?

[low idol]

It's most likely from the series above: adding the terms in the brackets gives $\frac{x^2}{(x^2-k^2)(x^2-n^2)}$, but you showed that an extra term comes out of decomposing the expression here.

[soft zealot]

MCAntDJC

[whole halo]

really, nothing?

I dont understand that sort of mentality but we can just focus on the problem then

[whole halo]

we already showed that the partial fraction decomposition does not work out

I tried to continue going with the approach and eventually got having to show that the thing on the left converges to pi^2/2, regardless of the value of x

needless to say this isnt happening

theres a few more other things we need to be talking about as well

[whole halo]

first, there already exists a clean way of finding S(x)

it uses the weierstrass factorization of sin(x), which you can search up

ln both sides, then d/dx both sides

this shows how cot is closely linked to a series like S(x)

and I imagine would be helpful in showing why those pi^2n and cot keep appearing in zeta(2n)

[whole halo]

with that, S(x) is already known to be

so if you wanted to find something more convenient, shouldnt you just work with the LHS instead of having to deal with the RHS

[low idol]

I assumed that I didn't know $S(x)$ was related to the cotangent, then showed it did via the differential equation for its relative $T(x)$

[soft zealot]

MCAntDJC

[whole halo]

we dont seem to be on the same wavelength at all here

what am I supposed to even do with a response like that?

[low idol]

I don't know, but I do know about the sine factorization; I just played with $S(x)$ as if I didn't know it.

[whole halo]

avoid combining discord and latex formatting like that, they dont mix well

going to ask, unrelated to the problem, are you aware that there is a major communication issue between us that would help things along if we were to fix it?

[low idol]

Yes

[whole halo]

in this message, are you aware that it was talking about the communication issue, and not about the problem?

it was a yes or no question

I wanted to look back and have us both figure out how we couldve communicated better, but Im going to take this silence as that youre not ready

there is nothing else for us to really point at right now, the mistake in the PD has been found, and the way forward after squaring T(x) as a series doesnt look promising

at the top here, the paper shows you had some more promise when the series wasnt squared

as soon as we squared it, we had to deal with this

this is my best attempt at simplifying it, as you can see its not ideal

at the very least itll involve things like S(k1^2) and S(k2^2) of some sort

it looks like you wanted to relate S(x) to a differential equation so that you can avoid getting into this? Idk

other than that, theres nothing else to point at, and we are done here

[low idol]

I can calculate series in parentheses, but it won't be pretty

[whole halo]

look at what the parentheses is connected to

those are each coefficients

the result youd get would just be a bunch of coefficients attached to the series, which in itself is too different from S(x) to really be of any use

it cant be immediately related to S or T since it appears to use x = k1, and S and T are both undefined for positive integer inputs

what Im thinking is we find a way around directly squaring T(x)

no other ideas though, and I dont see a way to even do that

[whole halo]

for quick reference, the closed form for this, cheating with knowing T(x) = a cotangent, is:

[low idol]

Try using $\frac1{k_1^2-k_2^2}=\frac1{2k_1}\left(\frac1{k_1+k_2}+\frac1{k_1-k_2}\right)$, then rewriting the sums as partial harmonic series

[soft zealot]

MCAntDJC

[whole halo]

wasnt that what I got earlier?

(also it took me way too long to realize why T(x) was so important:)

(its not just the cotangent thing, its got a really nice series representation)

[whole halo]

oh right, wrong image
it was supposed to include this one

I did try out using the partial fractions again but the harmonic series I got doesnt seem ideal

[low idol]

You should let $n\to\infty$ to make that last sum vanish (it's finite)

[soft zealot]

MCAntDJC

[whole halo]

I thought of the same, but that doesnt work

wait a minute

why didnt I think that would work, that does work

I was thinking since each term wouldve been a coefficient, we would need the initially finite sums at first

no idea what I was doing, but that just abou tsolves this

so much for throwing in the towel, lets finish this

we now have

WE'VE DONE IT

with this, we might not even need to know what zeta(2) is, we could just leave S(0)s everywhere then use the differential equation & the asymptotes of T to show that S(0) has to be -pi^2/6

[low idol]

You got it? Congratulations!

[whole halo]

and all it took was looking at the series a second time

and knowing where we were even going

[low idol]

Exactly! I was jubilant when I found out I tease out $\zeta(2)$'s value from all this

[soft zealot]

MCAntDJC

[whole halo]

you already found it some other way?

even looking at this section of the work, its wild we can find S(0) this way

[low idol]

For me, $\zeta(2)$ appeared as an unknown parameter: I had $T$'s form down to $A\cot(\sqrt{6\zeta(2)}x)$ (I don't remember $A$'s value, but it doesn't matter here)

[soft zealot]

MCAntDJC

[whole halo]

interesting, we have to had taken extremely similar paths to look for asymptotes this way

if you already had T's form down like though in advance, you couldve used it

well I got something completely off at first

Ill have to clean up the work and try this again

in particular, dont group things up as T(x)/x

[low idol]

I had found $T$ by solving its differential equation... and I may have then calculated $T(1/2)$ to get $\zeta(2)$

[soft zealot]

MCAntDJC

[whole halo]

well Ill have to clean up the work I have, but as far as I see it we're done here

[low idol]

We did it—it was fun and enriching for me my first time and I hope you had fun too

.close

[final saddle]

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[grand hound]

I currently have the 3rd last line as my answer but how did it jump to the 2nd last line??

[lilac moat]

its a typo

[grand hound]

damn!

[lilac moat]

cambridge has bad worked solutions at times

[grand hound]

but what abt the final answer?

how do I get that solution from what I have currently

[final saddle]

@grand hound Has your question been resolved?

[fast pulsar]

He factorized and put same denominator

[grand hound]

i'm struggling to see the factorising 😭

[dusky shadow]

just remove the (-) from the front of "1"

[grand hound]

ok

[severe canyon]

$-e^{-kx}$ is in common

[grand hound]

how should I approach question b?

[soft zealot]

Alberto Z.

[grand hound]

<@&286206848099549185> how to do part b :(((

[ivory bear]

weird

[ivory bear]

idk why i see this

[grand hound]

dw abt it

[severe canyon]

With the usual definition of the mean, i.e. the integral of x•p(x) from -∞ to +∞

[grand hound]

yeah idk how to antidiff it

i assume I have to use the answer i got in a

but idk how

[severe canyon]

Yes, you need that

You can apply it directly, your k is 1/λ

[grand hound]

so can I just switch 1/lambda for k and work with that and also set k=1/lambda on the side

ok yeah I did that and i got it

[grand hound]

how about part d?

nvm i got it

.close

[final saddle]

Channel closed

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[final saddle]

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.

[meager hedge]

What have you tried ?

[soft zealot]

Renato

[final saddle]

@gentle zephyr Has your question been resolved?

[final saddle]

@gentle zephyr Has your question been resolved?

[drowsy epoch]

[gentle zephyr]

I already proved reflexivity

and I have some doubts with symmetry

[final saddle]

@gentle zephyr Has your question been resolved?

[worldly mesa]

its of course symmteric

35

[gentle zephyr]

ye, but is hard to explain it in a proof

[worldly mesa]

no

just choose the same n

[gentle zephyr]

can you elaborate

[worldly mesa]

for 35 if fRg then f(n)=g(n)=1 for some n

then of course also g(n)=f(n)=1

so gRf

[gentle zephyr]

yes if and only if n = m

[worldly mesa]

you can choose your m

so you choose it to be equal

[gentle zephyr]

i mean, we can just assume they are not.

and use bijectivity thingy

idk if I am explaining myself correctly @worldly mesa

are you here Squee SQUEE?

[worldly mesa]

no need that

[gentle zephyr]

yes, that would be more appropriate

[worldly mesa]

its just that if f(n)=g(n) then g(n)=f(n)

bijectivity helps with the transitivity of the relation

[gentle zephyr]

dude

suppose n is not the same m

[worldly mesa]

m is not arbitrary

you can choose it

thats the point

[gentle zephyr]

then f(n) = g(n) = 1 = f(m) = g(m)

[gentle zephyr]

how would the proof would go like, then?

[worldly mesa]

you want to prove fRg implies gRf

[gentle zephyr]

can you give a rough sketch of your idea

[worldly mesa]

if fRg then for some n f(n)=g(n)=1

[gentle zephyr]

so we can prove that is symmetric

[worldly mesa]

to show gRf we need to show there is some m s.t. g(m)=f(m)=1

choose m=n and we are done

[gentle zephyr]

ye

can you put it all in one message, so we can continue?

[worldly mesa]

.

[gentle zephyr]

I just noticed it was a, there exists, if it would had been a forall, it would be different I guess

[worldly mesa]

.

[gentle zephyr]

you know what I mean?

[worldly mesa]

it wouldn't be so different

but yeah I understand what you missed

[gentle zephyr]

I mean, there is no need to use bijectivity here is what I mean

you understand what I am saying?

[worldly mesa]

yeah its symmetric either way

[gentle zephyr]

how to prove it?

[worldly mesa]

. this

we already talked about that

[gentle zephyr]

let me put it into words

@worldly mesa

you here?

[worldly mesa]

I am

[gentle zephyr]

@worldly mesa

[worldly mesa]

Hmmmm

First, in the line "Suppose gRf iff ...."

You should state instead that you want to prove gRf which means the there exists b.....

[worldly mesa]

You should dicard that iff

[gentle zephyr]

I WANT TO ASSUME X R Y AND GET THAT Y R X

[worldly mesa]

because when you say you "assume xRy iff exists ..." you assume the equivalence of those things

[gentle zephyr]

dude

its from the definition of the relation

[worldly mesa]

I know

But I help make your proof more accurate

[gentle zephyr]

?

can you elaborate how to do the proof please?

[worldly mesa]

You want to assume xRy

just this

[gentle zephyr]

ye

[worldly mesa]

so don't assume xRy iff exists a ...

[gentle zephyr]

and somehow, conclude that y R x

[worldly mesa]

yeah

[gentle zephyr]

any hints?

[worldly mesa]

so we assume xRy

then by definition there is an "a"

to show yRx we need to find some b

choose b=a

[worldly mesa]

good

[gentle zephyr]

@worldly mesa

[worldly mesa]

no need to ping

[gentle zephyr]

you here?

[gentle zephyr]

what now?

i got stuck

[worldly mesa]

what does it mean for gRf to hold true?

[gentle zephyr]

we cant asume g R f

[worldly mesa]

we don't assume it

we understand what it means

and we can show that instea

[gentle zephyr]

gRf iff there exists b in {1,...,10} s.t. g(b) = 1 and f(b) = 1

[worldly mesa]

yes

what b fits?

[gentle zephyr]

we cant asssume g R f

[worldly mesa]

we don't assume it

[gentle zephyr]

b = a

whats your point?

[worldly mesa]

my point is that by choosing b=a you proved "there exists b in {1,...,10} s.t. g(b) = 1 and f(b) = 1"

and therefore gRf

[worldly mesa]

hmmm.... you should phrase as, we want to show gRf or equivalently that there is a b s.t. g(b)=1 and f(b)=1

choose b=a and then continue

[gentle zephyr]

lets start from scratch

[worldly mesa]

no need to

its just that this first sentence in the second paragraph means something else then what you mean

[gentle zephyr]

@worldly mesa

this better?

[worldly mesa]

yes

[gentle zephyr]

ok, what about transitive property

is the relation transitive or not

no spoilers

[worldly mesa]

do you want me to tell yes or no?

[gentle zephyr]

no

did you already proved transitivity ?

[worldly mesa]

I know the answer yes

[gentle zephyr]

how?

[worldly mesa]

because I solved the question

I don't get your concern

[gentle zephyr]

how?

[worldly mesa]

you want help with showing transitivity?

[gentle zephyr]

ye

[worldly mesa]

first write what you need to assume

then write what you want to prove

then try to understand what approach you want to take

[gentle zephyr]

@worldly mesa you here?

[worldly mesa]

good

now what do you want to prove?

[gentle zephyr]

we want to prove f R h

[worldly mesa]

which means?

[worldly mesa]

good

what c do you think works?

[gentle zephyr]

a = b = c

[worldly mesa]

yes

but why does a=b?

thats what you need to answer

[gentle zephyr]

is hard

[worldly mesa]

do you have an idea?

[bold turtle]

there's something you're told about the functions here that you've yet to use

[gentle zephyr]

@worldly mesa @bold turtle you here?

[worldly mesa]

yes we are here

ok this is true

[bold turtle]

There's no need to constantly ping us (my message was literally just a MINUTE ago)

[worldly mesa]

but you wrote everything you did just to conlcude with the real argument

[gentle zephyr]

care to elaborate? i dont think I follow?

[bold turtle]

You can literally delete these lines and it has the exact same result

[gentle zephyr]

now what?

[bold turtle]

wdym "now what"

What's special about a and b here, like, at all?

We know they're equal, sure

But then why are either of them important here?

[gentle zephyr]

how do we conclude

[bold turtle]

ahem

[gentle zephyr]

is this proof perfect?

@worldly mesa you here?

[worldly mesa]

just c not b in that last line

but yeah its good

gj

[gentle zephyr]

what?

[bold turtle]

what?

[gentle zephyr]

ahh we need to take c = b

@worldly mesa

[worldly mesa]

yeah you wanted a c given the b

[bold turtle]

(I mean, the c IS b here so)

(we're done by that point)

[bold turtle]

Also, he is literally typing at the time you ping him - just don't

[gentle zephyr]

wat abt antisym

x R y, y R x <=> x = y

[worldly mesa]

this can't happen when the relation is symmetric unless the relation is the identity

[gentle zephyr]

dude

help ,me find, a, ounterexample

[worldly mesa]

choose different x,y such that xRy

[gentle zephyr]

[worldly mesa]

take f to fix all of {1,..,10}

and g to do that but send 9 to 10 and 10 to 9

[gentle zephyr]

ok

wait a sec

let me do a drawing

@worldly mesa

[worldly mesa]

yes

[gentle zephyr]

like this?

[worldly mesa]

yes

they both send 1 to 1

[gentle zephyr]

f = id

[worldly mesa]

but they are differnt

[gentle zephyr]

ye

this is our counter example

how lets show its not antisym

[worldly mesa]

it is transitive

[gentle zephyr]

edited

[worldly mesa]

I see

[gentle zephyr]

how do we show

its not antisym

[worldly mesa]

just present this counterexample

[gentle zephyr]

how?

[worldly mesa]

Lets take a look at f=id and g(n)=n if n in {1,...,8} and g(9)=10, g(10)=9

then f is not equal to g but fRg and since the relation is symmetric then also gRf

[gentle zephyr]

,, \ g : {1, \cdots , 10} \to {1, \cdots, 10} \ g(n) = \begin{cases} n &\text{ if } n \leq 8 \ 10 &\text{ if } n = 9 \ 9 &\text{ if } n = 10 \end{cases} \ f : {1, \cdots , 10} \to {1, \cdots, 10} \ f = \text{id}

[worldly mesa]

you can add more detail if you want

yes

just we usually don't put a \iff there

just a space

just a matter of notation

yeah good

[soft zealot]

Renato

[gentle zephyr]

is the identity function bijective?

[gentle zephyr]

ye

I think I follow

[worldly mesa]

of course

antisym not transitive

[gentle zephyr]

we would need to show f and g are bijective, but that is left as an exercise to the reader

[worldly mesa]

in homework this will not fly

depends on who grades the homework but yeah

[gentle zephyr]

homework is non graded and is optional

you do it if you want to do it

[worldly mesa]

ok so you are fine

[gentle zephyr]

what about equivalence classes

@worldly mesa you here?

[worldly mesa]

is it asking what is the equivalence class of Id?

what do you think?

[gentle zephyr]

i mean

[gentle zephyr]

we need to say 3 distinct elements of the equi cla of id

[worldly mesa]

ok, what is the condition to being in the same class?

what does it mean for fRId

[gentle zephyr]

they must be equivalent under the equivalence rela

[gentle zephyr]

here we go again

[worldly mesa]

fRId if f fixes some n

so you can use this g from earlier

and another one using the same idea

[gentle zephyr]

f R id <=> E x in {1,..,10} s.t. f(x) = id(x) = x = 1

[worldly mesa]

yep

oh yeah

I made a slight mistake xD

[gentle zephyr]

where

[worldly mesa]

here

what you said is right

fRId iff f(1)=1

[gentle zephyr]

ye

so there are infinitely many funcs that do the trick

bijective funcs

for that equi class

not inifinitely many but. . . like a lot

9! or something

[worldly mesa]

yes

[gentle zephyr]

the identity function is part of the equivalence class of id

[worldly mesa]

yes

since R is reflexive

[gentle zephyr]

so thats one

[gentle zephyr]

ye

another one is the one we defined earlier

@worldly mesa

you here?

[worldly mesa]

yes

its good yes

[gentle zephyr]

oops I made it backwards

@worldly mesa

you here?

[worldly mesa]

yes

what do you need

[gentle zephyr]

this three funtions are part of the equivalence class of id

[worldly mesa]

you are doing well

yes

[gentle zephyr]

now what?

[worldly mesa]

thats it?

you answered the question

[gentle zephyr]

the question was 36 aswell

[worldly mesa]

I think you should do that by yourself

its pretty much the ame

just a different relation

[gentle zephyr]

can you help me

its not the same

its similar but harder, ngl

[worldly mesa]

first try to understand what that relation means

then solve the question

[gentle zephyr]

please

[worldly mesa]

try it yourself

its good for practice

I cannot always help you with every questioon

[gentle zephyr]

its hard

[severe canyon]

We've told this to you hundred times, and still it seems you don't understand it... @gentle zephyr

[severe canyon]

But you haven't even started it

[gentle zephyr]

please

[bold turtle]

My guy

You've just worked through a similar question in excruciatingly more detail than would've otherwise been needed

You should, already, have an idea of how to approach 36

[gentle zephyr]

I tried doing the proof that the relation from 36 is reflexive, but idk if its correct, can I get some help?

[bold turtle]

yeah that's it

g o f = g o f is true because equality is reflexive

ergo, the relation R is reflexive

[gentle zephyr]

i tried to prove that the relation from 36) is symmetric, can I get some feedback?

[bold turtle]

Don't just plonk "I want to show that" in the middle; it breaks the flow of the proof

[gentle zephyr]

the otjer guy recommended dat

[bold turtle]

No, he commented that you check what you want to prove

So that when you do come across it, you can realise when your proof is done

You've already said gRh means g o f = h o f

That means h o f = g o f

[gentle zephyr]

.

[bold turtle]

Which means...?

[gentle zephyr]

yeah

[bold turtle]

Yh but you do this at the START of the proof

[gentle zephyr]

[bold turtle]

"Assume A. We want to prove that B. [PROOF]"

Don't do
"Assume A, i.e. [stuff] [half of a PROOF] We want to prove that B. [the other half of the PROOF]"

This just makes it harder to follow

[bold turtle]

WHICH MEANS...?

(Like, I'm just parroting what you've said; you literally have one line to finish this)

[gentle zephyr]

hRg

[bold turtle]

There we go

[Okay, I'm not gonna stay here any further; I've got CS work to get done ]

[gentle zephyr]

fuck my life

[opaque ember]

@gentle zephyr its ok to struggle but if you need to be handheld for so many similar problems, there's an issue with your learning method, and you need to fix it before doing anything else

[gentle zephyr]

wdym?

[opaque ember]

am i unclear?

[gentle zephyr]

wdym fix?

[opaque ember]

you need to find the issue with your learning method that makes you repeatedly get stuck on the same type of problem

[gentle zephyr]

the problem with proofs is that you dont know if they are correct unless you receive some feedback

is very different to what I have done in preuni, like, finding the area between two functions

[drowsy epoch]

if you dont know if your proof is correct, it's most of the times wrong anyway

[opaque ember]

the issue is not specific to this topic

[gentle zephyr]

also, some proofs look convincing but at the same time, are just plain wrong, you know what I mean

[opaque ember]

ive seen you post many similar questions over the past two years. the way you struggle with them has been the same

you struggle for a long time and learn a little something, but then you seem to lose all that progress when you do a similar problem

[trail crest]

Proof assistants

[gentle zephyr]

is not true imo, this has been only recently

or could be true, idk, preuniversity was very different

[opaque ember]

i remember when you posted the same type of linalg problems for a whole year

you had a pattern of struggle very similar to the one you have now

[trail crest]

Like the way you think i think you might be interested to learn some proof assistant like lean, rocq, isabelle etc...

[gentle zephyr]

back in preuniversity, everytime i tried a linear algebra problem I knew how to start but just, . . ., didnt managed to finish it entirely, like finish the exercise

now, is different sometimes I dont even know how to start

[gentle zephyr]

it was a little bit different, now, even if I finish the proof I dont even know if its true or not

[opaque ember]

this is what i observed from you every time

you struggle for a long time and learn a little something, but then you seem to lose all that progress when you do a similar problem

this is why i think there's an issue in your general learning method

[gentle zephyr]

this is my first few weeks in university, doesnt this happen to everyone? is very different from pre university

[opaque ember]

i think the issue is related to pattern recognition, applying what you learned from one problem to a similar problem

[gentle zephyr]

I mean

I think everyone in my class struggles with this, is just that some people convince themselves their proofs are good, but then when they send me pictures they have mistakes, because they receive no feedback from anyone, u know?

[trail crest]

May i ask what are you studying

[opaque ember]

it's normal to struggle but not for so long in the pattern that i described

[gentle zephyr]

as I said, this is the first few weeks of university, I am still learning and adapting my study mechanisms, I am open for feedback on it

[opaque ember]

i need to you to understand that the issues are not exclusive to university. my observations of you are from both pre uni and uni

the observations point to an issue with general learning method. ill restate:

you struggle for a long time and learn a little something, but then you seem to lose all that progress when you do a similar problem
i think the issue is related to pattern recognition, applying what you learned from one problem to a similar problem

so maybe my feedback is to pay more attention to the "big picture" of how you solve a problem. that way it's easier to apply the same method to a similar problem

[gentle zephyr]

isnt that the skill that I am trying to develop in order to pass my exams? the hability to apply what I learnt from one problem to another?

[opaque ember]

i see you getting caught up in the small details much more often than the big picture. part of ideal learning is having a good balance of attention to both small details and big picture

[trail crest]

I would argue that math is more about learning how to solve a problem (problem solving). And not about learning how to solve X, so that you can solve something similar to X.
(Just my opinion tho)

[gentle zephyr]

my point is, that hability you are describing, that applying past experiences in order to solve any problem you are thrown at, is easier said than done, is exactly the skill my professor wants me to learn, the issue is that I am a slow learner, but that is alright, no?

back at preuni, I managed to develop that skill by doing gazillion different exercises until something clicked in my head every time I saw a linear algebra problem

[opaque ember]

ofc it takes time. but i see you going through the same pattern of struggle so many times with very little progress, leading to my feedback on how you can improve your learning method

i have to leave now but i can tell you dont fully agree with my observations. as a first step to improvement i suggest reading your previous conversations with helpers where you spend a long time on a single problem. try to note the commonalities in your behavior and perhaps you'll see what i see

of course it's hard to be objective about yourself, so if you still don't agree with me, i'm sure there are plenty of helpers you can talk to who support my observations

[crimson bronze]

My impression is that a big problem you have is your fear of failing. You don't dare to try a problem on your own without someone explicitly telling you what to do. I think this is very detrimental to learning, because you're not forcing yourself to think. It's like trying to learn how to run but with someone constantly pushing on your back

@gentle zephyr I think if you spend 20-30 minutes on each problem, giving it an real, proper attempt, I think you'll be surprised how much faster you'll learn. The point isn't to solve the exercise on your own, the point is to force your brain to think deeply about the problem

[whole halo]

Interestingly, the last time I had renato do most of the problem on his own, he can handle it quickly

@gentle zephyr you should only ask for help once you run out of ideas, a lot of this is essentially ideas that you seem to already know or handle

[final saddle]

@gentle zephyr Has your question been resolved?

[ripe jewel]

.close

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[oblique estuary]

hey y'all, need help with this one, not sure why I got it wrong, will post my work

[oblique estuary]

pls @ me when someone pops in because the second problem I also got wrong and I assume it's for whatever the reason i got this one wrong is for

[oblique estuary]

right

[tame gust]

you are essentially using the difference of squares formula, so the signs underlined in green are GOOD, but the one with red is NOT GOOD

[oblique estuary]

okay because that was new but that's what my calc 1 teacher did so i was like "I guess we're doin this now"

[tame gust]

recall the formula is $(a-b)(a+b) = a^2 - b^2$, where in your case $a = \sqrt{4-v}$ and $b=\sqrt{7}$

[oblique estuary]

maybe I just misunderstood in class

[soft zealot]

Peter

[oblique estuary]

alr lemme redo the problem

[tame gust]

you had the right idea multiplying with the conjugate (the a+b term), but you accidentally changed the sign under the root which needs to stay fixed

[oblique estuary]

right on he changed the sign under the root too

ig maybe he messed up or somethin idk

[tame gust]

maybe if you show us the problem and solution we can discuss if that is right or wrong

[oblique estuary]

I mean it was in class

he gets them wrong sometimes

[tame gust]

might have been a mishap, happens

[oblique estuary]

alr so I got a question

how do I remove the (v+3) out of the denominatory then

or do I not

ig I could just not

but he told us not to do that

[tame gust]

we need to cancel the v+3 term on top and bottom so that we can calculate the limit

[tame gust]

and you remove it exactly the same as you did previously

[oblique estuary]

nevermind I do have to

[oblique estuary]

yep

[oblique estuary]

alr well one sec

ohhh I had made a mistake in my numerator

I see

alr lemme try my answer

well

I get -1/(0)

so that aint good

[tame gust]

can you show your solution

[oblique estuary]

ye

[tame gust]

they shouldn't be under the same root

[oblique estuary]

oh yeah

I just have shitty handwriting

mb

they aren't

[tame gust]

denominator should be $\sqrt{4-v} + \sqrt{7}$, again watch the signs as you should have gotten a plus

[soft zealot]

Peter

[oblique estuary]

well you see that's doggone confusin

no it's not

I'm just stupid

alr thx

[whole halo]

average math experience:

[oblique estuary]

FR

now then i got another question

[tame gust]

first, the exercise mentions that x>1

can you rewrite g(x) knowing that x>1?

instead of writing |x-1| could we write it differently knowing x>1?

[oblique estuary]

well obviously we can since you're asking

[tame gust]

im dead

[oblique estuary]

but not sure how 😭

[oblique estuary]

:)

[tame gust]

the absolute value is defined as |x| = x if x>0 and |x|=-x if x<=0

or

[oblique estuary]

right

[tame gust]

if instead of x we put x-1

[oblique estuary]

it'd be x-1

[tame gust]

|x-1| = x-1 if x-1>0 and |x-1|=-x+1 if x-1<=0
which is equivalent to
|x-1| = x-1 if x>1 and |x-1|=-x+1 if x<=1

[oblique estuary]

boom

[tame gust]

amazing

second question, what did you get for the conjugate of h

[oblique estuary]

it eas given but it would be 1

anyways I got that part right now

[tame gust]

the conjugate of h(x) = sqrt(x) - 1 is not just 1

[oblique estuary]

oh mb mb

but anyways I literally just put sqrt(x) - 1

and it took it

like

I put (x-1)/(sqrt(x)-1) for my answer

[tame gust]

alright, I'll just provide the steps how we got there

[oblique estuary]

since I multipled top and bottom by (1/(sqrt(x)-1)

[tame gust]

$$\frac{h(x)}{g(x)} = \frac{\sqrt{x}-1}{|x-1|} = \frac{\sqrt{x}-1}{x-1} \cdot \frac{\sqrt{x}+1}{\sqrt{x}+1} = \frac{x-1}{(x-1)(\sqrt{x}+1)} = \frac{1}{\sqrt{x} + 1}, \quad \text{when } x > 1$$

[soft zealot]

Peter

[oblique estuary]

so that's like

not what the book taight us to do

or rather

did for this problem

it started by multiplying numerator and denominator by (1/(sqrt(x) -1))

to get 1 for the numerator

[tame gust]

that is also a valid option, but it should be noted that sqrt(x) + 1 ends up being in the denominator

[oblique estuary]

right

wait

one sec

I got lost

too many homework assignments and problem open i swear

[tame gust]

it's ok, what's confusing you

[oblique estuary]

user error

aka I'm a dumbass

give me just one hot second

[oblique estuary]

so what problem is this even for

[oblique estuary]

this one

[tame gust]

this one

[oblique estuary]

alr

[oblique estuary]

okay that's awesome because it gave me my answer for the third part

which i was going to ask about

but like

okay so starting at 1/(x-1)/(sqrt(x)-1)

how do I get to 1/2

plugging in 1 for x at the end

[tame gust]

very quickly, why are we starting with this?

[oblique estuary]

that's what the previous answer to the second part was

[oblique estuary]

mb

missed a division sign

that'll change things

[tame gust]

oh, actually something funnier is happening

you simply rewrote the expression as

[oblique estuary]

sqrt(x)-1/(x-1)

[tame gust]

$$\frac{h(x)}{g(x)} = \frac{1}{\frac{g(x)}{h(x)}}$$

[soft zealot]

Peter

[oblique estuary]

aka h(x)/g(x)

ye

lol

[tame gust]

that's on the left yes

[oblique estuary]

ngl didn't see that until after I typed that out

[tame gust]

well, please check out this write-up fully and tell me do you understand everything that happens here

[oblique estuary]

I do, and with that

lemme see what I would do first

alr I figured it out, no need for an explanation

thx

aight should be all for now

.close

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[tawdry hatch]

There's a hint that was given

(Deliberately missing (e.g., firing in the air) is also a possible option!)

[summer sedge]

Do we assume A is completely self-interested?

[proper dagger]

you can enumerate the possible outcomes out of missing, firing at B and firing at C on A's first shot

[tawdry hatch]

ig?

[proper dagger]

but for this to work, there is one big assumption

[tame gust]

The answer to this question really depends on whether B and C play optimally, if they do, you can consider all possibilities of A winning along with the respective probabilities

[summer sedge]

Ok

[proper dagger]

we must assume that both B and C are not out for A's head

[tawdry hatch]

I'd assume they want to win?

[tame gust]

that's a fair assumption, but you should mention that in your solution

[summer sedge]

Yeah

Add that you assume everyone acts perfectly in their own self interest

But just calculate what the next terms would look like given each outcome

From there

[tame gust]

you might begin with enumerating the following possibilities as fox said

[tawdry hatch]

So assuming A misses, then B will fire at C because he's a greater threat right?

[proper dagger]

correct

[tawdry hatch]

okok I'll try

thanks guys

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[frank folio]

Heyo, Im working on an uniqueness proof and just wanted to make sure im doing it correctly and following proper wording and technique. Here is the problem:
https://cdn.discordapp.com/attachments/908076393198419988/1415529062834638878/image.png?ex=68c389ab&is=68c2382b&hm=d3634fd17f489c83c8fcb7357a5f16cbe2dc92cbf3b62e8fb5e5daed2ed65901&

[frank folio]

Here is my solution:

[soft zealot]

BOSS

[frank folio]

Existance:

[soft zealot]

BOSS

BOSS

[frank folio]

Uniqueness:

[soft zealot]

BOSS

BOSS

BOSS

[frank folio]

was there a more elegant way to do this?

[final saddle]

@frank folio Has your question been resolved?

[glossy zephyr]

I dont think that this is cleaner, but you could probably have shown that under addition the Complex form a Group and that 0 is the identity.

I dont think the process deviates a lot anyways

Since all elements from a group have only one inverse -> mostly known as an opposite here

[frank folio]

@glossy zephyr eh this class isnt in abstract algebra but I got your approach

other than that is it correct?

[whole halo]

yea

[frank folio]

@whole halo tysm

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[waxen fjord]

ok so if (this is about quadratic functions pre-calculus 11) - we have vertex (p,q) and i have a question that says: y=9-4x^2, is 9 p or q? i know the other one would be a value of 0 but it's just that this one is formatted a little differently than the usual vertex format y=a(x-p)^2+q

[junior token]

you can plot the graph and find out

[ornate tree]

Look, this is in the standard form of a quadratic function

[waxen fjord]

wha

the chapter thing says its vertex form

[junior token]

https://www.desmos.com/calculator/quvkxizuz1

[waxen fjord]

so

???

[ornate tree]

y=9-4x^2 is both in standard and vertex form

[waxen fjord]

oh

i didnt know that was possible

[junior token]

consider the position of the vertex. You can see from the picture that it is on the y axis, so the x-coordinate must be 0

[waxen fjord]

oh

ohh so like

its

(0,9)

[junior token]

yes

[waxen fjord]

WHAT ok what if i can't graph it cuz i dont have paper and it takes forever to draw a graph

[junior token]

You were asking this in a discord server so I assumed you had internet

and you could use desmos

[waxen fjord]

oh

that

[junior token]

well at least now you know that an equation can be in both standard and vertex form, and so in this case you can use either

[waxen fjord]

ok ok

tysm

[junior token]

or use both to check your answer

because it should come out the same both ways, so if it doesn't, then you know you made a mistake somewhere

[waxen fjord]

alr tyy

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[subtle folio]

so y

[subtle folio]

omg

guys

yelp

[junior token]

which parts have you done?

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@subtle folio Has your question been resolved?

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[nocturne frigate]

i dont get this

[nocturne frigate]

i got -e^(3 - x) - cos(3 - x) + sin(3 - x) + c

why is it positive cos and negative sin

[scarlet sequoia]

did you factor in that you're integrating sin and cos of 3 - x?

[nocturne frigate]

not really

what does that change

i would think it could be integrated like sinx

[scarlet sequoia]

well. Why did you put a minus in front of the exp term?

[nocturne frigate]

y = -e^(3 - x)
dy/dx = e^(3 - x) which is what we have in the integral

[scarlet sequoia]

ok... and did you find that by chance or...?

Just trying to find your reasoning for that

[stoic temple]

U sub

[scarlet sequoia]

so you used u sub to compute integral of e^(3-x)

but you didn't use it for the other terms?

[nocturne frigate]

i saw 3-x is the power, so using my knowledge of differentiating e^x, i knew that it would be negative since its a -x

[nocturne frigate]

idk what u sub is

these questions are supposed to be reverse chain rule

[scarlet sequoia]

ah sorry I thought you were the one that said u sub

[scarlet sequoia]

but so you computed dy/dx for y = -e^(3-x)

why don't you try for the cos and sin terms?

[nocturne frigate]

i did

y = -cos(3 - x)
dy/dx = sin(3 - x)

but this is wrong

[scarlet sequoia]

it is wrong.

[nocturne frigate]

idk why tho

[scarlet sequoia]

Are you not applying chain rule?

it's not -cos x you're differentiating

it's -cos(3-x)

[nocturne frigate]

bruh

im dumb

my bad

okay i get it now

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[cursive pilot]

Hey guys, I need some help with this:
Thought I had to do a polynomial division here, is that right? Since the top has x^3 and bottom x^2

[worthy wren]

u can try it

it is a good idea to use it here for the reason u mentioned

[cursive pilot]

I looked up a video on how to do that, and the guy in the video says the result has a form like _x^2 +- ^_x +- something
but I am not sure how that works. mustnt my result start with something like "2x" instead of "2x^2"? Because When I do the multiplication back 2x * x^2 = 2x^3,
but when the first x needs to be x^2 then we'd have 2x^2 * x^2 = 2x^4 which doesnt seem right

[worthy wren]

yeah u should get a linear term for the quotient

should be 2x indeed

2x-1

[tired walrus]

show the vid?

[final saddle]

@cursive pilot Has your question been resolved?

[cursive pilot]

https://www.youtube.com/watch?v=OdlYNZXjmWA
its in German, but as you can see, before even calculating anything he already had _x^2 _x __

[cursive pilot]

will try again in a sec, thank you

[tired walrus]

he was dividing a 3rd degree polynomial by a 1st degree polynomial

that's why he prepared the quotient to be degree 2

you're dividing degree 3 by degree 2

so your quotient will be linear

[cursive pilot]

ah I see

linear meaning ^1 is my biggest exponent yes

[worthy wren]

yes

[cursive pilot]

this is what I'm getting so far, how to proceed here? I dont get 0 as a result

[supple marten]

Or 2x-1+(2x)/(x^2-2x-15), yes

[cursive pilot]

like so?

how can I proceed with the integral now? Dont I have the same problem?

[supple marten]

integral of sum is sum of integrals, and the third summand has an obvious partial fraction decomp

[cursive pilot]

does that first part mean I can write the 2x-1 in front of the integral?

fraction decomp I use when the part below the line is a higher degree than the part above it yes

[supple marten]

int 2x-1+(2x)/(x^2-2x-15) dx = int 2x-1 dx + int 2x/(x^2-2x-15) dx, yes

[cursive pilot]

aah

then I use pfd on the lower part and proceed from there

I will try, thank you! Now to look up how to do the partial fraction decomposition :p

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[burnt maple]

How many 5 digits number are there whose sum of digits is 43

[desert mantle]

think about which digits you can even use

[burnt maple]

0 to 9?

[severe canyon]

Can all the five digits be less than 8?

And less than 9?

[burnt maple]

No

[tired walrus]

what's the biggest POSSIBLE digit sum for any 5 digit number

[burnt maple]

45

[tired walrus]

indeed

[desert mantle]

whats the biggest possible sum if one of the digits is a 1

[tired walrus]

(why the question mark?)

anyway, yours (43) is only two below that. so surely you can't really have that many different digit combos.

[burnt maple]

37

[desert mantle]

so can we have the digit 1 anywhere?

[burnt maple]

I thought something like this
Let number be x1x2x3x4x5
Now x1<9 x2<9, x3<9...
Let 9-x1=t1
9-x2=t2
.....
t1+t2+t3+t4+t5=2
So number of solutions are 6C4

Is this correct?

[desert mantle]

I hope you mean x1 <= 9

etc

[burnt maple]

Yeah

[desert mantle]

but yes

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[worldly mesa]

!status

[final saddle]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[civic cobalt]

so n = p * q and p <= sqrt(n) or <= sqrt(n)

[worldly mesa]

Or not and

7 * 2 for example

[civic cobalt]

is it possible to write a represenation of a prime number using symbols

like how i did with composite integer = p * q

[worldly mesa]

You can't decompose a prime as a product into smaller numbers

Thats the point of being prime

[civic cobalt]

yeah sorry that was a dumb question

[worldly mesa]

Note that when you wrote n=pq, p and q are not prime necesseraly

Anyway, do you see why at least one of p,q is <= sqrt(n)?

[civic cobalt]

for the 7 * 2 example, yes, but i feel like it wont always hold

[worldly mesa]

If both are bigger then can n=pq?

[civic cobalt]

wdym if both are bigger? like if p and q are larger numbers?

[worldly mesa]

If both p,q are larger then sqrt(n)

Then n=pq cant be possible can it?

[civic cobalt]

correct

its not possible

[worldly mesa]

Then at least on of p, q must be smaller or equal then sqrtn

[civic cobalt]

so lets say

p > sqrt(n) and q > sqrt(n)

this would imply that p * q > n, which isnt true. so atleast one of p and q needs to be <= sqrt(n)

[worldly mesa]

Yes

Now that you now that

What can you say?

You want a small prime divisor of n

[civic cobalt]

since n, the sqrt of it will always be larger or equal to a prime number that can n can be divisible by?

[worldly mesa]

Elaborate

[civic cobalt]

composite seem to either be divisible by 2 or 3 which are both prime, the sqrt(n) will always be greater than or equal to 2

[worldly mesa]

Ok ok hold on

Lets say p <= sqrt(nl

If p is prine you are done

Otherwise,...

[civic cobalt]

... the theorem doesnt hold?

but how do you prove p can be prime for any composite integer n

[worldly mesa]

P can be prime

But that not must happen

It can be composite

If it is composite it is divisable by a smaller prime r

And r <= p <= sqrtn

[civic cobalt]

ok so p or q is also composite, and since its composite it is divisble by a smaller prime r. this means that sqrt(n) is also divisible by a smaller composite int r

[worldly mesa]

No

Sqrt(n) is not an integrr

Most of the time

r | p | n

[civic cobalt]

what do the | mean

[worldly mesa]

Divides

[civic cobalt]

i have to go to class, i am going to close this channel and then repost later. thank you for the help

.close

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[tranquil pine]

Hi

[worldly mesa]

Hi

need help?

[tranquil pine]

yes please

lemme send the question

ive been stuck on this for a couple days

@meager hedge whats up bro

[meager hedge]

Hey ready to try again ?

[tranquil pine]

i was just thinking of @ing you 😂

[tranquil pine]

yes please thanks man

[meager hedge]

Awesome. Do you have a paper and pen ? It’ll help you organize your thinking

[tranquil pine]

yes i do

i wrote it out

The shape

[meager hedge]

Great. Now what have you done/understood so far ?

[tranquil pine]

so i figured that each cable doesnt really have a force of 9 pounds, because theyre pulling upwards not diagonally

but i just dont know how to express that

[meager hedge]

Actually

They pull diagonally

But the upward component is 9 pounds

Each cable pulls in its direction

[tranquil pine]

what i mean is see how all cables are tied up

wait

nevermind

@meager hedge

[meager hedge]

Yes

[tranquil pine]

can you visualy explain to me the 9 pound being the vertical component and not each cable separately

[meager hedge]

Look at T1, T2, T3. Each of them has a vertical and horizontal component

[tranquil pine]

true

i agree

[meager hedge]

Now the horizontals cancel out

And the verticals compensate the weight

This is the full written detailed explanation

[tranquil pine]

why?

[meager hedge]

Newton’s first law read the page and tell me what is not clear

[tranquil pine]

i dont know newtons first law

is that F=ma?

[meager hedge]

That’s the third. But you can use it too

If the object is at rest, then the forces add up to 0

So since the lamp is at rest, the forces that apply to the lamp add up to 0. So the 3 tensions + weight add up to 0

[tranquil pine]

@meager hedge the way im thinking of is it that each is pulling in the opposite direction

it’s like if you push a box with a certain force forward and i apply the same force backwards

[meager hedge]

It is true horizontally. That’s why they cancel out. But vertically, they all push up

What are you struggling with in this section ?

[tranquil pine]

right, is that what newtons law says

[meager hedge]

It is ! You can also apply sum F = ma, in this case a = 0 (no acceleration). So you get the exact same result ! Sum F = 0

[tranquil pine]

and mass here being?

[meager hedge]

We don’t care since a =0

What matters is sum F = m*0 =0

Do you see it ?

[tranquil pine]

yes

[meager hedge]

Ok so this part is ok ?

[tranquil pine]

i get the horizontals canceling out I guess

what are we really trying to find from the question

[meager hedge]

We are trying to find the tension vector, expressed as a function of L

[tranquil pine]

what does it mean to have a tension vector

as a function too

[meager hedge]

The tension is a force

It will depend on L (amongst other things)

[tranquil pine]

but what does this mean exactly

[meager hedge]

I feel like you’re struggling to grasp this. Do you wanna hop on a call ? So I can find easier examples and you get how this works ?

[tranquil pine]

okay ill see if i can in a bit thank you

thanks man

[meager hedge]

I’ll be available in 15 min

Keep me posted

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@tranquil pine Has your question been resolved?

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[valid atlas]

I don’t have time to adjust my proof but I’m wondering how to fix it

Also assume that for all a ∈ G ∖ {1}, a ≠ a⁻¹

[tired walrus]

G would be partitioned into k pairs and not n, by the way.

your proof has the correct general idea -- each element pairs up with its inverse, but with 1 getting paired with itself and |G| even, theres gotta be at least one more that has no pair save itself

[valid atlas]

Thanks

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

[tranquil pine]

My bad bro i was hella tired , mind if i send you a dm?

[meager hedge]

No it’s alright

[meager hedge]

Is this part clear ?