help-36

[trail crest]

But you did this which assumes x <= 0

[scenic junco]

I think this is what the textbook solution showed 😭 I’ll check in a bit

[honest tartan]

hi im new here can i try solving it

[edgy mauve]

Welcome to mathcord and yes u can help anyone u want

[honest tartan]

ok

[final saddle]

@scenic junco Has your question been resolved?

[scenic junco]

Any help is welcome 🙏

[honest tartan]

what question

is it the textbook

what number is it

[scenic junco]

Number 27

[honest tartan]

hold on i, should i find g inverse of f inverse and f inverse of g inverse

[final saddle]

@scenic junco Has your question been resolved?

[final saddle]

Channel closed

Closed by @scenic junco

Use .reopen if this was a mistake.

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.

[left trail]

Can someone explain what a vector projection is. My book describes it as such. But I don't see it's purpose. It just vector b multiplied by a special constant t. By special I mean that we placed emphasis on its derivation. But I still don't see it's point

[abstract bramble]

Cast a shadow of A onto B, where the light source is infinitely far away and perpendicular to B
that's the projection of A onto B

[left trail]

What does it mean for a shadow to be casted a onto b

I guess I also want to know what makes this so important that it has its own term because casting shadows on a vector doesn't seem to be an important matter

[white tiger]

@left trail .

[left trail]

I see all this. The book I was reading showed how to construct the projection. But then again I still don't see what makes it so important. It seems like a very niche thing.

[white tiger]

it has applications in physics

[final saddle]

@left trail Has your question been resolved?

[final saddle]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[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.

[gentle zephyr]

Show that if g:R->R is continous at x = a and the function f: R^2 -> R is given by f(x,y) = g(x) then f is continous at every point of the line (a,y), use this to prove that the following functions are continous in all R^2

seems simple but I cant find a way to prove it

more than simple, seems right, is just hard to put all the puzzle pieces together

from g : R -> R being continous at x = a we get this

the complicating part is translating that

f is continous at every point in the line (a,y)

no, wait I think I get it

this is a set

call it set D

[ivory bear]

is it reflecting this @gentle zephyr , where a is a general pt on the line y=a hence f(x,y) is continuous on every pt of line (a,y)

[gentle zephyr]

we need to prove that from the first claim that g is continuous at x = a

[gentle zephyr]

no

[gentle zephyr]

yes, sort of

like, we know f(x,y) = g(x)

from here is just a few steps more

not lateral limits exactly, look at the definition of continuity in multivariable

[ivory bear]

tramslate man

[gentle zephyr]

dude

[gentle zephyr]

Given $f : \mathcal{D} \subseteq \mathbb{R}^2 \to \mathbb{R}$ and let $(x_0, y_0) \in \mathcal{D}$ it is said that f is continuous on $(x_0, y_0)$ if $\lim_{(x,y) \to (x_0, y_0)} f(x,y) = f(x_0,y_0)$ and it is said that $f$ is continuous on $\mathcal{D}$ if $f$ is continuous on every $(x_0, y_0) \in \mathcal{D}$

[soft zealot]

Renato

[ivory bear]

betr

[ivory bear]

?

[gentle zephyr]

we are close

[ivory bear]

think think

[deep condor]

Lets double check our goal. What is left to prove here?

[deep condor]

this makes sense, but i dont see a way to prove $$\lim_{(x,y) \to (a,y)} f(x,y) = f(a,y)$$ without using $f(x,y)=g(x)$

[soft zealot]

haseeb

[deep condor]

my thought process is $$\lim_{(x,y) \to (a,y)} f(x,y) \overset{?}{=} \lim_{x \to a} g(x) = g(a) = f(a,y),$$ which is pretty much what you did in the screenshot above plus the critical equality at the start

[soft zealot]

haseeb

[deep condor]

the hardest part now is showing the 2D limit becomes the 1D limit. any ideas on how we could do that?

[gentle zephyr]

unsure

what I was thinking, I already wrote it down in the whiteboard and showed it to u

[deep condor]

fair enough lol, but that "want to prove" is probably going to hinge on this

[gentle zephyr]

I did a 200% gigabrain thinking for like 20 min and this the best I came up with

[deep condor]

hey, progress also entails ruling out the wrong things

[gentle zephyr]

wdym?

f(a,y) = g(a) by definition

[deep condor]

i meant this part

[gentle zephyr]

is it wrong by any means?

[deep condor]

not at all, but we will need some more work to get there

[gentle zephyr]

where are we?

as of now

[deep condor]

well we want $$\lim_{(x,y) \to (a,y)} f(x,y) = f(a,y) = g(a)$$

[soft zealot]

haseeb

[deep condor]

but because $g$ is continuous at $a$, this is the same as saying $$\lim_{(x,y) \to (a,y)} f(x,y) = \lim_{x \to a} g(x)$$

by the single-var definition

does that make sense?

[gentle zephyr]

g continous solely at x = a, careful

[soft zealot]

haseeb

[deep condor]

whoops, that is better

[gentle zephyr]

yes and this is equal to g(a) by def of continuity of single var

[deep condor]

so if we prove this, we prove that lim of f(x,y) = f(a,y)

[gentle zephyr]

ye

[deep condor]

i.e. that f(x,y) is continuous at (a,y)

[gentle zephyr]

just set up a equality after this, equal to g(a) and by definition of f(x,y) = g(x) we get that g(a) = f(a,y)

[deep condor]

that is exactly how, yes

[gentle zephyr]

we proved it

[deep condor]

well we still have to prove the starting equality

[gentle zephyr]

?

[deep condor]

we dont know this is true yet

unless you have a theorem that says $$\lim_{(x,y) \to (a,b) }g(x) = \lim_{x \to a} g(x)$$

[soft zealot]

haseeb

[deep condor]

but if not this is pretty ok to prove

[gentle zephyr]

how to prove it

[deep condor]

well, the way i did it was using epsilon-delta

but i think its better to discuss a hand-wavy proof, at least to start

let's use this as our "picture" of the two-variable limit

in the blue domain, $\norm{(x,y) - (a,b)} < \delta$ is represented by the (punctured) ball of radius $\delta$. This is the bottom red circle, D

[soft zealot]

haseeb

[deep condor]

with me so far?

[gentle zephyr]

yes

more like a ball, is a disk

[deep condor]

we say "ball" to refer to anything in n dimensions, but yes it is a (punctured) disc in the 2-variable case

now, we do this because f is a surface and not just a line

but if we put g(x) inside this model, it becomes a line/curve

gimme a sec to sketch what i mean

so the plane has basically become a line, because we're fixing y, and the surface has become a curve, the graph of g(x) like it would be in single-variable

in this case, do we still need to use balls/discs to approach (a,y)?

[gentle zephyr]

no

when y is fixed no

[deep condor]

exactly! why is that? / what can we use instead?

[gentle zephyr]

a segment of points?

like, idk

[deep condor]

well, what do we use in the single-variable case?

actually you were right you just said it interestingly

it would be |x-a|, there's only one way to approach a

which is along the blue line

all of this is to say that $\norm{(x,y)-(a,y)} = |x-a|$.

[soft zealot]

haseeb

[deep condor]

(for a fixed y)

[gentle zephyr]

intuitively the y-y gets to 0

if the limit exists

[deep condor]

and that's what the epsilon-delta proof hinges around

but here's a geometric idea of what happens when you "go down" a dimension

[deep condor]

so, can you try using this + the intuition to prove that $$\lim_{(x,y) \to (a,y)} f(x,y) = \lim_{x \to a} g(x)?$$

[soft zealot]

haseeb

[gentle zephyr]

using epsi delta?

[deep condor]

more like abusing it, but yeah

its not really an epsilon delta proof, it just starts with the definition

[gentle zephyr]

I mean I still find it hard ngl

assuming lim (x,y) -> (a,y) f(x,y) = L

∀epsi > 0, ∃ delta > 0 s.t.

[deep condor]

actually start from g

[deep condor]

i think it's a universal experience to find it hard :P but grappling with all of this makes it easier over time

[gentle zephyr]

assuming lim x->a f(x) = L

∀epsi>0∃delta>0s.t

|x - a | < delta

[deep condor]

you actually know more about L, because g is...

[gentle zephyr]

now add and subtract y

[gentle zephyr]

L = f(a)

[deep condor]

you can use g

but i appreciate the consistency

[gentle zephyr]

assuming lim x->a g(x) = g(a)

∀epsi>0∃delta>0s.t

|x - a | < delta

[deep condor]

ill leave you to write a lil bit and see if you can get to "f(x,y) -> f(a,y) as (x,y) -> (a,y)"

gonna go back on my word: you can write out the definition for f(x,y) so you know what your goal is

[gentle zephyr]

which is equivalent to ||x - a + y - y|| < delta for any y ∈ R

which is equivalent to ||(x - a) + (y - y)|| < delta for any y ∈ R

its not the same

idk what i am even doing

[deep condor]

not exactly, you dont wanna add and subtract

damn that's crazy. anyways remember norm takes in a point

not just a number

[gentle zephyr]

yes but sqrt is not

[deep condor]

again, definition for f(x,y) may help you with where to go

[gentle zephyr]

|x-a| = sqrt((x-a)^2) = sqrt((x-a)^2 + (y-y)^2) = ||(x,y)-(a,y)||

[deep condor]

you are in the math server. fork found in kitchen

[gentle zephyr]

@deep condor

[deep condor]

also #discussion is the general, this is someone else's help channel

[deep condor]

yep that is correct, so keep going to get to the definition for f(x,y)

[gentle zephyr]

the proof is almost done

at this point we just need the fireworks and stuff and the trumpets, the proof is almost done dude @deep condor

[deep condor]

see? wasnt so bad

keep going tho

[gentle zephyr]

|x-a| = sqrt((x-a)^2) = sqrt((x-a)^2 + (y-y)^2) = ||(x,y)-(a,y)||

[deep condor]

this still isnt the general channel, let this guy study in peace pls

[gentle zephyr]

|x-a| < delta then ||(x,y)-(a,y)|| < delta and thus, there exists an epsilon > 0 such that |f(x,y) - g(a)| < epsilon

[deep condor]

#discussion

[deep condor]

and g(a) is

[craggy plank]

no, let's move to #bots

[gentle zephyr]

g(a) = f(a,y) by def

[deep condor]

so what is $\lim_{(x,y) \to (a,y)} f(x,y)?$

[soft zealot]

haseeb

[gentle zephyr]

meaning we conclude that limit is f(a,y) :)

[deep condor]

:))

thus f(x,y) is continuous iff g(x) is (at x=a)

this should be enough to prove both directions, so try to write it down as a full proof

then the actual questions should be one-liners, using this "theorem" you proved

[gentle zephyr]

math is hard

[deep condor]

welcome to the life of a mathematician

<@&268886789983436800> hii if u dont mind :3

[deep condor]

but you did this proof pretty easily, id argue it was done more independently than the product of limits proof u worked on a little while ago, so you are actually progressing and learning :D

pretty soon this will be ezpz

[gentle zephyr]

that product of limits proof was a pain

[gentle zephyr]

I think so too, but we will see if prof agrees for the day of the exam

[deep condor]

well, try to put this all together into a proper proof, and make sure it satisfies both directions for equality

and if that turns out good, you will have it

(sorry if i bothered you omni, i was just really tired and didnt wanna deal with it ._.)

[cyan kayak]

no, you didn't bother me, I was just reading up and checking to make sure that there wasn't a subtlety that I missed.

[barren hound]

we exist to be bothered about this sort of thing

[deep condor]

hehe oki, do let me know if i can do something differently tho ^_^

and ty

[gentle zephyr]

ty to u haseeb

.close

[final saddle]

Channel closed

Closed by @gentle zephyr

Use .reopen if this was a mistake.

[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.

[warm python]

Let $a_1=1;a_{n+1}= \sqrt{3a_n}$ .Show this seqeunce converges

[soft zealot]

wai

[pliant shore]

1. monotonic?
2. bounded above?

[warm python]

yea, but how would I show it's monotonic

indunction, I suppose

[pliant shore]

yeah, exactly

[warm python]

I was trying to prove cauchy convergence for some reason 😭

Yea, I forgot MCT was a thing

[pliant shore]

if you ever take exams for RA you should have cheat sheets for ideas like these

strategies so that you can remember how you should attack a certain class of problem

cause yeah, the monotone sequence theorem is the one you need for these limit questions

[tired walrus]

this one even admits an explicit formula doesnt it

but i guess this would require you to know that the function x ↦ 3^x exists

which given your RA course is probably not kosher at all

[warm python]

I know, just ashamed I tried cauchy convergence instead

[pliant shore]

I mean from an RA lens the closed form is besides the point

[pliant shore]

that's okay!

don't beat yourself up for not knowing

[warm python]

yea, probably not for a while

[warm python]

tq

One more question, I'll try to get the idea this time

[final saddle]

@warm python Has your question been resolved?

[final saddle]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

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.

[ivory bear]

is it undergrad math?

[gentle zephyr]

ye this is my third week of uni

[tired walrus]

a or b?

[gentle zephyr]

a

[ivory bear]

[tired walrus]

the functions x^2 and e^x should strike you as rather easy to analyze

[gentle zephyr]

Given $f : \mathcal{D} \subseteq \mathbb{R}^2 \to \mathbb{R}$ and let $(x_0, y_0) \in \mathcal{D}$ it is said that f is continuous on $(x_0, y_0)$ if $\lim_{(x,y) \to (x_0, y_0)} f(x,y) = f(x_0,y_0)$ and it is said that $f$ is continuous on $\mathcal{D}$ if $f$ is continuous on every $(x_0, y_0) \in \mathcal{D}$

[soft zealot]

Renato

[gentle zephyr]

e^x is continuous?

,, \lim_{x \to 0^-} e^x \overset{?}{=} \lim_{x \to 0^+} e^x

[soft zealot]

Renato

[tired walrus]

yes, everywhere in fact

and by everywhere i mean at every real x

[tired walrus]

yes, these are both equal to e^0=1

[final saddle]

@gentle zephyr Has your question been resolved?

[final saddle]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[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.

[jagged flare]

i get the motivation of transforming it into a combinatorical model, but whats the motivation for it to be that?

[jagged flare]

can never understand where they get that from the problem

[final saddle]

@jagged flare Has your question been resolved?

[pliant shore]

I really don't see a way other than combinatorial proof

floor function is nasty

[tranquil pine]

yeah theyre asking why this thing specifically

[jagged flare]

first solution is algebraic

[tranquil pine]

you try to start from the simplest thing you see, the (2n+1 C n). This is just the number of ways of choosing n things from 2n+1.

Now you want to choose n things from 2n+1 but in a much funnier way. You try to look at how to interpret the LHS: you can see nCk and ((n-k)Cfloor(n-k)/2): it looks like you take k things from a set of n things, and you kinda do something with the remaining n-k things. So it looks like the LHS just deals with n objects instead of 2n+1. So you might think that the 2^k is there to fix this and somehow kinda double this n objects into 2n, but why should it become 2n+1? you dont know yet, but something that seems reasonable to do is to divide the 2n+1 into a block of n things, a single obect, and another block of n things.
Now you want to try to do the things in the LHS to this configuration:

first the (nCk): you can think of just choosing k objects from the first block of n. Then you do something else with the remaining n-k: you take half of them (if n-k is odd you take floor(n-k)/2). Now you try to understand why this gives you a way of actually choosing n things from the 2n+1: so far we have chosen k things then floor(n-k)/2, so it remains to pick ceil(n-k)/2, which is kinda weird if the only other piece left is a 2^k (by this i mean that it has a k, not a n-k) so the 2^k must have something to do with the k objects we chose at the start. At first (before doing this whole setup) what i thought of was that you could maybe color the k objects you chose, either black or white, something like that. You try to do something similar, keeping in mind that we actually only worked with the first block of n objects and didnt touch the single object in the middle and the second block of n.

[pliant shore]

ewwww

not even IMO wants you to do that lmao

[tranquil pine]

you can try to think the 2^k is there to do this: for each of the k objects you chose, you either take it from the first block, or from the second block. Now you started using the second block!
Now remember that we still have to choose another ceil(n-k)/2 objects. But the whole thing has to make sense: if with the 2^k(nCk) we chose objects that are either on the first block or the second block, with the (n-k C floor(blahblah) ) we should take another ceil(n-k)/2 which should be the corresponding ones from the second block. This is because we try to do different things with the two binomial coefficients, so that we dont overcount the number of ways.

So eventually you realise that what youre doing is choosing first the numbers that appear on one block and dont appear on the second, and then the pairs of those which appear on both blocks in the same position. And the single object in the middle is just there to fix the fact that sometimes 2floor(n-k)/2=n-k-1

[tranquil pine]

this was my thought process (continues in the following message), i solved it without looking at the sol you posted but it should be the same thing, they distinguish the 2 blocks of n by just saying one is boys and the other is girls

[jagged flare]

ohh and the ceil(n-k)/2 turns out to be the blocks of 2 that dont get chosen right?

[pliant shore]

well, they do get chosen

in the 1st block, each pair shares one ticket

[tranquil pine]

you have floor(n-k)/2 that is the number of pairs in which both the boy and the girl receive the ticket

[pliant shore]

but in the 2nd block, now each person gets one ticket

so it's like k * 1 + 2 * (n - k)/2 = n

[pliant shore]

but there's a floor function instead

[tranquil pine]

the ceil is just there to fix the off by 1

[jagged flare]

hmm i see

[tranquil pine]

like, floor(n-k)/2 + ceil(n-k)/2 = n-k

[jagged flare]

ooo

alright, thank you very much yaal! ❤️

.solved ❤️

[final saddle]

Channel closed

Closed by @jagged flare

Use .reopen if this was a mistake.

[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.

[vivid sundial]

Lol hello luminary

U could have asked me that

😔

[stone wagon]

start with u = sqrt(x)

or just x = u^2 (same thing)

[opal plume]

oh

[opal plume]

batao to

yaha bh1 answer kar skte ho, 1f dekh l1ya h to

dm krna zarur1 h?

.close

[final saddle]

Channel closed

Closed by @opal plume

Use .reopen if this was a mistake.

[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.

[lime crest]

Hi! im just curious about something
i think ive come across a theorem (i forgot the name) but it says if two 2d polygons have the same area then one can be cut into finitely many pieces and rearranged into the other. im curious if this holds for 3d?

[glossy nacelle]

pretty sure it does

actl maybe not..

[lime crest]

is there a proof? TwT

[glossy nacelle]

its the Wallace–Bolyai–Gerwien theorem

[lime crest]

OHH yea

[tranquil pine]

It's not true in 3d

[plucky rover]

This is false

[glossy nacelle]

hilberts 3rd porblem

[tranquil pine]

^

[lime crest]

whats that?

[plucky rover]

You need same volume and dehn invariant

[glossy nacelle]

like cube and tetrahedrom cannot

[tranquil pine]

(tbh I'm trusting him that it's the correct number)

Bruh bye bye WiFi conenction

[glossy nacelle]

cuz different dehn invarient

[tranquil pine]

Ok I'm back

[plucky rover]

It is third yeah

[tranquil pine]

Yeah

So sad

[lime crest]

ive never heard of hilberts problems😭 thank you tho

[glossy nacelle]

can search it up

[plucky rover]

It's okay, they're way beyond your pay grade lol

Just look it up yeah

[lime crest]

okay, thank you guys :3

ill close this now

.close

[final saddle]

Channel closed

Closed by @lime crest

Use .reopen if this was a mistake.

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.

[sinful knoll]

guys

[sinful knoll]

quick help

what is differential

for x it is delta x and dx

but for y its not delta y but dy which is dy= f'(x)dx

whats the idea of differentials

<@&286206848099549185>

[slate narwhal]

!15m

[final saddle]

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

[ivory bear]

differentials are just the denotion of slope of a function f(x), denoted by f'(x) (f'(x)=d(f(x))/dx)

[sinful knoll]

what

u know

ahmm

well the differential was used in finding the sq root of 16.4

i got confused

i always thought dy being very small change in y

[slate narwhal]

for future helpers: you might want to provide the original context of your problem

[final saddle]

@sinful knoll Has your question been resolved?

[ivory bear]

that is done using approximations, an entirely diff thing

comes uner the topic of applications of derivatives

[sinful knoll]

yuh yuh

can u tell

me i cant provide the pic

unfortunately

[final saddle]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

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.

[potent cobalt]

I'm stuck

[potent cobalt]

Q1

!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

[potent cobalt]

2

[potent cobalt]

<@&286206848099549185>

[thin cloud]

!15m

[final saddle]

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

[zenith cedar]

We can use similarity for the first part

[thin cloud]

chill man, you know the rule:p

[potent cobalt]

Sry

[craggy scroll]

[potent cobalt]

So how do I solve it

[zenith cedar]

Which similar triangles can you spot here

[potent cobalt]

SMQ and LQR

[zenith cedar]

What about snr and lqr

[potent cobalt]

Oh ya that too

Can you check my working if I did a step correctly by any chance

I knew only the first 3 btw then I did shi

[zenith cedar]

Q indeed is the midpoint of RN

[potent cobalt]

Oh 😯

[zenith cedar]

So you have already done the first oart

Part

[potent cobalt]

But I don't know the reason for it

That why it's a mid point

[zenith cedar]

Consider triangle SNR

We construct a parallel line

What does that tell us

[potent cobalt]

We gotta construct by our hands?

[zenith cedar]

No it is already in the figure in this case

LQ || SN

[potent cobalt]

Nah bro I am blind

[zenith cedar]

Wait let me make a new figure

[potent cobalt]

Can you tell me what is the converse of the midpoint theorem after this?

[zenith cedar]

Let ABC be a general triangle

And EF is parallel to BC

Now what can we infer about sides AB and AC

[potent cobalt]

They are parallel

[zenith cedar]

Have you done similarity

[potent cobalt]

What is that

[zenith cedar]

AB and AC are not parallel

[zenith cedar]

Oh

Wait let me reframe things a bit

[zenith cedar]

I wasn't aware about this

Let ABC be a general triangle
Now EF is parallel to BC
And let E be the midpoint of AB

What does that tell about F

[potent cobalt]

And half of it, right?

[zenith cedar]

That isn't really relevant for the first part

But yes it is half

So what do we know about F now

[potent cobalt]

It is a mid point

[zenith cedar]

Yes

[potent cobalt]

Of AC

[zenith cedar]

Now let us apply the same thing to triangle SNR

Can we spot a similar pattern here

[potent cobalt]

So what to do now?

[zenith cedar]

Look for a similar pattern in triangle SNR of your question

It is the same as the figure i mafe

@potent cobalt any progress

[potent cobalt]

Trying

[zenith cedar]

Alright

[potent cobalt]

Got it

Now to da second part

[zenith cedar]

You already told me the answer to this

[potent cobalt]

I did?

[zenith cedar]

^

[potent cobalt]

Oh

Can you tell me how do I write it

Like formally

[zenith cedar]

You can directly state that it is a corollary midpoint theorem for now

But a more formal proof comes from similarity

[potent cobalt]

Let me give you an example

My fair school notebook

@zenith cedar

[zenith cedar]

Uh you can say it like this

In triangle SNR

LQ is parallel to SN

L is the midpoint

By mbt

Is the midpoint of NR

You can write something similar for the other part

This isnt really that necessary as long as your argument works

[potent cobalt]

One more question

Where do we use cpct

[zenith cedar]

Congruent triangles

There aren't any here

It is mostly to prove things equal

[potent cobalt]

👍🏻 tysm for your help

.close

[final saddle]

Channel closed

Closed by @potent cobalt

Use .reopen if this was a mistake.

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.

[dapper hull]

!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

[opal plume]

2 ig

[dapper hull]

could you show your work if possible

[opal plume]

i substituted sqrtx

u

so u^2=x

now what

[dapper hull]

wouldn't say it's wrong but do you think it'll help ?

[plucky rover]

Well what do you have after you did that

[opal plume]

no idea

[plucky rover]

Oh sorry qimmah, I'll leave this to you

[opal plume]

tan^-1(u)

[dapper hull]

what is the integral you get after you did that subtitution ?

[plucky rover]

You should have more than just that

[dapper hull]

(even though subbing from the very start is quite shit imo it still works)

[opal plume]

gimme a sec pls

2utan^-1(u)

[dapper hull]

dx or du ?

[opal plume]

du

[dapper hull]

good now what do you think you should do next ?

[opal plume]

u can tell diff approach too, thats just what i thought

[dapper hull]

it's the same approach just a different order

[opal plume]

by parts ig , but idk what to do w tan^-1

[dapper hull]

you can't integrate it (at the moment) can you ?

then differentiate it

[opal plume]

derive tan^-1?

[dapper hull]

yes

[opal plume]

but why diffrentiate?

[dapper hull]

we want to do integration by parts no ?

[opal plume]

yh

one sec, let me process😭

[dapper hull]

why don't we do integration by parts from the very start wouldn't that be much easier ?

$\int \tan^{-1}(\sqrt{x}) dx$

[soft zealot]

qimmah

[opal plume]

ohh

how?

[dapper hull]

when doing integration by parts what notation do you use
u dv or f g' ?

[opal plume]

udv

[dapper hull]

ok, now it isn't obvious but isn't
tan^-1(x) = 1 * tan^-1(x) ? yea ? (I know it's sqrt(x) but give it a moment)

[opal plume]

hmm

[dapper hull]

we know how differentiate tan^-1(x) and also tan^-1(sqrt(x))
and we know how to integrate 1 dx yes ?

[opal plume]

yh differentiate but idk how to integrate

[dapper hull]

not even the integral of 1 dx ?

[opal plume]

i do

[opal plume]

i am talking abt tan^-1

[dapper hull]

yeah we don't know (yet) unless ofc you can find it using the same process anyways

u = tan^-1(sqrt(x)) , dv =1
what are the values of du and v ?

[opal plume]

v = x,

[dapper hull]

good

how about du ?

[opal plume]

du=1/(1+x)(2sqrtx)

u get it?

[dapper hull]

could be written better but yeah I understand you

now we know that
integral udv = uv - integral vdu yes ?

(where u and v are both functions of x)

[opal plume]

yh

[dapper hull]

ok sorry but I need to go AFK for the moment
but I think you should be fine afterwards

[opal plume]

ok thanks!

ill try

i am not fine

[dapper hull]

back

[dapper hull]

how did it go ??

[opal plume]

idk what am i doing

[dapper hull]

could you show me

[opal plume]

ill try this ques again later

[opal plume]

i dont have my phone w me rn

[dapper hull]

ok

[opal plume]

can i ping u later? when i reattempt this ques

[dapper hull]

don't ping me just ask it cuz I may be asleep

[opal plume]

oh ok

sorry

thanks!

.close

[final saddle]

Channel closed

Closed by @opal plume

Use .reopen if this was a mistake.

[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.

[wet hawk]

i need help

[wet hawk]

i’m not sure if this is right

[formal trail]

you can't just get rid of the absolute value like that

[wet hawk]

oh

how do i do it instead

[formal trail]

well there is a general rule for working with absolute value inequalities
[ \abs{\bsq} < \rsq \implies -\rsq < \bsq < \rsq ]

[soft zealot]

cloud

[formal trail]

but before we jump into applying that rule we should consider what sort of values an absolute value can produce anyway

[wet hawk]

okay

so the first step do i cancel out the 9?

[formal trail]

that is a good first step

but then we're left with
[ \abs{\bsq} \le -45 ]

[soft zealot]

cloud

[craggy plank]

No, you should ask yourself “what sort of value an absolute value can produce” in the first place

[craggy plank]

Oh, you’re going this way, I see.

[wet hawk]

wait i’m confused

[slate narwhal]

what do you know about the values an absolute value can produce?

[wet hawk]

nothing 🥹😞like the teacher notes are so bad

[formal trail]

just looking at the equality, is there any number whose absolute value equals -45?

[wet hawk]

uhhh

positive 45

[formal trail]

,w |45|

[soft zealot]

Results provided by WolframAlpha

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

Requested by cloudedhorizons

[formal trail]

absolute value is a function which can take in a number which can be either positive or negative (or 0) and always returns a positive number (or 0)

[wet hawk]

i see

[formal trail]

in other words, we wouldn't be able to find any number which, when put into an absolute value, returns a negative result

[wet hawk]

i see

so the answer is no solution

[formal trail]

yes, exactly

[wet hawk]

ohhh i see

thank you

would this be right

[formal trail]

it's partially right, but the rule we are wanting to use here is
[ \abs{\bsq} > \rsq \implies \bsq > \rsq \text{ or } \bsq < -\rsq ]

[soft zealot]

cloud

[formal trail]

you mostly applied that rule correctly but didn't flip the inequality for the second case

[wet hawk]

wait why would i flip it since i didnt divide a negative?

[formal trail]

it sort of is like dividing by a negative because we added a - sign to the right side

[slate narwhal]

3 > 2 but -3 < -2

maybe this helps explain why

[wet hawk]

ohhh

i seeeeee

so like this

[formal trail]

yes, that's correct

[wet hawk]

okayokay

how would i graph them

[formal trail]

do you know how to graph them individually?

[wet hawk]

nope i thought i did

they only thing i know is if there open or closed

[formal trail]

well the idea is to first graph the point, and then add an arrow indicating the rest of the points (to the right if it's greater than, to the left if it's less than)

[wet hawk]

so liek this

[formal trail]

yes

[wet hawk]

okay good

[final saddle]

@wet hawk Has your question been resolved?

Channel closed

Closed by @wet hawk

Use .reopen if this was a mistake.

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.

[iron ivy]

For symmetric, i solved it by stating that A = B implies A is not a proper subset of B, hence every element of B is in A and none more, further since every element of A is in B and none more, B is a subset of A, proving that B = A. I looked online and another proof went rather simply

• By definition, if A = B, then every element of A belongs to B and every element of B belongs to A
• By definition, that would be B = A
• Therefore, equality is symmetric.
  While I understand that the logic is correct and it pretty much says the entire thing as me, but there's a lack of rigour i feel, also it just seems so short to me to be marked correct in an exam. This might be a silly question but i was hoping if someone could give me a quick conformation on which approach is better.

[slate narwhal]

if the definition is given it's fine i suppose? it's not like the argument uses some invented definitions

i might be incorrect though

[iron ivy]

hmm that's true as well

i suppose i will go ahead and close this channel. The other approach is just better yea.

.close

[final saddle]

Channel closed

Closed by @iron ivy

Use .reopen if this was a mistake.

[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.

[full mist]

I have a question regarding limits. Is it possible to send a photo of the problem and my work so far?

[plucky rover]

Yes send it

[final saddle]

No need to ask “Can I ask…?” or “Does anyone know about…?”—it’s faster for everyone if you just ask your question! See https://dontasktoask.com/

[craggy plank]

just send :))

[odd seal]

this is the expected way

[full mist]

Number 2

Here’s my work so far

I feel like I should have one equation coming from the left and one equation coming from the right, but I also feel like the answer shouldn’t be infity or negative infinity

[plucky rover]

There's a mistake in what you've written

9+4 is 13, not 15

[full mist]

Would it not be 3 + 13?

[plucky rover]

$3^2+ \sqrt{16}$

[soft zealot]

Xavier 🌺

[full mist]

Which is what I had right?

[plucky rover]

Yes

Which is 9 + 4

Why would you get an extra 3

[full mist]

Oh right

[plucky rover]

(and even if it was an extra 3, you'd get 16 and not 15)

[full mist]

I see that mistake

[plucky rover]

Also 27-3+8 is not 16

[full mist]

32

So

My newest answer is 13/32

[plucky rover]

That's correct yeah

[full mist]

But shouldn’t I have two equations that split off? One that uses 3- and 3+?

[plucky rover]

Ah yes you can do it that way too

But this function is continuous

[full mist]

How did you know that

[plucky rover]

Oh you haven't done that

[full mist]

No I haven’t

Which is the part that I believe I’m missing

I just found the value

I don’t know if it’s continuous or not

[plucky rover]

In that case yeah you can do it using 3- and 3+

[full mist]

But how do I go about that

[plucky rover]

What have you done wrt continuity

[full mist]

Wrt?

[plucky rover]

With regards to

[full mist]

I don’t believe so, all of my work for the problem is there

[plucky rover]

No like in the class

[full mist]

Oh

Well that’s the part that’s confusing me

We can’t use l’hospitals rule and it all needs to be done algebriacially

[plucky rover]

I got a class to go to, sorry

Hope someone else drops in here

[full mist]

All good, thanks, I’ll thug it out

.close

[final saddle]

Channel closed

Closed by @full mist

Use .reopen if this was a mistake.

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.

[modern terrace]

Hii

[modern terrace]

Anyone know statistics?

[slate narwhal]

!da2a, please send your question

[final saddle]

No need to ask “Can I ask…?” or “Does anyone know about…?”—it’s faster for everyone if you just ask your question! See https://dontasktoask.com/

[modern terrace]

Huh?

[slate narwhal]

do you have a question you're stuck on?

if you have, please send it

[final saddle]

@modern terrace Has your question been resolved?

[tired walrus]

you should send your question(s) directly

[final saddle]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

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.

[serene tartan]

uhh

[crude rampart]

feel like im skipping over something

[serene tartan]

lemme try this

[crude rampart]

really simple here

and i cant figure out what it is

wait i think i see it

shouldve kept it like this instead of multiplying it out oops

[tired walrus]

that's generally a good idea yeah

[crude rampart]

WOOOPSYYYYYYYYYYYYYYYYYYYYYYYYYYYYY 😭

uh how do i close this

first time

sorry

[craggy atlas]

you wrote -4x instead of +4x

in the final answer

the denominator part

[crude rampart]

oh woops

thanks for catching that lol

[craggy plank]

type .close when you're done with the question

[crude rampart]

.close

[final saddle]

Channel closed

Closed by @crude rampart

Use .reopen if this was a mistake.

[fading geode]

.reopen

[craggy plank]

??

[fading geode]

What is the easiest way to prove any number divided by zero is infinity?

[craggy plank]

Claim another channel

just post your questions here

[serene tartan]

uhh limits maybe

idk im not a math major

[modern terrace]

Ok

[tired walrus]

well you were gone too much so your channel timed out and this one is taken now

so go take a new one and send your q there

[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.

[uncut topaz]

Can someone help me with this differentiation using chain rule

[solar crest]

chain rule states that

$\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}$

[soft zealot]

Flatus

[uncut topaz]

Yes

Letter b, this is the given

I'm kinda confused how to differentiate it using chain rule

[tired walrus]

does it say to find dy/dx?

this is kinda weird tbh

[uncut topaz]

Yes

I'm kinda confused also but he says find dy/dx

Can someone help

It's so confusing

[tiny gorge]

did you read about the chain rule?

not that you really need it here, but you could use it for the first two parts

[uncut topaz]

How

[crude rampart]

for $y = f(g(x))$

[soft zealot]

9k

[crude rampart]

$dy/dx= f’(g(x)) * g’(x)$

[soft zealot]

9k

[tiny gorge]

maybe you have seen it in the form $$\frac{dy}{dx} = \frac{dy}{du}\ \frac{du}{dx}$$

[soft zealot]

Bungo

[crude rampart]

ok

$y = 2u^4 +4$

[soft zealot]

9k

[crude rampart]

wait huh

did i mess up ? 😭

[uncut topaz]

Confusing right

[crude rampart]

i see

do you’re looking for dy/dx

[uncut topaz]

Yes

[crude rampart]

well for part b, both x and y depend on u

[uncut topaz]

Yes

[crude rampart]

so dy/dx = (dy/du)/(dx/du)

[uncut topaz]

Yes

Is this right?

[crude rampart]

so find the derivative of y

using basic rules

[crude rampart]

power rule and constant rule

yeah that’s good

[uncut topaz]

After this?

[crude rampart]

well u have dy and dx now

and you’re looking for dy/dx

so

substitute

[uncut topaz]

Ohhhh

Like divide?

Like 4u/3u²?

[crude rampart]

yes

now just simplify

[uncut topaz]

So yhe answer is 4/3u

[crude rampart]

yup

[uncut topaz]

That's the final answer?

[crude rampart]

yeah unless ur professor wants u to write it in terms of x

[uncut topaz]

How about c here

[uncut topaz]

Wdym

[crude rampart]

well $x = u^3 + 2$

[soft zealot]

9k

[crude rampart]

which means $u=(x-2)^(1/3)$

whoops

which means $u=(x-2)^1/3$

ugh how do i do this

$u = (x-2)^{1/3}$

[soft zealot]

9k

[crude rampart]

so this could be rewritten as $$4/3(x-2)^{1/3}$$

[soft zealot]

9k

[uncut topaz]

How did you get it

[crude rampart]

$x=u^2 + 3$

[soft zealot]

9k

[crude rampart]

oops

sorry

[uncut topaz]

I think you mean x=u³+2

[crude rampart]

$x = u^3 + 2$

[soft zealot]

9k

[crude rampart]

ok so first isolate the cube

so subtract 2 from both sides

$x -2 = u^3$

[soft zealot]

9k

[uncut topaz]

Then cube root both sides

[crude rampart]

yup

[uncut topaz]

To get (x-2)^1/3

[crude rampart]

$\sqrt[3]{x -2} = \sqrt[3]{u^3} =

u = (x-2)^{1/3}$

[soft zealot]

9k
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[crude rampart]

oops

whatever

u get the idea

[uncut topaz]

So 4/3(x-2)^1/3

[crude rampart]

if your professor wants it written in terms of x yes

if he didn’t specify that, then leave it as u originally had it

[uncut topaz]

He just didn't say anything, he only says find dy/dx

Hes crazy

How about this letter c

[crude rampart]

ok then just leave it as 4/3u and ur good

uhh lemme look at it

[slate narwhal]

,rccw

[crude rampart]

differentiate both sides with respect to x

dy/dx is basically just saying

“how does y change when x changes?”

when you take the derivative of the right side you have to take the derivative of the left side to keep the equation true

so take the derivative of both sides and lmk when you’ve done that

$x=y^4+1$

[soft zealot]

9k

[crude rampart]

what is $\frac{d}{dx}(x)$ ?

[soft zealot]

9k

[crude rampart]

the left hand side

@uncut topaz

[uncut topaz]

Wait the image doesnt load

[uncut topaz]

1

[crude rampart]

ok nice

now do the same thing to the right side

[soft zealot]

9k

[uncut topaz]

4y³

Ohhh

1=4y³?

[crude rampart]

yes, but remember that we are differentiating with respect to x.

so we must multiply y by dy/dx because y depends on x.

[uncut topaz]

Is this implicit differentiation?

[crude rampart]

yes !

anyways

[uncut topaz]

Should i use implicit differentiation even though our professor didn't discuss it yet?

[crude rampart]

1= 4y³(dy/dx)

[uncut topaz]

1/4y³ is the answer

He just jump straight to chain rule without discussing us the implicit first

[crude rampart]

no, remember u have to solve for dy/dx

$1=4y^3(\frac{dy}{dx})$

[soft zealot]

9k

[uncut topaz]

I will use multiplication in this right?

[crude rampart]

no, now you’re just trying to isolate dy/dx

[uncut topaz]

(4)(3y²)dy/dx+(y³)dy/dx(0)

[crude rampart]

no

[crude rampart]

recall this

here think of this

if we have $$1=3x$$

how do we solve for x?

[soft zealot]

9k

[uncut topaz]

Divide both sides of 3

[crude rampart]

so how do you solve for dy/dx here?

[uncut topaz]

Divide both sides by 4

[crude rampart]

close

[uncut topaz]

1/4=y³?

[crude rampart]

great job 🙂

[uncut topaz]

Then i will use cube root?

[crude rampart]

nope

that’s it

oh wait

[crude rampart]

no this isn’t the right answer, sorry

[lapis mesa]

this isn't right yet

[crude rampart]

think of dy/dx as the x

[crude rampart]

yeah i misread

my bad

[crude rampart]

@uncut topaz

[uncut topaz]

Yes

I'm listening

[crude rampart]

how do u isolate dy/dx in this instance

[crude rampart]

think of ur answer to this

[uncut topaz]

Divide both sides 4y³?

[crude rampart]

yes!

🙂

$\frac{dy}{dx} = \frac{1}{4y^3}

$\frac{dy}{dx} = \frac{1}{4y^3}$

[uncut topaz]

That's what i got earlier

[soft zealot]

9k

[uncut topaz]

This right?

[crude rampart]

…

i have a really bad misreading problem today…

sorry

[uncut topaz]

It's fine

How about this

[crude rampart]

anyways yeah

u got it

[uncut topaz]

D

It's just the same right?

I will use implicit differentiation?

[crude rampart]

sorry, i have to step away for a IRL emergency, need someone to take this over

but yes, implicit differentiation

ok nvm im back

[uncut topaz]

And the e is just normal chain rule right?

[lapis mesa]

could you send part e again?

[final saddle]

@uncut topaz Has your question been resolved?

[final saddle]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

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.

[craggy atlas]

Prove that sin (x+y) / sin (x-y) = (tan x + tan y) / (tan x - tan y)

i got the answer but how am i supposed to know beforehand that i have to divide both numerator and denominator by cos x cos y

[rustic wedge]

good question

[craggy atlas]

do i receive a good answer?

[eager shore]

because tan is sin/cos

¯_(ツ)_/¯

how else would you get tans in there

[craggy atlas]

hmmm

[valid carbon]

you can just as well go from the RHS too

[rustic wedge]

(arguably an easier question since its easier to see the sin(x+y) and sin(x-y) forming )

[craggy atlas]

that would take too long

i think

[valid carbon]

not at all

[rustic wedge]

nope

[craggy atlas]

oh

thanks

.close

[final saddle]

Channel closed

Closed by @craggy atlas

Use .reopen if this was a mistake.

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.

[whole halo]

.cloose whoops, unpinned the wrong message

.close

[final saddle]

Channel closed

Closed by @whole halo

Use .reopen if this was a mistake.

[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.

[low idol]

I have $T(x)=\frac1{2x}+\sum_{k\ge1}\frac x{x^2-k^2}$ and am stumped checking my PFD for $T^2(x)$:
\begin{align*}
T^2(x)&=\frac1{4x^2}+\sum_{k\ge1}\frac1{x^2-k^2}+\sum_{k,n\ge1}\frac{x^2}{(x^2-k^2)(x^2-n^2)}\
&=\frac{-1}{4x^2}+\frac{T(x)}x+\sum_{k\ge1}\frac{x^2}{(x^2-k^2)^2}+\sum_{k,n\ge1k\neq n}\left(\frac1{x^2-n^2}\frac{-n^2}{k^2-n^2}+\frac1{x^2-k^2}\frac{k^2}{k^2-n^2}\right)\
&=\frac{-1}{4x^2}+\frac{T(x)}x-\frac x2\sum_{k\ge1}\left(\frac 1{x^2-k^2}\right)'-\sum_{n\ge1}\frac{n^2}{x^2-n^2}\sum_{k\ge1k\neq n}\frac1{k^2-n^2}-\sum_{k\ge1}\frac{k^2}{x^2-k^2}\sum_{n\ge1n\neq k}\frac1{n^2-k^2}\
&=\frac{-1}{4x^2}+\frac{T(x)}x-\frac x2\left(\frac{T(x)}x-\frac1{2x^2}\right)'-\sum_{n\ge1}\frac{n^2}{x^2-n^2}\sum_{k\ge1k\neq n}\frac1{k^2-n^2}
\end{align*}
And I've calculated the last inner sum to be $\frac3{4n^2}$, which eventually leads to $$T^2(x)=\frac{-1}2T'(x)$$. But I recognize $T$'s form and it should instead satisfy $$T'(x)=-2T^2(x)-\pi^2/2$$

[final saddle]

@low idol Has your question been resolved?

[soft zealot]

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

[final saddle]

@low idol Has your question been resolved?

[scarlet sinew]

<@&286206848099549185>

[final saddle]

@low idol Has your question been resolved?

[whole halo]

@low idol in general, you can test if it’s correct by graphing each step then looking for the first time the graphs don’t match

have you done this before?

[low idol]

No.

[whole halo]

@low idol are you going to graph each line?

[low idol]

What do you mean?

[whole halo]

presumably this would solve your issue by showing when your work first acts incorrectly

[final saddle]

@low idol Has your question been resolved?

[brittle shadow]

Far out

[brittle shadow]

What are you stuck on?

[final saddle]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[low idol]

.reopen

[final saddle]

✅

[low idol]

Trying to find out why my PFD gives me the wrong result, but I also can't format the derivation properly here

[vital crag]

#latex-help

[final saddle]

@low idol Has your question been resolved?

[low idol]

No

[edgy mauve]

Click the cross

[low idol]

The bot last closed the question when I was sleeping.

[whole halo]

@low idol you here?

[low idol]

Yes.
I think the problem is from how I evaluated (\sum_{k,n\ge1}\frac{x^2}{(x^2-k^2)(x^2-n^2)})

[soft zealot]

MCAntDJC

[whole halo]

I tried graphing some portions of it and they dont appear to be completely identical

ending the terms at n (you shouldnt use n as a dummy variable, use something like j or l instead)

I still dont see where this came from

[low idol]

It was from the PFD of (\frac{x^2}{(x^2-k^2)(x^2-n^2)})
(There's also a typo in that expression, but the complier stopped working, so I couldn't correct it.)

[soft zealot]

MCAntDJC

[whole halo]

PFD?

partial fraction decomposition?

thats not at all how you partial fraction decomposit that

this is what I got instead:

have this be the default sum including the diagonal (k1 = k2) entries

take out the diagonal entries, so that the remaining ones cover k1 not equal k2

from here thres a choice, I chose to use symmetry to have k1 < k2

factor out the x^2 as well

whats left is $\frac1{(x^2-k_1^2)(x^2-k_2^2)}$

[soft zealot]

mtt

[whole halo]

from here, just consider subtracting $\frac1{x^2-k_1^2}-\frac1{x^2-k_2^2}$

[soft zealot]

mtt

[whole halo]

this would result in $\frac{k_1^2-k_2^2}{(x^2-k_1^2)(x^2-k_2^2)}$

[soft zealot]

mtt

[whole halo]

and so without needing to do any PFD with chance of mistakes, you can just conclude that $\frac1{(x^2-k_1^2)(x^2-k_2^2)}=\frac1{k_1^2-k_2^2}\left(\frac1{x^2-k_1^2}-\frac1{x^2-k_2^2}\right)$

[soft zealot]

mtt

[whole halo]

if you dont use this shortcut around PFD, theres another way that works for this specific case

heaveside coverup method, which then produces
[
\frac1{(x^2-k_1^2)(x^2-k_2^2)}=\frac1{(x^2-k_1^2)(k_1^2-k_2^2)}+\frac1{(k_2^2-k_1^2)(x^2-k_2^2)}
]

[soft zealot]

mtt

[whole halo]

either way, you should be able to tell that the numerator must be 1

[low idol]

So I should disregard the $x^2$ from the numerator when decomposing the sum? My calculations didn't produce any contradictions or divergent series, so I hoped someone here could find a flaw in them.

[soft zealot]

MCAntDJC

[whole halo]

you mvoe the x^2 out of the numerator, then do the PFD or so, then you can move it back in

[whole halo]

so youre saying you never tested your results?

[low idol]

Test them how? I feel it's not something I would use a graph to check

[whole halo]

you never asked

@low idol to be honest, what idea are you going for with this current approach?

[low idol]

I was trying to derive an expression with $x^2-k^2$ terms in the denominator, to find a relation between $T'(x)$ and $T^2(x)$

[soft zealot]

MCAntDJC

[whole halo]

thats it?

well in that case Ive hit a roadblock trying out what you have so far

are you allowed to use any other ways, such as simplifying T(x)?

if you can, you should post a screenshot of the original problem in case we have a faster way of going around this

[low idol]

Is this clear enough? Because of its width I don't think I can format it clearly here.

[whole halo]

right thats not answering any of the questions I asked, but that does help

[whole halo]

first, can you show a screenshot or a picture of the original description of the problem that required you to do all this

@low idol you realize its been an hour right

you gotta step up these response times, I cant wait forever

[whole halo]

yes or no question: can you show the original problem

you seem hesitant to answer a question like this

I need to know the original context or content of the question so that I can make sure that the approach youre taking is the best one

its been an hour and the mistake is still there

you arent even telling me a reason for these sorts of pauses

yes or no: can you show the original problem, so that I can read it too and so verify your approach

[low idol]

It's less a problem than a neat derivation gone awry when I thought using $T(x)$ would make it more elegant:

[soft zealot]

MCAntDJC

[whole halo]

neat derivation of what

is it a neat derivation of $\sum_{k=1}^\infty\frac1{x^2-k^2}$?

[soft zealot]

mtt

[whole halo]

yes or no

[low idol]

I thought it was a neat derivation of $4T^2(x)=-2T(x)-6\zeta(2)$, but I just remembered that it took up both sides of a sheet of paper

[soft zealot]

MCAntDJC

[whole halo]

thats not even on this paper, that goes way back

confirm: is $4T^2(x)=-2T(x)-6\zeta(2)$ the original derivation that you wanted to find a neat workaround?

[soft zealot]

mtt

[whole halo]

yes or no

[low idol]

It's the derivation that I eventually got, but I originally used $S(x)=\sum_{k\ge1}\frac1{x^2-k^2}$, and with it, I built $T(x)$ and hoped that using $T$ would lead to a sleeker derivation; but instead I got the mess at the beginning of my post.

[soft zealot]

MCAntDJC

[whole halo]

youre not answering the question I am asking

the question, again, is this:

all of these derivations are based on some fundamental problem

this problem is where you got all of this from

are you doing this for a school assignment? self-studying an alternate way to find a known result? the origin of this problem youve never said

youre just saying your progress and your steps already deep inside of the problem

you can see other people on here begin with the homework assignment theyre stuck on
with how loose the work currently is, it seems to me that youre trying to just explore around a particular result or goal

please tell me from what result, goal, or similar all of this is originating from

[low idol]

It's self-studying my own derivation of an established result

[whole halo]

and the established result is?

[low idol]

The differential equation for $T(x)$ (and, as a bonus, discovering $\zeta(2)$)