help-36

[whole halo]

hold on a sec

I have to pull up the context

here you go

[ruby crypt]

wow

[whole halo]

I just searched up "laser mirror" on google images, it might not even be a mirror for lasers

doesnt look like one either with that purple shade

but no one ever questions it

[ruby crypt]

yes same argument as "no one ever calls me vins here"

alr tysm bro

[whole halo]

yep

[ruby crypt]

u da real mirror

.solved

[whole halo]

dont get sniped

[final saddle]

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[wheat grove]

I have difficulty understanding something about conditional probability

[steep nest]

Please tell!

[wheat grove]

I understand that conditional probability is the probability of an event A happening that B is happened

[radiant igloo]

damn a lot of doubt channels so fast

[wheat grove]

this can be written as number of ways A and B both happen divided by number of ways B happen

[steep nest]

Yes that's correct

[wheat grove]

and the conditional probability definition is dividing the probabilities of these

and we reach the same thing after cancelling the sample space in numerator and denominator

but what does it mean to divide probabilities?

[steep nest]

to normalize the probability space back such that it equals 1

[wheat grove]

yeah I heard normalisation but I have no clue what it means

[steep nest]

Well take a venn diagram of A and B

[wheat grove]

Alright

[steep nest]

Now originally we have the entire space

A union B

but now after conditioning on B, we have only the circle of B

That's what given B means

[wheat grove]

alright

[steep nest]

and now the part of B where A happens (the probability)

is the small slit on the left side of B

which is P(A intersect B)

and now to have the case were we are only in the circle of B

we devide the probability by B

[wheat grove]

I understand

[steep nest]

P(B) is our total possible outcomes

and P(A intersect B) is our favourable outcomes

[wheat grove]

but why are we dividing probabilites?

it makes sense if the ven was number of outcomes

[steep nest]

P(A) = outcomes of A / all outcomes

[wheat grove]

but why probability? Isnt the definition of probability number of favourable outcomes by number of outcomes?

[steep nest]

Well we know B has happend

so our new number of outcomes is P(B)

[wheat grove]

isnt the new number of outcomes number of ways B can happen

and not probability of B?

[steep nest]

well

(number of ways A intersect B / total number of ways ) / (number of ways B / total number of ways)
total number of ways cancel out

So you're correct

[wheat grove]

yup I get that

I cleared that at the start

[steep nest]

And this is just P(A Intersect B) / P(B)

[wheat grove]

here

but what does dividing probabilities mean?

[vestal kestrel]

The best way to visualise this is looking at it in a tree diagram

[wheat grove]

how?

[chilly granite]

-# this might be very usefull https://www.khanacademy.org/math/ap-statistics/probability-ap/stats-conditional-probability/a/tree-diagrams-conditional-probability

[wheat grove]

ok let me check real quick

I really dont see what dividng probabilities has to do with tree diagrams :c

can someone explain

[vestal kestrel]

The respective branches shows the conditional probability

Take this as a simpler example, a coin toss

[wheat grove]

ok

[vestal kestrel]

The first toss will give a head or tail and so does the second toss

[wheat grove]

yeah

[vestal kestrel]

But note that the sum of the branches between the first event that happened and the events of the 2nd toss equate to 1

For example, after the 1st head, I have a probability of 0.5 to get another head or a probability of 0.5 to get a tail this time

Now if we want to find the probability of 2 heads, its just 0.5 times 0.5, straightforward enough

[wheat grove]

yeah but what does that have to do with dividing probabilities

[vestal kestrel]

Dividing probabilities is equivalent to removing the first set of branches

[wheat grove]

huh how

🙁

[vestal kestrel]

Example, lets say the 2 heads, the probability is 0.25 because 0.5 times 0.5

[wheat grove]

but why are you taking this example

this isnt even conditional probability

[vestal kestrel]

To do this, first we need heads which is 0.5, then another heads, which is a 0.5 on top of that

[chilly granite]

-# yes but wait from here we get conditonal probablity (its realted)

[wheat grove]

yup got that

[vestal kestrel]

That 2nd part, the 0.5 on top, is the conditional probability

And how we achieve that is we take the pathway (the 1st branch and 2nd branch), then we divide the first 0.5 (which removes the first branch), this leaves us with the 2nd branch, which is the conditional probability

[wheat grove]

hmmm

wait how does dividing by 0.5 remove the 1st tails branch?

I dont get what division does here?

how?

[chilly granite]

okay bassicly think of it this way ur essentiall reversing the proceess the probiblty you get 2 heads in a row is 1/2 * 1/2 so what is the probablity that you get heads the second GIVEN that u already gotten heads so by doing (1/2 * 1/2 ) / 1/2 YOU GET JUST 1/2

-# sorrry for interupting btw

[wheat grove]

this does make sense

[chilly granite]

oaky so one last thing- (als i am so sorry @vestal kestrel for interuptng you) but you know how to get the probablity of 2 events you multiply them?

[vestal kestrel]

Its fine

[chilly granite]

so wehn u want to remove one event as tzken said earlier you divide by it

[chilly granite]

-# okay sorry i will head out now please continue!

[wheat grove]

yes indeed

[chilly granite]

hence

[chilly granite]

this^

[wheat grove]

yeah i think ive understood

thanks alot both of you

.close

[final saddle]

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[chilly granite]

-# also really good exapmle tzkken mind if i steal this for later use?

[final saddle]

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[lime crest]

Find all functions $f$ from the naturals to itself such that [ f^m(n) +f(mn) = f(m)f(n) ] where $f^m = f\circ f \circ... f$

[soft zealot]

Copter

[lime crest]

i have that f(1) = 2 and f(m) = f^m(1), so every number can be expressed as some iteration of f(1)

idk what else tho

[ionic venture]

Did you finished the last problem?

[ionic venture]

Have you tried guessing the answer?

Normally we'd just try to approach it.

[honest gust]

a constant function could probably work here

[scarlet sequoia]

and what about m = 1, n any integer?

[lime crest]

f(1) = 2?

[lime crest]

yea, f(m) = 2 for all m

[scarlet sequoia]

oh yeah you got that

[lime crest]

my thoughts rn are like

case where f injective and not

if f isnt injective then f^a (1) = f^(b) (1) for a>b

then $f^{a-b}(f^b(1)) = f^b(1)$

[soft zealot]

Copter

[lime crest]

no idea ;-;

[ionic venture]

Does your natural numbers include 0

[lime crest]

no

[ionic venture]

cool

I think we can induct this

Alright I got something

(m, n) and (n, m) we get

$f^n(m) = f^m(n)$

[soft zealot]

Erebus

[ionic venture]

Not with your original idea of injectivity but something to keep in mind

[lime crest]

yea i have that too

[final saddle]

@lime crest Has your question been resolved?

[loud sundial]

.pin

[loud sundial]

If you need another pointer, ||try to show that the whole image of f is bounded|| for the non-injective case.

[final saddle]

@lime crest Has your question been resolved?

[compact dove]

There are two easy solutions:
f(n)=2, and f(n)=n+1

[lime crest]

let M be the maximum value of f, then f(a) = M for some a, and f(n) ≤ M for all nto get M^2 ≤ 2M ⟺ M ≤ 2

if there is some a with f(a) = 1 thrn f^a(a) + f(a^2) = 1 impossible for two naturals to sum to 1

hence in this case f(n) = 2 for all n

[loud sundial]

yeah the non injective case just gives the constant solution 2

you could also show that f(1)=f(2)=2 and proceed inductively

good

ok just do the injective case now

[final saddle]

@lime crest Has your question been resolved?

[final saddle]

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[urban raptor]

<@&268886789983436800>

[gaunt bison]

remove this user's kneecaps

[final saddle]

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[tulip coyote]

Did I get sniped

[vital crag]

yes i murdered you with a headshot. please proceed to heaven ☁️

[tulip coyote]

I've been packwatched, vengeance for all the souls of the MrBeast

[final saddle]

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[floral moth]

prove that $n x(1-x)^n$ tends to 0 as n increase to infinity, for $x\in (0,1)$

[soft zealot]

krvelty

[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

[floral moth]

okay so i searched bernoulli inequality and i think i can use this one to prove $(1+x)^r \leq e^{rx}$

[soft zealot]

krvelty

[floral moth]

and then i show that $\frac{n}{e^{nx}}$ tends to 0

[soft zealot]

krvelty

[floral moth]

ill brb

[late gazelle]

Since x is between (0,1), whatever we multiply it by (provided it’s greater than 1) will make our number smaller. And since n tends to infinity nx out the front will be positive and greater than 1. And for (1-x)^n, a number less than 1 multiplied by itself will continue to becomes less. So it should tend to 0 as n goes to infinity. Just by looking at the interval x can be on. [ofc this is not rigorous, but should be a good stepping stone]

[floral moth]

yeah i know this

im stuck proving it with delta and epsilon

and i cannot use l'hopital i think

yeah

[glossy zephyr]

btw you can argue that the lone term "x" is completely irrelevant to the limit

[floral moth]

also i think i can prove that the function is monotonous decrescent and limited inferior by 0

so it gonna converge to 0

?

[glossy zephyr]

Try evaluating the logarithm of the limit

Like, if you read it as f(n) -> 0
ln (f(n)) -> -infty

[floral moth]

maybe then i should show that 0 is the infimum

[floral moth]

yeah this can work too i think since the function is >0

[glossy zephyr]

which we can do cause f(n) is strictly positive

[glossy zephyr]

0 to be precise, yea

[final saddle]

@floral moth Has your question been resolved?

[glossy zephyr]

Kind of a spoiler: ||The proof eventually can be reduced to showing that 1/N is less than any positive real number for a suficciently big N, or, equivalently, that 1/N tends to 0.||

[floral moth]

.close

[final saddle]

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[stable viper]

How is this wrong

[willow cliff]

plug in x =3 and see what happens

[stable viper]

0 > 1

oh

hm

idk how to solve this otherwise tho

[willow cliff]

oh i see ur mistake

when you do the $x<0$ case, it becomes -(x-4)

[soft zealot]

TestTickler

[willow cliff]

then you proceed as usual and find the solutions (because you should be finding the roots of $x^2 - 7x+11$ in the end

[soft zealot]

TestTickler

[stable viper]

hm wait

[willow cliff]

but yes your first case is right, there should be no solutions if x > 4

[stable viper]

i thought we check if x - 4 < 0, so thne "(x-4)" is negative so then when we multiply to the other side the ineq sign flips?

[willow cliff]

yes $-(x-4) = 4-x$

[soft zealot]

TestTickler

[stable viper]

aaok so u get (3-x)(4-x) < 1?

[willow cliff]

yes

then solve as you did

i have ans if u needa check

[willow cliff]

hmm not quite

lets rewind a bit actually

the fact that $x>4$ makes it falls is true

[soft zealot]

TestTickler

[stable viper]

is this meant to havea negative -(3-x)(4-x)<1

[willow cliff]

no no thta part is right

and the roots you got are right

[stable viper]

oh cz i put the negative in

hm

[willow cliff]

thats right

the roots you got are right

just the interval you gave isnt

but back to what i was saying, $x=3$ makes it a false statement right. We saw that it gives 0>1, clearly false

[soft zealot]

TestTickler

[willow cliff]

if we plug in $x=3.1$, we get a false statement once again

[soft zealot]

TestTickler

[willow cliff]

x=3.01, same thing, and so son

so on*

now what happens if we plug in $x=2.9$?

[soft zealot]

TestTickler

[willow cliff]

do we get a true statement?

[stable viper]

im so confused

so we just testing for values around x = 3 from intuition

[willow cliff]

to begin with yes

because one of those roots you got doesnt work

[stable viper]

uh okay

[willow cliff]

just to give a quick summary:

• we found that right off the bat, x-4 > 0 ~ x>4 gives no solutions, so we look at x<4
• moreover, x>3 gives false statements, so then we look at solutions such that x < 3
• finally, solve the quadratic you got from the x-4 < 0 case and figure out which root satisfies the constraints we have (if any)

[final saddle]

@stable viper Has your question been resolved?

[stable viper]

.close

[final saddle]

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[spark agate]

Question 13 part b

[spark agate]

I’m confused cuz ik i can use

[runic needle]

,rotate

[spark agate]

But idk how to do b and my prof does it a diff way

[runic needle]

this is for an arithmetic progression

[spark agate]

Is there a formula i could use for b then?

[runic needle]

do you know what geometric progression is?

[spark agate]

I do not- it didn’t habe that in our videos

[runic needle]

what is ap?

[runic needle]

yes look it up

[spark agate]

Okok

[runic needle]

!done

[final saddle]

If you are done with this channel, please mark your problem as solved by typing .close

@spark agate Has your question been resolved?

[spark agate]

.close

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[stable viper]

I think I diffed the arctan weong by comparing to solns, but I’m not sure why it is wrong.

[stable viper]

Can someone check pls? Thanks

[final tangle]

you didn't applly chain rule when differentiating arctan(x/sqrt(3))

need to multiply by derivative of inner, i.e. derivative of x/sqrt(3)

[final saddle]

@stable viper Has your question been resolved?

[stable viper]

ohh oh my i keep forgetting that. thanks

.close

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[stable viper]

Where did I go wrong? It looks all good to me

[ornate knot]

not allowed to u sub? D:

cuz im pretty sure u = arc sin(x/3) would work

[stable viper]

didnt i do that

[ornate knot]

i didnt read properly , ggs

[ornate knot]

3sinu = x
dx = 3cos u

subbing back in

$$\int u \cdot 3\cos u \dd u$$

[soft zealot]

JustToPro

[stable viper]

what

[ornate knot]

and thats way simpler than what u made

yeah cuz u didnt use u sub , now that i read properly

[stable viper]

i mean the question says integration by parts

[ornate knot]

u just went straight to integration by parts

[ornate knot]

u cosu is solved using integration by parts tho

thats just a quality of life change i was describing

,w int^\frac{3\sqrt3}{2} _ \frac{3}{2} of 4 arcsin (\frac{x}{3})

[soft zealot]

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Requested by justtopro

[stable viper]

oh nvm i see my mistake

[proud cloud]

yeah that would be a much faster solution

[stable viper]

i dont think im allowed to do this

[proud cloud]

it technically is still using IBP

since you'd need IBP to solve ucosu

[brazen breach]

,w integral x/sqrt(9-x^2) dx

[soft zealot]

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Requested by nautilus2143

[brazen breach]

yeah you have a 1/2 there for some reason when it should have an /(1/2)

[brazen breach]

ping

[stable viper]

tyty

.close

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[floral moth]

i need to find the derivative of $(b - t a) (1 - \frac{1}{t} (1 + \sqrt{t^2 + 1}))$ for a,b positive and t negative

[soft zealot]

krvelty

[floral moth]

in respect to t

[onyx peak]

are you supposed to do it by hand? If not, just plug it into WA. If yes, I'd start by expanding it (probably a bit easier than applying product rule directly)

[bold turtle]

Well 1, presumably

[floral moth]

no, i tried to use a calculator but it gave me the incorrect answer

[bold turtle]

2, what have you trie-

[floral moth]

,ask simplify derivative (b-t*a)(1- 1/t ( 1+sqrt(t^2+1))),t

[bold turtle]

!show

[final saddle]

Show your work, and if possible, explain where you are stuck.

[floral moth]

its giving me this but

[onyx peak]

,w d/dt (b-t*a)(1- 1/t ( 1+sqrt(t^2+1)))

[soft zealot]

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Requested by methisalwaysright

[bold turtle]

No that's called "I couldn't use the calculator correctly" hate to break it to ya

[onyx peak]

this should be the right result

[floral moth]

😮

[bold turtle]

I mean it should have been a giveaway that it interpreted your ",t" as a misspelling of "/t", given this response

[floral moth]

nah

[bold turtle]

wait tbh that is slightly different

But the result should have been, yk, a differentiation

Not "I'm being asked to differentiate..."

[onyx peak]

,ask simplify derivative (b-t*a)(1- 1/t ( 1+sqrt(t^2+1))) wrt t

[soft zealot]

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[onyx peak]

maybe use wrt instead of ,t

or just yk, d/dt

also, in my experience the webpage interface is a bit more user friendly than the discord one

[floral moth]

,ask nsolve a(sqrt(t^2+1)-t)t^2 +b(-sqrt(t^2+1)-1) =0, t

[bold turtle]

bruh now you're asking a different question

[onyx peak]

,w d/dt (b-t*a)(1- 1/t ( 1+sqrt(t^2+1))) = 0

[bold turtle]

Just for clarification:

!xy

[final saddle]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[soft zealot]

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[floral moth]

look at this fucking graph bro

how can this calculater be so shitty

[onyx peak]

uh, your ab arent even negative over there

[bold turtle]

You have something with 3 variables tf are you expecting

[onyx peak]

if i understand it correctly

[floral moth]

a, b are positive constants

what varies is t

[onyx peak]

oh right, i thought you said negative

[bold turtle]

@floral moth

[onyx peak]

this one should hopefully wokr (assuming youre trying to find points with 0 derivative)

[floral moth]

okay wait

[onyx peak]

and fyi, this was my query

[floral moth]

isnt correct yet

strange

if b < a then it works

but if b > a it doesnt

yeah should be something wrong i did in the formulation

[onyx peak]

yeah, just noticed the same thin

my guess is that its bc of the square roots

here is the full thing WA gives

for a,b>0

[floral moth]

oh damn

[onyx peak]

apparently the last one works

weird tho

[floral moth]

yeah

thx

.close

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[onyx peak]

$t=\frac{b+\sqrt{2}\sqrt{\frac{b}{a}}\left(b-a\right)}{a-2b}$

[soft zealot]

MathIsAlwaysRight

[onyx peak]

@floral moth i managed to simplify it to a single expression that works for both

it was indeed caused by the square roots (the issue is that sqrt(a^2) is |a| and not a, so that causes a lot of possible sign errors)

<@&268886789983436800> (mrbeast)

was a spam

okay next one lol

$x=\frac{ab+\sqrt{2}\sqrt{ab}\left(b-a\right)}{a^{2}-2ab}$

[soft zealot]

MathIsAlwaysRight

[onyx peak]

possibly a bit nicer

[sand barn]

What's going on?

[onyx peak]

the goal was to find the point where the derivative of this (wrt t) is 0, but its solved now

[sand barn]

Ok

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[floral moth]

tysm!

[floral moth]

.close

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[timber leaf]

I have a question, is there exist a function such that continuous and bounded on $[0, \infty)$ but do not uniformly continuous on $[0, \infty)$? And how do I show that?

[soft zealot]

Nefer

[onyx peak]

well, to be uniformly continuous, you'd need |f(x) - f(y)| < epsilon for |x - y| < delta

try thinking of some functions which oscillate really fast (that'd make massive difference |f(x) - f(y)| for tiny |x-y|)

[timber leaf]

Talk about oscillating function i only think of sine and cosine

[onyx peak]

thats enough, you also know how to compose functions and stuff

[timber leaf]

but very fast meaning their period must be very small?

Yeah h o f or f o h

[onyx peak]

you'd want them to oscillate "infinitely" fast

it doesnt need to be just sin(constant * x)

it can be sin(f(x))

[timber leaf]

do i multiply a small constant in x?

Like sin(100x)?

[onyx peak]

that'd still be uniformly continuous

[stone wagon]

i mean $\sqrt{x}$ is not uniformly continuous on $[0,1]$ so take for example $f(x)=\min{\sqrt{x},1}$

or am i wrong

[onyx peak]

i think sqrtx is uniformly continuous on [0, 1]

[timber leaf]

Isnt sqrt x uniformly continuous?

[stone wagon]

at 0 too?

[onyx peak]

yeah

[timber leaf]

I think so

Should I go with sin(100x)?

[stone wagon]

oh i confused it with something else

sorry

[onyx peak]

|sqrt(x) - sqrt(y)| <= sqrt(|x-y|)

[stone wagon]

yes mathisalwaysright's idea is better

[onyx peak]

that is still uniformly continuous unfortunately

[timber leaf]

But isnt it oscillate very fast already?

[onyx peak]

it doesnt oscillate "infinitely" fast, its just sin squished horizontally by a factor of 100

[timber leaf]

How about something like sin(x^3)?

[onyx peak]

thats much better

[timber leaf]

Thats confusing

[stone wagon]

https://tenor.com/view/car-fast-car-speeding-driving-fast-driving-gif-6194660740054521025

[timber leaf]

I dont see much difference between them tbh

[onyx peak]

this is sin(100x)

[timber leaf]

sin(x^3)

It doesnt look much difference'

[stone wagon]

the argument of sin should probably be faster than linear

[onyx peak]

if you can bound |x-y| here, you can bound |f(x) - f(y)|

hence its u.c.

[stone wagon]

this is the answer

or sin(x^2) also works

[timber leaf]

But how do I prove it?

Suppose i found it but I also need to prove it by the definition also

[onyx peak]

here, due to the very fast oscillation, if you try to bound |f(x) - f(y)| by some epsilon, you can always go far enough to the right where the oscillations are really, really fast that they break your bound

[timber leaf]

Hmmmm

I can't understand a single word 😂

[onyx peak]

hm are you familiar with sequential defn of u.c?

[timber leaf]

not yet

[timber leaf]

I am only introduced to the early definition of u.c like this

[onyx peak]

that makes it a bit worse to work with

but well, for sake of contradiction, suppose it was u.c.

then for every epsilon > 0, there is a delta such that whenever |x-y| < delta, we have |sin(x^2) - sin(y^2)| < epsilon

[timber leaf]

Yes that is true

[onyx peak]

so in order to contradict this, we'd have to find epsilon, s.t. no matter what delta we choose, we will always be able to find x, y with |x-y| < delta, such that |sin(x^2 - sin(y^2)| > epsilon

looking at the graph, we can pick arbitrarily close x, y (so |x-y| arbitrarily close), such that |sin(x^2) - sin(y^2)| = 2 (so one is -1, the other one is 1)

[timber leaf]

So to unprove something, we find epsilon,

To prove something we find delta right

[onyx peak]

yeah, you just pick some epsilon and try to prove that there is no delta for it

[onyx peak]

nope, you have to disprove the existence of delta

[timber leaf]

Really?

Normally when i prove for continuity or differentiability i dont have to do that

[onyx peak]

U.C: forall eps, exists delta
Not U.C: exists eps, no delta

[onyx peak]

its because you disprove it

[onyx peak]

oh wait you said prove

yeah

you were right, sorry, i didnt read the "prove"

[timber leaf]

💀

[onyx peak]

anyway, do you get the intuitive idea for why its not UC? The idea is that you can always find two arbitrarily close x, y, for which f(x) and f(y) arent close. (more precisely, |f(x) - f(y)| = 2

that intuitive idea is gonna be very helpful in the proof

[timber leaf]

Yes I see it now

Will try it now

Thanks!

.close

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[onyx peak]

np

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[gentle zephyr]

how do i parametrize this surface

i am assuming cylindrical

(x,y,z) = (cos(t), sin(t), z)

[blissful meadow]

Since you know that $x^2 + y^2 \le 9$, you can pick $x= r\cos(t)$ and $y=r\sin(t)$, yes. What would $z$ be in this case given the constraint?

[soft zealot]

Azyrashacorki

[gentle zephyr]

@blissful meadow

[gentle zephyr]

wdym?

[blissful meadow]

Well you've done pretty much everything right up until you wrote down your P.

You're parametrizing a surface, so you should have 2 variables, not 3

[gentle zephyr]

what

[blissful meadow]

Like what you wrote with z = r^2(cos^2(t) - sin^2(t)) is right.

That's your third coordinate

It depends on r and theta.

[gentle zephyr]

ok

[gentle zephyr]

why don't we depend on z

[blissful meadow]

You already have z in terms of the other two variables.

Once you fix r and theta, z is determined.

[gentle zephyr]

i see

[blissful meadow]

So then your derivatives are wrong in the third coordinates, since you do have r and theta dependence in the third coordinates.

[gentle zephyr]

uff

[gentle zephyr]

@blissful meadow

[blissful meadow]

Might be unfun to compute the cross product's magnitude but yes.

[gentle zephyr]

no way

is there simpler route

[blissful meadow]

You can also parametrize as (x,y, x^2-y^2) if you think it's too complicated.

The bounds will be annoying, but once you've set up the integral you can always go back to polar.

[gentle zephyr]

what

[blissful meadow]

When your surface is the graph of a function, like z = f(x,y), you can always choose (x,y, f(x,y)) as a parametrization.

[gentle zephyr]

yeah

why did i used cylindrical then

[gentle zephyr]

it's not Minor

[blissful meadow]

Wdym? If you do use the (x,y,f(x,y)) parametrization then the cross product is easier to compute, so your integrand will be in terms of x and y, and then you can convert to polar inside your integral just like you can usually do for an integral.

[gentle zephyr]

is it legal

[blissful meadow]

Why not? You want the surface area, so you just find the suitable integrand. What you do after that is just computing a double integral as per usual.

[gentle zephyr]

@blissful meadow

[blissful meadow]

You didn't compute the whole cross product. Remember the cross product outputs a vector. What you put in the integral is the magnitude of that vector.

[gentle zephyr]

what

[blissful meadow]

What part is not clear?

You're meant to compute the "determinant" of the matrix you have written down.

You computed like one of the minors.

[gentle zephyr]

ah im so stupid

[blissful meadow]

-# Also you can still use the other parametrization if you prefer. It may be a good extra exercise to try it after this. You just need to use some trig identities to make it more manageable but it's far from being impossible (and not that long).

[gentle zephyr]

i don't like trig identities

[drowsy epoch]

Would u rather apply the chain rule on sin²(x)+cos²(x) if you had to differentiate it

[blissful meadow]

k is a vector, the coefficient is 1

[gentle zephyr]

i mean the basic trig ids is fine, but once we mess around with double angle and weird things it gets bleak

[gentle zephyr]

care to elaborate

k is a versor i think

[blissful meadow]

The point is your product results in (-2s, 2t, 1), not (-2s, 2t, k)

Like k is not part of the coordinates.

It's a unit vector

[gentle zephyr]

sure

[gentle zephyr]

now what

[blissful meadow]

Now go polar.

[gentle zephyr]

uff

[blissful meadow]

r is already positive, so no need to go from -3 to 3

Also don't forget the Jacobian factor that should pop out when you move to polar coordinates in an integral

[gentle zephyr]

oh shit

[blissful meadow]

Do you remember what it is?

[gentle zephyr]

of course

[gentle zephyr]

the problem is this nasty integral

might need to use hyperbolic shit

@blissful meadow @drowsy epoch

[blissful meadow]

It's not nasty at all.

[gentle zephyr]

hyperbolic sir

[blissful meadow]

You have an r trailing outside. What's the derivative of 4r^2 + 1?

[gentle zephyr]

8r

[gentle zephyr]

but i want the integral of the sqrt of that

[blissful meadow]

What I just pointed out is that you can just use a simple u-sub.

[gentle zephyr]

i don't get it

[blissful meadow]

What don't you get?

What u-sub do you think I'm nudging towards by asking you what the derivative of 4t^2 + 1 is?

[gentle zephyr]

dunno

if u = 4x^2 + 1

[severe canyon]

r, not x

But of course, that one!

[gentle zephyr]

if u = 4r^2 + 1
du = 8r dr
dr = du / 8r

[gentle zephyr]

@severe canyon @blissful meadow

[blissful meadow]

When you evaluate at 0 it’s just just 0 (after your u-sub)

But you’re pretty much there

[gentle zephyr]

@blissful meadow

[blissful meadow]

Yes

[gentle zephyr]

we good?

i still need to prove that the surface is smooth

@blissful meadow

[gentle zephyr]

my surface área should be this

in units^2

[blissful meadow]

How do you defined smooth surface? I suppose by the existence of a parametrization where the cross product you computed exists and is nonzero everywhere?

[gentle zephyr]

the parametrization

if S is smooth then it admits a regular parametrization

[gentle zephyr]

the parametrization needs to be regular

@blissful meadow

[blissful meadow]

Okay sure and what does that mean? How do you check it?

[gentle zephyr]

mmm

the parametrization needs to be c1

and injective

[gentle zephyr]

yeah the cross prod of the pdvs should be nonzero

that's the 3 requirements iirc

@blissful meadow

[blissful meadow]

So can you check those?

[gentle zephyr]

which parametrization

r(s,t) = (s,t, s^2-t^2)

[blissful meadow]

Idk the one you used?

[gentle zephyr]

this is c1

[gentle zephyr]

we also use cylindrical after cross product

remember when we multiplied by the jacobian

[blissful meadow]

That's a change of variable happening inside the integral. It didn't change the parametrization you had.

[gentle zephyr]

so what about injectivity

is it possible to prove it

[blissful meadow]

WHat does injectivity mean?

[gentle zephyr]

well

f(x) = f(y) <=> x = y

[blissful meadow]

Well more specifically f(x) = f(y) => x=y.

The other direction holds for any function anyway

So what does it mean applied to your parametrization

[gentle zephyr]

mmm

unsure

@blissful meadow

any ideas?

[blissful meadow]

You want to start from P(s,t) = P(s',t') and conclude that s=s' and t=t'.

[gentle zephyr]

well any idea how

[severe canyon]

...

[blissful meadow]

What is P(s,t)

What is P(s',t')

What does it mean for P(s,t) to be equal to P(s',t')

I shouldn't have to spell it out for you if you take 2 minutes to try/.

[blissful meadow]

Good

[gentle zephyr]

i mean

(x,y) = (s,t)

that should be enough

p(x,y) = p(s,t) => (x,y) = (s,t)

[gentle zephyr]

this is for R, -> R in R2 -> R3 it generalizes?

[blissful meadow]

Yes, it generalizes to any function R^m -> R^n.

$\vb{F}(\vb{x}) = \vb{F}(\vb{y}) \implies \vb{x} = \vb{y}$, where $\vb{F} : \R^m \to \R^n$, $\vb{x},\vb{y} \in \R^m$.

[soft zealot]

Azyrashacorki

[gentle zephyr]

alright so we good?

[blissful meadow]

Yes.

[gentle zephyr]

alright

and the cross product is never the zero vector in r3 because

the third component function is 1

im pretty sure @blissful meadow

[blissful meadow]

Yes

[gentle zephyr]

i appreciate your help

corki, thanks

thanks mate

.solved

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[crimson zenith]

<@&268886789983436800>

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[worldly spruce]

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[sour fiber]

Okay so

[sour fiber]

Basically

Oh dammit I have to translate

[slow coral]

tell the question first right off the bat

[supple jolt]

send the original one

in the original language

[sour fiber]

I know the answer now cause I got it wrong first but I don’t get why that’s the answer

The answer should be (120-2x)/2=a

Oh I cut out the question it asks write the area A for the function of x

[harsh stratus]

yes

if you use up x for the bottom portion of the rectangle

and another x for the top portion

[supple jolt]

and the perimeter is 120 m

[sour fiber]

The answer is basically saying the numbers that’s supposed to be where X is not but I don’t get ittt

[sour fiber]

How is that the area

Doesn’t that just say the unknown side

[harsh stratus]

how would you calculate the area of

[harsh stratus]

this rectangle

[sour fiber]

Öike

Like

X times 120-x/2

[harsh stratus]

yes

well actually 120 - 2x/2

[sour fiber]

Oh oh okay

[harsh stratus]

great there's the area

and sometimes in similar questions

they'll ask you what value x grants the largest area

[sour fiber]

Oooooh

[harsh stratus]

you can either graph the area and look for it that way

or set the derivative = 0

[sour fiber]

Oooh okay ty ty i understand now

[harsh stratus]

np 👍

[sour fiber]

Oh

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[ionic nest]

can someone help me ques 2.9

[ionic nest]

i notice that the groth of n^2 >> ln n so this is convegent

is there any approach to solve the problem that the numerator or the denominator growth much more faster the other

[winter grail]

You need to turn “(n^2) grows faster than (\ln n)” into an actual comparison

[clear moon]

it's generally not enough to compare growth rates, technically if you consider sqrt(n) / n, the denominator grows faster than the numerator but this series would still diverge

[soft zealot]

Dex

[rocky tusk]

ln(n) < sqrt(n)

[winter grail]

That idea is basically comparison test, but you need to compare it to a known series

[winter grail]

Turn the growth idea into an inequality or a limit comparison with a known series

[ionic nest]

yeah im looking for known series now

[clear moon]

integral test also works here btw

[final saddle]

@ionic nest Has your question been resolved?

[ionic nest]

@clear moon is this right

i stuck with -ln (n) / n so u use Lhopital

then the answer is a number which is convergent

[ionic nest]

im sorry but i cant find any known series

[ionic nest]

oh yeah thank u

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[severe cove]

i remember calculating determinants in highschool but this is so confusing for some reason

[severe cove]

it says that the determinant for row 1 would be a11C11 + a12C12 but then for the example, you'd do for both rows right?

idk what the minor is supposed to be bc it says the minor, Mij , associated with the element aij is the determinant of the (n − 1) × (n − 1) submatrix obtained from A by deleting row i and column j.

wait ok so i was trying to use the n > 2 formula, the cofactor expansion, but maybe this doesn't work for n = 2?

[clear moon]

it doesn't, because n = 2 is considered like the base case

the whole point of cofactor expansion is to reduce computing an nxn determinant to computing a bunch of (n-1)x(n-1) determinants

[severe cove]

ah okay

[clear moon]

for 2x2's you just memorize ad - bc

[severe cove]

okay cool

but for, lets say a 3x3 matrix, how am i supposed to know what the minor is?

[clear moon]

there are 9 possible minors for a 3x3, each of them comes from considering the 2x2 matrix you have left if you deleted one particular row and column from the original matrix

there isn't just like one minor

[winter grail]

You don't do both rows. For cofactor expansion, you choose one row or one column

[severe cove]

so do any of them work in calculating the determinant because C is found with the use of a minor

[winter grail]

So the (n>2) formula is mostly shown because that's where cofactors become more useful. For (2\times2), it still works, but it just turns into the shortcut (ad-bc)

[soft zealot]

Dex

[abstract bramble]

1 is the base case

it does work for n=2

[clear moon]

sure yes but practically speaking nobody cofactor expands a 2x2

[abstract bramble]

[clear moon]

i would hope...

[severe cove]

sounds like something a test would ask though

[abstract bramble]

1 2 3
4 5 6
7 8 9
the signs always alternate starting with positive
+1 -2 +3
then you multiply each with the minor of that cell
and ur done

for an nxn matrix you will need the n minors for each cell in the first row

[severe cove]

but what is the minor of the cell?

[winter grail]

cofactors work for calculating determinants, and minors are the pieces used to build the cofactors

[clear moon]

btw to cofactor expand a 3x3, you just pick a row or column to expand along (let's say i pick the first row). then i could move left to right along that row, remove that particular column i'm in, compute the determinant of the leftover 2x2 matrix after i delete my current row/column, and just account for whether i tag on a negative sign based on the (-1)^(i+j) term. if you repeat this for the three entries along the first row and add up your results, you get your determinant for the 3x3

[abstract bramble]

Read this?

the second sentence

[winter grail]

Minors and cofactors are both part of the determinant method

[severe cove]

for example, this one

[abstract bramble]

which row or column do you want to use

[severe cove]

so say for the number 5 in the first slot here, the minor would be -3 & 2 // 5 & 3?

because deleting row 1 and column 1?

[abstract bramble]

yes

[severe cove]

ah okay

[severe cove]

so you only calculate this for one row or column?

so like the first row and said minors

[clear moon]

yes

[severe cove]

ah okay

[winter grail]

Watch the sign on the (-2).

For the top-left (5), you delete row 1 and column 1

[soft zealot]

Dex

[severe cove]

and then find the determinant for each 2z2

[clear moon]

it's often useful (especially for big matrices) to expand along a row or column that has a lot of zero entries since you don't really need to compute the determinant of the corresponding minor for a 0 entry since the a_ij would be 0 and causes the product to be 0 anyway

<@&268886789983436800>

[winter grail]

yes

[severe cove]

yay

okay cool thanks guys

[severe cove]

ah okay this is good to know

thank you everyone have a nice weekend

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[harsh veldt]

So

Domain is a number that doesnt make the square root 0

And range is not making the square root = 0

But why

For the range here it doesnt only say 0 why there is 5 aswell

Or square root of 2

[tiny gorge]

doesn't make the argument to the square root negative

0 is fine

[harsh veldt]

Sorry didnt get it

Aaaa

So

Domain is not making the square root negative

??

[tiny gorge]

domain

range is the set of possible "output" values

domain is the set of allowable "input" values

[harsh veldt]

So my question was why 5 is there aswell

In the range

[tiny gorge]

what do you get when you plug in x=0?

[harsh veldt]

5

Yeaa

Thankk uu

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[harsh veldt]

.reopen

[final saddle]

✅ Original question: #help-36 message

[harsh veldt]

So the least possible outcome is 0 and the highest is 5 ? If i am not mistaken

I just wanna get the logic sorry for asking

[tiny gorge]

yes, that is correct. i would suggest plotting the function, it makes it easy to see what's going on

[harsh veldt]

Buttt that leaves me with a thing cant we make another possible outcome that is higher or less that those?

[winter grail]

The 5 or sqrt(2) is the maximum height/output value, not another forbidden value

[harsh veldt]

This is just an example like making x a number where u put it and it becomes 600 and when u add it with the 25 u get the 625 and square root it?

Or its not possible

[harsh veldt]

MAXIMUM HEIGHTTT

okaay

Yeaa range is height why i was not thinking of that

What is the fastest way pf solving this

[honest gust]

find critical points if you know them

and find where f(x) < 0

those points are not in f's domain

you could just factor x^2 + 3x - 4 if you want

then manually check where f(x) < 0

[compact laurel]

Factoring, and then using interval tests to figure out where were less than 0. Because where the function is less than 0, you can’t plug those into the square root

[harsh veldt]

I know factors but sometimes i just get stuck at it

[honest gust]

i'd suggest you also practice factoring with this one

take a bit of time to find what multiplies to -4 and adds to 3

it's a good recap

[harsh veldt]

This one is solvable but i have another question i dont even get it

Wait

[compact laurel]

Like a whole separate problem? Or a question that pertains to this problem?

[harsh veldt]

It was a whole seperate question

But i cant find it

I have been looking for it for hours

Yesterday i said i am gonna solve it today

[compact laurel]

Do you understand how to do this question?

[harsh veldt]

I marked it and i just cant find it

[compact laurel]

Or do you wanna go over it step by step?

[harsh veldt]

Yea u equal it to 0 ( equal or less than or equal to ??)

I think

Greater than or equal to zero

Then factor it

[compact laurel]

Well, here’s the idea, what values CAN’T you plug into a square root

[harsh veldt]

I send the -4 to the other side then take x from both sides

[floral zenith]

If √(g(x)) = 0
Then
g(x) ≥ 0

[harsh veldt]

• infinity to

[compact laurel]

Exactly. Negative numbers can’t be plugged into a square root. So, if you find where your inner function is greater than or equal to 0, you will find your domain

So, can you first figure out where your inner function is equal to 0

[harsh veldt]

Can u take me through it and i see the answer if u can cuz i have 3 mins left for exam and i wanna know how do i do it

[compact laurel]

Well, if you only have 3 mins, idk what I can really do

But the idea is you factor the quadratic, and set it equal to 0

[harsh veldt]

Its okay just i wanna see how

[harsh veldt]

Wait

It was equaling to zerp

Or greater than or equal to

Zero

Thank u bye bye i have to go actually thanks for ur help again

.close

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[tropic crest]

Tfw complex numbers

[compact laurel]

Well, yeah, but do you honestly think this question was in the complex domain?

[tropic crest]

(No, but i was joking when i said that)

[compact laurel]

What we really need are quaternions. lol

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[tropic crest]

Please do not post the same question in multiple different help channels

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[ornate token]

What do you mean?

And this channel is also occupied.

[tropic crest]

!occupied

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[crude sorrel]

ill use ai

[tropic crest]

!noai

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@lyric rose Has your question been resolved?

[final saddle]

@lyric rose Has your question been resolved?

[static raft]

Hi

Oh

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[pliant elk]

what is symmetric difference supposed to mean

[pliant elk]

in context of set theory

[honest gust]

it's the two circles without their intersection

by definition $A\triangle B=(A\setminus B)\cup (B\setminus A)$

[soft zealot]

chudcel

[honest gust]

this just completely removes all elements of $A$ that are in $B$ and the converse

[soft zealot]

chudcel

[honest gust]

in short it completely removes their intersection

it can be called symmetric because unlike $\setminus$, $\cup$ is commutative

[soft zealot]

chudcel

[pliant elk]

which makes $\Delta$ also commutative?

[soft zealot]

Mr. Smith

[pliant elk]

ah okay

.close

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[unkempt prism]

Let $p: X \to S$ be the map from $x \in X$ to the connected component of $X$, where $S$ is induced with the discrete topology. I want to show this is not always continuous.

[soft zealot]

ILikeMathematics

[unkempt prism]

@quasi rapids we discussed about this already

I was thinking about something like $X = {1, 2, 3}$ with $\mathcal T = {\varnothing, X, {1}, {2, 3}}$. But the problem is that then, $S = {{1}, {2, 3}}$ and so $p^{-1}({2, 3})$ for example is just ${2, 3}$ again which is open

[soft zealot]

ILikeMathematics

[unkempt prism]

No matter what topology I pick, it seems to always give this problem, it almost feels like the statement is true

[quasi rapids]

hm, i dont know how to get you to find the "easiest" counterexample

you want a space with the following properties: 1. The connected components are singletons; 2. these singletons are not open

and you have definitely heard of this space before

[unkempt prism]

Huh, but then why is the maximal connected component of any x in X not X itself

Since X cant be written as union of disjoint open sets

Because each of these are not open

Ah

We need open disjoint sets that partition X

Is this going to be infinite?

[quasi rapids]

yea

[unkempt prism]

So we can pick X = N right

How are we going to pick our topology now..

[unkempt prism]

Im not sure how to pick a suitable topology..

we cant just make everything open except singletons

that doesnt work

we need it to be "generated by this property" but thats just going to be messy and in the end we might not know how to work with it

[quasi rapids]

well, we have the subspace topology of R, but with that N is discrete

[unkempt prism]

yes

and we dont want that

[quasi rapids]

how about a subspace of R that isnt discrete

[unkempt prism]

Q?

[quasi rapids]

and what are the connected components of Q

[unkempt prism]

lots of singletons?

[quasi rapids]

what do you mean by that

[unkempt prism]

all singletons of Q

[quasi rapids]

right

and are those open

[unkempt prism]

No

[quasi rapids]

so thats our example

[unkempt prism]

Oh. How did you come up with this?

Isnt the most natural way to think about finite cases first

Like X = {1, 2, 3}

How do you know when to stop doing that and go infinite

[quasi rapids]

well in general finite topologies are quite weird to work with

[unkempt prism]

You mean they behave nicely?

[quasi rapids]

it depends on which behaviours youre considering

for things like "separation" they are horrible, only the discrete topology on a finite space is hausdorff for example

[unkempt prism]

ah

[quasi rapids]

in fact i dont know if connected components of finite spaces are necessarily open, but i dont think so

btw this behaviour "connected components are open" is equivalent to what we call being locally connected

its a rather fundamental property that most nice spaces fulfill

[unkempt prism]

well Q with the subspace topology of R doesnt sound too weird at first either

maybe it is

[quasi rapids]

well it appears so
So you could have kept on searching for examples of finite spaces with no luck

[unkempt prism]

ah, alright

Well thanks!!

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[bright river]

I rationalize and i got

√((1+2^10)+√(1+2^10+2^20)) =√a+ √b

i tried to do some algebra like making it into 2^10=a^2

so it becomes like 1+a^2+a^4 which is multiplication of (1+a^2+a)(1+a^2-a)

but i failed somewhere

[brazen breach]

consider first I think that 32=2^5=sqrt(2^10)

at the bottom there

2^20+2^10+1=(2^10+1)^2-2^10

[bright river]

It vanished when i rationaze

[bright river]

(2^10+1)^2 = 2^20+1+2^11 no?

,w (2^10+1)^2 = 2^20+1+2^11

[soft zealot]

Results provided by WolframAlpha

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Requested by neet420here

[brazen breach]

2^11=2 * 2^10=2^10+2^10

[brazen breach]

from this, you sould have sqrt(2^10 / (1+2^10-sqrt((2^10+1)^2-2^10))
let a=2^10 for simplicity for now and then multiply the conjugate of the denominator

[brazen breach]

you should then get a singular nonfraction

<@&268886789983436800>

[olive plinth]

how it supposted to equals that

[bright river]

@brazen breach

Wait

Actually we are at the same step

Which I wrote at the start

[bright river]

.

[bright river]

Here

,w I rationalize and i got

√((1+2^10)+√(1+2^10+2^20)) =√a+ √b

[brazen breach]

?

[brazen breach]

@bright river

[bright river]

Yes i have already solved out at the start of this question

Which is same

[brazen breach]

hm

[bright river]

So what next?

[brazen breach]

consider squaring it

[bright river]

I have wrote all of my efforts at the start of this question

[brazen breach]

you'd get 1+2^10+sqrt(1+2^10+2^20)=a+b+2sqrt(ab)

[bright river]

Sure

[brazen breach]

I'm realizing now I don't think it's expressible

as that

it's square root of odd in LHS and square root of even in RHS, perhaps a or b contain a radical

[bright river]

Wait wait

I solved out

[bright river]

1+2^10+sqrt(1+2^10+2^20)=a+b+2sqrt(ab)

1+2^10+2^20 =(1+2^10+2^5)(2+2^10-2^5)

1+2^10+1+2^10

2+2^11

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[ornate token]

Yeah, looks right.

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[onyx peak]

Is there a sequence an -> 0, such that sum an = prod (1 + an)?

[soft zealot]

MathIsAlwaysRight

[tropic crest]

Im tempted to say let a>1 then all of them would diverge

[onyx peak]

surely it cant be purely positive

[onyx peak]

i want them to exist and be finite

[tropic crest]

Bruh then idk

[onyx peak]

okay i found a boring example:
2, -1/2, 0, 0, 0, ...
2 - 1/2 = 3/2 = (2+1) * (-1/2 + 1)

is there something with no zeros?

[tropic crest]

(What about closed form ones)

[onyx peak]

like geometric sums, etc.. ?

i guess i could try some of those, maybe geometric with a negative ratio

Let $a_n = r^n$

[soft zealot]

MathIsAlwaysRight

MathIsAlwaysRight

[steep hare]

what is this???