help-10

[spice citrus]

oh, true. I've only heard of cauchy, not actively used it

This is also true btw $$f\left(zf\left(x\right)\right)=xf\left(z\right)$$, after applying cauchy

[warm shale]

Jelle

[spice citrus]

and cancelling f(y)

[median dome]

Yeah I already had that

.

[spice citrus]

oh, y = 0 is indeed easier

but this equation and cauchy together have the same information as the original equation

• injectivity I guess

[median dome]

Yes

How can we prove bijectivity?

[spice citrus]

z = 1

[median dome]

Ahh yeah my bad

We have that f is a bijection as well as an involution

[spice citrus]

yeah, (that means ff(x) = x, right?)

Oh, maybe we can prove f is bounded on some interval

Or proof by contradiction

[obtuse pebble]

@median dome Has your question been resolved?

[lucid sage]

Who has the whole question

[obtuse pebble]

@median dome Has your question been resolved?

[obtuse pebble]

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[frail depot]

I feel like what I have done is wrong. This is an assignment for maximizing the likelihood function with respect to theta as you can see. I’m trying to do that by using the log likelihood function, but I don’t think I’ve done that conversion right if that makes sense, I’d love to only get hints - so I can do the assignment myself

[robust raven]

you forgot to write on your photo the precise formula of the density function of the distribution, pelase add it and then display it

ok i can see now

[tardy epoch]

,tex .log rules

[frail depot]

Hahaha was about to say

[warm shale]

riemann

[frail depot]

Ye the first one

Riemann?

[tardy epoch]

You didn't calculate log(y) properly

[tardy epoch]

What

[frail depot]

Isn’t it the first one ur referring to

log ab equal log a plus log b

[tardy epoch]

What's y?

In your density

[frail depot]

I’m on an iPad to very difficult to write math

[tardy epoch]

Use keyboard

[frail depot]

Yeah doesn’t work I’ll send pictures

[tardy epoch]

Yea it does

[frail depot]

No, I am using voice to text speech

[tardy epoch]

Yea there's y

[frail depot]

Which is also why sometimes the things it writes is very weird

[tardy epoch]

You messed up here

[robust raven]

yes

$\log\mathrm{L}=\sum_{i=1}^{n}\left{ \log\frac{\theta}{x_{i}^{2}}-\frac{\theta}{x_{i}} \right}=\\=\sum_{i=1}^{n}\log\frac{\theta}{x_{i}^{2}}-\theta\sum_{i=1}^{n}\frac{1}{x_{i}}$

[tardy epoch]

The first term is right

[robust raven]

ok fogot log

[tardy epoch]

Why can't you use keyboard

[frail depot]

OK to be honest it’s because I have carpal tunnel syndrome in my fingers so I’m trying to not use them as much

[warm shale]

Joanna Angel

[frail depot]

And I just couldn’t be bothered telling because I know people will be very judgmental💀

[tardy epoch]

But you can write with a stylus?

[frail depot]

Well, yes, and I sort of have to because I have exam but typing on like keyboards and stuff is very bad if you have carpal tunnel syndrome

I study computer science and work as a junior software engineer so I type 24\7

[tardy epoch]

Oh you're in an exam?

[frail depot]

Yeah, this is an old exam set

So if it was up to me, I’ve just let my hands rest, but I don’t really have that option

I would’ve

Unless I wanna retake the course of course

But it’s easier for me to just send pictures so I’ll just do what I normally would just do it with my pen and send pictures

[frail depot]

What’s the difference between this and the first thing I sent except that I didn’t have like parentheses around the whole sum?

[robust raven]

you forgot about expoent

but log exp

removes itself

[frail depot]

Oh my God

Let me just write that up like you did that looks very delicious

[robust raven]

and that is all, if it comes to the hint

[robust raven]

correct?

[robust raven]

ok

now, use the necessary condition for the existence of a local extremum, you know what I mean

[frail depot]

Well, shouldn’t I take the partial derivative and set it equal to zero and solve for theta?

[robust raven]

yes but not partial, just an ordinary derivative one, in your case

$\frac{dL}{d\theta}=0$

[warm shale]

Joanna Angel

[frail depot]

Oh, I just assumed that since you had x and theta than it was partial

[robust raven]

you have one estimated parameter

x is not paramter

[frail depot]

Yeah, gotcha

Is this possible?

Or is that not valid?

[robust raven]

no

yo cant remove sum there

[frail depot]

Because it’s a different x sub i each time ?

[tardy epoch]

Didn't you just want a hint?

[robust raven]

$\text{only}\\\sum_{i=1}^{n}1=n$

[warm shale]

Joanna Angel

[frail depot]

Easy there, Tiger I was just asking whether what I was doing was legal or not

[frail depot]

Thank you

[tardy epoch]

You're the one who wanted to do it yourself

[frail depot]

Breathe

There are tons of other channels you don’t have to be in this one

[tardy epoch]

Just wanted you to know you're not doing it yourself

If you keep asking every step to be checked

[frail depot]

I appreciate the tip

[frail depot]

Well, because I was pretty sure that it was probably wrong, so I just wanted to ask before moving on with something that was most likely wrong

[tardy epoch]

That's just part of math

[frail depot]

Yeah, I get it. But it was a basic question there’s still a lot of things that I have to work out like how to derive the function then then I have to solve for the parameter so I can still solve the equation but I still appreciate your concern so thank you.

[obtuse pebble]

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[deft wharf]

I'm unsure of how to start solving this because of a language barrier with my professor and the notes don't really help either

[obtuse pebble]

@deft wharf Has your question been resolved?

[obtuse pebble]

@deft wharf Has your question been resolved?

[slim cove]

Which part are you stuck on so far, part a?

[deft wharf]

i just got a friend to help, ty tho, don't worry about it

idk how to close this channel

[slim cove]

no problem

.close

[obtuse pebble]

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[timid silo]

A boat moves relative to water with a velocity which is 1/n times the river flow velocity. At what angle to the stream direction must the boat move to minimize drifting?

[timid silo]

I’m unsure if the “the boat moves relative to water with velocity which is 1/n times the river flow velocity” means velocity of boat or velocity of boat wrt river?

[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

@timid silo Has your question been resolved?

[timid silo]

<@&286206848099549185> sorry for tagging but i would really appreciate some insights on this problem. i've been really lost for the last hour and a half

[jagged torrent]

Whoa

That’s not a lot

[placid oasis]

Go on

[jagged torrent]

Once it took me 5 hours and 25 minutes to solve one

[timid silo]

idk i just dont understand the wording

[timid silo]

this one?

[narrow needle]

for f?

so f(-2) is just asking what the value of f is when x = -2

[jagged torrent]

@placid oasis how did you get the pending G+ role

[timid silo]

im sorry but i think you meant to send that in a different channel

[narrow needle]

f' will be the slope of the function, which seems to just be a constant in many places because of there being straight lines

while the integral from a to b of f will be the area accumulated from a to b

[placid oasis]

Idk, I just joined in and got this role

[narrow needle]

which can be done geometrically

[jagged torrent]

Come on

[timid silo]

can you explain to me what this problem is saying

[placid oasis]

I have choosen the Graduate+ role in Channels & Roles

[jagged torrent]

Oh ok

[narrow needle]

f(a) asks for the value of of f at x = a f'(a) asks for value of first derivative of f at x = a integral from a to b of f(x) dx asks for the area accumulated from a to b of f

[timid silo]

what are you taking about?

how is this even relevant to my question

[narrow needle]

wrong problem

mb

[placid oasis]

The velocity of the both is 1/n times of the river flow velocity, it is the velocity of the boat

Sorry for long waiting, my wi fi is bad

No, it basically says that the boat is moving at that velocity, which is 1/n compared to the river, if the boat wouldn't move, if it would be still, it would move at the same velocity of the river.
(This is what I thought, idk if it is right)

[timid silo]

i am getting 90deg + tan^-1(n) as my asnwer

[placid oasis]

Oh man

Uhhh

[timid silo]

wait

[placid oasis]

I think I am still not at that level of helping, sorry

[timid silo]

oh no worries

thank you for your time

[placid oasis]

Np

[timid silo]

apparently the answer was supposed to be 90deg + sin^-1(1/n)

[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

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[empty pewter]

help how do I do this

[harsh remnant]

!status

[obtuse pebble]

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

[empty pewter]

1

[snow moth]

slope intercept form is y = mx + b so first you want to isolate y to one side

12x = -3y - 9
-3y = 12x + 9
y = -4x + 3

[empty pewter]

to isolate the y did you divide by -3?

wait but then how is this right

[hybrid stirrup]

isolate y just means put it on one side alone

so both of the things that were done were to isolate y

also penny made a mistake its y=-4x-3

if one side is divided by -3 the other side in its entirety is also divided by -3

she just forgot to put the negative for the second term on the right side

[empty pewter]

ohh ok so since the 9 is positive when we divide by 3 it became negative

that makes sense now thanks

can someone check on work on this rq

[hybrid stirrup]

yes

can you take a screenshot
so its easier to read

or a better picture

[hybrid stirrup]

no

u need to find the y intercept

u cant just put -1

the form is y=mx+b where m is slope and b is y intercept

then when uw ant specific x values u can plug in a y value to see what x is at that y value and the reverse also

[empty pewter]

what abt now

and then the line doesnt pass through the origin bcs b does not equal 0 right

[obtuse pebble]

@empty pewter Has your question been resolved?

[tepid sedge]

yes

[obtuse pebble]

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[meager citrus]

For the circled portion the answers says to factor out 2 but why cant i factor out 4?

[obtuse pebble]

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[smoky perch]

Um I was practicing a question within my textbook and I am unsure of how this answer was achieved specifically the second step as I understand everything after that

[safe haven]

difference of two cubes

[smoky perch]

oh ok sorry thank you very much

[true granite]

$a^3 - b^3 = (a-b)(a^2 + ab + b^2)$

[warm shale]

fede

[obtuse pebble]

@smoky perch Has your question been resolved?

[obtuse pebble]

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[opaque parrot]

Question 2. Ive found all the angles i can but idk how to get angle EOF

[eager niche]

Wouldn’t that just be 90-a ?

[opaque parrot]

no EOF not EOG

answer is 48 degrees but idk how

[obtuse pebble]

@opaque parrot Has your question been resolved?

[obtuse pebble]

@opaque parrot Has your question been resolved?

[obtuse pebble]

@opaque parrot Has your question been resolved?

[obtuse pebble]

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[timid silo]

A boat moves relative to water with a velocity which is 1/n times the river flow velocity. At what angle to the stream direction must the boat move to minimize drifting?

[timid silo]

i have the answer

its pi/2 + sin^-1(1/n) rad

but i have no idea as to how to tackle this

[obtuse pebble]

@timid silo Has your question been resolved?

[timid silo]

try start by drawing a diagram

i did

how do i continue

To minimise drifting, you should apply an equal and opposite force

as that would cancel out the drifting

[timid silo]

i dont understand

[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

@timid silo Has your question been resolved?

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[meager merlin]

part (b)

[median dome]

when is the argument of a complex number not defined?

[obtuse pebble]

@meager merlin Has your question been resolved?

[meager merlin]

When it is 0?

[median dome]

correct

and 1\z cannot ever equal 0

so the only domain restriction is when z=0 (as 1/z is not defined there)

[meager merlin]

okok thank you

[obtuse pebble]

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[hollow granite]

I don't really know where to start, I was gonna prove injectivity and subjectivity but that won't work as: Injective => Inverse isn't valid

[median dome]

you need to prove both sides

first prove that if f:A-> B is bijective then it has an inverse

then prove that if f:A-> B is invertible then it is bijective

the former can be proved by taking some $b\in B$ and noting that there has to exist a s.t $f(a)=b$ (by surjectivity)

[warm shale]

kheerii

[median dome]

We define a new function $f^{-1}: B\to A$ and let $f^{-1}(b)=a$

[warm shale]

kheerii

[median dome]

this works because f is injective, so a is unique

all you need to show now is that $fof^{-1}(b)=b\forall b\in B$ and $f^{-1}of(a)=a\forall a\in A$

[warm shale]

kheerii

[hollow granite]

I don't quite get the $fof^{-1}(b)=b\forall b\in B$ bit

[warm shale]

Josh

[median dome]

you don't get why we need to prove that or you don't get how we can prove that?

[hollow granite]

this all makes sense

[hollow granite]

I don't quite understand what its saying

surely $fof^{-1}(b) = 1$

[warm shale]

Josh

[hollow granite]

wait no it would be b

ohhhh i see

So youre acutally saying $\forall b fof^{-1}(b)=b \in B$

[warm shale]

Josh

[obtuse pebble]

@hollow granite Has your question been resolved?

[hollow granite]

?

[vast willow]

he is saying that you have to prove this

[hollow granite]

Oh so what I rewrote is correct

[vast willow]

yes

[hollow granite]

ok perfect, thanks

[hollow granite]

.close

[obtuse pebble]

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[median dome]

does $f(x^2)=xf(x)\forall x\in\mathbb{R}$ prove that f is a surjection? (Assuming f is not identically 0)

[warm shale]

kheerii

[median dome]

f is from R -> R

[civic sapphire]

I mean it seems like maybe we could construct such an f that only has non-0 values at +1 and -1?

Never mind that won't work

[median dome]

well why not?

f(1)=1 and f(-1)=-1

[civic sapphire]

Hmmm, you/I might be right.

[median dome]

and 0 everywhere else

hnnn

my thought here was that we can make x as large/as small as required, which means f(x) must take all real values

but that doesn't work if f(x)=0 somewhere

[civic sapphire]

If f is a surjection, you must f(x) = 0 for some value of x

How does that break things?

[median dome]

yes, f(0)=0

[elfin burrow]

doesn't the zero function satisfy this

[civic sapphire]

f not identically 0 is a condition

[elfin burrow]

oh your comment

[median dome]

comes out directly

[civic sapphire]

You know, I think that two valued function or some close variant should work as a counter example

[median dome]

Discord is bugging out bro

[median dome]

Yep

but if f(x)=0 for some non zero x then f(x^n)=0 for all x

basically I have the functional equation $f:\mathbb{R}\to\mathbb{R}$ such that $$P(x,y,z): f(x^2+yf(z))=xf(x)+zf(y) \forall (x,y,z)\in\mathbb{R}^3$$

[warm shale]

kheerii

[median dome]

$P(0,0,0)$ gives us $f(0)=0$ and then $P(x,0,z): f(x^2)=xf(x)$

[warm shale]

kheerii

[median dome]

one solution is $f(x)\equiv 0$; otherwise, we can easily prove f is injective using $P(0,y,z): f(yf(z))=zf(y)$

[warm shale]

kheerii

[median dome]

we also get $P(0,y,1): f(yf(1))=f(y))\implies f(1)=1$

[warm shale]

kheerii

[median dome]

this is where I got stuck

[obtuse pebble]

@median dome Has your question been resolved?

[median dome]

<@&286206848099549185>

[obtuse pebble]

@median dome Has your question been resolved?

[median dome]

.close

[obtuse pebble]

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[dark idol]

How do I solve this

[marsh geyser]

Do you know what kind of triangle is this?

If you answer this question I’m sure you will get the answer fast

[dark idol]

45 45 90?

[marsh geyser]

So what kind of triangle it is

[dark idol]

Isoceles

[marsh geyser]

Then, what is the value of the other side that’s not being asked?

If you have 2 sides of one triangle, do you know how to get the third?

[dark idol]

A^2+b^2=c^2

So 36^2+x^2

Or what

[tardy epoch]

this can work if you know what a, b, and c are

a, b are the sides of a right triangle and c is the hypotenuse

[dark idol]

36^2+36^2=x^2?

[tardy epoch]

correct

[dark idol]

50.9

[marsh geyser]

Yes

[tardy epoch]

,calc 36 * sqrt(2)

[warm shale]

Result:

50.911688245431

[dark idol]

How about this

[tardy epoch]

,tex .sohcahtoa

[warm shale]

riemann

[obtuse pebble]

@dark idol Has your question been resolved?

[obtuse pebble]

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[prime magnet]

Hey, can anyone explain why this converges to 0.75?

[prime magnet]

and why its equal to 1/4*3^-k

[robust raven]

$\text{The geometric series: }\\\text{ }\sum_{n=0}^{\infty}q^{n}\text{ }\text{ is convergent }\Leftrightarrow \left| q \right|<1\text{ }\text{ }\text{ then}\\\sum_{n=0}^{\infty}q^{n}=\frac{1}{1-q}$

[warm shale]

Joanna Angel

[prime magnet]

1/(1-1/q) is 1.5 no?

i dont really get it

[hidden compass]

No, it's 0.75

$\frac{1}{1 - q} =\frac{1}{1 - \left(-\frac{1}{3}\right)}=\frac{1}{1 + \frac{1}{3}}=\frac{1}{\left(\frac{4}{3}\right)}=\frac{3}{4}$

[warm shale]

Alberto Z.

[obtuse pebble]

@prime magnet Has your question been resolved?

[prime magnet]

oh my god thank you

[obtuse pebble]

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[orchid otter]

I need help with algebra

Ive gotten the final answer but in the wrong form

that is correct and is equal the the answer however on a calculator it doesnt show up in the form arootb

Ok

[high lily]

sin and arcsin cancel

[orchid otter]

Thats it#

thank you

suggested solutions

thanks

.close

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[pseudo raft]

Let n∈N. 2n participants, each participant shakes hands with other participants, but there are no three participants who all shook hands with each other (a 'handshake cycle' is not formed among them). Prove that the number of handshakes is not greater than n^2

[exotic walrus]

simple question

[pseudo raft]

need to proe using unduction

prove*

[pseudo raft]

induction

its my channel buddy

[exotic walrus]

oh mb

[obtuse pebble]

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[obtuse pebble]

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[frosty locust]

i’m not sure where to go with question b

oh nevermind i just did it

thx for ur help x

.close

[obtuse pebble]

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[slim summit]

What would the radius of the solid be?

[obtuse pebble]

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@slim summit Has your question been resolved?

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[viral star]

hey

[viral star]

lets say i had a 15x15x15 mm cube

what would be its area in cm³

[timid silo]

do you mean volume

[viral star]

yeah volume

basic math i know

[timid silo]

volume of a cube is a^3 where 'a' is one of its sides

your a here is 15

[viral star]

right

so the cube in context's volume is 15x15x15 mm

aka 3375mm³, right?

[timid silo]

,calc 15^3

[warm shale]

Result:

3375

[viral star]

if thats the case then how do i convert that volume to cm³

[timid silo]

yeah i guess

[viral star]

dont use i guess thats gay as hell

unironically that phrase is nasty

[timid silo]

okay and I'm not here to entertain your opinions on how i talk. Also don't use gay as an insult

[viral star]

its not an insult

[viral star]

but how do i do this

[viral star]

just giving you my honest opinion ykwim

[minor harness]

lol why on earth would you say this to someone trying to help you

[viral star]

i dunno bro

can u answer my question tho

[minor harness]

i can but I'm not gonna

[viral star]

okay

[gleaming saffron]

if u want to convert mm^2 to cm^2
u know that mm to cm is divide by 10
but since its ^2 u divide by 10^2

same process for ^3

[viral star]

so to convert 3375mm³ to cm³ all i do is divide it by 10

[gleaming saffron]

no

you would divide by 10^3

[viral star]

so 3375 mm³ = 3375/1000 = 3.375 cm³

[gleaming saffron]

yes sir

[viral star]

thank u

ppreciate it

[viral star]

unlike the way this bald headed french twat decided to respond

[minor harness]

lol you gotta chill

[gleaming saffron]

you can google a video on why that is but just think about squares for a sec and youll understand why

[dusk widget]

dont insult ppl lol

[viral star]

its a jokeeeee

[viral star]

okay, thanks!

[oblique glacier]

huh

the ping got deleted and it seems the user left?

strange

[gleaming saffron]

ing pinged u but idk why he left lol

[dusk widget]

i pinged, but seems like the issue was cleared up

sorry!

[oblique glacier]

yeah Baysel was being rude and i would have muted but they left so whatever

[minor harness]

person asking for help was being strange yeah

[oblique glacier]

ig ill close the channel

.close

[obtuse pebble]

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[pale drift]

please guys help me understand this please .

[obtuse pebble]

@pale drift Has your question been resolved?

[pale drift]

<@&286206848099549185>

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@pale drift Has your question been resolved?

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[solid cargo]

i have 0 clue how to solve this...

[obtuse pebble]

@solid cargo Has your question been resolved?

[vestal hull]

I might have an idea but it's quite not enough formal

The answer though should be b)

[daring rock]

notice $y = \sqrt{\sin(x) + y}$. Try finding $\frac{dy}{dx}$

[warm shale]

tatpoj

[obtuse pebble]

@solid cargo Has your question been resolved?

[solid cargo]

Wow thats genius thank you

[obtuse pebble]

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[river gale]

Hello, just saying I know the solution to this question, but can I ask about a math rule or something?

[river gale]

So the answer is 2^12, but I also found out that 8^4 is also an answer. While dealing with exponent questions, do you have to make the base small as possible?

[shut lagoon]

They're just equivalent.

8 = 2^3

[river gale]

Is it something like how a radical cant be a denominator

[versed stratus]

no, it depends on the question

at times the larger base is more convenient at times, the smaller base is

[shut lagoon]

In this case, I guess could just convert all of the choices in a common base, be it 2, 4 or 8, and then compare with whatever you got.

[river gale]

ic so theres no fixed method right

[shut lagoon]

No convention no.

[river gale]

Okay, thank you very much

.close

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[dusty drum]

Can someone please help me

[obtuse pebble]

Please don't occupy multiple help channels.

[pale drift]

why the hell no one answers my doubt/question !?

[dusty drum]

Don’t know man I’m also lost

[daring rock]

Draw in segments OF and OH. Then prove that triangles OGF and OGH are congruent.

Then, by CPCTC, you can show GF is congruent to GH

[dusty drum]

I have done this so far I’m not really sure if it’s correct tho

[daring rock]

looks fine so far

Is there anythng you can say about OF and OH?

[dusty drum]

Their congruent?

[daring rock]

They are, but why?

[dusty drum]

I’m not sure but I think it’s hypotenuse leg

[daring rock]

After you show OF = OH, then yes, you have Hypotenuse Leg

to show the triangles are congruent

but you need OF = OH first. They are they hypotenuses

Do you notice anything about OF and OH in relation to the circle?

[dusty drum]

All radii of a circle are congruent

[daring rock]

Bingo 👍

[dusty drum]

After that you would state that triangle OGF is congruent to triangle OGH because HL?

[daring rock]

Do you have enough information to say that?

[dusty drum]

I’m not sure

[daring rock]

In each of those two triangles, do you have a right angle, a congruent hypotenuse, and a congruent leg?

[dusty drum]

Yea

[daring rock]

Then yeah 👍

you have HL

[dusty drum]

And how would I prove that FG is congruent to GH?

[daring rock]

Do you know CPCTC?

[dusty drum]

I know a little bit

[daring rock]

It stands for
Corresponding Parts of Congruent Triangles are Congruent

So, once you proved the two triangles are congruent, any corresponding sides or angles of them are congruent

[dusty drum]

So because one of the legs are congruent I could say that for FG

[daring rock]

well, not exactly

[dusty drum]

Oh that makes sense now

[daring rock]

because the triangles are congruent, the legs are congruent

or like, whatever corresponding parts you want are congruent

[dusty drum]

Ok thank you so much man 🙏

[daring rock]

No problem 👍

[dusty drum]

Will keep that in mind

.close

[obtuse pebble]

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[bleak skiff]

hello, H is a subgroup of G. Order of G = 15. Prove that H is cyclic

[bleak skiff]

H != G

[shut lagoon]

It's Lagrange's theorem mostly.

What values can |H| take?

[bleak skiff]

1, 3, 5

[shut lagoon]

Right, if H has order 1, then it's just trivial right?

And if it's 3 or 5, then you have a group of prime order..

[bleak skiff]

ok thanks

.close

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[clear kraken]

What's the point in considering the power of 2's separately?

is it because this method modulo 2^k will not work

as the result will be -1 mod 2^k

[bronze mica]

it won't be -1 mod 2^k?

because 2 * -1 is not congruent to 1 mod 2^k

the issue is that 2^-1 doesn't exist at all

[obtuse pebble]

@clear kraken Has your question been resolved?

[coral nest]

$2^{-1}$ means the multiplicative inverse of 2 in $\mathbb{Z} / p^k \mathbb{Z}$

[warm shale]

LF

[bronze mica]

yes I know, and I'm saying -1 is not it when p=2

and then saying it doesn't exist at all when p=2

oh I have misunderstood and the claim is that 4a^2+9b^2-1 is -1 mod 2^k when we take b=0, which is true and indeed doesn't work

[coral nest]

oh ok sorry is misunderstood you

[clear kraken]

well their casework is like

not unique is it...

I feel like I could formulate an equivalent one by considering 3^k

and other p^k

[bronze mica]

if by non-unique you mean you could do the cases a different way, sure

but this way works, and since all you want is a solution there's no need to figure out whether this is the only way you can do it

[obtuse pebble]

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[wooden helm]

I spent hours on this please help

[wooden helm]

<@&286206848099549185>

[coral nest]

how did you get that?

[obtuse pebble]

@wooden helm Has your question been resolved?

[coral nest]

the number of elements in the lists is not constant

its (n-i+1)

[wooden helm]

yeah I did that

they just cancel out to 1/n

[coral nest]

right

but i dont think it works for n=3

[obtuse pebble]

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[coral nest]

do .reopen

[wooden helm]

.reopen

[coral nest]

where do you get that from?

[obtuse pebble]

✅

[wooden helm]

the size of each one decreases by 1 because the starting number moves forward like 2 to 4 to 6

[coral nest]

oh right sorry

that makes sense

i does work for 1 through 17

yes that makes sense

[wooden helm]

could you show me how to get the same answer because I don’t think it’s possible

[coral nest]

then what is the expected value in terms of X_i?

[wooden helm]

isn’t it just the sum of the individual expectations using linearity which is just their probabilities

[coral nest]

sure

[wooden helm]

@coral nest am i doing something wrong i have no idea how these are giving different answers

[coral nest]

im looking into it

what is your reasoning for that?

yeah that makes sense

let me check your sums

you should stop at n-1

because X_n=1

so add one to all that

i checked and you should get the same answert

[wooden helm]

wait sorry I don’t get it

[coral nest]

you are sure to have at least n turns

so X_n = 1

but up to n-1 this reasoning works

https://www.wolframalpha.com/input?i=(Sum[(n-i)%2Fn%2C{i%2C1%2C17}])%2B(Sum[(n-17)%2Fn%2C{k%2C18%2Cn-1}])%2B1

[wooden helm]

but that’s saying the probability of having n turns is 1

[coral nest]

at least n turns

[wooden helm]

yeah and it’s not possible to have more than n so isn’t it just n turns

[coral nest]

hm yeah i see how thats weird

right

maybe the indicator function should be at most n turns?

[wooden helm]

I don’t think that’ll make sense if we’re adding the probabilities of at most together to get the total

I don’t see why at least shouldn’t work

[obtuse pebble]

@wooden helm Has your question been resolved?

[coral nest]

@wooden helm are you still there?

you calculated the probabilities for X_i where X_i means it took more than i turns

for example if you lose after three turns

you said X_3 = (n-1)/n * (n-2)/(n-1) * (n-3)/(n-2) which would mean that you didnt pick 34

[obtuse pebble]

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[coral nest]

i.e. X_3 is observed to be 0

[coral nest]

thats why this is the correct sum

.reopen

[obtuse pebble]

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[spice chasm]

which leads to the conclusion that arcsinx = - arccosx

which isnt true

why

[versed stratus]

arcsin(x)+C=-arcos(x)

which is true

[spice chasm]

dont the C's cancel

[timid silo]

I’m p sure it’s true

[versed stratus]

no

the Cs are different

[spice chasm]

arcsinx + C=-arccosx + C

[timid silo]

Combine

[versed stratus]

replace one C with a D

[timid silo]

The Cs

You get the same thing

[obtuse pebble]

@spice chasm Has your question been resolved?

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[fathom flicker]

Does that work?

nvm

I am dumb af

.close

[obtuse pebble]

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[fathom flicker]

lmao

[obtuse pebble]

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[gleaming sleet]

Can anyone tell me how do i find the degree of (x+y)/(x-y)?

[gleaming sleet]

<@&286206848099549185>

.close

[obtuse pebble]

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[ornate turtle]

can someone help me with part c

[ornate turtle]

ill send what ive done so far

[sage geode]

Why is letting a^2 + b^2 = c^2 necessary?

Ah wait part c

I was reading part a

[ornate turtle]

i was trying to prove by contradiction but idk

[sage geode]

No need for contradiction

[versed stratus]

let A be an odd number of the form (2m+1) and B of the form (2n+1)

[ornate turtle]

alright

[versed stratus]

prove you can't express the sum in the form $(c+d)^2$

[warm shale]

Why am. I here

[versed stratus]

that would prove it isn't a square I think

[ornate turtle]

wait is that it

[versed stratus]

I think so

Could be wrong though, I'm bad at this stuff in general, let someone else confirm

[versed stratus]

did it work?

[ornate turtle]

yes but idk how to like conclude it

you get $4(m^2+m+n^2+n)+2$

[warm shale]

moe

[versed stratus]

Now proving it can't be expressed in the form (a+b+c.....n)^2 would be sufficent I think

where a,b,c...n are all integers

[ornate turtle]

i have 0 clue how to do that

[versed stratus]

I'd probably just say we can't express it in the form (a+b)^2 and thus it isn't a square

I may be wrong though

let someone else double check this

[versed stratus]

to complete the square this would have to be $(2((m+n)^2-2mn))+4(m+n)+2$ which isn't a perfect square

[warm shale]

Why am. I here

[teal ledge]

I think it can be solved with result of (a) and (b)

showing that a^2+b^2 is not divisible by 4 nor having 1 as remainder when divided by 8

as showing that not divisible by 4 shows that a^2+b^2 is not square of even number (from (a))
and not having remainder of 1 shows that a^2+b^2 is not square of odd number(from (b))

[obtuse pebble]

@ornate turtle Has your question been resolved?

[obtuse pebble]

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[fluid shuttle]

Hello. Could you please solve this deterative step by step, because i don't understand

[versed stratus]

,rotate

[shut lagoon]

Go at it term by term. It's power rule all the way.

[fluid shuttle]

How I need to start?

[shut lagoon]

Take the derivative of every term and add them up.

$$x^4$$

[warm shale]

Azyrashacorki

[shut lagoon]

$$-\frac{x^5}{5}$$

[warm shale]

Azyrashacorki

[shut lagoon]

And so on

Using the power rule $(x^\alpha)' = \alpha x^{\alpha - 1}$.

[warm shale]

Azyrashacorki

[fluid shuttle]

Alllright. It'a clear, but how for example solve the third (-4 root 4 of 3)?

[shut lagoon]

$$\sqrt[q]{x^p} = x ^ {\frac{p}{q}}$$

[warm shale]

Azyrashacorki

[shut lagoon]

Using this exponent rule

[fluid shuttle]

And that's the answer?

[shut lagoon]

Once you've computed the derivatives of all the terms and added them up, you'll have your answer.

If you want to check your work, feel free to use a smart calculator, like WolframAlpha with a query like "derivative of f(x)" where f(x) is your function

[fluid shuttle]

Okay it helped to my. How about fourth with franction? Where I should to put 6?

[fluid shuttle]

How to solve x^5 how I should to set?

[shut lagoon]

They're just coefficients, they stay there when you take the derivative.

[fluid shuttle]

So this answer would be -5x^4 × 5?

[shut lagoon]

The coefficient is 1/5, not 5

[fluid shuttle]

I need to see cleary vision answer, because in text is hard to me understand. Sorry

[shut lagoon]

$$-\frac{x^5}{5} = (\frac{-1}{5}) x^5$$

[warm shale]

Azyrashacorki

[fluid shuttle]

Now that is super. Cool thanks

Okay. Now I have learned how to solve with root. Please show me, for example the fourth deretative where should be number 6?

[shut lagoon]

For the terms where x is on the denominator, you can just change the sign of the exponent :

$$\frac{6}{\sqrt[3]{x^5}} = \frac{6}{x^{\frac{5}{3}}} = 6 x^{\frac{-5}{3}}$$

[warm shale]

Azyrashacorki

[fluid shuttle]

It's clear. The second one 2 / x^7. The answer is (2 / x^7)*x^5?

[obtuse pebble]

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[fluid shuttle]

No

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[fluid shuttle]

How to solve 2 / x^7

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[tender dove]

I'm studying the following series $\sum_{n=4}^\infty(\ln\ln n)^{-\ln\ln n}$

[warm shale]

Leo, ze FluffBøt

[tender dove]

I demonstrated that it is bigger than $\sum_{n=4}^\infty(\ln n)^{-\ln n}$ and tried using asymptotic comparison on this series with the harmonic series (as suggested by the book), but I get 0 as limit in the checking

[warm shale]

Leo, ze FluffBøt

[tender dove]

I used L'Hopital for such limit

$\lim_{n\to+\infty}{n\over(\ln n)^{\ln n}}=\lim_{n\to+\infty}{n\over(\ln\ln n+1)(\ln n)^{\ln n}}=\lim_{x\to+\infty}{e^x\over(\ln x+1)x^x}=0$, where $x=\ln n$, hence $n\to+\infty\iff x\to+\infty$

I sincerely have no idea of what series to use as comparison

Tried with Cauchy's condensation criteria, the integral criteria is off-limits, tried with normal comparison, root and ratio criterias are useless in this case

[warm shale]

Leo, ze FluffBøt

[tender dove]

Just noticed that L'Hopital was useless

lol

But still I don't know how to proceed

<@&286206848099549185>

[obtuse pebble]

@tender dove Has your question been resolved?

[obtuse pebble]

@tender dove Has your question been resolved?

[obtuse pebble]

@tender dove Has your question been resolved?

[tender dove]

Uh... anyone?

[bronze mica]

$$\frac{n}{(\ln n)^{\ln n}}=\left(\frac{e}{\ln n}\right)^{\ln n}$$

[warm shale]

Edward II

[tender dove]

Yeah but I get a limit of 0

And 1/n diverges

[bronze mica]

there is no 1/n here?

[tender dove]

the n on the numerator comes from the testing series of 1/n

I used asymptomatic comparison

[bronze mica]

oh I see

[tender dove]

So it's actually ${1\over(\ln n)^{\ln n}}\over{1\over n}$

[warm shale]

Leo, ze FluffBøt

[bronze mica]

could I see the hint / suggestion the book gave

[tender dove]

to use the harmonic series as comparison

just this

Apparently the series has to diverge, and geogebra seems to agree

[bronze mica]

I asked because I think this one converges

[tender dove]

Uh... it's possible

I mainly used that because $\forall n,(\ln\ln n)^{-\ln\ln n}\ge(\ln n)^{-\ln n}$

[warm shale]

Leo, ze FluffBøt

[bronze mica]

ok so you're not supposed to be using asymptotic comparison

or at least what I did doesn't use it

[tender dove]

Apparently so, but I tried other methods and nothing really stuck out

[bronze mica]

you do the same e^ln trick as I tried initially, and you get that

$$(\ln\ln n)^{-\ln\ln n}=e^{(-\ln\ln n)\cdot\ln(\ln\ln n)}$$

(I'm leaving intermediate steps out because I will cry if I have to latex it all)

[tender dove]

why greater or equal?

[warm shale]

Edward II

[tender dove]

Aren't they the same?

[bronze mica]

I got ahead of myself

[tender dove]

Ah ok

[bronze mica]

yes

equal

then use the fact that ln(x) < x-1 to get rid of that third ln which is annoying

and get started on inequalities

then do things with the resulting terms to obtain

$$... \geq e^{-(\ln\ln n)^2}\cdot \ln n$$

[warm shale]

Edward II

[bronze mica]

(that last one came from 'cancelling' an e^ln along the way)

this is in turn greater than just the exponential term

(you can probably get just that term at a variety of steps, I just think doing it at this one is clearest)

now ln(x) < sqrt(x), so we can get rid of the annoying square with another inequality, and then also ln ln n < ln n so overall we should get ... > exp(-ln n)

which is 1/n

[tender dove]

Uhm... lemme see if I can recreate it

[bronze mica]

oh wait

[bronze mica]

I forgot to get rid of an ln when doing this replacement

so using ln(x) < sqrt(x) gives > exp(-ln n) directly

alternatively the square of ln ln n is larger than just ln ln n and then use ln ln n < ln n to get the inequality with 1/n, either way on its own works

[tender dove]

Here I get this without the other ln

I get them only on the exponent

$(\ln\ln n)^{-\ln\ln n}=e^{-\ln\ln n\cdot\ln\ln\ln n}\ge e^{-\ln\ln n\cdot\ln\ln n}=e^{-(\ln\ln n)^2}$

[warm shale]

Leo, ze FluffBøt

[bronze mica]

oh

yeah that works too / better

(I used x-1 as the bound, in which you would end up with an extra term, but x is also an upper bound for ln(x) so this works fine)

[tender dove]

$(\ln\ln n)^{-\ln\ln n}\ge e^{-(\ln\ln n)^2}\ge e^{-\ln n}={1\over n}$

[warm shale]

Leo, ze FluffBøt

[tender dove]

@bronze mica I suppose this is what you meant?

[bronze mica]

yes

[tender dove]

Alright! Thank you!

[bronze mica]

I didn't clean up my inequalities to be only what was needed which is why my explanation was potentially confusing, sorry

[tender dove]

.close

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[tender dove]

Ah no worries

Thank you anyway

[obtuse pebble]

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[vocal obsidian]

woops sorry

[high lily]

you didn't simplify that -(4a-b) correctly

[vocal obsidian]

doesnt 4a cancel out?

which leaves u with ah - b - (-b)

[ashen dove]

-b-(-b) isn't -2b

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[proper sable]

what's the difference between a surface integral and a double integral? like if you're trying to find the flux across the region x^2 + y^2 <= 2 in the xy-plane, would that be a surface or a double integral?

[brittle bolt]

It's a circle i think so no need for integral just basic area formula

[proper sable]

i'm trying to find the flux, not the area

[brave bramble]

That's a vague question. There's differences between a surface and double integral, but you can convert one into the other.

So a double integral can solve your problem but you'd probably start with a surface integral

[proper sable]

isn't a surface integral across a surface?

is x^2 + y^2 <= 2 in the xy plane a surface?

[brave bramble]

With the ≤ it's a volume

[proper sable]

i said in the xy plane

it's not a cylinder

it's a circle

[brave bramble]

I mean yes, it's a surface

[proper sable]

is it a surface or a 2D region

[brave bramble]

It doesn't help that it's also a region

[proper sable]

so is it a dS or a dA

[brave bramble]

So a flux integral problem should start with a surface integral.

You will take that surface integral to a double integral

I'm slightly suspicious there is no difference between the two in this case. But there's differences in other cases

[wary vigil]

those are the same thing

dS and dA

The idea is that you take the dot product of the vectorfield at all the points on the surface with the normal at that surface

area is just a number associated with the 'size' of that surface

[obtuse pebble]

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[torpid grove]

find the are of that part of the cylinder $x^2 + y^2 = 2ay$ that lies inside the shere $x^2 + y^2 + z^2 = 4a^2$

[warm shale]

vuviCa

[obtuse pebble]

@torpid grove Has your question been resolved?

[robust raven]

use hint below:

$\text{Let }\text{ S}\text{ }\text{ be side surface of the cylinder then}\\\left| S \right|=2\int_{\Gamma}^{}\text{ }f\left( x,y \right)\text{ }dl\text{ }\text{ where }\\\Gamma:x^{2}+\left( y-a \right)^{2}=a^{2},\text{ }a>0\\f\left( x,y \right)=\sqrt{4a^{2}-\left( x^{2}+y^{2} \right)}$

[warm shale]

Joanna Angel

[torpid grove]

Yep than I go for polar coordinates

but than how

$ r = sqrt(2ay) $

changed $x^2 + y^2$ part with $r^2$

[warm shale]

vuviCa

[torpid grove]

and since $ r = sqr(2ay) $

I changed y = r sin(a)

so our r becomes 2a r sin(a)

[robust raven]

this way:

$\left| S \right|=2\int_{0}^{2\pi}\sqrt{4a^{2}-\left[ \left( \cos\text{}t \right)^{2}+\left( a\sin\text{}t \right)^{2} \right]}\cdot \sqrt{\left( \frac{dx}{dt} \right)^{2}+\left( \frac{dy}{dt} \right)^{2}}\text{ }dt\\\text{where }\text{ }x\left( t \right)=\cos\text{}t,y\left( t \right)=a\sin\text{}t$

[warm shale]

Joanna Angel

[obtuse pebble]

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[lapis bobcat]

i dont really get the answer to part (iii)

[obtuse pebble]

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@lapis bobcat Has your question been resolved?

[remote skiff]

R and S are identically distributed, you would expect that they both contribute to t equally, so t/2.

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[cyan merlin]

I have a problem where I'm supposed to find the number of spoonful's of water in the ocean, assuming oceans cover the entire earth and have a depth of 3km, and a spoonful is 18cm^3

[cyan merlin]

is it just the SA of the earth, times 3km and then divided by 18cm^3?

[eager niche]

What level of math is this?

[cyan merlin]

this is in my physics course lol, these are estimation problems and they are hurting my brain

so its not really math moreso just trying to make me think i guess

but im getting stuck

[eager niche]

But I think that’s not going to be right, try to visualize a sphere and how you would model the the water on top- what is the relationship you’ve been given?

[cyan merlin]

Okay thanks

[eager niche]

Do you see what I’m trying to say?

[cyan merlin]

yea, i basically have a sphere with 3km of water on top layer, with a total radius of 6400km, so i just need to find two volumes and subtract them

i think

[eager niche]

Yes exactly, do you know how to find the volumes?

[cyan merlin]

yea. i think im good from here

i can do calc but these conceptual things hurt my brain

thank you!

[eager niche]

Ah yeah, me too sometimes

[cyan merlin]

.close

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[shadow pecan]

ive gone ahead and proved that the sin of the right hand side is equal to the sin of the left hand side. when taking the sin of both sides i got 63/65=63/65
now all i need to do is prove that they're in the same intervals correct?
the best i could do is prove that for the LHS the interval is (0, 2pi/3) and for the RHS the interval is (0, pi/2)
is that enough to prove that the values are the same since sin is positive in both of those intervals since they occupy the first and second quadrant

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[austere whale]

hi everyone

[austere whale]

for no 5, how do u get tan 0 again 😭

and sec^2 0

[polar fossil]

well tan(0) is just 0

and $\sec^20 = \f1{\cos^20} = 1$

[warm shale]

hayley

[austere whale]

0 in tan 0 is tangent in 0 degrees right

[polar fossil]

yes, tan0 is tangent of 0 degrees

,tex .unit circle

[warm shale]

hayley

[austere whale]

is sec^2 x = sec x * sec

[polar fossil]

yea

[austere whale]

ohh cos^2 0 is 1?

[austere whale]

and is sec x = sec * x orno

[tardy epoch]

sec is short for secant function

functions take in a number and give back a number

In sec(x), sec is the function, x is the input number

Its definition is
sec(x) = 1/cos(x)

[austere whale]

ahh

i sont understand how cos^2 0 is 1

i understand how cos 0 is 1 but

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[stoic merlin]

Not sure what I’m doing wrong

[stoic merlin]

0 was also one of my possible answers but that was wrong as well

[inland matrix]

Line 2 should be
3(-4+h)²**+6** -(3(-4²)+6)

As the numerator

[stoic merlin]

oooh i missed that

okay thank you ill try that

[inland matrix]

Np

[stoic merlin]

Sorry still cant seem to get it right

Or wait wouldn’t each both sides give me an infinity answer?

[cloud willow]

you did not distribute the negative to the second term of f(-4)

and uhh idk why you're cancelling like that

you can't just cancel h like that

you're supposed to expand/simplify the expression first (hint: it WILL lead to a proper cancellation of h)

[stoic merlin]

ahh okay i'll retry again

[robust raven]

$\text{Analyse such an example if: }\text{ }f\left( x \right)=4x^{2}+5\\\lim_{h \to 0}\frac{f\left( -6+h \right)-f\left( -6 \right)}{h}=\lim_{h \to 0}\frac{4\left( -6+h \right)^{2}+5-\left[ 4\cdot \left( -6 \right)^{2}+5 \right]}{h}=\\=\lim_{h \to 0}\frac{4\left( 36-12h+h^{2} \right)+5-\left( 144+5 \right)}{h}=\\=\lim_{h \to 0}\frac{144-48h+4h^{2}+5-144-5}{h}=\\=\lim_{h \to 0}\frac{h\left( -48+4h \right)}{h}=\lim_{h \to 0}\left( -48+4h \right)=-48$

[warm shale]

Joanna Angel

[robust raven]

yours is an analogous one.

[stoic merlin]

-48 is not the correct answer either

[robust raven]

lol

[stoic merlin]

oh nevermind

[robust raven]

i wrote an exmple for you

not solution for you

[stoic merlin]

lmao just saw u used a different function

[robust raven]

analyse my example how to solve yours one

[stoic merlin]

will do

[robust raven]

ok

[stoic merlin]

I got it

Thank you!!

.close

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[cinder hedge]

An isolated farm house is located on the bank of a straight canal that is 200 m wide. Electric power for the house must come from a connection that is on the opposite side of the canal, 500 m downstream. If the cost to lay cable underwater is three times the cost of laying on land, what path should be chosen to minimize cost?

I'm not sure where to go with this. I have a diagram.

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.close

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[cinder hedge]

'

""

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[obtuse pebble]

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[coral willow]

what do you call the maths topic in which differential operators are transformed for use in local (eg cylinder coordinates) and physical coordinates (xyz)

[coral willow]

for example this formula transforms the gradient of a scalar field in local coordinates to gradient of a scalar field in physical coordinates