help-19

[quartz island]

?

[gritty oar]

can someone explain where is this inequality coming from, the context is brogains bourgains theorem. can provide more context if needed

hmm i suppose i will just dump it in case

[light agate]

whats lambda, mu, and epsiloj, and varepsilon that looks too big

[dapper canyon]

is it just mu = 1/16?

[light agate]

[gritty oar]

here are all the relevant bits

[gritty oar]

yes i am help me im stuck step bro'd on this one detail

[summer cradle]

help me im stuck step bro

[gritty oar]

oh

is it a chernoff bound

[light agate]

yeah thats what i was thinking too

what book is this btw, is it that gtm

[odd edge]

@gritty oar Has your question been resolved?

[gritty oar]

dk what gtm is but its just some random lecture notes i found online

[light agate]

oh i see cool

hope you figure it out

[gritty oar]

so its $P(X\leq (1-\epsilon) m/16) \leq min_t \frac{E(\exp(-tX))}{\exp(-t (1-\epsilon) m/16)}$

[clever fjord]

p norm reformed

[gritty oar]

$= min_t \frac{\prod_i^n E(\exp(-tX_i))}{\exp(-t (1-\epsilon) m/16)}$

[clever fjord]

p norm reformed

[gritty oar]

$= \frac{\prod_i^n 1-p(1-\exp(-t))}{\exp(-t (1-\epsilon) m/16)}$

[clever fjord]

p norm reformed

[gritty oar]

$\leq \frac{\prod\exp(p(\exp(-t) - 1))}{\exp(-t (1-\epsilon) m/16)}$

[clever fjord]

p norm reformed

[gritty oar]

$=\frac{\exp(np(\exp(-t) - 1))}{\exp(-t (1-\epsilon) m/16)}$

[clever fjord]

p norm reformed

[gritty oar]

$=\exp (m/16 (\exp(-t) -1 + t(1 - \epsilon)))$

[clever fjord]

p norm reformed

[gritty oar]

min at t = ln(1 / (1-epsilon)) ?

$= \left(\frac{\exp(-\epsilon)}{(1-\epsilon) ^{1-\epsilon}}\right) ^ {m/16}$

[clever fjord]

p norm reformed

[gritty oar]

take log of denom, taylor serise, expand terms and lower bound, exponentiate? and $\frac{1}{(1-\epsilon) ^{1-\epsilon}} \leq \frac{1}{\exp(-\epsilon + \epsilon^2 / 2)}$

[clever fjord]

p norm reformed

[gritty oar]

$(1-\epsilon) \ln(1-\epsilon) = -\epsilon + \epsilon^2 /2 + O(\epsilon^3)$

[clever fjord]

p norm reformed

[gritty oar]

ok yes

$\leq exp(-m\epsilon^2/36)$

[wanton bison]

brogains this is why you are the best

[clever fjord]

p norm reformed

[gritty oar]

unfortunately i cannot be the best because you are the best

https://tenor.com/view/pokemon-ash-hat-turned-around-throw-gif-8995789

$<\exp(-m\epsilon^2 / 48)$

[clever fjord]

p norm reformed

[gritty oar]

excellent

so for i <= m = 144log n, j <= t = log n

so

we require $\epsilon^2 > \frac{1}{\log e}$

[clever fjord]

p norm reformed

[gritty oar]

ok wait so

$S_{i,j}\subseteq X$ are the sets chosen for the frechet embeddings $f(x) = (d(x, S_{1,1}),\ldots d(x, S_{\log n, c\log n}))$

[clever fjord]

p norm reformed

[gritty oar]

and we get that

for all $x,y\in X, k = c\log^2 n, \frac{k d(x,y)}{b\log n} \leq ||f(x) - f(y)||_1 \leq k d(x, y)$

[clever fjord]

p norm reformed

[gritty oar]

so this embedding has distortion O(logn)

and is embedded into $L_p^{O(\log^2 n)}$

[clever fjord]

p norm reformed

[gritty oar]

so the probabilities that such a Sij works is bounded below by by constant time, and generating them is O(log^2n)

ok yes

[summer cradle]

👍🏻

[gritty oar]

so like

how the fuck do i solve max flow

with this shit

neck

[wanton bison]

you are so smart

[summer cradle]

oh i know someone who LOVES max flow

[gritty oar]

is it u

[summer cradle]

no

[gritty oar]

then who

[summer cradle]

bald guy from measure theory class

[wanton bison]

this is u

[gritty oar]

why does he LOVE max flow

[summer cradle]

LOL

[summer cradle]

not sure exactly. but he kept bringing it up when we talked about optimization

[gritty oar]

yeah dude

i shouldn't have started spectral graph theory

all i see is duality and i have to google 1000000000000000 things

[summer cradle]

i just felt my head and can confirm i am not bald

[gritty oar]

miss floyd fulkerson

[wanton bison]

https://tenor.com/view/hair-caressing-caress-love-couple-cuddle-gif-20005285

[summer cradle]

wish that was me

[gritty oar]

its me and ADONIS

[wanton bison]

wild

[summer cradle]

who is who

[gritty oar]

well adonis

are you a big spoon or a little spoon

[wanton bison]

lmaooo

[summer cradle]

answer the damn question

[light agate]

they aren't even spooning i dont see how this is relevant

[gritty oar]

who

[light agate]

^

[gritty oar]

ASKED

[light agate]

ok p norm

adonis you still havent answered

[wanton bison]

🚬

[gritty oar]

.

[wanton bison]

i dont know what spoon i am

[light agate]

its ok theres no shame in saying little spoom

[summer cradle]

i am little spoon

[wanton bison]

i think i would tend probably to big spoon

[gritty oar]

i was baout to make a joke

but hten i realziedthat im refomred

[wanton bison]

no you are reformed

so go ahead

[summer cradle]

well unreform yourself for a moment

[gritty oar]

no

fuck u kids and ure mums

[light agate]

so whats this lecture notes on

spectr graph theory

?

[wanton bison]

p daddy norm what do you wanna become later in life

[gritty oar]

yes

im permanently confused

[gritty oar]

dead

[light agate]

[gritty oar]

and by later i mean later today

[wanton bison]

good news, you will eventually

but i meant before death

[gritty oar]

dying

[wanton bison]

technically

you are dying every moment

[gritty oar]

everything is becoming so confusing with only extreme surface level knowledge in linear optimization

and normal graph theory is so easy

[wanton bison]

the p stands for pretty

[gritty oar]

that brings me enjoyement

[light agate]

i still havent studied graph theory somehow

[wanton bison]

oh no

[gritty oar]

will you

[light agate]

yes

but there are other things I'm interested in currently

[gritty oar]

am i one of them

[light agate]

yes

[gritty oar]

thx.

[summer cradle]

[wanton bison]

you are a little 🔪

[summer cradle]

🔪

[gritty oar]

[summer cradle]

:kuromiknife:

[gritty oar]

?

[wanton bison]

?

[summer cradle]

?

[light agate]

?

[gritty oar]

.close

[odd edge]

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[pulsar elbow]

h

[light agate]

i

[gritty oar]

g

[odd edge]

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[mystic saffron]

how do i discover the boolean expression by the table?

[mystic saffron]

oof

[noble obsidian]

when S = 1

look at those

(A, B, C) = (0, 0, 0)
(A, B, C) = (0, 1, 1)
(A, B, C) = (1, 0, 0)
(A, B, C) = (1, 1, 1)

What can you say about A

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[grave barn]

Pls help

[grave barn]

I did integration by parts twice alrea

Dy

.close

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[signal oar]

\begin{problem} Give any choice of $a, b, c$ so that the two lines [g_1: \mrm{1 \a \ 2} + r \cdot \mrm{b \ 3 \ 6} \quad \text{and} \quad g_2: \mrm{c \ 0 \ 3} + t \cdot \mrm{3 \ 1 \ 2}] intersect at one point. \end{problem}

[odd edge]

Please don't occupy multiple help channels.

[signal oar]

How do you do this?

[spiral basalt]

(g1-g2)(r,t) defines a plane
Does it contain 0 ?

[true garnet]

equate the 2 lines

try to solve the equations. Choose any a b and c such that the equations ar e consistent and linearly independent

in this case a is restricted

but u have to try to solve the equations to see it

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[mossy ruin]

yo

[mossy ruin]

cna someone give me the answer

im taking a test for math and i just got back from vacation and need the answers quick

[nocturne belfry]

<@&268886789983436800>

[mossy ruin]

is that

agains tthe rules?

[nocturne belfry]

yes

[boreal crag]

very

[mossy ruin]

oh ok

sorry

i didnt know that

mb

[odd edge]

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[nocturne belfry]

good luck on your test tho

[warped grove]

Yeah, no cheating allowed here

Good luck

[mossy ruin]

soryr mb

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[deep turret]

What is wrong with my answer to solving the integral of arcsin?

[wicked kestrel]

here in your 3rd line

your integrand contains u^-1/2, not u^1/2

[odd edge]

@deep turret Has your question been resolved?

[deep turret]

Wait the answer was my answer but without the 1/3

Wolfram alpha, integrate arcsin

,integrate arcsin

[tall veldt]

#help-28 message

[deep turret]

Thanks

Got it

[odd edge]

@deep turret Has your question been resolved?

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[outer coyote]

If a set $S \subseteq \mathbb{R}$ is open, then $\forall x \in S$, $\exists \varepsilon > 0$, $\forall y \in \mathbb{R}$, if $|x-y| < \varepsilon$, then $y \in S$.
\ \
(a) Write down the contrapositive of the statement.

[clever fjord]

poudel

[outer coyote]

how do I do this I can barely read what it says

is it

[summer cradle]

not open doesn't mean closed

you just write not open

[outer coyote]

is the contrapos I wrote correct tho

[summer cradle]

i didn't read it yet just that at the end

i'll read now

[outer coyote]

i havent learnt set notation yet i legit had to search up the symbols just so that I can type it

😭😭

[summer cradle]

no this doesn't look right

[outer coyote]

yeah thought so

can u run me thru

also im lowk confused there are like two statements stacked on top of each other

not sure how to treat them

[summer cradle]

so we wanna write the negation of $\forall x \in S$, $\exists \varepsilon > 0$, $\forall y \in \mathbb{R}$, if $|x-y| < \varepsilon$, then $y \in S$

[clever fjord]

chmonkey #1 simp

[summer cradle]

right?

[outer coyote]

yeah

[summer cradle]

that's what should go before the "then S \subseteq R is not open"

[outer coyote]

oh so should I be thinking of it as

[summer cradle]

NOT (for all x in S something is true)
means
there exists an x in S such that the something is false

[outer coyote]

if ..., then ( if then )

like the if then of the other statement is like in the then of the first

[outer coyote]

yeah I kind of follow

ill try again lmk if its write

right

[summer cradle]

so it should start like there exists x in S ...

[outer coyote]

yeah alr wait let me try lmao wanna see if I actually understand what u said

instead of getting spoonfwd

[summer cradle]

yea go ahead

[outer coyote]

ill take like two mins if thats alright

[summer cradle]

ping me

[outer coyote]

@summer cradle

[summer cradle]

that's closer

this isn't quite right though

[outer coyote]

the reason I put that

was cause

i thought i was negating

p --> q

ohh

andds

is it AND

[summer cradle]

yep

[outer coyote]

💀💀

yeah okay yeah thanks

then its right yeha?

[summer cradle]

yep

[outer coyote]

yeah alr tysm

think I get it

so like

since that if then statement was embedded in the overall if then

i just need to negate the embedded statment

and not worry abt doing the contrapositive of that

[summer cradle]

yea

[outer coyote]

yeah bet

tysm

[summer cradle]

np

[outer coyote]

ill close channel now

.close

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[craggy maple]

In Walter Rudin's "Principles of Mathematical Analysis", I want to know why he describes the relations of a rational p with q in sets A and B with the aforementioned conditions; with the identity highlighted in the attached screenshot.

[green elm]

this has been asked a lot in various forums, see for example this MSE question/answer:

https://math.stackexchange.com/questions/141774/choice-of-q-in-baby-rudins-example-1-1

the answer there is probably better than any i can give haha

[craggy maple]

thank you!

.close

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[grizzled summit]

im confused on how to get the areas for 1 2 and 3 without the area between curves formula, the desmos image is with my value for a which is -5

[grizzled summit]

.close

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[raw herald]

is it like this?

9 raised to 9 raised to 9?

it is a big number

[odd edge]

@mystic saffron Has your question been resolved?

[fickle coral]

Yes it is a big number, it may have around 370 million digits

[raw herald]

it seems like it's like this

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@mystic saffron Has your question been resolved?

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[balmy jackal]

can someone help with 10.b.

[frigid canopy]

is this a test?

[balmy jackal]

worksheet

online class

[frigid canopy]

okay

[balmy jackal]

i dont want to ask help on every question

[frigid canopy]

I think there's. typo

it should be g(h(x))

[balmy jackal]

yeah its cambridge

[frigid canopy]

does that make it easier

[balmy jackal]

my teacher says its bad

wait

he told me how to do it

.close

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[tidal barn]

need help with this, says find all maximal solutions

[north grove]

this is a homegenous differential equation

as f(kx,ky) = f(x,y)

put Y = Vx and solve it

[tidal barn]

i dont understand

[north grove]

just a minute, i will uplaod a picture

[tidal barn]

i can transform it into y'=.....

[north grove]

y' = root(x^2 - y^2)/x - y/x

[tidal barn]

yes

that

[mortal trench]

$$y' = \frac{\sqrt{x^2-y^2}+y}{x}$$

[north grove]

then put y = Vx

[tidal barn]

can you tell me how did you deicde that

moreover "why"?

this is incorrect

there should be +y

[clever fjord]

Mr. Gamer 🇵🇸

[north grove]

it is a homogenous differential equation, in these types of equations, inorder to solve them , we have to make it variable seperable , so we put Y = vx, where v is any real number

[mortal trench]

Let $y = v(x) \cdot x$. Then:

$y' = \frac{\sqrt{x^2 - v^2x^2} + vx}{x}$$

[clever fjord]

Mr. Gamer 🇵🇸
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[tidal barn]

and how did you know it is homogenous?

[north grove]

put x = Kx and y = Ky , if the k cancels out and there is no change in the equation, then it is a homogeneous equation

[tidal barn]

let me try to rewrite it and see if it cancels out

[north grove]

see it cancels out

[tidal barn]

v is what, real number?

[north grove]

yup

[tidal barn]

okay wait im writing it down and trying it myself

[north grove]

ok

[tidal barn]

meanwhile, can you explain to me in mathematical context why this substituion will solve it?

[north grove]

see, when we substitute Y = Vx, our equation is no longer homogenous, then new equation we get in the process is variable seperable which can be easily solve by integration

[north grove]

like this

[tidal barn]

and why can we put y=vx

y=vx means its a line

its a vector subspace

[north grove]

as we saw X, y = Kx,Ky (k cancels out), so putting y = vx also wont matter as in the end it will also cancel out, and not bring a change in the final answer
so we can only put V if it is homegenous and k cancels out

so like K , v will also cancel out

and not change the final answer

[tidal barn]

done, it holds.

[north grove]

that is just a way of writing the equation

[north grove]

nice

[tidal barn]

btw can you help me with more differential equations?

[north grove]

ya sure

i will be more than happy to

you can dm me or post it here itself

[tidal barn]

im calculating this one rn

[north grove]

okie

[tidal barn]

what if k is 0?

@north grove

[north grove]

k is just a variable with any real value we dont need to assign any particular value

[tidal barn]

but isnt it for all k apart from 0?

if k was 0, then the homogenous test wouldnt fit

[north grove]

ya, but we usually assume that k is not equal to 0

[tidal barn]

@north grove

[north grove]

this question is variable seperable

it is easier than homogenous

lemme just solve it and send a pic

see

the final ans is log y = x^4/4 + c

[tidal barn]

the function is defined on R, therefore it has solution on R

[north grove]

the solution to a differential equation is usally another equation itself, and it domain may change and it can may or may not be on R, that is immaterial to the solution

[north grove]

no idea whats this brother

[tidal barn]

do you know this one?

[north grove]

it is homogenous

put y = Vx

as k cancels out

if x = kx and y = ky

[tidal barn]

why not z=y/x

substitution

[north grove]

both mean the same thing , y = Vx means V = y/x

V , Z you can use any variable

[tidal barn]

v isnt a real number

v is some function of x

@north grove

[north grove]

oo yea yea , i forgot

sorry about that

[tidal barn]

i am still writing this down

[odd edge]

@tidal barn Has your question been resolved?

[odd edge]

@tidal barn Has your question been resolved?

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[normal sundial]

what is N? i dont get the explanation

[upper onyx]

i guess there is some useful information before "what is the value of N?" which you should share with us.

[odd edge]

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[normal sundial]

sorry

[odd edge]

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[vital warren]

can someone give me a step by step to find the direction please

[vital warren]

im having alot of trouble with this 😦

[frigid canopy]

is this. a test?

[vital warren]

no its a practic3 1uia

quiz

[frigid canopy]

so not graded, right?

[vital warren]

no they dont give me any answers when i get it wrong

so i cant even figure out what direction even is

[frigid canopy]

start by computing the magnitude

for that find the angle between the vectors

[vital warren]

okay

so the magnitude is 250

how would I find the angle between the vectors?

@frigid canopy

[frigid canopy]

what do you think the angle is

[vital warren]

would it be 90?

[frigid canopy]

yes

[odd edge]

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[opaque spindle]

why is this part equal to 2^n?

[leaden karma]

that's by the binomial theorem

$(1+x)^n=\sum_{i=0}^n\binom{n}{i}x^i$

[clever fjord]

kheerii

[leaden karma]

set x=1, then you'll have your result

[opaque spindle]

x is equal to one because you dont see it right?

[leaden karma]

yeah, sure you can think about it like that

[opaque spindle]

hmm okay

[dapper canyon]

x can be any value

just choose 1

[opaque spindle]

but i cant retroactively choose it to be one

[leaden karma]

no, particularly x=1 is what gives you the required result

if you set x to be something else you will get other results

[dapper canyon]

wdym

[dapper canyon]

this is true for all x

[opaque spindle]

I'm just thinking that if all i have is this

yeah im just thinking

[leaden karma]

yes it is, but PARTICULARLY x=1 is what is required to prove the identity used in OP's question

[dapper canyon]

yes i know

[leaden karma]

say you set x=2, then $3^n=\sum_{i=0}^n\binom{n}{i}2^i$

[clever fjord]

kheerii

[leaden karma]

this is true but it doesn't help us

[opaque spindle]

ah so yeah, it would be clearer if the one was stated $\sum_{i=0}^n\binom{n}{i}1^i$ because 1^i is always 1

[clever fjord]

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

[dapper canyon]

yep

[leaden karma]

yeah that's what you have to realise to use binomial theorem

[opaque spindle]

alright

[leaden karma]

this is a pretty common identity though so it's a good idea to memorise it

[opaque spindle]

okay

thanks both of you for the help!

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[dapper canyon]

np

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[silver crag]

For this question, I am supposed to "Express the volume of the solid described as a double
integral in polar coordinates". I got the r bounds easily, but I can't seem to understand why the theta bound is 0 to pi/2 and not 0 to 2pi, and also why the integrand is multiplied by r.

[silver crag]

This is what I thought the answer would be

[long fjord]

sorry bro i dont know the answer

[silver crag]

The answer is in the second screenshot

[split ice]

https://www.khanacademy.org/math/multivariable-calculus/integrating-multivariable-functions/double-integrals-a/a/double-integrals-in-polar-coordinates

[silver crag]

Well I mean I know what a polar double integral is

[split ice]

You saying why r come so thats why I send it

[silver crag]

I am just confused why the bounds would be 0 to pi/2 when the graph is a sphere and a cylinder

Because it seems like it should be 0 to 2pi

[silver crag]

...I know why r, I just don't know why it is r*sqrt(9-r^2) instead of just sqrt(9-r^2)

[split ice]

It is because graph is symmetrical around x and y axis

So he take volume in 1st quadrant and multiply it by 4

[silver crag]

Oh, I see, that makes sense

[silver crag]

So then the integrand is 2r*sqrt(9-r^2)? Where does the 2r come from?

[split ice]

There is 8 quadrant 4 up and 4 down that's why

And we use rdodr for double integral in polar form

[silver crag]

But doesn't 0 to pi/2 encompass two octants? Because it will contain both negative and positive z in that arc

[split ice]

You used surface integral here ?

How you come up with √(9-r^2)

[silver crag]

Because r^2=x^2+y^2, and x^+y^2+z^2=9

So z = √(9-r^2)

[split ice]

I think it is equal to z=+-√9-r^2

For uppar and lower half

[silver crag]

Ah, yes

Ok that makes sense

Thanks

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[split ice]

What we do after splitting

I think there is something more @silver crag

[silver crag]

.reopen

[odd edge]

✅

[silver crag]

Ok?

[split ice]

For uppar half we get 4 into our integration

But for lower half it's minus so it would be zero

[silver crag]

But we are looking for the total volume, so it doesn't matter if the volume is in a negative quadrant

[split ice]

No there should be minus somewhere in lower half which makes it positive

You should dig dip

I go to sleep now

Ok good night

[silver crag]

Good night

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[prisma wind]

anyone know how to solve the limit of (xy)/(x-y) as (x, y) approaches (0, 0)? either that or showing that the limit doesn't exist. i inputted it to a graphing calculator and it appears to be tending to 0, and i also set y = -x and surely enough it tend to 0. but i just cant figure out how to solve the limit itself

[prisma wind]

unless i chose a specific path it always results in 0/0 indeterminate

[latent quiver]

Yeah so it doesnt exist n its indeterminate

[prisma wind]

0/0 doesnt necessarily mean it doesnt exist i think

most of the time you can manipulate it in a way so that its not 0/0 but in this case i cant see a way to do that

[latent quiver]

there isnt a way to manipulate it thats y i am saying that

[prisma wind]

this is the graph

it appears to be tending to 0

if it doesnt exist then the way to prove that is to take two paths that result in different values but from what i can see all paths result in 0

which would mean it exists and is 0

unless theres something im missing and there is a path that tends to something other than 0

whats interesting is that theres some weird behavior near 0 though, it goes up and down in a weird way where the steepness seems to tend to infinity and yet at (0,0) it still seems to be 0

well i mean this problem was on the 2017 analysis 2 exam in my uni so theres probably a way to solve it since they wouldnt put an impossible problem

but still i cant find a way to either prove it doesn't exist nor prove it does exist

<@&286206848099549185>

[green elm]

maybe try to exploit the fact that the function blows up along the line y=x

you can't approach on the path y=x because the function is undefined there

but you can approach on a path that asymptotically approaches that line as you go to the origin

for example the path y = (1-x)x has this property

[latent quiver]

well it only exists when u take single limits or one is expanding and ther decreasimg

[green elm]

?

if i'm calculating correctly, the limit as you approach (0,0) along the path y = (1-x)x exists and is not equal to zero, so that should suffice to show that the overall limit DNE

(since obviously the limit if you approach along the x axis or y axis is zero)

[prisma wind]

phone ran out of battery just as i was about to send a picture hold on

damn it i accidentally tore a page in my calculus book

[green elm]

rip

literally haha

[prisma wind]

oh yea you're right its one

[green elm]

yea that's what i got

[prisma wind]

one of the weirdest functions ive ever seen, seems to have multiple values at 0,0

whatever guess that solves it

[green elm]

i wasted some time trying to do it in polar coordinates before i found this path

[prisma wind]

anyway ty

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[wanton bison]

Is $((-1,1), \circ)$ a group and we define $a \circ b := \frac{a+b}{1+ab} \text{ (for snosepsh: } a,b \in (-1,1)).\$
I managed to show that the operation is closed (again).
$\$
However I am stuck at showing the associativity. (apparently it is but I wanted to know if there is some trick)
$\$
Let $a,b,c \in (-1,1)$. Then $$ (a\circ b) \circ c = a \circ (b \circ c)$$

[clever fjord]

𝔸dωn𝓲²s

[wanton bison]

,w ((a+b)/(1+ab) + c)/(1+(a+b)*c/(1+ab)) = ((b+c)/(1+bc) + a)/(1+(b+c)*a/(1+bc))

[clever fjord]

Results provided by WolframAlpha

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

Requested by adoniszgks

[wanton bison]

ain't no way I have to evaluate this

[gritty oar]

[wanton bison]

I am gonna make a sandwich

[odd edge]

@wanton bison Has your question been resolved?

[wanton bison]

nvm+

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[drowsy root]

!status

[odd edge]

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

[green elm]

try multiplying and dividing by something that will make the denominator simpler

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@fierce wolf Has your question been resolved?

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[haughty dagger]

Hey. In my country we use derivative(y) = y'
But on english sources, I see it as dy/dx.
So here is my question.
When solving diferential equations of separable variables, how can y' (dy/dx) be split?

[haughty dagger]

There, shouldnt it be y'/y = x

[royal herald]

Same thing

y’ = dy/dx

[haughty dagger]

but on the step below

the y' is gone

the third step

Shouldnt it be 1/y * y'

[haughty dagger]

integral of this

[royal herald]

Yes they separated the dy/dx

So one side gets the dy

And the other side gets the dx

Look in step 2

That’s what they did

[haughty dagger]

I understant that they separated. But then the y' is just gone and we just primitivate the 1/y

and the x

[royal herald]

But look

[haughty dagger]

dont we need to primitivate the y' too

[royal herald]

There’s dy

Or dx

For each integral

If u keep it as y’, where will the dx come from?

[haughty dagger]

I see

So we split it so we can integrate each side

[royal herald]

I mean I think u could keep just y’ and have an ‘implied’ dx but I think the separation of dy and dx is good practice

[haughty dagger]

I think I understand

thanks man

hold on let me send a picture

[royal herald]

Am I wrong abt this @echo ginkgo

[echo ginkgo]

you just integrate both sides wrt x really

there's no implied dx

[royal herald]

Then it’s that dy is implied bc y is a function of x?

[haughty dagger]

Look here at 7

Left is how I usually solve and right is how this is suggesting

left is the correct final answer

[echo ginkgo]

$\int \frac{y'}{y} \dd{x} = \int x \dd{x}$

[clever fjord]

aPlatypus

[echo ginkgo]

and change of variables on LHS

but hey you end up with the same stuff in the end anyway

[royal herald]

Ah I see

[haughty dagger]

On the right side I dont

In the most recent picture

[royal herald]

Well u put y * dy instead of 1/y * dy

U needed to divide both sides by ‘y’

[haughty dagger]

you're right

alright thanks guys

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[lucid siren]

Hi everyone,
I would like to know if this loss of marks is justified in your opinion according to the mark scheme on the RHS.
Please have a look if you know basic linear algebra.
I feel like multiplying the new expression by the matrix again by A might not be necessary as we know that all the vectors are non-zero vectors, so the only linear combination which would satisfy the equation would be the one where c1 and c2 would themselves be 0, but I might be wrong.
Thanks

[lucid siren]

This is the official mark scheme in case anyone wishes to have a look as well

I feel like my reasoning alone should be enough to get full marks, but perhaps I am overlooking something, could someone please enlighten me

[odd edge]

@lucid siren Has your question been resolved?

[lucid siren]

<@&286206848099549185>

Guys I'm sure my question is not that hard Please

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@lucid siren Has your question been resolved?

[odd edge]

@lucid siren Has your question been resolved?

[mystic saffron]

Need help

[odd edge]

@lucid siren Has your question been resolved?

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[undone crypt]

How does the grafic change if is f^-1(x)

[undone crypt]

does it just reflects in the axis y?

[raw quiver]

nope

its the inverese function

do you have thee function itself?

[undone crypt]

No, just the graphic

[raw quiver]

think of 1/f(x)

i am not sure if there is another way to do it, didn't do inverse in a long time

g(x) = 1/f(x)
and try to draw g(x) from how you think it will go

[undone crypt]

Something like this?

[tawdry wagon]

Noooo

[raw quiver]

from what i remember inverese function domain is the image of the OG, and the image eis the domain of the OG

[tawdry wagon]

The inverse of f, f^(-1) is a function such that f^(-1) (f(x))= f(f^(1)(x))=x

[undone crypt]

I think i got it, samething but in diferent axis

[tawdry wagon]

Kinda but not really, i think your getting your terminology wrong

[tawdry wagon]

Yeah, generally if a function is invertible, the inverse is just a reflection across the line f(x)=x

[undone crypt]

thanks

[tawdry wagon]

Np

[undone crypt]

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[raw quiver]

given the series and a1, show that An is enclosed between 0,6.

How do i go around to proving that its enclosed?

I know that if An converges than An is inclosed

[random sigil]

probably by induction

[raw quiver]

do use induction, i first need to see if its monotonic

to know if i have a bottom/top limit

its also a recursive series

[random sigil]

are you asking about monotonic or bounded?

[raw quiver]

i first need to provee that is bounded

[random sigil]

and because a_n is in the bounded area then 6/a_n is greater then 1 and the whole fraction is <6 @raw quiver

[random sigil]

looks fine

[raw quiver]

i think so too, still insecure about induction and recursion

[random sigil]

maybe the wording isn't 100% but the logic itself is fine

wouldnt say קיים 0<a_n<6 more like assume for some n and then prove for n+1

[raw quiver]

induction proving is like this no?

1. base case
2. inductive step, show that there is a N that fullfills the induction,
   3.prove that it also fullfills for N+1

for step 2, the because wee have a base case, we know that atleast one number follows the induction rule

[random sigil]

its more of if it holds true for some n you have to prove its true for some n+1

it works because of the base case. 1->2 and 2->3 and so on

[odd edge]

@raw quiver Has your question been resolved?

[raw quiver]

This is the rest of the question

1. prove that the sereies montonic rising
2. provee that it converges
3. find the limit

[raw quiver]

couldn't explain it like this, but yeah, this is what i sortof understood

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[rapid reef]

The entire inequality is restricting sinx to the first and second quadrant

Take n=1 for example. On the left you get 2pi, so to make sure sinx remains positive, the left side must be 2pi + pi = (2+1)pi

I suppose it should be (2n+1)pi though

Since what they have right now fails for n=0

That would also fail for n = 0

Since you would want 0 < x < pi

2pi n < x < (2n+1)pi is what it should be, I believe

Then for n=0: 2pi(0) < x < (2(0)+1)pi

-> 0 < x < pi

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[violet peak]

what would be the answer

[proven beacon]

did you write the answers in blue pen

[violet peak]

yes

[proven beacon]

It seems good to me

[violet peak]

i used two point

[proven beacon]

Idk maybe use vertices

[violet peak]

which one

second or first

[proven beacon]

For vertices a b c d

[violet peak]

ok

thanks

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[winged turret]

Hi. I need some HELP
The question is
Three numbers are consecutive members of a geometric sequence. Determine those numbers, if their sum is 3/2 and is equal to the sum of their reciprocal values.

[winged turret]

PLEASE ping me when texting

[drowsy root]

you are missing an equation, you have 3 unknowns, so you need 3 equations

[winged turret]

Wait

Which one

b1+b2+b3=1/b1+1/b2+1/b3??

[drowsy root]

the first one is that theyre a geometric sequence

the second one is the sum is 3/2

and the third is that the sum of the reciprocals is 3/2

i dont see the first one

[winged turret]

b1,b2, b3?

Geometric sequence?

[winged turret]

I don't understand what you mean.

I apologise

[drowsy root]

ah wait, nvm i see it

you need b2=a*b1 and b3=a*b2=a^2*b1

[winged turret]

Seperately written?

[drowsy root]

they are part of a geometric sequence, that means, b2 is b1 multiplied by a constant, and so is b3 of b2

[winged turret]

Understandable

[olive sparrow]

A) $b{1}+b{2}+b{3} = \frac{3}{2}$ and
B) $\frac{1}{b{1}} +\frac{1}{b{2}} + \frac{1}{b{3}} = \frac{3}{2}$

[drowsy root]

wait, im a bit confused myself 😅

[clever fjord]

clonesolopros

[winged turret]

Yes

[winged turret]

Take your time

[drowsy root]

and $b_2=q\cdot b_1$ and $b_3 = q\cdot b_2 = q^2\cdot b_1$

[clever fjord]

Flappie

[winged turret]

Yes

Ik that

So i need to determine q

Right

[drowsy root]

yes

and b1

[olive sparrow]

so $A)-B) = \frac{b_1^2-1}{b_1} + \frac{b_2^2-1}{b_2}+\frac{b_3^2-1}{b_3} = 0$

[clever fjord]

clonesolopros

[winged turret]

May i ask how you got -

Minus

[drowsy root]

instead of adding like you did, subtract them instead

[winged turret]

So it's better to substract?

[olive sparrow]

like $b_1-\frac{1}{b_1} = \frac{b_1^2}{b_1}-\frac{1}{b_1}$

[clever fjord]

clonesolopros

[winged turret]

Yeah yeah

Just wait a second i think i got it

[drowsy root]

probably, sinc ethen you get = 0

[olive sparrow]

and well you can factor (b^2-1) into (b+1)(b-1)

so you get this:

$A)-B) = (b_1+1) (b_1-1)b_2 b_3+ (b_2+1) (b_2-1)b_1 b_3+ (b_3+1) (b_3-1)b_1 b_2 =0$

[clever fjord]

clonesolopros

[drowsy root]

i think we made a mistake somewhere

because b1=b2=b3=1 solves it

but that sum is 3, not 3/2

[winged turret]

I use

B1q

For b2

[olive sparrow]

wait when is 1+1+1 = 3/2?

[winged turret]

Are you sure that it's

1

For apl

All

Because we need q

[olive sparrow]

wait they are consecutive of a geometric sequence

[drowsy root]

its not, but if you plug in b1=b2=b3=1 then the (b1-1),(b2-1),(b3-1) terms all become 0

[winged turret]

Yes

[olive sparrow]

thats bc that is not a solution

[olive sparrow]

like for example: 1, 1/4, 1/9?

[drowsy root]

is b1=b2=b3=1 not a geometric sequence?

im pretty sure, all that is required is that b1=/=0 and b2=/=0

[winged turret]

Yk like b2=b1q
Arithmetic is b2=b1+d

[winged turret]

Yeah...i did not did requirements tho

[olive sparrow]

we missed the part where the numbers are of a geometric sequence xd

[winged turret]

So well... it's okay

[olive sparrow]

well I think this is solvable

the only "variables" we dont know are b2 and b3

in terms of b1 for now

[winged turret]

And q

We need q to determine b2, b3

[olive sparrow]

yeah

[olive sparrow]

you know theres a formula for summing, from "a" to "b"th number of a geometric sequence?

[winged turret]

Sn=b1*(1-q^n)/(1-q)

[olive sparrow]

yeah, well q^0 = 1

[winged turret]

Why

[olive sparrow]

its like you start from the 0th number aka start

[winged turret]

Okay

What then

[olive sparrow]

well I'd set: $b_n = b_1 \frac{q^n-q^{0}}{q-1}$

[winged turret]

Hm

q^3?

I got confused

[winged turret]

But that is kinda useles dont you think? We don't have b1 OR q

Is bad thing that from b1+ b1q +b1q² to determine b1

[clever fjord]

clonesolopros

[olive sparrow]

at the end you'll get (1+q+q^2)*b1 = 3/2

[winged turret]

Yeah

Can i go with that way ot

Or?

[olive sparrow]

yeah

[winged turret]

And it will be ok?

[olive sparrow]

I think so

wait we can determine q

just set b1 = 1

[winged turret]

We can do that?!

I am now solving algebra nightmares lmao

[olive sparrow]

well its like a function, you can fix one of the parameters while solving for the other xd

[olive sparrow]

well you can just also use the quadratic formula without setting the value? xd

well lets use: $b_1 (1+q+q^2) = \frac{3}{2}$

[clever fjord]

clonesolopros

[olive sparrow]

then we have a quadratic equation:

[clever fjord]

clonesolopros

[olive sparrow]

and this is the quadratic formula for q:

[clever fjord]

clonesolopros

[odd edge]

@winged turret Has your question been resolved?

[winged turret]

I did it

[olive sparrow]

when $b_1 = 1 \implies q=0.366$

[clever fjord]

clonesolopros

[olive sparrow]

what did you get?

[winged turret]

B1=2, B2= -1, B3=1/2

Q=-1/2

[olive sparrow]

that also works xd

[winged turret]

Thank you. You are very kind.

[olive sparrow]

😄

btw how did you know b1=2?

[winged turret]

I tried b1=1

But q isnt nice number

So i went with 2

[olive sparrow]

oh xD

[winged turret]

Kinda silly ig

[olive sparrow]

classic trial and error lol

[winged turret]

Thank you again AAA

[olive sparrow]

but do you want to know how the 2nd part was done?

[winged turret]

Second part? Where

[olive sparrow]

the 1/b1 + 1/b2 + 1/b3

[winged turret]

Well

1/2-1+2

=

[olive sparrow]

yea ik

[winged turret]

3/2

I mean

Just changing that

Values

Right

[olive sparrow]

I guess its kinda symmetric

[winged turret]

Yeah

3/2=3/2

Idk

[olive sparrow]

its an intresting problem tbf

[winged turret]

Yeah

I hope it doesn't comes on entrance exam lol

Gotta die

[olive sparrow]

lmao

just put random numbers surely it will work

[winged turret]

Okay

[olive sparrow]

at least you tried xD

[winged turret]

Yup!

Thank you

.close

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[fluid tundra]

that's a dubious way of doing this problem

[winged turret]

.open

.reopen

[odd edge]

✅

[winged turret]

Tell me

What is better way?

?

[fluid tundra]

a + ar + ar^2 = 3/2 --> a(1 + r + r^2) = 3/2

1/a + 1/ar + 1/ar^2 = 3/2 --> r^2 + r + 1 = 3/2 * ar^2

substituting one equation into the other equation, a(3/2 * ar^2) = 3/2 --> a^2 = 1/r^2 --> a = ± 1/r

testing a = + 1/r in the first equation, result is r + 1/r = 1/2 which is impossible

so a = - 1/r, leading to r + 1/r = -5/2, implying r = -1/2 and -2 leading to a = 2 and 1/2

[winged turret]

Second line

You multiply with ar^2 right

[fluid tundra]

yes

[winged turret]

Yeah now that is much easier

Thank you

[winged turret]

I understood

. close

.close

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[odd edge]

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[mystic saffron]

This is more me needing help with readin comprehension if anything but how do I interpret "the middle term is twice the product of their square roots"

because I'm confused at the "twice the product" part

5 * x = 5x and twice 5x is 10x.

I just got it.

ty

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[winter breach]

Can somone explain how to describe the region of the cone in cylindrical coordinates?

[winter breach]

This is the answer

[winter breach]

@placid atlas The angle is clear to me.

We're dealing with a cone, which has the base in the form of a circle.

The top of the cone is the apex. The base of the cone is the circle.

You mean the top of the cylinder?

So, in the first case, it's saying 3z/h ≤ r ≤ 3, which I don't understand the reasoning behind.

I can get behind 0 ≤ z ≤ h, it's just saying z is bounded by h and 0.

Yes

Sure

We know that r = 3 when z = h.

Yeah

h = 3/k?

Oh we solve for k

Okay, so if k = 3/h, and r = zk, then r = 3z/h

But how do we come to the conclusion that 3≥r, and r≥3z/h both hold?

3≥r comes from the fact that the radius is not greater than 3

Right?

Could you elaborate?

What's the region outside of the cone here?

Is it the triangle part?

@placid atlas

I'm not sure how you came to the conclusion it's about "the region outside the cone"

[odd edge]

@winter breach Has your question been resolved?

[winter breach]

Alright, thanks for the help.

.close

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[vocal whale]

Why is my way wrong

[wanton bison]

why 2 and not -2

[vocal whale]

That’s my question

[wanton bison]

mb

they explained it

[whole spire]

it's A(-2)=e^-(-2) = -e^2, not -e^(-2)

and don't forget the negative sign of e^(-4) at the end

[vocal whale]

I see

Thanks

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[glad wadi]

I know I have to factor the bottom but I'm not sure what I can do

yeah but I dont know how to factor it

like what factoring method

would i use

yeah

it doesnt factor nicely still

4 ( x^2 + x + 1/2)

it still doesnt factor nicely

[brave flint]

u can try this ques by completing the square and then it become one of basic itegral form

[glad wadi]

I tried but I dont understand it with this problem 😢

theres no good tutorials

i would have to put it in standard form right

first take the 4 out?

so would that be it factored?

yeah I messed up look at that now

[clever fjord]

..,,

[glad wadi]

yes

so I took the 4 out right

ohh so when i pull out that 1/2

okay

ill try agaibn

wait so what did I do wrong

i'm so confused

so you have the equation x^2 + x + 1/2

you move 1/2 to the other side right

[clever fjord]

..,,

[glad wadi]

yes I understand

but I dont know where I went wrong

[clever fjord]

..,,

[glad wadi]

okay yes but how do I get 1/2

like next time

if I have a problem like that

[clever fjord]

..,,

[glad wadi]

okay yes I understand when you foil it out but I'm asking what made me get 1 and what made you get 1/2

thats why I'm confused like what did I do wrong that made me get 1

[clever fjord]

..,,

[glad wadi]

yes I understand that but completing the square is an algorithm

what did I do wrong in the algorithm that gave me 1 instead of 1/2

like where in my work did I go wrong

[clever fjord]

..,,

[glad wadi]

okay

[olive sparrow]

correct