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[midnight plank]

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[leaden seal]

Hi, this is a physics question from the beam and cable lab

[leaden seal]

(hope you dont mind me asking here

my question is: how do we get 0.842 (the angle) where it says costheta at the bottom

[midnight plank]

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[midnight plank]

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[upper nexus]

How would you roughly sketch this diagram?

We are given a triangle ABC such that ∠BAC < 90◦.The point D is on the opposite side of the line AB to C such that |AD| = |BD| and ∠ADB = 90◦.
Similarly, the point E is on the opposite side of AC to B such that |AE| = |CE| and ∠AEC = 90◦. The point X is such that ADXE is a parallelogram.

[upper nexus]

it's just im not sure what "the opposite side" is referring to

like what is the "opposite side of the line AB"

[near cliff]

i think ABC is just an equilateral triangle

the more i look at this question

[upper nexus]

oh?

[shell wigeon]

C and D are on opposite sides of AB

B and E are on opposite sides of AC

I drew the perpendicular bisectors of AB and AC, and to make ADB = AEC = 90º, I just made BAD = CAE = 45º

[upper nexus]

oh thanks so much

[near cliff]

okay this made much more sense than what i was trying to visualise

[shell wigeon]

(I forgot to draw CE but you can imagine the right angles)

Then EX is just parallel to AD, and DX parallel to AE

[upper nexus]

yea

[shell wigeon]

Do you need help with a question about this?

[upper nexus]

well

yea

[shell wigeon]

... are you going to post it?

[upper nexus]

idk if im allowed to

tbh

[shell wigeon]

Why? Is it a test?

[upper nexus]

not exactly, it's just a problem to challenge me

[midnight plank]

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[lament flame]

I am trying my hardest...

[crimson wyvern]

Ok so which types of asymptotes do you feel you don’t understand?

Are you having trouble identifying them or having trouble providing the equation for them?

[crimson wyvern]

So for a vertical asymptote to happen, a function will shoot off to positive or negative infinity (or both) around the asymptote

[lament flame]

Hmmm

I see so...

[lament flame]

I think it is a positive right?

[crimson wyvern]

Positive what? Do you mean positive location, positive infinity?

[lament flame]

Ok positive 4

[crimson wyvern]

Yes! Now, what do you think the equation should be based on this?

[midnight plank]

@lament flame Has your question been resolved?

[lament flame]

X/x+4 +5

I dont know the numerator yet

[crimson wyvern]

ah, that's your function equation, but you just need the equation for the asymptote line

do you know what the general equations are for vertical and horizontal lines?

[lament flame]

The denominator can not be less then the numerator or equal 0

[crimson wyvern]

ok, let me get a visual aid

What is the equation of this blue line?

[lament flame]

Its 1/x-5 right?

[crimson wyvern]

you're trying to find the equation for the red function. You need to find the equation for the blue line

[lament flame]

Ohhhh

The vertical asymptote is 5

[crimson wyvern]

yes, but, it's specifically x=5

because that is the equation for a vertical line at 5 on the x-axis

[lament flame]

Ah....

[crimson wyvern]

yeah you have to specify because if you just write "5", that could mean either a vertical line at x=5 or a horizontal line at y=5

[lament flame]

Ohhhhhh

[crimson wyvern]

ok, so I think you have everything you need to answer this one correctly

[lament flame]

Oh

It said it was wrong.

[crimson wyvern]

what did you answer?

[lament flame]

Oh I missread

I dont know if I need to factor or if I dont and cancle for this one

[lament flame]

Do I cancle the 4x? It doesnt seem factorable

[crimson wyvern]

so, for vertical asymptotes, you need to find all the x-values that make the denominator equal to zero (with the caveat that you need to check and make sure the numerator isn't also equal to zero at this point, as that might do something else)

[lament flame]

Ohhhh

[crimson wyvern]

and because the denominator is linear (and not constant), it has to equal 0 somewhere, and also only passes through 0 once

[lament flame]

Ah

Alright I got x=-2/4 or 1/2

-1/2

[crimson wyvern]

yup

[lament flame]

Then we plug it in to numerator correct?

[crimson wyvern]

yeah

[lament flame]

Ok

So then...

Y=3

[crimson wyvern]

wait

what did you do?

[lament flame]

Plugged it in to numerator

[crimson wyvern]

you need to subtract that 2, not add it

nope my bad

I'm taking real analysis rn and apparently I still can't multiply two negative numbers correctly lmao

[crimson wyvern]

ok so your numerator is not equal to 0, and the denominator is, which means x=-1/2 should be a vertical asymptote

[lament flame]

Ohhhhhhh

[crimson wyvern]

for context, getting 0/0 somewhere on the function leaves a discontinuity, but not an asymptote. I will show an example

[lament flame]

So are there multiple meathods for achieving the asymptote?

Horazontal

[crimson wyvern]

like x/x is equal to 1 everywhere except 0

[lament flame]

Ohhhhh

[crimson wyvern]

but 1/x blows out to infinity because the numerator doesn't keep up with the denominator

that's why you want to check for 0/0

[lament flame]

Apparently there is a horazontal or oblique asymtote

[crimson wyvern]

yes. those are going to be quite different because they happen as you approach infinity or negative infinity on the x-axis instead of at a discontinuity

[lament flame]

What does it mean when they say num and denom both have a DEGREE of 1?

[crimson wyvern]

it is the power x is raised to

so like x+1/x^2+1 would have a numerator of degree 1 and denominator of degree 2

1/x has a numerator of degree 0 and denominator of degree 1

[lament flame]

Ohhhhh

So the coefficent is 4x/4x

So its an oblique then

[crimson wyvern]

nope

[lament flame]

Wha

[crimson wyvern]

well, be careful, because one of those terms should be negative

[lament flame]

Ohhhh

So -4x/4x

Thanks

[crimson wyvern]

so, since you are looking at it as x gets very large, what will happen?

[lament flame]

So wait- its... would it still be an oblique

[crimson wyvern]

ok, so since we are looking at very large x, we know x isn't 0 in this area

which means you can divide out the x

[lament flame]

Oh yeah

[lament flame]

So -4+1/4+2?

[crimson wyvern]

this is after removing the lower degrees

because they become negligible as x becomes very large

[lament flame]

I dont understand what you mean by that

When you say x becomes large

[crimson wyvern]

so take f(x)=x and g(x)=x+1. when x=1, g(x) = 2 and f(x) = 1. so, g(x) is double the height of f(x). Now take x=100. so g(x) = 101 and f(x) = 100. so the percent difference between them is now much less than double. as that x gets even larger, the percent difference "becomes negligible", or mathematically, approaches zero

[lament flame]

Ohhhhhh

[crimson wyvern]

this is why you only need to look at the highest degree. Because the highest degree of a polynomial is going to eventually blow out to infinity faster than all the lower degrees do, it's the only one that matters when you are looking at behavior as x becomes very large or small

(by small, i mean going to negative infinity here, not 0)

[lament flame]

Yeah

So how would I put this in as an equation?

[crimson wyvern]

well, you look at y= -4x/4x, and then just divide it out

[lament flame]

So -1

[crimson wyvern]

yeah

y=-1

[lament flame]

If it is a oblique it wants me to put it in slope intercept form?...

[lament flame]

Oh-

[crimson wyvern]

so it's horizontal, not oblique

you only get oblique if you end up with something in the form of ax^(n+1)/bx^n

[lament flame]

I apologize my head is just... I dunno today

Ok domain would be
(-inf, -1/2)U(-1/2, inf) right?

[crimson wyvern]

yeah should be

[lament flame]

... why is there a parabola here now...

[crimson wyvern]

my guess is the denominator is a quadratic function

since quadratics can pass through 0 twice

[lament flame]

Oh-

Ok uh- i was going to say the va's were -2,2

Hoazontal only has one

[crimson wyvern]

yeah that's correct

alright, I'm calling it for the night

[lament flame]

Oh shoot- thanks dude you helped me a lot

[midnight plank]

@lament flame Has your question been resolved?

[midnight plank]

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[steep ravine]

is this correct?

[steep ravine]

its for FSM finite state machines

[midnight plank]

@steep ravine Has your question been resolved?

[midnight plank]

@steep ravine Has your question been resolved?

[midnight plank]

@steep ravine Has your question been resolved?

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[agile root]

part c

[lyric charm]

c i, c ii, or c iii?

[agile root]

ii

[lyric charm]

p sure you want to do integration by parts

[agile root]

how do i choose u and dv in integration by parts

im always confused by that

[ashen kite]

LIATE

[agile root]

coffee?

[ashen kite]

Nope the rule for choosing u

[agile root]

what is liate

[lyric charm]

you want to make the log(x)^n into log(x)^(n-1) somehow

[lyric charm]

don't worry about that for now

[agile root]

im thinking differentiating but by chain rule we get a new term

[lyric charm]

try it anyway

[agile root]

nln(x)^n-1/x

[subtle zinc]

yes go on

[lyric charm]

paper

and do actually complete the ibp

in full

[agile root]

okay so integration by parts is:
uv - int(vdu)
i guess im gonna choose my u=ln(x)^n
so i get du=nln(x)^n-1/x
dv=x^2 so v=x^3/3
ln(x)^nx^3/3 - 1/2int(x^2nln(x)^n-1)dx

wait let me just write it out

[lyric charm]

write out the version for definite integrals

[lethal path]

also it's 1/3 multiplied by that integral

[lyric charm]

on paper not in discord

[tender pelican]

oh fun fact, ibp can be organized in a good manner using the D-I method

https://youtu.be/2I-_SV8cwsw?si=9vUoN0ZUVrcorXUl

[agile root]

like this

?

[agile root]

i have seen this method but my teacher doesnt allow it

[tender pelican]

https://youtu.be/xFk9cZYhFrw?si=nPt07OJnqY1b8ei4

lmao

[agile root]

it isnt that she doesnt believe in it

[tender pelican]

they are literally the same thing

[agile root]

she wants me to clearly show where im integrating

[lethal path]

the IB exams definitely allows it though

[lethal path]

oh that's a different issue then

[agile root]

is this correct?

[lethal path]

yeah to find v from dv, you better show your steps in the real exam too

[agile root]

my ipad died so um i used google docs

[tender pelican]

I do wonder what she would think of the video, but I learned integration on my own and currently am not up to integration in calculus, so I'll see what happens when I get up to it

[lethal path]

1. it helps to mention ln(e) = 1 and ln(1) = 0

2. it's better to write $\int_1^e \frac{x^3}{3} \frac{n \ln(x)^{n - 1}}{x} \ dx$ for clarity

otherwise, all good

[grand pond]

south

[agile root]

and now this integrand will just be n/3 I(n-1) right

okay i got the result

going back to the central issue

how do i choose u and dv?

[lethal path]

the key insight is: differentiating (ln x)^n is the way to make the power of (n - 1) appear in the integral

so it's a case of knowing what form you want the answer to be

writing it out as $\frac{1}{3} \left(e^3 - n \int_1^e x^2 (\ln x)^{n - 1} \ dx \right)$ helps a lot

[grand pond]

south

[lethal path]

and also, a common trick in papers is that they factor something out

so the 1/3 in front here

[agile root]

i see, so it is sort of like inspecting what i want or which ever term is harder to integrate per se

[lethal path]

yeah, that's the idea behind LIATE

though LIATE is not perfect and it 'fails' in many questions if you apply it in a braindead way

[agile root]

do u know a list of tricks we need for ib aa hl in integration

im so lost, i skipped almost all my calculus classes because i was super sick

[lethal path]

there's actually no tricks

[agile root]

techniques i mean

[lethal path]

IB will only give you integrals by substitution in the form $\int f'(g(x)) g'(x) \ dx$

[grand pond]

south

[lethal path]

otherwise, they will tell you exactly what to substitute

[lethal path]

(this is the reverse chain rule form)

[agile root]

here sub would be u=g(x)?

[lethal path]

I suggest practicing with the formula booklet also

[lethal path]

yep! du = g'(x) dx

[agile root]

oooh

[lethal path]

AAHL doesn't go as deep as you think for integration

[agile root]

which topic is the hardest

[lethal path]

there's also things like, if you want to integrate $\frac{x^2}{9 + x^2}$, that equals $\frac{9 + x^2}{9 + x^2} - \frac{9}{9 + x^2}$

[grand pond]

south

[lethal path]

it's subjective

do you find 3D vectors hard?

or how about complex numbers?

[agile root]

i find combinatorics very very hard

to a point where i have given up on that

[lethal path]

to be honest, I don't think it's worth your energy

[agile root]

WHAT

how did u get the insight to add and subtract 9

[lethal path]

if you practice you'll know

[agile root]

could we directly complete the square?

[lethal path]

that doesn't actually work, cause now you'll have -6x in the denominator

and that's even nastier

the key is to recognise the integral of 1/(x^2 + a^2) as a standard integral in your formula booklet

[agile root]

wow

clearly i need to practice a lot more

i think i understand this now, tysm ann, south, and a message!

.close

[midnight plank]

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[coral belfry]

,tex
hello, suppose we have the sequence $a_n = \frac{3n}{2n +1} $
\
and i have shown that
\
$ \frac{3}{2} \geq a_n \geq \frac{3}{2} + \frac{a}{n} $
\
\
where $a> \frac{-15}{4} $ but i havent actually proven anything more than that. is it enough to say that our sequence converges to 3/2? it seems rlly wrong to me bc im not sure if such "a" exists

[grand pond]

fijokazż

[umbral timber]

try taking the limit as n tends to infinity

[runic hamlet]

presumably you arent allowed to write a_n as 3/(2+1/n) and then move the limit inside the fraction?

[coral belfry]

yeah

btw the sequence is 3n/2n+5 my bad

[umbral timber]

then by the squeeze theorem you basically have a proof

[coral belfry]

yeah but i havent defined a so well

[runic hamlet]

well just pick any a so that the inequality works

a=-3 or something

[coral belfry]

hmm

oh wait it would just work?

[runic hamlet]

assuming the inequality you showed is correct, yes

[coral belfry]

okeoke ill continue it like that and see

thanks

.solved

[midnight plank]

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[cosmic jolt]

could I get some help please

[pearl hull]

which question specifically?

[cosmic jolt]

both

please

[torn compass]

isnt that as simple as resolution of forces

[cosmic jolt]

not sure I only js started mechanics

[torn compass]

u know what resolution of forces is tbh?

[cosmic jolt]

yh but I dont think the question is that simple

[torn compass]

it is tbh

unless if im missing smthing

[cosmic jolt]

bruh

[torn compass]

(idk if showing the image is giving the answer )

[cosmic jolt]

i dont get it icl

[torn compass]

which part

[cosmic jolt]

whats the mgcos theta part then

if mg is the weight

[torn compass]

breaking down a force into its x and y components , or resolution of forces

mg costheta would be the y component of the force

[cosmic jolt]

Yeah Im still new to this

[torn compass]

with respect to the block

[cosmic jolt]

.close

[midnight plank]

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[high wedge]

I have this, but I am struggling to make it to where I can have a palindrome in a palindrome, while maintaining the multiple of 3 I need

the goal is to make a context free grammar from the context free language provided

[crimson wyvern]

What course is this for?

[frozen talon]

I have no idea what this means

[high wedge]

theory of computation

[frozen talon]

Are you counting palindromes or something?

[high wedge]

or like discrete math 2

I think

[crimson wyvern]

It seems like a combinatorics problem yeah

[high wedge]

a grammar is basically a set of rules you can use to derive a string

so like

if I say
S -> A
A -> 00

I can do S -> A -> 00, which is just S => 00

[crimson wyvern]

I feel like this is something more in the realm of computer science than mathematics maybe? It seems very applied

[high wedge]

I need to make a grammar that accepts strings comprised of 0s and 1s which are palindromes, and are divisible by 3

[crimson wyvern]

What does divisible by 3 mean in this case?

[frozen talon]

what do you mean by "derive a string"

[high wedge]

so like 000 has a length of 3

[frozen talon]

the length of the string is a multiple of 3

[high wedge]

111 has a length of 3

111000 has a length of 6 but 6 is divisble by 3

etc.

[high wedge]

derive meaning can get to it via the starting rule and any combination of other rules in the grammar

[crimson wyvern]

So theoretically there is no upper limit on length?

[high wedge]

nope

infinite length is acceptable

[crimson wyvern]

Well that causes problems

<@&268886789983436800>

Scam

[frozen talon]

it doesn't if your goal isn't just counting palindromes, which it isn't

[high wedge]

so as a concrete example, the grammar for even length palindromes looks like this

S -> 0S0 or 1S1 or epsilon (empty string)

[frozen talon]

oh so this is a recursion thing

[high wedge]

yeah

[crimson wyvern]

What is the goal here exactly?

[high wedge]

create a set of rules such that by going from rule to rule you end up with a palindrome with a length div by 3

and you must start with the start rule

[frozen talon]

I think having two confused people asking questions is counterproductive

[crimson wyvern]

Yeah I’m going to have to leave here

[high wedge]

this problem feels hard to me because it's induction

[frozen talon]

you don't care about changing the alphabet from {0, 1} right?

[high wedge]

nope

a and b is fine

as long as it's binary

[frozen talon]

well, a palindrome is entirely determined by the first half of the string, as anything you append to the left must be appended to the right

• the middle character, if it exists

[high wedge]

yeah and in this case we will have a middle in a lot of cases

[frozen talon]

my first thought was to say something like S -> 000S000, 001S100, ..., 111S111

but that excludes cases where appending only three digits (disturbing parity, therefore) still results in a palindrome

e.g. "111" (odd) || "111" (odd) = "111111" (even)

[high wedge]

yeah this handles all cases of length 3 or less but

no more

what you had makes sense, but then you need S -> something

[frozen talon]

are grammars meant to let you derive satisfactory words from other satisfactory words?

[high wedge]

if you want it to represent a particular language then yes

like in this case, it must always end up in a satisfactory state eventually or else the grammar doesn't represent the desired language

but generally speaking, a grammar just needs to follow certain rules, satisfactoriness can only come from some constraint being set I suppose

[frozen talon]

ok, so grammars are not necessarily recursive, because this example generates palindromes of length 3

[high wedge]

yup

it can be but doesn't have to be

it's made to be expressive

[frozen talon]

is there a grammar-y way of expressing the reversal of a given string?

[high wedge]

I don't know, but that would probably help solve this 😹

[frozen talon]

can you access characters in a string in this setting?

and just to be transparent, I've actually never done this before, I'm not feigning this ignorance lol

[high wedge]

access wdym

[frozen talon]

maybe it's not useful. I was wondering if you're able to reference a character in a specific index of a given string

[high wedge]

no indexing

[frozen talon]

ok, sure

[high wedge]

hmm

so we do need recursion but

I don't see how to get that recursion while maintaining the divisibility by 3, for even length it was "easy"

here there is some extra conceptual leap

I had the idea of an intermediate variable Y, which can then go back to S

maybe I got it @frozen talon

[frozen talon]

I think that looks fair, cool

[high wedge]

yeah now I need to test it though

like 100010001 should be accepted as a palindrome right?

length of 9 and is same forwards / backwards

well anyways, thanks for the help, I do think it looks good now, I still don't really get it though lol 😹

.close

[midnight plank]

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[twilit field]

Unsure of where to even start, other than the norm is non-zero everywherre but 0

[spiral rock]

Consider two orthonormal basis for V w.r.t. to each of the inner products

[twilit field]

right

[visual tiger]

You only need to consider the orthonormal basis for the second inner product

also, weird that the exercise needs norms corresponding to inner products

why not just norms

anyways

[spiral rock]

Is it true for general norms?

[visual tiger]

I thought it was widely known

[twilit field]

this is from LADR 😭

in the chapter on inner products

[visual tiger]

Let's oblige then ig

[spiral rock]

I have used something like this once maybe

In ANT

[visual tiger]

(just know you could have shown this w/out requiring inner product norms)

let's leave this as a bonus exercise

[twilit field]

why is this

[visual tiger]

the answer lies in the proof itself

take v in V

and decompose it wrt said basis

wait lemme just check

[twilit field]

sure, but computing the norm from that is near impossible

[visual tiger]

you're not gonna compute the norm, you're gonna bound it

(hint: how do you bound ||u+v+...||?)

[twilit field]

Okay, I'll solve it in a norm-linear space then

[twilit field]

triangle inequality

quick sanity check , does. the concept of orthonormal basis make sense in a norm linear space

or something similar

[visual tiger]

norm-linear space?

[twilit field]

normed space

[visual tiger]

do you mean normed vector space

well no you need an inner product

[twilit field]

${\norm{\sum_{i=1}^{n} a_i e_i}}1≤ \sum{i=1}^{n} \abs{a_i} {\norm{e_i}}_1$

[grand pond]

wai

[twilit field]

does this work?

[visual tiger]

yes

[twilit field]

now under the same basis I decompose $v$ under the other norm

[grand pond]

wai

[visual tiger]

this will have no use

you're trying to show ||v||_2 is an upper bound

[twilit field]

cv_2

so it should work, right

[visual tiger]

sure but related to ||v||_2

by a constant factor

so it's useless to upper bound something that's already supposed to be an upper bound

imagine i want to show a <= b, and I show a <= c, b <= d

Showing c <= d will have no consequence on the proof

[twilit field]

got it

[visual tiger]

Here, the trick is to show that each term of the sum is smaller than (constant)*||v||_2

then the sum will be smaller than n * ...

[twilit field]

I was able to prove something similar I think
Let $e_1,e_2,\dots, e_n$ form a basis .
Then ${\norm{v}}1≤ c \sum{i=1}^{n}{\norm {a_ie_i}}_2$

[grand pond]

wai

[twilit field]

would this help us in anyway

[visual tiger]

it would

how did you prove that

[twilit field]

Okay, so here I'm assuming norms satisfy |v|=0 iff v=0

$\frac{{\norm{e_1}}_1 }{{\norm{e_1}}_2} = c$ (say) . We then chose a number bigger than $c$, say $c_1$. That then gives us ${\norm{e_1}}_1≤ c_1 {\norm{e_2}}_2$

[grand pond]

wai

[twilit field]

repeating and then replacing each c_1 ,with max of all c_i, we get the inequality

[visual tiger]

ok good

[twilit field]

now what we could do is, and I have an idea

[visual tiger]

So now, apply it to our problem, where e_1,...,e_n is specifically an orthonormal basis for <.,.>_2

[twilit field]

I thought we could directly do it for normed spaces?

[visual tiger]

we could if you want to

[twilit field]

I'd like to

[visual tiger]

ah, I guess you haven't seen compacity and borel theorem

[twilit field]

no

[visual tiger]

ok so unfortunately I think you need compacity to understand why it works for every norm

So here the proof we're gonna use still requires ||.||_2 to be connected to an inner product

but notice we haven't used that ||.||_1 is linked to an inner product

[twilit field]

I think I have a way around that

We know that $c \norm{ \sum_{i=1}^{n} a_i e_i }2≤ c \sum{i=1}^{n} \norm{a_i e_i}$.

[grand pond]

wai

[twilit field]

okay, let's do this, I guess the problem here will be getting a constant multiple

[visual tiger]

I know what you're going for, you forgot what I said above

[visual tiger]

.

[visual tiger]

ok

so now we have this, let's show $|a_ie_i|_2 \leq |v|_2$

[grand pond]

Raphaelisius Maximus MMIII

[visual tiger]

this way there's an "n" added to the constant, and we're done

any ideas

[twilit field]

this feels off, doesn';t this violate the triangle inequality

[visual tiger]

?

[twilit field]

I feel what you want me to prove violates the triangle inequality

[visual tiger]

How so?

What triangle inequality would be incompatible with what I'm asking you to prove

[visual tiger]

ok, but applied to what?

who would be "u" and "v"

[twilit field]

$\sum_{i=1}^{n-1} a_i e_i =v; a_ne_n=u$

[grand pond]

wai

[visual tiger]

ok?

[twilit field]

I feel the inequality has to be reversed

[visual tiger]

????

[twilit field]

here

[visual tiger]

nope

[visual tiger]

also, this would give us

[visual tiger]

keeping those notations

$|v|2 \leq |\sum{i=1}a_ie_i|_2 + |a_ne_n|_2$

[grand pond]

Raphaelisius Maximus MMIII

[twilit field]

okay, got it

sorry

oops

I kept inserting a summation in my head

[visual tiger]

yeah no there wasn't any

xd

it was just for one i

[twilit field]

no, sorry

[twilit field]

This is true as we're assuming the basis is orthonormal

[visual tiger]

this the idea to it

but prove it

[twilit field]

${\norm{a_ie_i}^2}2≤\sum{i=1}^{n}{ \norm{a_ie_i}^2}_2 = {\norm{v}^2} _2$

[grand pond]

wai

[visual tiger]

there we go

[twilit field]

ah cool

[visual tiger]

maybe don't use the same i for the particular term and the index of the sum

[twilit field]

so we stitch together the bit from the first part and this

[visual tiger]

now stitch this

[visual tiger]

with this

[twilit field]

We can select $e_i$ to be a basis of $V$, such that it's orthonormal under $\langle.,. \rangle_2$.
\ It then directly follows $\norm{v}1≤c \sum{i=1}^{n}{\norm {a_ie_i}}_2$. Further $ \norm{a_ie_i}_2≤ \norm{v}_2$. It then follows $ \norm{v}1≤c \sum{i=1}^{n}{\norm {a_ie_i}}_2≤nc \norm{v}_2$.

[grand pond]

wai

[twilit field]

does this work

[runic hamlet]

whats c

[twilit field]

a positive constant

[runic hamlet]

but what constant

[runic hamlet]

is this supposed to be a full proof or no?

[twilit field]

just the way to stitch together the two results proven here

so no

I can write a proper proof if needed though

We select ${e_1,e_2,\dots,e_n}$ to be an orthonormal basis under $\langle . , \rangle_2$. Under this basis $v= \sum_{i=1}^{n} a_i e_i$. Then there exists $c_1>0: \norm{v}1≤ c \sum{i=1}^{n} \norm{a_ie_i}_2$. Further $\norm{a_ie_i}_2 ≤ \norm{v}2$ so $\sum{i=1}^{n} \norm{a_ie_i}2≤ n \norm{v_n}$. It thus follows $c \sum{i=}^{n} \norm{a_ie_i}_2≤cn \norm{v_n}$. So $\norm{v}_1≤ nc \norm{v}_2$

[grand pond]

wai

[runic hamlet]

why does c exist

does its value depend on v?

[twilit field]

I've shown it does earlier

no

[runic hamlet]

then what is the point of the message you just wrote

I dont understand it

its not a proper proof

so what is it for

[twilit field]

this is the question

[twilit field]

It was a reply to @visual tiger when he asked me to stitch the two parts together

[runic hamlet]

but you stitched the results together here

[runic hamlet]

so what is this for

[twilit field]

You asked for a more comprehensive proof

[runic hamlet]

so this was supposed to be a proper proof?

[twilit field]

Isn't it comprehensive enough if I've proven all the required results previously

[runic hamlet]

what is the point of the message summarizing the previous results if I still have to read through this chat to find out those previous results

what is the point of that

either give me a proper full proof or just give me nothing

[twilit field]

Okay, will do

This is a mess

( typing it out)

I'll do it tomorrow, it's nearly 12 here

sorry

.close

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[wild wren]

hello, i have several questions from my textbook that i dont get quite understand, theyre word problems that are related to finding the derivative of when two functions/variables are multiplied or divided (idk correct terms since im translating everything im saying) like if f(x)= x^2 * 3x, the derivative would be (2x)(3x) + (x^2)(3)

anyway i hope i was able to clarify the lesson in my textbook, if it isnt too much, ill be sending 5 questions, i hope i dont need to send them one at a time, its fine if you can answer only 1 and then guide me with the rest

i downloaded the english version of my book to send these

[gleaming latch]

what are you stuck on for 25

[wild wren]

i cant understand what he wants me to decipher exactly

i understand 27 and 28 and im aware of how theyre solved

29 and 30 im lost but i know its something to do with derivative of something thats being divided

[wild wren]

like is it really that straightforward?

[gleaming latch]

yes theres nothing you have to do yourself they lay out all the steps

what is P'(t)?

[wild wren]

0.03 p(t)

[gleaming latch]

ok cool now find R'(t)

[wild wren]

im in a "memorized not understood" kinda mindset with that tho

[wild wren]

which is (q'(t)) (p(t)) + (q(t))(p'(t))

howd he find q'(t) and p'(t) seperately tho

and it was the percentage times the original function

is that a standard or is it exclusive to the question

[gleaming latch]

huh? they said find R'(t) which is what you said and you know what Q'(t) and P'(t) is

just plug it in

[wild wren]

alr then

[quasi goblet]

just replace q'&p' by their values from the last step

[gleaming latch]

idk man its written out in the question

dont overthink it

[wild wren]

29 and 30 are the real kicker for me

[gleaming latch]

we'll get to those

[gleaming latch]

memorised what

product rule?

[wild wren]

nvm im good with 25 to 28

its called product rule?

good to know

[gleaming latch]

R(t) = q(t) * p(t)
R'(t) = (q'(t)) (p(t)) + (q(t))(p'(t))

[wild wren]

whats the division version called?

[gleaming latch]

quotient rule

[wild wren]

.close

[midnight plank]

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[toxic nova]

I've been teaching myself linear algebra using Gemini and did a problem on paper, just want to know if I did my work properly. Here's the problem:

[toxic nova]

and my work:

[lavish venture]

yea

[toxic nova]

it's all good?

lets gooooo

thanks m8

.close

[midnight plank]

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[rigid turret]

Guys why am I getting negative time

[rigid turret]

In case those help

[agile root]

i believe it means that the event happened before your origin time

[toxic nova]

may i recommend using a pencil instead of a pen

[agile root]

your solution strictly decreases so there is no way to get it since the initial condition is P(0)=8

[midnight plank]

@rigid turret Has your question been resolved?

[midnight plank]

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[rigid turret]

wait

[rigid turret]

lol i probably should yea

[rigid turret]

so is my solution wrong ? I mean I'd assume that over time the disease spreads

[agile root]

your solution doesnt look wrong

your solution just means that if you went back in time by 0.159 days, you would have 100 cases

[rigid turret]

yea ill leave it i guess

my teacher sometimes makes bad questions

not sure why its the disease is decreasing but wtvv

[midnight plank]

@rigid turret Has your question been resolved?

[midnight plank]

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[last slate]

hi i need help

[last slate]

can someone explain this to me

i don’t understand how the strings work with this one

[cerulean oyster]

You know how R^3 works?

[cerulean oyster]

If you assume R^3 is a 3-tuple of 3 real numbers
And that R is the set of all real numbers, which I wont define cause thats complicating things too much

Then R^3 is a vector (a,b,c)

Which are elements from a set " {something something} "

well

If our set becomes {0,1}

Then {0,1}^3 is the set of vectors of 3 elements either 1 or 0

Which, from the computation side, is basically a string of bits

From there, x \in {0,1}^3 means that x is a 3-bit string
And x0 is a kind of notation abuse

But for this case, we assume x0 implies adding a 0 as the Least significant Bit (the rightmost)

[last slate]

yes

R x R x R

[last slate]

i dont get it bro

[cerulean oyster]

R is a set, right?

{0,1} is too.

A set to the square/cube/nth power is shorthand for the sets of vectors constructed from the original set.

$\mathbb{R}^3$ has $(a,b,c)$ for $a,b,c\in\mathbb{R}$

[grand pond]

∫ᴄ 𝐅·𝑑𝑟 = ∬ʀ ∇⨯𝐅 𝑑𝐴

[cerulean oyster]

${0,1}^3$ has $(a,b,c)$ for $a,b,c$ being an element from ${0,1}$

[grand pond]

∫ᴄ 𝐅·𝑑𝑟 = ∬ʀ ∇⨯𝐅 𝑑𝐴

[cerulean oyster]

So, either 0 or 1

R^3 has infinite elements since R has infinite elements too
{0,1} and its 3-tuples dont have infinite elements though

[midnight plank]

@last slate Has your question been resolved?

[midnight plank]

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[last slate]

but R is a set of lists

[last slate]

those bits are put in witout paranthesis

[cerulean oyster]

Nope, R is a set of numbers

{0,1} is a set of numbers too

R^3 is a set of Real Valued Vectors
{0,1}^is a set of 0 or 1 Valued Vectors.

[worldly geyser]

hloo

[cerulean oyster]

!occupied

[midnight plank]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[worldly geyser]

messi or ronaldo

[cerulean oyster]

Shut the hell up and go to #discussion 🥀

[worldly geyser]

did i make you angry

[scenic wyvern]

welcome to the server! please head to #discussion or #chill to chat.

[last slate]

set of lists of numbers..no??

[cerulean oyster]

By R i mean

$\mathbb{R}$

[grand pond]

∫ᴄ 𝐅·𝑑𝑟 = ∬ʀ ∇⨯𝐅 𝑑𝐴

[cerulean oyster]

Set of Real Numbers

[last slate]

yes but like

[cerulean oyster]

No, R is defined as a set of numbers

[last slate]

what do you call numbers in paranthesis

[cerulean oyster]

Set

[last slate]

(a,b)

like this

[cerulean oyster]

oh, thats a vector

[oak kelp]

you're looking for the term "ordered pairs".

[last slate]

Sure?

[oak kelp]

or "ordered tuples" in general.

[last slate]

yes

but also

[cerulean oyster]

Yes, a vector is synonym for ordered tuple.

[subtle blaze]

Does (a, b) mean the open interval

[subtle blaze]

Well not really but sure

[cerulean oyster]

okay yeah, but not in this case

[subtle blaze]

I’m asking lnrd

[last slate]

$$ R^3 = R x R x R = { (r1,r2,r3) : r1 ∈ R , r2 ∈ R , r3 ∈ R }

{0,1}^2 = { ? } $$

[grand pond]

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

[last slate]

bruh

anyway

[subtle blaze]

{(0, 0), (0, 1), (1, 0), (1, 1)}

[lyric charm]

badtex

if you want a cross for multiplication it's \times

[last slate]

how do you write that in definition terms

[lyric charm]

subscripts are _

actual braces are \{ and \}

the set membership symbol is \in

[subtle blaze]

${0, 1}^2 = {(a, b): a \in {0, 1}, b\in {0, 1}}$

[grand pond]

frosst

[lyric charm]

$\mathbb{R}^3 = \mathbb{R} \times \mathbb{R} \times \mathbb{R} = \{ (r_1, r_2, r_3) : r_1 \in \mathbb{R}, r_2 \in \mathbb{R}, r_3 \in \mathbb{R} \}$

[grand pond]

Ann

[last slate]

oh

[cerulean oyster]

Just in case

The big R is shorthand for a set

[lyric charm]

btw whats the original q

[subtle blaze]

It’s exactly what you would’ve thought it was

I didn’t do anything fancy

I just wrote exactly what you did but replaced some of the stuff with the new problem

[last slate]

i see

[cerulean oyster]

which cant be truly listed since there are infinite elements in it
meanwhile {0,1} is also a set, but with finite elements

[lyric charm]

infinitely many* and finitely many*

still tho uh

[subtle blaze]

One could imagine this if one wanted to

[subtle blaze]

Every real number is a list of integers in its decimal expansion

[lyric charm]

is this the original q?

[subtle blaze]

So technically you could say that R is a set of lists of integers

[last slate]

Are the elements (a,b) a random combination ?

from A and B

[cerulean oyster]

Nope

[subtle blaze]

You’d need to define what random means

[last slate]

letting A and B be sets

[lyric charm]

where are we at, actually. im a little confused

[subtle blaze]

(a, b) is used to denote any arbitrary element of A x B

[cerulean oyster]

Lets go over, do you understand what {0,1} is?

[lyric charm]

like i walked in to see some badtex that i fixed but what is the actual question you're asking or thing you're looking at right now specifically @last slate

[last slate]

im just trying to figure out definitions

[lyric charm]

of what

[last slate]

Initially it was with strings

[lyric charm]

what object are you looking at rn that you're trying to figure out the defn of

[last slate]

{0,1}^2

like how is the definition put in parenthesis when there wasnt any

[cerulean oyster]

(a,b) is an element from this set

[lyric charm]

{0,1}^2 is "formally" {(0,0), (0,1), (1,0), (1,1)} but commonly written as {00, 01, 10, 11} when talking about bitstrings

[subtle blaze]

What do you mean by parenthesis

Cos {} and () are different

[lyric charm]

parentheses are not some kind of magical beast with a life of their own btw

[cerulean oyster]

LMAO

[lyric charm]

we humans design our own notation

let's maybe step back a little; do you understand what a cartesian product is? @last slate

[last slate]

yes

[lyric charm]

ok

[last slate]

ill send the definition

[lyric charm]

then

{0,1}^2 is "formally" {(0,0), (0,1), (1,0), (1,1)}
do you also understand this?

[last slate]

look at the set on the right

We have ordered pairs

[lyric charm]

sure do.

[last slate]

But there isnt in the set on the right

[lyric charm]

ok, so your hangup is on how to quote-unquote "go from"
{(0,0), (0,1), (1,0), (1,1)}
to
{00, 01, 10, 11}

do i understand you correctly yes or no

[last slate]

Or rather why dont we have the second one

[cerulean oyster]

?

[lyric charm]

query unanswered

[last slate]

The paranthesis on the right depend on whah

what*

[lyric charm]

query unanswered

[last slate]

Uh

[last slate]

not really

im just hung up on how the definition is related to the product of the 2 sets

[last slate]

yeah

like different parts of the definition

[lyric charm]

alright,

"the definition". send it here please.

do you mean A × B = {(a,b) : a ∈ A, b ∈ B}?

[last slate]

yes

that

does this apply for every cartesian product of any 2 sets ever

[cerulean oyster]

The cartesian product of A and B A × B is a set constructed from two others.
Where all its elements are created from taking one elements from A, and one from B

All the elements from the cartesian product are equal to all the combinations possible.

[last slate]

yeah this is what i was asking

i was told no

[lyric charm]

by whom and where and when

do you remember the exact things that were said

[last slate]

.

[lyric charm]

you said "random"

they are not random

the elements of A × B are ALL POSSIBLE combinations of a member of A with a member of B

[cerulean oyster]

Lets say we have two sets A= {1,2,3} B= {4,5,6}
The elements from AxB are:

A×B= {(1,4), (1,5), (1,6), (2,4), (2,5), (2,6), (3,4), (3,5), (3,6)}

They simply cover all possibilities for mixing

[cerulean oyster]

with A = {0,1}, B = {0,1}
We end up with this.

And we use {0,1}^2 as shorthand for the cartesian product of that.

[last slate]

lets say we have R^3 = { (a,b,c) }

this means that (a,b,c) ∈ R^3

right

[cerulean oyster]

Yes

[cerulean oyster]

To be honest this is a bit of misnomer for the definition, but yeah, lets suppose

[last slate]

so why isnt R^3 x R^3 = { ( (a1,b1,c1) , (a2,b2,c2) … ) }

because (a,b,c) is an element

[cerulean oyster]

R^3 X R^3 has elements* ((a1,b1,c1) , (a2,b2,c2) )

[lyric charm]

you have an extra pair of brackets

[cerulean oyster]

But for ease we define it as (a,b,c,d,e,f)

[lyric charm]

however if you even need to treat R^3 × R^3 as a thing it is easiest to pretend it is the same as just R^6

it's kinda... not that deep ig

[last slate]

but thats not how its defined tho right

they do { (a1,b1) , (a2,b2) }

[last slate]

Because ur supposed to put the elements in the brackets and (a,b,c) is an element

[lyric charm]

are you just trying to think about this shit by yourself or are you trying to solve a problem with it

[last slate]

I don’t understand what ur asking

im tryna understand the definition

[lyric charm]

are you solving a homework question rn

[last slate]

no

[lyric charm]

ok so this is just idle musings

[last slate]

i am doing HW but not regarding thr R^3 question

[lyric charm]

so basically your question is "why doesn't R^3 × R^3 consist of fuckers that look like ((x1,y1,z1),(x2,y2,z2)) ie ordered pairs in which both coordinates are themselves 3-tuples?"

do i NOW understand you correctly?

answer yes/no only

[last slate]

Yes

[lyric charm]

right

ok well the answer to that is

formally it does, but we usually "flatten" that sorta nested tuple into just a basic 6-tuple for convenience

[last slate]

what does that mean?

[lyric charm]

it means we strip off all those layers of brackets and put the same six numbers in the same order into one big tuple

it means we pretend R^3×R^3 is just R^6

the difference between these sets is rather technical, bureaucratic and just doesn't matter 99.9% of the time

[last slate]

and does this thing exist in other sets

this kind of manipulation

at this point i cant trust the definition anymore or them

[sharp mauve]

It looks complicated

[last slate]

Lowkey

[cerulean oyster]

You can probably understand that R^2 x R^2 -> ((a,b) , (c,d) )
is obviously different in structure from R^4 -> (a,b,c,d)

[last slate]

right

[cerulean oyster]

And this will sound overcomplicated, but the connection between the two is a function.

But since those two structures, for us humans, seem so similar

We just consider them to be equal for practical cases.

[lyric charm]

generally A^n × A^k is identified with A^(n+k) in the same way

[cerulean oyster]

Its just an ordered list of 4 elements, it generally doesnt matter how we "bunch them up"

[cerulean oyster]

But again, the R^2 X R^2 -> R^4 isnt really special, and there are arguably more interesting ways to "re-structure" it.

[last slate]

Am i good with exactly following the definition

and taking the elements as they are

@cerulean oyster

[cerulean oyster]

Yes, thats the definition of Cartesian Product.

[midnight plank]

@last slate Has your question been resolved?

[midnight plank]

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[inland patio]

Let $(f_n)$ be a sequence of measurable, real-valued functions. Does it make sense to ask if $n\mapsto f_n(x)$ is measurable?\

Context; I have a sequence of exponentially distributed variables $U_1,U_2,\ldots$ and positive integer-valued random variable $N$. Then my book defines $U_i'=U_{N+i}$ and claims $U_i'$ is a random variable.

[grand pond]

psie

[midnight plank]

@inland patio Has your question been resolved?

[midnight plank]

@inland patio Has your question been resolved?

[spiral rock]

Ui' is a map from your probability space to R, not a map from N to R

[inland patio]

yes, but it is a composition of maps. Do you understand the maps it is composed of?

[spiral rock]

$\omega \in \Omega \mapsto (U_{N(\omega) + i})(\omega)$

[grand pond]

ExpertEsquieESQUIE

[inland patio]

Hmm, ok. That's the full map. Perhaps it is not a composition?

[spiral rock]

Its a composition, but you shouldn't view it as such

[inland patio]

Ok. I'm just curious how one would verify that this map is a random variable.

[spiral rock]

The inverse image of every measurable set in R will It like a big union and intersection of sets, which are measursble in omega

Something like based on where N(w) is fixed

Try it as an exercise

[inland patio]

Hmm, could you please elaborate a bit more? The $N$ in my original post is actually the Poisson process $(N_t){t\geq0}$ and I'm reading a proof of the fact that if $t>0$, and we set for every $r\geq0$, $N_r^{(t)}=N{t+r}-N_t$, then the collection $(N_r^{(t)}){r\geq0}$ is also a Poisson process. The author introduces $U_i'=U{N_t+i}$ and claims they are random variables, but without proof.

[grand pond]

psie

[inland patio]

I've been stuck for quite some time on this.

[spiral rock]

$$(U_{N+i})^{-1}(A) = \bigcup_{n \in \bN} \left ( (U_{N+i})^{-1}(A) \cap N^{-1}(n) \right)$$
$$ = \bigcup_{n \in \bN} \left ( (U_{n+i})^{-1}(A) \cap N^{-1}(n) \right)$$

[grand pond]

ExpertEsquieESQUIE

[inland patio]

Interesting, thank you very much. 👍 I have to carefully think about your equalities.

[inland patio]

By the way, we assume $\mathbb{N}$ is equipped with the power set, right?

[grand pond]

psie

[spiral rock]

Like the sigma algebra?

[inland patio]

Yes.

[spiral rock]

Well its just the lebesgue measure

And the borel sigma algebra

Which includes P(N)

[inland patio]

Ah, ok. 👍

[spiral rock]

N is a map from Omega to R, where R has the lebesgue measure

Nice

[inland patio]

Thanks a lot for the help! I'm closing here.

.close

[midnight plank]

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[coarse ledge]

given an ABC triangle, knowing that vct AB = (4, -2, 1); vct AC = (-1, -4, 0) and centriod G (-1, 1, 2/3). D(x; y ;z) is any points, prove that z(D) = 2

[coarse ledge]

I tried to use the centriod

the uhm..

since D is any point

vct DA + vct DB + vct DC = 3 vect DG

so that will be

A - D + B - D + C - D = 3G - 3D

or, A + B + C - 3D = 3G - 3D

but..

A + B + C - 3D + 3D = 3G

and because of that, there will be no D

but im finding D

can you all tell me what did i do wrong?

or.. teach me a way to find the z of point D?

[late rover]

Damn wth

Do you have a picture of the question

[coarse ledge]

uh uhh i do but its not in English

sorry for the horrible translation

[late rover]

Sure whatever language it is

Maybe someone will know

[coarse ledge]

okay

then ill send the pic of the question here i gues

[late rover]

,rccw

[late rover]

okay I have no idea what it says

[civic gazelle]

???

where is D???

go tell your teacher to fix the problem

[midnight plank]

@coarse ledge Has your question been resolved?

[coarse ledge]

D is any point

in the plane i think

[civic gazelle]

if D is any point then its z coordinate can be anything

[coarse ledge]

oh..

[coarse ledge]

.close

[midnight plank]

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[civic gazelle]

lạy

[midnight plank]

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[wind oxide]

in this question, why cant i simply equate pressures a depth of 36cm from the top of the U-Tube?

why must I equate the pressures at the interface?

[west iron]

Because the pressures at the interface need to be the same, or the interface will move

The pressures at a depth of 36cm on each side are not the same, so you can't equate them

[wind oxide]

o

i see

ty dreyuk

[west iron]

np

[wind oxide]

.close

[midnight plank]

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[gaunt nimbus]

.close

<@&268886789983436800>

[midnight plank]

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[scenic wyvern]

<@&268886789983436800>

[midnight plank]

Channel closed

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[blissful pier]

<@&268886789983436800>

[scenic wyvern]

oh that was super fast

[next rock]

bam

[blissful pier]

⚡

[dusk furnace]

.close (oh, already did)

[blissful pier]

.

[onyx tide]

Why ar there like 10 pings in every channel

In my too

[dusk furnace]

A spammer hit a lot of them.

[scenic wyvern]

cuz shit happens sometimes and we need help with those

[onyx tide]

Can you show what they send

[dusk furnace]

No. A series of images that look like they were attempting to recruit victims for an advance-fee scam.

[scenic wyvern]

in short, stuff irrelevant to the help channels

[spiral rock]

Something 2222?

[dusk furnace]

I'm not sure what 2222 means in this context.

[spiral rock]

Then no

I have a friend who got hacked recently

And the guy who went on his account spammed pictures similar to description you said

[dusk furnace]

Typically they're a sequence of three or four images of laptop screens that appear to show someone withdrawing prize money from the scam site. We ban them pretty much on sight, so if you're referencing some detail in the images, we won't see it before reaching for the banhammer. As soon as you see "[celebrity] is launching a crypto casino ..." there's no need to read further.

[midnight plank]

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[wind oxide]

what i dont get is

in the solution to this question

why is the pressure at the line marked p_atm + d_2yg + d_1xg

like where tf did the d_1xg term come from

[hybrid widget]

what's P, what's P_atm?

[wind oxide]

p is pressure of air inside the closed end

p_atm is, well, atmospheric pressure

[modern sapphire]

I think the pressure p should be at the red line level. The pressure is equal at both arms at red line level. On right side arm of the U-tube, its pressure is p, since only the trapped air exerts pressure. On the left arm, the pressure is exerted by the x-long column of d1 liquid + y-long column of d2 liquid + atmosphere

[wind oxide]

hmm makes much more sense

but clearly pressure at interface here isnt 0

so shouldnt i first let it be at equilibrium??

[modern sapphire]

well, you kinda have to assume, since it only makes sense for the measurement to be at equilibrium

[wind oxide]

bruh

[modern sapphire]

otherwise you need fluid velocities

[wind oxide]

well no like send it to equilibrum (left goes down by x right goes up by x stuff)

but yeah that isnt solvable either i think

[modern sapphire]

why would it? the trapped air column supports this configuration

thats why you get this pressure equation

[wind oxide]

but the p we derived is clearly not equal to d_2yg+p_atm which is interface pressure

theres a pressure gradiant so shouldnt it flow to equilibrium first?

[modern sapphire]

p would be p_atm + d_2yg + d_1xg this

[wind oxide]

yes yes

but

how can u assume

if its different like

is this an eg. of where i shouldnt use mah brain too much?

[modern sapphire]

honestly, I dont quite get what you are trying to say

do you wanna introduce shm of the interface levels here or smthn?

[wind oxide]

whts shm here?

[modern sapphire]

simple harmonic motion

[wind oxide]

wtf

god no

[modern sapphire]

since you dont think its in equilibrium, the liquid would oscillate

[wind oxide]

well it would eventually reach equilibrium right

[modern sapphire]

with the levels going up and down about the mean equilibrium position

[modern sapphire]

yes

[wind oxide]

and stay in that state right

[modern sapphire]

which I am assuming is already reached, coz you are clearly doing hs fluids

[wind oxide]

but clearly it isnt

like yes m doing hs

[modern sapphire]

yea, I meant steady state equilibrium position

[modern sapphire]

again, why do you say this? I see no reason for you to assume this is not stable

[wind oxide]

the pressure at the line originally marked (the interface) is not same on both sides

and i have verified this numerically

[modern sapphire]

can you post what all you did

coz in both, stable and unstable condition, the pressure should be equal at the interface

[wind oxide]

well i was confused so i ended up looking at the soln

so p = p_atm + d_1xg +d_2yg

putting values we find that

p = 2p_atm (nearly)

but p_atm + d_2yg is not 2p_atm

[modern sapphire]

ok, so you solved p = np ( ) and found n = 2

[wind oxide]

yes and then i checked whether the interface pressure was equal

which it clearly wasnt

[modern sapphire]

why tf would it be? 2P_atm is supposed to equal p_atm + d_1xg +d_2yg, not p_atm +d_2yg

[wind oxide]

because p_atm + d_2yg is pressure at the line on the left side, and p=2p_atm is the pressure at the line on the right side

no.?

we're talking bout the black line right

[modern sapphire]

I get what you are saying, but the pressure at black line on right side is not gonna be p

[wind oxide]

eh.?

what will it be then

[modern sapphire]

it would be less by d1xg

[wind oxide]

wth