help-13

[brazen wind]

because its multiplied by 9

[steep tinsel]

yes

a horizontal trasnlation wold have to do with the x right

as in

3^(x)

[brazen wind]

yes

[steep tinsel]

the x has to be tampered with

is it A?

[brazen wind]

one sec

[steep tinsel]

alrighty

[brazen wind]

ohh wait it says which is an equivalent form of h

so the answer will have to be equivalent to h

what can 9 be rewritten as

[steep tinsel]

3^2

??

[brazen wind]

right

so then h(x)= 3^2 * 3^x

and then since they both have the same base

of 3

you add the exponents

[steep tinsel]

2x

[brazen wind]

not multiply

add

[steep tinsel]

i mean

yea

sorry

crap

so B

i would need more poracice with that

[steep tinsel]

im gonna ask chat gpt to make a simolar question later

[steep tinsel]

thank you

[brazen wind]

yeah you're welcome

good luck on your test

[steep tinsel]

thank you sir

[brazen wind]

once you get to calc itll all start to make sense lol

ap calc is so fun

[steep tinsel]

are u sure abt that

[brazen wind]

yeah it has so many applications

opens up your mind a little

but it tanked my grade

[steep tinsel]

is this d.... i have no cclue

[brazen wind]

but best class i've taken in high school

[steep tinsel]

really

my friendfs hate it

[brazen wind]

maybe their teacher isn't good

mine taught at uc davis before so he's really good

[steep tinsel]

uc davis

i toured that scchool once

[brazen wind]

do you live in california

[steep tinsel]

no i went for a college expedition

[brazen wind]

oh ok nice

[steep tinsel]

yes

[brazen wind]

ok use the same exponent rules

to split it up

[steep tinsel]

well

4^-2 = 1/16 right

thats why i think its D

[brazen wind]

one sec lemme work it out

[steep tinsel]

after this question i have one final question then i will call it a night

and the last question is short

[brazen wind]

wait is this study guide from your teacher

[steep tinsel]

yes

she said

if i understamd study guide

test should be a breeze

or sum like that

there was 101 questions on the stufdy guide

[brazen wind]

wow

wait i don't know if im stupid or if none of these are the right answer

[storm seal]

Ur correct

[steep tinsel]

its D i checked the answer key

[storm seal]

4^2 is 16

Negative exponent is 1/16

[brazen wind]

right but the 4^x is missing

[storm seal]

All that to the power of x

[brazen wind]

it's not all to the power of x though

[storm seal]

U can split it as (4^2)^x

[brazen wind]

oh wait yes you're right

that's my bad

[steep tinsel]

its okay

[brazen wind]

I need to learn my exponent rules too 💀

[steep tinsel]

my llast question and i think its D

[brazen wind]

yeah that's correct

[steep tinsel]

great\

[brazen wind]

random stats question

[steep tinsel]

they give a lot of there

thank tou so much for ur help man

i appreciate i

it

[storm seal]

Tf was that bruh

[brazen wind]

yeah you're welcome

[storm seal]

Wrong gif

[brazen wind]

haha

[steep tinsel]

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[supple drum]

I need help with a few basic algebra problems

[willow sparrow]

leme see

them

[willow sparrow]

can you take it right side up tilting my head hurts

[supple drum]

Yeah

[willow sparrow]

you did number 1 right

[supple drum]

I can remember how i did number one. I completed that a few days ago and can’t remeber how

[willow sparrow]

you just made the equations and then used elimination

from your math a van carries 18

which seems kinda weird but i guess if thats what you got

[supple drum]

so where did the -2 come from in the work at the top? I’m a little confused on that

[willow sparrow]

you needed to make it so that 12b could have a -12b to cancel it out soyou multiplied the whole bottom one by negative 2

[supple drum]

Oohhh ok now i remember that

Yeah i fully understand that one now

[willow sparrow]

yeah you just need to figure out the people

do you need help on the othesr

[supple drum]

I might.

For the restaurant question i think i need help with

Question 3 of the second inage

[willow sparrow]

k lemme think

oh its like the first

since 4 people orderd the crab and gumbo whatever

its 4x+4y

or 4c+4g since you like to do that

and it equals 31

and then try to make the next one

[supple drum]

2c + g = 12.25

[willow sparrow]

yea

and then i htink its just elimination again

[supple drum]

Ahh ok. I guess i just needed a memory jog. I think i can take it from here

[willow sparrow]

ok cya

[supple drum]

Thanks

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[crimson sedge]

how to differentiate 1/log y wrt to x where y is dependent on x

[crimson sedge]

can anyone help me step by step

[fresh patio]

alright do you know how to start

[crimson sedge]

no

[fresh patio]

are you familiar with the chain rule

[woeful isle]

Quotient rule

[crimson sedge]

yes

[fresh patio]

or quotient rule also works

but i would use chain rule

basically, write the function you want to differentiate as the composition of functions you know how to differentiate

and then apply the chain rule repeatedly

[crimson sedge]

can we do this using substitution by any chance?

[fresh patio]

so in this case, ill label everything, let f(t) = 1/t, g(t) = log t, then you want to find d/dx f(g(y(x)))

[fresh patio]

not sure what you mean by this

[woeful isle]

Substitute what?

[crimson sedge]

substitute log y?

[woeful isle]

As what?

[fresh patio]

where? and how?

[crimson sedge]

as any variable

[woeful isle]

Oh

Could do thay

It would only make it longer

[fresh patio]

then i think that's what i just said

do you know how to differentiate f(g(y(x)))

[rain cloud]

i don't know if this is correct but this is how i would do it.

[woeful isle]

👏

Insane effort

[rain cloud]

i think my answer is correct

[woeful isle]

It is correct, yes

[rain cloud]

ah nice

[fresh patio]

yes seems correct

[rain cloud]

would prefer is prapti solves it herself though

[woeful isle]

Ye

[crimson sedge]

okay i think i understood

Is this correct?

[rain cloud]

,rotate

[rain cloud]

yes it;s correct

[crimson sedge]

Okay, thank you so much for helping everyone 🥰

[rain cloud]

weird question but are you indian prapti?

you can not answer if you want to

[crimson sedge]

yess

[rain cloud]

oh nice

[woeful isle]

I wonder what tipped you off

[rain cloud]

just wanted to confirm

i am also indian

[crimson sedge]

Ohh it's okay

How to close this channel?

[rain cloud]

.close

[crimson sedge]

okay thanks

.close

[cedar kiln]

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[round lava]

how do we do 1 (a)? I tried substituting root[(x-1)/(x+1)] as u and it simplified to 4u²/(1-u⁴)

[round lava]

how do we proceed further?

[mighty shuttle]

rationalise first

then proceed

[azure swift]

Perform partial fraction decomposition of ur new integrand

[round lava]

4/(1-u²) - 4/(1+u²)?

[blazing zephyr]

idts

[azure swift]

Hm, and then the numerator after simplifying back is (4+4u^2-4 + 4u^2)

So not correct

A factor of 2 too much

[round lava]

right, sorry

2/(1-u²) - 2/(1+u²), right?

[azure swift]

Yes!

[round lava]

so the second one is a standard form

what about the first one?

[azure swift]

Exactly, then do partial fraction once again for the first and you should be on ur way to the answer

[round lava]

okay, thank you!

.close

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[supple drum]

I need help with using a formula

[woeful isle]

No you don't

[azure swift]

Be more explicit

[void sand]

what specifically do you need help with?

[supple drum]

The formula is for apy and the formula is (1 + r/n)^n -1
Where n is 12 and r is 0.35
I get 0.41198 when using my phones calculators, but i’m unsure if that is correct

[lyric narwhal]

,calc (1+0.35/12)^12 -1

[wraith dagger]

Result:

0.41197997849137

[lyric narwhal]

looks good

[supple drum]

Ah i see. I typed the question into the google app ai and it insisted that the answer was something else

[twilit escarp]

That badboy was wrong then

[supple drum]

Is it possible that the ai was wrong, or is there different ways to calculate it?

[twilit escarp]

ai not so powerful sometimes

Specially in math

[lyric narwhal]

it doesn't matter how you calculate it

the answer will always be the same

[woven sphinx]

i'd like to see you 1v1 alphageometry on some olympiad problems

[twilit escarp]

"Sometimes"

[supple drum]

Gotcha. Thanks for clearing that up for me. That wraps up my question

[blazing zephyr]

can alphageo do olympiad problems

[woven sphinx]

"specially in math"

[lyric narwhal]

that's what it's made for

[blazing zephyr]

oh

[twilit escarp]

Yeah the avg one

[supple drum]

.close

[cedar kiln]

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[blazing zephyr]

how 2 use it

[woven sphinx]

wdym the avg one??? this is like talking about lebron's boxing abilities

[lyric narwhal]

i'm not sure honestly

you might be able to find the software online

[lyric narwhal]

https://deepmind.google/discover/blog/alphageometry-an-olympiad-level-ai-system-for-geometry/
the ai is open-source so you can download it and use it

[cedar kiln]

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[crimson sedge]

Having trouble with the cosine components

[mighty drift]

#advanced-analysis for fourier stuff

.close

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[tulip stream]

Wow!

[mighty drift]

this doesn't belong in the help channels

does it?

[woeful isle]

Nope

Too advanced

Cosine..?

Pfft never heard of it

[upper laurel]

You could just separate the integral line by line

[mighty drift]

you can compute the relevant integrals, sure
technically this is a simple sequence of integrals

[crimson sedge]

Yeah that is what I was doing

Was gonna send a pic until the channel got closed

[cedar kiln]

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[mighty drift]

<= is obvious
>= should feel intuitive

.close

[blazing zephyr]

what

[dreamy void]

he tried to convince a computer

[mighty drift]

fair enough

[mighty drift]

<= is fairly simple though at least I think

[blazing zephyr]

what

[mighty drift]

iff is a double implication

[dreamy void]

what

[blazing zephyr]

oh you mean like if n is divisible by a perfect square greater than 1 then sqrtn=sqrtx+sqrty has solutions?

[mighty drift]

ok writing it out it may not actually be obvious

[crimson delta]

(it is)

[blazing zephyr]

am i dumb what

[mighty drift]

ok I'll stop multitasking so hard

[undone halo]

Ok i think i have maybe an idea

i wrote up something in chat but idk if it proves it enough

[lyric narwhal]

Use the fact that x and y are integers to get some sort of contradiction

The backward direction is easy as people said

[blazing zephyr]

uh

[undone halo]

Ok sorry i forgot i had to ping

[blazing zephyr]

wdym

[undone halo]

ping4response

[blazing zephyr]

lmao

oh right

[undone halo]

Ok heres what i thought of i hope it works.

[blazing zephyr]

boo

[mighty drift]

indeed

[undone halo]

first lets say that n is divisible by some perfect square >= 1

[mighty drift]

I just forgot how I found it to be obvious

[undone halo]

then that means n has a factor lets call F which is a perfect square multiplied by some other thing ill call p

so n = Fp

in the equation it would look like

sqrt(x) + sqrt(y) = sqrt(Fp)

[crimson delta]

lets remember !nosols and all that

before we give too much

[mighty drift]

just write ||n = k m², x = k, y = (m-1)² k||

[blazing zephyr]

erm

oh whoops my brain aint mathing

x and y should be like p(F-q)^2 and p(q)^2 which always have solutions if we let q be 1?

so thats the backwards right

[mighty drift]

yes

[blazing zephyr]

im silly i wasnt processing shit

hm for forwards...

[undone halo]

if n is divisible by a perfect square greater than 1, then sqrt(x) + sqrt(y) = sqrt(n) has solutions for positive integers
and
if sqrt(x) + sqrt(y) = sqrt(n) has solutions for positive integers, then n is divisible by a perfect square greater than 1

right

[cedar kiln]

@blazing zephyr Has your question been resolved?

[blazing zephyr]

yeah ig

how do i prove the second one tho

[undone halo]

maybe you can try showing contradiction

like if n isnt divisible by perfect square then sqrt(x) + sqrt(y) cant equal to it

[normal cipher]

Why is everyone in a teacup

[blazing zephyr]

if we square it we get (n-x-y)^2=4xy
n^2+x^2+y^2-2nx-2ny-2xy=0

alternatively, $x+y+2\sqrt{xy}$ isnt divisible by a square larger than 1

[wraith dagger]

skissue.in.a.teacup

[undone halo]

1 moment

ill continue what i was saying

[blazing zephyr]

ok

[undone halo]

since F is a perfect square you can take F out of it

sqrt(x) + sqrt(y) = Fsqrt(p)

now since x,y can be any positive integer lets choose them to have some perfect square factor times p

[blazing zephyr]

uhuh

[blazing zephyr]

isnt it like this

[undone halo]

yea

now lets say hypothetically that n doesnt have the factor F (i.e. it isnt divisible by some perfect square)

then n = p

sqrt(x) + sqrt(y) = sqrt(p)

[blazing zephyr]

thats for backwards tho right

[undone halo]

Fuck

[blazing zephyr]

this was my idea

[undone halo]

ill just continue it then bec its not going to be useful anyway

choose x = Ap and y = Bp where A and B is a perfect square bla bla

Asqrt(p) + Bsqrt(p) = sqrt(p)
if it were Fsqrt(p), it would be easy (A+B = F)

but now its A+B = 1
since A>0, B>0, there is no way that A+B = 1 as A+B>=2. No sol

idk if that could be useful for the 2nd thing but im just throwing that out

[blazing zephyr]

you cant just say all x and y are factors of p can you?

[lyric narwhal]

I think this is overcomplicating it too much

we want to prove that $\sqrt{x}+\sqrt{y}=\sqrt{p}$ has no positive integer solutions, where $p$ is a square-free number

[wraith dagger]

kheerii

[lyric narwhal]

so assume a solution exists

then you get $x=p+y-2\sqrt{py}$ meaning $py$ must be a perfect square, and the only way for that to happen is if $y=pa$ where $a$ is a perfect square. Similarly, $x=pb$, where $b$ is a perfect square

[wraith dagger]

kheerii

[lyric narwhal]

but this yields $\sqrt{a}+\sqrt{b}=1$ and since $\sqrt{a}$ and $\sqrt{b}$ are positive integers this has no solutions

[wraith dagger]

kheerii

[undone halo]

this looks similar to what i did

[blazing zephyr]

hm

why is b a perfect square

[lyric narwhal]

the same as why a is a perfect square

[blazing zephyr]

a is a perfect square because py is a perfect square and y has to be pa so p^2×a is a perfect square, thus a is a perfect square right?

[undone halo]

x = p+y - 2sqrt(py)
y = p+x - 2sqrt(px) as well i think

[blazing zephyr]

o

[lyric narwhal]

y has to be pa because py is a perfect square

[mighty drift]

solution properly written
sqrt(x) + sqrt(y) = sqrt(n)
x+y + 2sqrt(xy) = n
||so xy is a perfect square||
||hence if you write x = a²b, y = c²d with b, d square-free, you have b=d||
||so x = a²b, y = c²b||
||hence n = (a²+c²)b + 2abc = (a²+c² + 2ac)b = (a+c)²b||
Done

[lyric narwhal]

ah this works too

[blazing zephyr]

ah alright thanks

[blazing zephyr]

mind if i use this on aops?

[mighty drift]

aops?

[blazing zephyr]

to submit as a solution

[dapper raven]

Don't use anything in here for anything other than helping yourself understand a concept

[mighty drift]

you're taking classes on https://artofproblemsolving.com/school ?

[lyric narwhal]

I think they're talking about posting your solution on a forum as an answer

[mighty drift]

I'd recommend adding a bit of detail maybe, but fundamentally I don't consider what I write here as mine

[blazing zephyr]

ah alright, cheers

.close

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[solid cedar]

I'm looking for problems that I could possibly take up research on in game theory specifically. I understand that any problem worth researching is probably being worked on, but a general idea would help.

[cedar kiln]

@solid cedar Has your question been resolved?

[solid cedar]

<@&286206848099549185>

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[vast junco]

Hello , How can i find the limit of this function as x --> pi/2 ?

[vast junco]

without dx/dy

[golden raven]

you need to take the derivative

there is no other nice way to do this

wait actually

[vast junco]

what about the Squeeze Theorem ?

[golden raven]

wait you can just plug in x=pi/2

that should work here

you see that the numerator goes to a positive value

specifically 1+sin(pi/2)=2

but cos(pi)=0

[golden raven]

this limit is actually not defined

[storm seal]

Isn't it infinity

[golden raven]

there is a positive and a negative limit

[golden raven]

from the negative side it is infinity, however, if we approach pi/2 from above it is negative infinity

[storm seal]

Ah

[vast junco]

How can I know when to simplify and where not to? For example, the limit (x^9-9)/( x-3) as x--> 3
will give you infinity if you don't simplify it
this might apply to this function also ?

[golden raven]

brb in 2-3 min

[vast junco]

tyt

[golden raven]

ok im back

um

okay so let us break this into cases

Case #1: Neither the numerator or denominator evaluate to 0 or infinity. Then the answer is just plugging in the value

Case #2: The denominator evaluates to 0 and is positive with the exception of 0. If the numerator evaluates to infinity or a positive number, then it is infinity. If the numerator evaluates to negative infinity or a negative number, then it is negative infinity

Edit: If the denominator is negative locally with the exception of 0, then a positive numerator implies negative infinity while a negative numerator implies positive infinity

Case #3: Both the numerator and the denominator evaluate to 0. In this case, you will need to use some tricks, such as L'Hopitals rule

does that make sense?

[cedar lance]

would be incorrect

circular reasoning

[cedar kiln]

@vast junco Has your question been resolved?

[cedar kiln]

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[lament hollow]

How did I screw up to get -bx instead of just bx?

[digital cliff]

im unsure what you mean

it may be that your got rid of the - in line 2

though im not sure that actually changes the result

maybe it should be x-... in each in the first line

rather than +

[lament hollow]

Hm

Yea

[digital cliff]

,w expand [x-(-b+sqrt(b^2-4c))/2][x-(-b-sqrt(b^2-4c))/2]

[wraith dagger]

Results provided by WolframAlpha

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

Requested by aldrnari_

[digital cliff]

yeah

that was it

[lament hollow]

Thanks

.close

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[potent dragon]

i just wanted to verify if the proof i wrote is viable for this question [this question is from Graph Theory with Applications by Bondy & Murty]
I apologize for the poor handwriting

[potent dragon]

for context epsilon is the number of edges, and nu is the number of vertices of the graph G. psi is the incidence function of the graph. not sure how widespread this notation is, so just thought i'd clarify

[cedar kiln]

@potent dragon Has your question been resolved?

[potent dragon]

<@&286206848099549185>

[cedar kiln]

@potent dragon Has your question been resolved?

[cedar kiln]

@potent dragon Has your question been resolved?

[worldly chasm]

@potent dragon the proof looks good to me.

Though perhaps you'd be able to use the pigeon hole principle on the second part.

[cedar kiln]

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[serene umbra]

from previous channel, on proving cos being strictly decreasing in the interval 0, pi using maclaurib series defn

[serene umbra]

so ig I've shown the result for [0, pi/2]

I js need to prove cos(pi/2 + x) = -cos(pi/2 - x)

to show the result for [pi/2, pi]

[dull oxide]

Have we shown that cos(pi/2)=0 yet?

[serene umbra]

nope

[dull oxide]

yeah that will be hard

at best, we can prove that there is some x between 0 and 2 such that cos(x)=0

To prove that it is pi/2, I never actually did.

[serene umbra]

Why do we need that for the formula

It is a particular equality

[dull oxide]

you could then use angle sum formula to prove the identity you want

[dull oxide]

It is not possible to prove this without first proving cos(pi/2)=0

[serene umbra]

I think we need to prove sum formula as well then

Which is difficult

[dull oxide]

Yes, we do

[dull oxide]

Yes, it is. Exactly why I said this would be a real challenge

I proved angle sum formula by showing that $y=\cos x$ was a solution to $y''+y=0$, but there's a lot of work to get us to where you want

[wraith dagger]

SWR

[dull oxide]

If we have not yet proven that cos(pi/2)=0, it also begs the question on how are we defining pi?

[serene umbra]

I think there are a couple of infinite sums for it, or we can use a limit

With like a regular polygon with increasing number of sides

[dull oxide]

I have a full set of notes on all of this stuff if you are interested btw. Unless you were more interested in trying to figure it all out on your own

[serene umbra]

I am definitely interested, but at the same time this is probably where I need to stop because I dont have enough preliminaries

If thats the word

[dull oxide]

it works as word

[serene umbra]

Can we evaluate the infinite sum with x=pi/2, just asking?

To yield 0?

[dull oxide]

That would be really hard, because how are we defining pi/2?

[serene umbra]

Pi is another infinite sum

Yea thats very hard

[dull oxide]

which sum did you want to use?

[serene umbra]

I dont know i havent learnt any i just know there’s a handful

[dull oxide]

Then yes, it would be possible, but I think very hard

That's not something I've done myself, but you ask a really good question

[serene umbra]

Yea i havent also learnt how to manipulate summations for infinite stuff

[dull oxide]

I can think of a way to do all of these things that you want. But I'll be honest, it's a ton of steps.

[serene umbra]

$\pi = 4\sum_{n=0}^{\inf}{\frac{(-1)^n}{2n+1}}$

Is one of them

Oops

[dull oxide]

yeah, that is the one I know best

[twilit escarp]

\inf

[dull oxide]

And it is based on the trig functions, so it may be your best bet

[twilit escarp]

And put {}

[wraith dagger]

violet

[serene umbra]

Better ig

[dull oxide]

$\pi = 4\sum_{n=0}^{\infty}{\frac{(-1)^n}{2n+1}}$

[wraith dagger]

SWR

[serene umbra]

I think it might be useful cause of the -1^N

It also appears in cos

[dull oxide]

good point

[serene umbra]

$\cos x =\sum_{m=0}^{\infty}{\frac{(-1)^m}{(2m)!}x^{2m}}$

[wraith dagger]

violet

[serene umbra]

If we put x=pi/2 uh

I don’t even know how this wil go

[dull oxide]

That doesn't look fun

[serene umbra]

It doesnt

Looks very messy

[dull oxide]

Yeah

I have another idea how to do it, but it revolves using the taylor series of arctangent explicitly, and I don't know it off my head, and am a little busy atm to deep dive into this

I can offer you my notes on what all I have done, and you can use that to research on your end if you are interested in expanding that

[serene umbra]

Sure

Definitely interested

Also we have chosen two adjacent terms in the expansion of sin x to show that cos x for x in [0, x] is strictly decreasing for x^2 < 6

Is there a way to improve this 6

By getting more terms

And then somehow find a way to keep getting closer to pi

[dull oxide]

wdym?

[dull oxide]

I'll DM it to you

[serene umbra]

Sure

[serene umbra]

We chose the first 2 and showed it is positive

What if we chose a different 2, or more than 2 terms

That increases the ‘6’

Until it gets to pi^2

[serene umbra]

Successively doing this to get a limit that tends towards pi^2

I’ll close for now

.close

[cedar kiln]

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[cedar kiln]

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[desert sparrow]

Can anyone help me with solving this?

[cedar kiln]

@desert sparrow Has your question been resolved?

[desert sparrow]

not yet

[prisma laurel]

This is better suited for the physics server https://discord.gg/physics

[dull oxide]

@desert sparrow, uhhhh. your pic is missing all of the variables in the problem text

[desert sparrow]

Yes that is the point

im supposed to make the variables

[desert sparrow]

But i hate physics!!

[dull oxide]

what?

[desert sparrow]

Yeah that is the point of the question

[dull oxide]

In here?

[desert sparrow]

our prof said the only formula we need is F = MA

[dull oxide]

all of the variables are mssing

consisting of a mass attached
See the extra space between "mass" and "attached"? I believe there should be an M there.

I can infer most of these missing variables, but what the heck why are they all missing?

And then this is just pure nonsense

I can't really infer anything here.

This isn't a physics problem, it's a mystery

[desert sparrow]

ok let me check the paper

[dull oxide]

bless

[desert sparrow]

Ye sorry you are right

here

[alpine geode]

what is the torque on the mass due to gravity?

[dull oxide]

much better

[dull oxide]

!status

[cedar kiln]

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

[desert sparrow]

T = -mgLsin(theta)

[desert sparrow]

7

[alpine geode]

and what is the rotational inertia of the system?

[desert sparrow]

dude i dont know

what is why im asking for a solution 😭

[dull oxide]

7? Then what is your current progress?

[desert sparrow]

I just know that F = ma

idk how to continue

[dull oxide]

F is your net force, so you must find the sum of all forces acting on your mass M

[dull oxide]

You kinda have one here

[desert sparrow]

i stole it from someone

i thought that person would help me further but it seems that they were just asking

i dont know how it came or anything lol

[dull oxide]

okay, then you need to consider what all forces are acting on your mass M

[desert sparrow]

Yes

what then

[dull oxide]

I'm asking you

what are the forces acting on M

[desert sparrow]

just gravity ig

[dull oxide]

Gravity is one, certainly.

But if only gravity were acting on M, then it would just fall straight down, but it is not fallingstraight down

So some other force must be there

[desert sparrow]

yeah idk

[dull oxide]

well, what do you think of that string attached to M? Do you think that might play a role in why the ball is not falling straight down?

[cedar kiln]

@desert sparrow Has your question been resolved?

[cedar kiln]

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[civic elk]

if i have a list of values, and the age they where added, like 1,2, ..., n, what would value/age tell me?

[livid hound]

still not sure what you mean by "age they were added"

[cedar kiln]

@civic elk Has your question been resolved?

[cedar kiln]

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[short otter]

why arent the bounds for phi in the triple integral 0 to arctan(R/h) instead of 0 to arctan(h/R)???

oops was missing this part

[cedar kiln]

@short otter Has your question been resolved?

[cedar kiln]

@short otter Has your question been resolved?

[cedar kiln]

@short otter Has your question been resolved?

[cedar kiln]

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[clear widget]

I don't understand how I can solve this, the lesson is called transformations of exponential functions and im supposed to write a function with a graph

[crimson sedge]

first graph gets translated 2 units upwards

So the y coordinate for each point is shifted two units upwards

[clear widget]

well i know that but how can I do it

[crimson sedge]

So f(x) + 2

So 3^(x) + 2

[clear widget]

the green words are the answer

[crimson sedge]

Can't exactly read the question lol

You've got the importance part cropped

I don't think that's the case tho

[clear widget]

this is all of it

[crimson sedge]

I think they want you to write a function for each transformation

[clear widget]

yeah

[crimson sedge]

The green just explains what the transformation is

That's not the answer

[clear widget]

the green is the answer

i took this from the answer sheet

[crimson sedge]

Tells you to write an equation based on the graph

The green is not an equation

[clear widget]

this one isn't the answer sheet

I want to know how I get 2 units up for A

what do I do

[crimson sedge]

I already told you

f(x) + 2

So 3^(x) + 2

He doesn't want you to get A 2 units upwards. A is a transformation of the original function f(x)

[clear widget]

ohh

[crimson sedge]

You can tell because the horizontal asymptote is y=2 but for the original function it's y=0

Hence the graph was shifted 2 units upwards

[clear widget]

could you show me that rq

[crimson sedge]

Show you what exactly

[clear widget]

how the asymptote is y=2

[crimson sedge]

Do you know what an asymptote is?

[clear widget]

its the line no?

[crimson sedge]

It's a line that the curve seems to keep approaching. In this case as x decreases it approaches the y value 2

[clear widget]

awe man im gonna fail the final exam

[clear widget]

Ohhh

[crimson sedge]

See the curve keeps getting closer to 2 when x keeps decreasing

[clear widget]

Wait im supposed to like find when the curve starts

[crimson sedge]

But it doesn't reach it

[clear widget]

Yeah

[crimson sedge]

A exponential function horizontal asymptote ab^x + c is c

In the case of the main function

3^x

a is 1 and c is zero

So the horizontal asymptote is 0

But in the new graph in part a

The horizontal asymptote is 2

So c = 2

And a remains 1

[clear widget]

ohhh

[crimson sedge]

Hence the equation is 3^(x)+2

Makes sense?

[clear widget]

Yeah

[crimson sedge]

That's good

[clear widget]

I'll continue from there thanks

[crimson sedge]

My pleasure

Good luck

[clear widget]

I'll need it

.close

[cedar kiln]

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[cedar kiln]

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[crimson sedge]

I am not entirely sure how to interpret the question

[crimson sedge]

When they say Find an equation of the line, do they mean y-intercept form?

[cedar kiln]

@crimson sedge Has your question been resolved?

[digital cliff]

any form

vector form would be the easiest here

though you can find m quite easily too if you want to do slope-intercept form

[cedar kiln]

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[serene basin]

for maximum likelihood estimation, suppose the distribution of the observations X is f_X(x), and that this distribution depends on theta. i am confused on what E_theta(X) is. i don’t think it’s a scalar because expectation should only give a scalar if theta is already fixed and you take the expectation w.r.t. X. so is it a random variable?

[serene basin]

the book seems to imply that this quantity is a numerical function of theta but i dont know why, an example would help

if anything, if you take the expectation w.r.t theta, then shouldn’t the variable be integrated out and you’re left with a function of X instead?

[cedar kiln]

@serene basin Has your question been resolved?

[serene basin]

.close

[cedar kiln]

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[indigo siren]

a,b,c are real num such that a+b+c=0 ,find all pair of (m,n) of natural num such that (a^(m+n)+b^(m+n)+c^(m+n))/m+n = (a^m+b^m+c^m)/m * (a^n+b^n+c^n)/n

[indigo siren]

my approach has been to put a = 0, b = 1 and c = -1

and i eventually concluded that both m and n have to be even

and got the relation mn = 2(m+n)

can u all tell me your approaches?

[cedar kiln]

@indigo siren Has your question been resolved?

[indigo siren]

heres a pic of the question

[cedar kiln]

@indigo siren Has your question been resolved?

[cedar kiln]

@indigo siren Has your question been resolved?

[cedar kiln]

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[graceful cedar]

This is just a practice question, I don't under this pattern/question at all, I need help please.

[brisk apex]

I’m guessing it’s the top left one

[graceful cedar]

Any logic?

[brisk apex]

I’m going on a whim here, but I think that the least 3 circles add up to the number of sides the next shape will have

[graceful cedar]

That's some complicated mechanic

[brisk apex]

Yeah

Maybe too complicated

Is it right?

[unborn mirage]

I think it's the bottom left cuz the sum of numbers seem to be increasing by 1 each time, and the amount of circles is the number of sides of the shape next it

[brisk apex]

Ooh

That seems way more plausible than my answer

[graceful cedar]

so how do you go from

to

[unborn mirage]

I don't think each term predicts the next, just that among the 4 choices only the bottom left follows the pattern (I know it's unsatisfying, but they really could put anything as a pattern in these sorts of tests, and I might also be wrong)

[graceful cedar]

In my opinion, I would exclude both of the top left and bottom left because the numbers are all spaced out a lot further.

[unborn mirage]

I don't think spacing has as much to do compared to patterns between numerical values, but if you don't have the real answer to compare it to we're all just guessing here

[graceful cedar]

my head hurts

I decided to continue anyway, the actual questions are no easier. I still need to understand these patterns.

[cedar kiln]

@graceful cedar Has your question been resolved?

[low holly]

Dunno if alrdy answered but i feel it's bottom left, simply a sum of numbers

[simple shale]

bottom left

N/S + number idicates how many shapes are at top/bottom

[cedar kiln]

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[leaden shadow]

Hello, I have a quick question about notation in derivation by perturbation: For a part of a project i am deriving shannon entropy equation, i did this by using perturbation with a small positive scalar and a small change to the function: so f(x) = f(x) + eqf(x)

I then treated the small positive scalar differently and cancelled it out by getting derivative as : Change in H (entropy) in the perturbation / q within a limit of q approaching 0. Since i am deriving it for a discrete distribution with area underneath distribution as 1 I considered qf(x) (perturbed function) to have area underneath it as 1 as well, and i used this to cancel it out at the end.

1. Is this approach justified and correctly stated?
2. This is a functional derivative, right?
3. If it is a functional derivative, I have to notate it differently, I am not certain how to do this particularly when deriving by perturbation, help greatly appreciated.

[leaden shadow]

Id be happy to provide more details, I could not find any help online so i am not sure whom to ask.

<@&286206848099549185>

do i tag again or what now? idk if should ask in other channels

[crimson sedge]

what's up? @leaden shadow

[leaden shadow]

im not sure about whether my approach to a perturbation is correct

im deriving shannon entropy and i perturbed but if i factor out small scalar, im not sure whether i can refer back to constraint for perturbed function

skipping the boring stuff

here i use earlier estabilished cosntraint that integral(f(x)) = 1 (probability underneath function is equal to 1) im not sure whether that applies to perturbed function

[crimson sedge]

tbh, I don't know so well. but y dont u use chatgpt?

I forgot those, I learned that long times ago

[leaden shadow]

gives unclear answers, and if challenged it just changes opinion

so i dont want to rely on it if i can

[crimson sedge]

it's good

u r right

perhaps it will take much time to solve that for me

[cedar kiln]

@leaden shadow Has your question been resolved?

[digital cliff]

do not use chat gpt, my lord, thats all i can help with here

[low holly]

wikipedia's page on this has a bit about this i think? but also your dx disappeared right?

[cedar kiln]

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[tired widget]

I need help with Part B here. Idk how to use a random number table.

[cedar kiln]

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[fair geyser]

.close

[cedar kiln]

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[void sand]

[fair geyser]

omg i missed 500

[cedar kiln]

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[urban condor]

I need help

[broken mist]

https://dontasktoask.com/

[urban condor]

L_infty is Banach prove or disprove

[crimson sedge]

elaborate

[urban condor]

Let f_n be a Cauchy sequence in L_infty

[crimson sedge]

whats the question?

[urban condor]

That

Prove or disprove the statement

[crimson sedge]

show linfinity norm is a norm confirm essential supremum exists prove triangle inequality use cauchy sequence completeness follows from convergence almost everywhere limit belongs to linfinity boundedness preserved check properties carefully

[small saffron]

hi

[urban condor]

I already have shown the infty norm is a norm, and that I just need to do show cauchy sequences converge

[crimson sedge]

show cauchy sequence in linfinity converges pointwise almost everywhere prove limit is essentially bounded since cauchy implies uniform boundedness check essential supremum of limit function is finite hence limit belongs to linfinity completeness follows from this argument

[cedar kiln]

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[cedar kiln]

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[crimson sedge]

can anyone help me find the particular integral by the D-operator method?

[crimson sedge]

replace d² with -a² in 1/(d² + a²), simplify the operator, and apply it to sec(ax).

it only applies to sin and cos

not sec

.close

[cedar kiln]

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[ebon radish]

How do I go about setting the constraints (creating the equations to solve this) for this problem?

[vestal bluff]

what would the triangle with the maximum area look like

where would it touch the circumference

[ebon radish]

wdym

[ebon radish]

one vertex would be touching the circumeference, the other two would be on the diameter

[sand osprey]

"one vertex would be touching the circumference" - at what point of the circumference exactly?

[ebon radish]

we need to find that

[vestal bluff]

intuitively where would it be

[red ingot]

ok

[ebon radish]

here?

[red ingot]

since the triangle is drawn inside a semi circle and one of its legs lies along the diameter that means 2 of its sides meet up on the boundary of the circle

because of the diameter condition

the base always is the same

we just find the max height

since area increases as height increases

max height is directly up from the center

which is the radius

0.5*1/2

= 0.25

M = 0.25

32sqrt3 M = 8sqrt3

[ebon radish]

The answer they gave was 9

[red ingot]

uhh

[ebon radish]

and they used a wierd method to achieve that

this

is there any other method to do this

[red ingot]

wtf

[sand osprey]

apparently, this solution is not correct

[ebon radish]

ah why

[sand osprey]

because the maximum area will be in this case

[ebon radish]

but they mentioned that the legs would be on the diameter and not the hypotenuse

[red ingot]

oh

[ebon radish]

else the max. area would be 1/4

[sand osprey]

if the triangle is inscribed into the circle, it is not possible

[red ingot]

exactly

[ebon radish]

the triangle here is not inscribed; it just is in the circle

[sand osprey]

draw the picture please

[ebon radish]

sure one sec

drew 2 cases

[sand osprey]

ah, I see

[ebon radish]

yeah

also it isn't necessary that both the vertices on the diameter should touch the circumference; but then I don't think we would get the max area in that case

am I right?

[sand osprey]

Then the solution of the picture is correct, except the small typo

but it does not affect anything

[ebon radish]

yeah

[sand osprey]

what exactly you do not understand in the solution?

[ebon radish]

I just wanted to know if there were any alternate solutions...

[sand osprey]

I doubt so...

[ebon radish]

hmmm

[ebon radish]

also why did they do $BC\times CE = AC \times CD$

[wraith dagger]

Flare

[sand osprey]

maybe come super sophisticated geometric reasonings, but I'm not aware

[ebon radish]

What property did they use here

[pastel vault]

power of a point

helps to know some circle theorems for this

well it comes from similar right triangles

[ebon radish]

oh

[ebon radish]

ah alr

[ebon radish]

oh

how do we know if those angles are the same

in this case

[pastel vault]

another circle theorem

angles in the same segment are equal

[ebon radish]

ah gotcha

i totally forgot about that theorem

[pastel vault]

which is an application of angle at the centre = twice angle at circumference

[ebon radish]

yeah

[pastel vault]

(you can move the point on the circumference around)

[ebon radish]

do you happen to know any another solution for that problem?

ty @sand osprey @pastel vault :)

.close

[cedar kiln]

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[pastel vault]

yeah

[ebon radish]

.reopen

[cedar kiln]

✅

[pastel vault]

the base length is the diameter so it is constant

just maximise the height

the max height must be the radius

[ebon radish]

but then they said one of the "legs" is on the diameter

[pastel vault]

well yeah

I mean they should have said sides on the diameter

otherwise the answer is wrong too

[ebon radish]

they did say that

[sand osprey]

no, have already been there:)

[lyric narwhal]

no way dedekind in the house

big fan of your work

[sand osprey]

@pastel vault here is the picture

[sand osprey]

thank you, thank you, glad to here:)

I thought it is going to be forgetten

[ebon radish]

what

[sand osprey]

nvm, google Dedekind:)

[ebon radish]

this dude?

[sand osprey]

yep

[ebon radish]

:o

@pastel vault you wanna say smth?

or should I just close this?

[pastel vault]

no

just close it

[ebon radish]

alr ty tho

.close

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[late ridge]

hiii Dms open

[proper mortar]

what is your question

[spare raptor]

what

[serene umbra]

???

[cedar kiln]

@late ridge Has your question been resolved?

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[regal sandal]

I have to prove a certain proposition in TU games (game theory) I can't find a solution, someone has some ideas?

[regal sandal]

V ~ V’ means that the games are strategically equivalent

[cedar kiln]

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@regal sandal Has your question been resolved?

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@regal sandal Has your question been resolved?

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[teal wraith]

chord of a contact of tangents drawn feom a point on the circle x^2+y^2=a^2 to the circle x^2+y^2=b^2 touches the circle x^2+y^2=c^2(a,b,c>0) prove that a,b,c are in a gp

[teal wraith]

i solved this via similar triangles

but need to solve by circles

by circles i mean coordinate geometry

[cedar kiln]

@teal wraith Has your question been resolved?

[teal wraith]

<@&286206848099549185>

[mighty shuttle]

Have you looked at MSE

[cedar kiln]

@teal wraith Has your question been resolved?

[teal wraith]

.

[teal wraith]

what

[cedar kiln]

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[wild zealot]

My solution was :

[mighty shuttle]

well

[wild zealot]

So I thought that any vales greater than 1 and less than 0 was the answer

[mighty shuttle]

what you want to do is find the minima of (x/(1+x))

[vestal bluff]

is there any restriction to a and x being integers?

[mighty shuttle]

or being positive

[wild zealot]

but appearently the answer is 0 and 1, I don't undertstand why my answer is not valid

[vestal bluff]

or rational

[wild zealot]

No, the book does not say anything about that

[vestal bluff]

is it not just 1 and 0

[vestal bluff]

according to this

[mighty shuttle]

,w minima of x/(1+x), if x>0

[wraith dagger]

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Requested by physicsrocks

[wild zealot]

Yeah, but I don't understand why my answer isn't valid

[vestal bluff]

whats your solution

[wild zealot]

This is my solution

[vestal bluff]

thats not a solution, how are you going to continue from there

[wild zealot]

Well, any positive integer A greater or equals 1 will be a problem because ax + a will be bigger than x

[vestal bluff]

try isolating x

[wild zealot]

wdym?

[wild zealot]

I isolated x here

Like, I understand the solution, but why shouldn't I continue the equation first like I did?

[vestal bluff]

you didnt isolate x

you should only have one instance of x

as the shbject

[wild zealot]

Hmm, how do I remove x from the right side? Can I divide x in both sides?

like x / x = a + a?

[vestal bluff]

bring all the x to one side

factor and divide

[wild zealot]

alright, so x = a / 1 - a

So the problem with my answer was not isolating x, now I can see that a can't be 1 because we would divide 1 by 0, but a can be 0 because 0 / 1 is 0 so x = 0

Where did i go wrong?

[vestal bluff]

and why is x = 0 invalid

go back to the original question

[wild zealot]

because there is a constant in the numerator that is 1

And we cannot divide by 0 here

[vestal bluff]

indeedd

so a cant be 0 and 1

[wild zealot]

Alright, that makes sense, but where did i go wrong continuing the equation?

[vestal bluff]

which equation

[vestal bluff]

this?

[wild zealot]

Yes

[vestal bluff]

it wasnt wrong it was just incomplete

[wild zealot]

Indeed, but here, when I completed it, 0 was a possible answer

Why?

[vestal bluff]

wym 1 is possible

[wild zealot]

0 actually

[vestal bluff]

a = 0 implies x = 0

[wild zealot]

because x = 0 / 1 is not a problem right?

[vestal bluff]

and x cant be 0 becayse of the division in the otiginal

[wild zealot]

Alright, so I think I understand now

Thank you

.close

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[inner solar]

what should be the first step?

[dusk finch]

I'd split it into sum of 2 seperate limits

[inner solar]

cosx-1 and x^2?

(with denominators)

[dusk finch]

yep

can you use lhop or taylor expansion btw?

[inner solar]

yup but how shold i use it there?

[dusk finch]

ah if you can use lhop then just do it

differentiate numerator and denominator

you might need to do it twice

actually once is enough

okay i take that back

you need to do it twice

[inner solar]

hm i managed to do it with only one lhopital

[dusk finch]

yeah, it might simplify nicely after that

[inner solar]

but ive got it now so thanks

[dusk finch]

2nd lhop would simplify it to constant / constant basically

[inner solar]

true

but deriving is hard haha

[dusk finch]

its actually possible entirely without lhop

[inner solar]

hm how?

[dusk finch]

$\frac{\cos\left(x\right)-1+x^{2}}{\left(x+\sin\left(x\right)\right)^{2}}=\frac{\cos\left(x\right)-1}{\left(x+\sin\left(x\right)\right)^{2}}+\frac{x^{2}}{\left(x+\sin\left(x\right)\right)^{2}}$

[wraith dagger]

MæthIsAlwaysRight

[dusk finch]

$\frac{x^{2}}{\left(x+\sin\left(x\right)\right)^{2}}=\frac{1}{\left(1+\frac{\sin\left(x\right)}{x}\right)^{2}}=\frac{1}{2^{2}}=\frac{1}{4}$

[wraith dagger]

MæthIsAlwaysRight

[dusk finch]

im ignoring the limits because im too lazy to type them

[inner solar]

ah yes i see it now

its the worst when you dont find that trick right away and then life's just misery

[dusk finch]

$\frac{\frac{\cos\left(x\right)-1}{x^{2}}}{\left(1+\frac{\sin\left(x\right)}{x}\right)^{2}}=\frac{\frac{\cos\left(x\right)-1}{x^{2}}}{4}=-\frac{\frac{1}{2}}{4}=-\frac{1}{8}$

[wraith dagger]

MæthIsAlwaysRight

[dusk finch]

the funny thing is that to see that trick, you need to be able to understand the expression intuitively on such a level, that you can estimate the limit without any computation

cosx - 1 and x^2 are both of quadratic nature

cosx - 1 is approximately -x^2/2

and sinx is approximately x

[inner solar]

xd yea so just knowing taylors will set you free