help-0

[mystic comet]

i mean in +infinite

[tardy stag]

$\lim_{n\to\infty} (-1)^n$ is undefined yeah (can you explain why?)

[ocean seal]

Hayley

[mystic comet]

it's always going to go from 1 and -1 depending of the n if it's pair or impair

so this should be 0 since the denominator is bigger

[mystic comet]

even or odd**

[tardy stag]

just because the denominator is bigger doesn't mean it'll be zero

eg $\lim \frac{n}{2n}$

[ocean seal]

Hayley

[mystic comet]

1/2

[tardy stag]

yeah not zero

does something similar apply here? does the limit even converge?

[mystic comet]

it doesn't since we have (-1)^n

[ruby current]

you can also add and subtract 0.5 from the numerator

[gray isle]

just because a certain component doesn't converge doesn't make the whole thing diverge

[tardy stag]

that's not necessarily true either, $\lim\frac{(-1)^n}{(-1)^n)}$ very much converges!

[ocean seal]

Hayley

[tardy stag]

the extra parenthesis is free

[mystic comet]

ohh i see your point

[ruby current]

you can also try approaching +infty by using the sequence of even numbers/odd numbers

[mystic comet]

lim to infinity

[ruby current]

i.e. changing variables to 2t = n or 2t + 1 = n

[mystic comet]

why not t=2n for example?

[gleaming ridge]

You could have from the start divided both the numerator and the denominator by n

Then take the limit, it's quick!

[mystic comet]

ah yes

[ruby current]

because you want to simplify (-1)^n

[vale crag]

you could also factor (-1)^n from the numerator and denominator

then squeeze it (théorème des gendarmes pour alitoo si tu connais)

[ruby current]

good method

[vale crag]

it's pretty much your idea tushar

[mystic comet]

yes i know it

his idea devided the squeeze in two pieces

i think

[vale crag]

$$\frac{n - 1}{2n + 3} \leq u_n \leq \frac{n + 1}{2n - 3}$$

[ocean seal]

_aplatypus

[vale crag]

then it's easy peasy

[ruby current]

gg

squeeze is always good when you have (-1)^n

[mystic comet]

I'm used to only use squeeze with cos and sin that's why i didn't notice it :/

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[lone heart]

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[tribal valve]

Find all complex numbers $z$ such that $|z-1|=|z+3|=|z-i|$.

[ocean seal]

dabbbingpotato

[tribal valve]

i tried replacing z with a+bi

[tardy stag]

this has a very beautiful geometric interpretation but algebraic should work as well

[tribal valve]

on a complex plane?

[median oar]

yeah

consider what it means for the absolute value of z - 1 to be some particular value

[tribal valve]

okay so isn't the point just moving left by 1 unit

and |z+3| would move right 3 points

and |z-i| would move down 1 point

[median oar]

no no

[tribal valve]

oh

[median oar]

let's maybe try consider what |z| means

[tribal valve]

magnitude right

[median oar]

what does |z| = c look like (for some real value c)

[molten pivot]

that's not always defined

[tribal valve]

isn't it just $\sqrt{a^2 + bi^2}$

[median oar]

good catch

[ocean seal]

dabbbingpotato

[median oar]

what does |z| = c (for some positive real c) look like

[molten pivot]

c has to be nonnegative

[median oar]

0 is degenerate and doesn't really matter in this case so i've chosen positive c

[median oar]

i mean what does |z| = 1 look like

what kind of complex numbers satisfy |z| = 1

[tribal valve]

1

[median oar]

1 does yes

what else

[tribal valve]

uhhhh

-i

[median oar]

what else

[tribal valve]

-1

[median oar]

anythng else?

what kind of thing has the property "has distance = 1 for every point"

[tribal valve]

circle

wait

[median oar]

if u plot every complex value that satisfies |z| = 1 you get a unit circle in the Re, Im plane

[molten pivot]

😮

[wind cloak]

Argand plane

[median oar]

yeah i know

[wind cloak]

ReIm plane sounds better though

[median oar]

it seems they dont know it

[wind cloak]

[tribal valve]

[median oar]

the Real/Imaginary plane is normally referred to as the argand plane or an argand diagram

[tribal valve]

oki

[median oar]

so if |z| = c (for some positive c) gives you a circle centered at the origin

[tribal valve]

right

[median oar]

what would |z - w| = c (for some positive c, complex w) give you?

[tribal valve]

and c would be the radius?

[median oar]

yes

[tribal valve]

it would give you a circle thats offset

[median oar]

offset by how much

[tribal valve]

w

[median oar]

ok

let w = 1

[tribal valve]

to the left

okay

[median oar]

now we have |z-1| = something

hey look that's the first part!

[tribal valve]

[tribal valve]

wait so its the overlapping parts of the three circles?

[median oar]

so we have some circle centered at 1+0i

exactly

[tribal valve]

:0

[median oar]

that's what hayley meant by "a very beautiful geometric interpretation"

[tribal valve]

it is indeed beautiful

better than the algebra solution

okay so three circles

one of them is centered at 1+0i

wait

-1+0i

another is at 3+0i

[median oar]

it's this one

[tribal valve]

and then the last one is 0-i

[median oar]

wait what

hold up

wait yes

you are right

sorry whoops

[tribal valve]

oh ok

-1+0i

-3+0i

and 0-i

[median oar]

wait what

[tribal valve]

huh

[median oar]

z = 1 gives you the center for the first one

because |1-1| = 0 so that's the center

,w graph x-3

[ocean seal]

Results provided by WolframAlpha

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

Requested by frosst#0

[median oar]

x-3 means you move it to the right

[median oar]

same here z-1 means you move to the right by 1

so the center is at 1+0i

-3 + 0i

and 0 + i

[tribal valve]

ah ok

i get it

(i have a doctors appointment in like 5 minutes do you think we could pick back up on this later its been a ton of help, if not thanks so much for your help so far)

[median oar]

have a think about it, i probably will be asleep by then

but there'll be others to help you

[tribal valve]

alright thanks so much for your help

.close

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[barren scroll]

can someone explain why they are sitting g(x)=2

ill screenshot original problem too

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[turbid root]

I am unable to do part b.

These are my workings.

[timber latch]

Seems like you solved it

Does the answer not match?

Wait what even is that method

The coordinates of c will be something of the form (x,0)

Since BC is 10 and lies on the x axis

It must be (x-10,0)

[turbid root]

yea it doesn't 😦

[turbid root]

the answer is (-5,0)

.close

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[median oar]

if $\underline \theta = (\theta_\mu, \theta_\sigma)^T$ and $X \sim N(\theta_\mu, \theta_\sigma)$

[ocean seal]

frosst

[median oar]

how do we know that $\mathbb{E}[(x-\theta_\mu)^4]=\theta_\sigma^2$

[ocean seal]

frosst

[naive valley]

is $\theta_\sigma$ the standard deviation of $X$?

[ocean seal]

Bungo

[median oar]

no it's the variance of X

[naive valley]

isn't there a factor of 3 as well

$3\theta_\sigma^2$

[ocean seal]

Bungo

[median oar]

oh there is

well

how do you get there

[naive valley]

offhand I'd think the moment generating function?

[median oar]

that means i have to go detour and learn mgfs first

D:

[naive valley]

i mean you could just brute force the math for the special case of the 4th moment

it'll probably be very grungy

[median oar]

theyve got this step here

then it goes here

[naive valley]

what's the left hand side of that first screenshot?

[median oar]

[naive valley]

ha, i call bs on their magically writing 3sigma^4 there

[median oar]

exactly

[naive valley]

the expression on the previous line is actually more complicated

[median oar]

where did it come from

[naive valley]

because it involves the non-central 4th moment

[median oar]

idk how we can evaluate E[x^4]

[naive valley]

exactly

the central moments are easier as usual

[median oar]

when in the previous case i had E[x^2] it was not too hard

i just expanded E[(x-mu)^2] = Var(x)

[naive valley]

here's a MSE answer that looks elementary enough (no MGF): https://math.stackexchange.com/a/1917666

[median oar]

and i just simply used the fact Var(x) is defined to be the 2nd central moment

ok lets see

FUCK

IBP

lmao

[naive valley]

twice, always fun haha

[median oar]

ooh that second answer is simplier

choosing a better ibp

[naive valley]

ah yes i just saw that one

yep

[median oar]

man whoever came up with this stuff is smart

[naive valley]

turns out easier than i thought

[median oar]

i guess i can do the same for x^3 term as well

ibp it with x and x^2 pdf

[naive valley]

well remember this is for the central moment (the MSE question assumes mean zero)

[naive valley]

so you don't actually need that complicated expression in your screenshot

[median oar]

oh

[naive valley]

doing the central moment directly is almost always easier

because you can assume it's zero mean

[median oar]

right right

oh i see what you mean

[naive valley]

then if you want the noncentral moment, you can do like:

Y = X + mu, where X is zero mean and mu is the mean of Y

[median oar]

doing E[(x-mu)^4] is the same as doing E[(x)^4] with 0 mean

so i can kinda just sub say Y = x-mu

[naive valley]

so E[Y^3] = E[(X + mu)^3] = E[X^3] + 3E[X^2]mu + 3E[X]mu^2 + mu^3

[median oar]

then go E[Y^4]

[naive valley]

where everything on the RHS are central moments

and similarly for the 4th moment

i once had this question on a midterm exam, still sticks in my memory for some reason

[median oar]

ah same idea but you have the letters the other way

[naive valley]

"express the nth moment of a random variable in terms of its nth central moments"

binomial theorem ftw

[median oar]

i just know mgfs are in next year's sem 1 class

[naive valley]

yep

[median oar]

this year they had an assignment question about a trinomial expansion

[naive valley]

oh fun

[median oar]

idk what i was doing but i tried my best to help the 2nd year students

by sheer will power and algebra without knowing what the symbols meant

[naive valley]

haha nice

[naive valley]

mgfs are cool because some distributions are harder than the normal one for computing their moments, but the mgf lets you do it in a nice albeit indirect way

[median oar]

mgfs are cool because they exist in physics too

[naive valley]

that's beyond my pay grade haha

but the idea behind them is similar to various tricks for computing tricky integrals as well

[median oar]

https://en.wikipedia.org/wiki/Moment_(mathematics) very cool

[naive valley]

ah yeah, physical moments that makes sense

defined basically the same way as the probabilistic ones

[median oar]

yeah

so you have the same mgfs

[naive valley]

yep cool indeed

[median oar]

they just mean different things

super cool

Hmm something happened

I’ve managed to lose the expectation on the first part

also the 3 disappeared??

[naive valley]

hmm, lemme check what you did

i don't really get your integration by parts

shouldn't it be:

hmm one moment i'm gonna write it out instead of trying to do it in tex

[median oar]

integration in tex

[naive valley]

ah here we go

instead of splitting it as y^2 and y^2 f_Y, try y^3 and y f_Y

[median oar]

OH

[naive valley]

lol

[median oar]

Ok now I got this

Still looks wrong

It looks closer than before I guess

i have a feeling my first ibp is wrong

FUCK

shouldn't the ∫yf_Y dy = 0

cos that's the expectation of Y

noooooo

(Also I’m so tired it’s 6am)

[naive valley]

hmm, let's be explicit here with the integration by parts:

$\int y^4 f(y)\ dy = \int y^4 c\exp(-y^2/2)\ dy$, where $c = 1/\sqrt{2\pi}$

[ocean seal]

Bungo

[naive valley]

let $u = y^3$, $dv = yc\exp(-y^2/2)\ dy$, then $du = 3y^2\ dy$, $v = -c\exp(-y^2/2)$

[ocean seal]

Bungo

[naive valley]

then integration by parts gives $\left.uv\right|{-\infty}^{\infty} - \int{-\infty}^{\infty} v\ du$

[ocean seal]

Bungo

[naive valley]

the first term is zero since exp(-y^/2) goes to zero faster than y^3 goes to infinity

so you're left with

$\int_{-\infty}^{\infty}3cy^2 \exp(-y^2/2)\ dy$

[ocean seal]

Bungo

[naive valley]

which integrates to $3\sigma^4$

[ocean seal]

Bungo

[lone heart]

@median oar Has your question been resolved?

[lone heart]

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[pale rock]

What is 362/20 equiv to

[torn elk]

it's equiv to a lot of things, you have to be more precise in your question.

[pale rock]

Ok what's 362/20 simplified as much as possible if that makes sense

[torn elk]

Do you know what simplified refers to?

When you simplify something, what you're doing is making it easier to work with.

Right now you have 362/20, which is a big fraction and hard to work with. The exercise is telling you to make an equivalent fraction whose numbers are smaller or even solve the fraction and use decimals.

[pale rock]

Like common denominators for example 5/10 is equiv to 1/2

[torn elk]

Yes.

Or writting as 0.5

That depends on what's the best way you put something to work with.

[pale rock]

I write fractions

[lone heart]

@pale rock Has your question been resolved?

[gilded vessel]

do you still need help simplifying the fraction

[pale rock]

No I figured it out by division but thanks anyways!

[gilded vessel]

alrighty

[lone heart]

@pale rock Has your question been resolved?

[lone heart]

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[hot bluff]

quick partial fraction decomp question. part b

[hot bluff]

that was my answer, but im missing a + 1

where does the 1 come from

[tardy stag]

PFD doesn't work if the numerator is the same degree or higher than the denominator

you need to do polynomial division first

[hot bluff]

ummm

[naive valley]

or you could write the numerator as $x^3 + 1 = (x^3 - 3x^2 + 2x) + (3x^2 - 2x + 1)$

[ocean seal]

Bungo

[naive valley]

which amounts to the same thing

[hot bluff]

is this right?

[naive valley]

yep

[hot bluff]

this would be the end of it, since i get a remainder that isn't easily reached with x^3 right?

[naive valley]

yes correct

[hot bluff]

so what does the result look like? 1 + 3x^2 - 2x + 1?

[naive valley]

1 + (3x^2 - 2x + 1)/(x^3 - 3x^2 + 2x)

[hot bluff]

ah

[naive valley]

now do partial fractions on the fraction

[hot bluff]

my man

thanks

[naive valley]

sure

[hot bluff]

.close

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[vapid steppe]

$\lim_{x\to 1}{\frac{xe^x-e}{x^2-1}}$

[ocean seal]

jashxdlol

[vapid steppe]

how to do

[plain flame]

are you allowed to use lhopital?

[vapid steppe]

no we havent learned it

im just wondering if i have to use my calculater to do it

since i looked it up and it turns out to be just e

[plain flame]

you can always check for values close to 1

yeah

its should be e

[vapid steppe]

so i guess there must be a algebraic way to do it

[vapid shuttle]

does 362 and 20 have any common factors?

[vapid steppe]

also i was wondering can u use lhopitals rule when x is approaching one of the infinities

[raven rover]

You could factor out an e and try a substitutiton?

[plain flame]

2

[vapid shuttle]

woa

sorry my screen was showing an old question apparently

[plain flame]

only for 0/0 or inf/inf

[vapid steppe]

i was wondering if i could use lhopitals to solve $\lim_{x\to -\infty}{\frac{3-2^x}{4-5^x}}$

[ocean seal]

jashxdlol

[raven rover]

No

[plain flame]

you can just plug it in

dont overthink it

[vapid steppe]

doesnt the denominater decay faster tho

[raven rover]

Doesn’t matter

[plain flame]

this is abuse of notation but $2^{-\infty} = 5^{-\infty} = 0$

[ocean seal]

heavy0201

[vapid steppe]

so when do i have to compare the dominant function

[proven leaf]

$\lim_{x\to-\infty}2^{x}=\lim_{x\to-\infty}5^{x}=0$

wouldn't that be better

[vapid shuttle]

no

[ocean seal]

XxMrFancyu2xX

[plain flame]

negative

[vapid shuttle]

yes

[plain flame]

yeah

[proven leaf]

ye i messed up lol

[plain flame]

thats better

[vapid steppe]

cuz like if u have polynomial/polynomial u have to compare the degrees

wait is that only when u get an indeterminate form

when u compare the dominance

[proven leaf]

happens all the time when I first boot up discord someone will be asking a question about circles and I say "use the quadraric formula"

[plain flame]

ngl i still havent figured this one out

[fallen verge]

You sure it isnt xe^x-e^x?

[vapid steppe]

this was a khan academy question and the answer was to approximate

[proven leaf]

l'hopitals?

[fallen verge]

Bleh

[plain flame]

well hes not allowed to so i am trying without

[proven leaf]

interesting...

[fallen verge]

Sure you can do numerical approximarion

[plain flame]

well yes the answer is e

[vapid steppe]

it isnt since the answer is e and not some random number

[plain flame]

but algebraically

[fallen verge]

The answer would be e/2 instead if this were the case not some "random number" lol

[vapid steppe]

the answer to the one i sent is e

[fallen verge]

I know

[vapid steppe]

i swear it wasnt that lol i wish i took a screenshot but this was a couple hours ago

[fallen verge]

I dont think its possible algebraically

[vapid steppe]

i mean the question was what are the steps to find this limit and the answer ended up being approximation

[plain flame]

naahhhh

it has to be

[vapid steppe]

and if it was the one u sent u could have done it algebraically

[fallen verge]

Yeah thats why i wanted to make sure

Exps dont play nice with polys tho

[vapid steppe]

since its xe^x maybe it has to do something with productlog

idk its fine

.close

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[frank quail]

When you have two well ordered sets with greatest elements but one's least element is removed and to find a contradiction we suppose we have an order isomorphism between them. How do you find the contradiction?

[molten pivot]

There is no bijective function between the two sets once you remove the endpoint

[frank quail]

I doubt that, the theorem is that there's no is order preserving bijection

[molten pivot]

How would you construct a bijection at all between [0,1] and (0,1]?

[frank quail]

I am pretty sure you can with uncountable sets

[molten pivot]

Try it. Lol

[worn fox]

they have the same cardinality so there is a bijection between them, lots in fact

[lone heart]

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[toxic elk]

Hello

From the last line i wrote, x = 1

And then we drew that on the graph

In red, x < 1, or x > 1

But which of those two are the answer?

And why

[lone heart]

@toxic elk Has your question been resolved?

[toxic elk]

<@&286206848099549185>

[rancid granite]

x<1 if you look at your graph you can see that the line for y=|x+2| is above the line for y=2x+1 for values of x to the left of x=1

[toxic elk]

okay okay

thanks

.close

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[dapper lynx]

I could use a hand with this...

[dapper lynx]

I can feel, on the tip of my frontal cortex, the direction this is pointing me in

But I'm a bit lost

[surreal meadow]

do you know how to, in general, show things are linearly independent?

[dapper lynx]

There can only be a nontrivial solution to av1 + bv2 = 0

[surreal meadow]

yes

[dapper lynx]

Sorry, more generally, there cannot exist a nontrivial solution to Ax=0 where the columns of A are the vectors in question

[surreal meadow]

so assume there exist a and b so that
a(2u) + b(u - v) = 0

show a = b = 0

[dapper lynx]

Oh

Simple enough, I suppose

Lemme try that

One thing that's sorta throwing me is that

In the context of av1 + bv2 = 0

a and b are known to be zero and only zero

[surreal meadow]

the direction of implication is
av + bu = 0
implies
a = b = 0

your goal is to rearrange terms so you reach

(some constant) * v + (some other constant) * u = 0

the conclusion will then be clear

[dapper lynx]

Aight

Thanks

[surreal meadow]

note that this comes from the independence of {u,v}

[dapper lynx]

Off the top of my head

[surreal meadow]

that works, i was hinting at going the other way though

[surreal meadow]

starting with this assumption

and arriving at a = b = 0

[dapper lynx]

Hmmm

[surreal meadow]

actually this doesn’t work i don’t think

ok nvm it should work

but yeah i was still trying to get you to go the other way

[dapper lynx]

I'm not entirely sure HOW it would work

I've proved that some constant times 2a fits nicely into this

[surreal meadow]

so

[dapper lynx]

But I'd somehow need to prove that a = b in order to wrap (v-u) into a single term

[surreal meadow]

we assume
a(2u) + b(v - u) = 0

then we get

2a * u - b * u + b * v = 0

(2a - b) * u + b * v = 0

what do we know about u and v

oh i had it slightly backwards sorry

let me edit

[dapper lynx]

K

I'm working my through this on my end, too

😛

[surreal meadow]

ok should be good now

[surreal meadow]

note this equation is of the form

c * u + d * v = 0

what do we know about c and d in this case

[dapper lynx]

Well, we know the must be zero by the definition of linear independence

[surreal meadow]

yes

so 2a - b = 0
and b = 0

the rest should be boilerplate

[dapper lynx]

Okay!

Thanks

.close

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[scenic wing]

$x^3-8$

[ocean seal]

putridplanet

[scenic wing]

how do you factor this

[long axle]

Do u know difference of cubes factorization

[scenic wing]

no

[oblique whale]

Hi people I need help about eigenvalues and eigenvector, what book do you recomend me to learn this topic?

[long axle]

#❓how-to-get-help

Use that image to help u factor @scenic wing

[scenic wing]

so

$(x-2)(x^2+2x+4)$

@long axle

[long axle]

ab = 8x ? U sure?

[scenic wing]

2x

[long axle]

And b^2 ?

[ocean seal]

putridplanet

[long axle]

Yep

[scenic wing]

and ab doesnt become negative because its x-2

[long axle]

Yea ab isn’t negative

[lone heart]

@scenic wing Has your question been resolved?

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[lapis fractal]

I need help with this geometry question:

[lone heart]

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'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[lapis fractal]

heres the accompanying construction

I dunno where to begin

[lone heart]

@lapis fractal Has your question been resolved?

[foggy current]

You should extend PK to A and QK to C

[lapis fractal]

I dont think PK can be extended to A

at least doesnt look like it can

I can however join A to K and K to C

Oh wait, I got the answer!

K is the circumcentre of ABC

thus AK = KB = KC

Hence proved

.close

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[vital oriole]

Rational Equation

[tacit arch]

What is the question

[vital oriole]

Solve for the Solution set

For 3x+6/x-1 = x+2/x

[tacit arch]

You're on the right track, you just left out parentheses

[vital oriole]

So far but i’m sure it ain’t right

[median oar]

There’s a bracket in (x+2) on step 2

And what happened to the = sign on step 4

[vital oriole]

Oh mb

all better?

[median oar]

Nope

You didn’t expand it properly

The reason you need the bracket is because it expands differently

[vital oriole]

O

[lapis fractal]

did you get the answer?

[vital oriole]

Still doing it

All good now

[lapis fractal]

Did you get x=2, 0.5?

[vital oriole]

Ah crap.

And I already submitted it

[lone heart]

@vital oriole Has your question been resolved?

[lone heart]

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[granite rapids]

Hi

[granite rapids]

Is my answer correct for (a)?

[vale wigeon]

6 times more lines than necessary

and in your quantifier-y statement the word "where" is unnecessary

$\forall n \exists p( p \text{ is prime} \land p>n)$

[ocean seal]

Ann

[vale wigeon]

you don't need to put any of this fluff around it

also your handwriting is mid

[granite rapids]

xD

[vale wigeon]

also you are not yet showing anything just restating it

[granite rapids]

so (a) is not correct

[granite rapids]

Is that better?

[vale wigeon]

it was correct all along just excessively wordy!!

erase everything except the quantifier-y statement

[granite rapids]

hm

i wont lose any marks for exessive words

i think

@vale wigeon

[vale wigeon]

suit yourself

the more you write the more chances you give yourself to make mistakes

[tacit arch]

Nobody can read your teacher's mind

[granite rapids]

What about b

I already got the answer but im not sure if its right

(b) To negate the expression in (a) and simplify it to resemble the logic used in the proof of Euclid's Theorem, we need to negate the quantifiers and change the logical connectives. The negation of the original statement would be:

Explanation:
∃n ∀p (p is a prime number → p ≤ n)
This negation can be read as "There exists a natural number n such that for every prime number p, p is less than or equal to n." This negation essentially claims that there exists a natural number n for which all prime numbers are less than or equal to n.

This negation contradicts Euclid's Theorem, which states that there are infinitely many prime numbers, as it suggests that there is a maximum prime number (represented by n) that bounds all prime numbers.

[granite rapids]

Is this correct for (b)?

[lone heart]

@granite rapids Has your question been resolved?

[lone heart]

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[safe tartan]

What transformation is applied when 2^x becomes 2^xsquared-4x+3

[upbeat hornet]

Do you mean $$ 2^{x^2} or (2^x)^2 $$

[ocean seal]

even order group => solvable

[safe tartan]

First

X^2 is the exponent

To two

[vale wigeon]

2^(x^2) - 4x + 3?

[safe tartan]

No the whole quadratic is exponent

So 2^quadratic

[lament forge]

$2^{x^2-4x+3}$?

[ocean seal]

bee [it/its]

[safe tartan]

Yep

[upbeat hornet]

You can complete the square in the exponent

[lone heart]

@safe tartan Has your question been resolved?

[lone heart]

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[vapid shuttle]

Let ${x_n}$ and ${y_n}$ be bounded sequences s.t. $x_n \leq y_n \forall n \in \mathbb{N}$ Then, show that $$\limsup{x_n} \leq \limsup{y_n}$$

[ocean seal]

austinu

[vapid shuttle]

I was hoping someone could check my proof for this. I will be typing it out below

$\limsup{x_n}=\lim_{n \to \infty} {\sup{{x_k:k\geq n }}}$ $\newline$ and $\newline$ $\limsup{y_n}=\lim_{n \to \infty} {\sup{{y_k:k\geq n }}}$ $\newline$ Then we want to show that $\newline$ $\sup{{x_k:k\geq n }} \leq \sup{{y_k:k\geq n}}$ $\newline$ $\forall k \in \mathbb{N}$

[rose sigil]

lots of typos?

[vapid shuttle]

Since $x_n \leq y_n \forall n \in \mathbb{N} \implies x_k \leq y_k \forall k \in \mathbb{N}$ $\newline$ Then $\implies \sup{{x_k:k\geq n }} \leq \sup{{y_k:k\geq n}}$ $\newline$ Then $\implies \limsup{x_n} \leq \limsup{y_n}$

[vapid shuttle]

should the limits be n -> inf?

or what are my typos

[rose sigil]

does this make sense to you?

[vapid shuttle]

it did, but now I see it should be n-> inf

[mortal trellis]

have you shown before that a<=b implies lim a <= lim b?

[vapid shuttle]

yes

and also that limits respect inequalities (if that is the right way to word that)

[ocean seal]

austinu

[vapid shuttle]

Without the typos now (hopefully that was the only typos?)

[ocean seal]

austinu

[naive valley]

since the whole content of the proof is to show that inf and sup respect inequalities, maybe it is worth elaborating slightly on that

[mortal trellis]

imo you don't spend enough time on showing that the supremums follow the inequality. by def you only know that x<=y pointwise (local) but you want to establish some global property

[vapid shuttle]

so

in the second textbox of mine

the step in between the end of the first line

and second line

I need to elaborate more there?

[naive valley]

i think you should

really just one more line

[mortal trellis]

why can't it be that eg x100 is the supremum of the xs and y200 is the supremum of the ys and x100>y200

[vapid shuttle]

is there where the bounded part comes in

[mortal trellis]

no

[vapid shuttle]

I see what you are getting at, but I'm not sure what to do about it Denascite

[mortal trellis]

the bounded part comes in when limsup is defined in the first place

[naive valley]

no, sups and infs respect inequalities even without the boundedness assumption

[vapid shuttle]

yeah I have proven this before

[naive valley]

i think taking an extra step when you go from x_n <= y_n to sup x_n <= sup y_n is warranted

like do one sup at a time

just to make it crystal clear why it's true

[vapid shuttle]

okay well

whatever the sup of x_n is

y_n at that n must be greater than or equal to it

so y_n necessarily has an element that is greater than or equal to the sup of x_n

[naive valley]

what do you mean "at that n"

[vapid shuttle]

whatever nth element

wherever the sup is

(how do we even know that the sup exists tbh?)

bounded isn't enough

we don't know that xn isn't nonempty?

[naive valley]

sup always exists in the extended reals

[vapid shuttle]

extended reals?

[naive valley]

how could it be empty?

[vapid shuttle]

oh right it is a sequence

nvm

[naive valley]

i mean you are given sequences haha

[vapid shuttle]

I was thinking sets

[naive valley]

extended reals = real line along with +inf and -inf

[vapid shuttle]

Okay Bungo

here is what I think

We have $x_k \leq y_k \forall k \in \mathbb{N}$

[ocean seal]

austinu

[mortal trellis]

its not?

[vapid shuttle]

I was thinking about sets

nvm about that

[mortal trellis]

reminder that the sup doesnt have to be attained for a certain value of n. it doesnt have to be a max

[vapid shuttle]

that's true

hmm

Okay

I think I got it

so

sup y_k is greater than or equal to all y_m

and y_m is greater than or equal to all x_m

[naive valley]

y_m is greater than or equal to x_m for that m

[vapid shuttle]

then sup y_k >= y_m >= x_m

fimfowmfwdaowd

okay nvm

hmmm

[naive valley]

but you're on the right track

that inequality holds for all m

[vapid shuttle]

which

[naive valley]

^

[vapid shuttle]

well

I was going to just insert sup x_k

in between the y_m and x_m

and then that would give what I wanted

[naive valley]

nope

can't do that

[mortal trellis]

well sup x_k could be bigger than the particular y_m

[vapid shuttle]

you guys are having fun with this aren't you

[mortal trellis]

yes

[vapid shuttle]

XD

[naive valley]

i am for sure

[mortal trellis]

but thats the hard part of real analysis

[naive valley]

you will too once it clicks for you

(probably)

[mortal trellis]

these little questions

[vapid shuttle]

hmm

is it to do with

[mortal trellis]

these are the questions you have to ask yourself when doing any proofs. are these inequalities true for all m? for a single m? can it be that its true for some and not true for others? and so on

lets take a step back and try to see the bigger picture. what is the definition of sup

[vapid shuttle]

least upper bound

[mortal trellis]

ok so it fulfills two things, its an upper bound and its the smallest one of those

[vapid shuttle]

yes

[mortal trellis]

how can you show that a sup is smaller than some other number

[vapid shuttle]

that number is another upper bound?

[mortal trellis]

yes!

and now think about what you have written above

[vapid shuttle]

so we want to show that

[naive valley]

yep look back at that inequality

[vapid shuttle]

sup x_k is an upper bound of y_k?

[mortal trellis]

no

remember what you want to show

[vapid shuttle]

sup x_k <= sup y_k

oh

do we show

sup y_k is an upper bound of x_k

?

[mortal trellis]

yes!

[naive valley]

and just for reference, recall that you have x_m <= y_m <= sup y_k

[vapid shuttle]

then there it is

^

[mortal trellis]

yes!

[naive valley]

that right hand side is just a number

it doesn't depend on m

[mortal trellis]

its easy to get lost in the inequalities. often its good to try take a step back like this to see whats going on

[vapid shuttle]

true

okay so

sup y_k >= y_m for all m

and y_m >= x_m

then

x_m <= y_m <= sup y_k

then sup y_k is an upper bound of x_k

then sup x_k <= sup y_k

[naive valley]

yes perfect!

[vapid shuttle]

then limsup x_k <= limsup y_k

[naive valley]

which is actually another application of the same technique

because as you probably know,

limsup is actually inf sup

[vapid shuttle]

[mortal trellis]

limsup is the smallest one of these supremums sup x_k for k->infty

"smallest" of course in the sense of inf, not necessarily attained

because the sequence sup x_k is decreasing

[naive valley]

and since the sequence of supremums is decreasing, the inf and the lim of that sequence are the same thing

[vapid shuttle]

why is hte sequence of supremums decreasing

[naive valley]

because you're taking supremums of smaller sets as n increases

[vapid shuttle]

nvm

shh Bungo

I got it

[naive valley]

haha sry

[vapid shuttle]

XD

[naive valley]

but the characterization of limsup as inf sup is often useful

because arguments that work for sup also work for inf with small modifications

that's the end of my soapbox for now, i have to go and wash dishes haha

gl with the rest

[vapid shuttle]

haha ty

goodluck with the dishes

[naive valley]

tx! cheers

[mortal trellis]

whenever you use any "obvious" implication in your proof, like you did in your line 2. think really about why its true. if you cant find an idea (you dont need to find the details yet), then you need to spend more time on that step

also the implication in line 3. you say you have proven it before, so it should not take long to think about the idea of that proof. if you cant, again spend more time on it. you will need that practice

[vapid shuttle]

Okay that is good advice

TY Denascite!

I revisted that proof

and I just proved the other way that they wanted of me too, that liminf xn <= liminf y_n much easier the second time around lol

[mortal trellis]

half the time when I speak with a friend and we are using some basic result of linear algebra or something, we will stop for a few seconds and think about why its true. you need to get in the habit of doing that

yes also good to prove the symmetrical version explicitly

[vapid shuttle]

Okay great

thanks alot

I should be doing that

studying is hard, and it is easier to ignore the difficult bits like that and kind of kid yourself

but you learn less that way

[mortal trellis]

yes

definitely

[vapid shuttle]

cya!

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[nocturne niche]

help pls 😦

[vale wigeon]

show progress?

[vale wigeon]

potentially thorny but ok

keep going

[vale wigeon]

...

did you confuse your own 1 for a 7?

[nocturne niche]

oh

mb

[vale wigeon]

also 3^[-2 log_3(1-x)] is (1-x)^-2 anyway

so you're going to be in medium-large amounts of pain and suffering with a 4th degree equation on your hand

[nocturne niche]

I don't know how to continue

[vale wigeon]

backtrack, divide both sides by 2 before incorporating the 2 into the log on the left.

get log_3(x+3) = -log_3(1-x)

and now raise 3 to the power of both sides to get x+3 = 1/(1-x)

[nocturne niche]

thanksss ^^

.close

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[distant frost]

is there anyone who is willing to hop in a vc with me to explain this

[distant frost]

i have to calculate the formula from the graph

but i don't fully understand how and this also applies for the parabola version

[distant frost]

what i mostly don't understand about this one is
i know that the formula is y = ax + b and B is the point where the lines intercepts the y axes but here i don't have a nice point on the y axes that intercepts it how do i calculate the b here?

[mortal trellis]

you can calculate the a first

a is the slope

rise/run if you heard that before

after that you can calculate b by plugging in one of the points

[distant frost]

yeah that would be y/x

[mortal trellis]

close but not quite

(difference in y)/(difference in x)

[distant frost]

ohh so i would first need to difference between 10 and 14 / differnce between -3 and -4
so then i would end up with 4/-1

[mortal trellis]

which is?

[distant frost]

-4

[mortal trellis]

yes

so a=-4

y=-4x+b

now you can plug in one of the points and solve for b

[distant frost]

how would i exactly do that?

[mortal trellis]

what does it mean for (-3, 10) to be on the line

[distant frost]

like in those are coordinates of the line so they are on the line at a given point

or more they indicate a point where the line is

[mortal trellis]

yes. and in terms of x and y, it means that x=-3 and y=10 satisfy the equation y=ax+b

so 10=(-4)(-3)+b

[distant frost]

ohhh

so then you would just solve it for b

[mortal trellis]

yes

[distant frost]

so then b would be 3 right

[mortal trellis]

no

(you can compare with the graph that the y intercept is not at 3)

[distant frost]

would i neet to multiply the -4 and -3

[mortal trellis]

yes

which gives?

[distant frost]

12

[mortal trellis]

yes

so 10=12+b

so b=...?

[distant frost]

2

[mortal trellis]

no

-2

[distant frost]

oh yeah since you kinda want to end with -2 = b right?

[mortal trellis]

yes

[distant frost]

so then the formula would be y =-4x -2

[mortal trellis]

and you can compare with the graph that the y intercept is at -2. so thats reassuring

yes

[distant frost]

thank you so much for the help

.close

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[grim berry]

HI

[grim berry]

hi*

i need help with uh

thi

s

[naive crystal]

Do you know distributive property?

[grim berry]

yeah but i forgot

can we go through it again

[naive crystal]

Well basically
a(a+b) = a×a + a×b

We distribute a

[grim berry]

wat about the 7'

[naive crystal]

We distribute 7a

[naive crystal]

This was just an example

If there is 45a(a+b), we distribute 45a

[grim berry]

yeah

i know the basics

[naive crystal]

Alr

[grim berry]

so lets do it step by step now

[naive crystal]

Ale

Alr

[grim berry]

7ax3a

21a

^2?

cause its a^1 + a^1

right

[naive crystal]

Yes

[grim berry]

21a^2b^2?

[naive crystal]

Yep

You distributited 7a to the first term... you need to distribute it to evry term

[grim berry]

ob so

ok*

21a^2b^2 is the first term right

[naive crystal]

Yes

[grim berry]

ok next term is 8ab

so

56a^2b?

[naive crystal]

You second term is not 8ab rather its -8ab

[grim berry]

ohhh

you right

[grim berry]

so its -56a^2b?

[naive crystal]

Yes

[grim berry]

alright

21a^2b^2 - 56a^2b - 7a?

[naive crystal]

Yes

,w expand 7a(3ab^2-8ab-1)

[grim berry]

rip

u gotta redo it

[ocean seal]

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[grim berry]

xd

oh what

it fixed ur edit?

how

[naive crystal]

Ye

See

[grim berry]

for this one

I got

-4b^2+3a^1b^1+1a^4

[naive crystal]

,w simplify (-16a²b⁵+12a³b⁴+4a⁶b³)/(4a²b³)

[ocean seal]

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Requested by _basudev#0

[grim berry]

so do i get it wrong

[naive crystal]

Ye

Re check it

[grim berry]

oof

[grim berry]

help pls

[naive crystal]

Wolfram is telling some other stuff so I though so it must be wrong dude, I didn't check manually

[grim berry]

oh word

[lone heart]

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[lone heart]

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[torpid kestrel]

I have 2 non-linear functions f and g that don't cross each other. How can I find the 2 points x_1 and x_2 on those functions that form a line with the shortest distance?
See example in desmos:
https://www.desmos.com/calculator/5r9f3b2mj3

[torpid kestrel]

I'm looking for the green line and not the purple one

In other words, for vectors v_1 = [x_1, f(x_1)] and v_2 = [x_2, g(x_2)], I wish to minimize the L2 norm v_1-v_2

[upbeat hornet]

have you tried setting the derivatives equal

[late mango]

what are the functions

[torpid kestrel]

So far, I've come up with the following loss function:
$$
L = (x_1-x_2)^2 + (f(x_1)-g(x_2))^2
$$

[ocean seal]

lucasyerz

[torpid kestrel]

$$
f\left(x\right)=\sin\left(x\right)
$$

[ocean seal]

lucasyerz

[late mango]

and g(x)?

[torpid kestrel]

$$
P\left(x\right)=x+\frac{\sqrt{2}}{2}-\sqrt{2\sqrt{2}x+0.5} \
$$

$$
g\left(x\right)=P\left(x-2\right)+1.1
$$

[late mango]

and g(x) = P(x-2)?

[ocean seal]

lucasyerz

[modern sedge]

Can't you just set it's gradient to 0? It will be a lot of painful work but it should work

[torpid kestrel]

you can see in the desmos example that I tried, but question is, derivate with respect to what?

[torpid kestrel]

maybe I should solve L for both x_1 and x_2.

[modern sedge]

Gradient is a vector, in your case it would be vector of derivative wrt x and derivative wrt y

both of these should equal 0

Lemme try it....

[torpid kestrel]

I just came up with that formulation, so haven't tried it yet, just figured there'd be a known famous solution. Anyway

[lone heart]

@torpid kestrel Has your question been resolved?

[torpid kestrel]

So after getting a numerical solution I get this red line:
https://www.desmos.com/calculator/ljxomc7rue

I'm not sure if there is a general solution for any function, but doing:
$$
L = (x_1-x_2)^2 + (f(x_1)-g(x_2))^2\
\frac{\partial }{\partial :x_1}(L)=0\
\frac{\partial }{\partial :x_2}(L)=0\
$$

[lone heart]

@torpid kestrel Has your question been resolved?

[mellow grail]

Couldn't you define d(x) to get the distance between f(x) and g(x) (using distance formula) and find the minima using derivatives

[harsh swallow]

that's the gradient

the distance formula is a function of x and y

and by doing the gradient and setting that to 0 you find a peak or a valley

[torpid kestrel]

$$
L = (x_1-x_2)^2 + (f(x_1)-g(x_2))^2
$$

$$
\frac{\partial }{\partial :x_1}(L)=0
$$

$$
\frac{\partial }{\partial :x_2}(L)=0
$$

[ocean seal]

lucasyerz

[torpid kestrel]

yeah, that's pretty much what I did, it's just that I was wondering that the solution for x_1 and x_2 can be writting as a function of f and g.

[harsh swallow]

you have the formulae

to do that

by just filling in f and g

into L

and then doing the derivative

you get the two equations you need to find x1 and x2

[torpid kestrel]

the f and g here are just example functions.

[harsh swallow]

yes but you can fill the functions you want to find in

[keen plinth]

https://www.desmos.com/calculator/yac0illz0o

[harsh swallow]

the problem is whenever there is no solution it might be funky

actually i think there's always a solution

[torpid kestrel]

Yup, exactly what I needed. Ty so much
$$
\left(f'\left(a\right)-g'\left(b\right)\right)^{2}+\left(g'\left(b\right)+\frac{a-b}{f\left(a\right)-g\left(b\right)}\right)^{2}=0
$$

[ocean seal]

lucasyerz

[torpid kestrel]

yeah, ts just a vector so there will always be a solution and if there are many solutions you can always introduce bounds.

[keen plinth]

note that this is actually two equations wrapped into one

[harsh swallow]

i wonder what happens if the solution is at a point where the derivative doesn't exist

[keen plinth]

[
f'(a) = g'(b) = -\f {a - b} {f(a) - g(b)}
]

[keen plinth]

this is what you actually want to solve

[torpid kestrel]

well, I'll be putting it into an optimizer, with an arbitrary f and g, so defining it as a single function is preferred.

[keen plinth]

sure

[2\textwidth] results may vary depending on if you do
[
(f'(a) - g'(b))^2 + \parens {g'(b) + \f {a - b} {f(a) - g(b)}}^2 = 0 \textqq{or}
(f'(a) - g'(b))^2 + \parens {f'(a) + \f {a - b} {f(a) - g(b)}}^2 = 0
]

[keen plinth]

if you're solving numerically

[torpid kestrel]

can I just take the loss average of the two?

or is that inefficient?

[keen plinth]

[keen plinth]

this is the only condition that matters

[torpid kestrel]

Ow, ok, nvm

[keen plinth]

you do need to make sure that you find a global min rather than a local min/max though