help-33

[marble summit]

no g : R->R does not necessary defined on R

[sinful flint]

Yeah

[marble summit]

c'est l'ensemble de départ pas forcément l'ensemble de définition

c'est mon problème

[sinful flint]

Can you translate it?

[marble summit]

g is the partial derivative

[still temple]

.

[echo bough]

let f be a partially differentiable function such that forall (x,y) dfx >0 and let X= .... , so X_ =.......

do they ask you to prove this fact about the closure?

or are we missing something?

closure= adhérence

[marble summit]

yes

if it is true

I think it is

sorry french

[echo bough]

f est continue?

[marble summit]

euh partiellement dérivable

donc continue non ?

[echo bough]

bah non

prend plein de fonction C1 sur R aleatoirement

et fait un millefeuille dnas la deuxieme dimension

sur chaque ligne on peut deriver selon x

[marble summit]

dans ce cas là non f n'est pas nécessairement continue

[echo bough]

genre si y est rationnel f(x,y) =cos(x)

sinon f(x,y)= sin(x)

donc ton argument de la boule ne tient pas

apres j'ai pas pris des fonctions decroissantes mais t'as compris

[marble summit]

bah je suis déçu mdr

l'inclusion de X bar dans A est fausse aussi alors

[echo bough]

tu fabriques un contre exemple?

[marble summit]

non je voulais montrer que c'était vrai

[echo bough]

bah oui mais c'est faux

mdr

[marble summit]

ah bon. ?

ah ok

oui il faut dire si c'est vrai ou faux

[echo bough]

je croyais que tu avais capté avec ce que j'avais dit au vu de ta reponse

en gros crée une discontinuité sur un ouvert selon y

genre si y>0 une certaine fonction

[marble summit]

ça veut dire quoi selon y. ?

ok

[echo bough]

si y<=0 une autre fonction

[marble summit]

et la dérivée par rapport à x doit être négative stric.

[echo bough]

ouais prend f(x) =-x et -x+1

sorry milo this devoluted into french maths

[sinful flint]

You have to prove that for the point (x,y) in R2 such that f(x, y)=0 every ball centred in (x, y) intersects X

Because for every set A we know that A is conatined in his closure

[marble summit]

it may be false

[sinful flint]

Contained*

[echo bough]

f is not continuous so there are counter examples

or i think there are

at least

[sinful flint]

Mhh why?

I may be wrong

[echo bough]

think about this

[sinful flint]

But I thought it was correct

[echo bough]

f(x,y) = -x if y>0 and = -x+1 if not

its partially differentiable

along x

[marble summit]

but the derivative is not strictly negative

ah si

pardon

ok

[echo bough]

and now look at the bit of line between (0,0) and (1,0)

all points on the line have y=0

so the function is -x+1

so the result is > 0 on that bit of line

so its outside the "definition" of the closure

bur those points are inside the closure of x

because any ball around those point intersect with points with y >0

and thus with f<0

[marble summit]

(tu peux le faire en français aussi stp)

[echo bough]

en gros etudie la fonction que j'ai donné

et nottament les points sur le segment

ouvert

entre (0,0) et (1,0)

prouve que ces points sont dans l'adherence de X

[marble summit]

on est d'accord que je dois trouver un point adhérent tel que f(x,y)>0. ?

[echo bough]

mais pas dans l'ensemble

donne

oui voila

mais qu'ils ont f(x,y)>0

exactemznt

[echo bough]

donc la fonction c ca

[marble summit]

en fonction de x alors

[echo bough]

hein?

bah c'est deux points du plan ca fait bien un segment

le segment il est sur l'axe x oui

si c'est ce que tu veux dire

en gros ta preuve aurait marche avec l'hypothese x continue je pense mais la ca permet de fabriquer des bizarreries

[marble summit]

en fait

ça correspond à quel points ça ? "entre (0,0) et (1,0)"

[sinful flint]

I tried a proof

Then I have to go sorry

[echo bough]

bah les points avec y=0

et entre x=0 et x= 1

[marble summit]

ok ok

oui je voi

[marble summit]

c'est ce que je voulais dire

[echo bough]

what you are trying to prove implies that you suppose f continuous

you prove that{x,y| f(x,y)=0} included in the closure

[sinful flint]

I just restricted the f along the x-axis

And the hypothesis says it's derivable

[echo bough]

partially differzntiable

each slice along x is differentiable

but that doesnt tell anything about the y direction

[sinful flint]

Yeah yeah, I meant the phi function is derivable

[echo bough]

as i said this fullfills the partially differentiable wrt x: f(x,y)= cos(x) if y is rationnal and f(x,y)= sin(x) if not

anyways your proof is probably good but it doesnt prove whats asked

[sinful flint]

Oh I see I should have added that the points (x, y) such that f(x,y) >0 cannot stay in the closure

[echo bough]

like in the counter example i gave, there are points inside the closure but with f(x,y)>0

yes

[marble summit]

il faut que je trouve une suite Xn et une autre Yn telle que f(Xn,Yn)<0 et telle que Xn->1/2 (par exemple) et Yn->0

déjà pour Yn je peux prendre la suite constante nulle

[echo bough]

pas besoin de suite, parle des boules autour des points

pour montrer qu'il sont dans l'adherence

[marble summit]

donc la boule B( (1/2,0), r) inter X n'est pas vide ?

[echo bough]

pour tout rayon r>0

oui

[marble summit]

ah oui

ok

[echo bough]

c'est le plus simple pour montrer qu'on est dans l'adherence en general

[marble summit]

dans cette boule il y a (1/2,-r)

[echo bough]

oui voila

[marble summit]

enfin -r moins un epsilon

[echo bough]

enfin +r/2

[marble summit]

oiui

[echo bough]

enfin ca depend de quel coté t'as mis les truc

mais attention faut que le coté "ouvert" ce soit le coté negatif

[marble summit]

et f(1/2,r/2)=-1/2

donc ce point est dans X

[echo bough]

et f(1/2,0) = 1/2

[marble summit]

ahhhh

moins non ?

ah oui

nan

ok

je vois

[echo bough]

en gros si tu met les bonnes inegalites stricte ou pas dans ta definition ca marche

[marble summit]

donc ce point est adhérent et pourtant l'image par f est positive

e comprends

je crois que c'est bon

merci beaucoup mec sérieux

j'aurais galéré pendant longtemps encore

[echo bough]

en gros la lecon a retenir: derivable partiellement ca veut pas di tout dire continu

il faut imaginer le cas le plus bizarre possible dans ce genre de questions

pour voir si ca va etre vrai

[marble summit]

yes

[echo bough]

t'aurais aussi pu prendre un truc du genre e(x) et -e(x) pour les fonctions pour avoir carrement que l'ensemble proposé soit egal a x

et soit un ouvert

et donc peu de chance d'etre une adherence

[marble summit]

t'es en quoi en maths ?

[echo bough]

et en fait la preuve serait encore plus rapide

[marble summit]

je veux dire en niveau d'étude

[echo bough]

puisque f est jamais egal a zero

[marble summit]

ok

[echo bough]

donc l'egalite avec x est simple

en fait e(x) marche pas

mais un truc avec arctan décalé

et retourner

puisuqe e(x) c'est croissant

je suis ingenieur la

ptet une these en informatique theorique bientot

mais pas sur

[marble summit]

t'as un master en maths ?

[echo bough]

non

j'ai un diplome d'inge et un master en informatique

[marsh citrus]

@marble summit Has your question been resolved?

[marsh citrus]

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[main mica]

Need help with 179 @rugged sierra If your not tired do you mind helping me with one more problem

[marsh citrus]

@main mica Has your question been resolved?

[main mica]

<@&286206848099549185>

[marsh citrus]

@main mica Has your question been resolved?

[main mica]

<@&286206848099549185>

[main mica]

<@&286206848099549185>

[tiny crown]

Same answer. Best of luck.

[main mica]

Yea i dont get your answer

[tiny crown]

All cool. See if you can attract someone else.

Or, you could try squaring it 🙂

[main mica]

I tried I can’t get it right

Can you write it on paper and show me?

How it would be

@tiny crown

[marsh citrus]

@main mica Has your question been resolved?

[tiny crown]

$LHS = \sqrt \frac{4+\sqrt(16-x)}{2} + \sqrt \frac{4-\sqrt(16-x)}{2}$

[elfin berry]

G. Spark

[tiny crown]

$LHS^2 = \frac{4+\sqrt(16-x)}{2} + \frac{4-\sqrt(16-x)}{2} + 2\sqrt\frac{4+\sqrt(16-x)}{2} \sqrt \frac{4-\sqrt(16-x)}{2}$

[elfin berry]

G. Spark

[tiny crown]

$LHS^2 = \frac{4+\sqrt(16-x)}{2} + \frac{4-\sqrt(16-x)}{2} + 2\sqrt{\frac{4+\sqrt(16-x)}{2} \frac{4-\sqrt(16-x)}{2}}$

[elfin berry]

G. Spark

[tiny crown]

$LHS = \frac{4}{2} + 2\sqrt{\frac{4^2-(16-x)}{2}}$

And so on... did I go wrong?

$LHS = \frac{8}{2} + 2\sqrt{\frac{4^2-(16-x)}{2}}$

That structure like (a-b) and (a-b) seems always to work out well when squared.

$LHS^2 = \frac{8}{2} + 2\sqrt{\frac{4^2-(16-x)}{2}}$

[elfin berry]

G. Spark

[tiny crown]

Any help? Maybe you can do the rest?
I suspect the probglem is cunningly crafted to make it work out ok.

But, check whether I goofed up, on the part I tried to illustrate witj.

[marsh citrus]

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[violet prawn]

is anyone able to give me the answer I already tried getting help but couldnt figure it out

[marsh citrus]

@violet prawn Has your question been resolved?

[marsh citrus]

@violet prawn Has your question been resolved?

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[marsh citrus]

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[azure bone]

Yo I wasn't taught how to get the midline

From an equation

Is it like

5

[gleaming pecan]

The midline is the horizontal centre line about which the function oscillates above and below. The midline is parallel to the 𝑥-axis. And the midline is affected by vertical translations.

[azure bone]

Does that mean it's 5

Cuz like the midline would normally be like y=0

But it's like higher by 5 fr

Omg I got it right

.close

[marsh citrus]

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[azure bone]

Idk wahr to do

Yo wait

Do I go down by 4.1

To the 6.7

Thing

Y=2.6

[past frigate]

Have you found the equation to the graph?

[azure bone]

Nah

Idk how to do that

Omg I got it right

Ez

Ez

Ez

🥱 🥱 🥱

[pseudo jackal]

Did you guess?

[tiny crown]

Nar, you are right. Maximum value - amplitude = mid value

[azure bone]

This is 3.5 right

[azure bone]

Bro skipped all my deductive reasoning 😭 😔

[tiny crown]

Yep

[azure bone]

Guys I got it right

I'm him

[tiny crown]

You are the man.

A problem can be made arbitrarily complicated.

[azure bone]

NVM I think I met my lucks end

I have no clue

[tiny crown]

No luck involved.

[azure bone]

Help

[tiny crown]

Ok, so the x-axis this time

Still easy

In phase terms, how far apart are the max & min points?

[azure bone]

Idk wat phase terms means

[tiny crown]

Or - What's a "period"?

[azure bone]

10.6

[tiny crown]

Was meaning in words.

[azure bone]

Like how far it goes until it like

Repeats

The cycle

[tiny crown]

Yes, in terms of maxes and miniums

or minima for purists

"repeats" huh? So returns to the same value?

[azure bone]

Uhhh

Idk

It starts the cycle again

How far it goes until it does that

[tiny crown]

Yep, so from a maximum - to the next maximum. repeats.

or minimum to the next mimimum

So they told you haw far from a max to a min.

[azure bone]

Yo wait

Like

π each time

so the period is 2π right

Wait

no

uhh

[tiny crown]

How far (on the x-axis) from the max to the mimimum?

[azure bone]

4π each period

[tiny crown]

Yep

[azure bone]

cuz 2π each min to max

[tiny crown]

Too fast for me.

[azure bone]

In terms of x

[tiny crown]

Yep

period = 4$\pi$

[elfin berry]

G. Spark

[tiny crown]

All good

[azure bone]

OKOK

isn't it like

Idk

[tiny crown]

Tell the bot to close when done.

Oh. more?

[azure bone]

2π/5π

2/5

[tiny crown]

As you said, how long (in x) until it repeats?

cos(theta) repeats after 2pi

[azure bone]

Uhh I'm just using this thingy fr

so 2/5 right frfr

[tiny crown]

so pick two, maybe x = 0 and then when the argument is 2pi bigger

[azure bone]

Wahr

[tiny crown]

5 pi (x+period) = 5 pi (x)+ 2pi

5 pi (period) = 2 pi

period = 2/5 - heh, you are right again.

[azure bone]

Ez

[tiny crown]

Tallent

[azure bone]

https://tenor.com/view/luke-skywalker-green-lightsaber-mandalorian-jedi-luke-gif-19627631

Me

Ty

.close

[marsh citrus]

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[muted glade]

If anyone can read what I did, could you try to see where I messed up?

[hollow oracle]

I think you made a sign error when factoring out the negative in the last line

[fallow wind]

the sign error is here

should've been $-\left(\frac{4x^{3}y^{7}+3x^{2}}{3y^{2}+7x^{4}y^{6}}\right)$

[elfin berry]

Combustion

[muted glade]

Ohh I see thanks

This is the right answer though so I’m still doing something wrong but idk what

[fallow wind]

it's the same as this

factor out x^2 from the numerator

then factor out y^2 from the denominator

[muted glade]

I’m not sure how to get the y^5 though

[fallow wind]

y^7 after factoring out y^2 becomes y^5

[muted glade]

Sorry I’m a little confused doesn’t the 3 have to have y as well to factor out the y?

On the numerator

[fallow wind]

oh wait yeah

good point

it's probably an error in the answer

[muted glade]

Aight thanks for the help

I hate these questions they’re always so long

.close

[fallow wind]

[marsh citrus]

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[marsh citrus]

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[marsh citrus]

@blazing sorrel Has your question been resolved?

[marsh citrus]

@blazing sorrel Has your question been resolved?

[still temple]

There will be infinite solutions

Because linear systems with more variables than equations

[marsh citrus]

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[marsh citrus]

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[still pumice]

How can the uniqueness of the solution f(x) = 1 - x be proven? I observed that f(x) = f(x + 1) + 1 numerically during my experimentation, but I couldn’t prove it using the conditions in the question. I’m guessing that if I could prove this, that would in turn prove that f(x) = 1 - x is the only solution, but I’m not sure. I saw induction being suggested online for such a problem, but I can’t really understand how that would prove uniqueness. Thanks in advance! :)

[marsh citrus]

@still pumice Has your question been resolved?

[marsh citrus]

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[void dagger]

what type of polynomial is 4a^2y and -x^3y^2 = 4x?

[marsh citrus]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

@void dagger Has your question been resolved?

[marsh citrus]

@void dagger Has your question been resolved?

[void dagger]

can anyone check my answers real quick

[marsh citrus]

@void dagger Has your question been resolved?

[still temple]

they look correct

[marsh citrus]

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[sharp mauve]

do you need to apply product rule aswell after chain rule?

[rain mango]

bro got that PHD math

[still temple]

what u have is perfectly fine

so no, no chain rule. you are done differentiating

[sharp mauve]

.close

[marsh citrus]

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[proper crane]

i ahve a question

[proper crane]

am i correct

[desert dirge]

seems so

[proper crane]

good!

[severe island]

lol where si the graph

[stark trail]

?

[hidden plaza]

[severe island]

if blue is the graph then how?

[stark trail]

Is the graph the blue?

[proper crane]

yes.

i think so

[stark trail]

Then no?

[desert dirge]

i presumed it was the black line going across

[stark trail]

Do you know the vertical line test

[native frigate]

the axis?

[desert dirge]

oh im blind as hell

thought the axis was lower then the line

[proper crane]

oh if blue is the axis then yeah i was wrong.

[white lion]

how did you define a function in class

[stark trail]

Blue isn’t the axis

[white lion]

cuz that's x = 5

[stark trail]

That’s not a function just because you can graph it as an equation

Functions have 1 output for every 1 input

Input x=5 to this

[earnest sage]

Solve for X

[stark trail]

The output can be anything

So clearly not a function

[stark trail]

!occupied

[marsh citrus]

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

[white lion]

yea for every 5 you get x 👍🏻

[proper crane]

i haft to be right on this right?

[stark trail]

What???

[white lion]

just joking

[hidden plaza]

f({5}) = R moment

[proper crane]

im confused on this one. is it c? if not then im clueless

[stark trail]

!onequestion

[marsh citrus]

It is suggested that you limit yourself to one question per help channel, opening a new one once your original question is answered and your original channel has been closed. This is to make your channel easier to follow for potential helpers and can bring attention to the fact that your question has changed.

[proper crane]

k

.close

[marsh citrus]

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[broken rover]

need help

[broken rover]

hello

[tall pewter]

[marsh citrus]

@broken rover Has your question been resolved?

[marsh citrus]

@broken rover Has your question been resolved?

[marsh citrus]

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[high igloo]

i got ||f(n)=1 + n^2 - n(n-1)/2 using eulers characteristic and also i got f(n)=f(n-1) + n from logic||

[high igloo]

for the proof bit im not completely sure how to approach it

[bleak ibex]

second statement is probably easier to prove, when you add a line, notice it crosses exactly n - 1 old lines, which means n regions, and all of them get split into 2 parts

[high igloo]

yeah

when they say find a formula would they be fine with something recursive

[bleak ibex]

well, you can deduce a formula from the recursion, it should be the first one if that method was correct

[high igloo]

oh cool

yeah

[bleak ibex]

best way is probably giving a formula and then using induction to prove it, with the above being the induction step

[high igloo]

right

thanks

[marsh citrus]

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[ivory marlin]

I'm somehow doing this wrong and I'm not sure. Can someone tell me.

I'm trying to find magnitude of the force on the 1nC charge

So what I did was first rotate this figure so the 1nC will be on the bottom right

Then i drew the forces, so for Force of 1 and -2, it will go to the right on the x axis because its attracted towards the positive

for force of 1 and 2, it will be repelled towards the southeast

So then i did my calculations, for force of 1 and -2, I got 1.8*10^-4 towards the right

for foce of 1 and 2, i got 1.8*10^-4 towards south east

Do components, 1.8*10^-4 x cos60 is 9x10^-5

1.8x10^-4 x sin60 is 1.56x10^-4

Since both the x components go to the right, i simply add them and get 2.7x10^-4

I do pythagorean theorem, so square the compoennts, and find the root, and I get 3.12x10^-4

but this is wrong apparently?

@sharp wing

[marsh citrus]

@ivory marlin Has your question been resolved?

[marsh citrus]

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[stark trail]

S is clearly a subset of Rn and f is >=0 and integrable on every compact subset of S, so I think this rewrite into an improper integral is fine, but I was wondering if I was applying this theorem correctly at that step?

Another point of contention for me was if I was applying Arzela's DCT properly.

[marsh citrus]

@stark trail Has your question been resolved?

[blazing pulsar]

Your writing looks a bit off to me, but I'm a bit too rusty to really comment on this

The obvious problem you have is that you are taking the limit as C approaches 1 from both sides, but you don't have convergence from the right.

The way I would set this up is to use Arzela's on the Geometric series to rewrite the integral as $\lim_{c\rightarrow 1^{-}}\sum_{n=0}^\infty}\int_0^c \int_0^c (xy)^n dydx$ and then just do the calculation.

[elfin berry]

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

[blazing pulsar]

The cheeky way to solve the problem is to just directly use basic calculus to show that the integral equals \zeta(2)

[stark trail]

we're told to solve it in this way

[stark trail]

also we don't need both integrals going to C

just 1

because at x=1 it is still defined

unless x=y=1

so it is only the second integral that needs to have C on top

[stark trail]

and then I did this, right?

[blazing pulsar]

I'm just telling you what I would do. I think it's easier to just cap them both off at c < 1 than to have to worry about the technicalities.

[stark trail]

What is different between what I did and what you're suggesting, besides the extra C

I'm not seeing it

[blazing pulsar]

Well, like I already said, the big thing is I am taking a left handed limit, the rest is just a cleaner presentation

[stark trail]

Ah okay

so just fixing it into a left handed limit

Did it seem like, if it were a left handed limit instead, that I applied Arzela's properly to rewrite the sum outside of the integral?

[blazing pulsar]

Well, I thought you needed the geometric series to be uniformly bounded too but it seems fine to me. It's been a while though

[stark trail]

we need it to be pointwise convergent

to use Arzela

I can show you the theorem

[blazing pulsar]

Yes, a uniformly bounded sequence of functions which converges pointwise

[stark trail]

it is uniformly bounded aswell

for each

k

the geometric series is

[blazing pulsar]

Sure, but you didn't say that.

[stark trail]

I did

or did I say it wrong?

I was meaning to say it here

[blazing pulsar]

That's probably fine, it just doesn't appear where I was expecting it to

On an unrelated note, what exactly is your degree you are studying? I'm confused because you seem to jump back and forth between asking first year problems and 3rd/4th year problems.

[stark trail]

Math

Do I do that?

I am taking 2 math classes and one is more basic it is called "Art of Problem Solving" so it is like a proofs & tricky questions class but nothing super involved. This class however is Honors Accelerated Advanced calc so it should hopefuly be more than 1st year questions

[blazing pulsar]

Oh okay

[stark trail]

Yeah

I've already done the basic calculus sequence

Anyways TY for your help

.close

[marsh citrus]

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[neat siren]

what I've tried so far

[frail orbit]

you're choosing for (0,3) to be on the the curve, but this is not necessarily the case

[neat siren]

I'm not?

[frail orbit]

maybe i'm just misreading

but the 'x' in point slope is different then 'x' of tangent/derivative

i think you are overthinking this

you know the equation of the tangent line

[neat siren]

yeah

[neat siren]

yes

but I'm just failing to derive a relation to find c

[frail orbit]

so you could set it equal to the curve...and this would give you where it 'touiches' the curve

[neat siren]

hmmm after that?

[frail orbit]

and you have that expression for the derivative and relating c...

[neat siren]

ok let me try

[frail orbit]

hmm...

[neat siren]

ok

[frail orbit]

ok i thought i was going insane for a sec but ye my approach works

[neat siren]

i solved it

[frail orbit]

:)

[neat siren]

but for the point of intersection

I'm getting two values

shit idk let me send

[frail orbit]

well only one makes sense

its -1 and 1, right?

but look at your function

:)

[neat siren]

yes

[frail orbit]

look at c/x+1

[neat siren]

ohhh yeah not defined at -1 oops

[frail orbit]

so it eliminates one one of them

[neat siren]

alr so I don't consider that case

yep

[frail orbit]

now you can plug in x=1 into the line

[neat siren]

so then I get f as 4

c*

yeah yeah

[frail orbit]

yes

:)

[neat siren]

thanks

.close

[marsh citrus]

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[frail orbit]

@neat siren

[neat siren]

yep yep

[marsh citrus]

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[fringe fulcrum]

Let V be an inner product space over R with orthonormal basis B = {v1, ..., vn} and let f: V->R be a linear transformation. Show that there is a unique vector w which is an element of V such that f(v) = <v,w>. Hint: Consider the vector Sum(f(vi)vi)

[fringe fulcrum]

I'm trying to prove existence by working with the formula, but can't figure out from there.

[marsh citrus]

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[vernal basin]

i have a question with double improper integral

[vernal basin]

i was asked to find $$\int_1^{\infty} \int_0^{\ln{x^2+x+1}} e^{-(|x|-|y|)} dy dx$$

[elfin berry]

omgatriple

[vernal basin]

well to find whether it's divergent or convergent

i drew the graph of what it looks like, don't mind the weird bit before the green line

but basically, y is bounded by f(x) where f is an increasing function

and x is unbounded

so i thought that it would be divergent, and i immediately tried to shorten the bound and simplify the question to get a smaller integral

but the answer key just started off with, we know this integral is convergent and we will do this via...

like how are u just supposed to have an intuition with this stuff? i'm actually so lost with it, not to mention i spent like 45 minutes on my divergent path, just to find out i lost the coin toss at the beginning

how do u know if ur wrong and there's a way or if there is no way

[marsh citrus]

@vernal basin Has your question been resolved?

[marsh citrus]

@vernal basin Has your question been resolved?

[marsh citrus]

@vernal basin Has your question been resolved?

[marsh citrus]

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[safe briar]

oh you managed to solve it?

nice

[vernal basin]

wait no but i was gonna ask another question

[safe briar]

enter .reopen please

[vernal basin]

.reopen

[marsh citrus]

✅

[safe briar]

thank you

what's your question

meanwhile I'm writing out some of my work here

I managed to deal with the inside integral as follows
\begin{align*}
\int_1^{\infty} \int_0^{\ln{x^2+x+1}} e^{-(|x|-|y|)} dy dx &= \int_1^{\infty} \int_0^{\ln{x^2+x+1}} e^{-(|x|-y)} dy dx\
&= \int_1^{\infty} \int_0^{\ln{x^2+x+1}} e^{-|x|+y} dy dx\
&= \int_1^{\infty} [e^{-|x|+y}]_0^{\ln{x^2+x+1}} dx\
&= \int_1^{\infty} [e^{\ln{x^2+x+1} - |x|}-e^{-|x|}] dx\
&= \int_1^{\infty} [x^2e^{x+1 - |x|}-e^{-|x|}] dx
\end{align*}

[elfin berry]

_Kookie

[safe briar]

now the inside is clearly increasing

therefore the integral will go to infinity

@vernal basin

[vernal basin]

hmmm but the integral is conergent

cuz u can do $e^{-x} e^y$ and then remove the $e^{-x}$ term from the inner integral and just do an integral $\int_0^{\ln(x^2+x+1)} e^y dy$

[elfin berry]

omgatriple

[vernal basin]

like i'm just wonedring how u get the intuition whether an improper integral is convergent or divergent

[safe briar]

e^{-|x|} obviously goes to zero as x -> infinity

but

$x^2e^{x + 1 - |x|} = x^2e^1 \to \infty$ as $x \to \infty$

[elfin berry]

_Kookie

[safe briar]

oh wait

hang on

your upper bound on the inner integral was $\ln(x^2 + x + 1)$?

[elfin berry]

_Kookie

[safe briar]

not $\ln(x^2) + x + 1$?

[elfin berry]

_Kookie

[safe briar]

in that case, this changes everything

[safe briar]

I was going by your original image here

the integral does converge now

because the double integral reduces to $$\int_1^\infty e^{-|x|}(x^2 + x) dx$$

[elfin berry]

_Kookie

[safe briar]

exponential with negative power will converge to zero faster than any increasing polynomial

furthermore, you can easily integrate this using integration by parts

so you'll have a definitive answer for this too that isn't infinity

[vernal basin]

sorry i mistyped

[safe briar]

no you're alright

[vernal basin]

but how would i, at the start when i just see the question, make a guess on which way to proceed in

i did understand eventually how it worked, but i just wasted a lot of time trying to show it's divergent since that's what it seemed like

[safe briar]

Honestly, if I were you, and I saw this at a glance, I would have no idea if it was convergent or divergent.

if I was forced to evaluate this integral myself

I would do it normally

start from the inside

work my way to the outside

[safe briar]

it was only after evaluating the inside integral to arrive at this

when I was able to tell if it was a convergent or divergent integral

If I was told to guess right off the bat without showing any working I'd evaluate the inside integral in my head

that's how unsure I was at the beginning

[vernal basin]

i see, i will try to kepe that in mind

improper integrals actually make me wanna die

but thank you for your help

.close

[marsh citrus]

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[frosty geode]

Having some trouble with this problem, its regarding Bayes Law/Theorem and Conditional Probability

[frosty geode]

The solution to 2 for P(A) is 0.58

I tried both the Tree Diagram and Bayes' Theorem methods, but I'm just getting stuck/confused

This is Bayes' Theorem/Law, which states its all good results occurring over all possible results that could occur

In this instance, the good event is B occurring given A, and all the events that could occur are the events that can occur for A

I guess what confuses me is, how do I actually do this in practical terms?

I've tried laying it out like a tree diagram, but I feel like I must be doing it wrong

I set it up similar to this

[marsh citrus]

@frosty geode Has your question been resolved?

[frosty geode]

<@&286206848099549185> So, I found the answer, but in case anyone in the future asks how you would solve a question like this using these formats of probabilities using like P(A), P(A|B), etc. The formula for Bayes' Theorem is:

$\frac{P(A|B)P(B)}{P(A)}$

[elfin berry]

Huntifer

[frosty geode]

.close

[marsh citrus]

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[marsh citrus]

@still temple Has your question been resolved?

[marsh citrus]

@still temple Has your question been resolved?

[marsh citrus]

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[marsh citrus]

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[marsh citrus]

Please don't occupy multiple help channels.

[main mica]

Need to find the x

[severe island]

it's power of a point or something

i can't really read your handwriting

is that 9 and 47

[main mica]

yea outside is 9

inside is 47

[severe island]

what is this?

[main mica]

the other ones outside is x

[main mica]

inside is x+72

i was writing fast mb

[severe island]

okay so basically 56 * 9 = (x+x+72) * x

from power of point

now you just solve for x

it's a quadratic

[main mica]

that wrong

right answer is 84

[severe island]

,w 9 * 56 = x(x+(x+72))

[elfin berry]

Results provided by WolframAlpha

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Requested by kannaaaaaaaaaaaaaaaaaaaaaaa

[main mica]

56??

its 47

[severe island]

is the right answer really 84?

[main mica]

yup

[severe island]

47 + 9 = 56

[main mica]

oh yea

[main mica]

why is 2x here

sould it not be x?

[severe island]

x+ (x+72) = 2x + 72

[main mica]

yup 6 is correct

[severe island]

it's CB * CA = CD * CE

[main mica]

72+6 is 78

and plus 6 is 84 and thats the answer

[severe island]

okay but that wasn't your question

your question was to find x

anyway

[main mica]

can you tell me the formula you used

[severe island]

I just did

[severe island]

Look at this

[severe island]

and this

[main mica]

yea i askd the question badly

[severe island]

CB = 9
CA = 9 + 47 = 56

CD = x
CE = x + x + 72 = 2x + 72

which is how i got: (56)(9) = (x)(2x+72)

then you solve the quadratic and discard the negative solution

[main mica]

i get it yup

i ot 1 more problem tho

got

[severe island]

[main mica]

wait 2 mins

Need to find the sides

@severe island

[severe island]

if that line passes through the center of the circle/triangle

then that line is a bisector

[main mica]

yea

it is forgot to say

[severe island]

so you have that the line is perpendicular to the side with length 18

and you can use angle bisector theorem i guess

[main mica]

so 10/8 eaquels what?

[severe island]

use pythag to find the outer edges

you have the outer edge of the triangle with 10 and 5x as side lengths:

,, \sqrt{10^2 + (5x)^2}

[elfin berry]

kanna

[severe island]

the outer edge of the triangle with 8 and 5x as side lengths would similarly be:

,, \sqrt{8^2 + (5x)^2}

[elfin berry]

kanna

[severe island]

now it's just applying the ratio of the angle bisector theorem and solving algebraically

[main mica]

so i solve thees first right

[severe island]

what's there to solve right now

[main mica]

this

[severe island]

nothing to solve there

,, \frac{8}{\sqrt{8^2 + 25x^2}} = \frac{10}{\sqrt{10^2 + 25x^2}}

[elfin berry]

kanna

[severe island]

now you can solve i guess

[main mica]

why do i need the sqrt tho?

[severe island]

doesn't that follow from pythag?

,,c^2 = a^2 + b^2 \implies c = \sqrt{a^2 + b^2}

[elfin berry]

kanna

[main mica]

oh ok

[severe island]

anybody i have to go

[severe island]

oh lol

u need to find the sides

bruh

[main mica]

wast all of that not the way?

@severe island

befour you go

.close

[marsh citrus]

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[cold rock]

I need help solving this

[cold rock]

I did this and got 17/55 and then arccos it but im getting a small decimal?

[native frigate]

that seems more complicated than it has to be?

one sec

oh i see

[native frigate]

what did you get

i did this and got ~72 degrees

[cold rock]

one sec

[native frigate]

cant wait for your answer to be in radians

[cold rock]

1.256?

[native frigate]

,w 1.256 radians to degrees

[mighty bramble]

that is correct

[elfin berry]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by tuzis

[cold rock]

So I might just have the wrong mode on?

[native frigate]

yeah whoda thunk

yes your answer is radians

[cold rock]

Well that explains the confusion

[mighty bramble]

yeah convert to degrees when needed

[cold rock]

Thank you

.close

[marsh citrus]

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[lofty fulcrum]

<@&286206848099549185>

[marsh citrus]

@lofty fulcrum Has your question been resolved?

[past spoke]

well you are right

4pi would make it 0

4pi > 9

but

that problem doesn't seem that much trivial

maybe the integral is supposed to be undefined there or something

[marsh citrus]

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[cobalt sentinel]

why does the sum equal to 1

[glass perch]

You can move out the term on the left and notice that the remaining sum is just exp u substituted with u= lampta*e^t

[cobalt sentinel]

but still, how does that sum to 1

[glass perch]

e^(-yt)*e^(yt) = 1

[cobalt sentinel]

Am I doing smth wrong here?

[glass perch]

Are you also in stochastic 1? 🙂

[cobalt sentinel]

this is intro probability, im taking stochastic processes next sem actually haha

[glass perch]

So that's just exp(ye^t-y) which is the same as they got

[cobalt sentinel]

oh yea thats true

[glass perch]

Happens

[cobalt sentinel]

yea its this that was throwing me off

[glass perch]

Just move the e^(-ye^t) out and the sum just becomes e^(ye^t) and then they cancel

[cobalt sentinel]

ah ok

yea that makes more sense

[glass perch]

I think you did it indirectly in (basically) the same way

[cobalt sentinel]

lol

[cobalt sentinel]

gotcha

thanks error

appreciate the help

[glass perch]

You're welcome

[cobalt sentinel]

.close

[marsh citrus]

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[marsh citrus]

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[boreal heath]

can someone check for me?

[marsh citrus]

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[boreal heath]

oh whoops

.close

[marsh citrus]

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[random sail]

So I have come up w the ceiling for the question being ⌈8/4⌉ = 2 and using the pigeonhole principle i concluded k = 9. Im not sure if my working or my answer is correct

[marsh citrus]

@random sail Has your question been resolved?

[cunning fiber]

yeah I got 9

I'm not quite sure what your logic is but

[random sail]

oh i omitted alot

from the text im sorry

[cunning fiber]

yeah I figured lol

nah you're good

[random sail]

so first i paired all numbers

such that the two selected numbers make a ps

so

[cunning fiber]

Does this mean you want me to check your logic?

[random sail]

no i was asking

whats the proper way to do it

without like cheeseing it

cause ngl i did the floor of ⌈n/4⌉=2

and thought

hey n = 8

[cunning fiber]

list out all the possible pairs of perfect squares

[random sail]

thus + 1 but im not sure if that makes snese

[cunning fiber]

there's like 5 or 6 iirc

not that many

then the construction is pretty trivial

[random sail]

the constructin is what i dont get

i got 5 pairs

but then what?

[cunning fiber]

can you list them out

I'm too lazy to find them again lol

[random sail]

np

1,4 1,9 2,8 3,12 4,9

[cunning fiber]

so we have numbers that aren't in any pair: 5, 6, 7, 10, 11
then we have numbers that are in one pair: 2,3,8,12
then we have numbers that are in 2 or more pairs: 1,4,9

see if you can figure out what to do from here

[random sail]

combination?

nCr

[cunning fiber]

🤷‍♂️

see if it works

[random sail]

im so sorry im rlly not good w this

ill try it out

[cunning fiber]

don't be sorry

as long as you're making a genuine attempt

usually the conversation is productive

it's only when you respond with only "idk" that ppl get annoyed

[random sail]

i c ty for the help

but to be sure

the answer is 9?

[cunning fiber]

yeah that's what I got

[random sail]

oh its worth noting that the numbers cant repeat in the set

ie 5,5 doesnt count

does that somehow change how we do it?

[cunning fiber]

I didn't use that explicitly

but maybe I used it subconciously

[random sail]

i mean u did obtain the right amt of pairs so theirs no doubt you probably read the question and j autopilot the restriction

[cunning fiber]

🤷‍♂️

[random sail]

ok wait i think i got it

so i used the pigeon hole principle

wait

nvm

ok so

i decided to

(|S-(A1∨A2∨A3∨...)| +3) + 1
(5+3)+1

[cunning fiber]

I did it a bit more caveman style

clearly you need 5, 6, 7, 10, 11

then putting the numbers in one pair

and then put it one element that's in 2 or more pairs

done

[random sail]

im sorry

why do we need 5 6 7 10 11 12 13 14 16

15*

i thought we excluded them

[cunning fiber]

you kinda need to put them there as placeholders

for lack of a better phrasing

[random sail]

i asked my peers

they got k = 12

.close

[marsh citrus]

Channel closed

Closed by @random sail

Use .reopen if this was a mistake.

[marsh citrus]

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[thorn marsh]

why does the answer use integration by parts?

[thorn marsh]

i subbed u for 2y and then just solved tan^-1 x = 1/(x^2 +1)

[past frigate]

Ye that is fine too

[thorn marsh]

but theres no way im getting an answer that has pi and 2ln here

[past frigate]

No wait

But to integrate tan^-1 x you need by parts

[past frigate]

That's the derivative

[thorn marsh]

oh wtf

💀

ok ty

[past frigate]

Lol

[thorn marsh]

.close

[marsh citrus]

Channel closed

Closed by @thorn marsh

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[pastel storm]

Is there a formula for the number of unique rationals that you can make only use integers from 1 to x

[sharp harness]

depends what you mean by "make"

...and "numbers from 1 to x", whether that's integers or rationals or reals or whatever will affect the answer

[pastel storm]

I was thinking integers

[pastel storm]

Like in form p/q where p and q are only integers from 1 to x

[marsh citrus]

@pastel storm Has your question been resolved?

[pastel storm]

<@&286206848099549185>

[unborn chasm]

Any open interval of the reals has countably infinite rationals

[pastel storm]

That is not the question I asked

[unborn chasm]

Oh I see what you meant now. There would be
$1 + 2\sum_{k =2}^x \phi(k)$

[elfin berry]

chencking

[unborn chasm]

Every coprime pair (m, n) corresponds to m/n and n/m unless m = n = 1.

(Here, $\phi$ is the Euler totient function)

[elfin berry]

chencking

[pastel storm]

is there a closed form forumla?

or nah

[unborn chasm]

https://en.wikipedia.org/wiki/Euler's_totient_function

No, there is not.

[pastel storm]

then this won't help me

damn

[unborn chasm]

Won't help with what? Are you trying to solvve a problem?

[pastel storm]

I was trying to find how many rational numbers there are (yes I know people say there are aleph null but that's not what I'm doing)

[sharp harness]

i feel like any way you do that is just going to be really arbitrary because of the reasons why actual infinite cardinals act so differently to finite cardinals...?

like why are you saying there are 2*inf-1 integers and not 2*inf+1, or inf because we enumerate them as 0, 1, -1, 2, -2, 3, -3, ...

[pastel storm]

2*inf-1 integers because for every natural there is a unique corresponding negative except for one case (which is 0)

[sharp harness]

if "inf/2" is meant to mean something different than "inf" then you have to either totally abandon anything resembling identity of indiscernibles, or say that the same set contains different numbers of things depending on what structure you put on it, and both of those are weird

[unborn chasm]

Is this a homework problem? The question doesn't really make sense. You seem to be taking the argument for sets being countably infinite and restricting to finite subsets to get some type of cardinality measure (which is not really mathematically sound but setting that aside).

The problem is the enumeration for the rationals isn't natural the way the enumeration for the naturals and integers are.

[unborn chasm]

Two times infinity is still infinity.

[pastel storm]

that is not true in all systems

[sharp harness]

in the first case, there are multiple distinct sets with different numbers of things that as far as i'm concerned are all "the set of integers", they are isomorphic rings

in the second case, it doesn't mean anything to ask "how many even natural numbers are there" without specifying what operations you're putting on the set of even natural numbers - addition? multiplication? comparison?

[pastel storm]

I have no idea what you mean by either but the second case seems way more wrong then the first

[unborn chasm]

Like what? I am not aware of any system where two times infinity is finite.

[pastel storm]

no one said it was finite

[sharp harness]

ok so if we take the set of natural numbers, and declare that even numbers are "positive" and odd numbers are "negative", and give it new + and * operations that act exactly the same as + and * of integers

this set still has only "infinity" elements and not "2 * infinity - 1", but it acts exactly the same as the "actual" integers in every way that actually matters

[pastel storm]

I mean I guess?

but yea there would still be less elements

[sharp harness]

well in that case, how do you know that this isn't how the set of integers is actually constructed?

maybe the "real" integers are this construction and there really are only "infinity" of them

[pastel storm]

because all naturals are integers but there is at least one integer that is not a naturals

[sharp harness]

...which one

do you have an example

[pastel storm]

-1 is not a natural

[sharp harness]

that's the natural number 1

(the integer 1 is the natural number 2)

[pastel storm]

what are you talking about

[sharp harness]

the natural number "1" and the integer "1" aren't the same object

[unborn chasm]

Anyway, the infinities there are why this doesn't make sense. You can't directly count infinities like that. It simply doesn't make sense.

For the relations you are getting at to make sense, you would need to restrict to elements in a set with finitely many integers. That would allow you to express the relations you are getting at.

However, then you run into what I said earlier - every open interval has infinitely many rationals.

[sharp harness]

there is a map from the natural numbers to the integers that sends the natural number "1" to the integer "1", and the natural number "2" to the integer "2", and so on

there is also a map from the natural numbers to the integers that sends the natural number "1" to the integer "-1", and the natural number "2" to the integer "-2", and so on

there's no reason to assume that one of these sends each natural number to itself and the other one doesn't

maybe neither of them do and in fact no integer is a natural number

[pastel storm]

Are you saying that the natural number 1 doesn't equal the integer 1

[sharp harness]

yes

[pastel storm]

because that would be a silly take

[sharp harness]

or it might not

it's possible that it does but how would we know

[unborn chasm]

He's pointing out they are different groups, and a map between the groups does not necessarily preserve the embedding into the reals.

You are implicitly looking at enumerations which drop any natural structure your group had.

[sharp harness]

there's also a 2x2 matrix that you could reasonably call "1"

the more common notation for it is $\begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}$

[elfin berry]

bee [it/its]

[sharp harness]

do you think that is equal to the integer 1?

[pastel storm]

I have no idea

[sharp harness]

so you're confident that the integer 1 is equal to the natural number 1, but don't know if the integer 1 is equal to the matrix 1

what's different in this case?

[acoustic gust]

hi

[pastel storm]

That I don't know enough about matrices to form an opinion

and that matrices are probably not numbers

[sharp harness]

what's a "number"?

[pastel storm]

I will not answer that

because that will bring us on another tangent that dosn't answer my question

[sharp harness]

...ok let's just look more directly at this question then

[pastel storm]

however I can show you this image I made about that

[sharp harness]

ok so here's a question

...nevermind i don't have a question apparently, i'm going to think more and then hopefully i will have one

[pastel storm]

bruh

[sharp harness]

ok wait just to check the notation convention we're using

is 0 a natural number or not

[pastel storm]

yea it is

[sharp harness]

ok so here's a question

if i take a set of things, and replace each of them with one other thing, is it possible that i end up with more total things than i started with

[pastel storm]

I would think not

[sharp harness]

so if i take the set of even numbers