help-39

[sterile python]

You made me sad layla 🥺

[random ermine]

sonion

im crine

who gave bro helpful 🥀

[sterile python]

God is

[cinder flower]

too bad

[sterile python]

Sigh.

[fleet ridge]

In slayla's thought bubble, it says she needs to be held. If you can offer her those services she might accept you

[vivid jetty]

Are we still focusing on solving the problem.

[fleet ridge]

Good point

@low kestrel did you need help in any other problem?

[vivid jetty]

@low kestrel !done

[wicked edge]

Supposedly we are.

[wicked edge]

You should have factoid first, then ping.

[vivid jetty]

Perchance.

[wicked edge]

You can see it didn't work.

[sterile python]

Pass

[fleet ridge]

Pass? What do you mean?

I really dont get it...

[lament dawn]

what is going on here 😭

[low kestrel]

@timelord tbh l have more trigonometry problems that l need help with

[sterile python]

If you know, you know

[lament dawn]

this is a psyop for me to do my homework

[lament dawn]

(i can help too!)

[wicked edge]

Guys, this is the 4th time for a redirection...

[fleet ridge]

I don't unfortunately

[random ermine]

no one knows twin

[fleet ridge]

Please share

[low kestrel]

alright but l have a requst

[sterile python]

She is too old for me, and too far away from me to hold. So I can't hold her.

[wicked edge]

Guys.

[wicked edge]

What is it?

[fleet ridge]

Supposed to be metaphorical

[low kestrel]

l'm requseting all the people who aren't willing to discuss trigonometry to leave please

if you're here to make friends l request that you dm each other privately thanks

ok moving on

[wraith jacinth]

go ahead!! we are listening

[low kestrel]

l just stated my request

anyways lets talk about the special angles in trigonometry

[fleet ridge]

This was the request @wraith jacinth

[wraith jacinth]

-# oh ah i though it was an acedmic request oops coniute dont mind me!

[lament dawn]

mm, what about them?

[wraith jacinth]

-# okay what would u like to know about this? why they occor or what are they or when do they happen?? or do u want to understand why its these angels?

[low kestrel]

well can you all solve cos(45- 30)?

[lament dawn]

also, special angles referring to the 15deg, 30deg, 45deg, 60deg, 75deg?

[fleet ridge]

Use cos(a-b)

Its a formula

[lament dawn]

we can! you use the cos(a-b) formula as timelord stated^^

[low kestrel]

special angles are only 3 , namely 45,30 and 60

[fleet ridge]

And 0, and 90

[lament dawn]

ah, fair enough, how ive learned it they expanded it to include 15 and 75 deg

[fleet ridge]

Maybe not special but still relevant

[wraith jacinth]

this might help-
-# also oops wrong reply dont mind me wjs

[lament dawn]

use the idea that cos(a-b) = cos a cos b + sin a sin b

[low kestrel]

so is there anyone who can use special angles to solve cos 720? it came in the end of term exam and mob guys were stranded

[vivid jetty]

Do you know about periodic functions? Such type of questions could be solved by using periodicity of a function.

[wraith jacinth]

well we can simplfy the angle first!

[fleet ridge]

This is basically it. Also there are formulas for cos(x+90k) where k is an Integer

[wraith jacinth]

this is bassicly because every 360 degress corect me if am wrong this circle repeats

[lament dawn]

cosine and sine have a period of 360deg

all this means is that if you have, say, sin(a) you can add or subtract whatever number times 360degrees to a, and still get sin(a)

[wraith jacinth]

so we would do atotal of two revolutions 720-360 and then 360 - 360

https://tenor.com/view/matematica-gif-7322246

[wraith jacinth]

this gif here might help and as u can see the angle (the exaxct nubmer is seen at the top)

notice how when we cross the 360 degree mark we starts over

so forexample if i gave u an angle of cos(370) is it the same as cos(10)

does that make sense so far?

[lament dawn]

||-# hey, sorry, i dont take unsolicited dms due to preferences and other reasons, but yes, i did make the guide myself and so on||

[low kestrel]

100%

[fleet ridge]

Yes I found a mistake in it

[wraith jacinth]

okay so u can imagne how removing the 360 from the angle can help!

so for example try it if i give u an angle of cos(400) = cos(x) can u tell me what we can use as x here instead of 400?

[fleet ridge]

I sent you the mostake. You dont have to respond if you dont want to

[low kestrel]

40

[wraith jacinth]

yup exactly!!

[low kestrel]

yayy!!

[wraith jacinth]

so if i give you an angle of cos(720) can u solve this now?

-# like cmon justopenur ownhelp channel atp or go to #serious-discussion

[vivid jetty]

You can remove or add 360° as much as you want, until you get whatever you want in parenthesis ( or something very close to it, or something you already know. )

[low kestrel]

cos 720 = cos 360 but using special angles how should l calculate that?

[wraith jacinth]

well here is the thing what does cosine and sine really mean?

[low kestrel]

cos = adjacent over hypotenuse and sine = opposite over hypotenuse

[vivid jetty]

You could remove another 360°, you do realise that?

[random ermine]

sohcahtoa

[lament dawn]

another good visualisation is using the sine and cosine graphs! this graph is in radians, and 2pi is 360degrees, but this shows that the output of sin(a+k*2pi) where k is some integer is basically the same as sin(a)

[wraith jacinth]

okay so for sin u can say it is the Ratio of how big the (the side opposite to the angel is COMPARED to the hyptonuse)

[low kestrel]

mhmm

[wraith jacinth]

so notice how when we increase the angel what happens to the hieght here?

[low kestrel]

but if we subtract another 360, hen we would get cos (o) is that acceptable?

[lament dawn]

using this idea, we can rewrite cos(720deg) as cos(a+k*360deg), and we know that that is also equivalent to essential cos(a)

[lament dawn]

yes!

[wraith jacinth]

wait wait- if u understand cos and sin u wll understand what this means

[wraith jacinth]

so here look at this notice how when the angle (on the left shown by blue green and light blue) gets bigger the (hieght) gets taller?

lmk if anything does make anysense at any point-

[lament dawn]

-# well, yes! cos 0deg is the same as cos 720deg!

[wraith jacinth]

-# they are essnetaily the same because if uturned around 0 degrres and 360 degrees and look at where ur looking in both cases it wouldbe as if u didnt move at all

[wraith jacinth]

so bassicly sin describes the hieght relative to the angel
this is why if u imagne increasing that angel until its 90 (Degrees) you would notice that both the hyptonuse and the hieght would be exactly = therfore once
But if u reduce the angel all the way down to 0 u will notice that the WIdth is what is = the hyptonuse now
this is why sin (90) = 1/1 so 1 and cos (90) ((remeber cosine describse adj which is width here)) would be 0

followning this logic can u guess what would be the width (cos) at angle 0 ?

[pearl pond]

@low kestrel Has your question been resolved?

[pearl pond]

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[daring bay]

yo so

[daring bay]

basically im trying to do 1.4

but somethings off

[summer imp]

The derivative of x^3 is not 3x

[ocean vale]

blud got hacked

[daring bay]

oh my god

what is wrong with me

but then can you just use uhh

wait

im thinking

do i use 0 and 4 as boundaries

but how do i actually get the triangle area

shit man

yes im
sure 0 and 4 are the boundaries, but

[summer imp]

If you find the point where they intersect you'll get the height of the triangle pretty much instantly.

The tangent line and g I mean

[daring bay]

yup but

i mean

[summer imp]

And then your base is between the two x-intercepts.

[daring bay]

oh

oh shit yeah

but

wait so i just integrate the lines only

[summer imp]

You don't really need to integrate anything.

The triangle's base is on the x-axis delimited by the x-intercepts of the lines, and the height of the triangle is given by the intersection

[daring bay]

wait ur saying i can just use a formula for triangles area

no way

[daring bay]

damn

man they just like to do some weird stuff

so 2*intersection/2

is my area

i got another problem i geniuenly do not understand them

so here i basically tried drawing it in

it does look like its the midpoint all the time but like

if you put 0 in then its 0

ill open a new channel

.close

[pearl pond]

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[stoic imp]

can i get help with exercise 10

[summer imp]

You need to show it's regular first.

Can you take its derivative and check that it's nonzero?

[stoic imp]

its an integral

you mean ftc1?

[summer imp]

You need to check that $\tilde{\sigma}(s)$ is regular.

[jolly parrot]

Azyrashacorki

[buoyant panther]

yes, you can use this theorem

[stoic imp]

how

[summer imp]

Consider the new parametrization they give you, differentiate it using the chain rule, compute the velocity.

[stoic imp]

but is an integral

[buoyant panther]

in your case g'(t) is literally the definition of this theorem, the derivative will simply be the function under the integral (the norm of velocity)

In other words:

[jolly parrot]

Modus

[summer imp]

Just use the chain rule. $(\tilde{\sigma}(s))' = (\sigma(g^{-1}(s)))' = \sigma'(g^{-1}(s)) \cdot (g^{-1}(s))'$. Now figure out what $(g^{-1}(s))'$ is with what Modus gave you.

[jolly parrot]

Azyrashacorki

[stoic imp]

can we start from the scratch I dont think I follow

[summer imp]

You need to show that the parametrization given by $\tilde{\sigma}(s) = \sigma(g^{-1}(s))$ is regular. To do that, differentiate it as above to figure out what the velocity vector is.

[jolly parrot]

Azyrashacorki

[summer imp]

Once that's done, the second part where you show the parameter gives the arclength is instant.

This will involve finding an expression for the derivative of $g^{-1}(s)$. There is a rule for differentiating inverse function you should know.

[jolly parrot]

Azyrashacorki

[stoic imp]

yes

inverse function theorem

[summer imp]

Good, so what's the derivative of g^(-1)(s)?

[blazing garden]

The GOAT!!

[stoic imp]

it should be 1/g'(g^1(s))

[summer imp]

Okay, and what is g'(g^(-1)(s)) given how g is defined?

[stoic imp]

thats sigma

t

[summer imp]

No

[stoic imp]

sigma hat

[summer imp]

Oh you mean the inside

Yeah

g'(g^(-1)(s)) is just ||sigma(g^(-1)(s))||

[stoic imp]

which is?

is the magnitud of sigma hat

[summer imp]

Well now that we know $(g^{-1}(s))' = \frac{1}{||\sigma'(g^{-1}(s))||}$

So now what is the velocity of sigma hat?

[stoic imp]

unsure

[jolly parrot]

Azyrashacorki

[summer imp]

What is sigma hat?

[stoic imp]

sigma circ g^-1

[summer imp]

So how do you differentiate the composition of functions?

[stoic imp]

⛓️

[stoic imp]

function composition rule for derivatives

[summer imp]

Yes

So what does that give

[stoic imp]

sigma' circ g^-1(s) cdot (g^-1)'

[stoic imp]

@summer imp

[summer imp]

Okay, so you have $(\tilde{\sigma}(s))' = \sigma'(g^{-1}}(s)) \cdot (g^{-1}(s))'$

[stoic imp]

yeah, what about sigma hat

[jolly parrot]

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

[summer imp]

And what was g^(-1)(s)?

[stoic imp]

there is some error with the latex @summer imp

[summer imp]

Yes it's still right.

[stoic imp]

Okay, so you have $(\tilde{\sigma}(s))' = \sigma '(g^{-1}(s)) \cdot (g^{-1}(s))'$

[jolly parrot]

Renato

[summer imp]

It's just an extra curly brace

[stoic imp]

what about it?

[summer imp]

Well this is meant to be the velocity of sigma tilde.

But you found an expression for the derivative of g^(-1)(s) earlier

[stoic imp]

$(\tilde{\sigma}(s))' = \sigma '(g^{-1}(s))\frac{1}{\norm{\sigma'(g^{-1}(s))}}$

@summer imp

[summer imp]

It's the magnitude in the denominator

But yes

[stoic imp]

wdym

[summer imp]

.

[jolly parrot]

Renato

[stoic imp]

I shee

what now?

@summer imp

[summer imp]

Well what is the magnitude of this vector?

[stoic imp]

I believe is normalized

[summer imp]

Yes, so it's constant 1.

Can it ever be 0?

[stoic imp]

how

[summer imp]

The magnitude is 1 always

It's normalized

[stoic imp]

right

[stoic imp]

no

[summer imp]

Right, so then that parametrization is regular.

[stoic imp]

what

[summer imp]

The velocity vector is never 0, so it's regular.

Now use the definition of the arclength to compute the arclength between 0 and s of this parametrization.

[stoic imp]

this shit is too hard

right???????

[stoic imp]

which parametrization thou

[summer imp]

Sigma tilde

The whole point is to show that the arclength of sigma tilde between 0 and s is s.

[stoic imp]

yeah

why did we normalized sigma tilde?

[summer imp]

We didn't normalize it. The velocity is normalized by definition

It's from how sigma tilde is defined

[stoic imp]

okay

anyways to find the arclength of sigma tilde between 0 and s

I need to differentiate sigma tilde once and find its magnitude

so the integral of the magnitude of the velocity of sigma tilde right

@summer imp

[summer imp]

Yes. You've already computed the derivative of sigma tilde though

[stoic imp]

when?

[stoic imp]

in here

?

but I need the magnitude of this right?

so $\int_0^s \norm{(\tilde{\sigma}(s))'} ds = \int_0^s \norm{\sigma '(g^{-1}(s))\frac{1}{\norm{\sigma'(g^{-1}(s))}}} ds = \int_0^s 1 ds = \left[s\right]_0^s = s$

@summer imp

[summer imp]

What is the magnitude of a normalized vector?

[stoic imp]

1

[jolly parrot]

Renato

[stoic imp]

@summer imp like this ?

[summer imp]

Yes like this

[stoic imp]

can we do a little recap?

@summer imp

[summer imp]

The way sigma tilde is defined makes it parametrize the same curve, but with speed 1.

Then it's regular, and by definition the arclength between 0 and s is s.

[stoic imp]

yes but can we rewind a little bit more

@summer imp

[summer imp]

What do you want to rewing to? That's pretty much all you did.

You computed the velocity of sigma tilde, found it was normalized and thus had constant speed 1.

[stoic imp]

yes

[stoic imp]

yes

we basically proved that g'(t) is always 1 for all t in T

@summer imp

[summer imp]

No, that sigma tilde' has speed 1 for all t.

g'(t) isn't 1 it's the speed of sigma'(t).

sigma'(t) isn't normalized

[stoic imp]

can we start from scratch?

it was a long amount of steps, first we used tfc1 on g(t) in order to find g'(t)

after that we get that g'(t) is the magnitud of sigma'

but where is sigma' defined

@summer imp

[summer imp]

Sigma is some parametrization.

[stoic imp]

it also says that g'(t) != 0 forall t

and then we find the derivative of g^-1

[summer imp]

There aren't that many steps.
You want to show sigma tilde is regular, so you compute its velocity using the chain rule and FTC1 for the factor that pops out.
This turns out to give that sigma tilde' is normalized, so has speed 1.

[stoic imp]

,, g(t) = \int_a^t \norm{\sigma'(\mathcal{T})} d\mathcal{T} \ g'(t)) = \frac{d}{dt} \int_a^t \norm{\sigma'(\mathcal{T})}d\mathcal{T} \ g'(t) = \norm{\sigma'(t)} \ (g'(t) \neq 0 \text{ , } \forall t \in \left[a,b\right] \implies (g^{-1})^{'}(t) = \frac{1}{g^{'}(g^{-1}(t)} \ (g^{-1})^{'}(t) = \frac{1}{g^{'}(g^{-1}(t)} \implies (g^{-1})^{'}(t) = \frac{1}{\norm{\sigma' \circ g^{-1} (t)}}

@summer imp

after applying fundamental theorem of calculus 1, then we are being told that forall t in T that g'(t) != 0 so we can use inverse function theorem

right?

[summer imp]

Yes

[stoic imp]

@summer imp

[lean totem]

estas en analisis 2 de la uba

[jolly parrot]

Renato

[stoic imp]

like this

[stoic imp]

anal 2 si

[lean totem]

jajajaja que coincidencia

[stoic imp]

recien empezando jeje

[lean totem]

es dificil, tenele paciencia

[stoic imp]

si?

anal 1 no era tan complicado, anal 2 es mas complicado?

[lean totem]

yy pega un salto

igual tene en cuenta que estos ejercicios son para que entiendas conceptos, no te van a tomar nada asi en un parcial

[stoic imp]

si claro

cual decis que es los temas que mas hay que darle bola

[lean totem]

esto era guia 1 si no mal recuerdo

[stoic imp]

y si, ando re atrasado mal

[lean totem]

siempre es primer parcial con algo de parametrizacion, segundo de green, tercero stokes y cuarto gauss

[stoic imp]

ando priorizando taller de calculo avanzado

[lean totem]

no se si seguira siendo asi

[lean totem]

banqué totalmente, yo estoy en datos pero se ve en analisis avanzado esas cosas

algunas nociones y demos de analisis 2 las vas a entender mas faicl sabiendo tca

perdon te re ocupé el canal

si queres hablame al dm

te dejo seguir

[stoic imp]

te mande soli

.close

[pearl pond]

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[valid nova]

is a negative number prime?

[valid nova]

like -5,-7 ?

[nocturne night]

technically not?

[valid nova]

well. i was solving an inequality, and they asked number of prime numbers satisfying that inequality. the boundaries are
$$(-\infty, -2.5) \cup (-2, 8)$$

[jolly parrot]

TheAstorPastor

[valid nova]

so i just have to include +ve ones right?

[nocturne night]

yep

[valid nova]

hmm

in some book, i saw them to be -ve

Hungerford's Abstract Algebra: An Introduction

i think

[buoyant panther]

the classical definition exludes negative integers (among other things, due to the factorization of numbers), so negative numbers do not enter

[nocturne night]

^

-ve's can be considered prime elements js not prime nums generally

[valid nova]

hmm

ok

.close

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[pearl pond]

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[warped violet]

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[wind lagoon]

Consider $\Phi ( r,\theta,\varphi ) = (x,y,z)$, the usual map from spherical to Cartesian coordinates. The domain of $\Phi$ is the $3$-cell $D=[0,1]\times [0,\pi]\times [0,2\pi]$. Is this map $C^1$? There's an example in Rudin's book where he integrates $dx\land dy\land dz$ over $\Phi$, but this integral is only defined for a $C^1$ $3$-surface.\

What makes me uncertain is that Rudin comments "... that the mapping is 1-1 in the interior of $D$ (but certain boundary points are identified by $\Phi$) ..." What does he mean by this?

[jolly parrot]

psie

[pearl pond]

@wind lagoon Has your question been resolved?

[wind lagoon]

.close

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[daring bay]

yo i hate fucking hate math

[daring bay]

how do i even approach the event D

[ebon skiff]

is this the entire problem

[daring bay]

no but like generally

ik the bernoulli formula

were allowed to use calculators for this but

[ebon skiff]

just send the entire problem

[daring bay]

im on 2.2 D

[ebon skiff]

whats P (salmon)

and P(salad)

[daring bay]

1/3 salmon

1/3 salad

[ebon skiff]

ok

so the first 4 choose salmon

so

(1/3)^4

for the first 4

[daring bay]

yeah

[ebon skiff]

now you hvae 16 people reimaining

and out of these 16

5 choose salad

so you can have something like

[daring bay]

so

[ebon skiff]

(s,s,s,s,s,notsalad,notsalad....)

or

(s,not s, s,s,s , nots, nots,...)

any combination

of 5

from 16

[daring bay]

so

2/3^9 + 1/3^7

?

[ebon skiff]

nah

[ebon skiff]

ok lets take this case

whats the probability for this

[daring bay]

1/3^5 + 1/3^4

[ebon skiff]

nope

1/3^5

[daring bay]

i mean other exponent

[ebon skiff]

this is correct

[daring bay]

huh

why

[ebon skiff]

ok

so

first of all there shouldnt be a +

only *

[daring bay]

oh yes right

[ebon skiff]

so

lets take this case

( salmon, salmon, salmon, salmon, salad, salad, salad ,salad, salad, notsalad, notsalad,...(in total 11 notsalads)

whats the probability for this

[daring bay]

wait its 4 salmon

[ebon skiff]

yeah mn

mb

[daring bay]

1/3^4 * 1/3 ^5 * 1/3^7

or no

[ebon skiff]

no

1/3^4 * 1/3 ^5 * this is correct

[daring bay]

what about the not salads

[ebon skiff]

whats P(notsalad)

[daring bay]

2/3

[ebon skiff]

yeah

and how many times does it appear

[daring bay]

7

[ebon skiff]

( salmon, salmon, salmon, salmon, salad, salad, salad ,salad, salad, notsalad, notsalad,...(in total 11 notsalads)

[daring bay]

oh shit yeah

11

times

do

so *2/3^11

?

[ebon skiff]

yes

1/3^4 * 1/3 ^5 * 2/3^11

but remember

this was jsut 1 case

( salmon, salmon, salmon, salmon, salad, salad, salad ,salad, salad, notsalad, notsalad,...(in total 11 notsalads)

we can arrange the salads in any way we like

not necessarily the first 5

out of 16

the salmon never changes though

cuz it says so in the question

so

only the 16 change

right?

[daring bay]

wdym only 16 change

[daring bay]

yep

[ebon skiff]

the last 16

of the total 20

change

cuz the first 4 are always salmon

[daring bay]

ohh

yep

[ebon skiff]

ok so basically

we need to find the number of all possible combinations

[daring bay]

oh hell nah

i hate this combinations shit

[ebon skiff]

#help-39 message

[daring bay]

but

so

there are like

so 5 times 16 maybe

something like that

or

[ebon skiff]

nah

[daring bay]

its something with ! right

[ebon skiff]

yeah

[daring bay]

but idk how

[ebon skiff]

hast du schon mal von "Binomialkoefizzient" geheort

[daring bay]

yeah

ja

oh

(n k)

[ebon skiff]

yeah

[daring bay]

so 16 5

[ebon skiff]

yeah

[daring bay]

is it just 16!/5!

[ebon skiff]

no

its

16! / (5! (16-5)!)

[daring bay]

oh damn

[ebon skiff]

https://de.wikipedia.org/wiki/Binomialkoeffizient

so you have this 1/3^4 * 1/3 ^5 * 2/3^11 and this many options 16! / (5! (16-5)!)

and you just multiply them together

so

1/3^4 * 1/3 ^5 * 2/3^11 * 16! / (5! (16-5)!)

,w 1/3^4 * 1/3 ^5 * (2/3)^11 * 16! / (5! (16-5)!)

[jolly parrot]

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Requested by masteroogway0599

[ebon skiff]

yeah

[daring bay]

oh damn

[ebon skiff]

so 0,257%

[daring bay]

i got a question like

did you just memorize this

or basically just think of it

[ebon skiff]

you dont usually need to memorize it

cuz you have the C button on the GTR

and you just plug in (16 5)

but yes i memorized it after using it for quite a while

[daring bay]

bo

[ebon skiff]

,w 16C5

[daring bay]

bro

[jolly parrot]

Results provided by WolframAlpha

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

Requested by masteroogway0599

[ebon skiff]

look

[daring bay]

do yk the vierfeldertafel

[ebon skiff]

yeah

[daring bay]

thats always conditional probability right

[ebon skiff]

um

not really

i mean

kinda

it depends what you mean

every square is P(something AND something else)

to get the conditional prob you need to take this square and divide it by the outside square

cuz the formula is P_B(A) = P(a und b) / P(b)

[daring bay]

damnn

[ebon skiff]

and you find P(b) on the outside

[ebon skiff]

https://tenor.com/view/duke-dennis-feels-the-aura-aura-duke-aura-farm-gif-12250327613032664645

[daring bay]

man i gotta learn some other stuff now holy shit

why did i study so late im so stupidd

[ebon skiff]

its ok bro dont think about the past

only the present

!done

[pearl pond]

If you are done with this channel, please mark your problem as solved by typing .close

[daring bay]

mb

.close

[pearl pond]

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[left anchor]

Can someone explain what reordering a sequence/series means and how it can affect the limit

[left anchor]

I saw a basic explanation on the fact that it doesn’t seem to affect the limit of a sequence because a sequence is essentially a set and sets don’t really care about the order of elements anyway

Now how is that different for a series

And how would I reorder a series to tend to any limit

I have a specific problem in mind but I’d like to figure out how everything works first
Maybe with simple examples

[pearl pond]

@left anchor Has your question been resolved?

[plush bramble]

just show your specific problem

[pearl pond]

Asking the actual question right away is more likely to get responses.

Asking "Can I ask...?" or "Does anyone know about...?" doesn't give people enough information to decide whether they can help, and answering can feel like a promise to help with the actual question, which they might find themselves unable to.

[left anchor]

This is the main part I’m interested in I was hoping it’d be clear enough

I can post the actual problem too but I wanted to make sure that I understand the concepts well enough in advance hence the questions

[plush bramble]

yes post the actual problem

[left anchor]

Working on it
Need to take photos of the necessary parts and explain

I have this series which converges to 0 but it is not absolutely convergent (is that the right term?)

All the problem asks for is reordering it to converge to 7 instead while maintaining the given structure from the photo

<@&286206848099549185>

[pearl pond]

@left anchor Has your question been resolved?

[pearl pond]

@left anchor Has your question been resolved?

[vagrant briar]

https://youtu.be/U0w0f0PDdPA?si=jXTM6fx0CnBMt1_r

Something like this?

[left anchor]

Yes
Changing the order of addends to get a different sum

[vagrant briar]

Maybe this video will help, or did u already watch this

[left anchor]

I’ve tried doing research by myself in the meantime and found the Riemann rearrangement theorem

[left anchor]

Not that specific one

[vagrant briar]

Its a great watch

[left anchor]

Now I’m still struggling to understand how I would do that for a specific series
The key concept of going back and forth by adding / subtracting numbers around the limit makes perfect sense

But how do I define the series to do that?

[left anchor]

I’ll open it up thanks

[left anchor]

.close

[pearl pond]

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[pearl pond]

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[inland ivy]

I've been struggling at (ii) for so long

So we get f(x+h) > f(x-h) for all x and sufficiently small h

but how do I go from this to f being strictly increasing?

I've been considering an approach similar to using the mean value theorem but it's not been working

[pearl pond]

@inland ivy Has your question been resolved?

[inland ivy]

<@&286206848099549185>

[rustic tendon]

I think I did this with contradiction

[rough stream]

f^s is just f', no?
So you just need to prove that f is strictly increasing when f' > 0

Which, I'd be surprised if that's not just a theorem in your book

[rustic tendon]

We can't assume differentiability though right

f^s existing doesn't imply f' existing

[inland ivy]

yeah that's the point

[vital crescent]

consider , on (0,1)

$f(x)\begin{cases}1 &x=\frac{1}{2}\
x&\text{otherwise}\end{cases}$

[inland ivy]

I've been trying to

[rough stream]

Ah the if. Didn't see it

[jolly parrot]

qwertytrewq

[inland ivy]

this is not continuous

[vital crescent]

this is a counterexample i think

[vital crescent]

oh shoot

missed that condition

[inland ivy]

yeah so continuity is critical here somehow

[rustic tendon]

I managed to find my old proof lol

Yeah there was some hand holding involved

[warm furnace]

what grade level math is this

[inland ivy]

it's first year uni

[rustic tendon]

I can give you a hint, assume for contradiction that there exist c, d in (a, b) such that c < d and f(c) > f(d). Then choose any y in between these function evaluations and define the set S_y = {x \in [c, d] : f(x) >= y}

[inland ivy]

what the blud

[vital crescent]

i think contradiction should work. Let f(a)<f(b)\geq f(c) with a<b<c. Find the maximum of the function on [a,c], and f^s should be weird there

[rustic tendon]

the continuity comes into picture when you show that ||f must possess an uncountable number of strict local maxima||

[inland ivy]

this is what I thought but you can't really say much about the symmetric difference even at the max

[inland ivy]

😭

[rustic tendon]

all I'll say is good luck bro

[vital crescent]

how so? eval at -h and h must agree no?

[inland ivy]

symmetric quotient always agrees, it's an even function

[vital crescent]

i dont see how f^s isnt going to be 0

oh wait

mb it could

[inland ivy]

if I take z = sup(S_y) then by continuity f(z) = y and f(z-h) >= y > f(z+h) for sufficiently small h

but then f(z-h) - f(z+h) > 0 which is a contradiction

does this work

[oak nebula]

okai youre kinda in the right way ig.

mvt does NOT work directly cuz f isn't necessary differentiable thr cheat code here s to assume for a sec that f has local max and if it did f^8 (x) would have < or = 0
which contradicts the given f^8 (x) > 0

so f cant have local maxima or minimum which forces it to be strictly mono. since f( x + h ) > f (x-h ) it has to be increasing ig

hope this helps

[inland ivy]

why does f having a local maximum force the symmetric derivative to be <= 0

oh my god i hate this fucking problem

[oak nebula]

at a local max the function has to stop right? and turn back down so if f(x) is a peak then f(x+h) and f(x-h) are both similar then the peak so when you subtract em in the sym derivative formula youll NEVER get a + result it will be < or = 0

since our f^8 =is strictly > 0 a peak is literally impossible it has no choice but to keep going up

[inland ivy]

@rustic tendon can you just give me the answer i've spent too long on this

[oak nebula]

does tha makes anysense?

tried my best 😭

[vital crescent]

consider x when x<0 and -1/2x when x>=0

0 is max

but the symmetric derivative is positive at 0

[inland ivy]

yeah f can decrease slower on the right than the left, so f(x+h) is still > f(x-h)

[vital crescent]

u have, in the derivative formula (-1/2h-(-h))/(2h), the limit is 1/4 i think

[rustic tendon]

why must f(z - h) >= y

[rustic tendon]

I just got left my room for a break

[inland ivy]

well for any delta there exists h < delta with f(z-h) >= y

but that's not enough i realise now

[rustic tendon]

see but you're getting there

[inland ivy]

wait it's still a contradiction right??

[oak nebula]

f(0) =0 then f(h)= -1/2h and f(-h)= -h ---> into the formula (-1/2h -(-)) /2h = (1/2h)/2h =1/4
so even with a shrp peak at 0 f^8 is still a 1/4 which s >0

@vital crescent

[inland ivy]

because you need f(z+h) > f(z-h) for ALL sufficiently small h

but you can always find arbitrarily small h with f(z-h) > f(z+h)

[vital crescent]

which is my point no? u claimed that at the peak it should be <0 or =0

[oak nebula]

but thats my whole point since ur examples symm derivative @ x = 0 turned out to be 1/4 which is > 0 it proves that x=0 isn't actually a local max in the context

a true local max would require f^8 (x) < or = 0
if the f^8 is strictly + everywhere a peak literally cannot exist lol

[vital crescent]

then how does f not having local max and local min (where they are defined in your context, not the actual local max/min) imply that it is monotone?

if u define critical points to be such that f^s at the point must be 0, then I feel like this is merely a restatement of the original question

[oak nebula]

its basic calc bro if a continuous function on an interval doesn't have any local max or min it literally can't turn around

if it starts going up and never hits a peak to turn back down

it has to stay strictly increasing

[vital crescent]

u r using the definition of actual local max/min

[vital crescent]

u said here the local max is "in your context"

this is my function

x=0 is, in the usual sense of local max and min, a local max

symmetric derivative is >0 at 0

but u claimed that it isn't a local max in your context, which I dont understand. If tou change the context of what local max/min means, youd have to reprove the statement that no local max/min implies monotone

[oak nebula]

mhm

[vital crescent]

so the issue sort of is that what if, every where, the function looks like this (it sure will be some weierstrass like function)

By the statement it can't exists but it's not immediately obvious

[oak nebula]

okai look im not changing the def of a local max but if s f^8(x) > 0 for all x then for any point x0 the function MUST be higher at x0 +h than at x0-h for small h this property alone makes a local max (where a point is higher than its neighbors on BOTH sides) impossible und in calca if a continuous function has no local peaks or valleys to turn it around it has no choice but to stay strictly monotone in this case increasing right ? its a standard result for symmetric derivatives so no need to overcomplicate the definitions ig

[vital crescent]

this property alone doesn make it a local max as illustrated by my example tho

i gave an example where it is local max ( in fact a global max) in the usual sense, but at 0, the symmetric derivative is 1/4

the f(x+h)>f(x-h) is always true using my function at x=0

u could have all these properties while not stopping f(x) being greater than every single one of those f(x+h)

If it were f' then sure all that you said is true. But it is f^s, and f^s does not directly compare f(x+h) with f(x), which is where the issue lies

[pearl pond]

@inland ivy Has your question been resolved?

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[leaden thunder]

Hello im struggling with finding which of those answers are wrong... I've looked over and I can't figure out which one is wrong. It says atleast 1 is wrong.

[arctic oracle]

you on your own

[leaden thunder]

how come?

[viscid shale]

Okay, so, do you recall the definition of a line integral over a vector field?

[pearl pond]

@leaden thunder Has your question been resolved?

[pearl pond]

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[leaden thunder]

.reopen

[pearl pond]

✅ Original question: #help-39 message

[leaden thunder]

not the full definition to be honest but like i used it for circulation and flux and masses. Since those are the formulas I have

wait no thats just for line intrgals... not over a vector field..

[viscid shale]

Thats how you do the calculation, visually, it has to do with tracing the dot product of the field with the direction of the parametrization.

https://en.wikipedia.org/wiki/Line_integral#/media/File:Line_integral_of_vector_field.gif

wikipedia has this quite intuitive gif.

So you go about the curve using r', evaluating the dot product at each point w/ the vector field, and finding the acumulation of this value.

Notably, you have to know how the dot product works

[viscid shale]

I want to believe youre familiar enough with it.

[pearl pond]

@leaden thunder Has your question been resolved?

[pearl pond]

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[fading nexus]

I’m not sure what to do for b and c

[timber cape]

,rccw

[errant solstice]

(hint: say, does x = 50 make sense for this problem? why (not)?)

[timber cape]

what function did you get for part A?

[fading nexus]

V(x)=40•12•x? Idk if that’s what you r looking for or not

[timber cape]

ouch thats wrong

[fading nexus]

I thought l•w•h is the formula for volume?

[timber cape]

yeah but we are cutting the lengths and breadths so they could turn to heights

[fading nexus]

Oh would it be 40-2x or smth? And then12-2x

[timber cape]

exactly

now what would be V(x)?

[timber cape]

can you visualise the maximum possible value of x we can take such that the box still has non zero volume?

[fading nexus]

Wdym visualize

[timber cape]

like for eg in the given diagram taking x=0 and gradually increasing x

[fading nexus]

I’m not really sure how to do that in my head

[errant solstice]

(or, if I may interrupt, you can try an extreme value like this and see what happens physically.)

[fading nexus]

I’m really confused 😭

[errant solstice]

in that case, sorry for interrupting. I'll let Itsuki continue.

[fading nexus]

I just don’t know how to approach b liek what math to do to get an answer

[modern talon]

can you have negative length

[fading nexus]

So it’s above 0 and not including so (0,_)

Oh wait

Could it include 0?

I don’t think so

Would it be (0,10)?

WAIR

[modern talon]

technically we could have 0 (flat, so not really a box)

[fading nexus]

(0,11.8)

[modern talon]

no

can you tell me the side lengths?

[fading nexus]

40 and 12

[modern talon]

no

[timber cape]

(reference image)

[modern talon]

40-2x and 12-2x

[fading nexus]

Oh yes

Wait

[modern talon]

correct

[fading nexus]

Can I just make them equal to zero to solve?

[modern talon]

yep!

[fading nexus]

So I got x=6 and x=20 but what one do I put in the domain

[modern talon]

so anything between 0 and 6 should work right?

because all the side lengths would be positive

[fading nexus]

Yea

[modern talon]

what about stuff past 6?

like 8, 10,... etc

[fading nexus]

It would make the side length to big

[modern talon]

could those work?

okay you're right that it wouldnt work

but for the wrong reason

[fading nexus]

To small?

[modern talon]

if i took something like 10, then 12-10(2) is -8, and we would then get a negative side length

i just want you to understand why stuff past 6 wouldnt work

anyways, we have our domain 0 to 6 now it looks like

can you carry on with the rest of the problem?

[fading nexus]

Yea but go c I’m stuck I know I need to isolate smth but u do not know what because there’s only x in the volume equation

[modern talon]

also, im sorry for taking over @timber cape , please carry on

you're both doing great

[timber cape]

its okay, sometimes i feel like observing more experienced people is better than just self trial and error

[modern talon]

no worries, you wanna take over for c? im gonna go back to studying, just taking a little break

[timber cape]

do you remember the concept of derivatives?

[fading nexus]

We haven’t done derivatives. I asked a question earlier and the person was surprised the way my teacher did it. We do it with like graphs and finding maximums and minimums

Not the same problem but yea

I know I’ll likely have to find the max on a graph

[timber cape]

yes you can try the same logic but the function here is a bit worse

-# it will work but will take forever

[timber cape]

if you want, i can show you the derivative approach at a glance

-# ill just show the derivative approach anyways

basically, the derivative or slope of a function y= f(x) (in this case V(x)) is equal to dy/dx

for a maximum/ minimum, the slope needs to make 0 degree angle with the horizontal(x-axis)

[fading nexus]

I caved and look at the answer key and I was overthinking it all I had to do was put the equation I already had into my calc and find the max 😭

[timber cape]

damn

[fading nexus]

I just need to get at least like a b on this test at this point I’m a senior I’m checked out

like I got 13 days left… college won’t reject me if I get a b in math

[pearl pond]

@fading nexus Has your question been resolved?

[pearl pond]

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[north talon]

Let $P(x) = x^{2026} - x^{2025} + ... -x+1$.Does there exists a sequence of integers $1 \le a_1 < a_2 <... < a_{2027} < 2027^{2027^{2027}}$ such that $\ (i)$ $a_i | a_j$ for each $1 \le i < j \le 2027 \ (ii) P(a_i) | P(a_j)$ for each $1 \le i <j \le 2027$

[jolly parrot]

Copter

[north talon]

i dont really have many ideas other than using a - b | P(a) - P(b) or factoring P(x)

[jolly parrot]

Copter

[north talon]

<@&286206848099549185> ;-;

[inner granite]

Maybe try finding $a$ and $b$ such that $P(a) \mid P(ab)$

[jolly parrot]

Erebus

[north talon]

hmmmm

[pearl pond]

@north talon Has your question been resolved?

[pearl pond]

@north talon Has your question been resolved?

[inner granite]

Do you perhaps know the source of the problem?

[north talon]

no

[wicked edge]

Are you doing question a, right?

[pearl pond]

@north talon Has your question been resolved?

[pearl pond]

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[winged pulsar]

I calculated.

A = 35°
B = 64,33°
C = 80,67°

a = 35
b = 55
c = 60,21

but apparently there are two solutions??? Explain please don’t y’all start to yap.

[winged pulsar]

It’s 3.

[sterile python]

What do you need to find, given a,b and angle A?

Would you mind translate the question?

[light helm]

have you been introduced to the ambiguous case for sine rule

[pearl pond]

@winged pulsar Has your question been resolved?

[winged pulsar]

I already founded it

[sterile python]

Great, any more questions?

[winged pulsar]

BUT THERE ARE FOCKING TWO ANSWERS.

[wicked edge]

What answer did you get?

Probably the other one was 180- your solution.

[pearl pond]

@winged pulsar Has your question been resolved?

[pearl pond]

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[swift spindle]

can someone epxlain this q

c)

[cobalt hinge]

what is p(x)?

[swift spindle]

i have no idea too

[cobalt hinge]

⁉️

no context?

[swift spindle]

no this is literally it

wait

[cobalt hinge]

nothing liem that?

[swift spindle]

nah they never gave p(x) befoer

oh, they might ltierally mean p * (x)

[cobalt hinge]

whaddafuq

oh it actually means f(x)=px…

why couldn’t they just write that bru

anyways

[pearl pond]

@swift spindle Has your question been resolved?

[split void]

horrible abuse of notation.

[pearl pond]

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[crimson haven]

I looked at the memo and I don’t understand why they kept the 90-3x

[light helm]

memo as in the stuff in blue?

[crimson haven]

No

I didn’t send it

Memo is the answer sheet

[light helm]

send it

Pretty much everything in the blue is wrong btw

[fleet ridge]

The first line in blue seems wrong. its supposed to be sin(90-3x) not cos(90-3x)

[crimson haven]

Why le

[fleet ridge]

Also you cant just divide by sine. you "cancel" it

[crimson haven]

*pls

[light helm]

complementary relationship between sine and cos

cos(t) isn't the same as cos(90 - t)

I looked at the memo and I don’t understand why they kept the 90-3x

Wdym

[crimson haven]

Huh

[fleet ridge]

Your going to have to be more specific. What are you confused about

[crimson haven]

What they said

[light helm]

90-x appears again in the following line
But what is your issue with that?
do you think it shouldn't be there anymore or that it should be something else

[crimson haven]

Yeah

[light helm]

Firstly, do you have any issue with the first line
sin(2x-20) = sin(90-3x)

[crimson haven]

No

[light helm]

For the next step
What do you think it should've been, and
what law/property are you applying

[crimson haven]

Divide the sin

To cancel

[fleet ridge]

Also you cant just divide by sine. you "cancel" it. This is done by taking the inverse of both sides

[crimson haven]

How pls

[light helm]

sin isn't something being multiplied

you can't divide by "sin"

[fleet ridge]

Its a bit complicated, but essentially:

sina=sinb
arcsin(sina)=arcsin(sinb)
a=b

This isn't always right due to the period of sine

[light helm]

they're applying the periodic property of sine to get the general solution

[crimson haven]

Wha

[light helm]

note that
sin(t) = sin(t + 360°)

you've done plenty of problems related to that this past week

or more generally
sin(t) = sin(t + k * 360°)

[crimson haven]

How would the inverse work pls

Can you show me

[light helm]

exactly what followed in their work, they split it into two cases

First case was a direct application of the above

[fleet ridge]

If sin(a)=b, arcsin(b)=a

[crimson haven]

I understand

Erm for the second one, why does k360 get affected by the -1

[light helm]

you mean doesn't get affected?

[crimson haven]

Yes

[light helm]

k represents some integer
could be positive or negative or 0

so when representing a general solution, the sign before doesn't matter

[crimson haven]

Oh

Ok

Ty both

Bye

[fleet ridge]

!done

[pearl pond]

If you are done with this channel, please mark your problem as solved by typing .close

[crimson haven]

.close

[pearl pond]

Channel closed

Closed by @crimson haven

Use .reopen if this was a mistake.

[pearl pond]

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[zinc zenith]

Can someone walk me through some problems

[random ermine]

!da2a

[pearl pond]

Asking the actual question right away is more likely to get responses.

Asking "Can I ask...?" or "Does anyone know about...?" doesn't give people enough information to decide whether they can help, and answering can feel like a promise to help with the actual question, which they might find themselves unable to.

[random ermine]

!status

[pearl pond]

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

[random ermine]

let's start with q1

[zinc zenith]

Ok

[random ermine]

quick

[random ermine]

?

[zinc zenith]

1

[random ermine]

...

-r^4 + 7r^3 - 2

how many terms does this have

[zinc zenith]

2?

[random ermine]

what are they

don't guess

[zinc zenith]

^4 ^5

[random ermine]

where do you see ^5

[zinc zenith]

I mean 3

Miss click

[random ermine]

right

no, ur wrong

-r^4 is 1 term

7r^3 is another term

[junior osprey]

Hello

[random ermine]

-2 is the last terms

make sense?

[zinc zenith]

Kind of

[random ermine]

how many terms are there

[zinc zenith]

3

[random ermine]

yes

now consider x^4 + 2x^3 - x^2 + x

what are the terms

[zinc zenith]

4, 3, 2 1

[random ermine]

no

you need to include the coefficient

try again

[zinc zenith]

^4

[random ermine]

https://tenor.com/view/whiplash-mad-angry-frustrated-throw-gif-6000995504968593846

NO

^4 is just an EXPONENT

not the TERM

[zinc zenith]

X^4

[random ermine]

yes that is 1 term

so 4 terms in total

[zinc zenith]

X^4

3

[random ermine]

now back to -r^4 + 7r^3 - 2

what is the degree?

[zinc zenith]

Degree?

[random ermine]

yes the questions asks for degree

degree = highest exponent

[zinc zenith]

Ty for the definition

r^r

4

[random ermine]

yes

[zinc zenith]

Yay

[random ermine]

now try the next question

gtg

someone else might takeover

[fleet ridge]

This is the original problem right?

[wraith jacinth]

h-

[wraith jacinth]

-# time lord how are you awlayws one step ahead lol

[fleet ridge]

Hehe. I have future sight

[zinc zenith]

Cool

[calm wing]

because he is literally time lord?? 😭

[wraith jacinth]

-# you truly are a time lord

[calm wing]

you fell right into that one xd

[zinc zenith]

…

[fleet ridge]

But still hat question are you stuck on

[zinc zenith]

Number 3 in the image

[fleet ridge]

Ok

So do you know how to group the terms

Essentially you need to group the terms with the same variable and exponent

[zinc zenith]

Let’s assume mo

Ahh

[fleet ridge]

Basically, you can group 9x and 4x because they have the SAME variable with the SAME exponent

[wraith jacinth]

so bassicly grouping terms is like grouping same type of objcests together like having 10 squares and 5 circles and 2 squares and 8 circles

[wraith jacinth]

so in total u have how many sqaures and how many circles?

[zinc zenith]

12

…

[fleet ridge]

-# would recommend expanding first

So @zinc zenith take the third question and expand the bracket

[zinc zenith]

My divice is about to die

Sorry to cut this short

[fleet ridge]

No problem. When your device is charged we can continue

[pearl pond]

@zinc zenith Has your question been resolved?

[pearl pond]

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[willow wren]

Hello guys. What is the point of finding the maximum of a function by finding the point where the derivative is 0 if you can just use the vertex maximum formula?

[sharp vigil]

the vertex maximum formula only applies to parabolas

[willow wren]

I see

Does a circle have two maximums then?

[sharp vigil]

a circle is not a function

[willow wren]

Aka two points where dy/dx = 0

[willow wren]

Alright

Thanks

[sharp vigil]

a point where the derivative is 0 can be a maximum, or a minimum, or neither

[willow wren]

I see

[willow wren]

So what else has a maximum?

[toxic fractal]

many continuous functions have a maximum

[willow wren]

Alr

[cinder flower]

sin has a max of 1

[pearl pond]

@willow wren Has your question been resolved?

[willow wren]

Got left on read but ig

[pearl pond]

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[valid nova]

hmm

[inland laurel]

How can I proceed further from this step

[valid nova]

{x} = x - [x]

right?

[inland laurel]

Yes

[valid nova]

x is an integer here,right?

[inland laurel]

Yes

[valid nova]

for example, take {4} = 4 - [4]

what is [4]

[inland laurel]

0

Wait

[valid nova]

um. [4]

[inland laurel]

4

[valid nova]

yeah

so {4} = 4-4

so {4} = 0

[inland laurel]

Yes

[valid nova]

so for {x) = 0, where x E I

right?

[inland laurel]

Yes

[valid nova]

now {x^2} must be 0 as well

[inland laurel]

Yes

[valid nova]

can you solve it now?

-x^2+3>0

[inland laurel]

Yes

[valid nova]

!done

[pearl pond]

If you are done with this channel, please mark your problem as solved by typing .close

[inland laurel]

.close

[pearl pond]

Channel closed

Closed by @inland laurel

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[sinful snow]

Why is it multiplying the others by 2 instead of 2x

Shouldn't it be 2x 6y not 2 6y

And 2x²