help-49

[lament knoll]

idk, u wanted convergence or divergence so i showed u how to prove it diverges lol

[frank iris]

Abel's test? I could be yapping.

[frank iris]

This?

[austere kraken]

didnt learn that

I'm just yr1 student, it should have some easy way to do this

[frank iris]

Abel's Rule may prove the first one diverges.

Idk.

[lament knoll]

i may have a counter example

u want to show S2n diverges and S2n+1 diverges -> Sn diverges ?

[austere kraken]

yes

[blissful rune]

Did you learn leibniz theorem?

[austere kraken]

yes

[lament knoll]

take $\sum \frac{(-1)^n}{n}$

[grand pond]

tm

[lament knoll]

S2n diverges

S2n+1 diverges

but the alternating harmonic series converges

[austere kraken]

u sure we are talking the same thing??

[lament knoll]

i think ?

[lament knoll]

that’s what u want no?

[lament knoll]

aaah i thought $\sum a_n$

[grand pond]

tm

[lament knoll]

fck

[austere kraken]

LOL

[frank iris]

To ln2 btw.

[frank iris]

哈哈哈

[lament knoll]

u want to prove it or u want to know if it’s true ?

[blissful rune]

i think he wants to prove that 1/(n-1) being divergent leads to the entire series being divergent or something like that

[austere kraken]

I think it's true and I have no idea how

[lament knoll]

i think it’s false but i have no counter examples..

[frank iris]

Limit values?

[blissful rune]

What is false?

I'm confused a bit

[austere kraken]

I'm asking this

[blissful rune]

Aren't these subsequences of Sn?

[lament knoll]

S2n and S2n+1 divergent -> Sn diverges

[austere kraken]

well u still need to prove it

[blissful rune]

If subsequence diverges then Sn can't converge as a sequence

Because a sequence Sn converges if and only if every it's subsequence converges to lim(Sn)

[frank iris]

Yes.

[blissful rune]

If Sn doesn't converge in X then it's said to be divergent

[blissful rune]

So infact you would only need S2n or S2n+1 be divergent

and that would imply that Sn diverges

no need for both to be divergent to conclude the latter

[frank iris]

Did we just state the same thing give or take?

[lament knoll]

if S2n and S2n+1 have two different limits we could conclude easily

by contrapositive

[blissful rune]

If Sn were to be a convergent sequence in a metric space X , then every it's subsequence must converge to the same point

Since S2n is a subsequence of Sn on the real numbers, if Sn were convergent then S2n would have to converge to the same value (both left and right implications hold)

Since S2n doesn't converge, then Sn can't converge either

by contradiction

[lament knoll]

yeah, so by contrapositive, if two sub sequences has different limits then Sn diverges

[blissful rune]

you only need any subsequence to have different limit

but yes

[lament knoll]

no, two sub sequences

[austere kraken]

so when we go back to this and say LHS diverges as n->inf so RHS diverge
S2n+1: similarly
then that orginal thing diverges
is that ok??

[blissful rune]

{pn} converges in X to p if and only if every subsequence {pnk} converges to p ==> not every subsequence {pnk} converges to p ==> {pn} doesn't converge in X to p. If {Sn} is convergent , then {Sn} --> p in R <=> the subsequence {S2n} converges to p in R. So if we prove that {S2n} diverges to infinity , then {S2n} doesn't converge to any p in R , hence {Sn} can't converge to any p in R, for if it did , then {S2n} would converge to something in R. which is a contradiction.

This wouldn't necessarily imply that the series {Sn} doesn't converge in extended reals

But {S2n} would indeed imply that {Sn} can't converge to anything in R , which is equivalent to saying that it's divergent by definition of the convergence of sequences on metric spaces.

[lament knoll]

ohio contradiction

[blissful rune]

nyehehehe

[lament knoll]

idk why but it seems strange lol

[blissful rune]

it isn't if you replace Sn with {an}

as in

think of the series as a sequence

since series will just be the limit of a sequence

[lament knoll]

ik ik

but idk

seems too easy

[blissful rune]

there are many powerfull tools in math

[frank iris]

What does?

[blissful rune]

Altrough i wouldn't call this anything particularily powerfull

[lament knoll]

isn’t it the other way around? u_n converges to p -> u_(n_k) converges to p

[blissful rune]

it's both ways

<=>

That this is obvious

[lament knoll]

ok so how u justify that the two infinities towards which S2n and S2n+1 diverge are the same 🤓

[blissful rune]

And doesn't require to think hard

[blissful rune]

U can't

U wont know whenever Sn diverges to -inf or + inf or undefined or amogus infinity

But it doesnt matter

Because

[frank iris]

They don't.

[lament knoll]

it matters

S2n and S2n+1 need to have different limits

to conclude Sn diverges

[frank iris]

They just need to diverge.

[blissful rune]

By definition of "divergence" due to daddy rudin it doesn't matter, therefore since rudin is always correct obviously this implies that it's divergent qed.

[blissful rune]

Remember how subsequences are defined

Increasing sequence {n_k} --> infty

[lament knoll]

ik ik

[blissful rune]

That the convergence of {p_n_k} implies the convergence of {p_n} , and that the convergence of {p_n} implies the convergence of {p_n_k} is clear. By contrapositive we have {p_n_k} not converging means {p_n} doesn't converge

[blissful rune]

You don't even need the left implication

[blissful rune]

You use this implication, apply contrapositive on this one

so then

[lament knoll]

i understand ur argument

but idkkkkkk

ask a professional in math discussion lol

[blissful rune]

u(n_k) doesn't converge to p --> un doesnt converge to p but since this holds for all real numbers p, therefore by rudin's definition of divergence, the sequence un is divergent

Rudin: a sequence pn is divergent if it doesn't converge in X

[lament knoll]

i know what is a divergent sequence pls 😭😭

[midnight plank]

@austere kraken Has your question been resolved?

[lament knoll]

hm ok

[austere kraken]

I got emotional damage within just a few hrs wtf is going on

[blissful rune]

rudin my beloved 😊

[austere kraken]

there are so many concept out of the syllabus I guess I just igonre anything and say S2n and S2n+1 diverges => Sn diverges is already good enough>

[blissful rune]

syllabus not introducing metric spaces?

Bad syllabus 😠

[austere kraken]

is this step ok? did that change anything in this sum?

[blissful rune]

Rudin > your syllabus

[austere kraken]

as n->inf

[blissful rune]

Rudin > abbot

Rudin > Spivak

Rudin > Tao

Rudin > Rudin

[lament knoll]

contradiction so rudin doesn’t exists

[blissful rune]

It does

grrr

Anyways

[lament knoll]

https://tenor.com/view/muehehehe-gif-4405977563994366182

[austere kraken]

this course is just to give us a tast of math next yr I'll get fked by Rudin

[lament knoll]

what is rhs ?

[blissful rune]

Yea but you don't need ilke

You don't need S2n+1 if you proved that S2n diverges

If S2n diverges then Sn will also diverge from R

[lament knoll]

ye

[blissful rune]

Similarily if you prove that S2n+1 diverges then Sn would also diverge from R

Unless you specifically want to analyze S2n+1 and S2n

in which case , if Sn diverges then they could literally be anything

well almost anything

but variety would be present

the set of subsequential limits would be closed so it wouldnt be a full freedom , but there still could be a lot of freedom

[lament knoll]

oh no topology

[austere kraken]

I guess this is the end thx guys

.close

[midnight plank]

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[blissful rune]

lmao

[midnight plank]

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[leaden seal]

Can someone please check my answer

have i shown full working?

[austere nest]

1a 1b and 1d look right in terms of methodology (im trusting the value for 2 is correct)

1c doesn't look right

you can't integrate x^4 and ln(x) like that if they're multiplied with each other

[leaden seal]

let me try again

would i use integration by parts?

[austere nest]

yep!

[frank iris]

Or just use Tabular Method.

[leaden seal]

hmmm

so far

[dusty portal]

Okay, so integrate dv/dx and calculate the result.

[austere nest]

yeah tabular is honestly just better

[dusty portal]

Good.

[austere nest]

Looks good

Just remember

To evaluate

Since it’s a definite integral

[leaden seal]

so this should be the final answer

[dusty portal]

Yeah.

[leaden seal]

Thanks

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[leaden seal]

can someone check this please

have I shown all my workings?

[chilly ginkgo]

in the first questions you have not substituted the x with u

[lethal steeple]

2a is wrong

[leaden seal]

Let me try again

[chilly ginkgo]

the remaining two looks fine to me

[leaden seal]

I don't understand can you be more specific

[chilly ginkgo]

after changing the limits, when you changed cos5x to cos u. You must also change the x that is multipled with cos5x

[leaden seal]

I've replaced itwith du/5

[chilly ginkgo]

no you have replaced dx with du/5

[leaden seal]

Like this?

[chilly ginkgo]

yeah

try it now you'll be able to solve it

[leaden seal]

do I multiply this

[chilly ginkgo]

yup

do you know hoe to integrate by parts?

[leaden seal]

Yh

[chilly ginkgo]

then solve it using that

[leaden seal]

Okay ill try gtg

[chilly ginkgo]

fine

[midnight plank]

@leaden seal Has your question been resolved?

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[tidal turret]

It is known that a molecule with covalent bonds formed by two elements A and B has a central atom of A. Additionally, the molecule is nonpolar. Indicate which of the following molecular geometries are compatible with this observation: A) tetrahedral; B) bent; C) pyramidal; D) trigonal planar.

[jolly plank]

C₃H₇COCL + C₂H₅NH₂----> C₃H₇CONHC₂H₅ + HCL

Trust me

[solid iris]

@jolly plank stop posting nonsense in others help channels

[jolly plank]

Sorry

[pearl idol]

btw you probably won't get much help on chemistry here

[tidal turret]

fuck my life

[midnight plank]

@tidal turret Has your question been resolved?

[midnight plank]

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[solid iris]

yes ibp. the integral isnt just sinu

[vast ginkgo]

i think you accidentally added the denominator in your second step after I =

coefficient should be -2/25

[leaden seal]

soshoukd it be this?

how do I do IBP

[vast ginkgo]

you have -2(1/5)(1/5)

[vast ginkgo]

you can just move the coefficients out of the integral

makes it a lot less painful

[leaden seal]

like this?

[midnight plank]

@leaden seal Has your question been resolved?

[vast ginkgo]

not sure how you got cosine in the denominator

[midnight plank]

@leaden seal Has your question been resolved?

[midnight plank]

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[scarlet flower]

Help

[scarlet flower]

how did -7pi/4 get there

also what is generally going on here

[fallow scarab]

those are from the constraints on x

and $\tan(t) = 1$ when $t = \pi/4 + n \pi$

[grand pond]

riemann

[fallow scarab]

where $n$ is any integer

[grand pond]

riemann

[scarlet flower]

I feel like the example used such a confusing way to do the problem

is it the best way to answer the question

[fallow scarab]

probably yea

[scarlet flower]

how are they just magically getting that fraction though I feel like I’m missing something

[fallow scarab]

alternatively, start with tan(t) = 1 and solve for all values t i showed you above

then equate t = 2x - 3 and solve for x in the solutions

[fallow scarab]

you need to be more specific

there are many fractions

[scarlet flower]

that’s makes a lot more sense to me

[fallow scarab]

these two are equivalent ways to do the problem, but your teacher showed you this specific way because they wanted you to learn this specific method

[scarlet flower]

What do they mean by “work down”

Work around

[fallow scarab]

they used this

[fallow scarab]

and then

[scarlet flower]

Oh I understand

Okay ty

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[steel coyote]

Could someone verify or check my answer

[lyric charm]

does not match mine

i think you miscalculated the area of the triangle part

[steel coyote]

is it -10?

5pi/2 -10

[lyric charm]

that sounds correct now

[steel coyote]

thanks!

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[wind oxide]

help please

[wind oxide]

me is getting 0.5 seconds

answer given is 0.33 seconds

but apparently the jee year it came in everyone did find the answer to 1/2s

now since its broad we dont care about the x component of velocity

as for the y

we r given that u_y rel = 4 => u_y=8

[dreamy lichen]

wdym by that?

whats u_y?

[wind oxide]

y component of our initial velocity

the rel means its relative to the

[dreamy lichen]

did you consider the angle?

[wind oxide]

lift

[wind oxide]

i dont need to?

i dont care about x or the speed we threw it with

i can just work with the y components of its velocity vector

[dreamy lichen]

you need to compute the y-component using the angle though

you just did 4 + 4

but the y-component of the velocity at which the ball was thrown isnt 4

[wind oxide]

that isnt what i was supposed to do? i just used relative logic

[wind oxide]

wait is it the speed that was 4?

[dreamy lichen]

not sure whether the 30° is taken from ground or from the vertical axis, but I think the diagram looks like this

speed = magnitude of velocity vector

[wind oxide]

ground it seems

[dreamy lichen]

Yeah, probably

[wind oxide]

ah so my actual speed (relative to the ground) is 8

so my u_y is just 4?

[dreamy lichen]

no?

[wind oxide]

because relative to a body moving with speed 4 its 4

[wind oxide]

why is this wrong now

[dreamy lichen]

this is just the throw

this would be the full thing

[wind oxide]

uh okay

hol up wtf

relative to the lift

the velocity vector of the ball is

(u/2 - 4)j + sqrt(3)u/2i

correct

[dreamy lichen]

relative to earth the velocity vector should be sum of these

and what's u?

[wind oxide]

u is the inital speed

[dreamy lichen]

oh okay

[wind oxide]

why would it be the sum

relative is of 2 with respect to 1 is v_2 - v_1

[dreamy lichen]

well if you are standing in the elevator and you throw the ball at speed 4m/s (relative to the elevator, 30° angle), then the vertical component is just gonna be 2

and then you add the vertical component of the elevator motion which is 4

so you'd get 4 + 2 = 6 m/s for u_y

or at least thats how I would interpret it

[woeful turret]

did u go inside the frame?

[wind oxide]

u mean the lift

[woeful turret]

yes..obviously

[wind oxide]

not really i am stupid when it comes to physics

[wind oxide]

hm this kind of makes sense

but

whats wrong with doing it

my way

[woeful turret]

ok just lmk if u need my help

[wind oxide]

kk

[wind oxide]

okay that does make sense

but now i need to understand whats wrong with my method

hmm

OH

got it

[dreamy lichen]

okay good lol

[wind oxide]

its not intuition but i do have a reason

[wind oxide]

this is correct

we r given this guys magnitude is 4

[dreamy lichen]

btw by adding 2 speeds you wont necessarily get the actual speed, cause the velocities might have different directions

so u = 4?

[wind oxide]

and while solving i made it into u/2j + sqrt(3)u/2i

[dreamy lichen]

wait isnt u given?

[wind oxide]

no we r given relative u

[wind oxide]

makes sense, ig?

[dreamy lichen]

oh okay and you are computing actual u?

[wind oxide]

no

it should be 12

[wind oxide]

yes

[dreamy lichen]

yeah, ignore that, i misunderstood what it means

[wind oxide]

u_rel is relative to lift, u is relative to earth

[dreamy lichen]

i actually dont know how to validate it because it feels like you're doing it in reverse

u is a vector btw? Or a scalar?

Cause velocity is a vector

[wind oxide]

u is the magnitude of the initial velocity vector (hence initial speed)

[dreamy lichen]

kay

[wind oxide]

hmm nvm even this seems to be wrong

[dreamy lichen]

well, i think its wrong

[woeful turret]

;/

[wind oxide]

because ur logic works with my workbooks logic

[dreamy lichen]

and initial speed would be of little use here

[wind oxide]

well if i have u, i have u/2.

[woeful turret]

the lift speed dosent matter btw @wind oxide

[wind oxide]

why?

[woeful turret]

when we go into the frame the lift speed becomes 0

[wind oxide]

yes i understand that

so okay yeah

[woeful turret]

what is the problem then?

[wind oxide]

ur right

[wind oxide]

i dont see where i go wrong

i said

the initial velocity vector of the ball

is

$\frac{u}{2}j + \frac{\sqrt{3}u}{2}i$

[woeful turret]

.

[grand pond]

rak³en

[woeful turret]

yes correct

[dreamy lichen]

the red vector is the actual velocity relative to earth. the angle is weird and the vertical component would certainly not be just u/2

[wind oxide]

why r u working relative to earth

[woeful turret]

@wind oxide do you know time period formula for projectile?

[wind oxide]

we r kinda supposed to work relative to each other

[wind oxide]

yess

u^2 sin^2/2g

or well

[woeful turret]

;/

that is maximum height

[wind oxide]

u_y^2/2|a_y|

oh shit what

oh fuck

its 2u_y/|a_y|

my bad!

[woeful turret]

no that is not for a projectile

yeah

2 uy/ a_y

[wind oxide]

sorry it takes a while to load shit in my brain

[woeful turret]

now can you tell me what is a_y in our case?

[wind oxide]

can we first resolve wtf is going on up there

[wind oxide]

this was correct right

[woeful turret]

resolve what

yes

[wind oxide]

just listen to me

yeah so now the velocity vector of the lift (INITIALLY) is 4j

so INITIALLY the velocity vector of the ball WITH RESPECT to the lift is

$\left( \frac{u}{2} -4 \right)j + \frac{\sqrt{3}u}{2}i$

correct?

[grand pond]

rak³en

[woeful turret]

no?

[dreamy lichen]

this already is wrt the lift

[woeful turret]

you cant combine the velocities of the projectile and the lift like that

[wind oxide]

how though

[wind oxide]

its relative velocity so i just did v_2 - v_1

[woeful turret]

and i already told you to go in the lift frame so that you wont have to consider the lift velocity

[wind oxide]

bro this is the velocity of the ball inside the frame no?

[dreamy lichen]

well it was said that it was thrown at 4m/s with respect to the elevator at angle 30°

[woeful turret]

it is not.

[wind oxide]

yeah so magnitude of velocity vector of ball with respect to lift = 4

[wind oxide]

why tho?????

the frame gets 0 relative velocity

[woeful turret]

^^^

okay first

can you tell me the acceleration in the lift frame

[wind oxide]

i dont even know tbh

uhh

12 m/s^2 down

?

[woeful turret]

yes correct

have you heard of pseudo force?

[wind oxide]

no

[woeful turret]

huh

[wind oxide]

why r we considering acceleration and forces...?

[woeful turret]

how did you get 12 m/s^2?

[wind oxide]

well because uhh

for the ball

the entire world is going 10 m/s^2 behind it while its at reast

so anything thats alr moving 2 m/s^2 upwards it would be 12 m/s^2 behind for it

and this is the relative acceleration of lift with respect to ball

[dreamy lichen]

its correct though in the relative to lift setup

[wind oxide]

and just reverse this to get 12 m/s^2 downards

[woeful turret]

ok so u applied the concept of pseudo force u just didnt know it..
anyway so now could you tell me what we have for the time period after substituting the acceleration

[dreamy lichen]

you will also need the vertical component of the ball velocity relative to the lift

[wind oxide]

i dont have u_y.

[woeful turret]

we will find that can u just tell me what we have till now dude

[wind oxide]

what if i just get the ball velocity relative to the lift and then half it

is that

a wrong logic

[wind oxide]

u_y/6

i thunk

[woeful turret]

yes

so now for u(y) they have alreaedy given us RELATIVE TO LIFT

so forget about the 4 m/s of the lift

and just resolve the given velocity in y direction

[wind oxide]

bro yoda

thats what

mathsisright did

and i understand

why its right

i need to understand

why I was wrong

[dreamy lichen]

I was working relative to ground, yoda is working relative to lift

[wind oxide]

huh didnt u do the same thing

kind of

yeh okay almost the same thing

just different frames

[dreamy lichen]

it's similar, because we both use somewhat conventional approaches

and i honestly have no idea what were you trying to achieve with that thing of yours

[wind oxide]

ugh

okay whatever

i shall quit on this approach

i guess?

[dreamy lichen]

Yeah, probably

[wind oxide]

ydzh alr

.close

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[cursive swan]

If x and y are the roots of x^2 - 3x + 3, find the value(s) of:

(i) x^3 + y^3
(ii) 1/x + 1/y
(iii) x^2/y + y^2/x
AND
(iv) x^10 + y^(-100x)

[fresh sparrow]

1. you are reusing x in two places, not cool
2. expression iv is not symmetric in x and y, it has two different answers depending on which root you take to be x and which to be y

and 3) $y^{-100x}$?? really?

[grand pond]

artemetra

[midnight plank]

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.

[obtuse basin]

For 1, 2, try rewriting the expressions in a different way (namely, ||factor|| for 1 and ||make the denominators|| for 2)

3 can be done using 1 and 2

[paper inlet]

the first 3 seem easy but what the hell is 4

[cursive swan]

Yes. (iv) is quite hard.

I don't know what the author of the book thought he was doing.

[obtuse basin]

Not to mention the roots are nonreal, so exponentiating makes it even more tedious

[cursive swan]

No wonder it's called "Problem.XYZ".

[cursive swan]

Yes.

[tawny apex]

maybe we express the roots in polar form?

$x=sqrroot(3) \cdot e^((pi/6) i)$

[grand pond]

nishi

[lethal steeple]

√3

[grand pond]

A Loner Bean

$x = \sqrt{3}\cdot e^{\frac{\pi}{6} i}$

[tawny apex]

yeahh

thanks

$y = \sqrt{3}\cdot e^-{\frac{\pi}{6} i}$

[grand pond]

nishi

[obtuse basin]

Whoops, didn't see the minus sign (nvm)

Just put it inside the {}

[tawny apex]

no no there would be 2 roots

one with the negative sign and other with the positive one

uhh and then maybe apply de moiver's theorem

it still wont be a good calculation tho

[midnight plank]

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[vocal flame]

I completely forgot tbh would only c be the Local max

[vocal flame]

Because obviously b d f are mins

And I think e would be global max

But idk about a

And ig f too

[lyric charm]

A is a local maximum and yes E is the global max

[vocal flame]

Thanks

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[wary trail]

How was the taylor expansion performed? I dont see where the sqrt comes in

[midnight plank]

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[wary trail]

.reopen

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✅

[frank iris]

f(x+h)=h*f'(x)+f(x)

[wary trail]

oh oh course

[proud ermine]

I think it is easier to see what is going on if you just view $L_a = \sqrt{f(x)}$ so that $L_a^{\prime} = \frac{1}{2\sqrt{f(x)}} f'(x)$. Then you know that $f'(x)$ can be written as $-2(a-1)\dot{q_1}\dot{q_2}$ and put it together.

Okay, I don't know what part of that fails to render.

[grand pond]

Hexicle

[frank iris]

Yes. Then combine it with f(x+h)=h*f'(x)+f(x)

[proud ermine]

Yeah, using substitutions tend to make things a lot easier to manage

[midnight plank]

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[wary trail]

thank you, I just had a brainfart then because I'd forgotten Taylor's Theorem on the spot

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[frank iris]

Did I use Taylor's Theorem?

[proud ermine]

You wrote a first order Taylor expansion of f(x+h), which I would not really call Taylor's Theorem but that was mostly splitting hairs.

I would prefer to write it in the form f(x) = f(a) + f'(a)(x-a) but I didn't want to explain the differences in notation because keeping track of that a and the a in the L_a would have been annoying.

[wary trail]

That's fine, I usually use f(x+h) = f(x) + hf'(x) anyways

thank you

Where did i drop the -

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[remote owl]

I understand I need to get a common denomenator but dont know what to do after

[gleaming latch]

what did you get for the common denominator?

[remote owl]

(v+6)(v-2)

[gleaming latch]

rewrite all the mini fractions so that their denominators are the common denominator

[dusty portal]

Great.

Now try and combine the fractions.

[rustic tartan]

Split the nomerator

[dusty portal]

$\frac ab+\frac cb=\frac{a+c}b$.

[grand pond]

;(

[remote owl]

ok so 8v+24/(v+6)(v-2)

or 8(v+3)

[dusty portal]

Yep.

[remote owl]

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[twilit field]

I'd like to prove that $\Q$ is countable by establishing a one one function from $\Q \to \N$

[grand pond]

What a wonderful world !

[twilit field]

Now abbott just does this visually, unless I'm missing something( I can share the proof)

[hard shard]

wheres that frosting gif

[twilit field]

what gif

I'm however, not convinced

[gaunt plume]

It's normally done visually by just drawing lines, but it can also be done symbolically

[twilit field]

More specifically, the issue I have is with the way he's showing the one-one correspondnace

[twilit field]

How would I do that

[gaunt plume]

Each rational number can be written as p/q for some unique coprime p,q where q is positive

So that gives an injection from p/q to N^2

Well, to Z * N

[twilit field]

yea

[gaunt plume]

Or actually, send it to N^3 where the 3rd variable represents whether it's positive or negative

Point is, there's an injection from Q to N^3

[twilit field]

We'd then need to establish a one-one function from N^3 to N

no?

[gaunt plume]

Then you can define an injection $(n_1, n_2, n_3)\mapsto 2^{n_1}3^{n_2}5^{n_3}$

[grand pond]

depression

[gaunt plume]

Which proves that $|\mathbb Q|\le|\mathbb N|$

[twilit field]

Cool

makes sense

Thanks

[grand pond]

depression

[gaunt plume]

And the other direction is fairly obvious

Then by the schroder-bernstein theorem, that implies that $|\mathbb Q|=|\mathbb N|$

[grand pond]

depression

[gaunt plume]

If you know what that is

[twilit field]

That I do not know

[gaunt plume]

That just says that if $|A|\le|B|$ and $|B|\le|A|$ then $|A|=|B|$

[grand pond]

depression

[gaunt plume]

Fairly intuitive but annoying to prove

[twilit field]

Got it

Thanks!

[gaunt plume]

Np

[twilit field]

Guess I'll get on with solving problems now

Show : If $A \subseteq B$ and $B$ is countable, then $A$ is either countable or finite.

[grand pond]

What a wonderful world !

[twilit field]

I was thinking of attempting to establish an injection from A to B

[gaunt plume]

I'm pretty sure that needs the axiom of choice

No it doesn't ignore me

[midnight plank]

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[twilit field]

I see

Okay

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[subtle peak]

What? Where did the extra s come from on the last line?

[subtle peak]

This is the solution key, btw

[thick granite]

Hello i dont know the steps for solving it but the answer should be x=0 and x=log4/log u btw

<@&286206848099549185>

[subtle peak]

.done

!done

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[midnight plank]

Please don't occupy multiple help channels.

[molten bay]

Hints

[midnight plank]

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[polar star]

i mean

u dont have any angles?

u need to resolve the vectors by components

then find magnitude / direction of the net force

[unreal nest]

idk what that means

[polar star]

have u learnt the parallelogram law?

[rich mirage]

Real

[unreal nest]

nope

[exotic stratus]

I am making some assumptions about the angles here based on the fact that the triangle is equilateral, These aren't completely out of the left field if diagram is an accurate representation of the original system

[paper inlet]

do you assume uhh F_2 is parallel to F_4

[polar star]

just use a protractor

[exotic stratus]

I think it is, because from the looks of it F_4 is extension of the angle bisector of that angle which would also be the side bisector of the side against it (because this is an equilateral triangle) and perpendicular to it too (again because it's an equilateral triangle)

[unreal nest]

im lost

[exotic stratus]

Have you worked with vectors before this?

[unreal nest]

a little

[exotic stratus]

So do you know why angles are even needed here?

[midnight plank]

@unreal nest Has your question been resolved?

[unreal nest]

for trig?

[exotic stratus]

Well yes we'll use trig but for dividing the force into vertical and horizontal components

[unreal nest]

i see

[exotic stratus]

Yea so using those can you do that

can you get the horizontal (parallel to base of triangle) and vertical (perpendicular to the base of triangle) components of the force

[unreal nest]

i think u can just assume f1 and f3 are pointing towards each other

and f2 and f4 are pointing in the same direction

maybe they just drew it badly

[exotic stratus]

I mean yea that's what I said in my diagram

[unreal nest]

ok

i still dont know whats goin on tho

[exotic stratus]

Oh so since the triangle is equilateral, all of it's angles must be 60 degrees

[unreal nest]

uh huh

[exotic stratus]

from there I just use a bunch of properties of parallel lines

and make a bunch of assumptions like F_2 is perpendicular to the side of triange

and that it's acting on the midpoint of the triangle

and that F_4 is the extension of the angle bisector of the angle it's tail is on, and from there it comes to be 120 + 30 from the both sides of triangles

and then F_1 and F_3 are just assumed to have just the horizontal component, the reason I added the angle for F3 was in case you also had to consider torque but that seems like a little far fetched in this scenario

[unreal nest]

im not really following

how do i find the net force

[exotic stratus]

Can you calculate the equivalent force of just F_1 and F_3, and then F_2 and F_4 as pairs?

[unreal nest]

is that just 21n and 2n

[exotic stratus]

Good

Now, seperate them into vertical and horizontal components

F_1 and F_3 just had horizontal component so ofc the resultant will also just have that

but F_2 and F_4 have both vertical and horizontal component

You have the angles

Determine the angle from the base

and from there use what you know about vectors to solve this

[unreal nest]

idk how to use the angles

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[potent fractal]

I am just wondering if YXB is a right angled and how do we verify that something does form a right angled triangle in 3d trig?

[tall lantern]

Nice username

[obtuse basin]

If given the coordinates of the vertices, it is possible to check whether an angle between two side lengths is a right angle by checking if the dot product of the vectors representing the side lengths starting at the angle is zero

If given the side lengths of the triangle, you can simply check if the Pythagorean theorem is satisfied

[tall lantern]

Sometimes it just LOOKS 90 degrees

But if you want a rigorous mathematical proof, the only way is to obtain the other 2 angles

Using other sides/other angles of corresponding figures

[midnight plank]

@potent fractal Has your question been resolved?

[potent fractal]

ah right that makes sense thanks

[potent fractal]

i see, thanks

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[bright bridge]

hey whats the best way to find all natural triple digit numbers which when divided by 3 give a remainder 2 sum?

[bright bridge]

i've tried using arithemtic progression but maybe theres a easier solution

[fresh sparrow]

wdym by "remainder 2 sum"?

[bright bridge]

like all the numbers that divide by 3 and give a remainder 2 sumed up

[fresh sparrow]

are you simply looking for triple digit $x$ such that $x\equiv 2 \mod 3$

?

[grand pond]

artemetra

[bright bridge]

yeah

[fresh sparrow]

that's kinda the solution

the smallest one is 102 and the largest is 999

so you have that its (999-102)/3

,calc (999-102)/3

[grand pond]

Result:

299

[fresh sparrow]

here you go

oh sorry

nvm

,calc (998-101)/3

[grand pond]

Result:

299

[fresh sparrow]

yeah it's still 299

but the first number is 101 and the last is 998

[bright bridge]

alright thanks

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[molten bay]

Hints

[obsidian relic]

$A = k*B$ , where k is a constant, and $A=l/C$, where l is another constant

[grand pond]

diptava

[obsidian relic]

then, $20 = k*25$ and $20 = l/18$

[grand pond]

diptava

[obsidian relic]

@molten bay do you see how you can proceed?

[molten bay]

Yes i can find values of constant

[obsidian relic]

yep exactly

[molten bay]

K=20/25

L=20×18

What should I do next?@obsidian relic

[obsidian relic]

i made a slight error

since it's given that A varies with B and C together, the equation looks something like this $A = (kB)(l / C)$

[grand pond]

diptava

[obsidian relic]

which becomes $A = kl * (B/C)$, then we consider kl as a new constant say m, so $A = m*(B/C)$

[grand pond]

diptava

[obsidian relic]

now if you solve for m, you will get the answer

previously we erroneously assumed independent variation, when B and C together influence A as per the question

@molten bay

[midnight plank]

@molten bay Has your question been resolved?

[molten bay]

20=kL 25/18

KL=20*25/18

[obsidian relic]

20 = kl * 25/18

[midnight plank]

@molten bay Has your question been resolved?

[molten bay]

I got B

Thanks

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[spice scroll]

how do i prove that

[spice scroll]

while

[gaunt nimbus]

doenst this diverge, waht 😭

[gaunt jetty]

[silver falcon]

yh

[spice scroll]

i tried using the last equality to show get a $1->inf \sum (-1)^k * b_n
where b_n is the 1-k/N+1

[silver falcon]

lol

[gaunt jetty]

diverges by alternating series test. There is nothing to do here

[spice scroll]

no

[woeful turret]

-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1 somehow becomes 1/2???

[gaunt nimbus]

nah

bros doing that -1/12 argument

[spice scroll]

this is Cesaro summation

[woeful turret]

ohhh

[spice scroll]

how do i prove this

[gaunt nimbus]

(-1)^k is non-converging

u cant find a sum

i have the wrong proof

wait

i found it

[grand pond]

k

k

[gaunt nimbus]

but this is wrong

because riemann rearragement theorem

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[twilit field]

I wish to prove that the sqeunces of the nth cesaro means of a given sequnce go to the same limit as the sequnce

[twilit field]

that is $y_n \to L$

[grand pond]

What a wonderful world !

[twilit field]

where $y_n= \frac{\sum_{i=1}^{n} x_i}{n}$

[grand pond]

What a wonderful world !

[twilit field]

so I have $\abs{ \frac{\sum_{i=1}^{n} x_i}{n} - L} < \varepsilon$

[grand pond]

What a wonderful world !

[twilit field]

or $\abs{\sum_{i=1}^{n} x_i- nL} < n \varepsilon$

[grand pond]

What a wonderful world !

[twilit field]

which doesn't really help

[simple field]

Move the L inside the sum then ||triangle inequality||

[twilit field]

I did think of that

so {x_i-L/n}

right

so I have $\abs{ \frac{\sum_{i=1}^{n} x_i- \frac{L}{n}}{n} } < \varepsilon$

[grand pond]

What a wonderful world !

[lyric charm]

hey may i suggest something that might at least make your life a bit easier notation-wise

prove it first for the special case L=0

ie prove that if x_n -> 0 then y_n -> 0 too

that should be easier i think

[twilit field]

lemme try that

I know that $\abs{x_n}< \varepsilon$

[grand pond]

What a wonderful world !

[twilit field]

we then have $\abs{ \sum_{i=1}^{n} \frac{x_n}{n}}< \varepsilon to be shown$

[grand pond]

What a wonderful world !

[twilit field]

we take a partial sum and use the triangle inequality

[simple field]

This isn't quite right

[twilit field]

that gives is $\abs{x} + \abs{\frac{x}{2}} + \dots + \abs{ \frac{x_n}{n}} < \varepsilon$ ( all converges by the algebra of limits to zero)

[twilit field]

yeah, I realise now

we we have $\abs{x} + \abs{ \frac{x}{2}} + \dots < \abs{ \sum_{i=1}^{n} \frac{x_i}{i}}$

now what

[grand pond]

What a wonderful world !

[twilit field]

I can't be sure that the inequality is less that \varepsilon

[quiet hinge]

have you not already asked this question once

[twilit field]

i know I've done this beofe yes

[quiet hinge]

also i have no idea what you are trying to do

this is practically incomprehensible

[twilit field]

Ann suggested I first tackel the case where it goes to 0

[quiet hinge]

i know what ann suggested, since it is also what i said last time

[grand pond]

What a wonderful world !

[twilit field]

I'm not getting an idea, I'll sleep over this

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[sour sierra]

.rotate

[sour sierra]

.rotate

,rotate

[sour sierra]

Why is it not 4x4/2 to find the amount of green
But it is 5x4/2 to find amount of green

[quaint seal]

what’s the longest continuous row or column of green you have displayed ?

[sour sierra]

4

[quaint seal]

also wait this doesn’t make sense

wait

[sour sierra]

Wait

I think I get why now

[quaint seal]

the arrow points to 4x4/2 to find green no?

[sour sierra]

Like to find the amount of green

Now I am more confused

Why is it not (5x4/2)-2

Bruh Idk

.close

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[midnight plank]

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[twin kernel]

this converges to 9/10 right?

[twin kernel]

i used ratio test

[quick nexus]

ratio test can find convergence, but cannot find what the series converges to

[twin kernel]

bruh what

so what am i supposed to do

[lavish venture]

it’s geometric

because you have a number raised to the power of n and nothing else involving n

$\frac{(-1)^{n-1} \cdot 3^{2n + 1}}{10^{n-1}} = -30 \left(\frac{-9}{10}\right)^n$

oh it’s 2n + 1

not 2n - 1

my fault

[grand pond]

knief

[lavish venture]

3^(2n) = (3^2)^n = 9^n

[midnight plank]

@twin kernel Has your question been resolved?

[midnight plank]

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[brittle grotto]

guys

[brittle grotto]

looking back on some earlier proofs

this one is for Gronwall's lemma. note that $v(t)\geq0$

[grand pond]

00100000

[brittle grotto]

and c is a constant

my question is

how does it follow that $u(t)v(t)\leq(c+R(t))v(t)$

[grand pond]

00100000

[brittle grotto]

i.e. how is it clear that $u(t)\leq(c+R(t))$

[grand pond]

00100000

[brittle grotto]

$u(t)\leq c+\int_a^tu(s)v(s)~ds$

[grand pond]

00100000

[brittle grotto]

because v(s) being lower can lower the value of the integral arbitrarily much, right?

WAIT I AM AN IDIOT

it's an assumption of the lemma 💀

💀 💀 💀 💀 💀

.close

[midnight plank]

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[gaunt gust]

I originally said that D1>D2 because the same momentum is transferred however trial 2 involves blocks with a greater mass which means the friction which depends on normal force will be greater and cause the blocks to decelerate more shortening D2

[midnight plank]

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[midnight plank]

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[grave crater]

Evaluate the double integral ∬_R (x² + y²) dA,
where R is the region bounded by the ellipse (x² / 4) + (y² / 9) = 1

[grave crater]

how do i do this

[fallow scarab]

Have you learned polar coordinates

[tepid nexus]

Ok I might be wrong here but

[grave crater]

its like how to find the point right

a method

oh

shit

thanks

wait arent you the same person that couldnt do the grade 8 question

the hell

[tepid nexus]

Yes but they taught me this one first

[grave crater]

wtf

@olive matrix can you ban this troll

[fallow scarab]

She obviously copied your question into chatgpt

[grave crater]

oh really

[fallow scarab]

Or some other ai engine

[grave crater]

is that right though

[fallow scarab]

Just ignore it

[grave crater]

cuz chatgpt can be wrong

tf

[olive matrix]

alright

[fallow scarab]

Example 15.3.2A
https://math.libretexts.org/Bookshelves/Calculus/Calculus_(OpenStax)/15%3A_Multiple_Integration/15.03%3A_Double_Integrals_in_Polar_Coordinates

Has the formulas for changing to polar coordinates

[midnight plank]

@grave crater Has your question been resolved?

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[grave crater]

thank you

[midnight plank]

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[vocal flame]

Have I lost my mind or is it 5pi/4 and 7pi/4

.rcc

[steep mirage]

.rcc

how u rotate in this

i forgot

[vocal flame]

,rcc

I think

[steep mirage]

,rcc

[vocal flame]

Broooooo

[steep mirage]

,rccw

[grand pond]

Couldn't find an attached image in the last 10 messages.

May 7, 2025, 5:52 AM

[steep mirage]

,rccw

[gaunt nimbus]

What have u tried

[steep mirage]

i did this weeks ago and i have forgot how to do this

[vocal flame]

Same

[vocal flame]

I just took the pi and 3pi/4

[gaunt nimbus]

Substitute k(theta) = -1 first

[vocal flame]

And went inbetween

[gaunt nimbus]

?

could u explain

[vocal flame]

I thought since it’s exactly half is one

And the amp is 2

The point of influx should be 1

And that would be in the middle of the the two points

Pi and 3pi/2

And then the other -1 would be inbetween 3pi/2 and 2pi

Not point of influx

I’m losing it

That would be the middle

[gaunt nimbus]

Let’s do it algebraically

[vocal flame]

Ya

[gaunt nimbus]

So -1 = 2sinx

So -1/2 = sinx

[vocal flame]

Yep

[gaunt nimbus]

Do u recall what x makes sinx -1/2?

[vocal flame]

60 maybe

Negative so

[gaunt nimbus]

-30 deg

[vocal flame]

Ok

One of the two bottom ones

[gaunt nimbus]

Yes

The two bottom ones that make 30 deg to the x axis, right?

[vocal flame]

So -30

330

And 180

[gaunt nimbus]

330 in rad is?

[gaunt nimbus]

180+30 =210

[vocal flame]

Ya

[gaunt nimbus]

So 210 and 330

Convert back to rad and should be fine

[vocal flame]

11/6

And 7/6

[gaunt nimbus]

There ya go

,w sin 7pi/6

[vocal flame]

Thanks

[grand pond]

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[gaunt nimbus]

,w sin (11 pi/6)

[grand pond]

Results provided by WolframAlpha

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Requested by nuclearinteraction

[gaunt nimbus]

👍

[vocal flame]

Bro this ap test bouta cook me

Thanks

[gaunt nimbus]

Ap pre calc?

[vocal flame]

Yuh

I think it’s easy tho cuz it’s newer

[gaunt nimbus]

Ah ic

Gl 👍

[vocal flame]

.close

[midnight plank]

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[midnight plank]

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[steep mirage]

Did I do this correctly chat

4.9m/s W

5.1 m/s E