help-43

[prime loom]

ohh k I found this

[urban flicker]

2

[winged lion]

.

[prime loom]

srry for wasting time lol

[urban flicker]

anyway, it's less then f(n)=n

that's enough

[prime loom]

f(n) <= n

yeah

so f(n) is decreasing

so upper bound is x_0,

[devout wraith]

i think its right

[urban flicker]

thank you @prime loom

[prime loom]

wait tho

my proof doesnt work for the 2

for 2 the bounds are sqrt(2) and x_0 I believe

[urban flicker]

how to prove that converges at sqrt(2)

[prime loom]

AM-GM?

[urban flicker]

uhmm

[prime loom]

$f(t) = 1/2(t + 2/t) >= sqrt(t * 2/t)

so thats sqrt(2)

[urban flicker]

$f(t) = 1/2(t + 2/t) >= sqrt(t * 2/t)$

[boreal girder]

Starland

[prime loom]

y

't's cancel

[urban flicker]

aha

and upper bound with n..?

[prime loom]

you can do it with derivatives

$f(t) = \frac12(t+\frac2{t})$, then
$f'(t) = \frac12(1-\frac2{t^2})$

[boreal girder]

AlmondAxis987

[prime loom]

for $t \geq \sqrt{2}, f'(t) \geq 0$

[stoic crescent]

Hello

[boreal girder]

AlmondAxis987

[prime loom]

and t is always greater than sqrt(2) as its lower bound

so f(t) is increasing

[urban flicker]

[prime loom]

ok etc (End of Thinking Capacity)

[urban flicker]

it has a lower bound sqrt2 and converges to sqrt2?

[prime loom]

Idk about the 2nd part

lower bound is def. sqrt(2)

[urban flicker]

uh okay

[prime loom]

<@&286206848099549185> can someone help this man and me out pls?

[mighty lance]

What question?

[prime loom]

Prove and find the bounds of the recursive sequence $x_{n+1} = \frac12(x_n+\frac2{x_n})$

[boreal girder]

AlmondAxis987

[prime loom]

we have found lower bound sqrt(2)

[urban flicker]

sure it's not increasing

it converges at sqrt2

[prime loom]

but this is correct as far as I can tell

[urban flicker]

compare x_n and x_(n-1)

[urban flicker]

mathematically correct, but it can not be applied on this recursive sequence

[prime loom]

ig

$f(t) -t = \frac{2-t^2}{2t}$

[boreal girder]

AlmondAxis987

[prime loom]

for all t>=sqrt2, this is negative lol

and t is >= sqrt2 as thats lower bound

so $f(t) < t$

[boreal girder]

AlmondAxis987

[prime loom]

is this correct?

anybody?

[winged lion]

I am sure you cannot just replace x_n with t because x_n can be any arbitrary sequence, right?

[prime loom]

?

[prime loom]

.

[fathom raven]

It does not matter if its replaced I think

[prime loom]

yeah

it starts with f(x_0) = x_1

then f(x_1) = x_2

and so on

and all x_n >= sqrt2

[fathom raven]

Yes

[prime loom]

so it is a monotinicly decreasing function

[prime loom]

from here

and its always >=sqrt(2)

so it converges to sqrt(2)?

@urban flicker I think this was the answer?

[urban flicker]

indeed

[prime loom]

ok cul

I feel a little dumb that it took me that much time tho

[fathom raven]

So with just that info is enough to say it converges to sqrt2?

[prime loom]

well thats yer questions solution done

[winged lion]

yea that's the issue, f(t) is not an independent variable, it's a recursion

[prime loom]

so?

[winged lion]

it doesn't prove anything

[prime loom]

I am using t to analyze the function and the restrictions on t placed in the analysis are placed on the recursion too

[prime loom]

Ok do you know what we have to prove?

[winged lion]

Induction would be a better choice to show possibly that sqrt(2) is a lower bound

[urban flicker]

lower bound sqrt2 and monoton decreasing
then it converges at sqrt2

[prime loom]

ig we are just doing induction

just not stating each step individually

f(x_n) < x_n

so x_n+1 < x_n

and x_n+1 >= sqrt2

so by induction x_n converges to sqrt2

also another way is to just set x_n+1 = x_n

in the recursion funxtion

[thorny kestrel]

if we suppose x_0 > 0, then clearly the sequence is positive. with AM-GM we find one lower bound to be 1.
then for t > 2, f - id is less than a strictly negative constant so the sequence is decreasing, less than x0, and will be less than 2 after a certain point
we can then show that x_n is strictly bigger than a constant C that is strictly more than 1
and when t is between C and 2, f' is < 1 thus proving both boundedness and convergence

x_n can be strictly decreasing & bigger than sqrt2 without converging to sqrt2

[prime loom]

oh so like it can converge to 1.5 and the conditions still hold

interesting

I have no idea how to find the limit of the sequence then

ohh lol I just solved by mistake

[prime loom]

here

this gives x = sqrt2

if you set f(x) = x

[fathom raven]

Idk I am new in how to do proofs

[prime loom]

I am not, but still I struggle

[fathom raven]

Thanks for discussing these topics it helps me explore the different views of everyone

[prime loom]

K Bai

OP close channel

*If youre done

@urban flicker

[compact pewter]

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[thorny kestrel]

you can only do that if x_n converges

[thorny kestrel]

ok mb it works

nvm nvm

.close

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[kind viper]

!status

[compact pewter]

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

[pseudo wasp]

Ehhh ig no idea

All I got was 18 * {11.....1(51 times)}^2

And I am somehw supposed to get 52nd digit

[kind viper]

well that's a good start actually

[pseudo wasp]

Okay....... any hints or patters?

I am just like 111....1^2 is always in the form of 123.........321

[kind viper]

try to look at the squares of 1, 11, 111, 1111 etc.

[pseudo wasp]

but ther than that idk

[kind viper]

why did you say "form" twice?

[pseudo wasp]

Ehhh cause doesn't it end with what it starts

like 12321

in this way

1234321

[kind viper]

no but why use the word "form" twice

[pseudo wasp]

so 123.....321

oh nvm that

[kind viper]

also consider looking at the expansion of $(x^n + x^{n-1} + \dots + x + 1)^2$ more generally

[boreal girder]

Ann

[pseudo wasp]

So......... eh........ well isn't this only true upto 9 1's

[kind viper]

and also expressing your number made of 51 ones as that but with n=50 and x=10.

[kind viper]

it only breaks down bc of carrying

[pseudo wasp]

okat....

ya

i know that..... :)

[pseudo wasp]

wait imma work on this exapnsion for 5 mins

see if I can find some pattern

Ok Just a hint

I tried to use Gp right

(X^n+1-1 ) /x-1

so I would get smth like eh.....( 10^51-1/9)^2

@kind viper

[kind viper]

GP will be kind of a no-go here sorry

also i am going to be slow to respond bc im cooking myself food rn

[kind viper]

the coefficients of this shit will be 1, 2, 3, 4, ..., n-1, n, n-1, ..., 4, 3, 2, 1
for any n

the 52nd digit from the right in any number is the 10^51 place

figure out how to account for that

and all the carries from places below

[pseudo wasp]

ehhh

[kind viper]

and do not forget about the 18 factor

[dusky quartz]

maybe try smaller cases.

i.e. 333×666
3333×6666
note any patterns you notice

then try generalizing it

[kind viper]

also this thing has missing brackets anyway and i will be a bitch about it

[pseudo wasp]

Bruh

I truly don't understand

[pseudo wasp]

I got this guys point

in that way it will be one

[pseudo wasp]

ehhhh

.close

[compact pewter]

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[prime loom]

@pseudo wasp

I can give a hint to solve the question you have above

if yu want, that is

ehh k whatever

[compact pewter]

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[west isle]

hi can I get some help? I’m unsure of my answer ☹️

[viscid canopy]

the total money he got = money from the merchandise + money from the sales tax

money from sales tax is 9.4% of the money from the merchandise

[west isle]

??

[viscid canopy]

what are his two sources of money?

[west isle]

1219.58 and 9.4

[viscid canopy]

no, the 1219.58$ is the total he earns from the two sources of money

what is his job?

[west isle]

artist?

[viscid canopy]

yes and what does he sell

[west isle]

pottery.

[viscid canopy]

good. So he sells pots and everytime he sells it, he taxes 9.4% money of the total sale. If he sells $100 worth of pots in a day, he earns an additional $9.4 from the tax

Does that make sense?

[west isle]

yes and no? that’s different from what I got when I solved it

[viscid canopy]

I was giving an example

What was your working btw?

[west isle]

no I know that, im just saying that I solved the problem and not sure if this is correct.

[viscid canopy]

ah, so what answer did u get?

[west isle]

1.00777 = x

and 9.47

because I had to solve for x

[viscid canopy]

what is x here

[west isle]

1219.58 = x + x ( 9.4)

[viscid canopy]

ah here is the problem

it is 9.4%, not 9.4 itself

9.4% = 0.094

9.4% of 100 dollars is 9.4 dollars, not 9.4*100 dollars

so the actual expression is: 1219.58 = x + x (0.094)

[west isle]

OH OH

I GOT IT

[west isle]

did I do it the other way around??

[viscid canopy]

you misinterpreted the meaning of percentage but the idea was correct

[west isle]

I got 1219.486

[viscid canopy]

for x?

[west isle]

yes

[viscid canopy]

,w solve 1219.58 = x + x (0.094)

[boreal girder]

Results provided by WolframAlpha

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[viscid canopy]

,calc 609790/547

[boreal girder]

Result:

1114.78976234

[west isle]

unless I simplify it

[viscid canopy]

hmm that is supposed to be x

[west isle]

erm

[viscid canopy]

how did u get 1219.486

[west isle]

I lowk don’t know unless i misplaced an number on accident when i did it

WAIT

I DID

oops

[viscid canopy]

did you figure out the issue?

[west isle]

yes

wait am I putting

1219.48/ 1114.789

like the full numbers for 1114.789?

or is it just 1114.78?

[viscid canopy]

for x, just 1114.79

because we are rounding to the nearest cent

[west isle]

o

this is what I have rn

[viscid canopy]

im not sure what you are doing here

1219.58 = x+0.094x
hence
1219.58 = 1.094x

Do you agree so far?

[west isle]

hey so

I am still fixing my equations.

[viscid canopy]

ah okay, let me know once u fix them!

[west isle]

1.094x??

where are u pulling that from 💔

[viscid canopy]

we are combining terms

as an example, what is x+x?

[west isle]

im overwhelmed.

[viscid canopy]

oh I apologize

have you seen stuff like x+2x = 5 before?

[west isle]

???

x+2=5 you mean or?

[viscid canopy]

yeah like in general

[west isle]

yeah I know x+2=5. not x+2x=5.

[viscid canopy]

okay so in algebra (the field of math we are doing right now), when you see something like
x+3x
we can combine the terms to get
x+3x = 4x
just like 1+3 = 4

with this logic, what would x+9x be?

[west isle]

umm

I don’t know, this is just overwhelming me more.

I think im going to stop here.

[viscid canopy]

alright

you may type .close

to close the channel

[west isle]

.close

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[mystic crater]

hey

[mystic crater]

I have the starting equation

5sin(2x)cos(2x) = sin(2x)

so then i divide both sides by cos(2x)

[native shard]

,av 82jz

[boreal girder]

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[mystic crater]

leaving me with

5sin(2x) / cos(2x) = sin(2x) / cos(2x)

which is

5tan(2x) = tan(2x)

which doesnt look right

[kind viper]

LHS is incorrect for one

and also you should be really careful dividing both sides by cos(2x) unless you wanna lose solutions

but also 5sin(2x)cos(2x) divided by cos(2x) does NOT give 5 sin(2x) / cos(2x).

[mystic crater]

yeah dumb error

[mystic crater]

why does this lose solutions

[kind viper]

whichever x's would satisfy cos(2x)=0 might be solutions but after this division they won't be.

[nocturne gate]

yea true u basically are considering tht cos2x is not the cause of the equality

[mystic crater]

why =0?

[kind viper]

also, third thing: i don't even think dividing by cos(2x) helps you at all.

[kind viper]

division by 0 no good

[mystic crater]

i was trying to get only one ratio in the equation

[kind viper]

you won't accomplish it that way.

[mystic crater]

so why would dividing by cos(2x)=0 not be an error but instead lose solutions

[nocturne gate]

tbh if u were trying to do tht u could have divided by sin2x

but dont

[mystic crater]

but this loses solutions too

possibly

[viscid canopy]

that is why, the optimal path is to take everything to one side and factorize

[nocturne gate]

yea thts why i said dont

[kind viper]

do you want a different example where solutions really are lost

non trigonometric, for simplicity

[mystic crater]

sure

[mystic crater]

like (1-5cos(2x))sin(2x) = 0?

[viscid canopy]

yups!

[kind viper]

take the equation x^2 = 7x

[viscid canopy]

this way, you get two factors that can possibly be equal to 0

[mystic crater]

ahhhhhhhh

[kind viper]

on the one hand, you can tell it has solutions x=0 and x=7 by a similar technique: put all terms on one side and factorize -- this gets you x(x-7) = 0.

on the other hand, if you give in to the temptation to divide by x immediately, you are left with x = 7 -- and now the solution x=0 is lost.

[mystic crater]

okay this makes sense

does this only happen when x (or the solution) = 0?

[kind viper]

rather it's when the expression by which you divide equals 0.

bc again, division by zero no good.

so funny business like this happens precisely when you divide by maybe-zero.

[mystic crater]

and if it doesn't then you won't lose solutions

[kind viper]

note that division by x, or another variable quantity that may be 0, isn't wrong in itself -- it simply requires knowing the consequences of doing such a thing.

[mystic crater]

interesting

[kind viper]

one counter-measure is to explicitly look at whichever x's might be lost as solutions, and note which ones among them really are solutions and write that down explicitly on the paper -- that way, they'll still be there and you'll still be able to put them back in the solution set.

[mystic crater]

.close

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[mystic crater]

.close

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[frigid furnace]

yo guys

[frigid furnace]

whats the smartest way to integrate 6arctan(8/w)dw

i mean

quicket

DI method?

[pseudo shale]

U sub

Then By parts

[frigid furnace]

I did

by parts

then u sub

.close

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[boreal girder]

;(

;(

[eternal pulsar]

So far so good?

[eager cipher]

can someone helpme wth something quick, its a really easy thing by the looks of it but i just have a little question

[short lantern]

ehh

id write it as a column vector of zeros

[short lantern]

!occupied

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[eager cipher]

why do you subtract 8ab from 15ab

oh sorry

[short lantern]

!occupied

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[eager cipher]

sorry sorry

[eternal pulsar]

Oh well

Too lazy

The problem comes when row reduce

[short lantern]

yes, you can see the sums of the columns are all 0

[eternal pulsar]

Ik that R1+R2=-R3

So we can "replace" the equation 3 with the equation w1+w2+w3=1

Right?

[short lantern]

wha

[slate sand]

Yea

Thats from the definition of a stochastic matrix

[eternal pulsar]

Oh ok

Good to know

[short lantern]

oh ok

[eternal pulsar]

Ok wait Lmao

[boreal girder]

;(

[eternal pulsar]

Ok so now aug should be

[boreal girder]

;(

[eternal pulsar]

Ow lag

Aaaaaaalright

So first row operation will be R2-2R3->R3

[boreal girder]

;(

[eternal pulsar]

Lets seeee

Next one will be 2R1+9R2->R2

[boreal girder]

;(

[eternal pulsar]

Good god

Y'all have any tips before i dive into this hell of a calculation?

Ok I feel like I did something wrong

Cause it's gonna result in w1=w2=w3=0

Sooooooooo

<@&286206848099549185>

[slate sand]

Did you try put it in a matrix calculator

[eternal pulsar]

Lemme se

,w rref {{-0.9, 0.2, 0.3, 0}, {0.2, -0.7, 0.4, 0}, {10, 10, 10, 0}}

[boreal girder]

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[eternal pulsar]

Oh well

💀

So da method is wrong

,w rref {{-0.9, 0.2, 0.3, 0}, {0.2, -0.7, 0.4, 0}, {0.7, 0.5, -0.7, 0}} with fractions

Gah

So

[boreal girder]

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[keen granite]

how did you get here?

mult and div by 10?

then shouldn't the bottom row be all 10s instead of all 1s?

[compact pewter]

@eternal pulsar Has your question been resolved?

[eternal pulsar]

,w rref {{-0.9, 0.2, 0.3, 0}, {0.2, -0.7, 0.4, 0}, {10, 10, 10, 0}}

[boreal girder]

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[eternal pulsar]

Same thing though

Aw

Damnit

You get the point

[eternal pulsar]

^

[keen granite]

well different matrices but it appears both are invertible

[
\begin{bmatrix}
-9&2&2\
2&-7&4\
10&10&10
\end{bmatrix}
]

[boreal girder]

PajamaMamaLlama

[eternal pulsar]

Yeah...

No clue

[keen granite]

wait do I suck at math?

[eternal pulsar]

This stupid questions and fractions

[keen granite]

,calc -9*lcm(2,9,10)

[boreal girder]

Result:

-810

[keen granite]

,calc -810/180

[boreal girder]

Result:

-4.5

[keen granite]

guess I do

[eternal pulsar]

wolfram won't give fraction output

[keen granite]

[
\begin{bmatrix}
-18&4&4\
18&-63&36\
10&10&10
\end{bmatrix}
]

[boreal girder]

PajamaMamaLlama

[keen granite]

[
\begin{bmatrix}
-180&40&40\
0&-59&40\
0&220&220
\end{bmatrix}
]

[boreal girder]

PajamaMamaLlama

[eternal pulsar]

Ugh

Well I got it

may or may not have asked gemini to do the brunt of the work

[keen granite]

alr yeah idk where this matrix came from but goodjob!

[keen granite]

[eternal pulsar]

I gaev up

There's no way shit like this will appear on the exam

[keen granite]

well wouldn't the full augmented matrix be:
[
\begin{bmatrix}
-9&2&3&0\
2&-7&4&0\
7&5&-7&0\
1&1&1&&1
\end{bmatrix}
]

[boreal girder]

PajamaMamaLlama

[keen granite]

why is that 1 in a_44 so far out?

[eternal pulsar]

OHHHHHHHHHHHH my lord

I'm dumb

[eternal pulsar]

&&

[eternal pulsar]

Just did row reduction on paper and it looks good

I guess I can do it but I make sill ymistake

[keen granite]

oh whoops

[keen granite]

happens

just make sure it doesn't during an exam

easier said than done tho

[eternal pulsar]

Yep

Aight cya, I'm VERY tired

.close

[compact pewter]

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[keen granite]

,ti ;(

[boreal girder]

The current time for exiled_hype is 06:33 PM (EDT) on Fri, 29/08/2025, the same as xxmrfancyu2xx!

[compact pewter]

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[livid spoke]

How do I do this

[bleak dock]

you know how to find the inverse of h(r) = 2r + 1 right

[livid spoke]

yes

[bleak dock]

what's the issue then

[livid spoke]

so i just do that for both?

[bleak dock]

the domains of both functions?

[bleak dock]

yes

and then note your domains, so for the inverse you will have $y \le 0$ and $y > 0$

[boreal girder]

south

[bleak dock]

h = y here

or actually, just find what h is when r = 0, so that would be 1 and 1 respectively, and that gives you the domains, cause you swap h and r

[livid spoke]

so the domains are not this?

[bleak dock]

no

cause you need the domain to be in terms of x surely

[livid spoke]

so what are the domains again?

[bleak dock]

figure it out from this

[livid spoke]

okay what is it, so i can see if i got it right

[bleak dock]

tell me and I'll tell you

[livid spoke]

wait wdym

[bleak dock]

if you've done it you can just say what you think the domains are

[livid spoke]

like 0-1/2 and cube root of 0-1?

[bleak dock]

well, no

okay so do you agree that (x, y) = (0, 1) is where both of those piecewise functions meet?

the functions break at x = 0

[livid spoke]

okay

[bleak dock]

yeah, and for the inverse functions

you just swap y and x

so that gives you which point?

[livid spoke]

1

[bleak dock]

well, (1, 0)

but the new x is indeed 1

[bleak dock]

so the new domains will be $r \le 1$ and $r > 1$

[boreal girder]

south

[livid spoke]

but how do you know this?

like where you got it from

[bleak dock]

you just sub r = 0 into 2r + 1 and r^3 + 1

and you get 1 and 1

[livid spoke]

ohh in the orginal thing

[bleak dock]

yeah

[livid spoke]

i thought i had to do r=0 on my new ones

[bleak dock]

yeah you could do it that way but I realised this shorter way

you could indeed solve for inverse of 1st function $\le 0$ and inverse of 2nd function $> 0$

it's not that much longer but still

[boreal girder]

south

[bleak dock]

it'd be repetitive actually

[livid spoke]

okay thanks for everything

[bleak dock]

no worries!

if you're done type .close

[compact pewter]

@livid spoke Has your question been resolved?

[compact pewter]

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[deft forum]

How do i do this

[deft forum]

To me it seems super unclear

[hushed magnet]

yeah its not well written imo

my interpretation: the clock in the mirror looks like its 6:45. so draw that, then mirror it, then read what time it shows now

[kind viper]

i think thats the only thing that makes sense yeah

[compact pewter]

@deft forum Has your question been resolved?

[deft forum]

Is that maths cuz I'm tutoring my lil bro's homework

This is ment to be a mock test so that pretty much can't be the solution

[hushed magnet]

why not

its definitely not any other subject

[deft forum]

Wait I solved it it's either 6:15 or 12:45

Depends on which axis which it didn't states

[hushed magnet]

neither one of those

where is the hour hand

[kind viper]

does your little brother know how a clock works

[prime loom]

why not?

is mirroring that high-level?

[compact pewter]

@deft forum Has your question been resolved?

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[native veldt]

can someone help me with this integral? I really cant find which convergent theorem i can use to bring the limit inside

[compact pewter]

@native veldt Has your question been resolved?

[native veldt]

I now have that the integrand goes to zero, but i don't know how to proceed.

[compact pewter]

@native veldt Has your question been resolved?

[sage sonnet]

<@&286206848099549185> 💔

[rotund sphinx]

https://math.stackexchange.com/questions/2023040/compute-lim-n-rightarrow-infty-int-01-dfrac-sqrtn-sin-x1n-sqrtx?noredirect=1

[native veldt]

But it's more to get the hang of changing the integral and limit from place, not just to solve it

[hazy yew]

Niemand kan bro helpen 🙏

[native veldt]

Ja ik denk da ik er gewoon mee ga moeten leven 😢

help jij mij

[hazy yew]

Ik kan je spijtig genoeg niet helpen 😔

[native veldt]

waarom niet

[hazy yew]

Ik heb daar niet de capaciteiten voor

[native veldt]

.close

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[strange pendant]

analize the existence of this limits and if they exist, say their value

[winged lion]

🐻‍❄️

[strange pendant]

take for example y = x and x = 0

we get

[winged lion]

(in case you wanna save time use polar coordinates)

[strange pendant]

it doesnt work

[strange pendant]

you mean (x,y) = (rcos(t), rsin(t))?

what is r then?

[winged lion]

r goes to 0

[strange pendant]

and t?

[winged lion]

t is arbitrary

u let it vary

letting t vary is like taking an arbitrary path

[thorny kestrel]

the numerator is y

[winged lion]

x²y

you can also simplify the denom

you can prove the limit with the squeeze theorem (use that sine and cosine are bounded)

[strange pendant]

[-1, 1]

[winged lion]

yea

[thorny kestrel]

Another way to prove this is using AM GM, though polar coordinates are probably the more general approach

[turbid surge]

Bro how are you already doing multivariable calculus ?

[winged lion]

yea and if you take the abs value and bound it you are done

[strange pendant]

what? I started university like 1 week and a half ago

anal 1, is multi var calc

[turbid surge]

I mean usually multi var calc is done after doing a calc 1 course with one variable idk

Ig its different for different unis

[strange pendant]

calc 1 course was what we covered in pre university math, idk

[turbid surge]

Oh

[strange pendant]

intro to mathematical analysis

[turbid surge]

Oh i see thats why

[strange pendant]

I am also taking this sem, algebra I which is intro to proofs, relations, basic set theory, functions, combinatorics, polynomials, and basic diophantic eqs, aswell as modular arithmetic

[winged lion]

crazy start

[strange pendant]

depends on the country if you covered calc 1 or intro to anal in pre uni, but I pressume it differs from across the globe

[strange pendant]

what now dude? how do I bound this thing?

cos^2 is bounded between 0 and 1?

[winged lion]

Yea

[strange pendant]

[winged lion]

,, \lim_{r\to0} |f(r,\theta)| = \lim_{r\to0} r\cos^2(\theta)|\sin(\theta)| \le ?

[strange pendant]

le?

[strange pendant]

ahh

[strange pendant]

1 i think

[turbid surge]

Good luck

[winged lion]

write it

[strange pendant]

thanks, u too dude, I saw you doing some exercises on modular arithmetic

[strange pendant]

but is also bounded from below by 0

the absolute value, no?

I am maybe tripping

[winged lion]

Yes

,, (0 \le) r\cos^2(\theta)|\sin(\theta)| \le r\to0

[turbid surge]

Yea I took abstract algebra this sem

[winged lion]

when i took it pre last, it took me

[strange pendant]

< or <=

[winged lion]

no

You still have r left

[strange pendant]

r tends to 0

lim r -> 0

[winged lion]

Yes

[strange pendant]

what about it?

[winged lion]

Why did you writr then 1

[strange pendant]

what about lhs of the ineq

why the absolute value ?

[winged lion]

Dont write limit

[strange pendant]

ok

[winged lion]

So we bound it from below with 0

Squeeze theorem says since lower and upper bound go to 0, then so must the middle term

[strange pendant]

its a completely different limit if we do that

[winged lion]

no

[strange pendant]

like, it would be simpler to argue that
since image of sin is [-1,1] and image of cos^2 is [0, 1] it follows that the image of cos^2(t)sin(t) is [0,1]

[winged lion]

You can bound it with 0

You dont have to take -1

[strange pendant]

but both things are equivalent

[strange pendant]

well

[winged lion]

I wrote the 0=< to make clear its not negative

If you resolve the abs value you get your function

Thats what I meant (I am cooking right now)

[strange pendant]

then what?

the problem is that I am having trouble concluding that its equal to 0

[strange pendant]

that is equal to 0

[winged lion]

Then what

[strange pendant]

but proving the original limit, is equal to

0

[winged lion]

Pull up the squeeze theorem

[winged lion]

g=-r
h=r
L=0

So f -> L as well

[winged lion]

Thats here

[strange pendant]

actually, I think that works

we dont need the absolute value on sin

[winged lion]

huh?

I just wrote the same thing shorter

I am on phone

[winged lion]

Remove the lim

[winged lion]

Do you see a lim in the inequalities

[strange pendant]

my bad dude

im cooking

some milanesa de pollo

[winged lion]

yes

[strange pendant]

I appreciate the help

how did you knew this was always zero?

also, cant i just use polar coordinates for all of this multivariable limits?

[winged lion]

not all

its like asking, can i always use a usub for integration

different tools for different problems

but when i see the x²+y² sure go with 🐻‍❄️

[strange pendant]

i appreciate it dude

.solved

[compact pewter]

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[turbid surge]

Given $\tau \sigma \tau^-1 = ( 1 2 5 4) \in S_7$
Given permutation $\tau \sigma^k \tau^-1$ is the inverse of $(\tau \sigma \tau^-1)^{40}$ find k

[turbid surge]

How do i do this

[boreal girder]

prograce

[turbid surge]

I know that ord($(\tau \sigma \tau^-1)^{40}$)=1 and ord($\tau \sigma \tau^-1$) =4 and that the permutation is odd from the previous parts

[boreal girder]

prograce

[kind viper]

$\tau^{-1}$ first and foremost

[boreal girder]

Ann

[kind viper]

also other

[turbid surge]

Im trying

[turbid surge]

What to do with it?

[kind crane]

should simplify $(\tau \sigma \tau^\inv)^{40}$ first

[boreal girder]

riemann

[kind viper]

was just correcting the most noticeable badtex on your part

gonna second riemann tho

[turbid surge]

Oh,

[kind viper]

alternatively

(tau sigma tau^-1)^4 = e

[turbid surge]

I could simplify it to (tau sigma tau^-1)^10 then

[kind viper]

simplify WHAT to that?

[turbid surge]

I think im looking at tau sigma tau^-1 as a whole when i should be looking at sigma=(1 3 4 6)

[turbid surge]

Simplify (tau sigma tau^-1)^40

[kind viper]

a^4=e does NOT imply a^40=a^10

[turbid surge]

Can I get a hint?

[turbid surge]

I tried again so: a^40=(a^4)^10=e^10=e ... idk

[kind viper]

yes

a^40=e indeed

[turbid surge]

So I need to find k such that $\tau \sigma^k\tau^{-1} = e^{-1}=e$

[boreal girder]

prograce

[turbid surge]

And since o($\tau \sigma \tau^{-1}$)=4 then k=4 ?

[boreal girder]

prograce

[upbeat gorge]

is this the smallest possible k?

[turbid surge]

Yes?

[upbeat gorge]

you seem a little unsure - can you back it up?

[turbid surge]

Since the order of the permutation is 4, then any smaller power would not give me the identity by definition

[upbeat gorge]

exactly, so you can be confident in your answer :D

[turbid surge]

Problem is my answer is wrong haha

Idk where it went wrong tho

[upbeat gorge]

ah.

[turbid surge]

The solution is supposed to be k=8

I'm quite lost on how to get that

[upbeat gorge]

well, i double checked by hand and also by theorem, and $\mathrm{ord}(1 ; 2 ; 5 ; 4)$ definitely seems to be 4

[boreal girder]

haseeb

[upbeat gorge]

is there more to the solution with k=8?

oh wait it's $\tau \sigma^k \tau^{-1}$, not $(\tau\sigma\tau^{-1})^k$

🤦‍♂️

[boreal girder]

haseeb

[upbeat gorge]

so we want $\tau\sigma^k\tau^{-1} = e$

[turbid surge]

Hahahah

[boreal girder]

haseeb

[turbid surge]

Oh right

[upbeat gorge]

can you simplify something here? perhaps make something reduce to e?

wait that also gives k=4

[turbid surge]

I think our answer is right, I just realized that 4 is not in the options,,

Sorry yes 4 is correct

8 is also correct and since it is in options

Wow that was a huge time waste haha

[upbeat gorge]

ohh of course oopsie

[turbid surge]

Any multiple of 4 is correct... so 8 is the answer here

Sorry I didn't realize 4 wasn't in the options

.solved

[compact pewter]

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[upbeat gorge]

. its on me too, didnt say smallest value

[eternal pulsar]

s

[eternal pulsar]

Ok

[boreal girder]

;(

[eternal pulsar]

Now I wanted to go by induction

[boreal girder]

;(

[eternal pulsar]

Hence true

Now we assume it true for n=k, and want to prove it for n=k+1, k>=2

The other stuff can be ignored since they will not be polynomial of degree k

[boreal girder]

;(

;(

[eternal pulsar]

So since it is true for n=2 and n=k+1, k>=2, it must be true for n>=2, hence proven

Is this good?

[compact pewter]

@eternal pulsar Has your question been resolved?

[azure vault]

yeah, just argue a bit more why "stuff" is a polynomial with degree <= k-1

and for constant coeff what did you do?

[eternal pulsar]

Oh oops

Well that's just f(0)=det(A-0lambda)=det(A)

[eternal pulsar]

Hmm, alright

Shouldn't be too hard

[boreal girder]

;(

[eternal pulsar]

Since there's n-2 entries with a lambda term, we have that the degree of the polynomial formed will be at maximum degree n-2=k-1

Since the same thing will happen for the other minors, the degree of stuff <= k-1

@azure vault is this good? I know its rough but I'm too lazy to elaborat emuch

<@&286206848099549185>

[eternal pulsar]

<@&286206848099549185> afraid

[azure vault]

yes this should be good

[eternal pulsar]

Typa timing lmao

[azure vault]

basically the minors from expanding

they are all k*k matrices

but they all get a row or column destroyed

where there was a lambda term

so only k-1 terms with lambda

[eternal pulsar]

awight

ezpz, cya

.close

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[flint wing]

i need help starting 2a iii)

[spark folio]

what did you get for i and ii?

like yes/no

[flint wing]

it is not injective and it is surjective

[spark folio]

good

so, you can find the trivial preimages for this right?

[flint wing]

like 2,4,8,16,...?

[spark folio]

hmm, not really

preimages are gonna be tuples

you might also wanna recall the rules of exponents

like for different values of k, the preimages gonna differ

[flint wing]

uh so

wat do i do

[spark folio]

lets start with some specific values of k

lets say you got k=2

what can you tell about its preimage(s)

[flint wing]

its (2,2 and 4,1)?

[spark folio]

yes

and uhh, format it better? like (2,2) and (4,1)

[flint wing]

{(2,2), (4,1)}

[spark folio]

what about if k=6?

no need to calculate the stuff exactly

I want what insight can you get from the values of k

[flint wing]

so we want the pairs (m,n) such that m^n = 2^6

[spark folio]

yes

so, how does that k affect the possible values of m and n?

[flint wing]

well like depending on k the amount of pairs u get are different i think

[spark folio]

ofc

[spark folio]

this is a small-ish hint

[flint wing]

are we trying to look like more generally ?

not like for specific k now

so i guess

like

we are solving m^n = 64

since output must be power of 2 then m has to be power of 2

[spark folio]

yep

and what about n?

[flint wing]

well if we let like m = 2^a so

m^n = (2^a)^n = 2^(an)

[spark folio]

well, yea, the final aim is to have a general case for the k's

[spark folio]

yea, thats a good step

[flint wing]

so 2^an = 2^6 if k=6

so therefore an = 6 and we have to find a and n that make an=6

we can have

a = 1, n = 6
a = 2, n = 3
a = 3, n = 2
a = 6, n = 1

[spark folio]

good, so this time you got 4 different tuples of possible preimages

and ofc, by now you must have realized how that depends on k

[flint wing]

like are u talking about if k is prime or something

[spark folio]

well, yea

thats a part of it

specifically, I am talking about the factorization of k

[spark folio]

this

each factor of k is gonna yield a different pre-image

[flint wing]

do we solve for n?

n = k/a

[spark folio]

how are gonna solve for n when you only know k and not a

[flint wing]

well like we need an = k right

so n = k/a

[spark folio]

imo going for prime factorization of k would make more sense

[flint wing]

uh do we need it if we're just looking at preimage?

[spark folio]

well, you did factorize k=6 as 2*3 and thats how you got the preimages no?

you put n=2 for one case, and n=3 for the other

and ofc n=1 and n=6 too

what are 1,2,3,6 if not the factors of k

[flint wing]

yes

[spark folio]

yea, so same way, you are gonna need the factors of k for any of its values so that you can write $2^k = (2^{a})^b$ and thats gonna get you $m = 2^a$ and $n=b$ yielding the preimages as $(2^a,b)$ where $k=a \times b$

[boreal girder]

Bacter14Fr0g

[flint wing]

ohh so (2^a, b) is our preimage

should we say b = k/a or it makes more sense to leave it as ab = k

[spark folio]

either way is fine

both statements mean the same thing anyway

[flint wing]

i see and if i needed to write a sentence about like the factors of k what should i say

[spark folio]

you can just write that the preimages depend on the factors of k, and that each of the distinct factor gives rise to a distinct preimage

and that trivially, since k = 1*k, you will always have at least two different pre-images for all k>1

[flint wing]

ohhhhh okay i think i get it now

thank you

[spark folio]

np

[flint wing]

.solved

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[compact pewter]

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[remote garden]

Can someone please help me with this question

[remote garden]

I just have no idea how to even start

[kind viper]

https://www.youtube.com/watch?v=7eyYNNUojEc

[ionic python]

!nosols

[compact pewter]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[ionic python]

and this looks like ai...

!nogpt

[compact pewter]

Please do not trust ChatGPT or similar AI tools for mathematical tasks, as they often generate output which "sounds correct" but has numerous factual or logical errors. Use of these AI tools to answer other people's help questions is strictly against server rules (see #rules).

[ionic python]

pretty sure no one will type latex without dollar signs

and not correct it

[kind viper]

<@&268886789983436800> AI answer

[ionic python]

and no one will add so many empty lines

[kind viper]

OP seems to be away

[ionic python]

probably watching tutorial

yep deleted

[remote garden]

Sorry, this isn't homework, this is just for revision for my upcoming test.

[remote garden]

yeah 😅

[ionic python]

but that's still not very good though

[remote garden]

ya

good thing i didnt see it lol

.close

[compact pewter]

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[compact pewter]

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[quartz yoke]

,rccw

[quartz yoke]

7. ii. and iii.

I dont know what to do

[potent berry]

ah yes 10th grade ncert

@compact olive do you wanna join here

mujhe maggi banani thi

[cursive harbor]

This is just basic conditional probability stuff right

[potent berry]

yes

12th grade ncert actually

[compact olive]

I wouldn't be the best at probability and PnC

Trust

[urban kindle]

oh these are fun! i have a deck of 52 questions

[cursive harbor]

Draw one

[compact olive]

In your own channel, diva

[potent berry]

we drew this

where is op

@quartz yoke

[quartz yoke]

Ok so how do i solve

[quartz yoke]

Thanks for this

[potent berry]

!status

[compact pewter]

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

[quartz yoke]

1

[urban kindle]

what is the probability that the lost card is a king without taking out any cards

[quartz yoke]

1/13

[urban kindle]

explain that first

[quartz yoke]

Oh 1 min

There are 52 cards and 1 card is lost, the probability of lost card is a king

So its 1/13

[urban kindle]

how many kings are there in the 52 cards

[quartz yoke]

Am i right?

4

[urban kindle]

given that you take out one king how many kings are there in the remaining deck

and how many cards are left

[quartz yoke]

3 king and 51 cards

[urban kindle]

within that, what is the probability of a random card being king

[quartz yoke]

1/17

Thats it?

Oh

[urban kindle]

do you know how to do part iii now

[quartz yoke]

Les see

1/425?

😭😭

[urban kindle]

wait

how did you arrive there

[quartz yoke]

3/51 x 2/50

Removed 1 card just like b4

[urban kindle]

wait why multiply

[quartz yoke]

2 cards to be drawn without replacing

Cuz 2 cards are drawn

[urban kindle]

they're drawn but you don't care abt the probability of that

you're just told that they're both aces

[quartz yoke]

Yes 1 ace is lost

We have 3 aces

Out of which 2 drawn are aces

[urban kindle]

how many aces do you have left after drawing 2, in the remainder of the entire deck including the lost card

[quartz yoke]

1

Since another is lost

2*

[urban kindle]

how many cards

[quartz yoke]

Including drawn + lost?

[urban kindle]

mhm

not drawn

everything not drawn

[quartz yoke]

51 excluding drawn

52 if no loss

First of all, i have a doubt in ii

How does probability of lost card being king relate to the kings after losing a card

[urban kindle]

wdym

[quartz yoke]

Like b4 losing it was 1/13

[urban kindle]

conditional probability - you're told that you remove a king from the cards

if the probability of losing any card was uniform before, it's still uniform now, except you take out a king

[quartz yoke]

Elaborate further please !

[west fern]

thats calss 8th ques guys

[quartz yoke]

Noo im so dumb

Could u please explain me even further 😭

[urban kindle]

consider the probability of losing any specific card, 1/52 at first

[quartz yoke]

Yess

[urban kindle]

after removing a king, that's 1/51 in the remaining cards

[quartz yoke]

1 or 3?

[urban kindle]

any specific card

[quartz yoke]

Ok

[urban kindle]

(because you did not lose the king that you removed)

there are 3 kings left

3*1/51 = 3/51 = 1/17

[quartz yoke]

Oh

[urban kindle]

the same with the 2 aces

you take out 2 aces—how many cards do you have left?

[quartz yoke]

50

[urban kindle]

how many aces remain

[quartz yoke]

1

[urban kindle]

there are 4 at first

[quartz yoke]

Then 2

[urban kindle]

2/50=1/25

do you get this

[quartz yoke]

Uhh

To some extent

[urban kindle]

don't consider losing a card until you have to calculate the probability

[quartz yoke]

Probability of which

The 2 drawn ace?

[urban kindle]

you're asked to calculate probability of lost card being an ace

[quartz yoke]

Ye

[urban kindle]

don't care about the drawn ace probability

[quartz yoke]

Then how do i proceed

[urban kindle]

you're not being asked to calculate the probability of drawing two aces

you take out two aces from the deck and want to find probability of the lost card also being an ace

[safe kelp]

In the bag of 5 red and 6 black ball whats chances of drowing the 2 red

[quartz yoke]

😭

[cursive harbor]

Are you trying to help him or asking your own question, I can't tell

[urban kindle]

2/11 but that's irrelevant

[quartz yoke]

5/11x 4/10 without replacing

[quartz yoke]

What 😭

I forgot the total

[urban kindle]

you want a probability on the lost card, not what you take out from the deck

[quartz yoke]

So could explain like how did u for ii

And tell what to take into consideration 1st and what not to

[urban kindle]

start here

[safe kelp]

No perhapsp(r) =5c2/11c3

[urban kindle]

ask in another channel

[cursive harbor]

Arpit please get your own help channel

#❓how-to-get-help

[quartz yoke]

Without removing we have 1/13 chance

After drawing 2 aces the cards are 50

Then the probability of ace being lost from 50 cards?

Or probability of ace being lost after 1 ace is removed then 2 cards are drawn

Which are also ace

[urban kindle]

you draw (remove) two aces

[quartz yoke]

Then caluclate lost ace?

[urban kindle]

calculate P(lost card being an ace | deck of 2 aces in 50 cards)

[quartz yoke]

1/25 simply

[urban kindle]

yea!