help-27

[lusty sapphire]

comes with time

[sturdy yew]

but B has to be real

[lusty sapphire]

I may not even be right

[sturdy yew]

which yr UG r u in?

[sturdy yew]

even is right

[lusty sapphire]

This is why I thought n must be even. You can have imaginary numbers properties with even-sized matrices, like $A^2=-I$ is possible for $n=2$

[woven radish]

SWR

[sturdy yew]

honestly i can't see how this is related to n being even

[gritty terrace]

ok i have an idea

can i say it?

[sturdy yew]

yess

but pls tell me its a high school level idea not a pg level one

[gritty terrace]

suppose $\lambda$ is an eigenvalue of $B$. then $\lambda^2 + \lambda$ is an eigenvalue of $B^2 + B$, which means $\lambda^2 + \lambda$ is an eigenvalue of $B$

[woven radish]

Pseudo (Cat theory #1 Fan)

[sturdy yew]

yes yes

[gritty terrace]

in other words, $\lambda$ being an eigenvalue means $\lambda^2 + \lambda$ is also an eigenvalue

[woven radish]

Pseudo (Cat theory #1 Fan)

[sturdy yew]

,.

[gritty terrace]

this is potentially bad becuase you're only allowed finitely many eigenvalues

[lusty sapphire]

I graduated forever ago

[sturdy yew]

oh damn

[gritty terrace]

like if 1 is an eigenvalue so is 2, and then so is 4 + 2 = 6, and then so is 36 + 6 = 42, ...

[sturdy yew]

ok sure

so we need cyclicity

basically

[gritty terrace]

yeah

[lusty sapphire]

I'm guessing you can only do this with complex eigenvalues

[sturdy yew]

thats true but its also kinda handwavy?

[gritty terrace]

yeah i think you can use this to show that $B$ cannot have any real eigenvalues

[woven radish]

Pseudo (Cat theory #1 Fan)

[gritty terrace]

i think this is correct

[lusty sapphire]

I suppose you can find cyclic eigenvalues by solving $a+b=\lambda$ and $a-b=\lambda^2+\lambda$?

[woven radish]

SWR

[sturdy yew]

where do those 2 eqns come from

[lusty sapphire]

No, that doesn't work

I have it backwards

hmmm....

[sturdy yew]

wait eigenvalues of B have to purely imaginary right?

[stone stump]

the obvious set of equations is x^2+x=y, y^2+y=z, z^2+z=t, t^2+t=w, ... until a^2+a=x. for some number of variables

[lusty sapphire]

I'm assuming $\lambda$ needs to be a complex number like $a+bi$, but I'm keeping it super generic and trying just $\lambda=a+b$ where $a$ and $b$ could be comples numbers

[woven radish]

SWR

[lusty sapphire]

this is better

for $n=2$, you can do just $\lambda_1^2+\lambda_1=\lambda_2$ and $\lambda_2^2+\lambda_2=\lambda_1$

[woven radish]

SWR

[sturdy yew]

oh nvm they wudve if there was no +x term rifht?

[lusty sapphire]

I think the right idea here is considering the eigenvalues need to be cyclic. Maybe the next question is to ask yourself is what happens to the cyclicity if n is odd

[gritty terrace]

well the thing is that if $n$ is odd

[woven radish]

Pseudo (Cat theory #1 Fan)

[gritty terrace]

||B has to have at least one real eigenvalue||

[stone stump]

you should try to show that for n=2 you cannot have real solutions. that approach immediately generalizes

[sturdy yew]

y wud that be a problem tho?

[gritty terrace]

because i think you can show ||B cannot have any real eigenvalues||

[spiral moss]

isn't 0 a possible eigenvalue?

which works in this example

[stone stump]

B is invertible

*nonzero real eigenvalue

[spiral moss]

why though?

[stone stump]

given in the problem

[spiral moss]

oh i missed it

[sturdy yew]

yea but how do u take it from here

[stone stump]

x^2+x=y, y^2+y=x. show that this has no nonzero real solution

[sturdy yew]

yea this is fine

[stone stump]

preferably not by bruteforce computing and instead by something that might generalize

[sturdy yew]

oh

[spiral moss]

||i think that using trace helps with showing no real numbers except 0 can be eigenvalues||

[stone stump]

oh yeah thats also nice

[sturdy yew]

using trace we get that tr(B^2) = 0

[spiral moss]

trB is sum of its eigenvalues, right?

[sturdy yew]

what

how 😭

[stone stump]

diagonalize

[sturdy yew]

yea when u diagonalize, the trace of the diagonal matrix is equal to the sum of the eigenvalues of the diagonal matrix right

and since say D = PBP^-1

D and B are similar

but wait

that still doesnt say anytg abt trB and the eigenvalues of B

[vast zenith]

the trace is invariant! super nice property

[sturdy yew]

so are the eigenvalues right?

[vast zenith]

hmm

what do you mean?

[sturdy yew]

eigenvalues of D and B are equal

i mean

[vast zenith]

precisely

[worthy raft]

Trace has the nice property that Tr(AB) = Tr(BA)

[sturdy yew]

ok so now tr(B) is equal to the sum of its eigenvalues

but this feels like its true for any matrix

also how does it help in showing thhat B cannot have real eigenvalues?

[vast zenith]

it is true for any matrix that is diagonalizable

[worthy raft]

So when doing PBP^{-1} you get that Tr(PBP^-1) = Tr(BP^-1 P) = Tr(B)

[vast zenith]

or actually for any matrix I think

[stone stump]

well jordan form is enough

[sturdy yew]

and how do we know if its diagonalizable?

[stone stump]

so as long as you are working over a field or integral domain you are good

[vast zenith]

yes true

[vast zenith]

there are some neat matrices that are diagonalizable

symmetric, orthogonal

but these are special cases

[sturdy yew]

so in this q how do we know B is diagonalizable?

[vast zenith]

in general you have to find a full basis of eigenvectors

[sturdy yew]

it feels like we've digressed very far

[sturdy yew]

n= 1 is not possible because the eigenvalue has to be zero

riight

[stone stump]

yes

[vast zenith]

this is an interesting problem

[sturdy yew]

for n = 2 the eigenvalues are imaginary

and thats completely fine

for n=3 if i show that there has to be atleast one real eigenvalue then this caase is not possible

as then the entire set wud diverge?

right?

i mean

if we have a real eigenvalue then the number of eigenvalues wud be infinite

as if x is an eigenvalue then x^2 + x shud also be an eigenvalue

[vast zenith]

they must both have the same set of Eigenvalues,

[sturdy yew]

yess

[devout snow]

@sturdy yew Has your question been resolved?

[sturdy yew]

<@&286206848099549185>

[tribal leaf]

Hi

What's the question

instead of typing csn you send like an image

[frosty crescent]

check the pin

[tribal leaf]

yeah

[sturdy yew]

we've moved away quite a bit from the original q

so lemme explain

[tribal leaf]

I mean

I read your explanation

and checked the pin

[sturdy yew]

ok'

[sturdy yew]

so the set of eigenvalues for B^2 + B and B are equal

[tribal leaf]

okay

soo

P^{-1}(B^2 + B)P = B

this is the question

[sturdy yew]

ok

[tribal leaf]

and you need to prove n can be written as any positive integer

Correct?

[sturdy yew]

i need to show that n has to be even

it cannot be odd

B is real and invertible

[tribal leaf]

Ohh okay

[tribal leaf]

Should've said it earlier

[sturdy yew]

mb

[tribal leaf]

without this I was getting n is both even and odd

so are you familiar with eigen values?

[sturdy yew]

yes

kinda

[tribal leaf]

B is real and invertible
eigenvalues of B are real or complex conjugate pairs

so

We can say

Lamda is an eigen value of b

[sturdy yew]

why complex conjugate pairs

?

[tribal leaf]

B is a real matrix

so

[sturdy yew]

ohhh because B is real and its characteristic eqn will have only real coeff

yesyes

mb

oof

wtf

[tribal leaf]

Okayy

[sturdy yew]

i think i got it

thanks

[tribal leaf]

okayy

lmao good job anyways

[sturdy yew]

thanks @stone stump n @tribal leaf

lmao

finally

😭

.close

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[stark mist]

Hello, I have a question regarding my college algebra assignment. I tried to use the remainder zero formula or what it was and got these. I asked my professor about it and he said that I need to use synthetic division to find the remainders of zero, but after doing them in my head I can’t find any values that equal zero, so I just want to check on if I’m right on this

[wise trout]

you are correct that there are no rational roots. why are you doing this type of algebra in college tho?

[stark mist]

I had to retake it cause I did it back in high school but not for dual credit so I gotta do the class again

[wise trout]

ok

[verbal wharf]

you are correct yes

[stark mist]

But thank you for clarifying!

[verbal wharf]

synthetic division is good to test but if they don't produce 0 remainder -> no rational roots

[stark mist]

Ah I see, thank you!

[winter patrol]

note that rational root theorem by itself only identifies candidates.
IF there are rational zeroes, it will be within the list generated

[stark mist]

So do I just close this now

[winter patrol]

you can close whenever you want

[topaz axle]

you don't have to, it'll close eventually

[stark mist]

Ah I see

Thank you for the help!

[winter patrol]

ideally you'd close when youre certain you're done

[verbal wharf]

welcome and gl with the rest !!

[winter patrol]

And don't require more assistance with the question

[stark mist]

Alright, thank you!

.close

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[gray prawn]

Helo

If i assume that every infinite sequence doesn't have any infinite subsequence which is monotonic

What are the consequences of it

(I am trying to prove that every infinite sequence has an infinite monotonic subsequence so I am trying to prove it through contradiction

I somehow constructed that i can find 3 terms following defination of monotone given that sequence is non monotonic
( Because if sequence is monotonic it is trivial))

[bronze void]

<@&268886789983436800>

[uncut crow]

this setup does not look right

not (every object has property X) means there exists an object that doesn’t have property X

[gray prawn]

As in

I want to prove that every infinite sequence has a infinite monotonic subsequence

For the sake of contradiction
I assumed that for any arbitrary infinite seq there is NOT a single infinite subsequence which is monotonic

[uncut crow]

the negation of that would be
there exists a sequence that has no monotonic subsequence

[gray prawn]

🤔

Oh damn

[gray prawn]

Forming a negation should be a common sense or there are rules laid for it?

[uncut crow]

hm both i guess

you can lay out rules and make it procedural

[gray prawn]

Yeah so, what are the consequences if i assume this so I can go further

[uncut crow]

or just practice and the ‘rules’ will be natural and make sense

[gray prawn]

......

@uncut crow hey

[uncut crow]

instead of doing this you could try to construct a monotone subsequence of an arbitrary sequence

[gray prawn]

Hmm 🤔

[devout snow]

@gray prawn Has your question been resolved?

[devout snow]

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[noble mango]

Are there any errors to my answer to part b?

[noble mango]

I'm sorry if the handwriting is a little hard to understand, i was writing fast.

[trail eagle]

Seems like a fitting counter-example

[noble mango]

Thanks for confirming, I have one more question relating to this topic but its a different question should I continue using this help channel or occupy another one?

[trail eagle]

Keep this one

[noble mango]

Alright, why did they use (x+5)^2 here for?

I really struggle with these types of prove questions, because I don't even know where to begin when writing my answer.

[trail eagle]

The way they wrote the proof I think makes it clear what they want to show, but I don't think it's a very natural thing you'd think about right away when seeing the problem.
I think the easiest way to go about it would be to expand (x+6)^2 as x^2 + 12x + 36 and try to separate 2x + 11 from it, so that you get x^2 + 10x + 25 + 2x + 11.
Then you'd spot that the first three terms are just (x+5)^2 and go from there.

Another way would be to show the equivalent statement that (x+6)^2 - 2x - 11 >= 0 for all x, which would probably even be more straightforward by expanding.

Again I think the solution you're given is kind of working backwards with the solution in mind already

[noble mango]

Okay you're right if i expand out (x+6)^2 and bring all the terms to one side I get (x+5)^2, but how do i go from there?

this is gonna sound weird but i'm not entirely sure what the question is asking

if (x+5)^2>0 hasn't that already proved that x is greater than 0?

[trail eagle]

Well (x+5)^2 >= 0 is true for any x. Anything squared is >= 0.

[noble mango]

yeah thats true

[trail eagle]

In general, one method of showing that a >= b is to show that a = b + (something)^2. This is what they're using in the proof

[noble mango]

a is (x+6)^2 and b is 2x+11 right?

[trail eagle]

Yes

[noble mango]

I see, do you have time for one more question? I don't wanna be a bother.

[trail eagle]

Go on

[noble mango]

where tf did a^2 come from in the answer sheet?

They didn't ask to square anything or imply to square anything and the original equation didn't have any square or square roots in it

[trail eagle]

They multiplied both sides of the inequality by a

[noble mango]

ohhh

to get rid of the fraction?

[trail eagle]

Yeah

[noble mango]

and we know (a-1)^2>=0 is true because anything squared is always positive?

[trail eagle]

Yes

[noble mango]

Geninually thank you so much dude @trail eagle

You're a life saver

.close

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[bitter dock]

This might not be a math question, but I need help searching for an article. The article I search for is "Jagers (1975): Branching Processes with Biological Applications". I somehow only found reviews to that article but not the article itself.
Additional: Do you have any general advice on finding articles like that?

[vital edge]

Are you sure about the author

The only article I can find with this name is Wu (2010)

[bitter dock]

I'm not, I just found it as reference in my professor's notes for possible bachelor thesis and wanted to check it

[vital edge]

The article you're looking for is simply titled Branching Processes

Or at least that's what it looks like

<@&268886789983436800> ??

[bitter dock]

Thanks, do you have link to it?

[vital edge]

https://onlinelibrary.wiley.com/doi/abs/10.1002/0470011815.b2a07005

I know this says 2005 but

This is the link I got from an article linking to the 1975 version

Not sure what's up with that

Maybe this is an updated version of the old article

[bitter dock]

Thank you very much

unfortunately my institiution does not grant access to it

[vital edge]

The emote I'm about to send has nothing to do with this channel

🏴‍☠️

I just really like this emote

[bitter dock]

.close

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[summer harbor]

i genuinely have no idea how to evaluate this

[mystic scarab]

Change of variable

(aka substitution)

I mean, it's not necessary but I believe it helps you see what to do next

[summer harbor]

example of this is basically u-sub in integrals right

u = particular equation in the integral

[last parrot]

Is this just finding the limits

[summer harbor]

yes

[last parrot]

Can you use algebraic manipulation, no?

[mystic scarab]

But you should have seen it also with basic polynomials (or other functions), such as solving x⁴ + x² - 2 = 0, where you usually let u = x²

[summer harbor]

y = x^2, y^2 =x^4

[summer harbor]

so in this case, x = t + 2

i assume,

[mystic scarab]

Yep

So now you have $\lim_{x\to ?}\ \frac{x}{\vert x \vert}$

[woven radish]

Alberto Z.

[summer harbor]

x approaches -2

[mystic scarab]

Careul 😉

[summer harbor]

right

[mystic scarab]

t → -2, but x = t + 2

So x approaches ...

[summer harbor]

0

[mystic scarab]

Awesome

Therefore, the limit has become $\lim_{x\to 0}\ \frac{x}{\vert x \vert}$

[woven radish]

Alberto Z.

[mystic scarab]

Can you see what we could do next?

In particular, we have an abs value which means we will have to study two cases, do you agree?

[summer harbor]

omg slr, i had to eat a bit

yes

cause abs val a negative and positive results in positive

[thick lotus]

yes

[summer harbor]

but how do we do that

[thick lotus]

do what?

[summer harbor]

how do we study this

[thick lotus]

oh

[summer harbor]

i get why but i dont know how if that makes sense

[mystic scarab]

Dw, I went eating too lol

Well, the two cases will be right- and left side limits

[thick lotus]

evaluate the left and right limit as x-> 0^- and x->0^+

[summer harbor]

but wouldnt that be 0/0 ?

[thick lotus]

-# off to you alberto

[mystic scarab]

Yes, but you can easily figure out the result, if you recall the definition of the absolute value

,tex .abs def

[woven radish]

Alberto Z.

[summer harbor]

so lim x -> 0- = -x, lim x -> 0+ = x

[mystic scarab]

Somehow yeah

$$\lim_{x\to 0^+}\ \frac{x}{\vert x \vert} = \lim_{x\to 0^+}\ \frac{x}{x}$$

[woven radish]

Alberto Z.

[mystic scarab]

$$\lim_{x\to 0^-}\ \frac{x}{\vert x \vert} = \lim_{x\to 0^-}\ \frac{x}{-x}$$

[woven radish]

Alberto Z.

[mystic scarab]

This is how you should write the two cases

And easily conclude, I guess

[summer harbor]

can we say then that the limit technically DNE because lim x -> 0- != lim x -> 0+ ?

[summer harbor]

right?

[mystic scarab]

Exactly! That's the correct conclusion, very well 💪

Because the limit is 1 from the right but -1 from the left

[summer harbor]

right

thanks !!

.close

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[silk panther]

hello, I need help in doing this Q

[silk panther]

so far, I already proved that the sequence is increasing, andd i just need to prove that it is bounded above

[willow helm]

Hello!

[silk panther]

then I can say the sequence converges

[silk panther]

hello good sir

im stuck in proving its boundedness from above

any idea?

[last parrot]

Use lim?

[willow helm]

Have you found possible limits?

[silk panther]

what

but the expression is like f(x+1) = (f(x) + 2) / 3

how to find limit

[last parrot]

If x_n converge L then x_n+1 also converge L

Let L=(L^3+2)/3

Then find from here

[silk panther]

but I wanna prove its boundedness first

from above

then find limit

[last parrot]

No

You find possible Limit first

Then you will find the boundness

[silk panther]

but shouldn't we prove that it converges first

[last parrot]

Thats what im saying

Assume it converge toward a limit

[silk panther]

are we doing induction

[last parrot]

Yes, but not now

[silk panther]

so ur saying

we find the limit L first, then find out its convergence...?

[last parrot]

Ye

[silk panther]

erm

if we already got the limit L, it must converge right

[last parrot]

No

Just find lim first lol

[silk panther]

im doing it

(L+2)(L-1)^2 = 0

so L = (-2,1)

oh wow there are 2 limits

[willow helm]

We don't find the limit

[silk panther]

what does this mean

[willow helm]

We find the possible values for limits

[silk panther]

yea its -2 and 1

[willow helm]

This simply tells you that if the sequence converges, the limit is either -2 or 1

We haven't figured out that the limit is -2 or 1, and most definitely not that it's both

[silk panther]

okay

[willow helm]

To prove boundedness

[silk panther]

so L must be 1

[willow helm]

No

We don't know that

[willow helm]

Simply compute the first few terms

You notice, as proven, they are increasing

But it seems like they are always above 0

And this is where we can use the possible limit 1

[silk panther]

sorry?

[last parrot]

Above

They are above 0 for xn

[willow helm]

Yes

My bad

[last parrot]

Baf

https://tenor.com/view/shikimori-shikimoris-not-just-cute-shikimoris-not-just-a-cutie-anime-anime-anime-girl-gif-26002808

Anyways

[silk panther]

the thing is, I have proven that the sequence is increasing, and given the possible limits are 1 and -2, and x_1 = 0.4, limit must be 1

[last parrot]

So it wont reach 1

You need to show 0<xn<1

[silk panther]

hmm I see

[last parrot]

Since for the first terms of x you see it go positive

[willow helm]

No

[last parrot]

Induction use here

[willow helm]

All that can be true, but still limit wouldn't necessarily have to be 1 (or -2)

[silk panther]

ah so base case would be finding x_1 and see if x_1 < 1 is true right?

[silk panther]

ok hold on now im confused

[last parrot]

Base case is 0.4

[silk panther]

yessir

[last parrot]

You find 2 L values, one is -2 other is 1, by intuitively find some x values, you see the sequence tend to go up positive, therefore, try to show 0<xn<1

Then by inductive, use the given formula

[willow helm]

You have proven it is increasing. You have found two possible limits, one of which must be the limit if the sequence converges. You haven't proven boundedness yet, hence you haven't proven convergence, hence the existence of a limit is still not proven

All you have to do is prove boundedness from above, you have proven boundedness from below by proving it is increasing and that x_1 = 0.4, hence all are > 0

The reason why we find possible limits is to prove this boundedness

[silk panther]

I SEEEE

[willow helm]

We find two values, -2 and 1

[silk panther]

I understand it now

[willow helm]

And what they mean is if the sequence reaches them, it stays there

So from that we know if the sequence ever reaches 1, it stays there

[silk panther]

wait lemme share my answer

[willow helm]

We don't pick -2 because see know all terms are > 0

So we work with 1

To show it is a good upper bound

[willow helm]

Once we do that, we have shown convergence

[silk panther]

so its like an educated guess

[last parrot]

Why dont you let possible limit be solved first?

[silk panther]

uhhhhhhh I wrote the first induction earlier before asking here

I haven't swapped it

does it matter tho?

[willow helm]

Kind of, but it's really more systematic

In practice you expect it to work

[silk panther]

I see I understand

yea need more practice

need to get used to this

sooo any suggestion for my answer? i will be diving to the last question

ill ask for help later haha

[last parrot]

Looks fine to me i think

[willow helm]

Yeah

[willow helm]

Sure thing

[last parrot]

My first year doing calculus too

[willow helm]

Best of luck! 🙂

[silk panther]

ur first year?

[last parrot]

Yes first year uni

[silk panther]

woahh dope

alr thanks yall, ill see u in a new channel 😏

.close

[devout snow]

Channel closed

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[willow helm]

[last parrot]

The smirk emoji👀

[devout snow]

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[silent hawk]

which method do i use to find the range of this thing

[last parrot]

Have you found domain yet?

[silent hawk]

i guess it will be R - {0}

[slow junco]

Use am gm inequality

[silent hawk]

how

[slow junco]

x^2+1/x^2>=2

Now add 4 both sides

It's [6,infinity)

[shy osprey]

!nosols

[devout snow]

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.

[silent hawk]

na dw its 6 to infinity cuz i looked in the soln

but just wondering how to reach there

[slow junco]

Am gm

[shy osprey]

jaisa ki @ daredevils_ldoe ne bola

[silent hawk]

i didnt understand how you used it

[slow junco]

Hn bhai

[shy osprey]

since x^2 cancel ho rha hai

wo apply karlo

[slow junco]

Bruh ( x^2 +1/x^2 )/2 >= √(x^2/x^2) which is 1

Multiply both sides by 2

[silent hawk]

oh

[slow junco]

Then add 4

[shy osprey]

AM>= GM for any two no.s is
(a +b )/2 >= sqrt(ab)

[silent hawk]

OHH

got it

[shy osprey]

Yahan par if you observe, tumko dikhega ki x^2 and 1/x^2 wala term multiply hokr cancel hota hai

But suno!

[silent hawk]

ye

[shy osprey]

AM GM inqequality is only for +ve numbers

We were able to apply this cuz x^2 and 1/x^2 (both are square isiliye postive)

[silent hawk]

ngl i never thought of using am-gm ineq

im 100% failing maths

[shy osprey]

always remember am gm and cauchy shwartz

Almost every qsn mein try karke dekha kar

Weighted am gm bhi works sometimes!!

[silent hawk]

whats weighted am gm

[shy osprey]

Abhi ineqality ka chapter nahi hua?

[silent hawk]

prob not taught

ill learn it myself

[shy osprey]

Oh okay then

aur kuch?

[silent hawk]

nothing for now

thanks

.close

[devout snow]

Channel closed

Closed by @silent hawk

Use .reopen if this was a mistake.

[silent hawk]

actually 1 q

.reopen

[devout snow]

✅ Original question: #help-27 message

[shy osprey]

sure!

[silent hawk]

so this was the original q

and range of h(x) is [6, infinity)

then why is the smallest positive integer taken as 7

and not 6

[shy osprey]

closed bracket nahi hai

[slow junco]

Bhai

[shy osprey]

open bracket hai 6 ke liye

[slow junco]

Am gm mei closed aata bracket

Kya hogaya

[silent hawk]

am gm mai closed hota hai

na

[slow junco]

Kuch glt kra hai

[shy osprey]

one sec i think catch hai

[slow junco]

Lagta hai

Aapne

[silent hawk]

na

100% sure sab correct hai

[slow junco]

Huh?

[vestal surge]

inequality ki condition ke according tabhi 2 hota hai jab numbers equal hoge

[slow junco]

How is that possible

[vestal surge]

idhar equal nahi hai

toh vo exactly 2 nahi hoga

[slow junco]

Are haan

[vestal surge]

2 se jyada hoga

[slow junco]

Kya baat hai

He got the point

[vestal surge]

isiliye open bracket hoga

[silent hawk]

ohh

shit

true

[silent hawk]

but vo ham equal consider nahi kar sakte kya?

[vestal surge]

no because x is in the base of log

[shy osprey]

wo nahi tha point!

[vestal surge]

in g(x)

[silent hawk]

crap

[slow junco]

Agar numbers equal hote jaise x^2 = 1/x^2 hota tbhi consider krte isko

[silent hawk]

no like i asked that why cant x^2 and 1/x^2 be considered equal here

i missed the minor detail

that x is in the base

of log

[slow junco]

The value of base of log cannot be 1 or -1

Broo

[silent hawk]

ye i missed that detail

[slow junco]

log(base x) (e)

Hai isme

To na hi 1 aayega na hi -1

[vestal surge]

he prob figured that by now

[silent hawk]

ye

tysm

yall

[silent hawk]

you also a jee aspirant?

[vestal surge]

jee 27

[silent hawk]

thx also btw if i keep this open for a bit is it okay

[vestal surge]

idk sure ig?

[slow junco]

Jee 26

Dera abhi

[silent hawk]

dang bro

gl

[slow junco]

Thx

[silent hawk]

april attempt ya adv?

[slow junco]

April

Jan mei percentile acchi ni aayi

[shy osprey]

#chill

[silent hawk]

dang gl

umm....

actually 1s

how do i find the range of this thing

[slow junco]

√(x+7) > 0

Now add 4 on both sides

And then change them to denominator

The inequality changes

[silent hawk]

got to this part

forgot to update it

mb

[slow junco]

It becomes 1/√(x+7) +4 < 1/4

Multiply by pi

Yea

Correct

[silent hawk]

now how do i find range of sin

tho

i basically found f(x) and 2f(x)

[slow junco]

The value of index of sin must be from (-π/2,π/2)

As it is it's principal branch

So -π/2< 2fx < π/2

[silent hawk]

whts principal branch

we've prob been taught by a diff term

[slow junco]

Huh?

[silent hawk]

cuz ive never heard/seen it anywhere before

[slow junco]

Maybe

Ho skta hai

Kuch alag batay ho

Range aagai ab -pi/4,pi/4

Sry closed

[vestal surge]

define 2fx and then just find sin values

thats your answer

[silent hawk]

ok wait brb

[slow junco]

Sin ki

To y belongs to -1/√2,1/√2

[thick lotus]

least value of f(x) is 0 as x-> infinity

so f(x) is never negative

[slow junco]

Yea

Missed that pt

[vestal surge]

i take that back, im wrong

[thick lotus]

me too

[vestal surge]

cool

[silent hawk]

um guys

i got this

but slight issue

the 0 should be in open bracket

why tho

umm what?

[thick lotus]

<@&268886789983436800> hope you good

[thick lotus]

when does the expression become 0

[silent hawk]

oh so the expression only tends to 0?

as x gets larger

?

but never becomes 0

[thick lotus]

yes

[silent hawk]

tysm

ill take a break now

.close

[devout snow]

Channel closed

Closed by @silent hawk

Use .reopen if this was a mistake.

[thick lotus]

np

[devout snow]

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[obtuse shell]

@dry robin

[cyan rose]

<@&286206848099549185>

[uncut crow]

[cyan rose]

You pinged a user instead of the role.

[obtuse shell]

💀

[shy osprey]

@obtuse shell you there?

[cyan rose]

Yes, he is with us.

[obtuse shell]

Yes

[shy osprey]

Okay do you still require help in this one?

[obtuse shell]

Yes

Please

[cyan rose]

What is the scale here? Is one face of the square equal to one unit? Then AB would be two units?

[shy osprey]

Right so first things first, could you send a cropped img of the qsn so as to protect your privacy!!

[shy osprey]

Yeah ig so

[obtuse shell]

I got this

I might be Wo]wrong tho i kinda suck

[cyan rose]

You need to crop it, and only show the question part. Remove everything else including the browser tabs and else.

[shy osprey]

-# for your privacy!

[cyan rose]

Alright, this is good enough.

[shy osprey]

Yeah see here, when you calc area of the left rectangle,

you cant just do 8*4

[obtuse shell]

Okay

[shy osprey]

Since the left part doesnt have that small chunk in the corner included

[obtuse shell]

Okay

[shy osprey]

so you would have to subtract the area of that small chunk BCD from the 8*4 area

[obtuse shell]

So 6 x 4 maybe?

is my triangle calculations fine?

[shy osprey]

Yeup!

[shy osprey]

Uhm no

okay lets do this step by step

whats the area of the chunk that you need to subtract?

[obtuse shell]

The little square?

[shy osprey]

-# i hope u understood which chunk am talking abt

[shy osprey]

Yeah!

with three vertices BCD right?

[obtuse shell]

Yes

[shy osprey]

Right well whats the area?

!nosols

[devout snow]

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.

[cyan rose]

Alright, apologies!

[shy osprey]

But no

[shy osprey]

no worried u joined tdy i see

[cyan rose]

Am I allowed to give them the approach by which I got the answer or do we need to do it in a step by step teaching way?

[shy osprey]

step by step so that the op understands it

[shy osprey]

well?

[cyan rose]

Alright.

@obtuse shell are you with us?

[devout snow]

@obtuse shell Has your question been resolved?

[devout snow]

Channel closed

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[devout snow]

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[late cave]

x^2 + y^2 = r^2 the equation of a circle centered at the origine.

For a circle of a radius 1 , determine the components of a vectr that origine is at the coordinate point of (0,0) and the extremity is at the trigonometric point P(pi/3)

[late cave]

P(pi/3) = (1/2 , sqrt3/2)

(1/2 - 0 , sqrt3/2 - 0) ?

Am i supposed to subtract (1/2,sqrt3/2) by (-1/2, -sqrt3/2) or (0,0)

[winter torrent]

P(pi/3) = (1/2 , sqrt3/2)

This is your answer. Points are vectors.

[late cave]

Oh ok thanks

.close

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[devout snow]

Channel closed

Channel closed due to the original message being deleted.
If you did not intend to do this, please open a new help channel,
as this action is irreversible, and this channel may abruptly lock.

[last parrot]

<@&268886789983436800>

[devout snow]

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[lost laurel]

I'm having trouble in (b)

[lost laurel]

$f_n(x)$ = $\begin{cases} \left(\frac{3}{2} \right)^n x & 0≤x≤1/3\ \left( \frac{3}{2} \right)^{n-1} x& 1/3≤x≤2/3 \ \text{ something} \end{cases}$

[woven radish]

Wai

[lost laurel]

The issue here is as n-> infty, doesn't the function on [0,1/3] diverge

[lunar harbor]

isn't $(3/2)^n x$ only valid for $[0, 1/3^n]$

or am I misinterpreting what your question is

[lost laurel]

hmm, I mean f_1 and f_2 have the same domains

like uh it's all defined on [0,1/3]; [1/3,2/3]; [2/3,1]

[lost laurel]

that's what I want to know too 😭

[woven radish]

Civil Service Pigeon

[lunar harbor]

edited

[lost laurel]

hmm, so because we want the middle third of each non-constant segment

I see

well, that would work for the middle segment, but what about the first bit

[lunar harbor]

The recursive step for $f_{n+1}$ references $[0,1/3]$, $[1/3,2/3]$, $[2/3,1]$, but the actual formula for $f_n$ is more granular than on just these three intervals.

[woven radish]

Civil Service Pigeon

[lost laurel]

Noted

[lunar harbor]

tbf it's quite late but I'm very confused on what your question is

is "first bit" referring to [0, 1/3]

[lost laurel]

yes

[lunar harbor]

$f_{n+1}(x)=\frac{1}{2}f_n (3x)$ is basically inheriting the entire "step" structure of $f_n$ then

[woven radish]

Civil Service Pigeon

[lunar harbor]

just scaled down to half the height and one third of the width

This is the graph of $f_1$, which has one "flat middle"

[woven radish]

Civil Service Pigeon

[lunar harbor]

when you construct $f_2$ from that, the original part stays flat (aka the middle $1/3$), but the recursive rule creates \textit{new} flat parts in the middle of the first and last thirds

[woven radish]

Civil Service Pigeon

[lost laurel]

so I flatten out the middle of the (3/2)x?

[lunar harbor]

yeah sure

[lost laurel]

mhm

I suppose I'll work on this

thanks, this is harder than I expected lol

[wicked turtle]

ooh cantor's function, one of my favorites

[lunar harbor]

yeah step back and just draw some of the smaller cases first

I think you're trying to juggle too much rn

[lost laurel]

Hi Bungo

[wicked turtle]

hey Wai!

[lost laurel]

I'll close this for now while I work out the details

.close

[devout snow]

Channel closed

Closed by @lost laurel

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[devout snow]

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[solid osprey]

Find the sum of all positive integers $n$ such that $1^3+2^3+\dots+n^3\equiv 17 \mod n+5$

[woven radish]

ihave<skissue>

[solid osprey]

ok so
n^2(n+1)^2/4=17 mod n+5
(n-(n+5))^2×(n-(n+4))^2=68 mod n+5
400=68 mod n+5
n+5|332
332=2^2×83, so >5 solutions are n+5=83, 166, 332 or n=78,161,327

n=78 and 161 works, but why does n=327 suddenly fail?

[lunar harbor]

n+5|332 is necessary but not sufficient

Consider what you need mod 4(n+5)

[solid osprey]

huh wdym

[lunar harbor]

In a rough sense, dividing by 4 in a congruence is only valid when gcd(4,modulus)=1

when n=327, 332 is divisible by 4

[solid osprey]

hmm

but why does n=161 still work

is it like it just so happens to work

[lunar harbor]

yeah

This one is infamous for trolling people because you had to plug back in to check

[solid osprey]

eugh thats terrible

actually can the check just straight up be 400/4=17 mod 332, and noticing that it just doesent work?

huh then why in the world does 161 work

[lean crater]

unrelated but i agree with your bio rankings

[solid osprey]

yeahhhhh

ok i forgot inotation doesent work in text and not all people have the same humor as my friend group

[lean crater]

unfortunately C will turn out to be the most useful in applied maths

[solid osprey]

ew

[lean crater]

wait it does work

[solid osprey]

thats what im saying

what the fuck is wrong with modular aritmethic

[lean crater]

for n = 161 the modulus is 166 right

4S = 400 + 166k
k is an even number represented by 2m
2S = 200 + 83(2m) => S = 100 + 83m

if m = 0, S is congruent to 100 mod 166
if m = -1, S is 17 mod 166

voila 161 works

@solid osprey

[solid osprey]

ok i think i see

thank you pigeon

thank you nefer

.solved

[devout snow]

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[lunar harbor]

[devout snow]

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[lost laurel]

Very confused here

So the ideals are sets sets of cosets here, right

so I'll first want to find something this is isomorphic to , in order to make my life easier

.close

[devout snow]

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[faint gorge]

Wai wait for other when Wai can help themselves

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[noble mango]

Why did he substiute -2 into the equation when there was already a x value inside it?

[noble mango]

There are two y values in this equation and one x value

thats kinda confusing me

If any of the helpers need context here's the video https://www.youtube.com/watch?v=7zkCj4vgBos&list=PLquyU_6YLv6di-RCY0DD_xbrkcD_TJuMO&index=5 the time stamp is around 5:00

oh nvm

he made a mistake

i kept watching and he realised it then

.close

[devout snow]

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[last parrot]

lmao

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[summer harbor]

how do i derive absorption law?

it just puts me back in a loop

[inland carbon]

v and ^ mean gcd and lcm? in which case p | p v q so p ^ (p vq )= p

[summer harbor]

no...

thats logic...

[inland carbon]

damn mb

[summer harbor]

v is disjunction and ^ is conjunction

[young spade]

btw when youre on paper, write v and "^" as:

$p \vee q \wedge r$

[young spade]

They go in line

[summer harbor]

right

[inland carbon]

maybe distinguish the cases p / not p if that's allowed

[summer harbor]

wdym

[inland carbon]

like a truth table

[young spade]

Yeah, proof by truth table is pretty straight forwards

given that you only have 2 variables

[summer harbor]

yeah

is it possible tho

to derive absorption law such that it is reduced to p?

[young spade]

I dont think you can do it with formal logic without appealing to itself

More like, with propositional logic or set theory

https://en.wikipedia.org/wiki/Absorption_(logic)

[summer harbor]

i see

so its inevitably

gonna be circular if i do it with js logic

[polar chasm]

it depends on what your axioms are

what laws can you work with?

[summer harbor]

just logical equivalence, so right off the bat

im using

distributive

and idempotent

[polar chasm]

hmm then you probably cant do much

[summer harbor]

yeah

[devout snow]

@summer harbor Has your question been resolved?

[devout snow]

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[mellow mulch]

Need help,

[mellow mulch]

this questions seems to be solved with help of products of determinant but how?

[dull parrot]

second determinant is just the determinant of the cofactor matrix

[mellow mulch]

ohhh... i thought i have to make a1 a2 and so on.

thanks bud.

[devout snow]

@mellow mulch Has your question been resolved?

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[crystal pumice]

how do i answer b

[willow helm]

!status

[devout snow]

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

[slow junco]

Have they mentioned minimum height or something

[crystal pumice]

i kindof got it now tbh

u get the variance of the original mean, 185. so u can get the Standard deviation of 190

nd its easy from there

idk how to close this thread

[supple knot]

!done

[devout snow]

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

[crystal pumice]

.close

[devout snow]

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[devout snow]

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[night rune]

i am having a hard time translating this

Elentis has produced 31 times the world average for the number of professional tennis players per 100,000 residents.

[thick lotus]

yh i also do not understand what is statement 1 and 2

[devout snow]

@night rune Has your question been resolved?

[small geyser]

I suppose the question asks you to pick statement 1 such that it implies a chosen statement 2.

at least, that's what a cold reading tells me.

[devout snow]

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[topaz axle]

it's 3→5

[devout snow]

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[fair canopy]

Can someone help me solve this?

[small geyser]

,rccw

[fair canopy]

No idea what to do

[shy osprey]

Any specific part or all?

[fair canopy]

All

Well I suppose all r same how abt we do 4th part?

[shy osprey]

Right, so the first step would be two identfiy the 'r'

[fair canopy]

Ye

[shy osprey]

r would indicate the max value of the expression right?

[fair canopy]

I know how to get r ye but I have no idea what this thing is

[shy osprey]

since sin() terms max value can only be 1

[fair canopy]

Ye

[shy osprey]

wait thats fine then how do you calc r?

[fair canopy]

I let 8 equal to r cos fi
And - 6 as r sin fi

Then square both sides and add both eq

Then take r Sq common

[shy osprey]

Okay yeah!
So what did you get r as?

[fair canopy]

10

[shy osprey]

Good!

[fair canopy]

Do we let - 6 equal r sin fi or 6= r sin fi

?

[shy osprey]

You can take anything

doesnt mattwr

[fair canopy]

Alr

[shy osprey]

if you take -6 case, you will just get phi as -

[fair canopy]

So wait lemme show u what I tried

Lemme send pic

[shy osprey]

Sure!

[fair canopy]

So if I let - 6 won't that change my answer in terms of sin (theta plus fi)

?

[shy osprey]

yeah it would

but

phi would be negative

so it would become theta + (-phi)

[fair canopy]

-phi? How

[shy osprey]

You can try!

[fair canopy]

Look

[shy osprey]

this is correct btw
now for the calc of phi?

[shy osprey]

Yeah am not denying this

[fair canopy]

Tan fie equals b over a

[shy osprey]

Am saying when you calc phi!

[fair canopy]

Ye

When calc tan fie equal b over a will I take b as - 6 or 6

[shy osprey]

<@&268886789983436800>

wow

tf is going on tdy

[fair canopy]

💀

This is not the solution bro

😂

[shy osprey]

whats b and a?

[fair canopy]

a is 8

B is minus 6 or 6 idk what to take

[shy osprey]

Oh right yeah

So see

[fair canopy]

Ye

[shy osprey]

if you take -6cos(theta)

then you would get a tan-1( -3/4)

which is same as -tan-1(3/4)

so thats how you get a negative phi

[shy osprey]

if you take positive as you did here

[fair canopy]

I solved this without using negative as

[shy osprey]

you just get a positive tan-1(3/4)

[fair canopy]

Oh ya

[shy osprey]

Yup correct

[fair canopy]

So calculation is fie is important here?

As the question jst asks for the sin theta plus minus fie form

[shy osprey]

and theta + (-tan-1(3/4)) is same as what you got here

[shy osprey]

as in?

[fair canopy]

While solving this q

Calculation of fie is needed?

[shy osprey]

phi is important yes

[fair canopy]

As the question does not ask for it

[shy osprey]

i mean you do need it for the answer no?

[fair canopy]

Ye

If I want to put it in that form

[shy osprey]

Oh u mean like exact value of phi?

[fair canopy]

Ye

[shy osprey]

Idts

but as far as i remember

3/4 gives 37 degrees

and 4/3 gave 53degrees

one of the somewhat standard values

[fair canopy]

Well thanks for clearing the doubt tho

[shy osprey]

Anytime!!

Well now would you like to try the others on your own first?

Or should we start with the others asw?

[devout snow]

@fair canopy Has your question been resolved?

[devout snow]

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[white pawn]

I dont understand conceptually vector lines and planes

[dull parrot]

what part are you confused about

[white pawn]

Like when do I use a point as a vector to descirbe it

its better in a differentq uestion

[dull parrot]

not sure what you mean

[white pawn]

why do they use A as the r vector

wtj

[dull parrot]

<@&268886789983436800>

[white pawn]

Anyway

in that question above

[dull parrot]

i think you mean direction vectors

which is different

[white pawn]

yeah why do they use A as the ditection vector

[dull parrot]

for this here?

[white pawn]

yep why cant they use b or c

to describe the plane

[dull parrot]

well it specfically notes it as $\vec{AB}$ and $\vec{AC}$

[woven radish]

MxRgD

[dull parrot]

so what we're saying is how do we get from point a to point b

[white pawn]

but above it has 3 points

[dull parrot]

and how do we get to point a to point c

[white pawn]

so cant tehy use bc and ba?

[dull parrot]

you can if you want to

but you'll get a different answer

[white pawn]

oh so its incorrect to describe the plane that way

[dull parrot]

ah no i meant, you get a different normal that's scaled differently

but it will be right

[white pawn]

ohhh I okay so yiu can use any of the points to find a normal and describe the plane

and each of the normals if used other points would have a scalar factor difference betwene them

not scalar factor difference like they are scalar apart

[dull parrot]

yes, specfically if you have three points $A(x_1, y_1, z_1)$, $B(x_2, y_2, z_2)$, $C(x_3, y_3, z_3)$. You can derive two direction vectors.

[woven radish]

MxRgD