help-49

[desert siren]

um i meant like

[lavish venture]

not piecewise?

[desert siren]

different definitions of the function at different domains

[nova pike]

Wait, sorry could you clarify more by this?

[lavish venture]

|x| has a turning point at x = 0 and yet it’s derivative is not zero at x = 0 (since it isn’t even defined at x = 0)

[desert siren]

yeah piecewise lol just looked it up, wasnt taught that word

derivative is mod(x)/x, not defined at x=0

[nova pike]

Ohhhh ok. Ahhhhhh, I was meant to say turning point of a differentiable function. I think that makes a difference

[lavish venture]

yea that’s different

[nova pike]

I mean it can at most be n-1, that’s true. But say for example we have a fourth degree polynomial. It cannot have two turning points im pretty sure, neither can it have 0 turning points, but it can have 1 or 3 turning points

[desert siren]

we have to take the maximum possible case though, so picking a specific example won't lead us to the result

[nova pike]

Sorry, could you clarify?

I’m sorry if I’m acting stupid, I’m still in pre-uni

[desert siren]

we have to take the maximum possible turning points

wait i misread your statement i thought you were saying that a fourth degree polynomial cannot have n-1 as its maximum possible turning points

i apologise

[lavish venture]

so it depends on if n is even or odd then

[nova pike]

Yes, that’s what I said before (in original message)

[lavish venture]

yea i misunderstood your question

lemme reread your solution

[nova pike]

I’m sorry if I worded my question badly. It seems that there are a lot of misunderstandings happening

[desert siren]

no it was my mistake not yours lol

about the proof for p(x) and ax^n

i dont know if there are other methods but i was taught this way:
take the limit of p(x)/ ax^n

it will come out to be 1

so for very high degree polynomials p(x) is approximately ax^n

[nova pike]

Ohh ok, makes sense

[lavish venture]

so you just need to show that the number of zeros of odd multiplicity of p’ are even/odd based on the parity of n

because that’s where the turning points occur

[nova pike]

Oh ok

Okay I think I’m good now

Thanks everyone!

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[desert siren]

you got your answer?

[nova pike]

I think so

[desert siren]

nice

[midnight plank]

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[neat canyon]

Find value of r so that this function has limit in x = -2

[neat canyon]

I dont understand if I just need to have the lateral limits the same

maybe not

[desert siren]

there's no need for finding the limit of the function for which x<-2

just the first two will suffice

put x= -2 in the second function to get your' reqd limit's answer

[neat canyon]

huh

in x-1?

[desert siren]

yeah since if you're taking limits the function is continuous

so it doesn't matter which one you solve the 2nd or 3rd

is that understandable?

[neat canyon]

yeah

ok so -3

is my limit

x = -3

[desert siren]

yep

now take the limit of the first function as x approaches -2

(that limit is -3)

so you'll find r

[neat canyon]

and just solve for r

[desert siren]

yep

[neat canyon]

r = 1

[desert siren]

you want me to check?

[neat canyon]

no need

thank you

[desert siren]

alr cool

yw

[neat canyon]

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[mint ravine]

Can someone please help me with this question? Thanks

[mint ravine]

This is AOPS btw NOT SCHOOL

so grading doesn't matter, thanks

[desert siren]

i just saw that message of yours lmao, why did you instantly close it after explaining

[mint ravine]

idk

i just need help with the problem

[desert siren]

well the hint looks like it could be useful have you tried that yet

[mint ravine]

yeah i did

[desert siren]

what'd you get

[mint ravine]

can i send a photo?

[desert siren]

yeah

[mint ravine]

sorry my wifi is laggy rn it will take a min

im tryna send it to my gmail

[desert siren]

well i'm not too familiar with what the conventional approach would be for these types of questions but if it were me i would do something like

first move from A to all adjacentR's

then move from R to adjacent C's

then from C to adjacent H's

[mint ravine]

yeah that's what i did too

finally it loaded into my gmail

[mint ravine]

okay here's the photo

please ask me if you have any questions (ik the handwriting is not the best)

[desert siren]

you did the other letters as well?

[mint ravine]

yeah i thought that's what you were supposed to do?

[desert siren]

no im asking if you did the same thing with A and C

[mint ravine]

yeah i did A and C

bro im so confused rn

[desert siren]

well your working seems right

for R

it's just a lot of working, nothing else to it just do the same thing for all of them

[mint ravine]

could you please clarify what "all of them" is

all the other letters i already did tho

[desert siren]

just A, R and C

not H since it's already at endpoints

[desert siren]

well then just add all the cases you got ARCH in

[mint ravine]

but there's only 1 option for A so why would that matter

[desert siren]

how is there only one?

you can move to both right and down

oh you mean as in there is only one A

[mint ravine]

yeah

[desert siren]

well yeah but you still need to make the cases for it

like all the different ways it can go to a R

[mint ravine]

oh ok

I got $\left(3\cdot2\cdot2\right)+\left(3\cdot2\cdot3\right)+\left(3\cdot5\cdot2\right)+\left(3\cdot5\cdot3\right)+\left(3\cdot5\cdot5\right)+\left(3\cdot5\cdot3\right)+\left(3\cdot5\cdot2\right)+\left(3\cdot2\cdot2\right)+\left(3\cdot2\cdot3\right)$

[grand pond]

RamenNoodleMan21

[mint ravine]

which is equal to 285

bruh it said it's wrong

[desert siren]

i didnt calculate but that number just seems too big at first glance itself

[mint ravine]

well i tried to include every possible choice

[desert siren]

i guess i'll give it a try

there's so many cases damn

[mint ravine]

yeah

[desert siren]

im getting 21

holdup sending my work

Idk if its correct theres so many cases

only the C to H displacements matter for our final result btw

[mint ravine]

no 21 is incorrect too

[desert siren]

lmfao

cooked

[mint ravine]

it's too small

[desert siren]

idk then man you should ask someone else perhaps

[mint ravine]

yeah

thanks for helping nevertheless

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[ornate bone]

it's not that far off

[midnight plank]

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[plucky prairie]

ok this is correct right?

i just wanted to clarify that when it says put in a+bi format this still counts even though its negative

[visual tiger]

looks correct

[visual tiger]

b is just gonna be negative, no big worries

[plucky prairie]

ok

i thought maybe it was an issue

thanks

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[static salmon]

can someone check my work under the line?

[static salmon]

this is the elipse equation im solving for

i didnt end up using the first equaition under the line sorry

[viral dagger]

im not sure where the /13 and /81.25 comes from, but without those its correct

[static salmon]

mhm

so wait

then you have to reduce it so they both get put over 325

and 25/325=1/13

and 4/325 = 1/81.25

[livid python]

Shouldn’t there be -225 in RHS ? before dividing

[static salmon]

what do you mean?

oh wait youre totally right

[livid python]

Most of these conic sections problems I saw have natural numbers so felt something is messed up

[static salmon]

yeah i got that

thank you!!!!!!!!

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[delicate flower]

my knowledge is very limited on this topic and i dont know alot of information.. how do i do these and what formulas do i use if i even use any

[night hawk]

Do you know about the unit circle?

[delicate flower]

Yes, thats what im able to use

I have a sheet of formulas too

[night hawk]

Do you know how to express tan in terms of sin and cos?

[delicate flower]

tan theta = sin theta/cos theta (?)

a quotient identity

[last slate]

yes

[delicate flower]

how would that help with these tho?

[brittle grotto]

understanding what cos and sin represent graphically can help you better understand how to figure these problems out. I wouldn't recommend drawing the unit circle and trying to re-derive them, but you should understand it to the point where you can go about doing something like that

this identity is useful because you're given a theta for problem 8

now, the problem reduces to determining the value of $\frac{\sin\theta}{\cos\theta}$

[grand pond]

00100000

[pearl hull]

[delicate flower]

i mean i have a unit circle drawn on a seperate sheet of paper with all the information written on it, which helps alot

[delicate flower]

i get that, but how would i find sin theta or cos theta?

[brittle grotto]

note that $-\frac{5\pi}{6}=2\pi-\frac{5\pi}{6}=\frac{7\pi}{6}$. This is because on a unit circle, the - represents the angle below $y=0$, which can be expressed by going ccw until your positive angle reaches that measure

[grand pond]

00100000

[brittle grotto]

excellent

wait why is texit she/her now lmao

anyways

[delicate flower]

you can take your time lol

[desert siren]

isnt this part all we need

[brittle grotto]

yeah, you could just say "plug that into your calculator" I suppose lol

[desert siren]

no

[brittle grotto]

but I'm gonna need you to use some of your imagination now @delicate flower

[desert siren]

tan 7pi/6 is tan( pi + pi/6)

[dreamy blade]

For better understanding I would recommend you to read on "Rotations & Reflections of Angles".

[brittle grotto]

wait a sec, is tan linear like that!?

well, not linear but

[delicate flower]

👍

[brittle grotto]

well anyways

[polar star]

periodic

[brittle grotto]

oh, it's not additivity. I see

[desert siren]

yeah

tan positive in 3rd quadrant

so its just tan pi/6

[polar star]

if tan was linear, all of math would change

[brittle grotto]

yes I didn't think it would be linear lol

[desert siren]

so just 1/root 3

[brittle grotto]

this works perfectly well I guess. I was gonna go for the intuition sort of route

[desert siren]

yeh sorry i forgot to put the brackets lol

[brittle grotto]

of thinking of moving the unit radius that angle

and then dropping the line down to form a triangle

and then noticing that it's a 30-60-90 triangle

[desert siren]

visualising is better with the unit circle

[brittle grotto]

and then getting the identities from that

[desert siren]

but for quick math ASTC works best

[brittle grotto]

and then using the tan identity

[delicate flower]

the quotient identity

[desert siren]

All- sin- tan- cos btw incase someones not familiar with the fullform

[brittle grotto]

yep

let me know if you're interested in me continuing, or if @desert siren's answer suffices for what you need it for

[desert siren]

you should learn the unit circle thing tori

will help you in the long term

[delicate flower]

i got it infront if me

i understand the

Negative numbers

and how they work

[delicate flower]

im gonna reread what was said again and try to solve it now, but thank you both alot

[brittle grotto]

of course. glad to be of help. good luck with the problem

[delicate flower]

i got my triangle, how do i know how to find sin and cos tho

[brittle grotto]

are you aware of the 30-60-90 triangle ratios?

if not, here's how you can derive it. First, construct and equilateral triangle of side length 2, and then draw an altitude of it

[delicate flower]

wait wait

s o/h and c a/h ?

[brittle grotto]

the two halves should be congruent by AAS (shared side, right angle, 60 degree angle)

then, the altitude bisects the side it connects to since the two segments it is bisected into are congruent (since they are corresponding parts of congruent triangles)

(i edited my message to use side length 2 for convenience)

then, the bisected segments are of length 1

[brittle grotto]

now, since it is a right triangle, apply pythagorean theorem

yep exactly

so now, your hypoteneuse needs to be 1, right?

[delicate flower]

oh correct

[brittle grotto]

you should be able to proceed from there on your own I think

[delicate flower]

because its a radius

rught

[brittle grotto]

yes, it is a radius of the unit circle centered at the origin.

[delicate flower]

so switch 1 and 2?

[delicate flower]

the pythagorean theorem

[brittle grotto]

definitely don't switch them

I'm suggesting that you divide both by 2, so that you can preserve the fact that your hypoteneuse should always be of length 1

[delicate flower]

ohh

[brittle grotto]

you can disregard this, because it was me telling you how you could reason that the 30-60-90 identity is true

[delicate flower]

well what im stuck on is how to find sin theta with the triangle

[brittle grotto]

well, this goes back to what sin is supposed to represent graphically

it will represent the signed size of the vertical part of the triangle you constructed

(with hypoteneuse of length 1)

it's signed because it needs to be negative if you're in a quadrant where y is negative

[delicate flower]

quad 3 is where only tan is positive so ya itd be negative

[brittle grotto]

yes, since sin and cos are both negative in that quadrant, right?

[delicate flower]

correct

[midnight plank]

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[molten bay]

-1,-w,-w^2 are singularities but 0 is also singularities if yes then why? Because of e^(1/z)??

[nova yoke]

z=0 is an essential singularity of e^(1/z), yes

what is w here?

[lyric charm]

presumably w = e^{2πi/3}

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@molten bay Has your question been resolved?

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[quartz hornet]

you are using l-hopital right?

why didnt you differentiate numerator but not denominator

[molten bay]

Ohh i made a mistake

Thanks

@quartz hornet

Can I do it by expansion?

,rotate

[desert siren]

Yes

[molten bay]

Where am I doing mistake?

[desert siren]

I dont understand how you went from the 2nd last step to the last one could you elaborate

[molten bay]

i expanded cosx and (1+a/x)^(1/2)

[desert siren]

Hm i forgot what the expansion of (1+a/x) sqrt would be, but from the second last step you can just cut out x^2

[molten bay]

Opps

I got a=4

Let me send it

[desert siren]

So you would be left with 2/sqrt (a+x)

Which is 2/sqrt a

So a is 4

[molten bay]

Yes

Thanjs

.clsoe

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[desert siren]

Yw

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[twilit field]

can you find f^{-1}

[molten bay]

Is this wrong?

i am stuck on the finding roots step

Should I square it or not?

Or the condition x>=-1 makes any effect?

[lyric charm]

yes, unjustified involvement of complex

you are working with real only

[molten bay]

So how will I solve?

(x+1)^2-√(x+1)=0

[lyric charm]

easy, (x+1)^3=1 gives x+1=1 only.

just don't put complex.

easy as that.

[molten bay]

Where is it mentioned?

The condition x>=-1

[lyric charm]

yes that condition

implies x is real

[molten bay]

Thanks

.closs

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[graceful ferry]

How do I find two sequence functions that converge uniformly on a segment I, and their product does not converge uniformly on I ?

[midnight plank]

@graceful ferry Has your question been resolved?

[wheat patrol]

@graceful ferry show me what youve tried

[graceful ferry]

I had 0 intuition on what the sequence function would look like so I just took a function that isn't uniformly convergent which f_n(x)=x^n on [0,1] and tried to get two sequence functions from it but every two I thought of were also not uniformly convergent on the segment

[wheat patrol]

id tell you think of a seqeunce function like any simple sequence x_n (for simplicity) where x_n which is uniformly convergent on I, now do you know the definition of uniform convergence?

[graceful ferry]

yes I do

[wheat patrol]

use that definition to find and proove that those two sequences are uniformly convergent in that interval I

[graceful ferry]

What two sequences

I need to find them in order to prove it

I can't find any that are uniformly convergent and idk how

(Where their product is not a uniformly convergent sequence funciton)

[wheat patrol]

can you come up with anyy example of a congergent sequence?

[graceful ferry]

f_n(x)=x^n on segment (0,1)

[wheat patrol]

well x^n converges but does not converge UNIFORMLY in the segment you can check using the definition

also sorry for late reply im at my work now

[graceful ferry]

IIt does uniformly on the Open segment (0,1)

But that doesn't matter this isn't my question

[graceful ferry]

I can show u proof

[wheat patrol]

show me i might as well be wrong

[graceful ferry]

Mb you're right

$f_n(x) = \frac{1}{x^2+n}$

[grand pond]

prograce

[graceful ferry]

This converges uniformly on all R

[wheat patrol]

okay i cannot check rn myself but can you check if you can confine it to a segment like [0,1]

also check another sequence, it could itself be the second sequence, where the multiplaction just squares it

[graceful ferry]

Ok

[wheat patrol]

these are the props you can use

to check for uniform convergence

[graceful ferry]

The sequence squared is still uniformly convergent

[wheat patrol]

on tip of my tounge one uniformly convergengt function i cant think of is 1/n

[steel crest]

whats a segment, a bounded interval?

i think Rudin uses that term to mean somethinh like that

[graceful ferry]

Just any domain idk I used google translate for the word

[steel crest]

ah okay

[graceful ferry]

Their product is still uniformly convergent

[steel crest]

!original

[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.

[wheat patrol]

yeah 1/n^2 is convergent

[steel crest]

קטע does mean segment but lmc if it means interval in math

[graceful ferry]

I feel like there's no such functions

[steel crest]

i think it does

are there any convergent sequences not uniformly convergent that you know of?

[graceful ferry]

Yes x^n on [0,1]

[steel crest]

so factor that into two uniformly convergent sequences

[graceful ferry]

I tried

It's not working

[steel crest]

maybe the answer will be more obvious if you consider x^{n+1} instead?

ah doesnt quite work

but i think this is the approach

lets try an unbounded interval

more room for pathological stuff

something that converges but not unifornly on all of [0,infty) let's say

[graceful ferry]

$f_n(x) = \frac{nx}{1+n^{2}x^{2}}$

[grand pond]

prograce

[graceful ferry]

this doesn't converge uniformly on all R

Maybe I can factor out two uniformly convergent functions from it

[steel crest]

oh looks promising

couldnt get it to work

[steel crest]

ah got one

found an example

think simpler

[graceful ferry]

Wait I found two functions

f(x)=x $g(x)= \frac{n}{1+n^2x^2}$ but interval has to be $[1, \infty)$ to prove g(x) converges uniformly which then works i think

[grand pond]

prograce

[graceful ferry]

Like x>0

Because then g_n(x)-->0 when n--> inf

[steel crest]

יופי

(f_n(x))=(x,x,x,x,...) to make it a sequence of course

[graceful ferry]

Oh yes

[steel crest]

https://tenor.com/view/the-goon-win-you-won-willy-wonka-oompa-loompa-fc-gif-14046847

[graceful ferry]

Thx

.solved

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[molten bay]

Integration (sinx+cosx)

[tawdry laurel]

is that what we need to find

[molten bay]

I want to find area

[tawdry laurel]

ok

Do you know how to integrate (find antiderivatives)?

[rancid vigil]

and what are you struggling with

[molten bay]

This is not where i am stucked

As you can see the limits

What will be the limits?

And how do we integrate

[tawdry laurel]

oh

[molten bay]

What oh

Just help quickly

😅

if I draw a line parallel to y axis

[tawdry laurel]

wait you want a synthetic way?

or want to find the antiderivative

[molten bay]

Bro sinx,cosx derivative is not a big problem

or integration

[tawdry laurel]

So what do you need help with

[molten bay]

Each integration needs limits

[tawdry laurel]

You want an indefinite integral??

[molten bay]

Brooo

.close

[midnight plank]

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[tawdry laurel]

the limits are up to you

... uh ok ig

[molten bay]

.reopen

[midnight plank]

✅

[molten bay]

@rancid vigil

You wanna say anything or help

How will you find the area of given graph and axis?

,w 0 to π integration sinx+cosx

[grand pond]

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[molten bay]

.close

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[dusty portal]

I am inspired

[dusty portal]

To do part i), do we just use integration by parts?

[quartz hornet]

looks like parts to me

try following your gut

[long dagger]

it is by parts

IBPeak

[dusty portal]

$\begin{matrix}\quad&\mathrm{D}&\mathrm{I}\+&\log(1-t)&\frac{\log(1-t)}{t}\-&\frac1{t-1}&\mathrm{Li}_2(t)\end{matrix}$

[grand pond]

;(

[dusty portal]

This is what my gut is telling me

Or did I screw it up

[quartz hornet]

overguting

[iron cradle]

any latest llm can solve it, dont waste time on it

[dusty portal]

I'm here to learn!

[quartz hornet]

check what you need to proof , and tally the first part

[dusty portal]

So log^2

[quartz hornet]

its there as it is

[dusty portal]

My brain is not braining

[quartz hornet]

and new term log x

[dusty portal]

Oh should I just

Oh

oh

[quartz hornet]

YEAHH

[dusty portal]

lemme cook

[quartz hornet]

make a brain baby

almost there

[iron cradle]

add ELI5 to your prompt.

[long dagger]

!nogpt

[midnight plank]

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).

[tawdry laurel]

bro was enlightened by flux

[dusty portal]

$\begin{matrix}\quad&\mathrm{D}&\mathrm{I}\+&\log^2(1-t)&\frac1t\-&\frac{2\log(1-t)}{t-1}&\log(t)\end{matrix}$

[dusty portal]

I'm not going to use AI lol

[quartz hornet]

lemme ELY5 , this is a grownup table 😍, if you wanna learn you can join

[dusty portal]

if it isn't clear enough already

[dusty portal]

Indeed

[grand pond]

;(

[dusty portal]

Ok so I have to make ln=log for this

ugh

OH FUCK THIS INTEGRAL?!

NOOOOOOOOO

Bruh I don't like this integral but whatever

[long dagger]

do u wanna do it or not

😭 make your mind cuh

[dusty portal]

$\log(t)\log^2(1-t)\bigg\vert_0^x+\int_0^x\frac{2\log(1-t)\log(t)}{1-t}dt$

[grand pond]

;(

[dusty portal]

I'm doing it

Trust

[quartz hornet]

bro is flexing his latex

geez

[dusty portal]

$I=\log(x)\log^2(1-x)+\int_0^x\frac{2\log(1-t)\log(t)}{1-t}dt$

[grand pond]

;(

[quartz hornet]

(doing great)

[dusty portal]

How else am I supposed to do it without paper

💀

[quartz hornet]

valid point

[dusty portal]

Ok now brain has frozen

Is this IBP again?

[long dagger]

Hint 2.4

[dusty portal]

💀

w hint

Now what to choose

differentiate log(1-t)log(t)?

$\begin{matrix}\quad&\mathrm{D}&\mathrm{I}\+&\log(1-t)\log(t)&\frac1{1-t}\-&\frac{\log(1-t)}{t}+\frac{\log(t)}{t-1}&-\log(1-t)\end{matrix}$

[grand pond]

;(

[iron cradle]

why is the group named help-49 like it reminds you to get 50 on that math test and not help-69

[dusty portal]

!redir

[midnight plank]

This channel is only for on-topic discussion. Please take casual conversation to #discussion or #chill.

[iron cradle]

ok

[dusty portal]

Ok so that looks wrong

[dusty portal]

I mean I guess it works...

Maybe it's meant to be

Oh I screwed up

$I=\log(x)\log^2(1-x)+2\int_0^x\frac{\log(1-t)\log(t)}{1-t}dt$

[grand pond]

;(

;(

[dusty portal]

Ok so I screwed up somewhere?!?!?!?!

[long dagger]

try and create a dilog

[dusty portal]

In the IBP?

[long dagger]

yes

[dusty portal]

uh

[long dagger]

specifically the du

[dusty portal]

$\int\frac{\log(1-x)}{x}dx=\mathrm{Li}_2(x)+C$?

[grand pond]

;(

[long dagger]

works

[dusty portal]

Oh

fuck I thought it didn't work

,w integrate log(t)/(1-t)

Too lazy

[grand pond]

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Requested by exiled_hype

[dusty portal]

.

Are we deadass

Aight so I was just being blind...

$\begin{matrix}\quad&\mathrm{D}&\mathrm{I}\+&\log(1-t)&\frac{\log(t)}{1-t}\-&\frac1{t-1}&\mathrm{Li}_2(1-t)\end{matrix}$

[grand pond]

;(

[dusty portal]

By the way do you like this integration by parts things that I am doing

with the LaTeX environment

ok I'm cooking

$I=\log^2(1-x)\log(x)+2\qty(\mathrm{Li}_2(1-t)\log(1-t)\bigg\vert_0^x+\int_0^x\frac{\mathrm{Li}_2(1-t)}{1-t}dt)$

.

[grand pond]

;(

[dusty portal]

I dont wanna \newcommand bro

whoa!

I think I see the simplification

$\log^2(1-x)\log(x)+2\mathrm{Li}_2(1-x)\log(1-x)-2\int_1^{1-x}\frac{\mathrm{Li}_2(u)}{u}du$

[grand pond]

;(

[dusty portal]

🙏

This is glorious

$\log^2(1-x)\log(x)+2\mathrm{Li}_2(1-x)\log(1-x)-2\qty(\int_0^{1-x}\frac{\mathrm{Li}_2(u)}{u}du-\int_0^1\frac{\mathrm{Li}_2(u)}{u}du)$

[grand pond]

;(

[dusty portal]

GLORIOUS I SAY

🗣️

$\log^2(1-x)\log(x)+2\mathrm{Li}_2(1-x)\log(1-x)-2\mathrm{Li}_3(1-x)+2\zeta(3)$ FROM OUR PREVIOUS DISCUSSION

[grand pond]

;(

[dusty portal]

PROOF BY FLUX BEING A NICE GUY

🗣️

@long dagger

Thank you bro

The next question looks cused

I'm gonna take a break

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[cursive swan]

?

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[main nebula]

I was just confused about the purpose of this question for my linear algebra class.

I didn't totally understand what I was being asked to do, but I feel like I already knew a b and c would all equal 0 once I got the matrix to rref. I still don't really understand what I was doing tbh

[steel crest]

it's well known that a polynomial cant have more roots than its degree, but this is showing it to you from a linear algebra perspective

[fallow scarab]

tells you parabolas don't intersect the x-axis more than twice unless it's the 0 parabola

[steel crest]

you can approach that theorem from different viewpoints

[main nebula]

Okay I see

So it’s like making me do the proof

[steel crest]

it's making you use linear algebra tools to view a problem as a linear algebra problem, even if you know the solution from somewhere else

you learned about bases yet?

[main nebula]

I did in high school 😅

That was like 10 years ago

[steel crest]

oh i assumed this was from a lin alg course

oh iut is!

[main nebula]

Yeah this is like

[steel crest]

anyway, you know that two polynomials in the same variable are identical if and only if the coefficients are equal , right?

[main nebula]

Yeah

[steel crest]

this can be written in linear algebra language in a very deep and profound way that implies that a polynomial cant have more roots than its degree

The space of all polynomials with complex coefficients in a variable t has a basis:

${1,t,t^2, t^3, t^4, \cdots}$

[grand pond]

gfauxpas

[steel crest]

try proving that two polynomials are equal iff they have the same coefficients by treating them as vectors written with this basis

fun exercise

... if youre looking for extra homework :P

[main nebula]

We haven’t talked about vectors yet, we started with solving systems with matrices

and Gaussian elimination

[steel crest]

ah okay

[main nebula]

but I’ll come back to this problem like next week I think

or two weeks

It does make sense now though, thank you 🙂

[midnight plank]

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[steel coyote]

anyone know the best way to solve this type of quesitons

[solid iris]

slope field must provide tangent lines everywhere along the graph of a solution

[steel coyote]

so how would i go about solving it

[solid iris]

plot solution and see which field provides tangent lines

[cedar pawn]

Tbh, you can kinda just see which one is correct

[steel coyote]

do i like visualize

like how od i visualize square root 4-x^2

[dusty portal]

You should just know that it’s a semicircle with radius 2, centre at the origin

Where y>0

[steel coyote]

how would i know that

[dusty portal]

x^2+y^2=r^2?

[steel coyote]

ohh just gotta rearrange the question

alr so what that be sf5

as the answer

[solid iris]

yes

[dusty portal]

Yeah

[steel coyote]

alr thanks

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[steel coyote]

can someone help with this im confused on the part it says its initally 300 pop

but the max capacity is 500 so wouldnt the rate of change be fastest at 250?\

[cedar pawn]

For part b?

[steel coyote]

yes

[cedar pawn]

It would

What do you get for dP/dt if you put in P = 250?

[steel coyote]

i think thats the answer but im confused on the inital population

how cn there be 250 pop when inital is 300

[cedar pawn]

Are we sure that 250/4 is the answer it wants?

[steel coyote]

no thats not the answer im pretty sure the answer si 250^2(0.001)

[cedar pawn]

That's the same thing

Assuming 250/4 is correct, it just says "the growth rate will never exceed", so I guess we're just putting an upper bound on the growth rate, even if it will never reach it. So the initial population doesn't matter.

[midnight plank]

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[narrow edge]

shouldnt we choose a person for the first seat in k ways?

cuz there are k seats in total

and then when one person is seated the no of seats for the next n should be k-1

so shouldnt the answer be k(k-1)..(k-n+1)?

[lyric charm]

no you have it backwards

we go down the seats in order and for each seat we look at how many people can sit in it

when we are on the first seat nobody has been seated yet so any of our n people can sit there

n is the number of ppl

[narrow edge]

ah

can u make the last one being (n-k+1) more intuitive?

like i get it but its not really intuitive

[lyric charm]

when we're filling the last of the k chairs

the remaining k-1 chairs are already taken

thus there are n-(k-1) people remaining at that stage

[narrow edge]

in how many ways would the k-3, k-2, k-1th chairs will be taken

[lyric charm]

i can't parse that

do you mean to ask how many options we would have to fill the (k-3)rd, (k-2)nd and (k-1)st chairs?

n-k+4, n-k+3, n-k+2 respectively

[narrow edge]

ohh

i get it now

ty

[narrow edge]

i dont understand why these are equal

[fallow scarab]

Use the definition of factorial on the right side

And cancel a lot of terms

[nova yoke]

or multiply the left hand side by (n-k)! and see what happens

[midnight plank]

@narrow edge Has your question been resolved?

[narrow edge]

i still dont see it

could u similfy it

[lyric charm]

n! is the product of all integers from 1 to n

dividing by (n-k)! destroys all the ones from 1 to n-k inclusive

the smallest one left standing is the next one after n-k, which is n-k+1

[midnight plank]

@narrow edge Has your question been resolved?

[narrow edge]

ohh so (n-k+1)(n-k+2)...(n-k+k)

i get it

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[molten bay]

√(2^x-5^x) domain

[molten bay]

Officially karna chahte ho agar toh

2^x-5^x>=0
2^x>=5^x

x(log2/5)>=0

x<=(log(2/5)^0

x<=1

But it doesn't satisfied at x=1

Which step is wrong?

[lyric charm]

x(log2/5)>=0

x<=(log(2/5)^0

wow you put like 5 newlines between these

anyway this is the wrong step

from log(2/5) * x ≥ 0 you divide by log(2/5) and flip the sign (because log(2/5) is negative)

and you get x ≤ 0

[molten bay]

log(2/5) is less than 1 so when i took it to right side inequality reversed no?

[lyric charm]

less than zero not less than 1.

comparison against 1 has no meaning here

you care about positive vs negative not <1 vs >1

[molten bay]

I think it depends on base

No on 0 and 1

[molten bay]

This is also wrong no?

less than 0 is complex

Which we don't care here

So 'wow' to you also you discovered a new hypothesis

[lyric charm]

no.

you confuse condition on x with condition on log(x).

also you didnt write any base for the logarithm anywhere so i assumed it was either e or 10 and both of these work equally well here we just need to be consistent

[molten bay]

Consistent at what point?

Logx when x<0 ?

How can we apply inequalities in complex?

[molten bay]

Could you explain this line means please?

[lyric charm]

consistent means we read your logs as log_e always or as log_10 always but not mix

[lyric charm]

i meant that when we multiply or divide both sides by a number in an inequality we care about whether this number is positive or negative

but we do not care if it's <1 or >1

in your case this number is log(2/5)

it is not 2/5 it is log(2/5)

log(2/5) is not same number as 2/5

[rose pilot]

i believe you have mistake even earlier
after 2^x>=5^x you should get
(2/5)^x>=1
so
log_{2/5}((2/5)^x)<=0
x<=0

[molten bay]

Yes (log2/5) would be a negative number

So inequality reversed

I got your point and mine too

I was telling about the base yes. Thanks again

And sorry for some silly words in middle of the chat

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[molten bay]

F(z)=e^(1/z)

[molten bay]

singularity at z=0

so here by laurent expansion

1+1/z+(1/2z^2)+...

here multiply of 1/z is 1?

so is it called residue at x=0 is 1?

It contains infinite isolated singularity so it is essential singularity

Hmm

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[graceful ferry]

how do I check whether the sum $\sum_{n=1}^{\infty}\frac{(n!)^2 \cdot 4^n}{(2n)!}$ converges or not

[grand pond]

prograce

[graceful ferry]

I tried ratio test but I got q=1 so it wasn't helpful

How can I possibly find a sum that is bigger than it which is easier to test for converges so I can use comparison test?

Or how do I do it in any way?

[wintry summit]

you can simplify first by dividing both numerator and denominator by (n!)

then you can do ratio test

[graceful ferry]

To get rid of the denominator ?

[wintry summit]

the denominator is just (1 * 2 *... * n * n+1 *... *n + n) = n! * (n+1)(n+2)...(n+n)

[graceful ferry]

Yea

What does multiplying and dividing by n! simplify

[wintry summit]

now youre just left with [(n!) * 4^n ]/ [(n+1)(n+2)...(n+n)]

then you can do ratio test

[graceful ferry]

Ok I will try

Idk how to continue tbh

[runic hamlet]

well thats not the ratio test

[wintry summit]

no ratio test

[tidal turret]

he said in the first message that the ratio test was inconclusive though because was equal to 1

[runic hamlet]

do the terms even go to 0

[tidal turret]

you mean the limit of the sequence as is?

[graceful ferry]

Bro omg mb

[tidal turret]

,w lim n to infinity ((n!)^2 * 4^n)/((2n)!)

[graceful ferry]

Yeah, and that's why I misunderstood for root test

[grand pond]

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[graceful ferry]

How would diving numerator and denominator by n! change the outcome I got which is q=1?

[rapid galleon]

ratio test fails? try raabe's test.. raabe's test fails? try logarithmic test. it will work in one of them surely

[graceful ferry]

I didn't learn raabe test or logarithmic test

Not in material of my course

There has to be another way ?

[runic hamlet]

.

[tidal turret]

this sequence diverges

[rapid galleon]

well as i see here it diverges, so maybe some comparison test

[graceful ferry]

What doe that mean..

Oh

[gaunt nimbus]

nth term test?

[graceful ferry]

Thta's what im trying to do but idk what to compare it with

[graceful ferry]

Ohh

[graceful ferry]

Ok so it's divergent by no.1

[tidal turret]

yes XD

i think so

[graceful ferry]

Yes

Thanks

[gaunt nimbus]

how do u show that the limit diverges tho

[graceful ferry]

My next problem

[tidal turret]

why did you deleted

[midnight plank]

@graceful ferry Has your question been resolved?

[tidal turret]

is it direct comparison?

[gaunt nimbus]

i made a mistake

again

[tidal turret]

is fine, we are getting there XD

[gaunt nimbus]

but i think

it can be found using combination functrion

[tidal turret]

I have some idea, why not direct compare it with
((n!)^2 . 4^n)/((2n)!) <= ((n!)^2 . 4^n)/(n!)

[gaunt nimbus]

ye

that'd work

[tidal turret]

now the problem is reduced to finding what does this limit approaches to

lim n to infinity of (n! . 4^n)

[gaunt nimbus]

it approaches infinity

[tidal turret]

yeah XD

[gaunt nimbus]

damn

😭

i was using am-gm

i really have tunnel vision problem

[tidal turret]

is fine, this one was really fun, also prograce is kind of my discord buddy so i tried extra hard

[tidal turret]

you had great ideas tho, and was really helpful aswell

[graceful ferry]

Oh nice

That was really simple

[gaunt nimbus]

ye

😭

[graceful ferry]

Thanks for helping guys

I was doing another question pff

[rapid galleon]

nope. the series on the right hand diverges, but it doesn't tell you that the series on left diverges. if you have an <= bn and bn diverges, it doesn't say an diverges. if an >= bn and bn diverges only then can you say an diverges.

[tidal turret]

damn i think thats right

this isnt over yet then

[gaunt nimbus]

we are doing nth term test

[graceful ferry]

What how? Comparison test is : if bn>=an and bn diverges then an also diverges

[gaunt nimbus]

its actually
[ \frac{(n!)^2 4^n}{(2n)!} \leq \frac{(n!)^2 4^n}{(n)!}]
since $(2n)! \leq n!$

[grand pond]

k

[gaunt nimbus]

so the thing still makes sense

wati

wait a sec

am i high

i am high

ye

u cant use that proof

[rapid galleon]

neh you have it flipped. just think lol a series bigger than your series is diverging it doesn't say shit, if you pick any convergent series there are many divergent series that are greater than the convergent series.

[graceful ferry]

Yes I am reviewing ur right

[rapid galleon]

I can't myself find a comparison, well i would've relied on tests to show the divergence as i have first told you.that usually sets the game.

[graceful ferry]

I would use them too but this is homework I will submit and I cant use tests I didnt learn

They usually teach rabbe and another test but this semester is shortened so we didnt learn them

[tidal turret]

idk

[graceful ferry]

Real

[gaunt nimbus]

chat

while trying to find other forms of my combination stuff

finally

my curiosity put to rest

@graceful ferry

[graceful ferry]

Thank you!!

[graceful ferry]

But since it's an approximation can u just substitute it ?

[gaunt nimbus]

yes for n approaching infinity

[graceful ferry]

Stirling approximation huh. For last measure I will use this but I will still try to think of another way

.solved

[midnight plank]

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[graceful ferry]

How do I calculate $\sum_{n=0}^{\infty}(-1)^n\frac{n^3}{n!}$ ?

[grand pond]

prograce

[tawdry ermine]

!status

[midnight plank]

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

[graceful ferry]

1

[tawdry ermine]

Try to write the first 5 terms first to infinity as a sequence

[simple field]

does this look like any series you are familiar with ?

[graceful ferry]

Does it look like e^-x?

Like the taylor expansion of it

[simple field]

yes

could write that out

[lyric charm]

yes and youre gonna need to fiddle around with it a bit

[graceful ferry]

$\sum_{n=0}^{\infty}(-1)^n\frac{x^n}{n!}$

[grand pond]

prograce

[simple field]

yes nice

can you think of any operation we could do to this get to get closer to our target series ?

[graceful ferry]

The n^3 bothers me

[lyric charm]

yeah there is a way to get rid of it

actually i recommend working with just \sum x^n/n! with the eventual goal of evaluating the thing at x=-1

here is something to think about

what if instead of n^3 you had n(n-1)(n-2) in the numerator

what could you do then

[graceful ferry]

differentiate e^x

differentiate the sum

[lyric charm]

no need actually

$\sum_{n=0}^{\infty} \frac{n(n-1)(n-2)}{n!} x^n = \sum_{n=3}^{\infty} \frac{1}{(n-3)!} x^n = x^3 \sum_{n=3}^{\infty} \frac{x^{n-3}}{(n-3)!} = x^3 e^x$

[grand pond]

Ann

[graceful ferry]

=(-1)^3e^-1

[lyric charm]

hold off on the evaluation at x=-1 until everything is actually sorted the way it needs t obe

do you understand the basic idea here

[graceful ferry]

I understand it

[lyric charm]

that if the numerator cancels nicely with some shit in the denominator to leave a lower factorial,

then you can rewrite things in terms of e^x directly-ish as i showed

[graceful ferry]

Oh but n^3 is not n(n-1)(n-2)

[lyric charm]

right

[graceful ferry]

Yes

[lyric charm]

so then you will perhaps want to express n^3 as a linear combination of:

• n(n-1)(n-2)
• n(n-1)
• n

and break the sum into 3 and do the trick on each one

[graceful ferry]

Ok

[graceful ferry]

How do I do that ?

[lyric charm]

$n^3 = An(n-1)(n-2) + Bn(n-1) + Cn$, solve for the constants $A$, $B$ and $C$

[grand pond]

Ann

[lyric charm]

there is a smarter way to do it if you're up for it

[graceful ferry]

Yes sure

[graceful ferry]

If there's a less work way

[lyric charm]

n^3 - n(n-1)(n-2) is going to not contain any n^3 term

so the coefficient on n(n-1)(n-2) has to be 1

and now you need to break the rest into a linear combination of n(n-1) and n

[graceful ferry]

Yes

[lyric charm]

figure out how much this "rest" is

[graceful ferry]

B=3 C=1

[lyric charm]

ok good

so now you can break the sum up as:

$\sum_{n=0}^{\infty} \frac{n^3}{n!}x^n = \sum_{n=3}^{\infty} \frac{n(n-1)(n-2)}{n!}x^n + 3 \sum_{n=2}^{\infty} \frac{n(n-1)}{n!}x^n + \sum_{n=1}^{\infty} \frac{n}{n!}x^n$

[grand pond]

Ann

[lyric charm]

(also trimming zero-coefficient terms here)

[midnight plank]

@graceful ferry Has your question been resolved?

[graceful ferry]

I got 1/e

Ty ann

.solved

[midnight plank]

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[midnight plank]

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[empty ivy]

is the answer A because when u equate lambda and mu it cancels out

[lyric charm]

mmm what do you mean by "it cancels out"

can you show your work

[empty ivy]

oh yep

[lyric charm]

ok so you got that the system reduces to one equation that expresses lambda in terms of mu (or vice versa)

yes, that means the two lines are one and the same.

[empty ivy]

wait they don't intersect if you find a value for lambda/mu and it doesn't work for the third equation right

i think i got that mixed up

[lyric charm]

if your system ends up inconsistent then yes that means the lines don't interact

but you arrived at no such thing

[empty ivy]

ohh

okay makes sense thank you

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[midnight plank]

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[graceful ferry]

Given $f(x,y)=\sqrt{\frac{y^2-1}{x^2+1}}$ I wanna find what shapes there exists between the contour lines of f

[grand pond]

prograce

[graceful ferry]

$\sqrt{\frac{y^2-1}{x^2+1}} = c$ \
$\frac{y^2-1}{x^2+1} = c^2$ \
$y^2-1=c^2(x^2+1)$ \
$y^2-c^2x^2=c^2+1$ \
now how do I find what shapes I can have based on the different values of c ?

[grand pond]

prograce

[barren loom]

Write (y-c x)(y+c x)=c^2+1 and consider the lines y= +-c x

[graceful ferry]

Can you give another tip ?

[midnight plank]

@graceful ferry Has your question been resolved?

[barren loom]

Sure, asymptotes

y^2/(c^2+1)-x^2/((c^2+1)/c^2)=1

[midnight plank]

@graceful ferry Has your question been resolved?

[midnight plank]

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[midnight plank]

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[safe onyx]

what am I doing wrong?

???

[old rampart]

lmao

just doesnt like you ig

[safe onyx]

so is the computer just dumnb

[fallow scarab]

extra 3

[safe onyx]

hmm?

[fallow scarab]

oh no you did nothing wrong

[safe onyx]

it wants the negative exponent first if I factored

ok thank you both

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[twilit field]

for T is injective -> T^m is injective , I was wondering if just saying compositions of injective functions is injective is enough, or if a proof is needed

[twilit field]

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[cobalt rover]

guys i gave an extremely difficult questions of Sequence and Series. Q) If (a+1),(b+1),(c+1),(d+1) is in GP then find the value of (b+c+2)^2 = _____ options:- (a) (a+b+2)(b+c+2) (b)(a+b+1)(c+d+1) (c) (b+c+2)(c+d+2)

.reopen

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[mint ravine]

Please help me on this question

[mint ravine]

I think the order of the sequence is ACACACAC

But I don't think there are any other sequences

But idk what to do with the "two seatings" part

<@&286206848099549185>

[round parcel]

It means that ABCDEFGH is the same as BCDEFGHA.

That's just the first one rotated, so it's the same.

[mint ravine]

oh ok

can you help me to solve the question tho

im stuck

[round parcel]

Well, you can pick a specific adult to start the sequence.

So, let's say A1.

[mint ravine]

A1, C1, A2, C2, A3, C3, A4, C4

[round parcel]

Yes, that's one possibility.

[fallow scarab]

!15m

[midnight plank]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[mint ravine]

A2, C1, A1, C2, A3, C4, A4, C3

[round parcel]

No, you want to start the sequence out with the same adult.

This makes it possible to ignore rotations.

[mint ravine]

Oh so the beginning has to be A1?

[round parcel]

Yes. This is because all rotations of a sequence are the same, so you just choose an arbitrary starting point for each sequence.

You could also choose like C3 to start it.

It doesn't matter. You just need to always start with a particular person to handle all rotations of the sequence.

Like ABCDEFGH is the same as DEFGHABC, so you might as well just write it starting with A.

[mint ravine]

So there are 3!=6 ways to place the A's

[round parcel]

Right.

[mint ravine]

And there is 4!=24 ways to place the C's

[round parcel]

Right.

[mint ravine]

and then we multiply them cause they're independent events

[round parcel]

Yes.

[mint ravine]

oh so the answer is 144 thanks

[round parcel]

No problem.

[mint ravine]

i got it wrong the first time cause i did 4!*4! so thanks for clarifying

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[round parcel]

You're welcome.

[midnight plank]

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[mint ravine]

Please help me on this question

[mint ravine]

I tried doing complementary counting

So 28 choose 3 minus X