help-13

[short heath]

Thank you for helping, i am not good as the other ppl in this server. i am in the 7th grade

thats why

Yeah

[crimson sedge]

No problem

[short heath]

Is it like a technique?

To solve math problems, like mine

[crimson sedge]

Yeah

[short heath]

Is it on youtube

i would like to learn about it

I want to see how it works

[crimson sedge]

I'm not sure, lemme search for you.

[short heath]

Ok

[crimson sedge]

Do you know how to solve linear equations in 2 or 3 variables?

[short heath]

3

[keen elbow]

ahh common core

[crimson sedge]

For these types of problems, you may use the concept of linear equations to solve em easily.

[short heath]

Ok, thank you for helping me

[crimson sedge]

If you are done, you may close the room by .close

[short heath]

.close

[cedar kiln]

Channel closed

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[cedar kiln]

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[crimson sedge]

Find the number of sides of a polygon which has 44 diagonals. How to solve this?

[crimson sedge]

Answer was given to be 11 but Is it possible for a polygon with odd number of sides to have diagonal? Like triangle has 3 sides, can we draw diagonals?

[dawn jetty]

how come?

a diagonal connects 2 distinct points with no common edge/side

thus creating a triangle

well at least it can create a triangle

[crimson sedge]

I'm not getting you

[dawn jetty]

in a triangle there are simply no such points so you can't define a diagonal

you're saying that a polygon of odd number of sides to hav ea diagonal

let me ask you this, what defines a diagonal?

[crimson sedge]

Answer to this problem is given to be 11. Im asking how can a polygon with odd number of sides have diagonal?

[crimson sedge]

Maybe a line segment which joins two opposite vertex of a polygon.

[dawn jetty]

2 opposite?

why necessarily opposite?

draw a pentagram

does it not have diagonals?

[crimson sedge]

https://www.cuemath.com/geometry/diagonals/ I've been taught that there must be 2 opposite vertices.

[crimson sedge]

I think no.

[fallen heath]

what.

[dawn jetty]

Oh okay I see

the use of the word Opposite is confusing

if you read the rest of the definition it says a diagonal is a line connecting 2 non adjacent vertices

Now consider a triangle

a triangle has no diagonals, because given the 3 points that form a triangle, every 2 points are adjacent to the third

[crimson sedge]

okay, so how many diagonals does a hexagon has? 3?

[dawn jetty]

well lets see

a hexagon has 6 sides

lets take 1 vertex
we can draw a diagonal to every vertex but the 2 adjacent to it correct?

that would give us 3 diagonals

[crimson sedge]

Yes

[dawn jetty]

now take vertex 2

[crimson sedge]

It will give 2 diagonals

[fallen heath]

can draw a diagonal to every vertex but vertex 1 and vertex 3 ( cause adjacent ), so 3 diagonals

what...

2..?

[crimson sedge]

From vertex 2

[wraith dagger]

Ansh

[crimson sedge]

I got 9 diagonals in a hexagon

[fallen heath]

Yes which ones?

[crimson sedge]

Wdym?

[fallen heath]

...

[crimson sedge]

From V1 and V2, i got 3 diagonals each, from V3, i got 2 additional diagonals and from V4, I got 1 additional diagonal.

[fallen heath]

Yes great...

Now, consider a heptagon :o

[crimson sedge]

I think I misinterpreted the word "Diagonal"

[fallen heath]

Count the diagonals :D

XD

Yes, I think just being able to count, would solve the problem for you..

[crimson sedge]

15?

[fallen heath]

Can you write them? or tell me how many you got from which vertices?

[crimson sedge]

Ohh wait I think i did wrong

[fallen heath]

From V1 and V2, 4, V3 -> 3, V4 -> 2, V5 -> 1, total 14 :o

[crimson sedge]

Yeah exactly

[fallen heath]

Generalize this setup for n, solve for a general : no. of diagonals in an n-gon, and equate to 44

:D happy solving

[crimson sedge]

How can I relate the number of diagonals given by the first and second vertex with number of sides or total diagonal?

@fallen heath

[fallen heath]

Counting?

[crimson sedge]

n-2?

Ohh wait... the actual question was to be solved using combination. How can I do that?

[fallen heath]

Pick any 1 vertex - times the number of diagonals that can be drawn from the vertex and consider the number of times each diagonal is repeated

that's a hint :o

[crimson sedge]

I didn't understand

[fallen heath]

in how many ways

can you

pick 1 vertex in the n-gon?

next

in how many ways

[crimson sedge]

nC1

[fallen heath]

can you

draw a diagonal from that 1 vertex that you got

within the n-gon

?

[crimson sedge]

I can draw (n-2) diagonals from that vertex

[fallen heath]

NO

you cannot

draw a diagonal from a vertex to itself

XD

[crimson sedge]

oh no, (n-3) xD

[fallen heath]

So, in how many ways can a diagonal be drawn within an n-gon?

smh, whichever article is used for "n-gon", my grammar sucks

[fallen heath]

hmm?

[crimson sedge]

Idk 😕

[fallen heath]

$\binom{n}{1} \cdot (n-3)$

wdym "idk"

[wraith dagger]

Ansh

[fallen heath]

.

[fallen heath]

.

[crimson sedge]

I always get confuse with "How many ways" :(

[fallen heath]

Okay so

you pick 1 vertex in nC1 ways

and

[crimson sedge]

Yes

[fallen heath]

draw a diagonal from it

in (n-3)C1 ways

so the total number of diagonals you get is

nC1 times (n-3)C1

no?

[crimson sedge]

Yes

[fallen heath]

NO! Lmao

Because, when you pick vertex 1, and draw a diagonal from it to vertex 5, say,

and later, you pick vertex 5, and draw a diagonal from it to vertex 1

you count this particular diagonal twice >_<

[crimson sedge]

Oh yeah 🥲

[fallen heath]

[crimson sedge]

We can select 2 diagonals from n?

[fallen heath]

So, you've counted every single diagonal out there, twice

[crimson sedge]

nC2

[crimson sedge]

Yeah

[fallen heath]

No 😂 why would you do that

[fallen heath]

so

2 times the total number of diagonals in an n-gon

equals

[crimson sedge]

Divide it by 2 😂

[fallen heath]

n(n-3)

So what is the total number of Ds in an n-gon?

[crimson sedge]

?

I'm not getting it, this chapter is really very confusing 😭😭

[fallen heath]

nC1 times (n-3)C1???

what's so confusing about that being n(n-3)

[fallen heath]

you don't understand how I got this?

[crimson sedge]

Ohh now got it.

Is there any property that nCr × mCr = nm?

[fallen heath]

XD whenever something like this happens... consider a short break to freshen up a bit

[fallen heath]

no lol 😂 nC1 = n

[crimson sedge]

Ohh 🥲

[fallen heath]

So anyways

[fallen heath]

.

[crimson sedge]

How can I conclude that? Is it n(n-3)?

[fallen heath]

Ehhhhhhhhhh

We just went through this

like 3 mins ago

[fallen heath]

.

[fallen heath]

.

[fallen heath]

.

[crimson sedge]

Is it the total number of diagonals in n sided polygon?

[fallen heath]

Yes

I'm asking for the total number of diagonals in an n-sided polygon

based on what you've understood so far about my explanation

[crimson sedge]

But why is it n(n-3)? Isn't it counting twice? You said earlier.

[fallen heath]

Thanks heavens

Do you get that part?

how the total number of diagonals counted twice is n(n-3)?

[crimson sedge]

Yes

[fallen heath]

then I'm asking

what is the total number of diagonals

[crimson sedge]

🤔

[fallen heath]

If the number of apples in my hand counted twice is 4, what is the number of apples in my hand right now?

[crimson sedge]

2

[fallen heath]

0 actually cause I'm typing, but you got it

[crimson sedge]

So n(n-3)/2?

[crimson sedge]

😂😂

[fallen heath]

Isn't it?

[crimson sedge]

Yes

[fallen heath]

So then the total number of diagonals in an n-gon is given as n(n-3)/2

now go ahead

use that formula wherever you want :o

[crimson sedge]

But ...

[fallen heath]

But.......................................

?

[crimson sedge]

n(n-3) is two times the total number of diagonals drawn from one vertex 🙂

[fallen heath]

No.

[crimson sedge]

[fallen heath]

(n-3) is the total number of diagonals drawn from one vertex

n(n-3) is the total number of diagonals you'd count, if you were to pick each vertex( in nC1 ways ), and draw a diagonal from it to the remaining vertices ( in (n-3)C1 ways )

[crimson sedge]

Okay 🙂

[fallen heath]

No it's not.... if you're not getting "why?"

Do you get why?

[crimson sedge]

I got it

[fallen heath]

Great then!

so

[crimson sedge]

Thanks a ton, you are really very good at explaining things. Ig you are a teacher. 🙂

[fallen heath]

the total number of diagonals counted by picking one vertex each and drawing a diagonal from it to the remaining vertices is n(n-3) but you then realize that you're actually counting each diagonal twice

so you deal with it simply by writing 2D_n = n(n-3)
so D_n must be n(n-3)/2

[crimson sedge]

Ok

[fallen heath]

Now equate with 44, n² - 3n - 88=0, and solve it

[fallen heath]

Uni student but :o I just know coz I've had exact same problems as you in the past

[crimson sedge]

n = 11

[fallen heath]

Great :D

Another way however

[crimson sedge]

Shukriya

[fallen heath]

was a pretty ... counting method

realizing the trend in hexagon : 3 + 3 + 2 + 1, in heptagon : 4 + 4 + 3 + 2 + 1, ..., in n-gon : (n-3) + (n-3) + (n-2) + ... + 2 + 1 = n(n-3)/2

[crimson sedge]

Yeah

[fallen heath]

but it'd be a lot better if you can just get used to thinking that nC1.(n-3)C1 and along those lines as you'll get more such problems like counting the non-colinear points and stuff

[crimson sedge]

Okay

.close

[cedar kiln]

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[marsh vapor]

if the tangents drawn from the origin to the y=x^2-3x+a parabola are perpendicular to each other, what is a?

[marsh vapor]

here’s my attempt at solving the question above:
let there be two lines y=mx and y=(-1/m)x. these two lines are perpendicular to each other, and they are tangent to the parabola.
therefore: delta=0 for both x^2-3x+a=mx and x^2-3x+a=(-1/m)x.

we get: (m+3)^2-4a=0 and ((-1/m)+3)^2-4a=0

we’re asked to find a, so let’s solve for m first: (m+3)^2=((-1/m)+3)^2

aaand I’m stuck

,w find m where (m+3)^2=((-1/m)+3)^2

[wraith dagger]

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[marsh vapor]

how do we find that m=sqrt10-3

<@&286206848099549185>

[crimson sedge]

I need help with my math @marsh vapor

[marsh vapor]

bro

[marsh vapor]

#❓how-to-get-help

read

[obsidian coral]

Don't try to take over an occupied channel

[crimson sedge]

Oops I thought it was under available, im blind clearly. Sorry

[marsh vapor]

no worries

[cedar kiln]

@marsh vapor Has your question been resolved?

[marsh vapor]

<@&286206848099549185>

[dire geode]

haven't solved it yet, but a first start is to expand the square, move all terms to one side, and then multiple the whole equation by m^2 to get a fourth order polynomial in m

[cedar kiln]

@marsh vapor Has your question been resolved?

[cedar kiln]

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[toxic carbon]

missed class today and i dont know what to do at all. teacher didnt give any notes. that first one is y = 105sinx btw.

[dawn jetty]

well do you know what amp and pd mean?

[toxic carbon]

amplitude and period from how im understanding it

that legit is all i know

[dawn jetty]

well yea obv but i mean like the meaning behind the word lol

[toxic carbon]

lmao nope

[dawn jetty]

well then I advise you to look up the definitions which are not hard to understand

and try to figure it out for yourself

[toxic carbon]

okay thanks

[dawn jetty]

It's honestly not too bad, you'll have an overall better learning experience this way

remember that cosine and sine are bounded and periodic functions when you approach it

[toxic carbon]

ight. i just didnt know how to start. thanks a lot!

[dawn jetty]

getting to know what amplitude and period of a function means is the place, good luck !

[toxic carbon]

thanks

.close

[cedar kiln]

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[gray vine]

I am asking for my sister: write an expression for the amount of matches in figure n, if figure 1 has 5 matches and figure 2 has 9 matches.

[dawn jetty]

figure n?

[gray vine]

Yeah

[dawn jetty]

so essentially a sequence?

a1 - figure 1
a2 - figure 2 etc...

[gray vine]

Yes

I think so

[dawn jetty]

what topic does it relate to?

[gray vine]

figure n would be something like 4(n+1) i assume, but i am 1) unsure about this, and 2) don't know how to explain this in an elementary level to a 13 year old

it's algebra

[dawn jetty]

4n+5 rather but yea

well its a simple case of an arithmetic sequence

[gray vine]

well then figure 1 would be 9 matches, but it's only 5

[dawn jetty]

fair enough , 4(n-1) + 5 would take care of the shift

but in any case this is simply asking you to construct an arithmetic sequence

and any such sequence is determined the the first term , a1, and the the difference between any subsequent terms , called d

[gray vine]

ok thanks

[dawn jetty]

in this case 4 would be the difference between each figure , and the first figure would be the other defining characteristic of the sequence

but then again if you're expected to solve it in a different way

if you want to explain it to your sister maybe show an elementary example where first you'd have 1 match to start with, that's how i'd do it with my nephew

[gray vine]

i'll do that, thanks

.close

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[cedar kiln]

Please don't occupy multiple help channels.

@stiff vigil Has your question been resolved?

[cedar kiln]

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[dusk gulch]

help needed, thanks

[cedar kiln]

@dusk gulch Has your question been resolved?

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[open agate]

How did he get from a to b:

[open agate]

I'm so confused

[jaunty mural]

I suspect context is needed?

[open agate]

it's from a midpoint circle drawing algorithm

https://www.youtube.com/watch?v=yfLm9nXsCvw&ab_channel=SundeepSaradhiKanthety

Im not understanding how he got from a to b

it's at around 9:50

Please someone help

[dawn jetty]

wait

that tick notation is a square?

((xk+1)+1)^2?

[open agate]

yeahh

lol

[dawn jetty]

oh, he just squared the terms then

((xk+1)+1)^2=(xk+1)^2+2(xk+1)+1

that's all there is to it

do you see it?

[open agate]

ok what about the other terms

[dawn jetty]

same for (y(k+1)-1/2)^2

he just used (a+b)^2 formula to square the expression.

[open agate]

Ah ok I get it now

[cedar kiln]

@open agate Has your question been resolved?

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[pastel grove]

I'm taking a paractice test for a physical cosmology class, and the question that popped up is throwing me a curve ball. The question is

Say that you have a system of two nonlinear ODEs. Under which conditions in terms of the eigenvalues of the characteristic matrix are the solutions stable?
I don't remember going over this at all, at any point. I'm hoping someone can explain it to me , and I can go over it.

[jaunty mural]

Interesting

Do you recall what eigenvector/value means

geometrically

[pastel grove]

I'm pretty sure. Eigenvectors of linear transformation are non-zero vectors that change by a scaling favor when that linear transformation is applied

I think that's right

[jaunty mural]

yes

$$A\boldsymbol{x} = \lambda\boldsymbol{x}$$

[wraith dagger]

Shuri2060

[jaunty mural]

So lambda is the stretching constant in the direction of eigenvector x

Now, for stability around the equilibrium

You need all the points around it not diverging

If we take the equilibrium as the origin

[jaunty mural]

Can you see what lambda must be?

note we have

$$\dot{\boldsymbol{x}} = A\boldsymbol{x} = \lambda\boldsymbol{x}$$

[wraith dagger]

Shuri2060

[cedar kiln]

@pastel grove Has your question been resolved?

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[gentle lintel]

can someone explain why we need the second condition

[jaunty mural]

Oh god.

I meant to draw something this morning 😂

[gentle lintel]

lol

[hollow osprey]

is this like

fancy notation for commuting the limit and the function?

cuz the second condition is simliar to "needs to be decreasing/increasing" around the functoin but im not sure why it cant be constant

[jaunty mural]

$$LHS = \lim_{x\to a}f(g(x))$$
$$RHS = \lim_{t\to\left(\lim_{x\to a}g(x)\right)}f(t)$$

[hollow osprey]

one more }

after the second limit

3rd***

before g(x)

lmao what is that

wait so i was right

[wraith dagger]

Shuri2060

[hollow osprey]

it is just fancy notation for interchanging limit / function

[jaunty mural]

@gentle lintel I don't know what argument you want to hear

we've been through them all.

[gentle lintel]

can you give an example of g(x) near a

[jaunty mural]

If g(x) is constant, we have a problem.

[gentle lintel]

except for constant functions

[jaunty mural]

ok?

we have g(x) constant on -1, 1

and everywhere else not

[gentle lintel]

alright

[jaunty mural]

Like what example do you want me to pull out, exactly

[gentle lintel]

uhh

[jaunty mural]

I can pull out some wacky ones if you want, but idk what to say

all that matters is g cant be constant near a

[gentle lintel]

if g(x) = 0 for x = 0 and x² sin 1/x otherwise

g(x) = 0 for x near 0

right

[jaunty mural]

no.

[gentle lintel]

no?

[jaunty mural]

,w plot x^2 sin(1/x)

[wraith dagger]

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[jaunty mural]

oh ok.

yes.

This appears to be an example of g that satisfies the equation but not (2)?

Possibly

[gentle lintel]

yea

[jaunty mural]

If you could give me an f that convinces me?

[gentle lintel]

not sure tho

[jaunty mural]

if g(x) = 0 for x = 0 and x² sin 1/x otherwise

f(x) = x

let me see...

lim fg = lim g = 0

lim g = 0

lim f = 0

@gentle lintel sure

I think.

You appear to have given an example that demonstrates the 2nd condition is not necessary, but only sufficient.

@celest seal can u check ^

if g(x) = 0 for x = 0 and x² sin 1/x otherwise
f(x) = x

This is interesting 🤔
I think a necessary and sufficient condition might involve non-locally constant but unsure

===
Well the thing is, for non equal cases to happen, I think we need f discontinuous

Oh no wait

[gentle lintel]

then remove the 0 for x = 0?

[jaunty mural]

Your choice of g

does not work for all f

I will choose

f(x) = 0, x not 0
f(0) = 1

Now lim fg will not exist.

Can you see why?

$$\lim_{x\to0}f(g(x))$$

[wraith dagger]

Shuri2060

[jaunty mural]

As our x gets closer and closer to the origin, fg will continue to switch between 0 and 1

fg(x) = 1 when sin(1/x) = 0; 0 otherwise

[gentle lintel]

huh

[jaunty mural]

Consider carefully what the composition of f and g is.

f(x) = 0 for x ≠ 0; f(0) = 1

g(x) = x²sin(1/x) for x ≠ 0; g(0) = 0

===
@fallen heath ^ check the above lol

It appears that condition on g is completely necessary, provided we don't know anything about f.

[cedar kiln]

@gentle lintel Has your question been resolved?

[gentle lintel]

lengthening the discussion lol

[fallen heath]

:o

[cedar kiln]

@gentle lintel Has your question been resolved?

[gentle lintel]

(i still dont get it)

(bruh)

[fallen heath]

I understand...
I think it would've been a lot easier to just present a rigorous proof from scratch

[gentle lintel]

nah im not hateful

[fallen heath]

Are you taking analysis courses?

[gentle lintel]

no

[fallen heath]

real analysis courses?

oh

[gentle lintel]

im self learning calc 1 with some notes from friend

but i know epsilon delta

[fallen heath]

Anyways, if you did prove squeeze theorem using epsilon delta... I'm supposed to assume you know the basics of limit conversion into epsilon delta

[gentle lintel]

yea

[fallen heath]

can you try writing everything said above, in the theorem.. with the idea of epsilon delta?

[gentle lintel]

which

which statement

[fallen heath]

[gentle lintel]

which line

[fallen heath]

the main limit L.H.S. = R.H.S. line after variable change

exactly what does that thing fundamentally mean :o

[gentle lintel]

wait

[fallen heath]

break it into three parts

and show me what's meant by each of them... fundamentally

[gentle lintel]

idk

[fallen heath]

1. L = lim f(g(x)) as x tends to a
2. M = lim f(g(x)) as g(x) tends to b
3. N = lim f(t) as t tends to b

^

[gentle lintel]

huh

not sure what you want

idk how to write when two limit expressions are equal

if lim f(x) = L sure

[fallen heath]

write those three according to the defintion of limit

[fallen heath]

Yeah, go ahead

[gentle lintel]

just sub in the defn

im lazy to write lol

[fallen heath]

me too >_< but then this conversation won't move ahead

[gentle lintel]

alright ill write it

,rotate

[gentle lintel]

something like this i guess

[fallen heath]

:o

Now forget the initial theorem >_<
And try equating the three of these implications using rigorous math

You'll come across some necessary conditions for that to happen :o

[gentle lintel]

not sure how

i havent equated limit forms

[fallen heath]

this?

:o

[gentle lintel]

im confused by what you mean about rigorously equating these

[fallen heath]

Like for example

consider what you got from lines 2, 3

[gentle lintel]

huh

[fallen heath]

here, in line 2, put g(x) = t, and you get : |t-b| < delta => |f(t) - M| < epsilon
so you can conclude that limit of f(t) converges to M but from line 3 it also converges to N, so M must be N

Is that okay?

[gentle lintel]

oh

wait

uhh… ok?

[fallen heath]

:o Not ok I guess

[gentle lintel]

wait no

i dont get it

[gentle lintel]

why M must be N?

[fallen heath]

what's so confusing about this?

[gentle lintel]

wait

i check note

for uniqueness of limit

[fallen heath]

because limit of f(t) can't be converging to M and N at the same time?

...?

[gentle lintel]

oh ok

yea

[gentle lintel]

okay

[fallen heath]

I want you, to similarly deduce a relation between L and M

scratch your head if you have to, but at least attempt something (ヘ･_･)ヘ┳━┳

[gentle lintel]

yea so

L = M = N?

whats next

[fallen heath]

but is L = M always?

or under certain conditions?

[gentle lintel]

idk, this is a guess

[gentle lintel]

i guess it’s similar to this one

[fallen heath]

Or is it?

[gentle lintel]

ejidjdoddididjxjxjdkddmfofktsd

[fallen heath]

the condition in 2)

says

specifically

that

$g(x) \neq b \forall x$ s.t. $|x-a| < \varepsilon$ is a necessary condition for L = M

[wraith dagger]

Ansh

[gentle lintel]

huh

we’re figuring out why we need the 2nd condition rifht

how does this help

[fallen heath]

huh?

You write the fundamental definition of the first limit L, you write the fundamental definition of the second limit M ... since you cannot, without assumptions jump from L to M, you have to study L, M to establish those assumptions...

Which part of this do you not understand?

[gentle lintel]

huh

‘since you cannot’

[fallen heath]

It's 'since you cannot without assumptions jump from L to M'

You don't understand that?

[fallen heath]

Can you trivially see L = M?

If not, you need to study under what assumptions can L be equal to M

[cedar kiln]

@gentle lintel Has your question been resolved?

[gentle lintel]

hi

[gentle lintel]

cant we apply a similar approach as we did to show M = N

[gentle lintel]

<@&286206848099549185>

[steep ingot]

hello

so what are we doing today

[gentle lintel]

understanding why we need condition 2

[steep ingot]

ok not to be that guy but what is this about

[gentle lintel]

wdym what is this about

calculus and limits?

[steep ingot]

ok thanks

i cant wrap my head around this right now ik its simple but its kinda hurting my head

<@&286206848099549185>

[dire geode]

Need a help-the-helpers role

[gentle lintel]

yea

[fallen heath]

Lmao

[gentle lintel]

had enough of the pings

[fallen heath]

Hmm

[gentle lintel]

also, i know to little too help in general

[fallen heath]

chromium, I need to wrap my things around this as well, for a bit as my basics in analysis classes aren't very well brushed up

so give me some time

for the time being

just consider the function you weree given yesterday

[gentle lintel]

can we use this

[fallen heath]

f(x) = 0, x \neq 0, f(x) = 1, x = 0

[gentle lintel]

seems to demonstrate better

[fallen heath]

NO

For any given value

[keen elbow]

damn, what did pings ever did to you? guessing you're a high schooler?

[fallen heath]

near 0

say x_k

g(x_k) would be x_k^2 sin (1/x_k)

and not 0, because x or sin x = 0 only at x = 0

Now, I skipped my topology class for a sweet nap coz I'm too exhausted from this week's continuous classes (ヘ･_･)ヘ┳━┳

Is it understandable that g(x) is not 0 near x = 0 but rather the limiting value of g(x) is 0

[gentle lintel]

i dont like that red thing on top of the app

[fallen heath]

if not, consider revising whatever you've done on your first few days of limit

[gentle lintel]

given any δ > 0

[fallen heath]

...

Okay go ahead... given any delta?

[gentle lintel]

we can always find h in (-δ, δ) such that g(h) = 0 right?

thats what near 0 means

i guess

[fallen heath]

:o

what is your f(x)?

[gentle lintel]

idk

but g suits the near 0 thing right

[fallen heath]

sure true for all x = 1/n(pi), but is that the only possible sequence near 0?

the solutions for g(x) = 0 would be x = 1/n(pi), n not 0, if I'm not wrong

[gentle lintel]

not sure what you mean

[fallen heath]

...

near 0 means

(0-∆, 0+∆)

and you're saying

1/n(pi) is the only possible numbers existing in that region

[gentle lintel]

????

no

[fallen heath]

Yes

[gentle lintel]

where

[fallen heath]

[gentle lintel]

im just guessing what near 0 means here

lol

[fallen heath]

so you know now that your guess was wrong

[gentle lintel]

since this is true, g(x) = 0 near 0

this is my interpretation

[gentle lintel]

this

[fallen heath]

x near 0 literally means x in (-δ, δ)

that's it

delta can be as small as possible

give any number, delta can be still smaller than that

[gentle lintel]

so second limit states that ∃ δ > 0 such that ∄ h ∈ (a - δ, a + δ) such that g(h) = b

right

[fallen heath]

Yes!

[gentle lintel]

thats literally my guess

[fallen heath]

?

[gentle lintel]

nvm

so x² sin 1/x doesn’t satisfy this right

[fallen heath]

∄ h ∈ (a - δ, a + δ) such that g(h) = b

and you're saying

∃ h ∈ (a - δ, a + δ) such that g(h) = b

:/

Anyways

[fallen heath]

Yes

As g(x) = 0 for x_k = 1/k(pi)

[gentle lintel]

so how do we go on about the limit part

[fallen heath]

you don't

first you have to settle on the part that you understand what near 0 means

[gentle lintel]

we’re trying to understand condition 2 right

[fallen heath]

or what near a means

[gentle lintel]

yea

[fallen heath]

Do you understand?

[gentle lintel]

yea

[fallen heath]

@gentle lintel

[gentle lintel]

huh

[fallen heath]

for this function, find the limits

$\lim \limits_{x\to 0} f(g(x))$ and $\lim \limits_{g(x) \to 0} f(g(x))$

[wraith dagger]

Ansh

[fallen heath]

g = f here

[gentle lintel]

what

[fallen heath]

[gentle lintel]

dne?

this is the example shuri picked

[fallen heath]

What are the limits?

[gentle lintel]

dne and 0??

not sure what you mean by g(x) → 0

[fallen heath]

wow

[gentle lintel]

isn’t it the same as this

[fallen heath]

wow

[fallen heath]

$\lim \limits_{g(x) \to 0} f(g(x))$ as in $\lim \limits_{g(x) \to b} f(g(x))$

[wraith dagger]

Ansh

[fallen heath]

Are we trying to do mental math here? Do you even have pen n paper down atm?

[gentle lintel]

yea

[fallen heath]

Then wdym by you're not sure abt this when we've just seen the same in the theorem and you seemed to have understood it just fine

[gentle lintel]

i translated it into a rigorous defn that algebraically makes sense

[gentle lintel]

here no

[fallen heath]

you mean, these two aren't the same? if g(x) converges to 0 instead of b, or simply if I put the value b = 0?

...?

Listen... if your book doesn't mention the proof of the theorem, and your teacher doesn't explain you that, you're not supposed to go ahead with it unless you have enough knowledge about the tools required

[gentle lintel]

it doesnt

it’s just there lol

[fallen heath]

So just understand what the theorem says

Use it in a particular example to see what's wrong

[gentle lintel]

i’m trying to know why theres such a specific thing

[fallen heath]

and continue

[fallen heath]

Don't

[gentle lintel]

because it comes up in disproving a common ‘proof’ of the same rule

[fallen heath]

Just giving an example should be enough for you at this level

which I did

[gentle lintel]

yea i still dont understand why it’s there

[fallen heath]

[gentle lintel]

just what it is

[fallen heath]

Yeah it's okay

In future, when you're taking analysis courses, you'd be more than qualified to attempt this question once again

and to discuss on it

[gentle lintel]

so this requires analysis knowledge???

inf sup etc

[fallen heath]

At least it requires involving sequences

[gentle lintel]

sequences are precalc

[fallen heath]

._.

[gentle lintel]

arent they

[fallen heath]

epsilon delta is also precalc?

[gentle lintel]

calc is based on precalc lol

[gentle lintel]

they’re introduced in precalc

[fallen heath]

(ヘ･_･)ヘ┳━┳

[gentle lintel]

general terms etc

does it involve some advanced knowledge that isn’t taught in calculus books?

at least in understanding this

if not i want to understand

[fallen heath]

Why are you so hell bent on ... okay smh

Give me a day 🤦‍♂️

I'll try to find a proof

"rigorous" proof

epsilon delta, etc.

but for now, I'm not sure about one thing

[fallen heath]

can you evaluate these limits?

[gentle lintel]

these?

[fallen heath]

f(f(x)) here is clearly 0 at x = 0 and 1 at x !=0 so no

the left hand limit is actually 1

as x approaches 0, f(f(x)) remains 1,

so the limiting value

is 1

as for the right hand limit however

At f(x) approaching 0, say f(x) = t approaching 0

lim f(t) is 0

and 1 != 0

so, for this particular example, even though 1st conditon holds

second one doesn't and the limit doesn't equal

[crimson sedge]

It is enough to Show(I'm sorry if I'm missing something here) that the thm won't be true if Second condition is false.Consider the constant function g(x)=b.Then this Change of Variable formula will hold iff f is continuous at b.Pick an f which is discontinuous at b and your statement will be false.So thus second condition is necessary

[gentle lintel]

huh

wdym

[crimson sedge]

Is this clear?

[gentle lintel]

no

[crimson sedge]

So if g(x)=b at every point x,then f(g(x)) is the constant function f(b) which is continuous.Thus Lim x goes to a (f(g(x))=f(b)

[crimson sedge]

The RHS of statement is lim t goes to b f(t)

So for that equality to hold we want lim t goes to b f(t)=f(b)

Or f to be continuous at b

[crimson sedge]

Is it ok now?

[gentle lintel]

wdym by this

[fallen heath]

:o understandable

[crimson sedge]

So we computed LHS of the statement and we got f(b) and RHS of it and we got lim t goes to b ,f(t) .Since LHS =RHS in given statement,this follows

[gentle lintel]

can you start over

i havent been sure what you’ve been referring to

[crimson sedge]

This was the first step

[crimson sedge]

Are you able to see this

[fallen heath]

Can you maybe take him on one of the VC channels?

[crimson sedge]

@fallen heath Did you understood this 😅.I can't do vc right now as I'm kind of close to the Library.If you understood this or if you have a different proof in your mind ,I think you can help him via VC.

[fallen heath]

Yes I did, and true, here, it's enough to prove that discontinuity of f at b is sufficient condition to follow with why second condition is important

and no ._. if I could, I would've already VC'd chromium

[gentle lintel]

what’s f and g again

qwq

[fallen heath]

g is a constant function

and we start by assuming second condition is not needed but the theorem still holds

assuming

we do the equations based on a function "f" and pick a particular case where f is discontinuous at b

and conclude with a contradiction as the theorem doesn't really hold in this case

so, we say the second condition is necessary 💀

[gentle lintel]

huh

[fallen heath]

Can't really help you there

it's just English

[gentle lintel]

so this is f

x² sin 1/x is g

?

[fallen heath]

nope

where the heck are you

Right now the context changed

we're talking about what sreedev proposed

earlier we were talking about what shuri proposed

Dude we can't do this unless you're paying attention ._.
I'm srry

Ping me after only if you've absorbed everything that's been said so far ._. including the two examples n your own example and what we did on those and whatever

[gentle lintel]

oh

@fallen heath

is this condition here to be aware of oscillating into DNE scenarios?

[elfin hemlock]

I don't understand this at all, you have been shown why 2nd condition is needed through some examples, what are you confused about?

Once you see a counterexample you have your reasoning why it is needed and you are done. So 2nd condition is clearly needed

[cedar kiln]

@gentle lintel Has your question been resolved?

[gentle lintel]

right..?

i just need to confirm thi

s

[jaunty mural]

wut is going on

[gentle lintel]

im nearly getting it!!

(amazing progress smh)

[jaunty mural]

I have an intuitive defn for limit

if it helps

[gentle lintel]

alright

[jaunty mural]

$$\lim_{x\to p}f(x)$$

[wraith dagger]

Shuri2060

[jaunty mural]

p being 'point'

For any open interval Y around f(p), no matter how small.

You are able to choose an open interval X around p so that f(X) = {f(x) : x in X} is in Y.

[gentle lintel]

this is my idea of defining this condition

[gentle lintel]

led me to this

[jaunty mural]

what does 'aware'

mean.

[gentle lintel]

g = x² sin 1/x

f = 1 for x = 0, 0 otherwise

you said lim fg dne at 0 right

[jaunty mural]

yh

[gentle lintel]

and here, one may wrongly apply change of variables

and lead to wrong answer

hence, we require this

since the limit oscillates into dne

[jaunty mural]

Well this is one example, I guess

while g oscillates

the oscillations get smaller and smaller

taking the composition makes that no longer true

The condition needs to be written in a way to dodge this case

But I think a solid reason can be explained through this visualisation of limit

==========
For any open interval Y around f(p), no matter how small.

You are able to choose an open interval X around p so that f(X) = {f(x) : x in X} is in Y.

[gentle lintel]

you make a visualisation?!

[jaunty mural]

Imagine the xy plane

Like draw a random function on paper and apply the above

[gentle lintel]

wait isn’t this literally the epsilon delta in layman’s terms

[jaunty mural]

Urgh I need to correct this

[gentle lintel]

holy shit

[jaunty mural]

That definition I just wrote is assuming f continuous

and yes

[gentle lintel]

ive been searching for this since i first learnt

i tried to understand it for 2 unholy weeks

[jaunty mural]

==========
If lim f(x) as x -> p is L. THEN we have:

For any open interval Y around L, no matter how small.

You are able to choose an open interval X around p so that f(X) = {f(x) : x in X} is in Y.

So like, try drawing on paper what this means

You give me an interval Y

I can give you an interval X

so that f(X) is contained in Y

And check continuous/non-continuous examples

The 2nd way of thinking about it is what some people like as well

Now, to consider this 2nd condition. How about drawing 2 separate graphs. One of f, the other g. However the y-axis of g graph corresponds to the x-axis of the f graph.

[gentle lintel]

alright

i’ll try to prove change of variables of limits i guess?

and see if that helps

idk

[jaunty mural]

Idk, I would use 2 pencils to point at what is going on with my limits

on each axis

[jaunty mural]

basically, you are going to fail f(X) in Y

In any of the counterexamples

f(X) is going to be miles away from f(p) because of the discontinuity

However much you shrink X

Due to the behaviour of G

(I'm describing lim fg, here)

[wraith crypt]

Stuff looks interesting, anything for me to help out with?

[gentle lintel]

exactly what was discussed yesterday

lmao

[wraith crypt]

LOL

[fallen heath]

:o

[ancient grotto]

👀

[crimson sedge]

anyone here good with exponential and logarithmic functions?

[ancient grotto]

get yourself a new channel if you want help @crimson sedge read #❓how-to-get-help

I am unable to get much out of the convo above

I don't see what you're stuck on exactly

but if you want a change of idea

[fallen heath]

Exactly XD
We've been there several times in last 36 hours

[ancient grotto]

I feel like sequential criteria would help here

[jaunty mural]

yes

[wraith crypt]

what exactly is the problem though?

[ancient grotto]

they want to understand a condition for a theorem from what I can tell :|

[wraith crypt]

i thought it was resolved yesterday

[jaunty mural]

but how do u propose to convince the sequential limit

is the same as continuous limit

if limit exists

[ancient grotto]

wdym by that

[fallen heath]

but the thing is, the book doesn't give a proof, and he's trying to prove the theorem ...

by sequence I suppose you mean defining a sequence x_n converging to x and then approaching this limit situation?

[jaunty mural]

thats what I mean.

[ancient grotto]

in sequential spaces you can define limits using sequences

and in this case that approach seems intuitive

to understand

[wraith crypt]

can we live in the real line for now HAHAHA

[fallen heath]

Yes... basics of real analysis :o

[jaunty mural]

u see, idk about getting into analysis here

[ancient grotto]

I am just proposing a change in idea I think they already have what they need to understand the theorem

[jaunty mural]

more analysis i mean

[fallen heath]

but I doubt he has the tools or the understanding / maturity of mind to be delving that deep atm when he's probably just trying to do precalc
heck whenever I was just given a theorem back then, I'd just understand it enough to stop myself from wasting my time over 60 questions long MCQ exam papers

[ancient grotto]

they've told me they're self studying so this approach won't make sense for them, I can see why they ask about these theorems here

[fallen heath]

Hmm ... Indeed

mb

[jaunty mural]

I think the thing we need rn

is an illustration

[fallen heath]

Yes

[jaunty mural]

I can try later if one of u dont

in the entire conversations today yesterday so far no ones given pictures lol. makes it hard imo

ig hard to draw on like desmos

[fallen heath]

Take them on VC :o

[wraith crypt]

in this case i don't even know what picture to draw

[gentle lintel]

i’ve no idea

[crimson sedge]

I don't think you have to .

Like you just need to show second condition is necessary

So give an example were second condition is false and because of that thm fails to work

[gentle lintel]

but are there other conditions

how do we show there are no other conditions?

[gentle lintel]

also, what if i insist a proof lol

[crimson sedge]

We can't and we don't know 😅

[gentle lintel]

there has to be a proof somewhere

[crimson sedge]

Like this may not be a perfect thm

[gentle lintel]

google gives calc 3 level proofs for mv limits

[wraith crypt]

well, yeah, we could always write a proof for the theorem, that's why it's a theorem

doesn't mean the theorem is the best possible, or there aren't other possible (maybe equivalent) conditions

[crimson sedge]

Yeah .Also sometimes it is better to take thms for granted untill you develop enough tools to prove it.

[gentle lintel]

so what does its proof require

beyond epsilon delta?

[wraith crypt]

at a glance it's just epsilon-delta, although i haven't actually tried writing it out on paper

[gentle lintel]

<@&286206848099549185>

proving why change of variables work

limits

[jaunty mural]

2 should be nsc on g

[gentle lintel]

wdym

[jaunty mural]

2 is a necessary and sufficient condition

on the function g

However, there are choices of f (continuous at the point) which make 2 unnecessary

This condition matters if f isn't continuous at the point

[gentle lintel]

i’m just concerned about proving this theorem

lol

not its conditions (which i’m sure will come up mid-proof)

[jaunty mural]

@gentle lintel u there?

[gentle lintel]

yes

[jaunty mural]

k i will try this visual representation

but its hard to do something like this with desmos

https://i.imgur.com/rpDjfMj.png

fg = blue 🔵
g = green 🟢

============

Green is like your x^2 sin (1/x) function (just the shape matters, not the actual function)

Blue will detect if green is 0 or not. If 0, output 0, if not 0, output something else.
Blue(0) = 0
Blue(not 0) = not 0

============

The numbers on the axes aren't relevant except for 0, more or less

ok?

[gentle lintel]

blue = fg you mean..?

[jaunty mural]

yes

Is that diagram ok?

[gentle lintel]

i guess?

[jaunty mural]

I'm then going to shift blue up (doesn't matter)

https://i.imgur.com/7ZO7jnp.mp4

Can you see that gif?

if you open it qual is better

[gentle lintel]

yea

it looks wacky

[jaunty mural]

Let's go back to the 'visual' definition of limit

======================
If lim f(x) as x -> p is L. THEN we have:

For any open interval Y around L, no matter how small.

You are able to choose an open interval X around p so that f(X) = {f(x) : x in X} is in Y.

If you require an open interval on the Y axis around 0 of whatever size

I can always pick an open interval on the X axis to contain the green graph

ok?

arghhh this is so hard to explain in words

[fallen heath]

O_O Imma save this chat as shuri lectures on limit visualization

[jaunty mural]

It would probably way easier

with a blackboard

cus u can point at the curve

@gentle lintel When you're back, you can think of discrete time as you're moving along the x-axis like this

$$f(x) = \begin{cases}0&x\neq0\1&x=0\end{cases}$$
$$g(x) = \begin{cases}x^2\sin\left(\frac{1}{x}\right)&x\neq0\0&x=0\end{cases}$$

[wraith dagger]

Shuri2060

[jaunty mural]

f(x -> 0) : [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, ...] (limit is 0)

g(x -> 0) : [1, 0, -1, -0.5, 0, 0.5, 0.25, 0, -0.25, 0, ...] (limit is 0)

(fg)(x -> 0) : [0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, ...] (limit is undefined)

f(x -> lim g) : [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, ...] (limit is 1)

========
The limit being L essentially means:
You give me an open interval on the y-axis around L
I have to find a cutoff point in the sequence where all the terms past it are contained in that interval

========

[jaunty mural]

Another way to think of it is if this is f

g determines how you approach the limit in the x axis

For a normal approach, you get arbritrarily close to x, but you don't actually get there.

However, if your g = x an unlimited number of times

Then your approach is going to continuously cross the point x, which is a problem here - your function will keep oscillating, and the limit won't exist.

Hence lim f(g) (and it won't be equal to the other way of evaluating the limit)

[gentle lintel]

yea i get all this

i guess

[gentle lintel]

it’s just this

[jaunty mural]

What I've shown is general.

It is quite a general argument for why condition 2 is necessary and sufficient if f is not continuous at that point.

[gentle lintel]

no thats not my focus anymore

[jaunty mural]

If f is continuous at that point, then it's clear you can do as the equation says.

Since there is no difference between taking the limit and just evaluating f

[gentle lintel]

i’m searching for a proof for why the change of variables work

[jaunty mural]

Your approaching method towards the x doesn't matter

as long as you are going towards it

That is what that theorem summarizes

[gentle lintel]

this is the intuitive way

i’m looking for a rigorous proof lol

[jaunty mural]

Then write down the epsilon delta defn's

for f and g

and it can be proven

[gentle lintel]

yea

[jaunty mural]

Triangle inequality is commonly involved in proving limit stuff (maybe not here)

[gentle lintel]

im not sure how to prove using epsilon delta

[jaunty mural]

breh

lim x-> a g(x) = b

So we write

$$(\forall\varepsilon_1\in\bR^+)(\exists\delta_1\in\bR^+)(\forall x\in\bR)(0<|x-a|<\delta_1\implies|g(x)-b|<\varepsilon_1)$$

If lim t->b f(t) = c, we write

$$(\forall\varepsilon_2\in\bR^+)(\exists\delta_2\in\bR^+)(\forall t\in\bR)(0<|t-b|<\delta_2\implies|f(t)-c|<\varepsilon_2)$$

[gentle lintel]

huh

[wraith dagger]

Shuri2060

[gentle lintel]

where is c

lol

[wraith dagger]

Shuri2060

[gentle lintel]

also, why epsilon1 epsilon2

and delta1 delta2

[jaunty mural]

If those statements are simultaneous

[gentle lintel]

just pick δ = min{delta_n}

[jaunty mural]

They are not necessarily the same

[gentle lintel]

size of epsilon doesn’t matter here either

[jaunty mural]

Then write the proof yourself lol

[gentle lintel]

idk just pointing things out

[jaunty mural]

It's not they don't matter

You justify they don't matter

There's nothing wrong with what I've written

As the starting statement

[gentle lintel]

it looks creepy but ok..?

[jaunty mural]

Now idk what's next, somehow you need to get t to be g(x)

I'm pretty sure you will want to set epsilon 1 = delta 2

$$(\exists\delta_1\in\bR^+)(\forall x\in\bR)(0<|x-a|<\delta_1\implies|g(x)-b|<\delta_2)$$

[wraith dagger]

Shuri2060

[jaunty mural]

By substitution, this is true.

[gentle lintel]

then t = g(x) then done?

[jaunty mural]

??????

hell no

[jaunty mural]

How do you make sense of 'for all g(x) in R'

[gentle lintel]

so can’t t = g(x) here?

[jaunty mural]

no???

$$\forall g(x)\in\bR$$

[wraith dagger]

Shuri2060

[jaunty mural]

What is this quantifier supposed to mean, exactly?

Think the next step is this

$$(\forall\varepsilon_2\in\bR^+)(\exists\delta_1,\delta_2\in\bR^+)(\forall x, t\in\bR)$$

$$[(0<|x-a|<\delta_1\implies|g(x)-b|<\delta_2) \land$$

$$(0<|t-b|<\delta_2\implies|f(t)-c|<\varepsilon_2)]$$

[gentle lintel]

uhh so what do we do

[wraith dagger]

Shuri2060

[jaunty mural]

This is all 1 line

no space with the bot

[gentle lintel]

alright

[jaunty mural]

Now we want this implication to go like this

We CAN say that

because according to condition 2

g(x) won't take on b

for epsilon small enough

Hence, we ARE able to combine statements

$$(\forall\varepsilon_2\in\bR^+)(\exists\delta_1,\delta_2\in\bR^+)(\forall x, t\in\bR)$$
$$(0<|x-a|<\delta_1\implies|f(g(x))-c|<\varepsilon_2)$$

[gentle lintel]

huh

[jaunty mural]

Then I chuck out unused quantifiers

Ah wait, I forgot to substitute

[gentle lintel]

confused starting here

[wraith dagger]

Shuri2060

[jaunty mural]

Ok let's talk here.

[gentle lintel]

why wouldn’t it work if g(x) was to take on b

at least in this algebraic sense

[jaunty mural]

Ok, the 3rd line

I want to use that implication

ok?

For t = g(x)

I cannot do this

With just the 2nd line

I only know that quantity is less than delta 2

I don't know it's greater than 0

However, I DO know it must be greater than 0 for sufficiently small epsilon_2

Because of condition 2

[gentle lintel]

which quantity