help-36

[chilly cloud]

Let $f_n$ be cauchy in $L^\infty$. Then for for all $\epsilon>0$, there exists $N(\epsilon)$ natural such that for all $n,m>N$, $||f_n-f_m||_\infty<\epsilon$

[soft zealot]

lilyenjoyer

[chilly cloud]

Carve out the bad set $E_n,m$ where this doesnt hold, and take the union on n and m, countable so it just doesnt hold on some measure 0 set

[soft zealot]

lilyenjoyer

[chilly cloud]

now work on complement

We have that ${f_n(x)}$ is cauchy (as a sequence of numbers)

[soft zealot]

lilyenjoyer

[mint orbit]

i dont think i understand this union

[chilly cloud]

now, let f(x) be the pointwise limit of f_n(x)

[mint orbit]

I mean what does this do to epsilon

we have no idea what the bound is anymore, do we?

or do we not care

[chilly cloud]

ok one thing at a time

the inequality doesn't hold on $E_n,m$ but this has measure 0

[soft zealot]

lilyenjoyer

[chilly cloud]

then, $\cup_n \cup_m E_n,m$ is still a countable union, so it has measure 0 as well

[soft zealot]

lilyenjoyer

[chilly cloud]

where n,m are naturals

[mint orbit]

no but i mean

what are the variables we are using N n and m

so for each epsilon, there is some N

[chilly cloud]

This is from definition of cauchy

yes

[mint orbit]

I know

but epsilon determines N

[chilly cloud]

yes

N(\epsilon) n depends on it

[mint orbit]

and N determines what m and n are allowed

[chilly cloud]

yes

and now we take a hammer

[mint orbit]

do these all have the same epsilon?

[chilly cloud]

and eliminate the set

where it doesnt hold

for all possible m,n

its measure 0, so it doesnt change anything

[mint orbit]

im worried about the set where it does hold

[chilly cloud]

ok wait

do u see a problem with anything i type so far ?

[mint orbit]

yes im asking about the bound

[chilly cloud]

which one

[mint orbit]

I don't understand how we can take a union and retain cauchy

[chilly cloud]

i havent done anything yet outside of state cauchy

Because the union is E_n,m

[mint orbit]

the union should destroy the bound

[chilly cloud]

is taken on a set of measure 0

[mint orbit]

I guess i dont understand what these sets are

I mean how one individual set is created

[chilly cloud]

Forget that part for now

[mint orbit]

We fix epsilon, N, m, n?

[chilly cloud]

I come back

to that part

[mint orbit]

its a big part lol

[chilly cloud]

yes

[mint orbit]

I don't get what you mean we destroy the place where bad things happen

I mean this union doesnt seem to accomplish that

it creates more problems

because now we cant talk about cauchy

but maybe we can and i just dont understand what the union is working on

[chilly cloud]

do you know what infinity norm is?

can you say how it is defined

for yo

you*

L^infinity norm

[mint orbit]

ugh maybe i should scribble

[mint orbit]

like in the context of a function? or as a metric?

[chilly cloud]

norm like

Does it look like this?

The way you have it

[mint orbit]

right

[chilly cloud]

yes

ok the important part here is

a.e. x

we ignore all sets of measure 0

right?

[mint orbit]

so when we say $||f_n -f_m|| > |f_n-f_m|$ this is fixed m n?

[soft zealot]

jan Niku

[chilly cloud]

yes if we qualify it that way

[mint orbit]

and then you say EXCEPT

on the set Zmn

[chilly cloud]

yes

but Z_mn is measure 0

[mint orbit]

so if you U {mn} Z{mn}

[chilly cloud]

sure

[mint orbit]

you get say Z

[chilly cloud]

we agree that this union is also measure 0?

[mint orbit]

yea

[chilly cloud]

and that for all m,n

[mint orbit]

so now you have Z

[chilly cloud]

outside of Z, the inequality holds

[mint orbit]

so if you exclude Z

then it doesnt matter really what m,n are

[chilly cloud]

exactly

[mint orbit]

because Z_mn is in there

okay

[chilly cloud]

thats the point

we want to work with infinitely many m,n later

so carving out Z is important

Can i move on

to the uniform convergence now?

[mint orbit]

sure, sorry

[chilly cloud]

its ok

its fun lol

[mint orbit]

this is the part that i was most confused on 😓

[chilly cloud]

i havent seen this in a while

oh

you get everything else?

or no

[mint orbit]

but i dont understand the rest either

[chilly cloud]

o

[mint orbit]

but this is what made me give up

[chilly cloud]

ok sure

[mint orbit]

and go home and eat burritos

for 1 hour

[chilly cloud]

SO now we're back to

f_n being cauchy

now, we fix x inside $Z^c$

[soft zealot]

lilyenjoyer

[mint orbit]

sure

[chilly cloud]

we have that $f_n(x)$ is a cauchy sequence

of real numbers

[soft zealot]

lilyenjoyer

[chilly cloud]

now, define $f(x)=lim_n f_n(x)$

[soft zealot]

lilyenjoyer

[mint orbit]

we dont know this exists

do we

[chilly cloud]

yes

reals are complete

[mint orbit]

cauchy isnt enough for convergence

[chilly cloud]

this is a sequence of real numbers

we fixed an x

[mint orbit]

its a cauchy sequence

[chilly cloud]

yes

[mint orbit]

but we dont know it converges

[chilly cloud]

cauchy implies complete

[mint orbit]

it may just be cauchy

[chilly cloud]

sorry

thats dumb typo

cauchy implies convergent

when you have a sequence of reals because R with standard metric is complete

[mint orbit]

I'd have to check this is true

[chilly cloud]

and this is a pointwise limit

[chilly cloud]

which part?

[mint orbit]

cauchy implies convergent

[chilly cloud]

ok wait

[mint orbit]

it doesnt seem like itd be true

[chilly cloud]

do we agree that

for sequences of real numbers

if a_n is cauchy, it is convergent?

[mint orbit]

I don't recall this as a fact it doesnt seem like itd be true

[chilly cloud]

wait

what

https://en.wikipedia.org/wiki/Completeness_of_the_real_numbers

this might help u

[mint orbit]

i mean it seems like okay log(n) would be one-

cauchy but not convergent

[chilly cloud]

no no

forget all functions

rn

[mint orbit]

but this is a counter example

[chilly cloud]

take a sequence of real numbers cuachy

there cannot be a counter example

to the reals being complete

[mint orbit]

log(n) is a sequence of real numbers

[chilly cloud]

its not cauchy

[mint orbit]

n=1,2,3,...

[chilly cloud]

log(n) is not cauchy

[mint orbit]

maybe i messed up my proof

i was arguing with someone about this earlier irl

[chilly cloud]

😭

i think

the most easy way to see this

is the cauchy construction fwiw

not the dedekind one, it hurts my brain

but its also in rudin i think, ch1 or appendix i forgot

but if you can accept that the reals are complete i can finish the proof of L^ infty complete

err just ping me if u want me to continue

illstill be on doing hw

😂

[mint orbit]

i need to check on what you are talking about

[chilly cloud]

sure sure

[mint orbit]

i dont believe it is true so i want to believe

[chilly cloud]

😭

ok wait

i can ask in the real complex channel LOL

[mint orbit]

im sure youre right

i dont belong in this class i dont know any analysis

[chilly cloud]

its ok lol

[mint orbit]

im just trying to do my best

[chilly cloud]

i dont mind im not an analyst no money in math

[chilly cloud]

ARE U A BELIEVER OR A NONBELIEVER 😭

[mint orbit]

sorry i am checking on things

[chilly cloud]

its ok!

https://en.wikipedia.org/wiki/Complete_metric_space THIS MIGHT HELP

the cauchy proof: https://math.stackexchange.com/questions/1186354/proving-the-usual-distance-metric-in-mathbbr-is-complete

[mint orbit]

always know im really reviewing some shit i shouldnt have forgot when author starts asserting stuff like 1>0

[chilly cloud]

its ok u got this

[mint orbit]

sorry im proving cauchy implies convergent

[chilly cloud]

well this is not true unless ur space is complete

but then its definition

the theorem is proving R is complete by construction

[mint orbit]

im not even sure what complete means tbh

but each thing in turn

[chilly cloud]

X is complete <=> (x_n cauchy => x_n converges)

for x_n sequence in X

[chilly cloud]

You cannot prove this in general

[mint orbit]

I'm trying to prove it for the reals

[chilly cloud]

oh ok

[mint orbit]

im stuck on a lemma

sorry

[chilly cloud]

its ok lol

ill be up for a while anyways doing hw

[chilly cloud]

did you get anywhere with it? no pressure btw its just the melatonin is kicking in sometime sooon

[mint orbit]

I proved cauchy is bounded but i'm trying to finish cauchy is convergent

i dont get the last line

[chilly cloud]

oh

it gets hairy

are you looking at the construction of R

from equivalence classes?

[mint orbit]

no im looking at a proof that cauchy implies convergent

[chilly cloud]

oh wait are you doing that thing

that divides in half nonstop

have fun lol

[mint orbit]

no im trying to follow this

im not doing anything cool

[chilly cloud]

oh

the proof i mentioned was of

bolzano weierstrass

[mint orbit]

no i cannot prove everything i dont understand

[chilly cloud]

ok yea lol mb

[mint orbit]

some stuff you just have to take at face value

[chilly cloud]

which line r u stuck on

yea

[mint orbit]

the second to last line

it looks like triangle ineq

but its not

|am - an| should be |am -ans -an + ans|?

but they inttroduce ell somehow

and remove a_n?

idk im probably like

i dont know enough analysis to complete this problem

[mint orbit]

idk assuming this is all true

is that the end of the proof you were talking about

i forgot where we even were now

[chilly cloud]

of which

uniform conv

uniform convergence?

yes it's nearing the end

if you take for granted that the reals are complete, you have it almost immediately

[mint orbit]

so we defined okay

we defined Z as the problem set

so on Z^c, if you fix some x in Z^c

then

[chilly cloud]

f_n(x) is cauchy, but these are just real numbers

so define f(x) = lim f_n(x)

[mint orbit]

then |fm - fn| < ||f_n - f_m||?

[chilly cloud]

uhhh

which metric do you have here?

the one induced by L_infty norm?

but i was gonna say

[mint orbit]

im trying to remember what we have on Z^c

its that if x is in $Z^c$ that $||f_n - f_m|| > |f_m - f_n|$ for all $m,n > N$

[soft zealot]

jan Niku

[chilly cloud]

we have that yes

[mint orbit]

okay

[chilly cloud]

then, we have an f

[mint orbit]

and this means that f_n is cauchy?

[chilly cloud]

yes

from that inequality

[mint orbit]

how do we know

oh i guess we assumed it was true

we dont have to show it

[chilly cloud]

Let e>0, choose n_0 such that ||f_m-f_n||<\e for all m,n>n_0

then we have the inequality so it is cauchy

then we define f(x) as the pointwise limit

[mint orbit]

what do you mean by inequality

[mint orbit]

youre talking about this?

[chilly cloud]

$|f_m(x)-f_n(x)|\leq ||f_m-f_n||_{\sup} <\epsilon$

[mint orbit]

did we have to do extra work for f_n to be cauchy

[chilly cloud]

no

its basically a given

[mint orbit]

but we had to do work to ensure it was true for this fixed x

okay

[soft zealot]

lilyenjoyer

[chilly cloud]

have you seen that

pointwise limit is measurable

then f is measurable

since $f(x):=\lim_n f_n(x)$

[soft zealot]

lilyenjoyer

[chilly cloud]

and we have that f(x) exists for all x since each f_n(x) is cauchy

[chilly cloud]

do you follow so far or no

[mint orbit]

im trying to tell if i know this fact or not

[chilly cloud]

o ok

[mint orbit]

but f is already measurable

[chilly cloud]

huh

[mint orbit]

or i guess this is just called f who the hell knows if thats what it means

im not sure why thats there

[chilly cloud]

we constructed such an f

but yes its measurable

its different from the f there

but tldr its nice either way

[mint orbit]

do we care how we know fn is measurable

or they just have to be

[chilly cloud]

f_n is in L^p

so its already measurable

by definition

[mint orbit]

but we havent shown completeness yet so we dont know that the pointwise limit is measurable

okay

[chilly cloud]

pointwise limit of measurables is measurables fwiw

[mint orbit]

im sure its true

[chilly cloud]

but to continue

[mint orbit]

okay so its convergent

and its measurable

so all thats left to show is that its essentially bounded

[chilly cloud]

yes so now take limit

from earlier: Let $\epsilon>0$, choose N so that when $n,m\geq N$, this holds $|f_m(x)-f_n(x)|\leq \epsilon$

[soft zealot]

lilyenjoyer

[chilly cloud]

then as $n\to\infty$ you have $|f_m(x)-f(x)|\leq \epsilon$

[soft zealot]

lilyenjoyer

[mint orbit]

hmmm

im not sure i follow

i may stop for tonight

[chilly cloud]

sorry im bad at explaining im just looking at my old lecture notes

but after this theres 1 line and ur done

[mint orbit]

its not your fault i just dont understand anything in analysis

[chilly cloud]

well, 2

[mint orbit]

it doesnt make sense why this would still be less than epsilon

we know its true for all n but why wont something else happen in the limit

[chilly cloud]

by construction ig

sorry thats not satisfying idk

[mint orbit]

i guess i will stop

or maybe im curious

@chilly cloud whats the last 2 lines

u dont have to tell me im not just gonna steal the answer

[chilly cloud]

its ok its fast lmao

[mint orbit]

in fact im gonna start over from the beginning as soon as you go to bed

[chilly cloud]

wel we have that

f_m converges to f in L^infty norm

[mint orbit]

?

[chilly cloud]

then triangle inequality

oh cause after we take

n to infinity

we have that for $m\geq N$, $||f_m(x)-f(x)||_{\infty}\leq epsilon$

[soft zealot]

lilyenjoyer

[chilly cloud]

earlier, we had this

but we have it true for all x in Z^c

so now we pass back to sup (infty norm) that way

and this is the L^infty norm so by definition we have that f_m goes to f in L^infty norm

[mint orbit]

and this is enough for convergence?

[chilly cloud]

yea its convergence in norm

[mint orbit]

but this isnt everything

[chilly cloud]

yea theres a short step after

[mint orbit]

we didnt show f is essentially bounded

also didnt we already show that the cauchy sequence was convergent to f

[chilly cloud]

yea its the last step

[chilly cloud]

wdym

[mint orbit]

wway back here

[chilly cloud]

we have this from

[mint orbit]

whyd we do all this extra stuff

[chilly cloud]

its to get that our f

is in L^infty

i.e. essentially bounded

[mint orbit]

but it doesnt show that

[chilly cloud]

the last step being triangle inequality

$||f||\infty \leq ||f-f_m|| + ||f_m||\infty + \epsilon < \infty$

$1$

[soft zealot]

lilyenjoyer

[chilly cloud]

test

[mint orbit]

you have a space after your final symbol

between $

[soft zealot]

lilyenjoyer

[mint orbit]

i think i get it

nm(f) <= nm(f-fm) + nm(fm) < eps + nm(f_m)

[chilly cloud]

ugh typos

[mint orbit]

but fm is essentially bounded

[chilly cloud]

sorry im on old laptop

yes

[mint orbit]

so f is essentially bounded

[chilly cloud]

yes

[mint orbit]

at least i understand the last step

maybe i will work backwards then

[chilly cloud]

the earlier step is the 'hard' one ig

where you pass to the reals

[mint orbit]

i still do not understand tbh

[chilly cloud]

to show that f(x) indeed does exist

[mint orbit]

i dont understand how we can use the reals here

but office hours tomorrow

[chilly cloud]

ask ur prof to explain how

the L^p completedness

[mint orbit]

i already know what she will say

[chilly cloud]

depends on the completedness of reals

[mint orbit]

she will say

[chilly cloud]

its key ingredient

[mint orbit]

you should review your analysis

[chilly cloud]

[mint orbit]

lol

sorry to keep you up mirin

[chilly cloud]

its ok

[mint orbit]

but

[chilly cloud]

my class not til 2

only midnight

[mint orbit]

u know theres a helpers vc

if u become helpful

[chilly cloud]

oh

[mint orbit]

we could pomo

[chilly cloud]

i only helped cuz i saw ur name

😭

sry

[mint orbit]

im in my last month of school lol

[chilly cloud]

i dont help normally lmao

[mint orbit]

ah

well

[chilly cloud]

oh me too

dang

[mint orbit]

thats okay

[chilly cloud]

[mint orbit]

we will always have the memories

[chilly cloud]

last 2 months for me but

[mint orbit]

[chilly cloud]

onto the real world ig 😔

[pliant shore]

Ay congrats on graduating

[mint orbit]

its not certain yet

each day it looks bleaker

but day at a time

[chilly cloud]

job hunt

[mint orbit]

yea that too

i mean

i have essentially no time

but thats okay

good luck on your finals if i dont see u mirin

💪

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[mint orbit]

@chilly cloud and thank you again

[chilly cloud]

NP HAVe fun

[paper timber]

hi guys

[mint orbit]

the channel will suddenly lock soon

just so you know

[paper timber]

I was wondering, i Send pics of problems in the chat maybe someone could give me the answers?

[paper timber]

Ohh okay

[mint orbit]

#❓how-to-get-help

get an open channel

[paper timber]

Ok, thank you

[final saddle]

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[neat flint]

someone mind checking my work?

[neat flint]

or would it be better to put -0.5

[marsh temple]

Your answer is fine, whether decimal is better depends on the software

[final saddle]

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[normal juniper]

can someone please help me with this

[muted wraith]

divide both lhs and rhs with sqroot 2

so you have 1/root2 + i/root2 on rhs

convert that to cis form

[normal juniper]

alr

[normal juniper]

1cis(pi/4)

[muted wraith]

yep

the general form is cis(2pi(k)+ pi/4)

then apply nth root on both sides

[normal juniper]

wait why though

if it was roots of unity I get the 2pi(k)

[muted wraith]

that is the general solution

pi/4, 2pi+pi/4,... all give the same value

[normal juniper]

oh ok

oh yeah that makes sense

[normal juniper]

so the answer is B?

[muted wraith]

noo

try again

[normal juniper]

wait what

oh shit

[normal juniper]

is it C??

coz then the modulus will be root 2

but from the original thing it is 1/root2 + i/root2

so when it multiplies it's going to become 1 + i

[muted wraith]

(cistheta)^1/n is cis(theta/n)

[normal juniper]

ooooohhh

[normal juniper]

D

[muted wraith]

yeah

[normal juniper]

ok that makes sense

thakns for the help

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[neon slate]

these definitions are from different books and one is about Rn and the other about complex numbers, but seems like either one could be used for the other set.

I was just wondering, would every accumulation point by these definitions also be a limit point and vice versa? My brain is new to thinking about sequences of points so not sure if i have the right intuition on this but seems like if you have an accumulation point you could pick successively smaller balls around it and picking points out of them to come up with a sequence of points that converges to that point. and i think also that if you have a limit point, you would need an infinite number of points in every ball around it since for at any point in the sequence, you would have an infinite number of upcoming elements within any desired radius of the limit point.

[trail mango]

yes

[neon slate]

😄

oops pinged wrong person lol

[trail mango]

nice ping

[neon slate]

so theyre equivalent?

[trail mango]

yes and your explanation is right

[neon slate]

i find the definition of accumulation points way more intuitive

what would be motivation to think of limit points in terms of sequences?

[trail mango]

not sure. sequences are nice to work with tho

[neon slate]

or wait, in definition of limit point, do all the points need to be distinct?

[trail mango]

no

[neon slate]

someone on stack exchange was saying this

[neon slate]

oh hm

i guess theyre using a diff definition

[trail mango]

{1} doesn’t have any limit points or accumulation points with your definitions

[neon slate]

right

[neon slate]

but you would need to end up with an infinite number of distinct points in the sequence right

cause otherwise if there were only a finite number, you could figure out the closest point's distance from the supposed limit point and there'd be no more points in the sequence if you pick epsilon < that i guess?

[trail mango]

yea

[neon slate]

so when youre building these sequences of points, youre basically stringing points one after another in whatever random fashion you want, the sequence of points converges if for any epsilon > 0 you can find some N so that after N all the points in the sequence afterward are within < epsilon distance from L right?

[trail mango]

you could say if z is a limit point of S then there is a sequence in S converging to z with all elements distinct and not equal to z

but there will also be sequences that don’t have all elements distinct

[neon slate]

yeah i think that makes sense, cause you could just eliminate duplicates from whatever existing sequence i guess?

[trail mango]

yea

[trail mango]

building for what?

[neon slate]

oh uh

i dunno if thats the right word, just imagining these sequences, and basically was just having a bit of trouble with what a sequence of points even is, but i guess just literally any list of points lol

that goes on forever

and convergent ones i guess are special cause they become bounded by their distance to something?

but it could literally just be random points forever right? 😄

[trail mango]

yea. a way to look at what a sequence (in S) is, is that it’s a function from the positive integers to S

[neon slate]

righto

btw layla

today i fell asleep at 8pm

and woke up at 2 lol

guess i was just tired

[trail mango]

WTF

#「helpers-lounge」 message

[neon slate]

thx for help btw that was helpful ^^

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[trail mango]

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[past bronze]

can sshouldnt it be D

[zealous reef]

what seems to be the problem

[past bronze]

its decreasing then increasing

so point of inflectio

i learned is f'(x) is under the x axis, then f(x) was decreasing

[zealous reef]

how would you define point of inflection

[past bronze]

concave up to concave down

or from negative to positive

[zealous reef]

concave up to concave down or vice verse doesnt mean the slope goes from negative to positive

[final saddle]

@past bronze Has your question been resolved?

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[tired snow]

Help?

[final saddle]

@tired snow Has your question been resolved?

[static plinth]

so since the rod is in equilibrium, we can choose our pivot point anywhere to make it more convenient

i would suggest putting the pivot point on point A, so the normal force at point A would be canceled out

actually your last line is very close to the correct answer

double check the lever arm you’re multiplying by for R_p

[final saddle]

@tired snow Has your question been resolved?

[tired snow]

I dont know the distance A-P

[static plinth]

we can use trig for that

[tired snow]

Wait

[static plinth]

also be careful that it is 45cm not 45m

[tired snow]

Ou thanks

This?

[static plinth]

yep

but we want the hypotenuse here

[tired snow]

Yeah

[static plinth]

actually you kinda labeled the adjacent side wrong as well

[tired snow]

Why

[static plinth]

you can only do that when the hypotenuse is given

[tired snow]

Oh

[static plinth]

since cos(15)=A/H

but we don’t know H

we only know O

[tired snow]

I need 2 sides

[static plinth]

so we have to relate sine with the opposite and hypotenuse

and solve for the hypotenuse

[tired snow]

How do I do that

[static plinth]

$\sin(15)=\frac{.45m}{h}$

[soft zealot]

y0shi

[tired snow]

Only 1 side is given

[static plinth]

how do we solve for h

[tired snow]

Oh that

I thought something else

[static plinth]

recall that sine relates the opposite and the hypotenuse

oh

were you going to use tangent?

to find the adjacent

and then use pythagorean to solve for hypotenuse?

[tired snow]

Messed up my bad

[static plinth]

alright all good

[tired snow]

Yeah its 25 now (shown)

As for part b if we do upward forces equal downward forces?

[static plinth]

we can

as long as you make sure to split up R_p into components

or we can use torque again

and place our pivot point elsewhere

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[rigid plinth]

Is there supposed to be 2 sets of answers to this q? The answer key says its only r=8, theta=1/4

[languid yoke]

Well r=1 and theta=16 doesn't really make sense

since if r=1 the area of the full circle is pi which is less than 8

Also keep in mind that theta won't be higher than 2pi.

[rigid plinth]

Oh good point, was just at a loss cz when i plugged in the values it was equal to 8

Thanks

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[wispy trellis]

What is name of this formula/property??

[vapid basalt]

pythagorean identity

the proof is pretty simple

[wispy trellis]

Same like this what is for tan cot sec and cosec?

[vapid basalt]

those are other trig identities

unrelated to this one

[wispy trellis]

Okay thanks

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[heavy grove]

I determined a is 1 but that’s it

[heavy grove]

11

I tried solving for the horizontal asymptote of came up empty handed

[vapid basalt]

for c you have 2 possibilities

[heavy grove]

Yep

[vapid basalt]

either the function is undefined at that point or you have a global extremum at -1

[heavy grove]

I tried lhopital because I think it can apply here?? I got 1/2bx+c and so 2bx + c = 1 and if they can only be 1 and 0, c must be 1, and b must be 0.

Is that right?

[heavy grove]

What point

[vapid basalt]

i had the wrong idea lmao

my bad

i need to sketch it gimme a sec

[heavy grove]

I think I need to solve for the horizontal asymptote and compare it?

But when I solve it normally it’s 0

Ignore x->0 that’s supposed to be inf

[vapid basalt]

i have a better idea

first reasoning for why a=0

[heavy grove]

Why would a be 0?

It’s 1

[vapid basalt]

i meant b sorry

[neon slate]

yeah that seems correct, you can lhopital the limit as x -> inf and you need that limit to equal 1

[heavy grove]

So how can there be two of them there?

0 and 1

[neon slate]

so 1/(2b+c) = 1
well b is either 1 or 0, but if b is 1 then this fraction is < 1 so b = 0

[heavy grove]

Right

And c must be 1

[neon slate]

wait no 😄

[vapid basalt]

if b =1 it would have a vertical asymptote

[neon slate]

its 1/ (2bx+c) i guess

[heavy grove]

Yes

But what about when I take it normally

[neon slate]

so that limit would be 0 as x -> inf if x isn't 0

[heavy grove]

By dividing by x^2

I get 0

Why don’t I get 1?

[neon slate]

take what normally?

[heavy grove]

The horizontal asymptote

Solving for it normally it’s 0

[neon slate]

what does "take the asymptote normally" mean?

[heavy grove]

A different method of

[neon slate]

i am not understanding what youre trying to say

[heavy grove]

Ig

[heavy grove]

This

[neon slate]

so x-> inf?

[heavy grove]

Yes

The top becomes 0

And I can’t make it 1

[neon slate]

what do you mean becomes 0

as x-> inf, top -> inf

[heavy grove]

It approaches 0

[heavy grove]

Yes, but dividing top and bottom by x^2 it approaches 0

That’s the way I was taught

[neon slate]

wut

[heavy grove]

Wdym

[vapid basalt]

@heavy grove you sure you took the limit correctly?

[neon slate]

but then both top and bottom approach 0

so youre back where you started

[heavy grove]

No?

0/b

[neon slate]

and how do you know b isn't 0?

as it turns out thats the case

[heavy grove]

Mmm

That’s a good point

So is lhopital the proper method here

[neon slate]

yea

[vapid basalt]

yeah

[heavy grove]

Is that probably what the wanted

[vapid basalt]

not explicitly

[heavy grove]

What other method was there

[vapid basalt]

no this is the method but i assume they want you to show that you know why it works

ie you approach a point with the slope 0

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[final saddle]

@fierce osprey Has your question been resolved?

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@fierce osprey Has your question been resolved?

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@fierce osprey Has your question been resolved?

[muted wolf]

what's the doubt in this actually?

u already found out the radius and height for the minimum cost, u just need to substitute the radius and height in the expression of their respective areas and find out the area and multiply with the cost of the material corresponding to the area, tada u found out the minimum cost

[fierce osprey]

how do I do the minimum accuracy

[muted wolf]

here

[fierce osprey]

thx

[muted wolf]

[fierce osprey]

!close

[muted wolf]

.close

ig

[fierce osprey]

.close

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[magic loom]

i have a theory question

[magic loom]

this is sin(x) with domain [0, 2pi]

if you calculate the area from 0 to pi it's 2

but the area from pi to 2pi is -2

but that doesn't rlly make sense does it

cuz it's an area

[formal trail]

integrals don't give the area, they give a "signed area", which says that area above the x-axis is positive and area below the x-axis is negative

[magic loom]

if my goal is to calculate the area then should i just put a negative before the integral sign

like this

O(W) just means surface

[formal trail]

if you have a function that is negative for an entire interval, then putting a negative in front will give a positive area

[magic loom]

yeah

so i think in this case adding the negative would give the correct area

since from pi to 2pi it's completely under the x-axis

[jolly tapir]

well just enclose it in absolute value then

[magic loom]

my book isn't a fan of that so i asked about the adding negative thing instead

[jolly tapir]

taking the absolute value works for all f(x) when you want to take the absolute area

as opposed to - which only works for this instance

[magic loom]

yeah

idk why they won't do that in the book

maybe it'll come later

i'll keep it in mind

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[formal trail]

when we do absolute value, it's not practical to integrate. so we use the definition of absolute value [ \abs{f(x)} = \begin{cases} +f(x), & f(x) \ge 0 \ -f(x), & f(x) \le 0 \end{cases} ] which comes out to the same approach

[soft zealot]

𝕔loud

[magic loom]

thanks a lot

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[normal juniper]

can someone please help me with this question

[normal juniper]

I'm pretty sure it involves using modulo but i'm not sure how

[hollow iron]

Try squaring both sides

[normal juniper]

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

[hollow iron]

No

[normal juniper]

wat

[hollow iron]

Try again and do it carefully

[normal juniper]

ohh crap

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

[hollow iron]

Yup

So what does x^2 + 1/x^2 equal to?

[normal juniper]

-1

[languid whale]

I might suggest multiplying x on both sides of the equation and get a quadratic which you can solve.

[hollow iron]

No don't do that

Alright so x^2 + 1/x^2 = -1

Try squaring both sides again..

You might notice something

[normal juniper]

ok so I assume it always equals to -1

[hollow iron]

Well not always, but for some powers it does

Specifically, powers of 2

[normal juniper]

right

so I find the power of 2 thats closest to 2016

?

1024 is the closest power

[brave shoal]

2048 is closer, is it not?

[hollow iron]

Yes it is

[normal juniper]

oh thats true

i thought it couldnt be above

so what now 🤔

[hollow iron]

Hmm

Actually x is just a third root of unity is it not

[normal juniper]

wait why

[hollow iron]

x + 1/x = -1
x^2 + x + 1 = 0

[normal juniper]

where'd the 1/x^2 go

[hollow iron]

What do you mean

Forgot the squaring thing, there's a simpler way

x^2 + x + 1 = 0 implies x^3 = 1

[normal juniper]

(x+ 1/x)^2 = x^2 + 2 + 1/x^2

[hollow iron]

You can multiply both sides by (x - 1)

Yeah forget that

[normal juniper]

alr

[hollow iron]

So x^3 = 1 and you can easily use that to find the value of the expression you want

[neon slate]

?

[normal juniper]

how do you get that though

[hollow iron]

.

The solutions to x^2 + x + 1 = 0 are two of the three solutions to x^3 - 1 = 0

The only one missing is x = 1

[normal juniper]

ah ok

[neon slate]

...no kidding you're introducing that solution by multiplying by (x-1) on both sides, but what does that have to do with the problem at all?

[hollow iron]

Because x^2016 = (x^3)^(672)

[normal juniper]

so the answer is a?

oh wait its 2

so b

[hollow iron]

Yes

[normal juniper]

oh ok

I just did the multiply both sides by (x-1) on paper and it makes sense

Thanks for the help

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[undone idol]

help

[lucid marsh]

pi/2 +(pi/4-pi/6)

If i understand the question correctly

[undone idol]

why is this?

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[faint condor]

Could I get some hints on how to approach this, I’ve got the two equations by themselves in vector form but don’t know how to go any further

[terse siren]

try to write the equation of the line in terms of xyz

[faint condor]

How do you do that?

Just a hint please

[tiny gorge]

your line equations have three rows
one for each of x, y and z

[terse siren]

yes

x = -3 + 7 * lambda for example

[faint condor]

Okay, thank you

[terse siren]

now i'd try writing the equation as x + ay + bz = 0

making it independent from lambda

[faint condor]

Got it @terse siren thanks for your help

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[terse siren]

np

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[tame thicket]

Can someone help me with this contest challenge?

[tame thicket]

I have finished this problem and I got it wrong. Can someone tell why?

I think the error can not come from the solving equations part because I have double checked

Probably the problem is in establishing the equations

[zenith pollen]

yea the equations get solved correctly, and I think the 5 degrees might be 30 degrees instead, if you mean the angle between numbers on a clock

[tame thicket]

Oh yes, the hour hand moves 360/12 = 30 each hour

Now the new system is
270 + x/2 = 6x + 360t - 360
6x = 270 + x/2 + 30t
solving I got t = 12/13

yes this is indeed the correct answer

thank you for help

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[hoary oxide]

yo

[hoary oxide]

I thought in symbolic logic, "AND" has precedence over "Implications"

[lucid marsh]

Thats why they put the brackets there

[hoary oxide]

so shouldn't the AND execute first (thus implicity parantheses over the Fat and Happy), and then the conditional?

[lucid marsh]

To emphasize why this is wrong

[hoary oxide]

OHHHH I missed those

thank u hired

< 3

.closed

wut

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[magic sparrow]

@hoary oxide hello

[hoary oxide]

omg Austin

HEYYYY

[hoary oxide]

I am becoming an expert in symbolic logic 😎

[magic sparrow]

Very nice

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[rugged tide]

i'm at a loss here

[static fractal]

are you asking how the integral on line 2 evaluates to that?

[rugged tide]

no I drew an arrow with ??? on what i'mc onfused on

like what next

[iron dagger]

you isolate y

what is the inverse of the e function?

[static fractal]

move the minus to the other side and take logs

[lucid marsh]

Negate both sides and take the ln

[rugged tide]

ik isolate y but how do I do the thing to do that

[tawdry storm]

logs are the "opposite" of exponents

ln is base e

[rugged tide]

ohh

[tawdry storm]

Just remember, what you do to one side, you have to do to the other

[iron dagger]

more specifically the inverse of e, meaning that taking ln(e^x) = x

[rugged tide]

oh

Thanks all

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[vital spruce]

just wanted to verify since not much solved examples online, if i want to solve the tsp, can i define first two points as farthest point from middle as first point and closest point to that as second one?

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[mortal horizon]

Can someone check my logic? I think it’s 21, since for the number of distinct digits in this type of palindrome there are 2 options, so 2^n, but we have to subtract 2 since it can’t be all 8 or all 9. 2^11 - 2 = 2046, and 21 is the smallest length palindrome with 11 distinct digits

[zenith pollen]

that's right

[mortal horizon]

Ty

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[wanton spindle]

Stuck on part 1, I can do part 2 by reduction of order but I need an existing solution which I am having trouble coming up with from part 1

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[wanton spindle]

<@&286206848099549185>

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[final saddle]

Please don't occupy multiple help channels.

[sinful fox]

a+b = 15

a-b= 4

a(radius of larger circle) b(radius of smaller circle)

than find the both radius

calculate their areas and substract them from rectangles area

[restive copper]

will be worked like a normal equation then?

[sinful fox]

yes

[restive copper]

I worked it now however I got confused cause since there's no - on neither 15 or 4 it become 19

[pliant shore]

Yeah, so 2a = 19

Do you see how that happened?

[restive copper]

ok i gotta sleep

[pliant shore]

Cool ok have a nice night

[restive copper]

im missing so many obvious steps bruh

[sinful fox]

have a nice night

[restive copper]

i meant i need rest

but yeah i understand it now

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[slim saddle]

how many ways are there of choosing 3 different numbers in increasing order from the numbers 1,2,3,4,...,10 so that no two of the numbers are consecutive?

[trail mango]

there might be a better way but i have a stupid suggestion which is to count the number of choices that don't include 10 first

and to translate it to a combinatorics problem: I THINK choices that don't include 10 correspond to ways to place down two-length "blocks" onto the list 1,2,3,...,10

[vital surge]

this q is stars and bars problem so

[trail mango]

well it already is a combinatorics problem

[slim saddle]

wait wat

what r two-length blocks?

[trail mango]

this corresponds to the choice 1,4,6

[slim saddle]

as in the numbers 1,4,6?

[trail mango]

yes

[slim saddle]

isn't that consecutive tho?

[trail mango]

1 and 4 and 6 are not consecutive

the blocks are what prevent the choices from being consecutive

[slim saddle]

right

so what do i do from there

[trail mango]

it is not too hard to count the ways you can place the blocks but i don't recommend doing it like this on second thought

[ancient basalt]

to make the first choice we can choose from 1-6 i guess

[slim saddle]

mhm

[ancient basalt]

i have no clue how to go further

maybe if we choose 1 then we can choose the second number from 3-10

if we choose 2 then 4-10

(i have no clue if this will even work)

[trail mango]

if you want to count cases like that you should use a visual aid like this

that was going to be my not so nice way of doing it

[slim saddle]

wait so is the # ways to place the blocks 9 ways

[trail mango]

should be more than that

[slim saddle]

can the blocks be separate? i.e. (1,3)

[trail mango]

(1,3)?

[slim saddle]

like place 1 and 3 together as one block

[ancient basalt]

hmm alright

[slim saddle]

im like actually lost

<@&286206848099549185>

[ancient basalt]

is this an olympiad question?

[slim saddle]

k just found the solution it's 10C3-9C1*8+8

[slim saddle]

nope its a practise test question

[ancient basalt]

oh alr

[slim saddle]

👍

[trail mango]

that's (10-3+1) choose 3

it seems they did not count it very elegantly if they wrote the answer as 10C3-9C1*8+8

[slim saddle]

oh-

[slim saddle]

so its just 8C3? Why?

[trail mango]

https://math.stackexchange.com/questions/20132/how-many-ways-can-r-nonconsecutive-integers-be-chosen-from-the-first-n-integ there is an explanation here it's a bit complicated tho

[final saddle]

@slim saddle Has your question been resolved?

[final saddle]

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[delicate fiber]

Im getting 1/e^2 when it should be 1/e, anyone see a mistake?

whole q btw

[misty torrent]

As you multiplied by 2 in this equation both sides

you wrote integral 0 to pi directly because of part ii

but for right side it is 2 * integral 0 to pi/2 right?

And you did..

@delicate fiber

[delicate fiber]

Ohhh crap

Thats why

Thanks

.close

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[final saddle]

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[glass ember]

hello

[glass ember]

if riding a toy car for A minutes cost B dollars. How many mintues would you be able to ride for C dollars?

[robust fulcrum]

see

[pliant shore]

The assumption there is that it's linear

[robust fulcrum]

do this with unitary method