help-38

[kindred jungle]

d = a?

because then its less than or equal to a

[willow urchin]

from two numbers how can we pick a number smaller than both?

[kindred jungle]

but then what if its bigger than b

[kindred jungle]

subtracting them?

MIN

MIN

MNIN

d = min(a,b)

[willow urchin]

lol

[kindred jungle]

im so smart

is that right

[willow urchin]

ya

ok we want d<=sqrt(e/6) and d<=e/2

[kindred jungle]

so d = min(sqrt(e/6), e/2)

[willow urchin]

ya

scratchwork done 🙂

[kindred jungle]

wait but

thats different from what my teacher got

[willow urchin]

their stuff looks annoying af

i saw discriminant and noped out

[kindred jungle]

yeah 😭😭😭

so like

would he still mark this correct?

[willow urchin]

u need to reformat the scratchwork as the proof

and if every step is clear there should be no prob

if they take points off let me yell at em 🙂

[kindred jungle]

HAHAHAH fr

hang on lemme see if i can

write it

cuz i still dont know how to lay it out correct

ly

also this answer i got here

i learnt that from youtube

and thats like another different answer

ive been soo confused

[willow urchin]

i hate starting at ..<e bc its so inflexible

[kindred jungle]

is this okay?

[kindred jungle]

ahh i see lol

[willow urchin]

this is good scratchwork but its not the proof

[kindred jungle]

oh

wait lmme see if i can wrtie the proof then

[willow urchin]

the proof format is

let e>0

pick d=…

if 0<|x|<d then

(chain with justification on each line that needs it)

[kindred jungle]

what ab in this case

where delta is the minimum of the two

[willow urchin]

u need to justify the step ..+..<e/2+e/2 using that

[kindred jungle]

can we not do like

[willow urchin]

its like

[kindred jungle]

"suppose sqrt(e/6) < e/2"

and then the other way round

[willow urchin]

no

[kindred jungle]

oh

[willow urchin]

d=min so d<=.. and d<=.. so ..<e/2 and ..<e/2

[kindred jungle]

but we've chosen d = min(sqrt(e/6), e/2) so how can delta be less than/equal to either if we're choosing delta as being the minimum of the two

is it possible for u to write me the proof part so i have an idea

[willow urchin]

didnt we do this already

if d=min(a,b) then d<=a and d<=b

[kindred jungle]

yeah but

we chose d to be the minimum of a and b

so like

if d = min(a,b)

and a<b

then d=a

if b<a

then d = b

no?

[willow urchin]

theres no need to think of cases…

[willow urchin]

just follow this logic

[kindred jungle]

okayy

so

[willow urchin]

u just have to fill in the ..

[kindred jungle]

what do i write after this?

even thsi

[willow urchin]

just paste the chain

[kindred jungle]

ohh ok

so

[willow urchin]

reread our convo to piece it together

[kindred jungle]

like this?

oh wait lemme js

like this?

[willow urchin]

perf

[willow urchin]

fixed

[kindred jungle]

so i dont need the bit after 0<|x|<d

[willow urchin]

no bc thats what u wanna prove using the chain

[kindred jungle]

rightt because we assume |x| < d and we try to prove

so wait

all of this stuff

this is js workign out

and its wroth no marks

[willow urchin]

scratchwork

[kindred jungle]

but the proof itself has all the marks?

so why dont we do bprp's method

hang on let me write it

[willow urchin]

e d be funky like that

if the problem is prove limit then we need the formal proof format

if its just “find good d” then scratchwork is fine

[kindred jungle]

bprp does it like this

then he's like "what do you do to turn delta into epsilon"

and you choose delta = e/3

and he writes the chosen delta after lol

which iw as like

it feels like cheating 😭

[willow urchin]

well hes doing scratchwork in a way so he can easily make it formal later

but thats impossible to do for any remotely harder limit

[kindred jungle]

ohhh fair enough

also EVERYONE ive seen on youtube does quadratic limits like

like this (if you ignore the bad working out)

[willow urchin]

ik some ppl like breaking down absolute value but i hate it

[kindred jungle]

ohh i see

but if we choose delta to be js about anything why dont we choose a really big delta

or a really small delta

whatever suits

because that doesnt prove "approach"

rgiht

thats honestly so weird

[willow urchin]

delta must be small enough so ..<e happens

thats the whole point of e-d

[kindred jungle]

righttt

im gonna try a square root limit and see how i go

its hte next q

and this is how mr professor did it

and heres what i got

completely different answers

[willow urchin]

@kindred jungle looks good except in the proof id show the multiplying by “1”

but yeah see how easy life is if u dont break down absolute values?

[kindred jungle]

HAHAHAH YEAH 😭

thanks man

wait also

oh hang on thatd never happen

mmm 1 sec

[kindred jungle]

what do you do after this?

[willow urchin]

$<3\delta/|x-2|$

[solid kiln]

ロケットジャンプ

[willow urchin]

u must find lower bound of |x-2|

[kindred jungle]

does that mean

|x-3| < d
-d < x-3 < d
-d+1 < x-3 <d+1
so lower bound is -d+1?

[willow urchin]

recall my hatred of breaking absolute val

we want |x-2|>=..

hint, reverse triangle ineq

[trim jolt]

@kindred jungle Has your question been resolved?

[kindred jungle]

Ooo

[kindred jungle]

What's reverse triangle ineq😭

|x-2| >= ||x| - 2|?

That means

[willow urchin]

gotta be smarter

[kindred jungle]

||x| - 2| <= |x-2|

[willow urchin]

|(x-3)+1|

[kindred jungle]

||x-3| - 1| <= |x-2|

And |x-3| < d

So ||x-3| - 1| <= |d-1|?

I'm confused😭

[willow urchin]

its ur first time so ill do it

[kindred jungle]

Okk

[kindred jungle]

Why is ||x-3| - 1| = 1-d?

[trim jolt]

Channel closed

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Use .reopen if this was a mistake.

[kindred jungle]

.reopen

[trim jolt]

✅

[willow urchin]

oh should be >

@kindred jungle

[kindred jungle]

Why is that the case tho?

[willow urchin]

if d<=1

then |x-3|<d<=1

[kindred jungle]

|x-3|-1 < d-1 <= 0

||x-3|-1| < |d-1| <= 0

Then what

[willow urchin]

overcomplicated

gotta go, make sure u totally get this before continuing

[kindred jungle]

Okkk

I'll think abt it lol

Thankss

[willow urchin]

@kindred jungle ima take a sec before i go

i was dumb

ah wait i thought my new thought is easier but it doesnt work

stick to this one

ok peace

[kindred jungle]

HAHAH okk

tyy

[trim jolt]

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[thorny epoch]

after the thinking I did I have it figured out that one must make one of those kinds of pattern of summation where the other term gets canceled till there are 2 terms left (supposedly first and last)

[dull pilot]

SHEESH

[trim jolt]

@thorny epoch Has your question been resolved?

[thorny epoch]

<@&286206848099549185> rescue me 🥀

[proud wave]

I can't see C

[proud wave]

Is C) 4?

Coy Idk what C choice says. Can you retake?

[thorny epoch]

c) 2 d) 4

[thorny epoch]

the answer is not 4

answer : ||2||

[proud wave]

Uhh welp

[simple panther]

Here's my solution:

Take $\frac{k\ k!}{2^k}$ common out

[solid kiln]

Suika

[simple panther]

Numerator:

$k(k+1)k!=k(k+1)!=((k+2)-2)(k+1)!=(k+2)!-2(k+1)!$

[solid kiln]

Suika

[simple panther]

Take out the LCM:

$\frac{(k+2)!}{2^k}-\frac{(k+1)!}{2^{k-1}}$

[solid kiln]

Suika

[simple panther]

Sum from k = 1 to 2010:

$\frac{2012!}{2^{2010}}-\frac{2!}{2^0}$

[solid kiln]

Suika

[simple panther]

Substituting this summation value in, we get 2!/1 = 2

[simple panther]

Take out the LCM, more like: split the fraction

[dapper swift]

exactly and the name for this in English is a telescoping sum/series

you can collapse something like this so that you are only left with the two ends

[dapper swift]

next time don't give full solutions until the person has shown some working

[simple panther]

Okay

[trim jolt]

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[idle parcel]

Need help with 3 statics (physics) questions

[idle parcel]

pls help

.close

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[covert path]

I understand that f''(c) = 0 implies that there is a possible point of inflection at c

but when you find a local max of f' at x = c

isnt that just saying take the derivative of f', which is f''

and set it equal to 0

i just dont understand this possible point of inflection stuff

could somebody explain it to me

[covert path]

is it something to do with the behaviour of a local minimum and local maximum?

that maybe the case of a local minimum is different?

[trim jolt]

@covert path Has your question been resolved?

[west birch]

Hi there

[azure pulsar]

!noans

[trim jolt]

The purpose of this server is to help you learn, not to hand out answers. Do not ask someone to give you the answer directly.

[west birch]

Oh shit imma delege fast

Delete

Did it dissapear?

[azure pulsar]

i dont even think u gave the right answer, also, delete the img

[west birch]

Can u help then?

[west birch]

Why isnt it right?

[west birch]

I want to know

[azure pulsar]

polynomial isnt restricted to highest power only

it could be x^3 + 11x^2 - 7 lets say

poly means multiple

[azure pulsar]

first, do you know what an inflection point is and what it looks like?

[covert path]

yea its where concavity changes

[azure pulsar]

ok, what does it look like

here ill show you an image so you can visualize it

[covert path]

I dont know how to describe it but the graph changes shape

[covert path]

yea

[azure pulsar]

ok good

lets go with elimination now

a says local max. is that true?

[covert path]

i get the answer by elimination

but im just trying to understand why its a) if i coudlnt use process of elimination

D and E are very obviously not true and B is not true because the graph of f''(c) could bend up at one of its roots, which also then disproves C)

hence its A)

[azure pulsar]

well is a true for all curves? could it also be perhaps a local min??

[covert path]

thats what im asking

the asnwer key implies that its true for all curves

[azure pulsar]

look at the img i sent. try to visualize the graph of that deriv

draw it out it possible

[covert path]

it'll look like a parabola with concave up

right

hold on

[azure pulsar]

then what does that mean

[covert path]

there's a local minimum

[azure pulsar]

ok, so is that what a says?

[covert path]

no

[azure pulsar]

then can it be a?

[trim jolt]

Channel closed

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[covert path]

.reopen

[trim jolt]

✅

[covert path]

yes it could be a

what are you trying to get at

[azure pulsar]

how so, you just said that its a local min?

[covert path]

literally just reverse the end behaviours of your graph

[azure pulsar]

well local max =/= local min

[azure pulsar]

im tryna get u to think abt how u proved a statement and to eliminate it based on that proof

[azure pulsar]

you just claimed that for this image, f'x at c, is a local min.
a) says it is a local max

[covert path]

first of all the question says a) must be true and you're trying to say a) is false?

second of all the image clearly shows that it could be a local max too

[azure pulsar]

yes but thats not generalized

[covert path]

and your proof is not generalized either

[azure pulsar]

it is a counter example, counter examples arent generalized

[covert path]

ok sorry

go on

.close

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[green lava]

if f : (a, b) -> (c, d) is stricly monotone, bijective and continuous, can we always extend it continuously with f(a) = c, f(b) = d

[clear cloud]

i don't see why this couldn't work

[trim jolt]

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[supple robin]

I need help with 33b

[supple robin]

I cannot imagine the angle

Can someone guide me?

[trim jolt]

@supple robin Has your question been resolved?

[supple robin]

<@&286206848099549185>

[supple robin]

<@&286206848099549185>

[warm trellis]

which one

[supple robin]

The angle between planes BFH and BCH

[warm trellis]

theres a cube

[warm trellis]

you cut off a pyramid can you imagine that

[supple robin]

Ya

[supple robin]

Emm

[warm trellis]

nope?

[supple robin]

Are you expressing this

[warm trellis]

nope

i meant the acute angle

[supple robin]

So is the angle in it?

[warm trellis]

which direction are you looking from

b to c

or c to b

[supple robin]

B to c

Just like the figure

[warm trellis]

im considering

c to b

if you want b to c

flip it over

did you understand?

i can draw it the other way if you wanted

[supple robin]

Understanding

[supple robin]

Angle FBC?

Maybe I can more clearly in this way

[warm trellis]

fbc would be

pi-theta

[supple robin]

Maybe I should construct FC

[warm trellis]

if that helps

that would beoh wait

i made a mistake

in the diagram

how are two points b

wait my entire thing is wrong

ill redraw

my bad

[supple robin]

Yes

[warm trellis]

this was right

you're right

my bad

[supple robin]

Don’t worry

[supple robin]

.close

[trim jolt]

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• After 15 minutes, feel free to ping <@&286206848099549185>.
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[mild cosmos]

for part a, is the following proof correct:\Proof: $\rho_E (x)=0\iff \inf_{z\in E}d(x,z)=0\iff\forall\varepsilon>0\ \exists z\in E\ \text{such that}\ 0\leq d(x,z)<\varepsilon\iff x\ \text{is a limit point of}\ E\iff x\in\bar E?$

[solid kiln]

pirateking0723

[mild cosmos]

note: \bar{E} denotes the closure of E

[errant vortex]

sounds right to me
the third equivalence is a bit more complicated than the other 3, but to be fair it's pretty intuitive

[mild cosmos]

i agree with you

but it follows from the definition of a limit point, ie $x$ is a limit point of $E$ if $\forall\varepsilon>0\ \exists z\in E\ \text{such that}\ d(x,z)<\varepsilon$

[solid kiln]

pirateking0723

[mild cosmos]

alright tysm

.close

[trim jolt]

Channel closed

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[mild cosmos]

have a great day

.reopen

[trim jolt]

✅

[mild cosmos]

for the first part of this exercise

isnt it sufficient to just show that ρ_F(x)>0 for all x in K ?

[wraith hinge]

how do i open

[mild cosmos]

#❓how-to-get-help

you go to any channel under "Math Help (Available)" and post your question there 👍

[lilac flame]

what's rho_F

[mild cosmos]

so it goes like this : So first of all $\rho_F(x)\geq 0\ \forall x\in X$. Now $\rho_F(x)=0\iff x\in\bar{F}\ (\text{from the previous exercise})\iff x\in F\ (F\ \text{is closed})$.\But $x\in K\implies x\notin F\implies \rho_F(x)>0\implies\inf_{z\in F}d(x,z)=\delta\ \text{for some}\ \delta>0\implies d(x,z)>\delta$

[mild cosmos]

mb i forgot to mention this

[mild cosmos]

it was defined here originally

[mild cosmos]

this exercise is the previous exercise

here it is

[solid kiln]

pirateking0723

[trim jolt]

@mild cosmos Has your question been resolved?

[dull island]

yes

[mild cosmos]

then is my attempt correct ?

[mild cosmos]

this one

[dull island]

you lose me before the inf

[willow urchin]

inf doesnt preserve strict inequalities, thats a fatal flaw

[mild cosmos]

ah yes you are right mb

[mild cosmos]

but then you can just choose any number 0<δ_1<δ

so it can be corrected like this right ?

[willow urchin]

not really

[mild cosmos]

why ?

[willow urchin]

in general all we can say is inf preserves nonstrict ineqs

lets do this

what does rho_F continuous on compact K imply?

[mild cosmos]

i agree with this

but what i am saying is this : we know that d(x,z)>=δ, now choose δ_1 such that 0<δ_1<δ

[dull island]

1/n has inf 0 but is never 0

[mild cosmos]

then d(x,z)>δ_1 no ?

we can find such a δ_1 because otherwise we would have δ=0

[dull island]

find such a delta such that 1/n > delta > p

[mild cosmos]

this implies that ρ_F(K) is compact

[mild cosmos]

i dont follow your point

[willow urchin]

xetrov says u can have inf=0 but not have positive lower bound

so u cant guarantee delta this way

[mild cosmos]

ah yes

[willow urchin]

lets continue my way

[mild cosmos]

now i get the flaw in my way

tysm

[willow urchin]

ok so its conpact but its a subset of R. anything special to say?

[mild cosmos]

closed and bounded

[willow urchin]

yes and we’re looking for a lower bound of rho_F

so it has a min (on K)

there exists $k_0\in K$ st $\rho_F(k)\ge\rho_F(k_0)$ for all $k\in K$

[solid kiln]

ロケットジャンプ

[willow urchin]

agree?

[mild cosmos]

we are in this exercise no ?

[willow urchin]

yes is smth unclear?

[mild cosmos]

how did we conclude that ρ_F(K) is a subset of R in this case

[willow urchin]

def of rhoF

[mild cosmos]

isnt d(x,z) another metric on X ?

i am assuming that d(x,z) is a metric on X

since nothing said that it is not on X

and the only metric space found in this definition is X

[willow urchin]

$\rho:X\to[0,\infty)$

[solid kiln]

ロケットジャンプ

[willow urchin]

u should see this from def of rho

[mild cosmos]

wait nvm i am stupid

ρ is a metric so its range is definitely a subset of R

mb

[willow urchin]

well d is the metric

[mild cosmos]

that too

[mild cosmos]

ok so back to this

[willow urchin]

so now its clear?

[mild cosmos]

yes

[mild cosmos]

and now ρ_F(k_0)>0 since k_0∈K

[willow urchin]

true but can u justify it more

brb orderin pizza

[mild cosmos]

k_0∈K implies k_0 not in F implies k_0 not in the close of F implies ρ_F(k_0) neq 0 (from the previous exercise) implies ρ_F(k_0)>0 (ρ_F(k)>=0 for all k in X)

[mild cosmos]

np take your time

enjoy your pizza

i gtg for a bit to eat too so brb

[willow urchin]

@mild cosmos looks good

now can u see a good delta?

[trim jolt]

@mild cosmos Has your question been resolved?

[mild cosmos]

I am back

[mild cosmos]

Give me a sec to think about it

Well ρ_F(k_0)>0 so any number 0<δ<ρ_F(k_0) works no ?

For example ρ_F(k_0)/2

@willow urchin

or ρ_F(k_0)/3 or ...

[willow urchin]

yep

[mild cosmos]

But then I have a question

[mild cosmos]

Isn't this essentially the same ?

[mild cosmos]

If I take this into consideration

[mild cosmos]

By doing this

[dull island]

here it haven't shown that delta>0

[mild cosmos]

Take δ_1=δ/2 for example

[mild cosmos]

How didn't it

"ρ_F(x)>0"

[dull island]

the inf

[willow urchin]

as we said ur argument doesnt hold in general

we only got the result by using compactness of K and closedness of F

[dull island]

@willow urchin sorry to ping you again but can you just check that my solution is what you meant

[willow urchin]

yes xetrov

[mild cosmos]

"x in K implies x not in F implies ρ_F(x)>0" shows that inf>0

[mild cosmos]

I am trying to find why this is the case

Why is compactness essential here

How does my argument not hold

[dull island]

the not in F is true by the closure being the set

[willow urchin]

thats up to u to explore

bc the problem follows up by asking why it fails on noncompact sets

[mild cosmos]

That doesn't relate to the compactness of K

[mild cosmos]

Hahahahah

[dull island]

but you could have a sequence of points getting infinitely close to a point in F

[mild cosmos]

If you are saying that there can be a sequence of points in K that converge to F that's not possible

Ah wait

[dull island]

nice

now argue the inf thing out

at the end of the day, you used similar arguments

[mild cosmos]

Wait shouldn't K be closed for this to be true

[dull island]

I'll draw an argument with where you were heading out

lol

[mild cosmos]

Take your time

When you are back ping me

so i will write my problem rn so that when you are back you can clearly see where i am stuck and what i am saying and then you can correct me/give me a hint to let me find my mistake

we know that $\rho_F(x)=0\iff x\in\overline{F}$, so $x\in K\implies x\notin F\iff x\notin\overline{F}\iff\rho_F(x)\neq 0\ \text{but}\ \rho_F(x)\geq 0\implies\rho_F(x)>0\implies\inf_{z\in F}d(x,z)>0$ and we know that $d(x,z)\geq\inf_{z\in F}d(x,z)$ so let $\delta=\frac{\inf_{z\in F}d(x,z)}2$ then $d(x,z)>\delta$

[solid kiln]

pirateking0723

[mild cosmos]

so from here where exactly is my mistake

[willow urchin]

rho>0 => inf>0

[mild cosmos]

how is this the mistake, rho=inf no?

[dull island]

Wait hold on

[mild cosmos]

ρ_F(x)=inf d(x,z) by definition right ?

so if ρ>0 then inf>0

[willow urchin]

im just skimming bc im doing smth else rn, lemme read slower

[dull island]

Can you send the 4.20 and 4.21 again please

[mild cosmos]

sure

[mild cosmos]

np take your time and if you are busy then dont bother yourself with this

[mild cosmos]

here there is the definition of ρ and part a is useful in 21 (at least i used it in the argument that i want to be checked)

[dull island]

Okay yes, the bit in purple is true

[mild cosmos]

let me read this

[dull island]

It’s complement is open so why not

[mild cosmos]

but what does inf ρ_F(k_n) have to do with anything

i only need ρ_F

i didnt mention inf ρ_F anywhere

maybe i am failing to see your point

[dull island]

Im struggling to write it down

Hold on

[mild cosmos]

but ρ_F(x)>0 means that inf d(x,z)>0

because ρ_F(x)=inf d(x,z)

inf d(x,z)>0 implies d(x,z)>=0

[dull island]

But these rhos can get infinitely close to zero

[mild cosmos]

ig this is the point you are trying to show ?

[willow urchin]

oh i see ur error

[mild cosmos]

but any ρ_F>0

[willow urchin]

its very subtle

[dull island]

Can you spoiler it

[willow urchin]

$d(k,f)\ge\rho_F(k)>0$

[solid kiln]

ロケットジャンプ

[willow urchin]

that IS right

[mild cosmos]

yes

[willow urchin]

and u think u can just put delta=rho/2

but delta must be indep of k,f

[dull island]

This was my point

[mild cosmos]

yes for example .

[willow urchin]

ok no spoiler bar then

[mild cosmos]

I see

[dull island]

Sorry haha

[mild cosmos]

but still

[dull island]

I’m currently taking topology so this is good practice for me

[mild cosmos]

since d(k,f)>0

and its inf>0

then there exists a delta

[dull island]

It’s inf is zero for that one k

[mild cosmos]

regrardless of its form

[willow urchin]

the claim is exists delta forall k,f

what ur proof does is forall k exists delta

[mild cosmos]

ohhhh

thats it

now i fully understand my mistake

[willow urchin]

death by quantifiers 🙂

[mild cosmos]

[mild cosmos]

i wouldnt have noticed this mistake even if i was left for 100 years

[mild cosmos]

ohhh thats nice

i am doing this as an exercise in rudin's pma

[willow urchin]

i was still eating pizza so i didnt have enough attention to catch it earlier haha

[mild cosmos]

hahahha

[dull island]

Back to complex analysis for me

See ya

[mild cosmos]

cya

good luck in your mathematical journey

[willow urchin]

solve some pdes using conformal maps for me

[mild cosmos]

cant blame you because the taste of pizza takes away the concentration

[willow urchin]

esp when its this pizza

[mild cosmos]

good luck (although i am not sure what conformal maps are )

[mild cosmos]

🔥

[mild cosmos]

back to solving some more exercises and then jumping into a new chapter in rudin for me

tysm both of you

great help and great insights

have a great day/night

.close

[trim jolt]

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[willow urchin]

peace dude

[dull island]

I’m currently doing Cauchy Taylor’s theorem because I haven’t revised all year eek

[willow urchin]

taylor boring

pde fun

use conformal maps to solve laplace’s equation
$$\grad^2T=0$$
in the upper half plane with boundary conditions
\begin{itemize}
\item $T(x,0)=0$ for $x<-1$
\item $T(x,0)=1$ for $x>1$
\item $T_y\br{x,0}=0$ for $|x|<1$
\end{itemize}

[solid kiln]

ロケットジャンプ

[willow urchin]

thats the most fun stuff i did in complex

[dull island]

Most fun I found is residue theorem

Perhaps rouches theorem

[willow urchin]

residue calc is very fun at first but then they get samey

every time u end up estimating terms that vanish for large R or small eps

but its great for practicing big o notation

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[last vale]

I'm having trouble finding the sine equation.

The answer I came up with is y = 3sin(3 ( x - pi/6 ) + 2

I think I've found:
A = 3
D = 2

I'm having trouble finding B and C.
I think that the sin graph shifts to the right pi/6 units.
So if the phase shift is pi/6, B = 3?

I plugged my answer into desmos but the graph doesn't look right. Any help is appreciated 🙏

[trim jolt]

@last vale Has your question been resolved?

[simple panther]

Yes

C = pi/6 and B= 3

No, C is pi/2 - pi/6

Sin graph starts by going up, not down

[simple panther]

How did u get the phase shift as pi/6?

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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[ionic cargo]

how do i do this 😭 i have no idea how to do the unit circle shading crap

[trim jolt]

@ionic cargo Has your question been resolved?

[lime sphinx]

what have you tried?

[ionic cargo]

i have isolated the sin and cosine

but dont know what to do from there

[lime sphinx]

what are the max and min values sin(theta) can take?

[ionic cargo]

These two?

Sine is -root2/2

[lime sphinx]

I was thinking about the bounds on the sine wave max = 1 min =-1 to get you to have a feel for the function

[ionic cargo]

oh

the way the website is trying to teach to get the answe is like this

but i dont know how to shade it

this is an example problem so the answer is going to be different than the one im trying to solve

[lime sphinx]

tbh i ever learned this this way, but i can show you my way

[ionic cargo]

im down

[lime sphinx]

my first step would be to do a sketch like i did above

[ionic cargo]

alr

[lime sphinx]

then identify the two theta interval [—] i put on it in blue

where sin is greater than sqrt(2)/2

then we need to identify the 2 ends of the interval that are not -pi or pi

for this i would use the trig circle

[ionic cargo]

i think i get it

[lime sphinx]

then since I know some value on the unit circle, I would draw this

to get the 2 missing end of the intervals

o h there is an error

[lime sphinx]

do you get this part?

[ionic cargo]

still trying to comprehend it

[lime sphinx]

you have to know by heart some values of the trig circle

[ionic cargo]

i check the answer of the question and its showing this

[lime sphinx]

namely the first quadrant

[ionic cargo]

yeah i have probably like 2/3ds of it down but we get to have a cheatsheet for the test

so im just gonna put it down

im wondering

is the green part of the graph the intervals -pi < theta <pi?

cause it doesnt look like it

or im just really dumb

[lime sphinx]

I got the -3pi over 4 but we still need to tackle the other constraint on theta the cos(theta) <= -1/2

[ionic cargo]

okay

[lime sphinx]

to get the 2pi over 3

[lime sphinx]

but now i’m wondering if it would be better for you to learn it this way since they probably expect you to spit out this on your test.

[lime sphinx]

I don’t think so

[ionic cargo]

yeah i honestly think its better to do the shading thing cause we only get like 2inchx3inch boxes to show our work

[lime sphinx]

it look pretty straigthforward imo y = sin(theta) and x = cos(theta) so that indicate how to do the red and blue shading

[ionic cargo]

yeah i got that also

im just confused on the green

oh shoot

i see it nvm

wait so for the -pi < theta < pi intervals

this might be a really dumb question

[ionic cargo]

that makes a lot more sense ty

[lime sphinx]

then you get the angle with the unit circle

[ionic cargo]

how do you get -3pi/4? i didnt really understand the equation you showed earlier

[lime sphinx]

the values of sine and cosine repeats after each increment of 2pi to theta. Since we want thetha between -pi and pi you can do 5pi/4 - 2pi=-3/4pi

but what i did earlier is I exploited the pattern with the minus signs in the unit circle and my understanding of how the sine wave relates with the unit circle

[ionic cargo]

i see

im going to try another problem and see if i get it

@lime sphinx would this be the interval?

[lime sphinx]

yep

[ionic cargo]

im mean i got it

but the question was quite easy

so im just going to try another one

ok for somereason its giving easy questions

but for the one you showed me how to do, when are you supposed to use the [] and the U thing

@lime sphinx

[lime sphinx]

U is the union of sets, it’s a collection of objects containing everything in the first interval and everything in the second one

[lime sphinx]

do you know interval notation?

howto use []

[ionic cargo]

i know what it does but when are you supposed to use the brackets and the parenthesis

wait im dumb

you use the [] if its a less than or equal to right

or greater than or equal to

[lime sphinx]

if 1 <= x<= 2, then x is in the interval [1,2]

[ionic cargo]

yeahhh i understand now

and you use the U if it goes outside of the interval and goes back in right

ok i udnerstand how to do these problems now

but im not gonna like

lie*

[lime sphinx]

because we had -pi <= theta <= pi making the solution sets two disjoints intervals even though they don’t look disjoint on the trig circle

[ionic cargo]

i still didnt comepletly understand how you got the -3pi/4 still

wouldnt [-pi,5pi/4] work?

[lime sphinx]

because 5pi/4 is bigger than pi and most importantly because it’s wrong, a bunch of point in this interval don’t respect the conditions like 0 is in there and cos(0)=1 which is not less or equal to -1/2

[ionic cargo]

😱

i c

[lime sphinx]

it all boils down to what are sine and cos fondamentally

[ionic cargo]

alright i get it

finally

thanks

sm

anyways if i were to get it wrong on the test i can still get some credit depending on how far i got into the problem

[lime sphinx]

like just saying for your intuition cos is by definition the “length” (but it can be negative) of the blue line. can you imagine in your head the white dot rotating on the circle to the left (theta getting bigger) making the blue line shorter and then eventually negative when theta becomes bigger than pi/2.

https://www.pinterest.com/pin/607352699725790017/

[ionic cargo]

wait thats so cool

[lime sphinx]

i mean that’s how i got the -3pi/4. I made the white dot move in my head clockwise (thetha getting smaller). from my graph i knew it would be past -pi/2 so in the 3rd quadrant then i used the unit circle like at 5pi/4 because since is the same there

ima go now gl with your test

[lime sphinx]

ty

[trim jolt]

@ionic cargo Has your question been resolved?

[trim jolt]

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[hybrid field]

dont advertise in this server.

[tulip violet]

<@&268886789983436800>

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[carmine mauve]

!status

[trim jolt]

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

[viscid wagon]

2

i have some steps

if we say Red triangle has sides a,b,c
then green triangle has sides a,a,b

ratio of area of R/G is (1/2(ab)) / (1/2(bh)) = a/h

not sure how to get height of green triangle

[carmine mauve]

does this help?

[viscid wagon]

kind of

so it would be pythagorean thm with sides (1/2b, h, a)

[carmine mauve]

yep

[viscid wagon]

ok so i then get

R/G = a/h = a / sqrt(a^2 -b^2/4)

so i need b in terms of a or vice versa

[viscid wagon]

oh dang i didnt see that

[carmine mauve]

and here

[viscid wagon]

c = 2a

[carmine mauve]

yep

[viscid wagon]

i got 2

in the end

R/G = 2

[carmine mauve]

[viscid wagon]

are you able ot help with another problem

[carmine mauve]

yep

[viscid wagon]

What is the next date (after today) such that all of the 8 digits used to represent it are different and none of them is zero?
Note: we are writing the dates as DD/MM/YYYY, e.g. 05/12/2000 is 5th of December 2000.

i am getting no solution

[carmine mauve]

you know where the 1 needs to go

[viscid wagon]

yeah

[carmine mauve]

DD/1M/YYYY

[viscid wagon]

has to be december

and then u cnat use any days

[carmine mauve]

so you also know where the 2 goes

DD/12/YYYY

what's the smallest year, greater than 2025, with no 0s?

[viscid wagon]

2111

[carmine mauve]

with no 1, 2, or repeating digits

[viscid wagon]

3456

idk

[carmine mauve]

yep

so we have DD/12/3456

you can do the rest 🙂

[viscid wagon]

so its not possible right

because DD has to be 78

or something

[carmine mauve]

ah, that just means we need to use one of the 3, 4, 5, or 6 in the day

because, yes it's 78 is too big

so, what's the smallest day with no 1, 2, or repeating digits?

[viscid wagon]

30

[carmine mauve]

or zeros XD

[viscid wagon]

03

lol

yeah

okay im just gonna put DNE

[carmine mauve]

*you're right, btw, there will never be a next date with your requirements.

do you need to say why?

[viscid wagon]

no i dont

[carmine mauve]

cool 🙂

[viscid wagon]

do you want to see more problems

[carmine mauve]

do you want help with more problems?

[viscid wagon]

i have just started one so not sure yet but i can send if you want to look

does this picture look right

[carmine mauve]

looks right to me

lots of algebra in this one

[viscid wagon]

yeah im not even really sure how to start

[carmine mauve]

I would start py picking an origin point (any point will do), then use the lengths and angles given to determine the other points.

Also, label the unknown lengths so you can do algebra with them

[viscid wagon]

wdym by determine other points

like determine angles

[carmine mauve]

so, let's say B is at the origin. Then C would be at (7,0) because it's 7 units in the x axis from B

keep going for A and D

[viscid wagon]

oh

hmm

i can find slopes for ab and cd

[carmine mauve]

If we define length AB = a, then A would be at (a cos 30, a sin 30), and you should know cos 30 and sin 30

[viscid wagon]

i see

yeah let me try

[trim jolt]

@viscid wagon Has your question been resolved?

[viscid wagon]

still solving

[carmine mauve]

np take your time 🙂

https://www.desmos.com/geometry/cfsu7ggt7m

[viscid wagon]

i think i have all the points as coordinates

[carmine mauve]

cool

[viscid wagon]

in terms of one of the lengths

DC

[viscid wagon]

i got 7 - DC

as the distance EF

any idea how to keep going

[carmine mauve]

what do you have for the points?

[viscid wagon]

for E and F?

[carmine mauve]

A,B,C,D,M,N

[viscid wagon]

A (3Y/2, sqrt(3)Y/2)
B (0,0)
C (7,0)
D (7 - Y/2, sqrt(3Y/2)
M (7/2,0)
N ((7+Y)/2, sqrt(3)Y/2)
E (3/4Y, sqrt(3)Y/4)
F ( 7 - Y/4 , sqrt(3)Y/4)

where Y is DC

[carmine mauve]

Do you mean Y is the height of line AD?

[viscid wagon]

did u mean to delete this

just in case Y is the line DC

[carmine mauve]

yes

the length of line CD

[viscid wagon]

yeah

oh i think i know

i make small right triangles with E and A

[carmine mauve]

I think you've got cos 60 and sin 60 the wrong way around

for D

no, ignore me

I got it wrong

[viscid wagon]

gotcha

yeah i was sus bc it came out looking too clean

[carmine mauve]

D is good

how did you get A?

[viscid wagon]

i set the heights equal to each other

[carmine mauve]

Or did you mean A = (3X/2, sqrt(3)X/2) where X = |AB|?

[viscid wagon]

i had (sqrt(3) X /2, X/2) where X = AB

then set x/2 = sqrt(3) Y / 2

[carmine mauve]

yep 🙂

[viscid wagon]

i have AB = sqrt(3) Y
AD = 7 - 2Y
DC = Y

[carmine mauve]

A is good

I think you've got them 🙂

sorry it took a while XD

[viscid wagon]

yeah for me too

maybe im misising someting

but everything is based on Y so how to find it

ohh

i forgot they gave us NM

[carmine mauve]

yep 🙂

[viscid wagon]

got EF = 4

i have one more actually if oyu are still free

its a bit weird one

[carmine mauve]

I don't thin you need to worry about EF?

Oh, it's find EF

i'm a dumb dumb

[viscid wagon]

yea

haha no ur good

[carmine mauve]

So, what did you get for MN in terms of Y?

[viscid wagon]

i got Y

for MN

[carmine mauve]

cool

[viscid wagon]

do u have time for one more q?

[carmine mauve]

ye

[viscid wagon]

the task is to explain why this proof is wrong

i know it has something to do with the 7s being double counted

[carmine mauve]

yep, I also get EF = 4

[viscid wagon]

yay nice

[carmine mauve]

are any of the statements wrong?

[viscid wagon]

i think only statement 5 is technially wrong

but its like super wrong

but i am having trouble explaining it

[carmine mauve]

it's 4

||we also don't subtract the numbers beginning with two zeros, or three zeros||

ah shit, I've just given away the answer

[viscid wagon]

well

[carmine mauve]

sorry

[viscid wagon]

i think that there are probably a lot of answers

but that is definitely one of them

because the answer i got is 7623

[carmine mauve]

it may also be wrong for other reasons, but it's at least wrong because of that

[viscid wagon]

yeah

[carmine mauve]

how did you get that answer?

[viscid wagon]

but he asked us to give as many as possible

i found 5 digit numbers with only 1 or 0 7s

and subtracted

but this proof is going the other way so im having trouble conceptualizing it

but its really impressive u saw that so fast

i ddint even think abou tit

[carmine mauve]

he may want you to say something allong the lines of
subtract the erroneously counted numbers with a leading zero
subtract the erroneously counted numbers with two leading zeros
but now we've over-subtracted numbers with only one leading zero, so add them back in
subtract the erroneously counted numbers with three leading zeros (there's only one of these)
but now we've over-subtracted numbers with two leading zeros, so add them back in
but now we've over-added numbers with one leading zero, so remove them again

[viscid wagon]

is this how to fix the solution?

i think he mainly wants us to explain some logical errors

[carmine mauve]

it's a classic problem in combinatorics

[viscid wagon]

i know there is one main error which is something about numbers being double counted

[carmine mauve]

ah yes

let [7] notate our fixed position for a 7.
Then a[7][7]7e and a7[7][7]e are the same number, but counted twice

actually, it's counted three times

[viscid wagon]

yeah

but i dont understand that exactly

so this is even before the subtraction

right

[carmine mauve]

he probably over subtracts for the same reason

[viscid wagon]

wait i thought it was under subtracting

because we dont subtract numbers that start with 00

[carmine mauve]

it's both, by different ammounts

[viscid wagon]

why is there over subtracting?

[carmine mauve]

because 07[7][7]e, 0[7]7[7]e, and 0[7][7]7e are the same number subtracted three times

[viscid wagon]

ohh

i see

and in the original counting

the more 7s the more overcounting

like 17777 is counted for each pair of 7s

[carmine mauve]

they're over-counted, and over subtracted. but we don't over-subtract the same number as we over-count because we over-count repetitions beginning with 7, but don't over-subtract them.

[viscid wagon]

oh boy

[carmine mauve]

yep

17777 is counted 6 times

[viscid wagon]

gotcha

[carmine mauve]

77777 is counted over-counted by 9 times, but only over-subtracted by 5 times

[viscid wagon]

you wouldnt subtract it right

bc no leading 0

[carmine mauve]

oops

[viscid wagon]

1. The same numbers are overcounted (ex: 1777 is counted 6 times based on which 2 7s are picked)
2. Doesn't subtract numbers beginning with multiple leading 0s
3. Oversubtract for the same reason as 1. ex(07777 is subtracted 6 times)

this is what i wrote so far

[carmine mauve]

so it's never over-subtracted at all

[viscid wagon]

07777 is over subtracted right

[carmine mauve]

yep 07777 counted 6 times and subtracted 6 times for the same reason

[viscid wagon]

gotcha

[carmine mauve]

but 77777 is counted 9 times and never subtracted

[viscid wagon]

yeah

makes sense

ok im gonna stick with these 3

[viscid wagon]

this should be 10 rihgt

5c2

[carmine mauve]

5C2 = 10

yes

[viscid wagon]

bet

okay thanks a lot man

[carmine mauve]

5P2 = 20

Or have I got it the wrong way around

[viscid wagon]

oh

[carmine mauve]

nope, right way around

[viscid wagon]

im not sure which one it is

for this case

[carmine mauve]

nCr means combinations, where order doesn't matter
nPr means permutations, where order does matter

for example 7_1, 7_2, c, d, e and 7_2, 7_1, c, d, e are the same combination, but different permutations. (where 7_1 is chosen before 7_2)

[viscid wagon]

i think the order does matter in this situation right?

ohh

the order you choose them in

[carmine mauve]

yep

[viscid wagon]

so its still combinations

for this counting

[carmine mauve]

nCr is the number of ways to select r things from a set of n things
nPr is the number of ways to arrange r things from a set of n things

[viscid wagon]

i see

if i think about it too hard it gets confuing

[carmine mauve]

yr XD

[viscid wagon]

loll

[viscid wagon]

thanks for helping man lmk if theres any way i can like

give you a good review

lol

usually the helpers are not so helpful

[carmine mauve]

the correct way is to
1.a count exactly 2 sevens in a 2 digit number
1.b count exactly 2 sevens in a 3 digit number
1.c ...
2.a count exactly 3 sevens in a 3 digit number
2.b count exactly 3 sevens in a 4 digit number
2.c ...
3. ...

[viscid wagon]

u lost me xd

is this how you can correctly subtract

from 10,000

[carmine mauve]

we're not subtracting. we're correctly counting

[viscid wagon]

why do you have to go 2 digit, 3 digit, etc

you can count exactly 2, 3, 4,5 7s right

for just 5 digit number

[carmine mauve]

avoids the problem of leading zeros

[viscid wagon]

i guess im confused if we count exactly 2 sevens in a 2 digit number

how do we then take that to our count for 5 digit number

[carmine mauve]

we don't need to

we never count 2 digit numbers again

[viscid wagon]

so once we add up this count

we are still not subtracting it?

[carmine mauve]

nope, no subtraction

[viscid wagon]

im kind of confused

so this number tells us all the 7s up to 5 digit number

all the counts added up

[carmine mauve]

yep

[viscid wagon]

but we only are counting 5 digit numbers

[carmine mauve]

we count

• how many 2 digit numbers contain 2 sevens (only one)
• how many 3 digit numbers contain 2 sevens
  1. how many 2 digit numbers contain 2 sevens, multiplied by the number of non-zero, non-seven leading digits we can choose
  2. plus the number of 2 digit numbers containing 1 seven (which counts 3 digit numbers starting with a seven)
• how many 4 digit numbers contain 2 sevens
  1. how many 3 digit numbers contain 2 sevens, multiplied by the number of non-zero, non-seven leading digits we can choose
  2. plus the number of 3 digit numbers containing 1 seven (which counts 4 digit numbers starting with a seven)
• keep going all the way up to 5 digit numbers...

[viscid wagon]

oh

i see

but then you have to add cases of more than two 7s

which makes it even worse

[carmine mauve]

we don't need to subtract anything. we do it recursively

[viscid wagon]

yeah

i guess its just very long

[carmine mauve]

Let $N(n,r)$ be the count of $n$-digit numbers containing exactly $r$ of a digit.
\begin{itemize}
\item If $n < r$, then $N(n,r) = 0$
\item If $n = r$, then $N(n,r) = 1$
\item Otherwise, $$N(n,r) = 8 N(n-1,r) + N(n-1, r-1)$$ Because there are $8 N(n-1, r)$ $n$-digit numbers containing $r$ of our digit, and don't start with that digit or a zero (to avoid overcounting), and $N(n-1, r-1)$ $n$-digit numbers starting with our non-zero digit containing $r$ of our digit.
\end{itemize}

[solid kiln]

Shuba

[carmine mauve]

Hopefully that isn't too complicated

[viscid wagon]

ig im kind of confused about the solution

like why we are doing it this way

also are you good at differential equations