help-49

[wary thorn]

oh wow, you're in your last year now?

Congrats

[twin ridge]

Nah, not last year.

Just a senior.

[past mantle]

i see letters and numbers and variables

[wary thorn]

[twin ridge]

My major requires me to do 5 years.

[past mantle]

congrats

[wary thorn]

What are your plans, if you don't mind me asking

[past mantle]

get the money bro

[wary thorn]

we can go DM if you'd like too

[twin ridge]

Same as everyone in my major lol

[wary thorn]

teach?

[twin ridge]

Yeah.

Haven't decided what grade yet.

[past mantle]

my plans is to work with solar pannels is what my brother wants me to do

[twin ridge]

I really liked 8th grade, but I haven't figured out what high school is like yet.

[wary thorn]

Become a mathematician instead

[past mantle]

oh hell yea

[wary thorn]

I think whatever grade you pick you'll do great with

[past mantle]

if you graduate and get a better one you can make a 6 figure to 8 figure job

[wary thorn]

My mom teaches high school geometry

[past mantle]

thats whats up

[twin ridge]

If $x_{i+1}<x_i$, then that means that $\frac{1}{2}x_{i}+2<\frac{1}{2}x_{i-1}+2$?

[twin ridge]

I think I'd be content with high school geometry. I actually really loved geometry when I took it.

[past mantle]

i enjoyed alegbra

[twin ridge]

Most of that was due to a really good teacher, but the subject itself is also fun.

[wary thorn]

you must make them read Euclids elements

[past mantle]

geomentry sucked kinda

[twin ridge]

Geometry seems very teacher dependent.

[past mantle]

yea true

[twin ridge]

Because I had a really old dude with a monotone for geometry in high school and I hated it

[grand pond]

Narutoes

[twin ridge]

I can just take the 2s out then? Or do I factor something out

[wary thorn]

Let me just lay it out

x_(i+2) = 1/2 (x_(i+1)) + 2

[twin ridge]

Normally with induction I can highlight a section that was our original statement

[wary thorn]

try using the inductive hypothesis on this

remember it was that x_(i+1) < x_i

[twin ridge]

So I have the original x_(i+1) in there

[wary thorn]

which is less than x_i, by assumption

so

[twin ridge]

Right.

[wary thorn]

x_(i+2) < 1/2 (x_i) + 2

does the RHS look familiar now?

[past mantle]

you guys have a goodnight

[wary thorn]

goodnight

[past mantle]

its late here

[twin ridge]

I'm not goin to bed unfortunately.

[wary thorn]

its 3:25am in india bitch

[twin ridge]

Is it actually?

[wary thorn]

no

[twin ridge]

That doesn't quite math

[wary thorn]

it's just my catchphrase

[twin ridge]

What a catchphrase.

My catchphrase is "If it doesn't concern me and it doesn't affect me, idgaf so don't tell me."

[wary thorn]

anyways we get this haha. simplify the RHS and you're done

[twin ridge]

Simplify the RHS meaning isolate x_i?

[wary thorn]

no

it should look familiar

1/2 (x_i) + 2

what's it look like

[twin ridge]

It's x_i+1

[wary thorn]

mhm

so

x_(i+2) < x_(i+1)

which is what we wanted to show

completing our inductive step

[twin ridge]

Did we jump from the x_i+1 to the x_i because it's known that it's less?

[wary thorn]

yes we used our inductive hypothesis there

[twin ridge]

So now we've proven it is montone, and we proved it was bounded earlier.

And we have the limit of 4 so we're done and I just gotta type this so it makes sense.

That's where most of my homework points get lost. I know how to do the problem I just don't know how to write it.

[wary thorn]

Something that you never get used to is that proofs aren't written how they are solved

its almost always in the reverse order

the outline I'd use for writing this proof would be like

• prove its monotone decreasing
• prove its bounded
• therefore by monotone sequence theorem it has a limit, call this L
• solve for L

which is the opposite of what we did 😓 but it is a much smoother read

Hello Flip, any doubts?

[frozen talon]

sorry for cutting in but I'm curious how you demonstrate monotonicity here

[wary thorn]

Induction

[frozen talon]

I just went through the motions myself and I had to use boundedness to show for monotonicity

[wary thorn]

You can do it by induction

[twin ridge]

We proved that for every term in the sequence, the previous one is larger than it.

So it's always decreasing.

[wary thorn]

Indeed

[frozen talon]

well you're kinda sweeping the issue under the rug I think, I just want to know why you know that $\frac{1}{2}x_n+2\leq x_n$

fug

[twin ridge]

same.

[twin ridge]

story of my life

[grand pond]

Flip

[twin ridge]

By assumption

[frozen talon]

you can't use assumption here, the indices agree

[wary thorn]

The base case is x_2 < x_1.
Then we assume x_(i+1) < x_i
Then we prove that this implies x_(i+2) < x_(i+1)
This proves that for all n>=1
x_(n+1) < x_n

which is monotone decreasing

[frozen talon]

I know

[wary thorn]

Alright then there is no issue

[twin ridge]

The induction step here being that since x_(i+2)<x_(i+1), x_(i+2)<x_i

[frozen talon]

"since x_(i+2) < x_(i+1)" suggests that you're assuming what you want to show

[twin ridge]

I don't know nearly enough about proofs to know if I'm allowed to do that or not. I get very mixed answers whenever I ask.

[frozen talon]

I'm sorry I probably glossed over context here and being distracting, I just thought it was important

[twin ridge]

It could be.

[wary thorn]

Oh no, I see what flip is saying

we used that x_(i+1) < x_i to replace the x_i on the RHS

but

this makes the LHS >= to what is now on the RHS

not <= which is what we wanted

you have to argue differently

[frozen talon]

I ended up showing that (x_n) is bounded (below) first

[twin ridge]

We know it's bounded below because x_n is never negative, so it can't go under 2

[frozen talon]

because, spoilers, knowing that the sequence is bounded by a particular number implies this portion

[wary thorn]

yes yes

Thank you sir Flip

good catch

[twin ridge]

I didn't quite catch what he caught.

[frozen talon]

I can try to explain

restarting the induction, we know that x_1 = 8 and x_2=6 < 8 = x_1, so that's good

we assume inductively that x_{i+1} < x_i for some index i

the goal is then to show that x_{i+2} < x_{i+1}

but x_{i+2} = 1/2x_{i+1}+2, so we need only verify that 1/2x_{i+1} + 2 < x_{i+1}

[twin ridge]

Right.

x_{i+1}=1/2x_i+2 though

[frozen talon]

yeah, and here I just realized that I'm dumb and consequently embarrassed lmao

[twin ridge]

I'm just permanently dumb.

[frozen talon]

you can use boundedness to show that this inequality works

but as you say, x_{i+1} < 1/2x_i + 2, so we need only verify that 1/2x_{i+1} + 2 < 1/2x_i + 2

but of course this is true lol

[wary thorn]

?

[frozen talon]

because you're assuming that x_{i+1} < x_i, so this follows by multiplication and addition stuff

like, you halve both sides and the ineq. still obeys

[twin ridge]

Did I write this correctly? I am not great at writing induction proofs so I always look for input.

[frozen talon]

you add 2 and the ineq. still obeys

[wary thorn]

I thought my sir was losing it for a second, but we're all good

[twin ridge]

If it makes you feel better, I just lost the game so I'm also losing it.

[frozen talon]

I am losing it and trying to get it back together, I apologize

[wary thorn]

State that you're going to prove by induction that x_n is monotonically decreasing at the beginning

Then say the base case is...

Then say, "For our inductive hypothesis assume that..."

"doing algebra then leaves us with"

Do the algebra

if you want to be certain not lose points

[twin ridge]

How do I explain that I'm doing algebra?

[wary thorn]

by showing it

[twin ridge]

Our inductive hypothesis is the xi+1<xi, or xi_2<xi+1?

[wary thorn]

the first

[frozen talon]

"by preservation of order under (addition / positive scalar multiplication)"

and by substitution

[wary thorn]

At that point just show the algebra lmao

[frozen talon]

well yeah

[twin ridge]

So the *2 and the -2?

[frozen talon]

I'm just saying this is the supplement to saying "by algebra" when coupled with showing said algebra

[wary thorn]

If I had to grade someones work for this limit and they said by preservation of order under ... I would find it very funny

they wouldn't lose any points though of course

[twin ridge]

How can I go from x_i+2<x_i+1 to 1/2x_i+1+2<1/2x_i+2 to have the proof still flow?

[frozen talon]

it depends on where you are

[wary thorn]

Have you actually done the algebra yet?

[twin ridge]

A lot of my wording is clunky.

[frozen talon]

not geographically I mean mentally

[wary thorn]

And are you writing this in Latex?

[twin ridge]

Well this is the substitutuion before the algebra, no? And yes, LaTex.

[frozen talon]

it looks like word

dang

[twin ridge]

I get extra credit for using LaTex.

[frozen talon]

based

[wary thorn]

I'd just separate the algebra into lines like
$$ equation 1
$$ equation 2
...

ok ignore the render

u get my point

[twin ridge]

I just split screen it so I can read both.

[wary thorn]

$$2x+5=10$$
$$2x=5$$
$$x=\frac{5}{2}$$

[grand pond]

Austin

[wary thorn]

Like this

is how I would show the algebra

[wary thorn]

I just use overleaf

[twin ridge]

I don't always have connection at school since they're reworking the networks.

[frozen talon]

you can use the align environment to make the equality signs 'align'

[wary thorn]

ah

[twin ridge]

How do I do this part though?

Like from a wording perspective.

[wary thorn]

I recommend not using any words at all, and writing it like I did here

[frozen talon]

it'll read a lot nicer if you do the logic in exact reverse to how I was trying to explain it earlier

[twin ridge]

I don't have to say that I'm substituting the definition?

[wary thorn]

Or if you ever use the inductive hypothesis, put text in between those steps and say "By the inductive hypothesis"

[frozen talon]

so you start with "suppose x_{i+1} < x_i (for some natural i)" then proceed

thinking in reverse, you want this to eventually look like x_{i+2} < x_{i+1}

and to get there, you need only halve both sides and add 2

identifying that 1/2x_i + 2 = x_{i+1} is what I think of as substitution

[twin ridge]

Does this flow better and make sense?

Or should I say "Therefore for all n... xn+1<xn by induction

[frozen talon]

"If x_{i+2} < x_{i+1}" is where you lose the game

you can't assume what you're trying to prove is true

[twin ridge]

I lost this game long ago.

[frozen talon]

because otherwise what's the point lol

I assume I'm right and there's nothing left to be done

[twin ridge]

I wasn't really given any tools on how to write induction proofs so all I know about them is what I've managed to learn from this disc.

[frozen talon]

word

[twin ridge]

I am aware I am wrong, I just need the tools to not be wrong. Or at least less wrong.

[wary thorn]

its not "decreasing... and subsequently monotone"

just say its monotone decreasing

[frozen talon]

wording could also be monotonically decreasing?

[wary thorn]

ye

just not what you wrote

can't be that

[frank wolf]

Hmm so

[wary thorn]

😂

[twin ridge]

mono dec.

[frank wolf]

Do you know how to prove that if $x_{i + 1} < x_i$ then $x_{i + 2} < x_{i + 1}$?

[grand pond]

Pseudonium

[wary thorn]

I hope he does at this point

[twin ridge]

I thought I did, but apparently not.

[frank wolf]

Hmm ok so

Let me try something...

[twin ridge]

I already wrote what I thought the answer was.

[frank wolf]

So, you can view $x_{i + 1} < x_i$ as a function

[grand pond]

Pseudonium

[frank wolf]

Do you see how?

[twin ridge]

Aren't they both functions?

[wary thorn]

they're both sequences

[frank wolf]

Yes, so

$x_i$ is a function

[grand pond]

Pseudonium

[frank wolf]

You plug in a natural number

[frozen talon]

sequences are just functions on natural numbers

[frank wolf]

And you get out a real number

[wary thorn]

this is unnecessary

[frank wolf]

It's a function $\mathbb{N} \to \mathbb{R}$

[grand pond]

Pseudonium

[wary thorn]

he was done wasn't he

I didn't check his proof but I assumed we were just critiquing the wording

[frozen talon]

the logical flow is reversed

[twin ridge]

I was told the middle of my proof was wrong

[frank wolf]

Right, I see, I haven't read most of what's come before

[frank wolf]

$x_{i + 1}$ is also a function

[grand pond]

Pseudonium

[frank wolf]

You plug in a natural number $n$

[grand pond]

Pseudonium

[frank wolf]

And you get out the real number $x_{n + 1}$

[grand pond]

Pseudonium

[wary thorn]

Well I think we have enough helpers here so I am going to peace out and have faith in the rest of you all to clear this doubt

[frank wolf]

Again, a function $\mathbb{N} \to \mathbb{R}$

[grand pond]

Pseudonium

[twin ridge]

Historically, hasn't always panned out that way lol

[wary thorn]

sir flip will ABCD and he has backup if needed

[frank wolf]

But we can also view $x_{i + 1} < x_i$ as a function

[grand pond]

Pseudonium

[frank wolf]

It takes in a natural number $n$

[grand pond]

Pseudonium

[frank wolf]

Computes both $x_{n + 1}$ and $x_n$

[grand pond]

Pseudonium

[frank wolf]

Compares them to each other

And spits out TRUE if $x_{n + 1} < x_n$, and FALSE otherwise

[grand pond]

Pseudonium

[frozen talon]

so my main critique with the writeup is that the logic is exactly reversed: as it's written now, you've shown that if the claim is true, then the hypothesis is true. it's not what we want: we want the hypothesis to prove the claim

[frank wolf]

So it's a function $\mathbb{N} \to {\text{TRUE}, \text{FALSE}}$

[grand pond]

Pseudonium

[frank wolf]

Does that make sense?

[twin ridge]

I've been told this like 20 times this entire week, so I'm thinking I just have to completely relearn how to write proofs.

I haven't had a teacher tell me I was doing it wrong so I've done it this way all year.

[frozen talon]

one trick when writing I used to do a lot was having a skeptic in my head as I wrote

as a reader, I immediately protested when I read "if x_{i+2} < x_{i+1}"

[twin ridge]

So how would you revise it to make it so I don't do that?

[frozen talon]

completely reverse the whole thing

[twin ridge]

I am very unsure what that means.

[frank wolf]

If you want, I have my suggestion..

[twin ridge]

Always open to suggestions.

Worst case I just don't use it but multiple viewpoints always good

[frank wolf]

.

[frozen talon]

by reversing it, I mean reversing the logic

[twin ridge]

I still don't know what all that entails.

[frozen talon]

so if you read what you have right now, if you read it bottom to top, it looks salvageable

[twin ridge]

So start by saying the sequence is monotonically decreasing?

[frozen talon]

no, sorry

[twin ridge]

I hear "reverse the logic" and I write all the sentences from back to front.

Or are you saying, go from x_i+1<x_i and try to build that into the x_i+2<x_i+1?

[frozen talon]

your steps are effectively this:
\begin{gather}
x_{i+2}<x_{i+1}\
\implies\frac{1}{2}x_{i+1}+2<\frac{1}{2}x_i+2\
\therefore x_{i+1}<x_i
\end{gather}

[grand pond]

Flip

[frozen talon]

but of course, this isn't what's desired

however, this logic is reversible

(3) implies (2), and (2) implies (1)

[twin ridge]

Gotcha

Lemme try to write that

So I have x_i+1<x_i

And then if I write that into the recursive thing, does the x_i give me a x_0?

Or do I leave it as x_i?

[frozen talon]

no, the indices did nothing wrong

we do not eliminate them

actually in this context we don't even have an x_0, it starts with x_1 lol

[twin ridge]

Well now I'm thoroughly confused

You just said to reverse it

Oh

I see

[twin ridge]

?

[frozen talon]

yeah, that looks much more functional

[twin ridge]

Any syntax weirdness?

[frozen talon]

yeah, I never write $a(x<y)$ personally

[grand pond]

Flip

[frozen talon]

idk why I texed that lmao

[twin ridge]

Austin said to "do the algebra"

I was just gonna jump to it

[frozen talon]

the manipulation works, and whether you want to provide full justification to it is up to you and what you think is more beneficial for you

[twin ridge]

Can I just say "Multiplying both sides by 1/2 then adding 2 gives us ..."

[frozen talon]

yeah basically

you may consider citing some kind of substitution going from (2) to (3)

you know the culture of the class better than I do

[twin ridge]

It's a very intro level course, and the teacher isn't particular about syntax.

[frozen talon]

one thing I would do, and it's incredibly minor, is to briefly derive showing x_2 = 6

[twin ridge]

Me being critical of my syntax is just me wanting to write proofs better personally.

[frozen talon]

that's a good mentality to have

[twin ridge]

So like

[frozen talon]

yeah exactly

and it doesn't need its own sentence or anything either, just shove the calculation in there when you say x_2 = 6

it's small but it's nice

[twin ridge]

Did I mess up any logic by trying to shorten it?

[wary thorn]

its equivalent to

not equal to

[twin ridge]

What's the difference?

[frozen talon]

exactly what I looked at lol

[twin ridge]

In this realm

[frozen talon]

equivalent is like equal but reserved for logical statements

[wary thorn]

there's no equivalence relation for two inequalities 😓 idk it just isn't something we say about them

2<5 isn't equal to 0<3

[twin ridge]

It is now.

[wary thorn]

😂

[twin ridge]

2<5 is equal to 0<3 by TrustMeBro.

[frozen talon]

actually I'd like to retract my statement

[twin ridge]

Too late, we in there now.

[frozen talon]

we don't say logical statements are equal, but we do say they are "equivalent" if they imply each other

[wary thorn]

[frozen talon]

2<5 isn't the same content as 0<3, but they imply each other

so they are equivalent, but not equal

[wary thorn]

2<5 <=> 0<3

very poor example yet a true one

[frozen talon]

or more commonly, contraposition yields equivalent statements

hmm there's an example but I forget it

[wary thorn]

let's not get into it

[frozen talon]

it's for the best

[wary thorn]

has the limit doubt been cleared?

[frozen talon]

I think we've only been talking about proof writing, and specifically for monotonicity

[wary thorn]

ah yes. after making that tweak, just show its bounded and then conclude by finding the limit. I have faith in your ability to write up both of those steps

Flip sir are you doing any math of your own?

[frozen talon]

mmm not really, I'm out of school at the moment

I've been transcribing some notes from a ring theory class, struggling to read an algebraic topology book, and doing amateur research and programming lol

[wary thorn]

ah so it sounds like you are doing math of your own after all

nice

[frozen talon]

very very slowly

[wary thorn]

I dread topology

[frozen talon]

I'm effectively nocturnal

[wary thorn]

I like graph topology but it isn't real topology

I can hide behind the combinatorics

[frozen talon]

and I do like 3 hours max a day

[wary thorn]

better than most

[frozen talon]

that seems like the standard take on topology from someone who likes combinatorial thingies

[twin ridge]

I didn't look much at the syntax, just wanted the ideas on the page first.

[wary thorn]

yah i mean looks acceptable

[frozen talon]

it's a little handwavy at the bounded below bit

[wary thorn]

I think its fine

[twin ridge]

I just need to prove there exists a lower bound.

For it to be bounded below.

[wary thorn]

you've effectively proven the lower bound is 0

[twin ridge]

I could just go with 0, that works.

[wary thorn]

and then unnecessarily expanded on that by saying its then 2

[frozen talon]

oh hey this ties in with my shenanigans earlier on

in this context, I claim that monotonicity implies boundedness (below)

[wary thorn]

does anyone like topology? 😂

[frozen talon]

I like the idea of topology, and some of the logic in early basics is pretty fun

[twin ridge]

I have to take Combinatorics next semester.

[wary thorn]

its so cursed

jordan curve theorem shouldnt exist

that is my take

[frozen talon]

see you know more than I do I think

this is why I say I like the idea of topology and not that I like topology lol

[wary thorn]

haha

[frozen talon]

combinatorics is fun, counting is hard

[wary thorn]

If you want a fun math recommendation you should read Pearls in Graph Theory Chapter 10

[twin ridge]

Counting is subjective.

[wary thorn]

yes

[twin ridge]

Considering I took French in high school and their counting system is whack.

[wary thorn]

recently I had to do some partitions

very simple looking questions but

[twin ridge]

So now I just gotta do this process 4 more times for the rest of this homework.

[wary thorn]

complicated answers

[frozen talon]

partitioning is illegal here. they locked me in a cell

[twin ridge]

Although the next one is x1>1 rather than equalling.

[wary thorn]

How many ways can you write 210 as a sum of 19 positive integers all even and >= 10?

https://tenor.com/view/right-to-jail-jail-parks-and-rec-right-away-fred-armisen-gif-16039907319194806297

[twin ridge]

Is this one increasing?

[wary thorn]

sir just plug in 2 values

[frozen talon]

isn't there some connection between partitioning into even clumps and standard partitioning?

[wary thorn]

yes

its a nice problem to think about

I can give hints

[frozen talon]

gah it's terrible that I have to think about it again lol

I wish I could rember

[wary thorn]

if all 19 numbers have to be >= 10 then its the same as partitioning ||210-190|| into even integers greater than or equal to ||0||

[twin ridge]

So it's decreasing but losing acceleration

[frozen talon]

oh that's not what I was tackling first, I was tacking the evenness lol

[twin ridge]

Is the limit 0?

[wary thorn]

then partitioining ||20|| with 19 even integers >= ||0|| is the same as partitioning ||10|| into 19 ||integers not necessarily even|| >= ||0||

and then there's ||42|| ways to do that. which is just a gross answer for such a simple question lol

[twin ridge]

The limit is 42?

[frozen talon]

that makes sense

[wary thorn]

no haha

we got off topic

sorry sir

I suggest you try this problem on your own first, and then if you find yourself stuck after 10 minutes, open a new doubt channel

[twin ridge]

It is decreasing tho, right?

[wary thorn]

no

if x1=1 then x2 = 1.5

[twin ridge]

But if x2=1.5, then x3=1.33333333333

[wary thorn]

no

and the question is asking you to prove it monotone for all x1 > 1

so

[twin ridge]

This problem does not calculator well

[wary thorn]

it certainly cant be both

increasing and decreasing

for different choices of x1

[twin ridge]

So it is increasing?

[wary thorn]

well it cant be both

[twin ridge]

Why'd you tell me it wasn't at the beginning then?

[wary thorn]

i told u it wasnt decreasing

sir

[wary thorn]

[twin ridge]

You said this, but I had already done that and saw it was increasing, so I thought you saw something I didn't.

[wary thorn]

no I just didn't think you needed a sirs confirmation for that doubt is all

[twin ridge]

I need all the confirmation, I am so confused.

[wary thorn]

yes it is increasing

now go attempt the rest

[frozen talon]

wait, I don't think it is increasing

[twin ridge]

Is it similar to the first one?

[wary thorn]

[frozen talon]

your emotes scare me lol

[wary thorn]

lmao

yeah its decreasing

sory

x >1 means x cannot equal 1

lol what am I doing

[twin ridge]

Bruh I'm so confused

[wary thorn]

its increasing for x<= 1

[twin ridge]

So this thing?

[wary thorn]

[wary thorn]

ye

[frozen talon]

you have died and there are two doors, with one guard in front of each door. one guard always says increasing, one always says decreasing

[twin ridge]

Because we started with x1=2 at that point.

[frozen talon]

you must go to heaven

[wary thorn]

Alright

it is decreasing

now go forth

[twin ridge]

What is my base case in this situation? It doesn't stipulate a universe of discourse for x_1

[wary thorn]

you are on 15 minute timeout

from my help

after that if you are stuck ping me

[twin ridge]

I can't move forward without an answer to that question because I'm already fully stuck

[wary thorn]

https://tenor.com/view/im-doing-this-for-your-own-good-im-doing-this-for-you-this-is-all-for-you-youre-the-reason-im-doing-all-this-kalama-epstein-gif-16761813

https://tenor.com/view/terminator-i-am-not-authorized-to-answer-your-question-not-answering-question-cant-answer-arnold-schwarzenegger-gif-27410901

[twin ridge]

This helps me none

Is my base case 1.1? 1.01? 1.001?

Is it a complex number?

[wary thorn]

Sir..

please

its a sequence still

[twin ridge]

I'm genuinely so confused here.

[wary thorn]

think about it

for 15 minutes

on your own

no questions

[twin ridge]

Bruh

[gloomy scaffold]

what are y'all doing 💀

[twin ridge]

This literally is just making insanely anxious.

[wary thorn]

x1 is arbitrary

[twin ridge]

Can you please just answer that one.

[wary thorn]

but the sequence still starts at 1

its a sequence

of natural numbers

[twin ridge]

How does it start at 1 if it's greater than 1

[wary thorn]

please dont give the solution

[wary thorn]

x1, x2, x3, x4, ...

[gloomy scaffold]

oops my bad

[wary thorn]

the sequence is indexed by the natural numbers

[twin ridge]

Don't I need a value for x1?

[wary thorn]

But that is not what you are inducting over

[twin ridge]

How no?

[wary thorn]

in the previous problem we didnt start inducting at 8

[twin ridge]

We didn't?

[wary thorn]

x1 = 8 in the previous problem

but our induction started at n=1

of course we didnt

go reread your solution

[twin ridge]

I thought we said our base case was 6<8

[wary thorn]

x1 > x2

[twin ridge]

Can I just say that?

[wary thorn]

..

[twin ridge]

Without providign a base case?

[wary thorn]

the base case is x2 < x1

[twin ridge]

Just because?

[wary thorn]

because 6 < 8

[twin ridge]

That was for the first one though

[wary thorn]

yeah

now do the second one

[twin ridge]

That has actual numbers.

[wary thorn]

in the same way

[twin ridge]

Don't I need an actual number for a base case?

[wary thorn]

yes but you will find that the actual value of the number does not matter, given that x1 > 1

for this problem

[twin ridge]

But it's a base case

[wary thorn]

yes and the base case requires showing that x2 < x1

[twin ridge]

I've never not had an actual number for my base case.

[wary thorn]

Okay well now you do

i don't know what else to say about it

[twin ridge]

But I can't do that

[wary thorn]

its like a variable

you can

[twin ridge]

What's telling me I can do that?

[wary thorn]

try working out the algebra

[twin ridge]

But I need an actual value or it's not a base case

[wary thorn]

compute x2, in terms of x1

[gloomy scaffold]

yeah that's kinda the point of induction

[wary thorn]

x2 = 2 - 1/x1

[twin ridge]

x2=2-1/xn

[gloomy scaffold]

there doesn't have to be a numerical base case

[wary thorn]

yes

use that to prove that x2 < x1

given that x1 > 1

[twin ridge]

This is what I'm stuck on. I've only ever done 3 inductions in my entire life so my sample size isn't nearly large enough to do things I've never done.

[gloomy scaffold]

ah

[frozen talon]

you could define x = x1 and proceed, if it helps

[gloomy scaffold]

well you can start trying now!

[twin ridge]

So our base case is x2<x1 because x2=2-1/x1?

[wary thorn]

Well now we've explained how
you need to prove that
x2 < x1
you have that
x2 = 2- 1/x1
So you need to prove that
2-1/x1 < x1
Use the fact that x1 > 1 to prove this, and that is your base case.
I don't think any more explaining is needed, I think you just need to try these steps

[gloomy scaffold]

yeah I agree

@twin ridge just do the math see where it leads you

[twin ridge]

I'm just trying to establish my base case I can do the rest.

I just have no idea how to form that idea into words.

[gloomy scaffold]

your base case is a variable

[wary thorn]

I just formatted the idea into words

right here

[twin ridge]

My base case is x2=2-1/x1?

[wary thorn]

When did I ever say that?

[twin ridge]

I'm infinitely more confused than I was 5 minutes ago.

[wary thorn]

Reread the explanation then

[gloomy scaffold]

I have no idea where you got that from

[twin ridge]

That is what I drew from the explanation.

I don't know how else to read that.

[wary thorn]

it starts with "you need to prove that x2 < x1"

in the first sentence

so that is your base case

[twin ridge]

He said "you have that x2=2-1/x1"

[wary thorn]

yes you do have that from the definition of the sequence

[twin ridge]

My base case is x2<x1?

[gloomy scaffold]

^ do it step by step here please 😭

[gloomy scaffold]

no, prove it first

[twin ridge]

Bruh what the actual shit is my base case here

is it 1.1?

Idk why I can't just get a straight answer

[wary thorn]

I have given you a straight answer multiple times

[twin ridge]

I've been told a billion things of what it isn't.

[wary thorn]

The base case is proving x2 < x1

[gloomy scaffold]

use x1 as your base case, seriously

[twin ridge]

Huh

[wary thorn]

x1 is arbitrary

[gloomy scaffold]

yeah

[twin ridge]

How does that work

[wary thorn]

Re read above where I explained how it works

[twin ridge]

Do I isolate x1?

[wary thorn]

Get x2 in terms of x1

Yes

[wary thorn]

Which is why I wrote this as the next step

[gloomy scaffold]

that's isolating x2

(not x1)

[wary thorn]

I figured that’s what he meant, no real difference

[gloomy scaffold]

but yes same idea

wait aren't most inductions done without a number anyway? (I think I'm missing something here)

[wary thorn]

The base case is typically a numerical equality

Or inequality

[gloomy scaffold]

ah maybe I've got some problems with my understanding, haven't actually learnt this properly yet, might be doing it right with different terms

[twin ridge]

This may be true, but all of the ones I've done have only been with, so I have negative clue on how to do without.

This is my first year of doing this type of class.

I think I went down the wrong rabbithole here. I don't know how to isolate x1 any more. All of my basic algebra skills have flown out the window with all my confusion.

I may just need to stop here. I don't think I'm figuring the algebra out on this one any time soon.

[gloomy scaffold]

@wary thorn sorry I hope you don't mind me asking but for the first part shouldn't it be limited to x1>4 if not x2>x1

[wary thorn]

Are we trying to show monotone decreasing

I haven’t been keeping up

[gloomy scaffold]

I believe so

since you start at 8

[twin ridge]

Well I'm not, but the problem is.

I gave up a while ago.

[wary thorn]

If we’re trying to show it’s monotone decreasing you need to show
x_(i+1) < x_i
For all i
So I’d say the base case is x2 < x1

[gloomy scaffold]

yeah that's what I thought so you do then need to limit it to above 4

if not you will get monotone increasing (also approaching from wrong direction of qn)

[twin ridge]

I'll probably come back to this tomorrow when I'm able to do basic algebra. Thanks for the help.

.close

[midnight plank]

Channel closed

Closed by @twin ridge

Use .reopen if this was a mistake.

[twin ridge]

Although looking at this thing Idek if I'd be able to do that under normal circumstances.

It doesn't even look isolatable.

The longer I stare at that thing the more I'm convinced I'm getting punked

[midnight plank]

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[inland summit]

The complete and exhaustive range of values of 'x' satisfying the inequality |2x+ 1| > x, is

[midnight plank]

Please don't occupy multiple help channels.

[sweet wing]

!status

[midnight plank]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[inland summit]

1

[sweet wing]

whenever you see absolute values, you have to think about the positive and negative part

the positive part is removing the absolute value

and the negative part is taking the negative of whats inside the brackets

[sweet wing]

this is a bit weird phrasing, i cant think of the correct term, but you get the point

[inland summit]

Not a problem bro

Yes I do

[sage helm]

square both sides

[sweet wing]

shhhhh

you could also square it, yes

[sage helm]

but there are scenarios where you create extra solutions

so be careful

its always better to piecewise break the absolute value

[inland summit]

I will try it brothers thank you so much

[idle phoenix]

why not consider the values when (2x+1) is negative an (2x+1) is positive then find the intersection of the sets of the solutions

[inland summit]

I tried it and got -infinity to -1/2

But it was wrong according to the answer key

[sweet wing]

you mean $-\infty<x<-\frac12$?

[grand pond]

Flappie

[inland summit]

S

Q. 56

[sweet wing]

,rotate

[sweet wing]

can you explain how you got here?

[inland summit]

It took 2x+1 as +

So 2x+1>x

Subtract x on both sides

Oh shit

X>-1

[sweet wing]

you look at the absolute value

and break it up into two parts

the first part is where the inside is positive

e.g. 2x+1>0

which is for x>-1/2

and we take that as our condition

x>-1/2, so 2x+1>0, so 2x+1=|2x+1|

we can then reduce the absolute value to just its value

solve |2x+1|>x with x>-1/2

[inland summit]

I don't know how to do this

[sweet wing]

$|2x+1| = \begin{cases} 2x+1 \text{ for} x>-\frac12\-2x-1 \text{ for} x<-\frac12\end{cases}$

[grand pond]

Flappie

[sweet wing]

does this make sense?

[inland summit]

Ya

[sweet wing]

to solve this, we look at every case

[inland summit]

-x<a<x

[sweet wing]

lets start with the first one

$|2x+1|>x\iff 2x+1>x \wedge x>-\frac12$

[grand pond]

Flappie

[sweet wing]

you can solve this

you know how to solve 2x+1>x

[inland summit]

S

[sweet wing]

can you solve this please?

[inland summit]

X > -1/2

[sweet wing]

no

$2x+1>x\x+1>0\x>-1$

[grand pond]

Flappie

[sweet wing]

yes?

okay

so the bound that we get here

is less strict

[inland summit]

K

[sweet wing]

$x>-1\wedge x>-\frac12\implies x>-\frac12$

[grand pond]

Flappie

[sweet wing]

agree?

or disagree

okay

so we have solved the first case

now we look at the second case

which is $-2x-1 > x \wedge x<-\frac12$

[grand pond]

Flappie

[sweet wing]

can you try and solve this one please?

[inland summit]

X < -1/3

[sweet wing]

indeed

and what can we simplify $x<-\frac13\wedge x<-\frac12$ to?

[grand pond]

Flappie

[inland summit]

What

[sweet wing]

you can simplify it to a single <

[inland summit]

X<-1/2

[sweet wing]

exactly

so, in our first case, x>-1/2 and in our second case x<-1/2

which is all x

[sweet wing]

(to be precise one of them should be <= or >=)

[inland summit]

Ok

[sweet wing]

does all of this make sense?

,w graph |2x+1|>x,x=-1/2

[grand pond]

Results provided by WolframAlpha

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Requested by flappie.

[sweet wing]

wtf.....

,w graph |2x+1|>x

[grand pond]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by flappie.

[sweet wing]

finally

[inland summit]

So the answer is (- infinity, -1/2) ,right brother

[sweet wing]

no

its $\mathbb{R}$

[grand pond]

Flappie

[inland summit]

-infinity to +infinity?

[sweet wing]

look at the graph

[sweet wing]

yes

if you look at the graph

you see the notch

[inland summit]

Ya

-1/2

[sweet wing]

yes

so we look at the case where x<-1/2

and then we look at the case where x>-1/2

and we take the union of that

lemme give you a harder one

[inland summit]

K

[sweet wing]

$|2x-4|>x$

[grand pond]

Flappie

[sweet wing]

this one is a bit harder

[inland summit]

Is this correct

?

[sweet wing]

no

you made the mistake in the cases

|2x-4| = 2x-4 or -(2x-4)=-2x+4

you have to take the negative of the whole statement

[inland summit]

I have taken it as -2x+4

I ts my handwriting

[sweet wing]

ah okay, sorry

the x<0 is wrong

[inland summit]

I made it look like a.

[sweet wing]

x>=0

aswell

its at a different x value

[inland summit]

Oh ya

I have interchanged the cases

Sry abt that

[gray lance]

@sweet wing @inland summit did you ever realize it was wrong from the start??
It should be x<2, not x<0

also idk how to type the bigger and equal sign so

[sweet wing]

what?

thats what i commented

[sweet wing]

right here

[gray lance]

I meant to erase your name but I accidently included it

[sweet wing]

but i didnt want to give the answer right away

[gray lance]

sorry

alright i gtg

cyaaaaaaa

[sweet wing]

its oaky

ive made similar mistakes before

[inland summit]

So it is x>=2 and x<2 ?

[sweet wing]

yes

you are looking for 2x-4>0

which is for x>2

[inland summit]

Then 2x-4>x

X>4

Right?

[sweet wing]

yes

[inland summit]

But how can we prove when x<2

[sweet wing]

so then we take the other case

x<2

-2x+4>x

[inland summit]

It can be + or - right

[sweet wing]

and solve again

[inland summit]

X<4/3

I got it now

[sweet wing]

okay, so whats the interval?

[inland summit]

(4/3, 4)

[sweet wing]

almost

[sweet wing]

here you got x<4/3

[sweet wing]

and here x>4

so how can 4/3<x<4?

[inland summit]

(4, 4/3)

[sweet wing]

come on

you know thats not the answer

[inland summit]

(4, infinity)

[sweet wing]

yes and what else?

[inland summit]

U(-infinity,4/3)

[sweet wing]

exactly

there you go

[inland summit]

I drew the number line brother

[sweet wing]

,w graph |2x-4|>x

[grand pond]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by flappie.

[inland summit]

Thank you brother

For your time and patience

[sweet wing]

no problem

i just hope that you understand it

and if not

keep asking

[inland summit]

I understood it bro

I have another doubt

[sweet wing]

just ask

[inland summit]

Here should I do the same and take different cases of x in terms of the absolute values

?

[sweet wing]

its a bit harder since you have some more intervals

but there is probably a smart way to do ti

but i cant see it

[inland summit]

This is a pyq of worlds second toughest exam bro

[sweet wing]

i would split it into 3 intervals

[inland summit]

I can see two of them

[sweet wing]

which do you see?

[inland summit]

One using x+2 and another using 2x+3

Right

?

[sweet wing]

i see 3, $-\infty<x\leq -2$,$-2<x\leq -\frac32$, and $-\frac32<x<\infty$

[grand pond]

Flappie

[midnight plank]

@inland summit Has your question been resolved?

[midnight plank]

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[sweet wing]

whats the question

[twilit field]

differentiate it I guess

[last slate]

find dy/dx

[twilit field]

try bringing all variables to one side and then differentiating

$x^3=sin(u)$

may help

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

$(y^3)=sin(t)$

[grand pond]

ƒ(Why am. I here)=I don't Know

[last slate]

how did you get this idea?

[twilit field]

The powers are messy

so I thought a trig sub may help

doubt it does though

[sweet wing]

yes i see that, but whats your question

[last slate]

oh

[last slate]

status 1

[sweet wing]

!status

[midnight plank]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[sweet wing]

1

this?

xd

[last slate]

yes

xdf ?

[sweet wing]

okay

[sweet wing]

typo

[last slate]

oh okay

[last slate]

it does

[dull yoke]

@last slate which class is this for

[twilit field]

$cos(u)+cos(t)=a^3 (sin(u)-sin(t))$

[grand pond]

ƒ(Why am. I here)=I don't Know

[last slate]

12th

[dull yoke]

like calc1, calc2 ?

[last slate]

no idea

ig calc 2

[twilit field]

in asia afaik that really isn't a thing

[twilit field]

really?

[sweet wing]

you can use implicit differentiation

$\frac{d}{dx}(\sqrt{1-x^6}+\sqrt{1-y^6}=a^3(x^3-y^3))\\frac{d}{dx}(\sqrt{1-x^6}+\sqrt{1-y^6})=\frac{d}{dx}(a^3(x^3-y^3))$

[last slate]

but i don't see how it leads to any of the options

so i didn't go any furthur

[grand pond]

Flappie

[last slate]

yes,3 min

I am doing it

Got it

thanks everyone!

.close

[midnight plank]

Channel closed

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Use .reopen if this was a mistake.

[twilit field]

[sweet wing]

howd you find it?

[last slate]

more like i understood the solution

should i send it ?

.reopen

[midnight plank]

✅

[last slate]

$cos(theta+phi/2)+cos(theta-phi/2)=a^3 (sin(theta+phi/2)-sin(theta-phi/2))$

[grand pond]

Cnidarian

[last slate]

wow

okay

now

this would be simplified to

[sweet wing]

$\cos(\theta+\phi/2)+\cos(\theta-\phi/2)=a^3 (\sin(\theta+\phi/2)-\sin(\theta-\phi/2))$

[grand pond]

Flappie

[sweet wing]

so, trig substituion, yes?

[last slate]

yes

oh wait

mb

[last slate]

simpifying this

[sweet wing]

ah okay

i see

no need to write it all out

[last slate]

Okay

[last slate]

this is wrong btw

[sweet wing]

i just cleaned up what you wrote

:)

[last slate]

well mine is wrong too

[sweet wing]

okay, i get the gist of it

[last slate]

$cos(\theta)+cos(\phi)=a^3 (sin(\theta)-sin(\phi))$

[sweet wing]

i was just curious what you used to eventually solve it

[grand pond]

Cnidarian

[last slate]

hard to get the idea of trig ssubstitution

[sweet wing]

these should be squares, no?

[last slate]

nope

there is a square root

[sweet wing]

ohhh, right

xd

[last slate]

.close

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[midnight plank]

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[last slate]

For the first station
Since there are 19 more stations in front of it we need 19 tickets
For the second
There are 18 stations so we need 18
and so on

so the total no of tickets required would be
19+18+17+.....+1= 190

But the answer is 20P2 = 380

[twilit field]

,w 19*20

[grand pond]

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[twilit field]

yup, so 20 choices for station 1

[last slate]

where was i wrong ?

[twilit field]

19 for station 2

[last slate]

19 for the first

because the station itself would not come under the count

[last slate]

And why is it 20P2 ?

[junior flower]

what exactly would the route look like?

[last slate]

i am guessing its a straight line

[junior flower]

no i mean like

the visiting order with your way

[last slate]

i didn't get you

[junior flower]

ok nvm that

there are 20 stations

for each two stations a and b we need a ticket from a to b and from b to a, right?

[last slate]

Since the question says "every station after" isn't it just 1 ?

if a comes first then a to b or if b comes first then b to a

[last slate]

or did i interpret this wrong ?

[junior flower]

the question is poorly written so i’m choosing to ignore the word ‘after’ because that makes the answer 20P2

[last slate]

Okay

but how is it 20P2 ?

[junior flower]

“the number of tickets required in order that it may be possible” like bruh what

who writes that

[junior flower]

how many ways to choose 2 stations?

[last slate]

20P2

no

20C2*

[junior flower]

well sure with order

and you do want order

because you want tickets from a to b and b to a to be considered distinct

alternatively, 20C2 * 2

[last slate]

oh

i get it

Thank you

.close

[midnight plank]

Channel closed

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[junior flower]

i don’t really know what ‘after’ is supposed to mean

it does sound like what you described earlier but probably not supposed to be that

[last slate]

Okay,I will report it

[last slate]

Say the stations are the numbers on a number line
for station 1
all the stations "after" would be 2,3,4,5,...
but i think that is what gives a different answer

[junior flower]

yea

[midnight plank]

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[grave locust]

hello, can someone help me understand why this is true?

i dont really get it

[main current]

I'll use ' for "inverse".

(AB)' = B'A'

That's most of it right there

[grave locust]

yes

and for the B

because i would expect B transposed and inversed * A to be =

but they changed the order of the inverse and transposed

[main current]

Did have to look it up to be sure, yes the inverse and transpose do commute

https://math.stackexchange.com/questions/340233/transpose-of-inverse-vs-inverse-of-transpose

[grave locust]

ah ok

that makes sense

thank you very much

.close

[midnight plank]

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[midnight plank]

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[bleak pier]

A coin is tossed six times then the probability of obtaining heads and tails alternatively is?

[bleak pier]

Total outcomes 2^6

Favourable 2

So 2/64 =1/32

[full marlin]

seems right

[bleak pier]

I see

.close

[midnight plank]

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[midnight plank]

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[severe glade]

Can someone explain to me why my answer is wrong for this equation of a circle question?

[hard shard]

you have two x variables

should be (x-x1)^2 + (y-y1)^2 = r^2

[midnight plank]

@severe glade Has your question been resolved?

[severe glade]

I’m confused what does that mean for my answer ? @hard shard

[hard shard]

look at how you wrote your circle equation

you put (x-x1)^2 + (x-y1)^2 = r^2

you have two x variables instead of an x and a y

[midnight plank]

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[wild glen]

I am having trouble justifying my answer for b). Isn't positive orientation by definition outward? I am struggling to find a clear definition.

[wild glen]

Surface Integrals

The textbook gives this example, and the last sentence is partially where my confusion is coming from

[pearl idol]

No, the positive orientation of the cylinder is outward in that example since n in that example points outward

You should look at the vector n you found in part a and check which direction it points in

[wild glen]

So the vector I get is -cos(v)i+sin(v)j

So would that be inward because it has a negative term?

What exactly am I checking, the direction compared to r?

[pearl idol]

think about what happens when for example u=v=0

then the point is r(0,0) = (1,0,0)

and n(0,0) = (-1,0,0)

so it would be inward because it points from r(0,0) towards the center of the cylinder (x=y=0)

[wild glen]

How do I know it points to the center?

[pearl idol]

Do you know how to plot a vector

If you imagine the vector n(0,0) being based at the point r(0,0), then the start of the vector will be (1,0,0), and it will point in the direction (-1,0,0), so it will end up at (0,0,0)

you can try this for a couple other points too just to make sure you understand, like for example v=pi/2 and u=1

[wild glen]

Why is that the start of the vector though? And if it points in the direction of -1,0,0, whouldn't that be out of the surface?

[pearl idol]

r(0,0) is the point on the surface that n(0,0) corresponds to

Can you try drawing the surface sigma and labeling r(0,0) and n(0,0) and showing me?