help-49

[subtle blaze]

It’s not your first day here you know how we work

You post your question along with any work you’ve done so we know how far along you are

[tidal turret]

as I said, I only verified the base case

because when trying to assume n holds and then doing the n + 1 Idk if I can do it

[tidal turret]

I am . . . stuck

but is fair, if you dont want to help me, is okay, I am not only stuck I am also kind of like disappointed with how things have went this week, I have so much classes daily and I get back to home just to sleep or do homework, sorry it feels like I am demotivated sometimes or non enthusiastic

university life has hit me hard, like killed all my motivation I had from before, topics are covered very fast, too many classes, too many people in my classes, from monday to friday my classes end at 10.30pm and I get home by 11pm

is not like I am venting, is just that I understand you despise me, I also despise myself, things have been rough this past weeks, and as I said, I have had like a complicating time just not giving up, at this point I just want to deenroll and do something else or something, is hard when things dont go well at the start

[crude compass]

It's not that we don't want to help, but I'm too tired right now to get into this. My guts tell me that the proof must be similar to that of the binomial theorem, if you want to get into it. You can ignore the weird binomial notation.

[grand pond]

Sam

[tidal turret]

well, combinatorics are covered later in this class, it might still be possible without them

[crude compass]

Yes, my point is that the binomial theorem uses methods to prove that might translate to your case, but do feel free to ignore the combinatorics and treat them as some magical coefficient that just works.

And I might be wrong, but it is a good example of proof by induction in general.

Also, do take breaks from working. You can't solve problems efficiently when you're tired.

[fast pawn]

Do you need help with proving the general formula or the binomial one

[crude compass]

Nothing to do with binomial, check the pin for the initial problem.

[fast pawn]

So just the classic yes?

[crude compass]

All I was saying is that I feel, at a first glance, that both proofs are similar. I'm going to sleep, so I won't really be helping more than that.

[fast pawn]

Okay, @tidal turret i can relate, so

Where are u getting stuck

[tidal turret]

proving using induction

[fast pawn]

Okay

[crude compass]

Something to do with geometric series formula by induction.

[fast pawn]

Do you understand how induction works?

[tidal turret]

yes

[fast pawn]

Great

U got base case I understand right

[tidal turret]

ye

n=1

[fast pawn]

We proving this?

[tidal turret]

well

[fast pawn]

I mean, that just first look looks way harder than the classic, but still doable

[tidal turret]

we need to use this to prove that is equivalent to the classic geometric formula

I mean, from this sum we need to derive the geometric formula

[fast pawn]

I think it should go with something along

Proving that this holds true

And then setting b to some arbitrary value

[tidal turret]

that works ig

[fast pawn]

So, would u know how to prove this

Or need some clues

[tidal turret]

a little bit of help is appreciated

[fast pawn]

Okay, u obviously assume this holds for n

And for n+1 what you’re gonna want to do

[tidal turret]

,, a^{n+1} - b^{n+1} = (a-b) \sum_{i=1}^{n+1} a^{i-1}b^{n-i}

[grand pond]

Renato

[fast pawn]

I mean

There is certainly a way to do it

By multiplying the sum out with the a b

And showing that the terms cancel out

But there has to be a more clever way

[tidal turret]

what?

[fast pawn]

Think of the sum

What it represents

If u went to write out everything by hand

[tidal turret]

cant I divide by (a-b)?

[fast pawn]

U would go like

[fast pawn]

Wouldnt help really

[tidal turret]

dude

[fast pawn]

Just change the sum into

[tidal turret]

,, \sum_{i=1}^{n+1} a^{i-1}b^{n+1-i} = \frac{a^{n+1} - b^{n+1}}{a-b} \ \sum_{i=1}^{n+1} a^{i-1}b^{n+1-i} = \sum_{i=1}^{n} a^{i-1}b^{n-i} + \frac{a^{n+1} - b^{n+1}}{a-b} \ \sum_{i=1}^{n+1} a^{i-1}b^{n+1-i} = \frac{a^n - b^n}{a-b}+ \frac{a^{n+1} - b^{n+1}}{a-b}

[tidal turret]

a cant be equal to b

[fast pawn]

(ab^n-1 - b^n) + ……. + (a^n - a^n-1* b)

[fast pawn]

Thats fine

[fast pawn]

Isnt even induction lol

[fast pawn]

Eventually showing that all terms cancel out except for -b^n + a^n

[tidal turret]

we need induction

[fast pawn]

This proof works just fine as well

But if your strict on induction

This would be

😂

Well

But does this move you somewhere?

I think this is the type of induction where u just add and subtract the same term

I see a slight mistake however

If the sum upper bounds is

n+1

The b should have n+1-i in the exponent

[grand pond]

Renato

[fast pawn]

I suppose ab^k probably helps

[tidal turret]

wait

,, \sum_{i=1}^{n+1} a^{i-1}b^{n+1-i} = \frac{a^n - b^n}{a-b}+ \frac{a^{n+1} - b^{n+1}}{a-b} \ \sum_{i=1}^{n+1} a^{i-1}b^{n+1-i} = \frac{a^n - b^n+ a^{n+1} - b^{n+1}}{a-b} \ (a-b)\sum_{i=1}^{n+1} a^{i-1}b^{n+1-i} = a^n +a^{n+1} - b^n - b^{n+1} \ (a-b)\sum_{i=1}^{n+1} a^{i-1}b^{n+1-i} = a^n(1 + a) - b^n(1 + b)

[grand pond]

Renato

[fast pawn]

I think these operations will probably make you spin in circle

Okay I think I got it

[tidal turret]

I need to use induction dude

[fast pawn]

With induction

Okay

So

U begin the inductive step with

a^n+1 - b^n+1

[tidal turret]

do you have paper?

can you write it in latex?

[fast pawn]

I can write it down on a whiteboard

[tidal turret]

sure

feel free to ping me dude

[fast pawn]

Sorry, took me a while

[tidal turret]

how did this happen

[fast pawn]

Okay, so

I was thinking that I could use a term that I add and subtract to make it easier

And something ideal would be

[tidal turret]

how did this happen

[fast pawn]

That would leave me with the original a^n + b^n

So I could use the early assumed formula

[tidal turret]

oh right

ye ye

[fast pawn]

Substituting the formula back

[tidal turret]

ye

inductive HIPO

[fast pawn]

Thats the smart word for it ye

[tidal turret]

how did it happen

[fast pawn]

Okay

So

The a

U put it into the sum formula

As its essentially just a bunch of a^0b^n etc

And u make it

a^1 b^n

a^2 b^n-1

[tidal turret]

how did this happen?

[fast pawn]

Well, u essentially add the term into the sum

a^0*b^n

Oh wait

[tidal turret]

this last step is not correct my guy

we. . . are close though, maybe

[fast pawn]

Yes

It is correct I believe

The first and last term are correct

As it stands

The first term being a^0 * b^k

And last one being b^0 * a^k

[tidal turret]

?

[fast pawn]

Its right

[tidal turret]

elaborate

[fast pawn]

Uh, well

U add one element to the sum

b^k* a^0

Currently the sum goes from a^1 * bk-1 to a^kb^0

[tidal turret]

it doesnt make any sense

[fast pawn]

What exactly?

[fast pawn]

Well

It is verifyable easily

You just adjust the indices based on the term we add

[tidal turret]

it doesnt make any sense whatsoever

[fast pawn]

Hm?

Well, at least

Point out the exact part

I can try and help

[tidal turret]

it went from a sum to a multiplication

literally impossible

[fast pawn]

Wdym

It just added one term into the sum

Wait lets get this straight

Is it the first row thats confusing?

Or the second one

[tidal turret]

dude

,, \sum_{i=1}^{n+1} a^{i-1}b^{n+1-i} \neq \sum_{i=1}^{n} a^{i}b^{n-i} + b^n

[fast pawn]

I mean

It works

Shall I try explaining the indices more clearly?

[grand pond]

Renato

[tidal turret]

a^n . b^0 is the last term

you are tripping hard here

[fast pawn]

I dont know what to say

Does the last term change with the adjustment?

Just write out a few terms

First and last

[tidal turret]

your claim I dont think its true

[fast pawn]

What do you think is the first term prior to the adjustment

And after

[tidal turret]

first term is a^0. b^n, last term is a^n . b^0

[fast pawn]

Good

[tidal turret]

from the lhs

[fast pawn]

RHS now

[tidal turret]

lhs, first term is ab^(n-1) last term is a^nb^0

[fast pawn]

Gooood

Now

When we add b^n a^0

Doesnt it become the one on the left?

[tidal turret]

no

[fast pawn]

😭

[tidal turret]

you need to add the last one

a^n . b^0

[fast pawn]

Which side is missing that one please?

[tidal turret]

the rhs

[fast pawn]

U sure?

[tidal turret]

you add a^n . b^0 to the rhs and it becomes lhs

[tidal turret]

yes

[fast pawn]

So u say the summation on the right does not include a^n* b^0?

[tidal turret]

oh, my bad

[fast pawn]

Is aight

Just made me reconsider my education for a second there

[tidal turret]

maths surely are tricky

[fast pawn]

So

All good?

[tidal turret]

ye

[fast pawn]

Ggwp

[tidal turret]

math is tricky my guy

[fast pawn]

🤨

[tidal turret]

u from america?

[fast pawn]

No

Europe

[tidal turret]

i appreciate the help

[fast pawn]

No problem

[tidal turret]

i lowkey had a good time with this one

[fast pawn]

Appreciate that

So did I smelling the markers

Surely enlightened my morning

[tidal turret]

the fk?

[fast pawn]

On the whiteboard

[tidal turret]

you get taught basic proofs at hs level there?

[fast pawn]

Uhhhh

We did

Induction is not a standard here generally

Until uni

I am CS tho so I get to use mostly that

[tidal turret]

i see, fair enough

i appreciate the help jurj

[fast pawn]

No problem, maybe someday again

[tidal turret]

ye, have a good day, fella

.solved

[midnight plank]

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[midnight plank]

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[vale charm]

what is the TSA

[vale charm]

for this

[fiery onyx]

dyk how to find tsa?

[vale charm]

yep i got 1690.84 but idk if that is correct

[fiery onyx]

okay walk thru how you did it

[vale charm]

ok

hold on

yo why is that different fom my caculator

the desmos calcualtor

[buoyant linden]

$\frac{2\cdot\pi \cdot 14.65^2}{2} + 2\cdot 14.64 \cdot \pi \cdot 16.2 - 29.3 \cdot 16.2$

[grand pond]

Flatus

[buoyant linden]

so people can see it betteer

[vale charm]

so is my answer right

[buoyant linden]

i didn't get your answer when i subbed it into my calculator

can you explain

how you got your solution

[fiery onyx]

why did you subtract 29.3x16.2 ?

that's the area of the base it should be included in the tsa

[vale charm]

desmos shows 1690.784 but my calculator said 1690.84

[fiery onyx]

so it's open at the base?

[buoyant linden]

wait it says open on one end

wouldn't that mean

one of the semicircles

is gone

[vale charm]

yep its openeed at the base

[buoyant linden]

oh jokes?

[fiery onyx]

idk are you in radians?

[buoyant linden]

wouldn't change a thing?

we're not dealing with trigonometric functions

[fiery onyx]

idk lmao

anyway

[vale charm]

wait so is my cacultor right or desmos

or do i have to change the settings somehow of my calculator

[fiery onyx]

well if it's open at the base subtracting the base doesn't give you the tsa? you just dont include it in your answer

and instead double whatever you get

im pre sure

it's kinda like, you're subtracting air?

unless you did the working out differently

[buoyant linden]

jessberry

kinda random but js a heads up

discord keeps tagging you as a likely spammer for me

[fiery onyx]

but I wld find the area of the long stuff on top, then the 2 half circles

and then double it

[vale charm]

so i just ignore it

[fiery onyx]

well that's js how I would do the question

if it makes sense

[vale charm]

thats suppose to be the answer

[fiery onyx]

ok so let's go thru it

so dyk how to find the area of the long bit on top?

[buoyant linden]

oh yeah

i got that

there is no open side btw

[fiery onyx]

omg 😭!

[vale charm]

oh me flipping lord

[buoyant linden]

i assumed no open sides and got that answer

would you like me to run you through the solution or guide you to the solution

[vale charm]

so i add the rectoangel instead of subtracitg

[buoyant linden]

yes

you wouldn't ever subtract it though

if it was open

[vale charm]

wait lemme figue this out

[buoyant linden]

you just leave it

cuz it was never added anyways

[fiery onyx]

no it just isn't included

[buoyant linden]

yeppers

except in this case

you do need it

[fiery onyx]

Yuh

[vale charm]

?

[fiery onyx]

okay just try do it knowing that it's actually closed now

[buoyant linden]

so basically

add the half circles faces

add the rectangle

add the curved face

[fiery onyx]

bc I think you know how to do it

[vale charm]

but isnt it asking for half the cyclinde r

[buoyant linden]

essentially, yes

you could do it that method

find total surface area of a cylinder

half it
then add a rectangle

[vale charm]

ok ok ok

lemme try

oh me goshh

i got it

[buoyant linden]

noice

[vale charm]

so we caculate the whole cylinder, divided it by two and add the rectangel

but it said one open end

am i tripping

[buoyant linden]

yep

[buoyant linden]

question's trippin

[vale charm]

SO WHY THE HECK

DO WE NOT MINUS

[lapis ruin]

Hi, I'm new here. Anyone want to take the IMO test here?

[vale charm]

BUT PLUS IT

[vale charm]

no thanks

[lapis ruin]

🙂 really

[vale charm]

ye dude. i am doing math

[buoyant linden]

wait what

[fiery onyx]

Let’s see the whole question maybe

[fiery onyx]

Well it’s a “closed”half cylinder

I think it’s a misprint

See how it says “cylinder with” then just stops

I think it’s mesnt to be “cylinder with one open end”

[vale charm]

R WE FR RN

R WE FRRRRR

[fiery onyx]

Close half cylinder one open end doesnt make grammatical sense lmao

[fiery onyx]

Rip..

[vale charm]

its ok we still figured it out

THANKS

.close

[midnight plank]

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[fiery onyx]

so instead of multiplying by 5 I multipled by 10, but won't i get the same answer anyway?

[buoyant linden]

hey jess 😭

[fiery onyx]

hey again !

oops just realised i didnt simplify 4 to 2

but it's still wrong anyway

[buoyant linden]

hm

[lyric charm]

4+72 isn't 80

[buoyant linden]

ah lol

[lyric charm]

sloppy arrow, sorry

[fiery onyx]

omg!

Well that’s embarrassing.. thanks 😅

.close

[midnight plank]

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[lyric charm]

<@&268886789983436800> baldi trolling again

[midnight plank]

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[prime hornet]

I guess that tweet did draw in some attention

[midnight plank]

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[proud abyss]

Problem Statement: Given a string s, return the longest palindromic substring in s (without using dynamic programming). how do i start with this one?

[ashen sequoia]

i would try choosing each letter/pair of letters as center of palindrome and check whether if you expand it in both sides, would it be palindrome

[proud abyss]

time complexity here would be O(n^2), right?

also would doing it recursively an easier option? or no?

[midnight plank]

@proud abyss Has your question been resolved?

[proud abyss]

.close

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[twilit field]

I'm trying to prove this sequence converges

Now calculations hint at the limit being 1

so I first want to prove the seqeunce is positive

[dreamy lichen]

Why if I may ask

[twilit field]

Well, I thought it would come in handy to prove the seqeunce is monotone decreasing

Anyhow's I have proven it is positive , so I'll move on to that

[dreamy lichen]

I'd start by sketching the proof of it being monotone decreasing and then you'll know what will be handy and what will not (in this case, i dont think it'll be very handy)

[twilit field]

hmm

I mean $a_{n+1}-a_{n} = \frac{1}{a_{n-1}}-\frac{1}{a_{n}}<0$. Which follows directly from the inductive hypothesis that $a_{n-1}<a_{n}$

[grand pond]

wai

[dreamy lichen]

Oh, I guess this works

[twilit field]

I now have to show the infimum of the sequence is 1$

Let there be a greater lower bound( let' call this $l$, let it be the supremum of (a_n)

[grand pond]

wai
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[west portal]

what level math is this

[dreamy lichen]

seems like basic real analysis, so like first (or 2nd) year uni

[west portal]

seems complicated wow

[dreamy lichen]

you probably mean infimum btw

if you want a hint, lmk

[twilit field]

yea

[lyric charm]

is it actually that?

,calc 2-1/1.5

[grand pond]

Result:

1.3333333333333

[lyric charm]

,calc 2-1/(4/3)

[grand pond]

Result:

1.25

[lyric charm]

ok, no CE from the first 3 terms, it looks like.

[twilit field]

So $\forall \varepsilon \exists a_m : l<a_m < l+\varepsilon$

[dreamy lichen]

I think that wai's argument is correct, base case is trivial and the induction step seems to work, given that the sequence is positive

[grand pond]

wai

[twilit field]

Right so far?

[dreamy lichen]

it might work, but it's not the simplest approach

[twilit field]

what;s another approach

[dreamy lichen]

consider what will happend with the equation a_(n+1) = 2 - 1/an for very large n, given it converges

and specifically, what happens to a_n+1 and an

and what happens in the limit

[twilit field]

the difference goes to 0

[dreamy lichen]

(an) goes to 0?

[dreamy lichen]

Yeah, so in the limit they'll be equal

[twilit field]

So you're essentially saying argue for cauchy convergence

[dreamy lichen]

if you expand lim |a(n+1) - an| = 0, you should be able to find the limit

[dreamy lichen]

u know it converges, dont you? It's monotone and bounded from below (since its positive)

you only need to find the limit

[twilit field]

I don't get it?

[dreamy lichen]

Well, we know that the sequence (an) converges (it's monotone decreasing and bonuded from below by 0)s

so we only need to find the limit

say the limit is a. What I claim is a = 2 - 1/a (because in the limit, an+1 is the same as an)

[twilit field]

sure, but I have to prove that

[dreamy lichen]

to rigorously prove it, you can use this

replace an+1 with the defn

and apply few limit theorems

[twilit field]

$\abs{a_{n+1}-a_n} =\abs{2-a_n-\frac{1}{a_n}}$

[grand pond]

wai

[dreamy lichen]

we can even get rid of the absolute value and just write an - an+1, since its decreasing

[twilit field]

sure

so 2-a_n-1/a_n

[dreamy lichen]

then take the limit and apply algebraic limit theorem, until everything is in terms of lim(an)

then you have an equation and you can solve for lim(an)

[twilit field]

how( sorry , just really confused)

I don't know this is 0, do I

[dreamy lichen]

lim 2-a_n-1/a_n = 0

[dreamy lichen]

you do

[twilit field]

how

[dreamy lichen]

the sequence converges, therefore its cauchy

[lyric charm]

a_n is not yet known to converge is it

[dreamy lichen]

that's one way

[twilit field]

It's monotone bounded

[dreamy lichen]

we know its bounded from below (wai proved its positive) and we know its monotone decreasing

[twilit field]

Sure, it's cauchy, but how does that tell us , this is 0

[dreamy lichen]

the difference of consecutive terms must tend to 0

[twilit field]

I'd have to prove that then

[lyric charm]

if $\lim a_n = L$ then $\lim (a_n - a_{n+1}) = L - L$

[grand pond]

Ann

[dreamy lichen]

a_n+1 is the same sequence just shifted, so it must converge too and converge to the same value

lim(an - an+1) = lim(an) - lim(an+1) = L - L

[zealous schooner]

Why is [ a = \lim_{n\to\infty} a_{n+1} = \lim_{n\to\infty} 2-\frac1{a_n} = 2 - \frac1{a} ] not rigorous enough?

[dreamy lichen]

thats another way to prove it without refering to it being cauchy

[grand pond]

kheer257

[dreamy lichen]

(an) = (3/2, 4/3, 5/4, ...)
(an+1) = (4/3, 5/4, ...)
Do you understand why these have the same limit?

[twilit field]

yea

[dreamy lichen]

okay so we can use this

a = lim an = lim_an+1

thats how the first step is justified

[twilit field]

yea, then the limit is 1

[dreamy lichen]

yeah, right

[dreamy lichen]

the 2nd step is just defn of an+1 and the third one is algebraic limit theorem lim(2 + 1/a) = lim(2) + lim(1 / a) = 2 + 1 / lim(a) = 2 + 1/a

this is how recursive limits are usually computed btw, so this approach is worth remembering (and understanding ofc)

[twilit field]

I know this method 😭

I just need extreme rigor and formalism for this course

[dreamy lichen]

I see, well, do you now understand how that method is rigorous? Each step can be justified by some limit theorem

alr, great

[zealous schooner]

This is rigorous enough

You can unpack the definitions to make it rigorous

[twilit field]

I mean we do still have to do cauchy convergence in class, but should be fine soon

[dreamy lichen]

this doesnt need cauchy convergence, it was just the first and simplest justification that came to my mind

lim(an+1) = lim(an) is really all you need (and this is obvious, its the same sequence, just shifted, it can be said it follows from lim(subsequence) = lim(sequence) if you know that theorem)

[twilit field]

thanks

[twilit field]

yea, I've proven that

One more question

Checking the convergence of this

the limit is 3/2

[dreamy lichen]

u sure?

[twilit field]

uh, no

I forgot this was recursive

I'll first determine if it's increasing

Well, it is increasing

the question is how to prove it

Let $a_{n}>a_{n-1}$. Then $4+3a_{n}>4+3a_{n-1}$

[grand pond]

wai

[twilit field]

And $3+2a_{n}>3+2a_{n-1}$

[grand pond]

wai

[dreamy lichen]

the issue with this is that you'd need 3 + 2an < 3 + 2an-1 in order to put those together

[twilit field]

yea

I

I'm thinking

I could break it down a bit

[zealous schooner]

If you think it’s increasing just try to prove a_{n+1} > a_n directly

It’s especially easy when a_{n+1} is defined purely in terms of a_n

[dreamy lichen]

you might need to prove a lemma or two with induction in the process, but it's prolly the easiest way

[twilit field]

I see

okay, I think I can do it from here

Thanks!

.close

[midnight plank]

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[zealous schooner]

Find a domain in which (4+3x)/(3+2x) > x, and then prove a_n always lies in that domain

[midnight plank]

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[idle dirge]

Hello!

[idle dirge]

How do I solve this one?

I tried e^ln

And tried to fish some L hopital

But I got stuck

[lyric charm]

the obvious first step imo is to put $t := \ln(x)$ so that you get a limit as $t \to 1$

[grand pond]

Ann

[idle dirge]

Alright, let me continue

Now what?

Also did this

<@&286206848099549185>

[cobalt heart]

salut romania

[midnight plank]

@idle dirge Has your question been resolved?

[idle dirge]

Salut boss!

[midnight plank]

@idle dirge Has your question been resolved?

[midnight plank]

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[umbral scroll]

How to calculate lim (x->+oo) e^(-x) * ln(x)

[umbral scroll]

I have no idea

I tried nothing

[alpine gyro]

$\lim_{x \to \infty} \frac{ln(x)}{e^x}$

[grand pond]

Good

[alpine gyro]

What do you expect when you solve the limit?

[spare beacon]

just think e^x grows way faster than x

[sick fossil]

are you required to show your work, and can you use L’Hopital?

[midnight plank]

@umbral scroll Has your question been resolved?

[midnight plank]

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[safe shale]

can anybody help me find a, b and c ?

[wheat patrol]

what did you try until now

[safe shale]

everything is on paper

everything i tried

[wheat patrol]

last question, is there any constraint given like, a,b,c is any real number or restricted to integers?

[safe shale]

it's an 8th grade problem, so no. and i see now it asks me to find the angles as well

i don't know how i should approach this

[dusty portal]

Is it asking you to prove the inequality?

[safe shale]

no, just: find a, b and c (the sides of the triangle) and the angles of the triangle

[wheat patrol]

if its 8th grade then I am assuming all a,b,c are integers, so in that thought process atm just plug in whole number a,b,c in the second equation and keep the combo which satisfies the triangle inequality and you might see a pattern

[safe shale]

i see a pattern but i have no idea how to work it out from under the radical

[civic gazelle]

find the lowest value of each term

[safe shale]

so for sqrt (a-6)^2+4 the lowest value would be 2

for a =6 ?

[civic gazelle]

you'll be able to work out the lowest value of the expression and see something quite peculiar

[civic gazelle]

ye continue

[safe shale]

then b = 3sqrt3, c = 3 remains 2+1+4=<7 true then it's a right triangle

and the angles are 90, 60, 30

[civic gazelle]

^

what would be the lowest value

[safe shale]

7

no ?

[civic gazelle]

yeah

it has to be >= 7

but the problem asks us to find values for a,b,c so that it's <=7

which means only one thing can happen

[safe shale]

wait

why does it have to be bigger then 7

[civic gazelle]

because we were trying to find the lowest value for the expression

[safe shale]

please keep explaining, maybe it will click if i see the whole picture

[civic gazelle]

I think you would agree with me that (a-6)^2 >= 0 right?

[safe shale]

yes

[civic gazelle]

do the same thing for the other ones

you'll get the whole thing >= 7

[safe shale]

and from that where do i go

[civic gazelle]

you notice that the problem asks us to find a, b, c so that the expression is <=7

[safe shale]

yes and a, b and c cannot be other then 6, 3sqrt3 and 4 because the expression will be bigger then 7

[civic gazelle]

which means the only thing that can happen is if the expression is exactly equal to 7

[safe shale]

we found the lowest value which is 7 so a, b and c cannot change the values because the expression will be bigger then 7 and that contradicts

ok i got it

thank you very much

[civic gazelle]

ye no problem

!done

[midnight plank]

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[safe shale]

.close

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[analog torrent]

I dont know how to do this at all question 8

[zenith snow]

Let's do the easier one first

c.

Do u understand the concept of position vectors

[analog torrent]

Yes

But in hexagon no

I can do the triangles,parallelogram

This is the first time I came with a hexagon

[south gorge]

isnt a, b, and vector AB just a triangle though

[zenith snow]

Js consider that AOB is a triangle and find AB

[analog torrent]

Ok lemme try

[south gorge]

Would FA, BD, and AC not just be equal to AB?

[zenith snow]

Also magnitude of a and b r prolly supposed to be same

[zenith snow]

That's magnitude,that would be equal

But the vectors differ ig

[south gorge]

its a regular hexagon, so all sides are equal though

[zenith snow]

But only for FA and AB

[analog torrent]

Is AB
b-a?

[zenith snow]

BD and AC r not sides

[south gorge]

Oh I see

[zenith snow]

Yes

[south gorge]

you're right

[zenith snow]

Magnitude is equal,yes but direction isnt

[south gorge]

so BD is just 2AB and AC is just 2AB aswell

[analog torrent]

So 2×(b-a)

Is BD?

[south gorge]

AB is the vector from A to B on the outside

lowercase a and b are their own vectors respectively

[analog torrent]

Ok

[zenith snow]

Honestly i dont think this is right

[analog torrent]

Im sp confused

[zenith snow]

Cuz the direction of vectors matter as well

[analog torrent]

I have the answer sheet

[zenith snow]

Care to share?

[analog torrent]

Yea sure

FA=b
BD=-b-a
AB=b-a
AC=b-2a

[zenith snow]

Yeah thought so

[analog torrent]

So far AB is correct

[zenith snow]

Yes

Alr let's move to FA

[zenith snow]

Notice that FABO is a parallelogram

[analog torrent]

Yess

[zenith snow]

So u can say that the parallel sides r equal

[south gorge]

b-a makes sense

[analog torrent]

Like this right

[south gorge]

I forgot that direction matters 😞

[zenith snow]

Yes

[zenith snow]

Issok

[south gorge]

If you draw the tail end of a from the tip of b and then connect tail of b to tip of a

that just gives AB right?

[zenith snow]

The direction of FA would be from F to A

[analog torrent]

So FO+OA?

[zenith snow]

Yes if they were free vector

[analog torrent]

Ohh waiy

FA = b

Cus OB=b

[zenith snow]

Yes both in case of magnitude and direction

[south gorge]

you can make them free vectors for the sake of finding AB value

right?

[zenith snow]

That would be the case if O wasnt mentioned

They rnt free here

They r bound

[south gorge]

you can always stick them back together after finding the values

[analog torrent]

OOHH I GOT AC TOO

[zenith snow]

If u r told that the position of A with respect to O is a
Then how can u say that the position of A with respect to B is also a

[zenith snow]

Yeahh,good work

Lets do BD then

[analog torrent]

Oki

OK YAYEEE I GOT IT TOO

Thank u very muchhh both of yall

[zenith snow]

Good one!

[analog torrent]

Thanks!have a nice dayy

.close

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[midnight plank]

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[tardy bloom]

Number of odd integral solution of
x+y+z=21

[runic hamlet]

so all of x,y,z are odd?

[tardy bloom]

Yes

[dapper monolith]

Any other condition

[runic hamlet]

set x=2a+1 etc

[dapper monolith]

Like postive integers?

[tardy bloom]

Yes

[tardy bloom]

No
All integers

[dapper monolith]

Isn't it infinite then

[shell wigeon]

It is

[chrome vessel]

z=21-x-y, pick your favorite odd integers x,y, you get a z

[shell wigeon]

Or find the number of integer solutions to a+b+c = 9

[dapper monolith]

"Favourite odd integer"

💀

[tardy bloom]

Is it infinite?

[dapper monolith]

Yes

Unless more constraints r applied

[tardy bloom]

Then ig, positive odd integer

[runic hamlet]

wdym, guess

[chrome vessel]

then you could follows what denascite said

[runic hamlet]

do you have a question in front of you

[tardy bloom]

No, I had a test and a question like this appeared

Don't have the exact question now

[dapper monolith]

So u r asking it by memory

[tardy bloom]

Yeah

[dapper monolith]

Was it distinct integers too?

[tardy bloom]

No

[dapper monolith]

Yea than do it by what denascite said

[tardy bloom]

I'll do it now, thanks

Wait I have another question

There are n objects out of which p are alike. How many ways are there to arrange these n objects in a circular combination?

.close

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[neat sun]

Is the fact that ""false statement implies true statement" is a true statement" a convention thing, or is there some reason why

[prime hornet]

the way I like to think about vacuous truths is to use an explicit example and then think about when the statement is a lie

[radiant roost]

i think it's fine to consider it convention

[prime hornet]

for example, let our P => Q statement be "if I go outside today, then the world will end tomorrow"

now, when is that statement a lie?

okay well

if I go outside today and the world ends tomorrow, I told the truth when I said this statement, agreed?

and if I go outside today and the world doesn't end tomorrow, I told a lie

but now, if I don't go outside today, and the world doesn't end, I didn't lie!

[neat sun]

Yes

[prime hornet]

for all we know, the world could've ended if I went out

[prime hornet]

so the statement cannot be false

hence it's true

likewise, if I don't go out today and the world still ends, I also didn't tell a lie

[neat sun]

Yeah

[prime hornet]

so it's not false there either

[neat sun]

Hmm

[prime hornet]

hence true

[neat sun]

I was thinking about it

In context of the fact that

We can't do a proof like:

1. Start from statement to be proven
2. Develop by deductions
3. Arrive at a true statement

[neat sun]

Does this relate to the vacuous truth scenario

?

[prime hornet]

hmm, I don't think that's related?

I could be wrong, but I don't see how it is

[neat sun]

I mean why exactly is this kind of proof not valid

[prime hornet]

well I mean, you can prove that -1 = 1 doing this

assume -1 = 1 and square both sides to get 1 = 1

the latter is a true statement, so the former must be too by this logic

hello arohi

[neat sun]

Yeah, and this is the case because we can start from a false statement and go to a true statement

Like

Idk how to put into words

[prime hornet]

but you can prove all sorts of wrong theorems if you allow this, is my point

[neat sun]

Yeah

But why is thatbthe case

[prime hornet]

if this is a valid proof technique for a true statement, then you also get consequences like -1 = 1

[neat sun]

That we can go from a statement that is always false

To a statement which is always true

Through deduction

We can't do the thing in the inverse case

We can't go froma true statement to a false one

[prime hornet]

hm, I'm not sure if I know the answer to that

if anybody else is watching, feel free to chime in

[midnight plank]

@neat sun Has your question been resolved?

[midnight plank]

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[junior wedge]

I am stumped on where to start with this one, but I think it involves the formula (vector) = manitude(cos(theta))i + magnitude(sin(theta))j

if if m = 4000

CA + CB magnitudes = m?

[midnight plank]

@junior wedge Has your question been resolved?

[sharp coral]

they don't add up to m in magnitude

in order to be in equilibrium, all the forces (CA, CB, the weight) should vector sum to 0

[brisk iris]

@last slate this could be a good exercice for you

[midnight plank]

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[last slate]

Hi, have a maths exam tommo, looked thru an old test anf saw this, was wondering what the correct way to approach this would be( proper to a gcse/igcse markscheme)

[last slate]

<@&286206848099549185>

[midnight plank]

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

[last slate]

ah shit mb my ass read that as 10😭😭

[ebon nebula]

bro why do I have helper role

[last slate]

😭😭

[ebon nebula]

like I didn’t even choose the option to help people

(Well one time I had to help someone because no one was answering but then I figured out the answer was right..)

[last slate]

Lmaooo

[ebon nebula]

yeah nah.. if I have this exam I’m cooked

Im lucky I live in a other country than you

[last slate]

..😭😭😭 shiii

[near cliff]

,rotate

[grand pond]

Couldn't find an attached image in the last 10 messages.

Sep 14, 2025, 8:24 PM

Couldn't find an attached image in the last 10 messages.

Sep 14, 2025, 8:25 PM

[near cliff]

bro my wifi

[ebon nebula]

yuhhh, I mean.. other people can help 🤷

[near cliff]

,rotate

[ebon nebula]

bros like: OHHH MYYY PCC—

[last slate]

🙏🙏

[ebon nebula]

[last slate]

Gtg gonna showr rq

[near cliff]

(x+1)(4x - 3 - 4) + 4(3x) = 40

decompose the figure into two rectangles

[near cliff]

simplify from there

[last slate]

Wym decompose?

[near cliff]

break apart

[last slate]

Ohh

Kk tysmm

.close

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[last slate]

indeed

[midnight plank]

Please don't occupy multiple help channels.

[last slate]

i unintentionally claimed this channel too

[vast cipher]

type .close

[last slate]

.close

[midnight plank]

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[last slate]

yo

[last slate]

these are my teachers notes but shouldnt for positive be (-infinity,5) not 5]?

[wicked flint]

what am i looking at

[small jasper]

yes

[last slate]

?

[last slate]

also for increasing

should it be (-3,2.5) or [-3,2.5]?

[wicked flint]

im having a stroke reading this

[wicked flint]

()

interval of increase is a concept

[last slate]

but why

[wicked flint]

youre going infinitly close to the point

BECAUSE

at the point

the roc is 0

[last slate]

-3 is at 0

[wicked flint]

so its not

increasing or decreasing

[last slate]

and you increase from 0 to 150

[wicked flint]

man what

youre going infinitly close to the point

because at the point

the roc is 0

[last slate]

at 0 or 150?

[wicked flint]

the rate of change

is 0

[last slate]

for both?

[wicked flint]

bro what

[last slate]

for both points

is the roc 0?

at 0 AND 150

[wicked flint]

if its not increasing nor decreasing

the rate of change is zero

[last slate]

ok so if it was also asking for decreasing

i would always do ( )

[wicked flint]

yeah

you cant include the point because

[last slate]

never [ ] with decreasing or increasing

balright thank you

[wicked flint]

your just going close to it

[last slate]

and positive and negative is always ( ) ( ) too

assuming the x is real for any real number

[wicked flint]

uhh no

wait

[last slate]

and assuming it goes through the x axis

[wicked flint]

lemme read the graph

oh wth hey civil

[small jasper]

Increasing means $x_1<x_2 \implies f(x_1)<f(x_2)$. Take that as you will.

[grand pond]

Civil Service Pigeon

[last slate]

yeah 0<150

but apparently it has to be 0.00000001 to 149.9999999999999

[wicked flint]

youre including the point 0

[last slate]

how

0 is not positive

and not negative

[wicked flint]

wait

are you

i dont even know

lemme check

[last slate]

like for this example it should be (-infinity,5)

(-infinity,5], not that

because 0 isn't a positive numebr

alright i think i got it

[wicked flint]

yeah

i think ur teacher just did it wrong

[last slate]

also if the local min and local max is known as those

wouldn't they also be the absolute min/max?

[wicked flint]

no

this is within a specific part

you only have part of the graph

[last slate]

ok

[midnight plank]

@last slate Has your question been resolved?

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[last slate]

im confused how they came of witht hat equation with that description

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[odd solar]

Do you know what an inflection point is?

[odd solar]

@last slate if you still need help, open a new channel

.close

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[last slate]

hey

[last slate]

i solved this problem but i have a question abou tit

[rich yew]

’tis the question?

Mr?

[last slate]

oh

[rich yew]

are thou present?

[last slate]

what if it was

that instead?

i get out to solve that using coefficent comparison

but i dont get how to do if if it is a bigger polynomial

like that

[midnight plank]

@last slate Has your question been resolved?

[humble wolf]

method too slow

use synthetic division

[last slate]

i kjnow but our teacher wants us to use that method

idk why

[midnight plank]

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[midnight plank]

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[uneven minnow]

i used the product rule for absolute value to get that f times sinx is 3, but cant go past it

[uneven minnow]

∣f(x)sinx∣≤∣f(x)∣⋅∣sinx∣<3⋅1=3

[spiral rock]

do you think there are any more upper bounds you missed?

[uneven minnow]

i dont believe so.. i dont know what to do with the four

[spiral rock]

there is a useful known inequality here

does anything come to mind?

[uneven minnow]

f(x) is less than 3?

oh or sin less than or equal to 1

[spiral rock]

this part you already handles well

something about handling the absolute value and the 4 etc

[uneven minnow]

can i do abs value of four plus abs value of −f(x)sinx?

[tender dew]

that is the idea, but think about how you would go from |4 - f(x) sinx| to |4| + |-f(x) sinx|, what very common and famous inequality for absolute values will you use?

[uneven minnow]

The one where if you add them, you also add the inequalities?

[tender dew]

that is known as the triangle inequality

[uneven minnow]

Ah I see

[tender dew]

|a|+|b| >= |a+b|

[uneven minnow]

But then how would I get the collective inequality from that? All I get is

[uneven minnow]

This

[tender dew]

well you use the inequality

can you tell me what a and b would be for the terms |4 - f(x) sinx|, |4| + |-f(x) sinx|

[uneven minnow]

a is 4 and B is -f(x) sinx

[tender dew]

yup, plugging those in you get your desired inequality

[uneven minnow]

And f(x) sinx after taking abs value

[uneven minnow]

How so?

[tender dew]

Let's write what we have so far

[grand pond]

Peter

[tender dew]

this is obvious right?

[uneven minnow]

Yes

[tender dew]

ok, we want to simplify i.e. bound the RHS further, |4| = 4 so no complications there

|f(x) sin x| you bounded before

what was its bound?

[tender dew]

^

although the <= sign should really be =

are you following so far?

if something doesn't make sense we can go back to it, just mention it

[uneven minnow]

Where did the negative sign go? Did you get rid of it because it's in the absolute value signs?

Why is it equal to 3?

[uneven minnow]

Question says it's less than 3

[tender dew]

correct

[tender dew]

I never said it's equal to 3?

[uneven minnow]

Oh I see now

The f(x) sinx is equal to the next part

[uneven minnow]

Yes

[tender dew]

perfect, well by bounding |f(x) sinx| you have a bound for the original expression

I'll write it out more neatly

give me a second

[uneven minnow]

Alright thank you

But the 4 still counts towards the bound, right?

[tender dew]

\begin{align*}
|4 - f(x) \sin x| &\leqslant |4| + |-f(x) \sin x| \tag{triangle inequality} \
& = |4| + |f(x) \sin x| \
& = |4| + |f(x) | |\sin x| \
&\leqslant |4| + |f(x)| \cdot 1 \tag{since $|\sin x| \leqslant 1$} \
&< |4| + 3 \tag{given $|f(x)| < 3$} \
&= 7
\end{align*}

[grand pond]

Peter

[tender dew]

absolutely (no pun intended)

[uneven minnow]

If f(x) is less than 3, why would we use it and not a lower number? Because we are finding the upper bound?

[tender dew]

overall considering all inequalities above, we have
|4 - f(x) sin x| <= |4| + |f(x) sin x| <= |4| + |f(x)| < 4 + 3 = 7 or |4 - f(x) sin x| <= 7

[spiral rock]

f can be as close to 3 as you want

[tender dew]

precisely because of that, we are trying to get as close to the upper bound as possible

[uneven minnow]

Ohh, okay thank you!

[tender dew]

it sounds contradictory because we are trying to find the lowest UPPER bound

but we need to make sure that the expression |4 - f(x) sin x| at the end of the day is bounded by this value for all x

[spiral rock]

exactly! we know 7 is good

we need to show the others are not

[uneven minnow]

would 8 still count though?

[spiral rock]

yes

[tender dew]

yes

[uneven minnow]

\

[tender dew]

7 is the best (lowest) we can do

anything above that also works

[uneven minnow]

ohhh

[spiral rock]

to do that we need a case where the function is as big as possible

[uneven minnow]

right that makes sense

[tender dew]

it's a bound, but it is not the optimal per say

however the question asks

all bounds

not just the tightest one

[uneven minnow]

ahhh got it now

Thank you!

[midnight plank]

@uneven minnow Has your question been resolved?

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[uneven minnow]

.close

[midnight plank]

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[jade magnet]

I is the converse

II is the contrapositive

and I'm pretty sure III is equivalent to I

well thats what i thought anyways

but apparently it isn't

i thought for something to be sufficient it means the converse is true

i picked F but the answer was G

[lyric charm]

"only if" in isolation...

"A only if B" must be read as "if A then B".

i can tell you as somebody whos done, like, Serious Math™ that fucking nobody says just "only if".

you only ever hear it as part of "if and only if"

[midnight plank]

@jade magnet Has your question been resolved?

[midnight plank]

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[midnight plank]

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