help-10

[trail cloak]

Ooop

Tension toward B will be more than tension toward A

[fluid meadow]

dont think so

answers say tension is the same

and it would be because the string is inextensible

[trail cloak]

Oh wait

Yeah you're right

[obtuse pebble]

@fluid meadow Has your question been resolved?

[obtuse pebble]

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[late flare]

E and f please

[obtuse pebble]

Please don't occupy multiple help channels.

[gleaming ridge]

,rotate

[late flare]

There is no dot or circle

How could I know

[gleaming ridge]

What's the question?

[late flare]

Its written

[gleaming ridge]

oh e and f?

[late flare]

Yes

[gleaming ridge]

e is easy, you got e correct?

I mean f is almost the same

Do you know continuity?

[late flare]

None kf them i got them correct

[late flare]

Nope

[gleaming ridge]

So, do you the meaning of a limit?

[late flare]

Yea

Approcahes x value

But doenst intersect

[gleaming ridge]

intersect?

[late flare]

Explain please hiw to solve

[late flare]

Touch

[gleaming ridge]

So, the limit of a function is basically the value a function takes when the input approaches some value

In simpler terms it is how the function behaves very very close to the input but NOT the input

So, in (e), it says the input approaches 4

[late flare]

I know this

[gleaming ridge]

now notice that there's no circles or dots anywhere near 4

[late flare]

Thats why its confusing to me

[gleaming ridge]

which means that we can go as close to 4 as possible

In practice this means we are almost there at 4

Lemme think of a better way to express it

[late flare]

Can u draw maybe?

[gleaming ridge]

Drawing won't help actually

Btw, how did you do (a) part?

[late flare]

Its obvious from the dot

Its 3

[gleaming ridge]

okay, what about part (b)?

[tacit shadow]

hlo

[late flare]

Same way

[tacit shadow]

I saw limits and I thought I could help

[late flare]

Just please tell me how to find it when i dont have circles or dots

[late flare]

Please do

[tacit shadow]

you want e and f

so its basically asking

if we get as close as possible to four, what will be the limit of that?

@gleaming ridge Am I correct?

[late flare]

There is no dot??

How do i know

It wont exist then

[tacit shadow]

no

tell me the f one

[gleaming ridge]

Well, I am trying to understand your infatuation with the dot. There are no dots in real life functions

Do you just see the dot and mark that as the answer?

[late flare]

Yes

[tacit shadow]

what will be the function of 4 at x = 4

[late flare]

Oyherwise how do i know?

[tacit shadow]

?

[tacit shadow]

see the graph

[late flare]

4

[tacit shadow]

good

[late flare]

If thats the answer then e and f is the same

[tacit shadow]

now what will the limit of that be as x approaches 4

yes good

@gleaming ridge hes correct right?

[gleaming ridge]

So, for a weird analogy when you see no dot every point is a dot

[late flare]

So i basically place a dot

To find the answer

[gleaming ridge]

Kinda like yes

[late flare]

And x= 4

[tacit shadow]

yup

you can do that

[gleaming ridge]

yes so for x approaching to 4, you can place a dot at 4

[late flare]

Whats the difference between x=4 and the limit then

If they give same value

[tacit shadow]

we are asking what happens as the value approaches

[gleaming ridge]

Nothing, that is the point of the exercise

[tacit shadow]

so we are getting closer and closer to 4

but you will get the same answer

[gleaming ridge]

The function is continuous at 4 hence limit and the value at 4 will be the same

[tacit shadow]

in this case, atleast

[late flare]

Thanks alot

[gleaming ridge]

That is the definition of continuity

[late flare]

I still havent took ot

[tacit shadow]

what is the d one?

[gleaming ridge]

Yeah, most likely these exercises are there to motivate the continuity

The very basic definition of continuity is that you can draw the graph without lifting the pencil

Which you can do at x=4 but not at x=2

[tacit shadow]

cause the graph ends there?

[gleaming ridge]

I hope you can see that you'll need to lift your pen/pencil cause there is a jump at 2

[tacit shadow]

even I havent studied continuity, so im learning somthing too

[woeful dew]

HOW TO GET GOOD AT MATHS

[late flare]

@gleaming ridge

13 plz

also this

lim -1 right side

D and E please of 15

<@&286206848099549185> i need help asap plz

[solemn trout]

For D, x<pi.
For E, x>pi.

@late flare

So, use the part of the piecewise definition that satisfies those inequalities.

[obtuse pebble]

@late flare Has your question been resolved?

[late flare]

what is pi

like how do i evalute it

is it 0 or 1

or how do we take limit for it]

@tacit shadow mind helping?

[obtuse pebble]

@late flare Has your question been resolved?

[timid silo]

3.14

[obtuse pebble]

@late flare Has your question been resolved?

[obtuse pebble]

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[ruby skiff]

is the formula not a=theta(r)?

where did the theta/2(pi) come from?

[nocturne minnow]

Percentage/ratio of the angle/sector of the entire circle

[warm canopy]

the formula comes from the work done there

[obtuse pebble]

@ruby skiff Has your question been resolved?

[obtuse pebble]

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[obtuse pebble]

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Channel closed

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[haughty schooner]

Im having trouble understanding why we would use nCr instead of nPr for this example:

https://cdn.discordapp.com/attachments/903430332324405288/1058482945737891971/image.png

[tardy epoch]

Fyi this channel will close

[haughty schooner]

ok thanks, and srry

[tardy epoch]

All good. Just didn't want you to waste time and lose messages

[obtuse pebble]

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[sturdy blade]

I don’t understand this, haven’t learned it but it was given to me

[sturdy blade]

<@&286206848099549185>

[obtuse pebble]

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.

@sturdy blade Has your question been resolved?

[proper thistle]

Wait

[quaint orbit]

whats up?

[sturdy blade]

I need help

<@&286206848099549185>

Anyone?

[livid canopy]

@sturdy blade What is your question exactly?

[nocturne minnow]

It gives you step by step instructions on what to do

[sturdy blade]

The thing is there are backsides and frontside flips but I have no backside or front side

[nocturne minnow]

Then label the sides

[obtuse pebble]

@sturdy blade Has your question been resolved?

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[timber plinth]

given that $x^2+y^2 \geq 2xy$, prove $$(x^2+1)(y^2+1)\geq x(y^2+1) + y(x^2+1)$$

[timber plinth]

this is proving to be far more annoying that anticipated

the RHS factorizes to $(x+y)(xy+1)$

[warm shale]

RedJive

[haughty coyote]

My first guess would be to expand and simplify

[timber plinth]

and i tried AMGM which didnt really work

[timber plinth]

no simplications possible

full expansion leads to $$x^2y^2+x^2+y^2+1 \leq xy^2+x+x^2y+y$$

[warm shale]

RedJive

[timber plinth]

now we do have $xy^2+x+x^2y+y = (x+y)(xy+1)$

which is just something i found

[warm shale]

RedJive

[haughty coyote]

I don't think this is possible

The LHS is a polynomial in x^2y^2 which is of degree 4

[timber plinth]

oh?

[haughty coyote]

For x,y big enough it gets larger than the RHS which is of degree 3

[timber plinth]

hold on let me read the quesitona agin

[haughty coyote]

And the constraint always holds because it's the simple AMGM, a rewriting of (x-y)^2 >= 0

[timber plinth]

aw fartnuggets

i wrote the wrong one

[warm shale]

RedJive

[timber plinth]

but yeah i tried amgm

[haughty coyote]

This can be rewritten as (y^2+1)/y >= (x^2 + 1)/(x^2 - x + 1)

[timber plinth]

wait what

oh

[haughty coyote]

Then it suffices to prove you can slide >= 2 >= ibetween them

Though that makes no use of the AMGM unless it's in the details so probably not the intended proof

[timber plinth]

no AMGM

im not too sure on how to see 2 works

[haughty coyote]

Ngl I desmosed it

[timber plinth]

why is it that (y^2+1)/y necessarily >2

[timber plinth]

oh

[haughty coyote]

But good old function analysis should work though that's painful

[timber plinth]

ok so for some context

this was on a fench middle school math oly

[haughty coyote]

Middle school lmao ok

[timber plinth]

¯_(ツ)_/¯

someone sent it to me

[haughty coyote]

What level of olympiad ? POFM ?

[timber plinth]

uh no idea

i can ask

brb

ok well

i kinda lied

not really

this is the original problem

[haughty coyote]

What age is this

[timber plinth]

¯_(ツ)_/¯

no idea

[haughty coyote]

13-15

Late middle school basically

[timber plinth]

yah

i guess

8-9th grade

[haughty coyote]

Weird to say 8th-9th when these aren't french grades, but anyways

[timber plinth]

oh

well

bros lying to me i guess

idk that person

i just do math

[haughty coyote]

French middle school is 6e -> 3e

[timber plinth]

yeah wtf

💀

probs translated to american grades

or not

no idea

[haughty coyote]

High school is 2nd -> 1ere -> Terminale (final)

[timber plinth]

ok nevermind what the hell is wrong with the french education system

[haughty coyote]

It counts down to graduation

[timber plinth]

OH

anywyas

i dont see how one would do this without desmos

¯_(ツ)_/¯

or amgm

which wasnt used i guess

also the x^2+y^2 >= 2xy part

wasnt used at all

[haughty coyote]

The first one is >= 2 by AMGM

[timber plinth]

oh

[haughty coyote]

So maybe it's intended

[timber plinth]

the second one is

[haughty coyote]

Just gotta figure out the 2nd one

[timber plinth]

yeah but also coming up with that at all is like

damn

[haughty coyote]

Olympiads

[timber plinth]

¯_(ツ)_/¯

[haughty coyote]

the 2nd also follows from the AMGM

In a slightly more subtle way

[haughty coyote]

But I'll concede that figuring out that this form is the way to go requires some mathematical maturity

[timber plinth]

idk i always hated amgm even when i did oly

anyways

thanks

much appreciated

[haughty coyote]

Btw

[timber plinth]

hm?

[haughty coyote]

Do you know why I knew this had to work with this rewriting ?

[timber plinth]

why

are u like. a mopper or smt lmao

[haughty coyote]

Mathematical maturity/experience/intuition question

[timber plinth]

oh

[haughty coyote]

What's a mopper?

[timber plinth]

are u from the US?

the US math oly system ends with MOP (not counting IMO)

[haughty coyote]

You might have guessed that I'm French

[timber plinth]

oh

💀

i thought u like

googled that

LMAO

[haughty coyote]

I googled the age for grade 8 and 9

[timber plinth]

did u do IMO

👀

[haughty coyote]

No. Stopped at the national level, didn't really do Olympic math

[timber plinth]

oh

still cracked

[haughty coyote]

It's not really something I practiced so I never did better than 30th nation wide

[timber plinth]

damn wtf

30th nationwide

[haughty coyote]

Somehow

[timber plinth]

here thats MOP level cuz mop takes 40 kids

¯_(ツ)_/¯

[haughty coyote]

US is also 5x the population

[timber plinth]

oh

good point

i did (technically) top 100?

i say that because usamo takes 200 and median score is 0, and i got a 1

so

¯_(ツ)_/¯

anyways yeah

youve been real helpful and i really appreciate it

[haughty coyote]

Median at 0 sounds stupid

[timber plinth]

¯_(ツ)_/¯

[haughty coyote]

Like this is so inefficient

[timber plinth]

well

they go from 200 -> 40

[haughty coyote]

To actually compare people, you then them to have a lot of points

[timber plinth]

plus usamo score counts for 50% of the IMO team

the other 50% is tst

at MOP

[timber plinth]

theres plenty of points to go around

theres always a handful that get 20+

which u can score

the usamo is formatted exactly like the IMO

theyre also (intended) to be of equal difficulty

but yeah thanksss

.close

[obtuse pebble]

Channel closed

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Use .reopen if this was a mistake.

[timber plinth]

<3

[obtuse pebble]

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[wicked crown]

I have a question on when a funktion diverges, exact question follows:

[trail musk]

Function

[wicked crown]

yes, sorry let me switch the device , one second

so S is supposed to be a language, but I think that doesnt matter too much

now I wanna find a S such that f does not have a limit

and afaik for that I'd need to find S such that f diverges, there are no other ways to find such a S that come to my mind

but I dont really get an idea what S could meet this requirement

maybe someone can provide me some intuition, some inital thought that might be helpful

I thought about the cases where S = {0,1}^k where k>=n, that should lead to the intersection {0,1}^n and that gives me ( 2^n / 2^n ) = 1
and also
S = {} would result in 0/2^n = 0

And from what I can think of, the cardinality of the intersection of S and {0,1}^n would always be inbetween 0 and 2^n , which shouldnt cause any diversion in my eyes

[timid silo]

ok so, what kind of divergence are you aiming for?

[wicked crown]

basically I am just trying to find S such that f(S) does not have a limit

I am very sorry but lim is not my strongest discipline

so I might say stupid stuff sometimes

[timid silo]

$s_n = \frac{\abs{S \cap {0,1}^n}}{2^n}$

[warm shale]

rbit ✨

[timid silo]

hmm

[wicked crown]

$s_n = \frac{\abs{S \cap {\left 0,1 }\right^n}}{2^n}$

damn

[timid silo]

well i guess you get the idea

[wicked crown]

yes !

[timid silo]

so, first, do you think theres any way s_n can diverge to infinity?

[warm shale]

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

[wicked crown]

there certainly has to be one because the exercise I am trying to solve requires me to prove that

should even be a decidable language S

[timid silo]

but i dont think it can, how about making it so that
s_n goes 1, 0, 1, 0, 1, 0, ?

making it diverge by cycling

[wicked crown]

something like that was the only thing that came to my mind

from this basic example of (-1)^n

yeah

but I couldnt find a way to construct S so that S_n would behave like that

let me think , maybe I can find one way

[timid silo]

|{0,1}^n| / 2^n

what is this equal to?

[wicked crown]

I'd say 1

[timid silo]

yes

and |{}| / 2^n is 0 of course right?

[wicked crown]

yes right

I agree

[timid silo]

so .. can you find some infinite union for S, to make it behave like that?

[wicked crown]

I'll think about it real quick, I actually got an idea

[timid silo]

$S = \bigcup_{n=0}^{\infty} ???$

[wicked crown]

wait that would be |N \ {0} ?

[warm shale]

rbit ✨

[timid silo]

like this then?

[wicked crown]

so N ?

[timid silo]

doesnt really matter

[wicked crown]

I am sorry I cant follow with this one

I got an idea but it was quite different

[timid silo]

oh you mean this notation?

[wicked crown]

no the notation is familiar to me

but like

wouldnt that lead to {0,1,2,3,4,5,...}

but S is only in {0,1}*

or like, what kind of numbers are n

[timid silo]

no no you gotta put something into the question marks

[wicked crown]

ohhh

I thought the question marks were like "do you understand"

lol sorry

[timid silo]

oh haha

[wicked crown]

ok give me a moment

its 0^n

?

no

no would always end in |S|=1

ahh my brain is too slow, but I feel like I can find it

[timid silo]

well you want
$S \cap {0,1}^n$
to always alternate between {0,1}^n and empty set right?

[wicked crown]

yes

damn wait

no ahhh man , I think I got it but then I realize it doesnt make sense `:D

[timid silo]

hmm what should this sequence of intersections look like then?

for n = 0, 1, 2, 3, ...

[wicked crown]

ok one moment, its just that I am a bit slow

but I can solve it I am convinced

ok so I just noticed that I had a bit of a misunderstanding of what {0,1}^n would look like

but now i think that maybe we could say that for all even n we just take S = {0,1}^n and for all odd n we say S = {0,1}^n-1 or n+1

but I would have to construct this term such that it would actually behave like that that

[timid silo]

ok that idea leads to the right direction

[wicked crown]

:D

[timid silo]

why not make it empty when n is odd?

[wicked crown]

that also possible

just so that there is no intersection between {0,1}^n and S

but

let me think how to make these sets the empty set then

ok so to not waste any of your time:
I struggle with finding a way to construct the term the big union is applied to such that we differentiate between even and odd n's

I could construct a simple function and put that as the term

but idk if that would be valid

and also

[timid silo]

well one tip is to just use 2n for the union

[wicked crown]

for that function I would need to make a case check

hmm ok thats something that came to my mind but I didnt know hwo to execute properly

one moment

[timid silo]

its very simple actually

[wicked crown]

I feel like if I thought about it for some time I'd find it, but right now, as I said, I really dont wanna take too much of your time

and I feel like an idiot for not seeing it lmao

[timid silo]

well what if
S = {0,1}^0 U {0,1}^2 U ... ?

[wicked crown]

damn

now I see

[timid silo]

so what would you put into the big union?

[wicked crown]

then for every odd n the intersection of {0,1}^n and S would be empty, becuase S doesnt contain them

yeah then I'd say S = U {0,1}^{2n}

[timid silo]

yeah

[wicked crown]

I completely missed how the big union would act

I knew it was there

but I missed that it would lead to a set that skips the odd n

omg thank you so much, that did not just help me with this problem, I actually gained some useful insides

thank you so much, I really appreciate your efforts to help me a lot !

[timid silo]

np

[wicked crown]

.close

[obtuse pebble]

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[obtuse pebble]

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[solemn dew]

$\frac{a^{\frac{3}{2}}}{\frac{b^3}{\frac{a^{-1}}{b^2}}}$

[warm shale]

DarkSyde

[lost tree]

$\frac{a^\frac{3}{2}}{b^3\div\frac{a^{-1}}{b^2}}$

[warm shale]

SilverSoldier

[solemn dew]

.close

[obtuse pebble]

Channel closed

Closed by @solemn dew

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[obtuse pebble]

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[timid silo]

Which topic is this?

[timid silo]

Induction? It does not look like a high school problem.

[sharp pecan]

induction for sure

A neat trick here is: to compare ||L(n+1)-L(n)=R(n+1)-R(n) where L is left side and R is right side||

[timid silo]

Nice name you have.

[sharp pecan]

thanks

[timid silo]

Can you set the constants to be any terms?

[sharp pecan]

sure, any term that isn't double

[timid silo]

Lastly, do you think this is a high school problem?

[sharp pecan]

Well, if you know what induction is you can solve it

I guess the random variables are a bit confusing at first, but if you know how to deal them from big equation systems that should be ok

[timid silo]

I'm actually sending it to a student and I think it is like tougher than a usual test

Slightly tough is fine but I think the difficulty for this one would be a little too much so that's why confirmation needed

[sharp pecan]

Yes, I really hope this is one of those 'very hard exercise to get bonus points in the next test' thingies

The Summation sign with 2n intervals sure is brutal for induction

[timid silo]

Icic thanks

.close

[obtuse pebble]

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[rotund vector]

can someone explain how this isn't square root of 45?

[haughty coyote]

It is

But then you simplify it because it contains 3^2

[rotund vector]

then why is the answer 3 x square root 5

[haughty coyote]

And we prefer having things outside of square roots if possible

[rotund vector]

?

so how would you simplify it?

[haughty coyote]

sqrt(45) = sqrt(3^2 * 5) = sqrt(3^2) sqrt(5) = 3 sqrt(5)

[rotund vector]

I'm confused

[river galleon]

write sqrt(45) in terms of prime and composite numbers, you get sqrt(9) x sqrt(5)
sqrt(9) is 3, so you are left with 3 x sqrt(5)

[rotund vector]

oh I understand

[rotund vector]

thanks

[river galleon]

no problem

[tawny fog]

$√45=\sqrt{9×5}$

[warm shale]

Arnab Pal

[tawny fog]

What's square root of 9?

@rotund vector

You have to solve it here 💀🔪

[obtuse pebble]

@rotund vector Has your question been resolved?

[obtuse pebble]

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[gleaming ridge]

Show that if f: X -> Y is injective then there exists g: Y -> X such that gf is the identity map.

[gleaming ridge]

My problem with here is the question itself. I know how to do it but unless f is surjective the whole of Y won't the range of f

Hence, how can we define g from the whole Y?

[gilded needle]

you can set g to be anything you like on the part of Y that is not in the range of f

it won't make any difference because when you form the composition gf, that part of g's domain will never be used

[elfin burrow]

should probably assume that X is nonempty

[gleaming ridge]

Yeah I understand that which is why it's more of a technical nitpick cause we aren't restricting the domain of g to range(f)

[elfin burrow]

no you're not

what Bungo is saying is that you don't need to restrict the domain of g if you send everything in Y that is not in range(f) to a random fixed element of X

[gleaming ridge]

Ah, okay so for everything in Y - Range(f) can be sent to any point in X cause it's of no practical use

Thanks!

[gleaming ridge]

True, this is an algebra textbook so empty sets haven't even been defined and hence neither empty map

Anyway, I understood

[elfin burrow]

abstract algebra?

[gleaming ridge]

Yes

[elfin burrow]

nice

[gleaming ridge]

.close

[obtuse pebble]

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[gleaming ridge]

@elfin burrow were you gonna say something?

[obtuse pebble]

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[elfin burrow]

that the sets are usually assumed to be nonempty in these proofs

[gleaming ridge]

I guessed so, although they should have specified it clearly

Anyway, I'll close this channel to free it up

[elfin burrow]

.close

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[obtuse pebble]

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[frozen acorn]

hi

[frozen acorn]

$(1+e^x)/(1-e^x)$

[warm shale]

M.WALI

[frozen acorn]

how do i find the dy/dx of this?

[kind hawk]

Quotient rule

[limber quartz]

Same as always

[frozen acorn]

i know its quotient rule

[limber quartz]

You might like product + chain rule better for this one

[frozen acorn]

wdym?

[limber quartz]

$=(1+e^x)(1-e^x)^{-1}$

[warm shale]

Disorganized

[frozen acorn]

i see

ill give it a try and tell you the results

[limber quartz]

Same result though

Also exponentials are super easy to deal with anyway

[frozen acorn]

iv been trying to do it with quotient rule but it just dosent seem to work

or maybe am stupid

[limber quartz]

I assure you...
...it works.

[frozen acorn]

then am stupid

[limber quartz]

No

[frozen acorn]

yes

[limber quartz]

NO

[frozen acorn]

YES

[robust sleet]

Yes

[frozen acorn]

wait no

[limber quartz]

$(uv)' = u'v + uv'$

[warm shale]

Disorganized

[limber quartz]

$\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}$

[warm shale]

Disorganized

[limber quartz]

@frozen acorn ?

[frozen acorn]

am sorry

am a disappointment

[timid silo]

quotient rule is just product rule+ chain rule so i can assure you quotient rule works lol

[limber quartz]

Geeze, relax man

What is the derivative of a constant?

[frozen acorn]

1

[limber quartz]

No

1 is a constant. What is the derivative of 1?

[frozen acorn]

0

[limber quartz]

Yes

All constants have derivatives of 0

What is the derivative of e^x ?

[frozen acorn]

same

e^x

[limber quartz]

What is the derivative of e^(2x) ?

[frozen acorn]

2e^(2x)

[limber quartz]

(Actually beyond the scope of this problem)

Yes

Ok do
d/dx (1+ e^x)

[frozen acorn]

ok

[limber quartz]

You did both parts of this already @frozen acorn

[frozen acorn]

i did?

[limber quartz]

It's the sum of the derivatives of both terms

[frozen acorn]

ohh

d/dx of (1+e^x) is e^x

[limber quartz]

Differentiation is a "linear operator"

Yes, that is it.

You can do this whole thing now.

I really wanted to read from @robust sleet

I'm sure they had something profound to say

[frozen acorn]

and am scared to say something

[limber quartz]

Ask questions

Questions = good

@frozen acorn

[timid silo]

Thats the derivate

[limber quartz]

They did this part earlier

Well, apparently my patience is terrifying

Good luck @frozen acorn

[timid silo]

What's the question?

[obtuse pebble]

@frozen acorn Has your question been resolved?

[frozen acorn]

.close

[obtuse pebble]

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[toxic hollow]

Is this a common result used when the events are independent?

[obtuse pebble]

@toxic hollow Has your question been resolved?

[obtuse pebble]

@toxic hollow Has your question been resolved?

[obtuse pebble]

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[gleaming ridge]

.reopen

[obtuse pebble]

✅

[brave bluff]

yes this is the definition of T_1 and T_2 being independent (conditioned on C)

[obtuse pebble]

@toxic hollow Has your question been resolved?

[obtuse pebble]

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[toxic hollow]

.reopen

[obtuse pebble]

✅

[toxic hollow]

Doesn’t C also have to be independent with T1 and T2?

[obtuse pebble]

@toxic hollow Has your question been resolved?

[obtuse pebble]

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[tawdry meteor]

Can someone help me solve matrix b)?

[tawdry meteor]

so basically

i have to solve for a,b and c

i know i whould be making the determinant

i have to find the values of a b and c that are equal to 0

the second image is the solution to the question

which i don't understand

[spiral knot]

which matrix ? b) or c) @tawdry meteor

[tawdry meteor]

b)

oh hey

it's you

[spiral knot]

\begin{align*}
\matr[v]{1&a&a²\1&b&b²\1&c&c²}
&=\matr[v]{1&a&a²\0&b-a&b²-a²\0&c-a&c²-a²} \\
&= 1 \matr[v]{b-a&b²-a²\c-a&c²-a²}\
&= (b-a)(c²-a²) - (c-a)(b²-a²)\
&=(b-a)(c-a)(c+a) - (c-a)(b-a)(b+a)\
&= (b-a)(c-a)[c+a - b-a]\
&=(b-a)(c-a)(c-b)
\end{align*}

[warm shale]

Mehdi_Moulati

[spiral knot]

which is the same answer in your image

[spiral knot]

hey, how are you @tawdry meteor ?

[tawdry meteor]

oh so it's just the determinant

[tawdry meteor]

afraid of failing maths

:c

[spiral knot]

Just try to solve the problems as much as you can

and everything will be right

[tawdry meteor]

that's the problem

i get distracted so easily

and stop doing maths :c

.close

[obtuse pebble]

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[night sonnet]

$$ f(x) = x^2 , f(0) = 1 $$
I want to use the epsilon delta definition to prove its not continuous in 0.
So let $$ \epsilon = \frac{1}{2} $$ and then $$ \forall \delta > 0 \exists x= ... : ∣ x - 0 ∣ < \delta $$ and $$ ∣ f(x) - 1 ∣ > \epsilon $$

But when I e.g. pick $$ x=\delta/2 $$ I can find some delta where it fails, I would appreciate some input or a nudge into the right direction

[warm shale]

aabb

[timid silo]

why did you choose epsilon yourself?

also how is f(0) = 1?

[night sonnet]

just the way the function is defined

and I chose epsilon because thats how you negate continuity, no?

[timid silo]

ohh, not continuous, i see

[night sonnet]

Yeah my problem is however I choose my x in relation to delta, I can find some value for delta where the second inequality breaks

[timid silo]

well, what if you just ignored delta at first, and just choose some constant x so that
|x² - 1| ≥ 1/2
which x could satisfy that?

[night sonnet]

any x?

from 2 onwards

for example

0-sqrt(0.5)

[timid silo]

that would work, yes, but thats a rather big x, can you find a small x?

[night sonnet]

(0-sqrt(0.5)) meaning anyvalue in that interval

[timid silo]

oh right

so how about x = 1/2?

then
|(1/2)² - 1| = 3/4 ≥ 1/2

[night sonnet]

but dont you have to pick your x in relation to delta?

what if delta is smaller than 0.5?

[timid silo]

now as you already said, it can be any value in that interval, even as small as you want to

so if delta is smaller than 0.5, what could we simply do to satisfy |x| < delta?

[night sonnet]

pick a smaller x?

[timid silo]

yes, like you already did with x = delta/2, and it works for small values of delta right?

[night sonnet]

okay so you think delta x=delta/2 holds for any delta?

[timid silo]

it holds for small delta, but if you do something like delta = 10 it breaks

[night sonnet]

it breaks for delta 2 too

[timid silo]

so if delta is large, it would be nice to switch back to x = 1/2 right?

[night sonnet]

Okay so I pick my x from some set, so that it can meet every condition?

[timid silo]

well yeah, we want the following:

if delta is small, choose x = delta/2
if delta is large, choose x = 1/2

now how can you make this more precise?

[night sonnet]

something like x=min(1/2, delta/2)

[timid silo]

yeah thats it

[night sonnet]

great, I wasnt aware you can do that, but it makes sense

[timid silo]

using the min function is a nice thing to keep in mind for these problems

[night sonnet]

apparently haha,
Okay I think I can tie everything together, is my x good enough defined as is now?

[timid silo]

$\exists \epsilon > 0: \forall \delta > 0: \exists x \in \bR: |x| < \delta \text{ and } |x² - 1| \geq \epsilon$

[warm shale]

rbit ✨

[timid silo]

this is what we want right?

[night sonnet]

indeed

[timid silo]

then yes,
any x ≤ 1/2 satisfies the right inequality and any x ≤ delta/2 satisfies the left inequality, so combining that gives
x = min(1/2, delta/2)
and we have both then
x ≤ 1/2 and x ≤ delta/2

[timid silo]

ok we should exclude x=0 from this

[night sonnet]

yeah true

Okay then everything is covered, no?

[timid silo]

right

[night sonnet]

great, thanks for the help!

If you dont have anything else to say, i will close the thread

[timid silo]

no problem

[night sonnet]

.close

[obtuse pebble]

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[ruby skiff]

yo so my tutor showed me how to do this problem but how is log(3) and (x+1)log(2) added with each other?

are they not supposed to multiply?

[dim bobcat]

He used the property log (ab)= log a + log b , in above pic

Then log a^b = b log a

[ruby skiff]

oh k

[ruby skiff]

yo man could u also tell me about this one?

like this one

[dim bobcat]

What problem do you have in it?

[ruby skiff]

i do not understand how the log(3^x + 15) would be solved

how would u bring the x of log(3^x + 15) to the bottom so it looks like this? xlog(3+15)?

like is the goal not to bring the x in front?

[obtuse pebble]

@ruby skiff Has your question been resolved?

[tall tusk]

@timid silo @timid silo

rbit didnt friend snow

[timid silo]

hey my alt

[tall tusk]

weird

heyhey

[obtuse pebble]

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[static beacon]

https://gyazo.com/e39817a48bdd73419f1be286963293a5

[elfin burrow]

combine the fractions

[static beacon]

Yea but i dont even know

what im doing

like there is no big N

what is big N

[spiral knot]

A number for which all numbers greater than it satisfy the equation

[static beacon]

so like

am i just re-arranging that equation

for small n

[elfin burrow]

lim n to infinity of (4n-5) / (9n - 100) equals 4/9

[static beacon]

Yea

i know tha

[elfin burrow]

this is because the 4n and 9n terms dominate

so you are seeking a large enough integer N such that the difference between (4n-5) / (9n - 100) and 4/9 is extremely small

[static beacon]

by extremely small u meant 10^-6?

[spiral knot]

yeah

[static beacon]

thanks

.close

[obtuse pebble]

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[frail cargo]

is this the right system of equations?

[timid silo]

the RHS of the second eq should be 2 * 18 for the total cost

[frail cargo]

so it woul be 3x+1.5y=36?

[timid silo]

yeah

[frail cargo]

ok thx

[nocturne minnow]

No I don't think so

You want and average of $2 per lb

[frail cargo]

yes

[nocturne minnow]

So the LHS of the second equation would be divided by 2, for average, and the RHS would the total weight... Oops coincidence that the total weight was 18, my bad

It would be 2 * 18

[timid silo]

huh?

[frail cargo]

ok

[timid silo]

If x and y are the weights of chocolates and nuts respectively (in pounds) in an 18 pound bag

then x+y = 18, first equation

if, the bags have an average cost of 2$ per pound. Then an 18lb bag will cost on average 2*18 = 36lbs.

but also the cost of an 18lb bag will be 3x+1.5y

hence 3x+1.5y=36, second equation

[frail cargo]

i understand now thanks

.close

[obtuse pebble]

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[final girder]

Imma need a lil help

so firsy

factorise

-(x^2+14x) +9

for the first equation

then

complete da square

so - ( ( x + 7)^2 -40 )

then expand

[nocturne minnow]

You forgot the ^2

[final girder]

yea just pretend its there

so

[nocturne minnow]

Yeah... no

[fierce lagoon]

Im blind

[nocturne minnow]

$-( ( x + 7) -40 ) \neq - ( ( x + 7)^2 -40 )$

[final girder]

-(x+7)^2 +40

[warm shale]

dldh06

[fierce lagoon]

Communication is important and some people just want it there

[final girder]

this is correct right

therefore c is -7 and d is 40

and the coordinates of the turning point would be the inverse of c

[fierce lagoon]

c is not -7

[final girder]

fuck me

c is 7

i got ahead of myself

but

coords of turning point

is -7, 40

[fierce lagoon]

The turning point is (-7, 40) yes

[final girder]

and c is 7 d is 40

alr

i think

i forgot the extra bracket here thats why

Wowzers

[final girder]

my mans

[nocturne minnow]

I was gonna say that equation wasn't right

[final girder]

its -40

right?

wait

wait

no

[fierce lagoon]

Oh I did a dummy

[final girder]

its a maximum graph so the y value would be positive right

[nocturne minnow]

No

[fierce lagoon]

,calc -(-7)^2 -14(-7) + 9

[warm shale]

Result:

58

[fierce lagoon]

Idk how I got 40

[final girder]

huhhh

so 49 x -1

-49

minus

man

[fierce lagoon]

I mean I used a shortcut to get it but

[final girder]

I am not gonna lie to you

my mind is boggled 7

Can I do this one step by step

first

factorise

so -(x^2 +6x) +1

then complete square

so

• ( (x + 3)^2 + d

to find d

[fierce lagoon]

$$\begin{align*}
-x^2 - 14x + 9 &= -(x^2 + 14x + 49 - 49) + 9 \
&= -((x+7)^2 - 49) + 9 \
&= -(x+7)^2 + 49 + 9 \
&= -(x+7)^2 + 58
\end{align*}$$

[final girder]

do 1

[warm shale]

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

[final girder]

merci

1 minus (3)^2

so 1 minus 9

-8

[nocturne minnow]

Are you saying d = -8?

[final girder]

yes

[nocturne minnow]

Because no

[final girder]

fek

wait

what would d be then

oh no

i know what youre saying

im not saying d is 8

-8

let me cook

• ( (x + 3)^2 -8 )

so now

expand

-(x+3)^2 + 8

[nocturne minnow]

That's wrong still

[final girder]

how

[nocturne minnow]

-(x^2 +6x) +1
-(x^2 +6x + 9 - 9) +1
-((x+3)^2 - 9) +1

[final girder]

why is there -9 and +9

[nocturne minnow]

Because when you complete the square, you do (b/2)^2, right?

[final girder]

yes

[nocturne minnow]

So (6/2)^2 = 9

Because you add 9, you need to balance it by subtracting 9

[final girder]

wouldnt this just be equal

[final girder]

to this

[nocturne minnow]

No

[final girder]

fml

[nocturne minnow]

Distribute that negative on the outside

What do you get?

[final girder]

-(x+3)^2 + 9 + 1

bro youre confusing me

[nocturne minnow]

And 9 + 1 =?

[final girder]

wait wait wait

[final girder]

in here

would + 9 - 9

be D ?

[nocturne minnow]

No

[final girder]

wait wait wait lemme explain

[nocturne minnow]

The answer is still no, it's not going to be D

[final girder]

bro i dont know wtf is going on

i feel stupid

[nocturne minnow]

You are skipping too many steps

[final girder]

okay

can we start with the original question

[nocturne minnow]

$$-x^2 - 6x +1$$
$$-(x^2 + 6x) + 1$$
The next step, you went straight to $-((x + 3)^2 + d$
You went straight to plugging in a d variable instead of doing it one step at a time

[warm shale]

dldh06

[final girder]

okay

so first we find c

[nocturne minnow]

Yes that is completing the square but since you skipped steps, it threw you off

[final girder]

which would be coefficient of x /2

so is c 3?

[nocturne minnow]

$$-(x^2 + 6x) + 1$$
Let's start there

[nocturne minnow]

Not quite

[warm shale]

dldh06

[nocturne minnow]

The proper equation is (b/2)^2

[final girder]

??

i have always been taught c = coefficient of x/ 2

in the form (x+c)^2 + d

[nocturne minnow]

That's when you jump to that form

[final girder]

and d = constant - c^2

[nocturne minnow]

The key is, you need to understand how to process it one step at a time

$$-(x^2 + 6x) + 1$$
When you complete the square, you do $(b/2)^2$ to get
$$-(x^2 + 6x + 9)$$

[warm shale]

dldh06

[nocturne minnow]

Because you added 9, you also need to subtract 9, to keep it balanced

[final girder]

is this specifically for equations in the form

-(x+c)^2 + d

[nocturne minnow]

$$-(x^2 + 6x + 9 - 9) + 1$$
Is what you end up with if you completed the square

[warm shale]

dldh06

[nocturne minnow]

Then you factor $x^2 + 6x + 9$ to be $(x + 3)^2$

[warm shale]

dldh06

[nocturne minnow]

$$-(x^2 + 6x + 9 - 9) + 1 = -((x + 3)^2 - 9) + 1$$

[warm shale]

dldh06

[final girder]

yes

but

i think im beginning to understand but

[nocturne minnow]

This is you trying to memorize things

And not fully understand the step by step process

[final girder]

it because i do it with the other way

with equations not in this form and get it right

but apply the same process when its in this form

and get it wrong

[nocturne minnow]

Sure d = constant - c^2 but you are forgetting that there was a negative factored out

So I'm pretty sure the more appropriate equation is d = constant - a * c^2

[nocturne minnow]

Notice how this form isn't what you have
You have a negative in front, this form -(x+c)^2 + d is what you have

[nocturne minnow]

That equation will apply if it is in this form (x+c)^2 + d

[nocturne minnow]

Pretty much this is the key point

[final girder]

so when theres a negative factored out

we do it like this?

[nocturne minnow]

You would use this to find d

The general form is $a(x + c)^2 + d$

[warm shale]

dldh06

[nocturne minnow]

When you were taught this, a = 1

Hence why you could do this d = constant - c^2
With no mistakes

[final girder]

so when its in the form -(x+c)^2 + d

[nocturne minnow]

But what if a wasn't 1, what if it was -1? Like in the current problem. That changes things

I best suggest this as the general equation
d = constant - a * c^2

Because right now you are only factoring out -1

Like $$-x^2 - 6x +1$$ to $$-(x^2 + 6x) + 1$$ was just factoring a negative

[warm shale]

dldh06

[nocturne minnow]

But what if it was $-2x^2 - 12x +1$?

[warm shale]

dldh06

[nocturne minnow]

You will still factor but when you factor it'll be $$-2(x^2 + 6x) + 1$$

[warm shale]

dldh06

[final girder]

wait wait wait

c is 3 and d is 10

[nocturne minnow]

Long story short, it's best if you know this because this works for all the cases

[final girder]

right?

[nocturne minnow]

For the original problem, yes

[final girder]

im cooking

Video on website told me this

did the exact same method detailed and got 3 and 10

wait

man how the fuck is the video from the site wrong

lemme show you my working

$$ -x^2 -6x +1 $$

[warm shale]

Homebound

[final girder]

okay

[nocturne minnow]

It's not, it took in account for the negative sign that was factored

[final girder]

$$-(x^2 +6x -1)$$
$$-((x+3) - (3)^2 -1)$$
$$-((x+3)^2-9-1)$$
$$-((x+3)^2 -10)$$
$$-(x+3)^2+10$$

[warm shale]

Homebound

[final girder]

this should be right

[nocturne minnow]

Yes that is a proper method

[final girder]

wait

this is kinda like your method

[final girder]

this

wait no its not

its just

c ^ 2 - constant

so its normal

I think im on the brink of locking it in

OH NO

[nocturne minnow]

Here is the key point, you learned d = constant - c^2, that does not work for all cases of completing the square

[final girder]

wait

it IS your method of finding D

since a is -1

we do -1 x (3)^2 - constant

which is essentially your method right?

because -1 x 3 squared is -9

and -9 -1 is -10

right?

[nocturne minnow]

That would be related to the second to last line here

[final girder]

oh wait

yes

im confusing multiplication and addition

its - +9 so it becomes -9

[nocturne minnow]

If you wanted to go straight to this form, then d = constant - a * c^2 is more appropriate

[final girder]

alright

one more

sorry for taking up so much of your time

i know how to do this one

first

put in form

$$-x^2 -8x -25$$

[warm shale]

Homebound

[final girder]

this is correct, right?

no wait

8x positive

so

$$-x^2 +8x -25$$

[warm shale]

Homebound

[final girder]

then factorise it in terms of -1

$$-(x^2 -8x +25)$$

no

25 should be positive

[warm shale]

Homebound

[final girder]

then

complete square

$$-((x-4)^2 - (-4)^2 + 25)$$

[warm shale]

Homebound

[final girder]

so then

$$ -((x-4)^2 - +16 + 25)$$

[warm shale]

Homebound

[final girder]

$$-((x-4)^2 - 16 + 25)$$

[warm shale]

Homebound

[final girder]

so then

$$-((x-4)^2 + 9)$$

[warm shale]

Homebound

[final girder]

so then

$$-(x-4)^2 -9$$

[warm shale]

Homebound

[final girder]

@nocturne minnowmy friend is this correct?

[nocturne minnow]

Submit and see

Process looks fine

[final girder]

x would be 4

y -9

Got it right