help-10

[final girder]

Trusting the process is insane

.close

[obtuse pebble]

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[nocturne minnow]

And doing one step at a time

Notice how you did one step at a time instead of skipping multiple steps

[obtuse pebble]

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[final girder]

there is more hard math to be done

[final girder]

In any case

this would mean that

$$-(x+3)^2 - 2 $$

[warm shale]

Homebound

[final girder]

so we must get this back into its original form

first

lets get d back into original form

so

wait

step by step

just going backwards

$$ -((x+3)^2 +2) $$

what now

[warm shale]

Homebound

[final girder]

would i just

wait

am i overthinking

do i need to just expand and simplify

[teal turret]

Expand and simplify w this

[final girder]

yea

[teal turret]

There’s no difference between those 2 expressions

[final girder]

figured that much

so -6 and 7

no

[teal turret]

?

So what do u think it is now

[final girder]

dont know bruh

got a new question

[teal turret]

Bruh

[final girder]

Same principle applies though

thats how this site works

[teal turret]

Yea I figured

[final girder]

$$-(x+4)^2-5$$

[warm shale]

Homebound

[final girder]

expanding

fuck it

lets put it all in brackets

$$- ((x+4) (x+4) -5)$$

[warm shale]

Homebound

[final girder]

wait

i dont need the whole think including -5 in brackets right? @teal turret

cause thats only multiplying the (x+c)^2 term

not the -5 constant

[teal turret]

It’d have to be +5 if u wanted everything in parenth right

[final girder]

id rather not

that sounds like a lot of work

[teal turret]

lol

[final girder]

so can i keep -5 out the brackets?

[teal turret]

U have to unless u change it to a +5

So yea

[final girder]

which i am NOT doing

[teal turret]

Yea

Not “can” but “I must”

[final girder]

so -8 and -11 is what i got

[teal turret]

No

[final girder]

Fml man

where the HELL am i going wrong dude

[teal turret]

So did u keep the -5 out of the parentheses

[final girder]

yea

got

$$-(x^2 +4x +4x +16 ) -5 $$

[warm shale]

Homebound

[teal turret]

Yep

Keep going

[final girder]

brackets first right?

okay then so

[frigid robin]

whats hommebound?

[final girder]

Wait no

im going to simplify it

[teal turret]

This is the same part u got wrong from the last problem, this step

The distribution of the -

[final girder]

$$-(x^2 +8x +16) -5$$

wait wait wait wait

[warm shale]

Homebound

[teal turret]

Yep

Keep going

[final girder]

now

-x^2

-8x

-16

$$-x^2 -8x -16 -5$$

[warm shale]

Homebound

[teal turret]

Yep

Keep going

[final girder]

$$-x^2 -8x - 11$$

[warm shale]

Homebound

[teal turret]

👎

[final girder]

how

[teal turret]

The c term

What’s -16-5

[nocturne minnow]

-16 - 5

[final girder]

im a dud

-21

[teal turret]

Ye

[final girder]

bruh

such a simple mistake

[teal turret]

yep

[final girder]

2023

I ENDED 2022 DOING A MATHS QUESTION

THIS IS THE YEAR THE PROJECT GOES BAMNG

LETS GOOOOOOOOOOOOO

[teal turret]

European

Happy new year

But ur in the future atm

[final girder]

.CLOSE

.close

[obtuse pebble]

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[frigid robin]

nice

[teal turret]

do u have a math question

lol

[thick oracle]

😂

[obtuse pebble]

@frigid robin Has your question been resolved?

[obtuse pebble]

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[frigid robin]

no

[fierce lagoon]

Bruh

[wooden cipher]

whats your question?

[nocturne minnow]

• Show your work, and if possible, explain where you are stuck.```

If you don't have question, then close the channel and open a channel when you actually have a question

[fierce lagoon]

Send this man to the dungeon where he will rot for all eternity!!!!

[frigid robin]

Nop lease

Oh sorry

never mind

I have resolved my own question

[nocturne minnow]

Then close the channel since you're not properly using it

[fierce lagoon]

.close

[obtuse pebble]

Channel closed

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[frigid robin]

how to close?

[fierce lagoon]

This has to be the biggest bruh moemnt

[frigid robin]

im new

Im very sorry

.close

[frigid robin]

How do I close the channel?

[little warren]

it's closed

just leave it as it is

[nocturne minnow]

It was closed by someone else

[frigid robin]

Oh ok

[nocturne minnow]

If you see this, you are good

[frigid robin]

But it still has my name next to it

[little warren]

dw it will close

[frigid robin]

Ok

[nocturne minnow]

It takes a few minutes to fully close

[frigid robin]

oh ok thanks guys

[obtuse pebble]

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[desert finch]

when f(1/n) = 1 and f(-1/n) = -1 , show that the limit f(x) when x approaches 0 doesnot exist

how do i do that exacly ?

[fierce lagoon]

What happens when n -> infinity

[desert finch]

it approaches 0

[fierce lagoon]

So essentially when you take |n| -> infinity, you get f(0) = 1 and -1

But then

Like

Hell breaks loose then

[lament tartan]

There is a discontinuity yes

So when you take the limit from both sides

It doesn't match up

I.e. the limit doesn't exist

At x = 0, the point of the discontinuity

[desert finch]

so can i ask why are they using sequences here ?

[fierce lagoon]

Prob to do with squeeze

[desert finch]

so let me get this right

[quaint glen]

Its a known result that if lim_{x \to a} f(x)=L, then lim_{n \to \infty} f(a_n)=L for any sequence a_n that goes to a

So, if we can find two sequences that go to zero along which the function has different limits, that would contradict this result

[desert finch]

could u type this one in latex , i donot really umderstand it

[quaint glen]

Its a known result that if $\lim_{x \to a} f(x)=L$, then $\lim_{n \to \infty} f(a_n)=L$ for any sequence $a_n \to a$

[lament tartan]

Just take the limit from both sides, of f(x) as x approaches 0, by taking the inner limit as n approaches infinity

[warm shale]

Sneaky

[lament tartan]

A discontinuity is defined as where limits as approached from both sides are unequal

[desert finch]

oh okay i understand now thank you very much both of you

[obtuse pebble]

@desert finch Has your question been resolved?

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[untold cave]

Is the following a correct approach to $(\cosh(x+iy))^i$?

$$\cosh(x+iy) = \cosh x \cos y + i \sinh x \sin y$$\

Thus, in polar form
$$\cosh(x+iy)=\sqrt{\cosh^2 x \cos^2 y + \sinh^2 x \sin^2 y} \ \text{exp}(i\tan^{-1}(\tanh x \tan y)).$$\

So $$\ln(\cosh(x+iy))=\ln(\sqrt{\cosh^2 x \cos^2 y +\ i \sinh^2 x \sin^2 y})+i\tan^{-1}(\tanh x \tan y).$$

Therefore we can use $(\cosh(x+iy))^i=e^{i\ln(cosh(x+iy))}$ and that should be it.

[warm shale]

𝙹o

[obtuse pebble]

@untold cave Has your question been resolved?

[brave bramble]

Should be what? What's the question?

[trail cloak]

The question is about whether the method used has any flaws or not

[obtuse pebble]

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[raw coral]

is this correct

[nocturne minnow]

Do you know how to tell if two lines are parallel or perpendicular?

[raw coral]

i changed it to y=mx+b

if it was parralel it will be the same

if perpendicular it will be inversed

its none

thanks

.close

[obtuse pebble]

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[obtuse pebble]

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[robust sierra]

I’m kinda stumped with the following proof. I understand the idea and how it works and all, but I just can’t seem to wrap my head around where the (b-a)/3 value came from for epsilon. It just seem extremely random, how could I have come up with it if I had been coming up with the proof myself?

[viral blade]

Probably doing the proof backwards

Maybe even starting with just arbitrary delta and seeing which epsilon it guarantees

[unreal musk]

Weird way they've done it there tbh

[gleaming ridge]

The point here is that if they are not the same you want a value of epsilon that would result in a contradiction since you assumed a != b
choosing e = |b - a|/something makes sense

Although, the further proof is weird ngl

[robust sierra]

Alright thank you guys

they did give a second proof for that one iirc

using the triangle inequality

.close

[obtuse pebble]

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[valid crown]

I'd like to have a formula for function along a curve "if doable", for example cosine along exponent, where cosine is non deformed in respect to exponent, has same amplitude and period

[valid crown]

I came up with this

[dark stirrup]

It would have to be parametric

[valid crown]

I can measure length of a function for constant period

[dark stirrup]

this is a differential geometry question as you need to know the curvature of the exponent

[valid crown]

and create perpendicular in any point to map points correctly

but here is a little problem

for any x, I can easily map only previous points

at this point, is this still possible to map properly a point for x without iterative methods?

[valid crown]

so, make a tangent, then perpendicular and mapping

but that doesn't guarantee a formula in terms f(x)

[dark stirrup]

You almost certainly cannot

Hence why it needs to be parametric

[valid crown]

almost?

[viral blade]

e^x gets more and more vertical

So you want a wavy line going straight up that passes the vertical line test?

[lost tree]

like this?

[valid crown]

no

Like I drew

here cosine is symmetrical to y=0 axis

I want it to be symmetrical to y = e^x curve

your only holds for constant period

[versed turret]

They want cos along exp

[viral blade]

e^t arc length differential is sqrt(1+e^2t) hmm

and slope is e^t

x = t + cos(t)sin(arctan(e^t))
y = e^t + cos(t)cos(arctan(e^t))

this won't be good but maybe it's failures can be improved

[valid crown]

I try to put it in desmos

[valid crown]

length function is taking much time to compute

[versed turret]

Don't know how useful it is, but the function's average along exp will equal exp

[viral blade]

length function?

For your solution I assume

Do you think t = ln(tan(u)) is too optimistic lol

if it is try integral calculator .com maybe

[valid crown]

https://www.desmos.com/calculator/xtwjm1lmbg

got any clue what is the syntax for n(m) ?

[viral blade]

You can't do derivatives like that

$g(x) = \f d{dt}f(x)$ maybe

[warm shale]

monikanicity

[valid crown]

derivatives were computed correctly

as I checked

[viral blade]

Desmos doesn't recognize the prime symbol

[valid crown]

it does, look for tangent a normal with a for slider

[tardy epoch]

Did you freehand this

[lost tree]

no

[valid crown]

it's from n(m) ?

[lost tree]

what is n(m)

i used geogebra for this

[valid crown]

oh ok

[wild swallow]

looks kinda scuffed

[valid crown]

Whoa

I knew beyond certain steepness it's no longer a function

[viral blade]

need smaller period ig

bendy sine curve

[lost tree]

make it e^x/5 or something

ig

this is too vertical

[wild swallow]

thats only going to postpone the vertical

wont make a difference

[lost tree]

yeah but

[valid crown]

Could you share a code please?

[wild swallow]

i mean you could technically wrap any graph around it

[lost tree]

at least itll look kinda less vertical near the origin

[wild swallow]

heres a funky wave

[tardy epoch]

Make one that says 2023

[wild swallow]

im not about to compute the fourier coefficients of that

[untold scaffold]

wave that wavy wave with more waves

[viral blade]

Make an anime girl next

[wild swallow]

https://www.desmos.com/calculator/1t527gkhlz

[valid crown]

this one is static to just e^x

I'll try to generalize

[wild swallow]

its for efficiency

giving desmos the arclength formula isnt good for its health

[valid crown]

yeah

[viral blade]

Particularly computers are bad with double integrals lol

Which, nesting the function twice you were somehow doing I think

idk

[valid crown]

welp, I got also this from some time ago, it doesn't hold constant amplitude but it works fast

https://www.desmos.com/calculator/tjmgf6bja4?lang=pl

nontheless I couldn't reverse engineer this to find out why it doesn't work for linears (y=x) and fix amplitude

[wild swallow]

sine along sine

[viral blade]

Nah u need to get rid of the f(t) in the g(t) expression lol

Idk maybe not

That screws with the period ig

[wild swallow]

[valid crown]

ok, I finally repaired those 2 parametrics

finally works for all cases

https://www.desmos.com/calculator/sivn40tdme

[wild swallow]

https://www.desmos.com/calculator/c0mluctqlb

[viral blade]

bored

[wild swallow]

[viral blade]

i have an idea

[valid crown]

circular function and fourier series for "one to rule them all" ?

[viral blade]

I deeply regret this

is my expression just too long

nah ok ik the problem

[valid crown]

meh, just add step by step

[viral blade]

ur gonna despise me after this

I promise

[lost tree]

i just realized this was wrong

[valid crown]

can't wait to

[valid crown]

I pasted reworked formula

[lost tree]

this is crrect

[valid crown]

here

[wild swallow]

[lost tree]

oh

[valid crown]

almost

[tardy epoch]

e^(1/10) lmao

[wild swallow]

this already calculates the required graph in general

[valid crown]

your period is not constant

[lost tree]

oh

[valid crown]

I denoted fe as function along which curve is drawn

fc drawn curve

fl length of fe

[valid crown]

call me dumb but where do I enter 2nd function?

[viral blade]

its so beautiful

https://www.desmos.com/calculator/yinfer1vjw

[wild swallow]

what second function

[wild swallow]

[versed turret]

.close?

[valid crown]

[tardy epoch]

Bookmark this to the server

[viral blade]

nah someone needs to @everyone for this

[valid crown]

welp, thanks to everyone invested, happy new year

[tardy epoch]

How long do desmos links last for

[valid crown]

.close

[obtuse pebble]

Channel closed

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[viral blade]

252 million years

[wild swallow]

forever

[obtuse pebble]

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[rapid cove]

how do i solve this: 2cos(x)+2cos(2x)=0

[sage geode]

Use double angle identity for cosine (preferably the one where sine is absent)

[rapid cove]

👍 thx

[obtuse pebble]

@rapid cove Has your question been resolved?

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[timid silo]

what are the steps to graphing this?

[warm shale]

_____

[obtuse pebble]

@timid silo Has your question been resolved?

[viral blade]

What happens when you plug in +-2 for x?

[timid silo]

<@&286206848099549185>

[cloud berry]

do you know how to graph sec(x)

[timid silo]

I know how it looks

[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

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[hollow sail]

Hi can someone help me with b5

[plain stag]

what about it are you having trouble with?

[hollow sail]

Wait nvm I solved it alr

.close

[obtuse pebble]

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[hexed stirrup]

hi i need help

[hexed stirrup]

z-7+7xy-xyz

anyone help me out

i have doubt in last step

[dim bobcat]

lol isn't that the same question which you asked earlier

[hexed stirrup]

where you get

[hexed stirrup]

yes it is

[dim bobcat]

Ask the doubt directly

[hexed stirrup]

i didnt understood the last step

wait a min

[dim bobcat]

Which step

Show

[hexed stirrup]

$(z-7)+xy(z-7)$

[warm shale]

Rock Man

[hexed stirrup]

how this would come as

$(z-7) (1-xy)$

[warm shale]

Rock Man

[hexed stirrup]

@dim bobcat this step i didnt understood

[dim bobcat]

You know about this property that ab+ac = a(b+c)?

[hexed stirrup]

no

its factoriztion

[dim bobcat]

Dosen't matter

[hexed stirrup]

i dont know about the formulas

[dim bobcat]

Answer what i am asking

It is regarding your doubt

Can we write ab + ac as a(b+c)?

[hexed stirrup]

yes

yes

[dim bobcat]

Then we did the same thing

In above step

[hexed stirrup]

uh?

[dim bobcat]

(z-7) is acting like a there

[hexed stirrup]

oh

[dim bobcat]

Common in both terms

[royal shard]

Shouldnt it be +xy?

[hexed stirrup]

can you plz do it on paper and post it here

the pic of the solution

plz

[dim bobcat]

Yea he wrote it wrongly

It was -xy

[royal shard]

Ah ok

[dim bobcat]

Ok

Wait

[hexed stirrup]

okay

@dim bobcat done solving?

[dim bobcat]

Was telling other one about something

Sending wait

[hexed stirrup]

ok

[dim bobcat]

ab+ ad can be written as a(b+d)
Just like it
z-7+7xy-xyz
=(z-7)-xy(z-7)
=(z-7)(1-xy)

[hexed stirrup]

okay so in this solution where i have doubt is

as you wrote z-7-xy(z-7)

i didnt understood how did you write xy(z-7)=1-xy

how this?

@dim bobcat

[dim bobcat]

I took (z-7) common just like i took a common in above one

I didn't write 1-xy in place of xy(z-7)

[hexed stirrup]

my doubt is how can we solve

xy(z-7)

[dim bobcat]

We don't solve anything

[hexed stirrup]

to get answer as (1-xy)

[dim bobcat]

We take common

Because we have to factorise

There's no equation here that we have to solve

You agree that ab+ ad= a(b+d)?

[hexed stirrup]

i dont even know bcuz we just started the chapter like 2 days back

!

[dim bobcat]

It's basic maths

Nothing to do with your chapter

[dim bobcat]

?

[hexed stirrup]

i dont even got to understand something

sorry to say this

[dim bobcat]

Ok

ab+ad= a(b+d)

This is called distributive property of addition

Which is used above

[hexed stirrup]

okay

.close

[obtuse pebble]

Channel closed

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[hexed stirrup]

.reopen

[obtuse pebble]

✅

[dim bobcat]

lol

[hexed stirrup]

i need it for sometime

[dim bobcat]

What else do you have to ask bro

[hexed stirrup]

if i may get another doubt

[dim bobcat]

Ok

That one clear?

[hexed stirrup]

i may get bcuz i am currently solving

[hexed stirrup]

yeah my mom explained it to me

[dim bobcat]

What mistake were you making earlier

[hexed stirrup]

not knowing how to solve

[obtuse pebble]

@hexed stirrup Has your question been resolved?

[obtuse pebble]

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[wild orbit]

31/pi is the final answer, can someone help

[obtuse pebble]

@wild orbit Has your question been resolved?

[stable rain]

,rccw

[stable rain]

sry whats the integral

im only seeing a phi but why does it end with dphi dtheta

[wild orbit]

Its a spherical integral

[stable rain]

eh

idk but

i guess the integral is not dependent on theta so we can just treat it as a constant 2 pi outside

we can split the integral into 2 bits but idk how to do the um

cos 3phi sin phi

probably convert cos 3phi to trig with just phi

then int

[obtuse pebble]

@wild orbit Has your question been resolved?

[wild orbit]

Would u get the answer as 31 pi/6

[stable rain]

idk

[wild orbit]

I dont understsnd this ,can u like wtite it down or smth?

[stable rain]

no idk how to do it

im just saying u probably hv to do something similar

[wild orbit]

👍

[stable rain]

probably like

sin(A+B) = ...

and

[wild orbit]

Looka like someone else have to come in , i eill wait for some time...

[stable rain]

cos)A+B) = ...

[wild orbit]

Oh

[stable rain]

eh

yea mayb try it

[wild orbit]

Hmm,what about integration by parts?

[stable rain]

also u can just try integral calculator to check

er

idk mayb it works too

[wild orbit]

Yea i used symbolab

[stable rain]

just google integral calculator

convert it to a single integral fisrt

or idk

[wild orbit]

For phi everywhere ,i used x ,cos i didnt have the symbol

And im getting 17 pi/3

[stable rain]

anw idl weird trig integration

idk how to do

gl

n im out

[wild orbit]

I shuld be getting 31 pi /6

[wild orbit]

👍

[wild orbit]

Have fun

[stable rain]

thanks

atb

[wild orbit]

Will wait some time to see if anyone else will try to help ,otherwise will close the channel ig

[wild orbit]

Hmm

.close

[obtuse pebble]

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[cobalt kiln]

https://cdn.discordapp.com/attachments/423244559682764800/1058545751745114222/image.png

[cobalt kiln]

I have done letter b, can anyone check my proof? Btw, it should be a+c/b+d, there's an error

[obtuse pebble]

@cobalt kiln Has your question been resolved?

[obtuse pebble]

@cobalt kiln Has your question been resolved?

[icy badge]

@cobalt kiln how did you justify that a/b < c/d?? Your proof is okay apart from this. Check for the case that c/d < a/b

show that for both of these inequalities a + c/ b + d is between a/b and c/d

[cobalt kiln]

@icy badge You don't need to justify it, you begin with it as your definition/assumption. And yes, I was thinking the same thing about whether I should do c/d < a/b to show that it goes both ways. Did you see the second part of my paper?

I think I will say as a note that if you do c/d < a/b, you end up with a+c/b+d also between them, With the same exact procedure as the first one

[icy badge]

then, great. Make sure you state that you assume that a/b < c/d not just "such that" it sounds like you are changing the problem

[cobalt kiln]

Oh ok, yes. English is not my first language and I don't usually write proofs, so this is really my first time doing this. What about the second part, is it any good? Thanks

[icy badge]

looks good to me, great job

[cobalt kiln]

ok thanks, btw wait a minute just going to check something

[icy badge]

what kind of proof did you do?

you should know

like a direct proof? contradiction? contrapositive?

[cobalt kiln]

This is contradiction here

[icy badge]

im testing u

[icy badge]

very nice

[cobalt kiln]

It seems to be the only oen coming to my head

lol

[icy badge]

that's fine

[cobalt kiln]

Meaning, I don't know all other proof techniques

Btw, I did c/d < a/b and it also works. Although, if you establish that a/b < c/d is set in, do you really need to put the inverse? I mean, this would be changing the number line to something else, even though it also works. I think one would suffice no?

[icy badge]

yes, that suffices. Just say that by symmetry, the case or the assumption that c/d < a/b leads the same result regardless

you have to cover all cases.

you covered the a/b < c/d case, but what a bout the c/d < a/b case?

just say it checks out using same argument but different variable names

[cobalt kiln]

Yes, I guess I lose nothing doing such a thing also

Btw, I did letter c, do you want to check it?

[icy badge]

no thanks

[cobalt kiln]

Ok, thanks again for the help btw

[obtuse pebble]

@cobalt kiln Has your question been resolved?

[dark stirrup]

In part b, you wrote ad-bc=+1

But it's given that it's +-1

[cobalt kiln]

The minus one is done at the end, I gave -1 also

The -(ad-bc) part

[dark stirrup]

Nah that's not how it goes

ad-bc=+-1

It could be either

Neither should be disregarded

[cobalt kiln]

wait im not following you

[dark stirrup]

It's possible that ad-bc=-1

[cobalt kiln]

oh oh, it does not have to give +1

ok*

[dark stirrup]

That's what +-1 means

Yes

[cobalt kiln]

ok, so writing +- instead should do the trick then right

[dark stirrup]

Literally read out loud as "plus or minus 1"

And then -+ for -(ad-bc)

[cobalt kiln]

The minus is already given in the second part, so I only have to write the + to take into account this also

[dark stirrup]

Yes

Write the plus under the minus for the second one

[obtuse pebble]

Channel closed

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✅

[cobalt kiln]

oh i understand now why you inverted the signs

it's because - * +1 and -1 * -1, which gives at the end -+1

@dark stirrup Is that right?

[dark stirrup]

Yes exactly

[cobalt kiln]

ok thanks my friend!

Btw, did you have time to read my solution to c or not yet?

[obtuse pebble]

@cobalt kiln Has your question been resolved?

[timid silo]

i will close this channel in a few minutes due to timeout

ok

.close

[obtuse pebble]

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[obtuse pebble]

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[obtuse pebble]

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[surreal grotto]

\begin{align}
&\text{Prove that,}
&\frac{(x^2+1)(y^2+1)(z^2+1)}{xyz} \geq 8\
&\end{align}

[warm shale]

Mushfiqur

[surreal grotto]

For any positive real numbers

[rigid lintel]

with positivity constraints on x, y, z right

$x, y, z > 0$

[warm shale]

Gijs

[slate zephyr]

Proof first that ((x²+1)/x) >= 2

[obtuse pebble]

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[cobalt kiln]

https://cdn.discordapp.com/attachments/423244559682764800/1058545751745114222/image.png

[cobalt kiln]

Can someone check my proof for letter c

https://cdn.discordapp.com/attachments/903430333096132618/1058794598111117332/IMG_0589.jpg

[dark stirrup]

mediant, not median

[cobalt kiln]

yes sorry, I did not change the paper since I last sent it to you

It's the same one

[dark stirrup]

"first" mediant implies multiple. I'd write "given mediant" instead.

[cobalt kiln]

What I meant by "first mediant" was the one which you obtain by doing a+c/b+d, because you can still subvide your line into other mediants

[dark stirrup]

Yeah I know what you meant

But saying "first mediant" implies that a/b and c/d have multiple mediants

It's just a clarity thing

[cobalt kiln]

oh ok, I'll clear this then

other than that, is there anything else?

[dark stirrup]

What word is this

[cobalt kiln]

its numerator is

[dark stirrup]

what about here?

[cobalt kiln]

a distance

I write like this btw lol

[dark stirrup]

gotcha

How do we know that d1 is non-zero?

[cobalt kiln]

wait a minute let me think about this

yes so, as far as I remember, since this is a distance, which is located between a/b and c/d, you know it is going to have some quantity, right?

If it was zero, this means it would be equal to a/b or c/d

when we assume e/f is between those two fractions

[dark stirrup]

It's weak wording, mathematically, but it makes sense.

yes

[cobalt kiln]

How would you write it in words?

[dark stirrup]

Something like this

[cobalt kiln]

yes ok

I should have specified e/f > a/b and e/f<c/d

and done that it stays superior to 0 in mathematical terms

[dark stirrup]

Your proof is correct up to this point

This indeed is a true statement

But it does not mean that f has no upper bound

Showing what the lower bound is does not mean that it has no upper bound

[cobalt kiln]

oh ok, thanks for mentioning this. To be honest, I did consider this case but I was not sure how to interpret my result. I assumed my original statement "disappeared"

btw, just to be clear I'm following, you'Re saying that f < b+d is my upper bound, right?

[dark stirrup]

yes

As a hint

Consider c/d-e/f

You let d1=e/f-a/b

So let d2=c/d-e/f

It must be true that d1+d2=c/d-a/b

[cobalt kiln]

Ok, thanks, I'll try to derive the rest of the proof by myself and hopefully my head will "click"

[dark stirrup]

good luck

[obtuse pebble]

@cobalt kiln Has your question been resolved?

[cobalt kiln]

Ok, so I obtain this final result. I think the first one, -f <=d+b can be disregarded since we are only working with positive values of f here in general, based on what the problem stated

So we only take a look at the second part, with the positive values, and we obtain a contradiction

@dark stirrup is my reasoning correct?

[obtuse pebble]

@cobalt kiln Has your question been resolved?

[dark stirrup]

looks good

you're finally done

@cobalt kiln

[cobalt kiln]

So, I was right to exclude the first one based on what I said?

[dark stirrup]

yes

[cobalt kiln]

Thanks again

man, what a problem this was

Take into account it is my first time doing this

[dark stirrup]

You can generalize to consider negative fractions, but it's just the same result with extra work

Yeah it was a non-trivial problem

[cobalt kiln]

While I'm happy I solved it, I'm kinda realizing that this pushed and showed the extent on my knowledge

So I'm not sure of how to feel about this, besides that I need to work on myself

Btw, like I already told you, I did not check the paper which I sent you guys after the first time

So everything depended on what my head gave me and what i remembered

And yes, I inspired myself from it, but this is not a problem as long as I don't look at it while doing the problem and I furnish the detials of my steps, right?

The reasoning

[dark stirrup]

self-learning is a self-discipline

This is a hard problem algebra wise

doable, but hard

[cobalt kiln]

At the same time, I feel like I have grown from this and learned a lot, just from one problem. Btw, do you still want to see my final paper for letter c or you trust that I shall write a good thing?

[dark stirrup]

Nah it all looks good

I trust you put it together right

[cobalt kiln]

ok thanks again. Btw out of curiosity, do you study math at uni?

[dark stirrup]

I graduated

a while ago

[cobalt kiln]

undergraduate you mean?

[dark stirrup]

yes

But I work now

Out of school

[cobalt kiln]

I always wondered something when stuyding math at school

since you have to take a number of classes, and you end up piling stuff, how do you manage to find all these tricks in such a short amount of time?

How did you "survive"

like, one week you have to solve some problems, and obviously next week you also shall have to do so

[dark stirrup]

experience

[cobalt kiln]

Do you really manage to find the tricks or you jsut have to take some hits sometimes?

[dark stirrup]

"doing" is the best teacher

[cobalt kiln]

yes, but you have 1 week for example to solve x problem, with the other classes

[dark stirrup]

I mean, there are no set "rules" to solving problem

But, there are patterns that can be learned

How well you learn them really depends on the person

Some people have a talent

Some don't

If you're doing it in your free time, you have the capacity to develop it

[cobalt kiln]

So you indeed have to learn some patterns to make this an effective thing

[dark stirrup]

Yes

And the more you do, the more patterns you learn

It'll help your problem solving

[cobalt kiln]

yes, thanks. At the same time, the cosntant time pressure is really an unpleasent thing

I'm "scared" that this will kill my joy

[cobalt kiln]

what did you mean here?

[dark stirrup]

If you're learning math on your own, you should be able to develop a talent for it since you seem to enjoy it enough

[cobalt kiln]

ah yes, I get your point. Even someone who has a thing for something has to work to become better

nothing comes for free

It was really pleasant talking with you, and you were right to stand on your position in the face of adversity. This is my original error when doing this problem : I considered the upper-bound option, but since I wanted to really solve this problem, I let my bias disregard this option

I was self-deceiving myself

[dark stirrup]

It happens to all of us

I've been confidently wrong before so I know the feeling

[cobalt kiln]

I'm really happy you still did not go with the flow, since this would have put me in the error. Truth is better

Thanks again and good night!

[dark stirrup]

good night

[cobalt kiln]

.close

[obtuse pebble]

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[median thicket]

i need help

[median thicket]

Hello?

[meager mantle]

hi

ok so

mnp congruent tus

angle mnp = tus

mnp = 180- (142+24) = 180-166 =14

2x-50 = 14

or 2x = 64

su = np (cpct)

2x-y= 13

64-y = 13

y=51

[obtuse pebble]

@median thicket Has your question been resolved?

[earnest elk]

Archi don’t give out the whole solution

[obtuse pebble]

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[delicate lion]

In this example, we have 2 free variables in the Ax = b. But why x2 and x4?

Thank you

[gleaming ridge]

Basically, the variables corresponding to the pivots are what forms the space of solutions rest form the free variables

I am not sure if this is just convention or if there's an actual theory behind this

[delicate lion]

I don't know why x1 and x3 are pivots too.
If pivot means a diagonal components that contains non-zero values,
In this example, you can add row 1 to row 2, then component(1,1) and component(2,2) will be non-zero pivots.
That means column 1,2 is pivot column

[gleaming ridge]

I think by definition pivot means the first non-zero entry in row reduced form

So, the corresponding pivot variables are x1 and x3

[delicate lion]

I see. That's clear. Thank you @gleaming ridge

[gleaming ridge]

sure

pls close the channel

[delicate lion]

.close

[obtuse pebble]

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[queen mica]

Can someone hint how to start this

[versed turret]

In general, if you have a function, how would you try finding its maximum/minimum points?

[jaunty goblet]

u can find the minimum point

[kind hawk]

Derivative

Set it equal to 0

[queen mica]

and solve for x or?

[kind hawk]

Yes

[queen mica]

Okay i'll try that

Im not sure how to solve for x, as i end up with x^3=-250, and i cant change it so that it would =x as i cant have negative square roots

[kind hawk]

Show your work

[versed turret]

You need to find the derivative first

[queen mica]

oksy

[violet kiln]

step 1

[queen mica]

i got, f'(x) = 2x-250x^-2

[violet kiln]

set it to 0

and solve

tell me what u get

i can help you

if you want my help

[queen mica]

Okay

[queen mica]

when i set it equal to zero, i got this

[violet kiln]

show me

oh

since you cant have negative answer

there are no critical points

that's that

you done

[queen mica]

?

[violet kiln]

x has to be greater than 0

right?

[queen mica]

Yes

[violet kiln]

wait

there is A critical point

you can have a critical point where there is an asymtote

since the derivative is undefined at 0

that's your critical point

[queen mica]

I tried finding the derivative

[violet kiln]

can you give me f(x)

[queen mica]

But i ended up with f'(x) = 2x-250x^-2

[violet kiln]

whats f(x)

x^2 - 250ln(x)?

[queen mica]

yea

[violet kiln]

ok

let me graph it

there are 2 critical points

let me solve

ill let you know when i'm done

[queen mica]

okay

[violet kiln]

the 2 critical points are x=5sqrt5 and x=0

the derivative is 0 at 5sqrt5

try it

[queen mica]

huh

[violet kiln]

is it x≥0 or x>0

[queen mica]

0>

[violet kiln]

so x=0 is NOT a critical point

[queen mica]

Okay

[violet kiln]

so the only critical point is x=5sqrt5

[queen mica]

How do you know that

like how would i find that

[violet kiln]

so this is my process

f'(x)=2x-250/x right?

[queen mica]

yes

well

[violet kiln]

now we set it to zero

2x-250/x=0

[queen mica]

the x goes up i think?

to make -250x^-2

[violet kiln]

no no no

[queen mica]

no?

[violet kiln]

you ADD 250/x on both sides

that will yield 2x=250/x

[queen mica]

Ok hold on

f(x)'=2x+250/x

[violet kiln]

NO NO NO

f'(x)=2x-250/x

you said that didn't you

[queen mica]

yea yea

[violet kiln]

so we set 2x-250/x equal to zero

let's divide by two

so we get x-125/x = 0

add 125/x on both sides

x=125/x

multiply by x

x^2=125

take the square root

x= √125 or -√125

we exclude -√125 because it's less than 0

we can simplify x=√125 to x= 5√5

that's what you get

[queen mica]

Ohhh

That makes much more sense

[violet kiln]

you had a simple addition mistake

or you were using the wrong function

[queen mica]

But do i leave it at 5root5

[violet kiln]

yes

[queen mica]

Probably

[violet kiln]

you were getting f'(x)=2x+250/x

which implies f(x)=x^2+250ln(x)

but you said f(x)=x^2-250lnx

so you were differentiating incorrectley

[queen mica]

i wasnt sure how to move the x without making it negative, i get it now

you made it much clearer

[violet kiln]

it was fun helping you

[queen mica]

Just one last thing,

[violet kiln]

👍

tell me

[queen mica]

The answer scheme says "x=5 minimises the area" im not sure what that means

[violet kiln]

can you show me the book

the answer scheme

[queen mica]

Thats all the teacher put

[violet kiln]

i think your teacher is referencing the area under the curve

can you show me the question

no. 12

[queen mica]

?

[violet kiln]

ok let me see

i got confused

nvm sry

let me take a look of this

take the derivative

set it to zero

A'=2x-250/x^2

set it to 0

2x-250/x^2 = 0

divide by two

x-250/x^2=0

x=250/x^2

x^3=250

x=cbrt(250)

simplify the cube root to get x=5cbrt(2)

no i made a mistake

when you divide by two you get x-125/x^2 = 0

let me restart

A=x^2 + 250/x

A'=2x - 250/x^2

set it to zero

2x-250/x^2 = 0

divide by 2

x-125/x^2 = 0

add 125/x^2 on both sides

x=125/x^2

multiply by x^2 on both sides

x^3 = 125

take the cube root

you get x=5

because the cube root of 125 is 5

so that's the answer

x=5

[queen mica]

Yes! that makes so much sense

Thanks a bunch, was so confused

[violet kiln]

excellent

glad i could help

[queen mica]

.close

[obtuse pebble]

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[young solar]

The result is supposed to be infinite but I get 0 instead
And I have no idea where the error is

[cold linden]

i found the mistake

[slow cairn]

The limit doesn't even exist

[cold linden]

-sqrt3 times sqrt 3 is not -9

[cold linden]

and then if you put 2 in the variable x you get a 0 in the denominator

[young solar]

OH

but then what about the numerator?
It still becomes 0/0

[cold linden]

yeah it means you the answer to the limit is still hidden in this formula

[young solar]

It’s because I am supposed to find the point of discontinuity

[young solar]

I should have a number that is not 0 in the numerator so that lim n/0 = infinite
But I already rationalised and,,,I am not sure how else to rationalize it

[little warren]

limited developments and bam good bye limits

[young solar]

maybe I should have multiplied the numerator and denominator with rad(x-2) instead??

[young solar]

But I still would get 0/0 in the end…

[cold linden]

no because you still have 0 in the numerator

exactly

[young solar]

I have no idea on how else to solve it though

The other solution would be to simplify something but
They are radicals

😭

[tawny fog]

I got a result where numerator is 3 and denominator is 0 then it can be equal to infinity right?

[cold linden]

yes and how did you exactly do it?

[tawny fog]

L'hôpital

Got this result

And √(x-2) is 0

But this is undefined

[tawny fog]

$\frac{\dv {x} (\sqrt{x-2})}{\dv {x} (\sqrt{x²-1}-\sqrt{3})}$

[warm shale]

Arnab Pal

[tawny fog]

Could simplify that first

[cold linden]

i wonder if you could do it in a different way than l'hôpital

[tawny fog]

Well I don't know about any alternative methods actually

Only this thing popped into my mind

Seeing one hint

[cold linden]

like what if l'hôpital doesn't work

[tawny fog]

It should work right

Cause this is 0/0 situation

[cold linden]

in this situation yes

but what if you got a limit where l'hôpital used doesn't work?

[tawny fog]

Well then there would be so many complications

You could simplify

The expression

Or you could apply some theories of series

Like tylor series

Another great example will be this

$\lim_{x\to \infty}{(1-\frac{a}{n})^n}=e^{an}$

[warm shale]

Arnab Pal

[tawny fog]

I am bad at tex

To proof that you have to use series

And there's much more

[tawny fog]

Well @young solar I think you should use this

L'hôpital

To get your answer

[young solar]

I can’t

I still didn’t do it in the math program
So I can’t use things my math teacher still has to teach

[tawny fog]

Well I am 14 not I am learning all of these stuff so I don't know that advance calculus sometimes if you can't use L'hôpital then it would be complicated@cold linden

[tawny fog]

Ohh 🥲

[cold linden]

have you learned this online or?

[tawny fog]

Online

[violet kiln]

where is the x term in the limit

[tawny fog]

n tends to infty

[atomic bobcat]

use sequence?

[tawny fog]

Yupp

[young solar]

Yeah I know :, )
But my math teacher dislikes if we use derivates when

1. he still needs to teach us
2. these exercises are “””simple””” so they are meant to be solved and used without derivates

A classmate of mine did all the exercises with hopital and the teacher said that those exercises don’t count because they were supposed to be done without derivates😭

[cold linden]

so no derivates?

[atomic bobcat]

yeah you should avoid l hopital if you are learning to solve questions

[young solar]

Yeah 😢

[cold linden]

i don't know how to else solve this

[young solar]

🥲

In the meanwhile can I ask you where did you learn derivates and hopital?

This is the only exercise that I couldn’t do so far
So maybe I should at least solve it like you did and then ask the teacher how else to do it (so that at least I can say I did solve the exercise somehow😭)