#help-10
1 messages · Page 37 of 1
okay so
Now ill sub 0=0
we got all these
Get rid of some terms
yes
the ones you want to get rid of
are the xy, yz, xz terms
using the fact that they add to 0
hmmm
if you
delete teh 48xy and the 24yz
you also gotta delete 12pxz as well
oh wait
you did
A wat
yeah
Now i will yeet out xz
you cant just delete the first 3 terms
the only things you can delete
are these
not the same
Oh

hmm yes
alright
Train my brain
inequality
What is that
its the inequality that says
Damn
and the arithmetic mean is always >= the geometric mean
Is that a olympiad thing
O
its a pretty olympiad thing
have i 
i dont want you stalking my name
do you really believe @stable rain

alright
we only need AM-GM for two numbers
so lemme restate
this one i think we can even prove
I dont think i can but yes
lets prove it
HM
so
Direct proof?
agreed?
Yes
Nani
agreed?
Yes
so we expand
agreed?
Yes
Wut
wait shit
Lmao
now agreed?
OH
and then
alright so when does equality hold
equality holds when this equality holds right?
Wait but how the heck did u know we needed sqrt(a) instead of just a
I seriously dont get how u see it
Yes
in other words
thats when equality holds right
Ohhhhhhhhhh
when a = b
Yesh
over here
Yes
AM-GM for 2 numbers
i cant remember if i know the proof for n numbers
its probably jensens
but we dont need it
Yes
Wat
cool
now the idea is
the LHS
is a sum of square numbers
if we can show that
its gonna be too large
then itll never be equal to the RHS
Wait how does that help
But i thought we wanna find a solttution
yeah but
we gotta show that
there is no solution
to find when solutions start appearing
because at some point
it goes from there being no more solutions
to there being more solutions
we need to figure out when and how that happens
here
Yes
cuz squares
Ue
hmmmm wait
Wot
i think i missed a case in my solution
Ó
actually hmmm
Huh
because
the idea is we want to apply AM-GM
actually we might be fine
because it just doesnt matter lol
alright
Whats happening
lets just continue
Oki
how do we produce an xz
no no
we need to produce an xz
using AM-GM
so that we can compare sizes
because comparing x^2 and z^2 against xz is weird
this right
so
o oops
i do tat
yes
but simplify the square root a bit
and then move the 2 over to the other side
LOL

you did forget your z
and also
unfortunately
we dont know that x and z are positive
My 2 is also gone
lets fix it
Yes
okay cool
Tatata
Almost?
cuz we dont know x and z are positive
Yes
but x^2 and z^2 certainly are
Yes
actually this is still right
its just weaker than this
Yes
cuz |xz| >= xz
Yes
We replace
this one
what do we get
lol okay maybe i was a bit misleading
wot
should not have said sub in
we actually want to create a bound on the LHS from below
so we have
LOL
huh
36
lol
this
if we can show that the LHS is too large
to be equal to the RHS
then they wont be equal right?
Yes
we're now doing that
so we have a lower bound for the LHS right here
the LHS must be at least as large as this
this LHS has the 16y^2
WHY IS THIS QUESTION SO HARD
Wtffff
we need
to convert the y^2
into xs and zs now
to do that
we use the other equation
so
Wait
Where do i write this
yes
Then now
this
So another computation?
HEH
we're gonna sub it into the y^2 eventually tho
Where did u eben get this
what
this
so
i divided all the constants out in my working
so i didnt have the 4s and 6s and 3s flying around
okay
im like following and not following a the same time
i dont get why are there so many new computations
i wrote this
so what is it squared
sub in
nooooo
;-;
thats not an inequality either
it just
cant be subbed in
like theres no relation right no
between the inequalities we're deriving
and the original equation
not yet
yeah but
we cant sub an inequality into the equality like that
well we can
but
you gotta do it differently
pok done
nice
that was the whole inequality chain
yes
bruh i legit forgot what we were tryna find
cus we did so much more stuff
we are trying to find the value of p
yes
yes
alright cool
yes
so then
kind of braindead
what if we add in
but we will finish this
WHAT
from here
so we actually have
Thats how long weve spent on this
this is whats changed
Do i have to write this out
since here
Somewhere
okay good
now is where it gets a bit fucky
because i didnt consider this case in my original proof
lemme see what happens
possibly

aaa
@feral sentinel Has your question been resolved?
@feral sentinel Has your question been resolved?
@wild swallow HI
what is the question?
oh it was the old question?
we are like in the middle of doing it
ye
do u mind
helping me
with another qn
just curious tho, what does it meant by more than one set of solutions?
i think its that there exist another solution which isnt the trivial solution
like x,y,z = 0
ah right then
uh, you might not want to do that tho
oh why
multiple occupied channels stuff
also, i think i have a solution for this
maximum channel occupied is just 1
oh wot
yes
one channel at a time
im gonna ignore that there is a message above this one
yesyes

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Can someone walk me through this pls?
@quiet dome Has your question been resolved?
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Can anyone help me explain how "S(n) = n/2 [2a + (n-1)d]" works?
I assume you mean for AP
Yes
Wouldn’t it be easier for you look for some videos on YouTube
Well let S_n = (a)+(a+d)+….(a+(n-1)d)
Well yeah also that
Wait, what happens after (a+d?)
a+2d
(a+2d)?
OH I GET THE (N-1)D NOW.
It could save us so much time
But why 2a?
Go ahead..
A tutorial explaining and proving the formulae associated with arithmetic series.
VISIT MATHORMATHS.COM FOR MORE LIKE THIS!
Follow me on www.twitter.com/mathormaths, and like www.facebook.com/mathmathsmathematics to stay up to date with tutorials and examination walk throughs. You can always request your own videos!
I mean the thumbnail itself has all the required details
Oh ok, I got it! Thank you!
,tex \np
Perhaps close if done
deep is all about the closing today

I’m very active now

e4e5
No d4 best opening 
e5 
then bring out the bishop
e5 

Someone.close
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✅
ig ppl with helpfull can close channels now
helpers can't .close anymore?
ya
.close
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✅
they can
aight guess it's everyone then
even deep
Need to have helpers role
ping simulator
I am very active and thats what i want 😌

very active takes alot of patience tbh
No
does it
will miss my blue name tho
for me ya
Snow did it in less that 28 days lol

must be very active then

Yeah that very is like VERY
OO
It took me ig 2 months yeah

deep is very active now
Now i need kawaiiCat
welcome to no life role
lol


Lol bot only shows ur name in very active @timid silo
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Did we close this thing?
o
nope
.close
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can a improper fraction be a rational number
sooo?
for example?
what's your definition of imporper fraction
and what's your definition of rational number
idk
i am dumb
i would guess like 1 is rational
pie is not
1.333333333 is
1.2432534123124325 is not
1.252525 is
why don't you think
1.2432534123124325 is not
isn't rational
also doesn't answer the questions i asked earlier
This is rational
It terminates
so it would be irratiol
Rational fractions have decimals that terminate;
Irrational keep on going forever
you could view terminating decimals as decimals with a repeating trailing 0
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Hi, I don't understand this definition of a limit.
more specifically I get lost at the part after: x element of domain(f)
the upsidedown v is an and symbol, so its saying that if x is in the domain and x is delta away from a, then f(x) is epsilon away from L
wait so '0 < | x - a | < delta' means what exactly?
and where does the 'a' come from?
and why absolute value?
its basically saying the distance between x and a is less than delta
the absolute value just says it doesnt matter if x>a or x<a
so a is the like asymptote
and x is just a point you plug into the formula?
and delta is some line close to the asymptote?
if that's right, what about the part afterwards?
| f(x) - L | < e
its saying that as x gets closer to a, then f(x) gets closer to L
okay so wait
this is what im imagining rn
where in this picture does 'e' go?
and why are 'L' and 'a' different things?
L is on the f(x) axis, and a is on the x axis
ohhh okay
lemme draw a pic
what about the difference between 'e' and 'delta'?
great !
(disclaimer, its gonna be bad lol)
yes cause this is really high quality xd
ahhh, okay
if we are within delta of a on the x axis, then we are within epsilon of L on the y axis
Gigabyte
why is this like
why isnt it for example
$\forall \delta > 0 \exists \epsilon > 0$
that means that if I give you any positive epsilon, you can always find delta so that this image holds
Gigabyte
we want f(x) to get close to L
yes
we can always get close to a (by assumption)
so we want to prove that f(x) can be epsilon close to L, not x delta close to a
yeah, but idk if im doing a good job at explaining it to you lol
you're doing great so far!
thx 😀
uhm, im not following this entirely though
what would exactly be wrong about doing it the other way around or for example 1 sec
let me get all other possible permutations(?)
$\forall \delta > 0 \forall \epsilon> 0$
$\forall \delta > 0 \exists \epsilon> 0$
Gigabyte
$\exists \delta > 0 \forall \epsilon> 0$
$\exists \delta > 0 \exists \epsilon> 0$
can you like say what's wrong about them
same :p
we are looking at the definition of $\lim_{x\to a}f(x)=L$, ie, we want to show that f(x) gets close to L when x gets close to a. $\forall\epsilon\exists\delta$ does this.
$\forall\delta\exists\epsilon$ would be saying we can make x close to a as long as f(x) is close to L, which I guess says something about the inverse of f(x). The following, the limit doesnt exist but this permutation of exists/forall works
Toby
this is because "exists" means we can make a choice of epsilon
"forall" means it needs to work for any epsilon, in particular very small ones
this clears up a lot of confusion !!
so here, for any small delta (ie we want to get as close to a as possible), we can choose an epsilon (such as the one drawn) so that f(x) is close to L
so i get why that permutation is logical, but why would the other ones be incorrect?
like actually mathematically incorrect
so right now it says "for each delta we can pick any epsilon" and P(x) should hold
why would "for each epsilon we can pick any delta" and P(x) should hold be exactly wrong
$\exists\delta\forall\epsilon$ would be too strong
Toby
ie it says that there is a single delta that works for any epsilon
(normally, delta depends on epsilon)
oh so there exists a single delta for all possible epsilon
and this there exists a single epsilon for all possible delta
no, otherway. thats for all possible delta, there exists an epsilon (possibly depending on delta)
$\forall\delta\exists\epsilon$
Toby
yup
(logic hurts my brain :p)
same
okay wait so i also study computer science so
this is like a for loop and if statement right
for (delta) if (epsilon) P(x)
this is the correct one
yes for the for, not sure about using if
i think if is right cause logic is the foundation of programming languages im pretty sure
for (epsilon) if (delta) P(x)
you said this was too strong right? but would it be incorrect?
ig this is psudo code for the limit definition
set delta=delta_epsilon
for x in (a-delta,a+delta):
if not(|f(x)-L|< epsilon):
print("incorrect limit :(")
break
print("correct limit :)")```
and your question is about the order of the top two lines
isnt set delta=delta_epsilon always the same?
set delta=delta_epsilon
for epsilon in positive reals:
for x in (a-delta,a+delta):
if not(|f(x)-L|< epsilon):
print("incorrect limit :(")
break
print("correct limit :)")```
can you do this?
there exists delta for all epsilon would mean the limit exists, but it does not have to hold if the limit exists (ie converse doesnt hold)
ig im emphasising that delta depends on epsilon there
oh yeah cause delta comes from x and epsilon from f(x)
thats what you mean right
and L is never reached by the graph, but a is always accesible
if f(x)=x and a=0, then you cannot find a delta that works for all epsilon
f(x) = x does not have an asymptote?
your definition works for continuous functions too
continuous functions also have limits
like a precise definition
how does f(x) = x have a limit?
$\lim_{x\to a}x=a$
Toby
yeah but that doesnt say anything
why would you take a limit of a function that can reach everywhere
sure, but it still satisfies the defition
fucking math
xd
this really does improve my thinking though
think about $f(x)=\begin{cases}x&\text{if }x\neq0\\pi&\text{if }x=0\end{cases}$. Then $\lim_{x\to 0}f(x)=0\neq \pi=f(0)$
Toby
then $\lim_{x\to 0}f(x)=0\neq \pi=f(0)$
Gigabyte
why $0 \neq \pi$
Gigabyte
what did mean with that
writing math formulas (in latex) is so cool btw
best part by far
Gigabyte
you said $f(0) = 0 \neq \pi$
Gigabyte
is this the same?
I said $\lim_{x\to 0}f(x)=0\neq \pi$
Toby
yeah so you plug in 0 for the x right?
no
and then you get f(0)
huh
thats only true for continuous functions
whenever we see the limit symbol, we have to use this definition
(it just so happens that for continuous functions we can substitute and skip the definition)
wait lemme copy your message 1 sec
think about $f(x)=\begin{cases}x&\text{if }x\neq0\pi&\text{if }x=0\end{cases}$. Then $\lim_{x\to 0}f(x)=0\neq \pi=f(0)$
Gigabyte
wait why does it equal 0
i dont get it
like x approaches 0, but never becomes it
so shouldnt it be f(x) = x != pi?
yeah, the limit only cares about what it approaches, not what it becomes
then why does it equal 0
cause it never becomes 0 right







