Hmm
#help-23
pseudo scroll
What's tough
grim plover
say by some constant k
pseudo scroll
Oh are you trying to come up with a formula
grim plover
What's tough
nthin it was a general remark
grim plover
Oh are you trying to come up with a formula
yeah
desert pasture
so ${x+y}= (x+y)-[x+y]$
flat frigateBOT
ƒ(Why am. I here)=I don't Know
grim plover
well yeah
desert pasture
${x}+{y}= x+y -([x]+[y])$
flat frigateBOT
ƒ(Why am. I here)=I don't Know
grim plover
use $\{$
desert pasture
notice that if$ {x} + {y} \geq 1 , [x]+[y] \neq [x+y]$
flat frigateBOT
rak³en
Compile Error! Click the 
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(You may edit your message to recompile.)
grim plover
notice that if$ \{x\} + \{y\} \geq 1 , [x]+[y] \neq [x+y]$
no render
grim plover
notice that if$ \{x\} + \{y\} \geq 1 , [x]+[y] \neq [x+y]$
well if [{x} + {y}] exceeds 1 by k, then [x] + [y] + k = [x+y] right?
flat frigateBOT
ƒ(Why am. I here)=I don't Know
grim plover
Yeah i understand that
grim plover
well if [{x} + {y}] exceeds 1 by k, then [x] + [y] + k = [x+y] right?
is this true tho?
(and obviously, k is a non-negative integer)
desert pasture
well if [{x} + {y}] exceeds 1 by k, then [x] + [y] + k = [x+y] right?
try with a few examplers
like 1.9 and 0.9
desert pasture
desert pasture
well if [{x} + {y}] exceeds 1 by k, then [x] + [y] + k = [x+y] right?
I mean this is true
desert pasture
k will change dpending on teh number though
desert pasture
yes
tardy oxide
desert pasture
!occupied
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desert pasture
<@796939110815629322> ?
yes
copper pewter
**ƒ(Why am. I here)=I don't Know**
If {x}+{y}≥1 then [x]+[y]=[x+y]+1
you can just casework this probably
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obtuse bobcat
obtuse bobcat
I don't know how to solve this and the person who was helping me went offline I think
low vector
!status
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5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above
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median flare
Pls help me with this question I have no idea how to move forward
safe radishBOT
Pls help me with this question I have no idea how to move forward
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median flare
<@&286206848099549185>
noble rose
?
polar minnow
wuts h
noble rose
So I can understand.
polar minnow
look
radius = 7
which is r
height = 11
which is h
now input it
into the formula
they give in the picture
noble rose
Okay.
brittle crown
into the formula
exactly
polar minnow
<@&268886789983436800> @noble rose is trying to get help in other help channels
idk sry for the ping
brittle crown
<@&268886789983436800> <@1104133323766370457> is trying to get help in other he...
is that illegal
polar minnow
probably?
polar minnow
asking for help
its that hes invading other peoples help
lik
u can see that thich channel's for lost
not
scemer
brittle crown
its that hes invading other peoples help
wdym invading
polar minnow
bro
every channel
is meant
for someone's individual probelm
once they're done
u can come and look for help channel thats open
or post in forum
noble rose
brittle crown
oh sorry idk how this server works
polar minnow
yes
brittle crown
im new
noble rose
ok.
median flare
bro this is my channel
<@&286206848099549185> pls help
Question is pinned
<@&286206848099549185>
acoustic holly
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lean otter
hey guys i need help with shell's method, more calculating integral but yeah, so using shell's method i have to solve the following equation, y = 2 - x^2 , y = 0, x = 0. its rotating on y axis
safe radishBOT
hey guys i need help with shell's method, more calculating integral but yeah, so...
lean otter
now i know that formulla for shell method is 2pi integral from lower bound to higher bound (h * r) dx
but i dont know how to actually integrate im having trouble
i know that lower bound is 0 and that upper bound is square root of 2
i connected that its becuase you get 0 when you plug that value into y
but i odnt know reasoning behind it
i also dont know what would go into integral
i think it'd be (2-x^2 * r) and i dont know what r would be here
quiet plume
If you're doing shells about the y axis like this, r is just x. So (2-x^2)*x
lean otter
If you're doing shells about the y axis like this, r is just x. So (2-x^2)*x
oh is it always x?
and how would i integrate that?
i think itd turn into 2x - x^3
and then 2x would turn into x^2 and x^3 would turn into (x^4)/4 right?
so itd turn into x^2 - (x^4) / 4
and thne i just plug the lower nad upper bounds in
and thats that?
quiet plume
oh is it always x?
Shells are done by taking a fixed distance from the axis of rotation, so if you rotate around y, r=x and if you rotate around x, r=y
quiet plume
and thats that?
Yes
lean otter
Shells are done by taking a fixed distance from the axis of rotation, so if you ...
alright so generally its gonna be
value of y multiplied by x
and lower bound of integral is 0 if the y=0 and x=0
and upper bound of integral is wherever y intercepts?
quiet plume
Yes, in this case. It can depend on the construction/ shape, but you can always draw it out and figure out what your h and r are
lean otter
oh alright i get it now
i was confused by integratin but it cant be too bad
thank you kind pumpkin man
.close
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tacit atlas
Any help?
We will look at the L-3 proof system, which is obtained from the normal proof system by omitting (A3) from the group of axioms. Also, we will mark each proposition with A+ the proposition obtained from A by omitting every instance of the negative conjunction.
A'. Prove that if A is proved in the system L -3 then A+ is a tautology
B. Infer that the L -3 system is incomplete.
the axioms -
$A_1 :A\to (B\to A)
A_2:(A\to (B\to C))\to ((A\to B)\to (A\to C))
A_3:(\lnot B\to \lnot A)\to ((\lnot B\to A)\to B)$
flat frigateBOT
mtr123
rustic goblet
Any help?
We will look at the L-3 proof system, which is obtained from the nor...
you might find better luck asking this in #foundations
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carmine saffron
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carmine saffron
hi I have to determine if this series converges using limit test
so I guess this is actually a calc 1 question bc I just don’t know how to calculate that limit
what about the root?
oh nvm
the highest power in the denominator right?
got it
what was that technique called again?
nvm i think it was smth like forced factorization
thanks!
.solved
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burnt shoal
can someone clarify the solution with like an illustration thanks
burnt shoal
idk where the 9 came from
swift hare
ok
lets say the bottom coind is facing up what are the possibilites for the next coin
burnt shoal
its all up
swift hare
yeah
burnt shoal
so wouldn’t you need to multiply by 2 (up or down) not 9?
swift hare
no because lets say the bottom coin faces down down
how many options are there for the next coin
swift hare
what about DUUUUUUU
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rocky heath
m^3-2m-4=0
safe radishBOT
m^3-2m-4=0
blissful osprey
m^3-2m-4=0
u need to simplify?
valid jetty
m^3-2m-4=0
u can see that 2 is a factor
chrome aspen
u need to simplify?
I assume we’re finding the value of m
rocky heath
u can see that 2 is a factor
how do i know that tho
valid jetty
how do i know that tho
u substitute and find out
chrome aspen
how do i know that tho
Are you solving for val of m or simplifying
rocky heath
i think im solving
rocky heath
u substitute and find out
is there a way to find out why 2 works
blissful osprey
umm firstly u need to break down the initial m^3 first
rocky heath
or i have to substitute in random values
blissful osprey
is there a way to find out why 2 works
to simplidfy the 2m and 4
rocky heath
to simplidfy the 2m and 4
ok
chrome aspen
f(1) = 1^3 - 2(1) - 4 = 0
Doesn’t equal 0 so move to 2
f(2) = 2^3 - 2(2) - 4 = 0
Works out to 0 so we know one factor is m +2
arctic raven
is there a way to find out why 2 works
you can verify 2 works by substitution, and you'd think to try 2 on your own by the so-called rational root theorem, if you're interested
blissful osprey
f(1) = 1^3 - 2(1) - 4 = 0
Doesn’t equal 0 so move to 2
f(2) = 2^3 - 2(2) - 4 = ...
ah this works too
mental dove
you can verify 2 works by substitution, and you'd think to try 2 on your own by ...
well I think you can solve the equation without using it
chrome aspen
Okay
arctic raven
well I think you can solve the equation without using it
for sure, I just thought I'd mention it as a bonus fact
mental dove
you can do it using splitting as well
mental dove
for sure, I just thought I'd mention it as a bonus fact
alright
safe radishBOT
(m+2)(m^2-2m+2)
The purpose of this server is to help you learn, not to hand out answers. Do not ask someone to give you the answer directly.
blissful osprey
(m+2)(m^2-2m+2)
yes this one
mental dove
no
chrome aspen
Well that is what (m-2)(m-2) is
mental dove
Well that is what (m-2)(m-2) is
no
chrome aspen
wdym no bro
mental dove
wdym no bro
😭 do the math
mental dove
but please don't give away the answers
valid jetty
is there a way to find out why 2 works
as far as i know there isnt any definitive way to solve higher than 2 degree equations
sometimes u substitute smtimes u simplify and go on like that
chrome aspen
hm
blissful osprey
Well that is what (m-2)(m-2) is
the m^2-2m+2 will be simplify as (m+1)^2+1
chrome aspen
(m+2)(m^2-2m+2)
am I right in saying you can’t factor past this?
acoustic holly
hi what is the question
chrome aspen
Well I mean the start should be correct, (m+2) works as a factor if I’m not wrong
acoustic holly
uh
blissful osprey
Well I mean the start should be correct, (m+2) works as a factor if I’m not wron...
yes the firsst one was right
acoustic holly
changeit into x^3 - 4x + 2x - 4 = 0
not factorize l
and now factorize that
x(x^2-4) + 2(x-2) = 0
blissful osprey
changeit into x^3 - 4x + 2x - 4 = 0
would it be the same if u simplify it too? the factorization
blissful osprey
would it be the same if u simplify it too? the factorization
from the initial eq m^3-2m-4 = 0
acoustic holly
x(x^2-4) + 2(x-2) = 0
factorize it
x(x-2)(x+2)+2(x+2) = 0
i guess its good enough to solve
factorize that again and solve it
( m = x)
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inland hawk
It has been reported that men are more likely than women to participate in online auctions. In a recent survey, 65% of respondents reported that they had participated in an online auction. In this same survey, 45% of respondents were men and 38% were men who had participated in online auctions. What is the probability that a respondent selected at random is female and has never participated in an online auction?
safe radishBOT
It has been reported that men are more likely than women to participate in onlin...
inland hawk
how to solve?
safe radishBOT
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inland hawk
.reopen
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inland hawk
It has been reported that men are more likely than women to participate in online auctions. In a recent survey, 65% of respondents reported that they had participated in an online auction. In this same survey, 45% of respondents were men and 38% were men who had participated in online auctions. What is the probability that a respondent selected at random is female and has never participated in an online auction?
neat fable
what have you tried
safe radishBOT
@inland hawk Has your question been resolved?
inland hawk
what have you tried
Honestly nothing really works what I do, I have no clue to begin what I have is P(Female and never participated)
low vector
try making a table
inland hawk
Btw answer should be 0.28
inland hawk
try making a table
Ok lemme try
low vector
columns male and female, rows participated and not participated
or vice versa ofcourse
inland hawk
I did it and I come to 0.38
low vector
can you show your working?
inland hawk
low vector
the problem you have is that the total doesnt add up to 1
all the probabilities should add up to 1
inland hawk
All 4 add up equal to 1?
low vector
yes
inland hawk
Oh I thought each row and Colomn had to be 1
low vector
if that was the case, then in a 2x2 case it would always be symmetrical
which doesnt have to be the case
you can think of it like this: theyre buckets
every person fits into one of the 4 buckets
the total of all the buckets is the total of all the people
inland hawk
Oh yeah I see
low vector
Oh I thought each row and Colomn had to be 1
this can easily be disproven with 3 people, 1 female who participated, 2 males, one which participated and one which didnt
low vector
yes its correct
low vector
yeah that looks right
low vector
do you realise how you did it?
inland hawk
You also don’t give the answer right away you are great
inland hawk
do you realise how you did it?
Yeah
Most important for me to find it was all add up to 1
I didn’t knew it before
inland hawk
Nice seems hard are you in university ?
low vector
yep
inland hawk
Oh well pretty tough then
low vector
well, its adjusted ofcourse
for me its doable
but for others probably not so much
but i have the same with different subjects
inland hawk
but for others probably not so much
Yeah true you must be great
And do you like statistics the most?
Or more algebra etc
inland hawk
And physics is it hard?
low vector
it depends
inland hawk
Cus I actually maybe am going to do physics
low vector
one course is easier than the other
and also depends whether you like the topic
at least, thats how it is for me
low vector
Wait lemme check rq what we will see in physics
physics exam or physics at uni?
low vector
youll get multiple topics
inland hawk
Not the. Ourse
inland hawk
youll get multiple topics
Yeah well I’m actually bit scared because I didn’t have alor of physics yet
You think it’s doable with not to much knowledge? Or can I adjust myself?
low vector
its built up in a way such that youll understand if you have high school knowledge of physics
inland hawk
Oh okay cus I will have that too
inland hawk
but im guessing its gonna be similar
Let’s hope so xd
low vector
the only requirement my physics study has is that you need to have had physics in high school
since in high school you couldve had the possibility to drop it after a certain point
inland hawk
Oh okay I see
The most scared I am is for math because we need to know like 200 proofs for theory
And it’s oral exam
low vector
like a pre-exam?
or whatever you call it
idk what you call it
but a small exam you have to do before just to kind of test you
to make sure you can succeeed
inland hawk
No like you the math examination but 50% is theory which is like 200 proofs oral exam and 50% exercises
inland hawk
but a small exam you have to do before just to kind of test you
Yeah the thing is I’m going to switch from course so I already had the math but I’m scared for the proofs of theory
low vector
No like you the math examination but 50% is theory which is like 200 proofs oral...
is it actually hard proofs or a proof you could think of on the spot?
inland hawk
And I have no clue how I’m going to know all of that
inland hawk
is it actually hard proofs or a proof you could think of on the spot?
Wait,lemme tell u
I will check
Its about all highschool math but mostly of last year
Like integrals and differentials and binnimium of newton
Limits like everything from high school
But like I mean really everything 😂
The document is 70 pages of just proofs
low vector
thats okay
inland hawk
Lemme send it
I mean how to learn all of this?
This is the only thing that scares me
To learn all that theory
low vector
most of it you probably already know
cuz this is mostly high school exam stuff
and by mostly i think all
inland hawk
Oh okay but you think it’s doable?
inland hawk
Yeah a bit
low vector
so, could you tell me what the conjugate of (3-4i) is?
inland hawk
3+4i I think
inland hawk
By differentiating?
inland hawk
now, hwo do i find extrema of a function?
Or yeah by finding min and max
low vector
Or yeah by finding min and max
yes, but how do i find those
you told me differentiating
but then what do i do with f'(x)?
low vector
exactly
we differentiate and set it equal to 0 to find the extrema
now, how do i find if its a maximum or minimum?
inland hawk
No put the x value in f(x)
low vector
By putting the x values in f’’(x)
continue
inland hawk
And then checking if it’s lower, equal or higher then 0
low vector
exactly
so
i have a point
i know its an extrema
and ive found that f''(x)>0
is it a minimum or a maximum
inland hawk
Minimum I think
low vector
exactly
see, you know this
lemme find smth else from the doc
hmm, i do see some stuff thats not high school level
inland hawk
Which?
inland hawk
Like the first pages?
inland hawk
Which page?
low vector
or determining convergences of series
low vector
Which page?
page 63, theorem 16
or l'hopital, i had that 1st year, but no way i wouldve had that in high school
inland hawk
page 63, theorem 16
Yeah I see
The thing is I don’t think many of them are really hard but it’s just a lot
low vector
yeah, and it can be overwhelming
but if you look at the stuff you need to learn for Wiskunde A or Wiskunde B, its also a lot
yet, you still know almost all of it
inland hawk
Yeah that’s true
I’ve seen almost all of it
It’s just the recognizing which will be the hardest for me
To not confuse stuff
low vector
youll be fine
try to not stress too much during the oral exam
thats where a lot of people struggle
they freeze up or cant think
low vector
it depends
- how much you already know
- how good you are at maths
- if you have the will to study it
inland hawk
- I would say 70%
- normal level I think
- I’m motivated yes, but I don’t like to learn to much on 1 day
low vector
then you could probably learn it in a few days
if you spend like 3 hours per day
and whenever you get confused or dont understand smth you can just open a channel here and ask for help
anyway, its already way past 6 so i have to go eat dinner :)
inland hawk
Okay thank I might go for it then
low vector
goodluck
low vector
Thank you, have a great dinner
you should say "eet smakelijk" instead
inland hawk
Are you Dutch? 😂
Your name seems Dutch
Oh wow handig; nu kan il Nederlandse opgaves sturen en je pingen
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jade comet
hello
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cunning pasture
<@&286206848099549185>
gusty trench
so 7inf-7inf = 0
infinity - infinity is an indeterminate form
you absolutely cannot conclude that this is 0
consider the function f(x) = x^2 - x, then consider its limit at infinity, you'd get an infinity - infinity form but clearly x^2 - x grows off to infinity
try multiplying both the top and bottom by the conjugate
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gusty trench
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gusty trench
so write your limit as (sqrt(49x^2 + 8x) + 7x) * (sqrt(49x^2 + 8x) - 7x) / (sqrt(49x^2 + 8x) - 7x) which simplifies to (49x^2 + 8x - 49x^2) / (sqrt(49x^2 + 8x) - 7x) which simplifies to 8x / (sqrt(49x^2 + 8x) - 7x)
this is something you can use l'hopitals rule on
and this will be how you get your answer
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balmy loom
You have to check in which one A and B are independent (statistics). I don't really know how to start this question.
safe radishBOT
You have to check in which one A and B are independent (statistics). I don't rea...
potent osprey
Start from the definition of independence
balmy loom
f(x,y)=f(x)*f(y) ?
potent osprey
Well yes, but for events this becomes $P(A \cap B) = P(A) P(B)$
flat frigateBOT
DAILI
potent osprey
$P(A)$ is just the area of $A$ divided by the total area $4$ (assuming uniform distribution on the square)
flat frigateBOT
DAILI
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balmy loom
Is checking this relation enough to conclude they are independent?
Nevermind, I think I got it.
.close
blissful frost
hey guys!
I really suck at math so I got a question wrong and I've been given the answer process but I'm not sure how even how they got the answer?
what happened to the power of 2 between the two sections I highlighted?
it's probably some really simple answer but like I said I really suck at it </3 thank you so much to whoever explains it though!!
safe radishBOT
hey guys!
I really suck at math so I got a question wrong and I've been given th...
inner parrot
they factored out an rcos(θ) on the right hand side
gusty trench
2r^2cos^2 is the same as (r cos) * (2r cos)
blissful frost
I'm losing it how did I not see that
thank you guys
literally had my eureka moment
.close
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bronze yoke
prove that for a in R, if a>=0 then √a <= (a+1)/2
safe radishBOT
``prove that for a in R, if a>=0 then √a <= (a+1)/2``
bronze yoke
so my thought process with this is that √a >= 0
and I got (a+1)/2 >= 1/2
if both √a and a are bigger than or equal to 0, that'd mean that √a >= a, but a <= (a+1)/2
actually that doesn't make sense
this question is breaking my mind
actually it makes sense
a <= (a+1)/2
so is it good enough for a proof to write it in a similar fashion?
odd crest
umm am gm
toxic stratus
,w a <= (a+1)/2
flat frigateBOT
odd crest
arithmetic mean of two positive real numbers is greater than or equal to the geometric mean
take the arithmetic mean of a,1
bronze yoke
oof
odd crest
take the arithmetic mean of a,1
It's a+1/2
and take the geometric mean
That's just sqrt(a)
bronze yoke
no
bronze yoke
uhhh do you know what geometric series are
what's related to the sigmas and all that stuff?
odd crest
well it's just a progression
toxic stratus
basically, you should start with the inequality [ (\sqrt a - 1)^2 \ge 0 ]
flat frigateBOT
bronze yoke
we have skipped those in this math course (I will be studying it later on) so I should be able to do this without prior knowledge of series
odd crest
Yeah this works fine^
toxic stratus
because after you expand it, you'll get what you want
bronze yoke
because after you expand it, you'll get what you want
a - 2√a + 1?
toxic stratus
= 0
bronze yoke
oh then I can get it to (a+1)/2 >= √a
very clever
but how would I figure out to start with that
odd crest
uhh you kinda don't, you learn from this procedure, since it's kind of tough to conjure that
toxic stratus
i mean
you kinda do
you just shuffle the terms around in the original inequality
bronze yoke
wha
toxic stratus
and see that you get what i wrote
toxic stratus
(a+1)/2 >= sqrt(a)
multiply through the 2
a + 1 >= 2sqrt(a)
a - 2sqrt(a) + 1 >= 0
just simple algebra
bronze yoke
snoseph answered what I didn't understand
toxic stratus
a - 2sqrt(a) + 1 >= 0
this is literally it though
odd crest
oh okay
toxic stratus
like
you do the algebra
and it's the natural conclusion
there's nothing nontrivial about it
bronze yoke
it does not at all feel like a natural conclusion for me but I guess I just need more experience lol
I can see the thought process
I would start doing something weird like a + 1 >= 2sqrt(a) => a >= 2sqrt(a) -1
and noticing that a - 2sqrt(a) + 1 is fitting for (a-b)^2 is also pretty hard to spot when you're used to seeing a^2 - 2ab + b^2
the brain just goes "oh it doesn't start with a^2"
toxic stratus
i think you just need more experience with algebra
bronze yoke
i think you just need more experience with algebra
you're 100% right
I didn't have a lot of time to gain a lot of experience but the next months will be more than enough for me to gain that much needed experience
anyways, thanks a lot
.close
safe radishBOT
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somber cape
I am confused on this problem
safe radishBOT
I am confused on this problem
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safe radishBOT
@patent gazelle Has your question been resolved?
humble oak
what does it mean when a derivative does not have a closed form solution?
humble oak
please mention me if you respond!
merry swift
@humble oak it just means you cant write it using just normal operations like +, -, etc, well as commonly known functions such as ln, sin, constants, and the variables. It may still be perfectly well defined but we just dont have a good way of writing it
humble oak
can you give an example perhaps? or is it hard to find such examples?
median vigil
do you have an example of where you encountered that phrasing?
humble oak
i encountered it in ML
merry swift
The funny thing about derivatives is that using the main functions we're familiar with and ways of combining them, we will always come up with a closed form solution, so such functions will never have a closed form themselves.
humble oak
The funny thing about derivatives is that using the main functions we're familia...
that makes sense
but it is still really intriguing
median vigil
not necessarily the derivatives themselves but also solving the "set it equal to 0 and solving for the parameters"
even simple transcendental equations often lack closed-form solutions
humble oak
oh okay
so how do you maximise/minimise such expressions?
here they used EM algo in ML, but what do people do in math?
gusty trench
this isn't saying that the derivatives don't have closed forms, just that the solutions to them being equal to 0 don't have closed forms
for example, f(x) = x^2 + sinx has f'(x) = 2x + cosx and clearly that derivative has a closed form, but there isn't a closed form for solving 2x + cosx = 0
humble oak
so a closed form is kinda like defining a set?
sin x = 0, x = n(pi) is a closed form
but you cannot define a closed set for 2x+cosx = 0
is that what it is?
gusty trench
more like you can't really algebraically solve for the solutions so they have to be approximated numerically
solar hazel
‘closed form’ isn’t even a well defined thing
you might as well say ‘nice form’ instead because that roughly means the same thing and is more descriptive
solar hazel
if you aren’t going to add any more info
yea but like
then a nice simple piecewise representation may not be a “closed form”
under that definition
you know what, i hate the term ‘closed form’
added to my mathematical irk list
humble oak
so basically you cannot find maxima minima directly here
solar hazel
what does directly mean
solar hazel
sure
humble oak
so how do these functions get maximised/minimised?
solar hazel
there are numerical methods for this
solar hazel
gradient descent is a simple one
humble oak
here they are using an algo called EM algo, but i wanna know how it is done in math
solar hazel
but not too robust
humble oak
cool, got it, thanks all of you!
solar hazel
also you may be able to approximate solutions to the derivative = 0 equation
but again it would be numerically
trail otter
I need to determine convergence of series
[\sum_{n=0}^{\infty}z^n\frac{\arctan (n^n+1)}{n^{\alpha}\ln(n^2+7)}; \ \alpha\in\mathbb{R}.]
I already know that radius of convergence is 1 so i can rewrite it into
[\sum_{n=0}^{\infty}e^{i\phi n}\frac{\arctan (n^n+1)}{n^{\alpha}\ln(n^2+7)}; \ \phi\in [0,2\pi )]
What should I do next? What test should I use?
safe radishBOT
I need to determine convergence of series
\[\sum_{n=0}^{\infty}z^n\frac{\arctan ...
safe radishBOT
@trail otter Has your question been resolved?
trail otter
<@&286206848099549185>
safe radishBOT
@trail otter Has your question been resolved?
grim plover
**Slowaq**
Yes
safe radishBOT
@trail otter Has your question been resolved?
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narrow grove
does an identity element need to be a member of the group?
safe radishBOT
does an identity element need to be a member of the group?
narrow grove
E e € G or A e € G ?(dont know latex sorry)
peak estuary
well both of these say that it needs to be in the group
narrow grove
but not always
peak estuary
but its the first
narrow grove
why
peak estuary
definition of group
peak estuary
the backwards E means "exists"
narrow grove
it says that there can be some e's which are not a subset of the group G
peak estuary
"there exists e in G such that ..."
narrow grove
how would it be any different if it was upside down A?
peak estuary
"for all e in G ..."
narrow grove
i know the literal meanings but im not sure if i could understand them properly
can you give me an example?
peak estuary
of what exactly
I mean this issue doesnt seem to have anything to do with groups or identity elements
narrow grove
of the difference between upside down A and inversed E
narrow grove
I mean this issue doesnt seem to have anything to do with groups or identity ele...
it does
kind of
peak estuary
no
narrow grove
of the difference between upside down A and inversed E
in this example
peak estuary
well I mean the A means "forall" and the E means "exists". those meanings are quite literally
narrow grove
OOOOOH
aight
i dont 💀
ok i get it now thanks
so, one more thing, if it was a backwards A, would it mean that all of the elements in group G are identity elements?
peak estuary
yes
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hot breach
fuck
solar pike
why did you multiply 5 by 6?
hot breach
because you have to
for something idk but
it's just a step someone told me
because you x 2 by 6 you have to x 5 by 6 as well
solar pike
Yeah, well you should try to understand better what you are doing
solar pike
It seems like someone else is about to instruct you, so I'll leave it to them 🙂
Well you want both fractions to have the same denomnator so you can add
hot breach
okay i understand that
solar pike
this is called "common denominator"
so you can achieve this by manipulating the second fraction
How can you manipulate the second fraction to have the same denominator?
hot breach
you just make it 12?????
solar pike
By multiplying the numerator and the denominator by the same number
proud storm
you only need to mulitply one side by 6 because you're trying to get them both at the same denominator
which is 12 for this example
solar pike
Yeah, well, when you multiply the numerator and the denominator by the same number, you don't change the fraction
because you basically multiply by one
Can you see this?
@hot breach
solar pike
When you have the same denominator, you can add the numerators
solar pike
5 + 30 + 9?
where did it come from?
Look, you should change the second fraction right?
by multiplying by both numerator and denominator by 6
you get $\frac{9\times6}{2\times6}$
hot breach
oh
flat frigateBOT
Cain
narrow grove
what grade maths is this?
hot breach
hi my name is riah and i'm 25 and i never fucking learned how to do math
^ that's me
that's it
solar pike
Oh god
hot breach
i could write u a novel but i can't do equations so here i am
failing basic math in college
so let's try this
hot breach
i'm confused off top because what the fuck can u multiply 2 by to get 5
solar pike
2.5, but you could modify both
proud storm
Just multiply them together to get a common denominator of 10
solar pike
Just multiply them together to get a common denominator of 10
That means you modify both fractions, and each fraction is modified by multiplying both numerator and denominator by a different number (to get a denominator of 10)
hot breach
so 2 x 5 = 10
3 x 10 = 30
1 x 10 = 10
so we have 30/10 + 10/10 = 40/10
40/10 is the answer?
solar pike
no
why 3x10?
@hot breach you must understand what you are doing!
It's not a magic
hot breach
the steps are not CLEAR
solar pike
Let me try to help you out again
proud storm
here lets simplify it a bit. If the denominator is 10, what would the numerator be if the fraction must be equal to 1/2
proud storm
you're not changing the value of the fraction but making it easier to work with
solar pike
You can't modify fractions as you like, you must do things that don't change the value of the fractions to get the right answer
Is this clear?
solar pike
The way to do this is by multiplying by 1
multiplying by 1 doesn't change the value
by we can write 1 in many ways
hot breach
okay
solar pike
$\frac{a}{a}$ is 1
flat frigateBOT
Cain
hot breach
yes
solar pike
so we can multiply by a/a to get a common denomiator
so that we can add the fractions
now we need to get a common denominator of 10
How would we change the denominator to be 10 for the first fraction?
3/5
we multiply by ....?
@hot breach
hot breach
2
hot breach
how can you tell me that 2 x 2 = 1
2 x 2 is 4 and 2 x 5 is 10
so i don't understand any longer
hot breach
2 divided by 2
solar pike
yes
proud storm
you mulitply both numerator and denominator by 2
solar pike
so $\frac{3}{5} = \frac{3}{5} \times 1 = \frac{3}{5} \times \frac{2}{2} = \frac{3 \times 2}{5 \times 2} = \frac{6}{10}$
flat frigateBOT
Cain
solar pike
Follow this closely ^
hot breach
okay
solar pike
This is the way you may rewrite fractions without changing their values
now do the second fraction
rewrite it to have a denominator of 10
solar pike
the second one is 1/2
hot breach
but you said rewrite it
hot breach
oh okay
winged pecan
1/2 * 5/5 = 5/10
solar pike
oh what the fuck
hot breach
LMAOOOOOO
winged pecan
oh what the fuck
wtf 🤨
winged pecan
ur right
hot breach
that makes sense
winged pecan
but yeah anything multiplied by 1 is itself so u can use it to ur advantage to manipulate things to the way u want it
hot breach
so now we have 6/10 and 5/10
solar pike
and what operator between?
hot breach
solar pike
no
hot breach
it becomes positive??
hot breach
ofc i do
winged pecan
well if it's plus, then a/b + c/b = (a+b)/c
hot breach
i have to leave but do you guys think i'm going to pass the rest of my exams :c
solar pike
Hopefully
hot breach
the next one is on exponents and factoring and other stuff but i'm restudying this module so i can do a make up exam
i think this was the first module and i was like oh yeah that'll be easy
but i made a 40 on it the first time i took it
after studying really har
d\
safe radishBOT
winged pecan
gl
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clever ridge
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winged pecan
the question asks u to sketch the graph but u don't really need to to answer this ig
sin and cos in particular do not have absolute maximums
winged pecan
they have local maximums and minimums
clever ridge
they have both
lime dust
<@&268886789983436800>
winged pecan
they have both
so where would u say the absolute maximum for cos x or sin x would be
white umbra
Tourette's is so not an excuse for this, go educate yourself about it
clever ridge
mm would appreciate if someone would help me solve f'(x) = 0
everyone scrambles out the door
white umbra
sin and cos in particular do not have absolute maximums
(every local maximum of sin or cos is a global maximum tho, no?)
clever ridge
yes
idk y he said that
but i am trying to prove critical points
it is literally written in my book
winged pecan
x = pi / 2 + pi * n ?
winged pecan
3pi / 2 would also work
winged pecan
there are infinite solutions
clever ridge
so those are the local min and maxes and absolute min and maxes
thx i got it
i think..
winged pecan
but getting critical points isn't really enough, u then have to prove that they are minimums and maximums
clever ridge
i did by plugging them back into the original equation
winged pecan
every local maximum and minimum occurs at a critical point but not every critical point is a local maximum or minimum if i recall correctly
clever ridge
im not sure abt the second part
mm
yea i think u can be rigfht
but in this case they are all local min and max for infinity
clever ridge
how do you get the derivative of f(x) = |x|?
winged pecan
get a piecewise function ig
|x| = x for x >= 0 and -x for x < 0
i don't think |x| is differentiable at x = 0
definitely not
winged pecan
prove what
winged pecan
it's only not differentiable at x = 0 isn't it?