#help-41
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someone please give me the answer to this question
anyone 😭
@keen quartz Has your question been resolved?
oof that's hard to do without paper
thjank yto
8sin(12t/pi − 9) maybe
no wait that's not right
+55
that looks right
maybe
could it be upside down?
no that's worse
this but 55
OH MY BAD
I DIDNT SEE YOUR MESSAGES
IM SORRY
is it +55 instead of 47
@sinful sapphire
IT WAS RIGHT
TYSM
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Bonus question help please
I literally tried but I couldn’t even make a system of equations
@manic spindle Has your question been resolved?
if the sum of 2 numbers is 5.5, what is their mean?
(T1 + T1 + d)/2
don't think of it as an arithmetic sequence
if I have two numbers a and b, what is their mean in terms of a and b?
(A+b)/2
5.5
great, so the mean of the two numbers must be 5.5/2
Yes
yeah and that is enough information to set up the linear equation actually
the "inserted" part is confusing I know
Wait
But the thing is there’s not only 1 arithmetic mean
There’s “k” amount of means
yes, but all of those means are equal to 5.5/2
Liek the sum of them?
wait, do you have the answer for the bonus question?
Yeah 12
Wait so does (a+b)/2 give u the value of the sum of the means
no, that just gives you one of the means
this
no, all of the means are equal to 5.5/2
no
mean 1 = 5.5/2
mean 2 = 5.5/2
mean 3 = 5.5/2
....
mean k = 5.5/2
So all the means are just the same?
But isn’t it an arithmetic series so all the means have a common difference?
no, you're overthinking it
it's just k * 5.5/2 - 21 = k
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need help with this question, this is my attempt for the first direction and need help on how to approach the other
@mortal vault Has your question been resolved?
@mortal vault Has your question been resolved?
@mortal vault Has your question been resolved?
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Tw: business math 💀
How to get the total number of payments (n)
Ive done the 4 × 10 since the time is 10 years and its quarterly
Since theres the word semi-annually will it be 4 × 10 × 2 😭
Or is it just 40
Am I overthinking it
wait what do you mean by n?
isnt that just 40? There are 40 payments involved, the interest rate/compounding/etc is irreleavant
number of patments is just 40
the semi annually affects the interest not the numver of payments
Thanks
👍
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✅
Same problem
How do I get the interest per period
Its r/m
So it should be 2%/2 right
Or is there something im not seeing 😭
This wasnt explained properly by my teacher so I have a lot of questions 💀
i think so (its been a while since i did financial math)
Thank you 
Alright
Im finding the future value now
Is this correct

<@&286206848099549185>

ano ba yan anteh?
wait madali lng yan may notes em
English only fo
Fatingin
huhu sorry i misplaced my notes
awh
I was just confused on finding the total number of payments

@celest sun
What about the total number of payments here
what's quarterly
Its where m=4
sorry i dont remember how to do that
yes
remember the compound annualy or quarterly or whatever only applies to interest, not number of payments
If its compounded ____ it -
YES
Thank you
sorry cant help with finding fv tho 😞
So if its compounded whatever it will only apply to i and n where it is i/m and t(m) respectively

Confusioj #1 solved ig
Thank you kristy
ig so
hurray
🔥🔥
like quarterly means 4 months
This is investment,annuities and loans right
Yes 
HOL UP
If its 7 x 12
Then i ( interest rate per period ) should be?
idk anymore
mhmm
Interest rate should just be divided by 12
what do u think semi annually means?
isnt it by 6
bc semi annually
Whats the purpose of compounded semi anually then
Oh ye
😭😭😭😭
I was only doing 12 cause he did 7 x 12
Wait so
semi anually means its compudned every 6 months (aka half a year)
our qn says u make payments per MONTH
In payments its per month and in interest its different
so u wanna change this from per 6 months to per one month, thus u divde the 8.5 by 6
Oh
the qn states were paying for 7 years right, and we make payments each month thus u get 7x12 payment installments
Uhuh
the compounded semi anually basically indicates the kinda like extra accumative value u own per 6 months
essentially ur interest from the qn tells us it accumates by an additional 8.5% per 6 months
but sicne were paying per one month
we want to change the compound interset from per 6 months to per one month
thus we divide 8.5 by 6
and rhats ur interst for ur qn
to solve and stuff
Cus
u want everything to be in terms of only per one month
Its monthly and then the semi annually
So its 8.5% / 6

Wait i got this let me process the information
OHHHH
I get this now
Okay thank you again kristy
yussss
yayyyy
nwwws
If it said quarterly instead then it will be every 3 months and we gonna do 8.5% / 3

Ig
Gonna close this now
.close
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Hey, i need help with probabilities calculations
!da2a
No need to ask “Can I ask…?” or “Does anyone know about…?”—it’s faster for everyone if you just ask your question! See https://dontasktoask.com/
I have an urn with 13 black balls, and 7 white balls,
i draw them one by one and sort them in a line until i run out of balls in the urn.
A. what is the probabillity that the last 4 balls in the line are black, if its known the first 3 ones are white.
B. GIVEN that 4 black balls are last in line, and 3 white balls are first in line, what is the probabilty that the first 6 balls in line at White?
i have solved for Question A, but i am having trouble understanding B,
i have calculated in A where:
A = the time where 3 white balls are drawn first,
B = the time where 4 black balls are drawn last.
and then did P(B|A), which gave me the answer
but for question B, i am unsure on how to approach it, is it like this:
C = the occurence where first 6 balls are white ?
and then P(C|AB)
or just take the "Starting position" where i have only to sort 13 balls, where 4 of them are white and 13 of them are black, and i need the first 3 ones to be white
its my first time doing homework in Probability and i am unsure on how to approach this
Do you know how many way it could satisfy the arrangement for Question B?
How
you are left with only 1 white ball, and 13 black balls
First six are fixed, you only have one more white ball going around other 9 balls
there are only 14 ways to rearrange them in different ways
Last 4 is fixed to be black
ohh thats for exclusvily 6 white balls
9 including the 4 fixed black balls
basically 10!/9!
Think 10 more second
10 ways
yeah, 12!/3!*9!
hmmm
is it, not P(A|B) in this situation?
since its a GIVEN state?
and not "if a then b?"
this is why i am confused, i am unsure of what to actually look for
P(A|B) means probability of A happening, when we already know that B has happened
i am still a bit confused
in question A they said if its known that, that implies A|B
but in B they explicitly said given
Here already give |AB
so given the state the first 3 are white and last 4 are black
i think you're confusing your A and B sets with the A and B questions
Okay, what are you calling A, what are you calling B, and what are you calling C?
A - 3 white balls at the beginning
B- 4 black balls at the end
C- 6 white balls at the start
Okay. So you calculated the amount of arrangements that have A and B happening, correct?
yeah
can you please mark it for me?
i have already calculated P(AandB)
the text is in hebrew, but essentially, this is what i did for the first question
writing is a bit funky, but i hope its enough to understand
yeah that looks fine
so you're only missing the amount of arrangements C
you can work easier if you fix the balls at the start and end
that does not sound right
well, if given the first 6 ones are white
we have 1 white and 13 black remaining
so there are 14 ways to sort them
yes, but you also have 4 black at the end
remember, you have the first 3 whites and the last 4 blacks fixed
so should my C be 6 white and 4 black at the end?
so you can only move the 4 white and 9 blacks remaining
That is C|AB as you defined
yeah, C|AB
and out of those arrangements, only the ones that have 3 whites at the start (fixed) are "successes".
So the valid cases would be the arrangements of 1 white and 9 blacks, while your total arrangements would be the arrangements of 4 white and 9 blacks
and that would be your P(C|AandB) as defined
so starting point is AandB
3 white, 4 black
and what are teh chances to get from AB to C
essentially
yeah, you can work with the probabilities of both, or you can work with the number of cases of both
usually, the number of cases is easier unless you already have the probs computed
so, how many arrangements of 1 white and 9 blacks do you have?
so your prob would be those 10, divided by the 13!/4!9!
yeah,
and is that equal to P(C|AB)?
or are we calculating something different
i am just confused defentionally
It is
i see
i think that i need to seperate my intuition from this
but thank you very much everyone!
much appriciated
as you defined them, it is
@clever sage Has your question been resolved?
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!status
What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
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
3/4
I am only well acquainted with the disk method
So I can check that
However you could wait for someone to come along and help you with the shell method
Well I was required to use online calculator for this for my project
But then the disk and shell method do not compute the same answers
Huh
That’s interesting
I’ll just let someone smarter come along and help you with this. I am really quite unaware of the shell method
I even double checked with my friends the equation for the disk and shell and still isnt the same
Do you know someone who can help?
@mint nacelle (Sorry for the ping senõr) is quite a smart fella. He may be able to help
He’s helped me multiple times before
Thank you man :)
can you not ping me for random questions
Noted. Sorry
I will not do that ever again
Could you help me out with a disk and shell method question
be active enough in help channels
I'm working on it
okay so you have $x = \left(\frac{37}{10} y \right)^{1/3}$
south, just south
yeah
the bottom horizontal line is y = 7.297 approximately
Just as a background info( With this equation when revolving around Y axis creates a cup and I'm supposed to calculate the disk and shell method for it)
,w integrate 2 pi (y - 7.297)(4 - (10/37 y)^(1/3) from y = 7.297 to y = 17.297
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jesus
Shouldn't this be calculated using disk method?
huh it's not working
you did it using the disk method though right
Yeah disk in terms of y
and shell in terms of x
Disk method gave me 397 with some decimal value
wrong
While shell method gave me 265 with some decimal value
can you explain this?
so each of these horizontal lines represents a shell
it has a radius, so that's the distance from y to the bottom horizontal line, y = 7.2973
Also since its revolving around y axis, shouldn't disk method be in terms of y?
it also has a length, so that's the distance between x on the function and x = 4
no, the disk method revolves around the x axis
But in the question I provided it revolves around y axis?
.
oh shit
Yeah I did it again and I'm still getting different answers for both disk and shell
ok give me more time
what calculations did you do?
For the disk method I put in terms of y with limits from 7.297 to 17.297
For shell I put the same equation *2pi x with limits from 3 to 4
no, what integrals did you get exactly
You mean the answer?
How i set it up?
yes
give me a minute
Yeah so
Disk method
shell method
@mint nacelle are they correct?
Yeah I'm here
wouldn't that just be $\pi\int_{7.2973}^{17.2973}\left(4-\left(\frac{37}{10}y\right)^{\frac{1}{3}}\right)^{2}\ dy$
south, just south
but that gives the volume as 9
Why is it 4-the y equation?
you want to find the horizontal distance between the curve and the vertical line x = 4
that's your "height"
but horizontal
Isnt x=3 and 4 limits to use for x?
no
you were correct originally
when you revolve around the y-axis using the disk method
the integral is in fact dy
What does it mean if disk and shell method answers are different?
Like which one do I need to consider then
they should be the same answer ofc
But idk my equation isn't giving the same answers
okay now check this graph
this should be for the y-axis being the axis of revolution
I thought you didnt need to put subtract 4 or 7.297 since they're just boundaries you need to calulate area for
they are super important
please take a look at my annotations
Oh damn
You literally solved what I was thought was unsolvable
Ty so much man
You have no idea how much you helped me
no worries!
also like 4 - x is also the height when you do it with the disk method
I see
the only difference is that we use $x = \left(\frac{37}{10}y\right)^{1/3}$
south, just south
You literally helped me so much
and for the washer method, we use $y = \frac{10}{37} x^3$
yeah that's how it works
alr no problemo!
Ty so much once again
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south, just south
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Solve the following sets of equations by graphing them on a rectangular corridinate system. Check your answers by substituting them back into the original equations.y=x+4 and y= -x+2
go on desmos.com and graph those two equations
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Can someone help with the 3rd question
This is what I did (the one I have cut)
@split sail Has your question been resolved?
<@&286206848099549185>
What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
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
@split sail Has your question been resolved?
do you still need help?
so you can actually find the smallest angle on the triangle arctan(32/75) and that angle will be an alternate angle
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Can someone explain the annihilator method to me please?
I don't feel like I understand it, factoring operators feels wrong and theres nothing intuitive about it
do you have a problem you're working on?
I can make a random one but its really just the concept I'm confused about
$9y'' - 4y' - 12y = x - 6$
Nathan
In mathematics, the annihilator method is a procedure used to find a particular solution to certain types of non-homogeneous ordinary differential equations (ODEs). It is similar to the method of undetermined coefficients, but instead of guessing the particular solution in the method of undetermined coefficients, the particular solution is dete...
I'm looking for an explanation than examples
I just dont get the concept
what line exactly
$\frac{d}{dx}\frac{d}{dx} = \frac{d^2}{dx^2}$
riemann
it's not more complicated than that
How would you do $$(D^2 + 5D + 6)y = (D+2)(D+3)y$$ in normal notation though
Nathan
$D = \frac{d}{dx}$
riemann
Sure but I mean exanding it in that format and factoring
use this 2-3 times
Nathan
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nope it's not saying this
Whats it saying?
$\left(\frac{d}{dx} + 2\right)\left(\frac{d}{dx}+3\right)y$
Nathan
$\left(\frac{d^{2}y}{dx^{2}}+5\frac{dy}{dx}+6y\right)$
Nathan
Again this is no clearer to me
It didnt pass up the fact that its factoring operators
this?
I'm quite literally factoring operators
correct the derivative is not multiplication
dy/dx * dy/dx is not d^2y/dx^2
this isn't multiplication either
it's an operator acting on an operator
yes I know
but its still having me write it as a binomial that is expanded through multiplication
or factored through division
its abuse
I cant say that $$5^2+5 = (^2 + 1)5$$
it makes no sense
Nathan
nobody writes it like that, but you can if $f(x) = x^2$ and $g(x) = 1 * x$ so that $(f + g)(5) = f(5) + g(5)$
riemann
so this is what a composite?
If you understand that $\frac{d}{dx}$ is an operator, then you can understand that $\frac{d}{dx} + 3$ is also an operator with 3 being the function $g(x) = 3 \cdot x$
riemann
So can I say that $$ D^2 + D + 1 $$ is like $$ D^2(f) + D(f) + f $$ such that we can define it as $$L(f) = D^2(f) + D(f) + f$$ and so $$ L(y(x)) = \frac{d^2y}{dx^2} + \frac{dy}{dx} + y$$
Nathan
Is this right
Yea
Okay...
and so
$$ D^2 + 2D + 1 = (D+1)(D+1)$$ and if we put $$ L(y) = D^2(f) + 2D(f) + (f) $$ then we can say that it is the same as $$(D(f)+f)(D(f)+f)$$?
Nathan
But then that wouldnt make sense
because expanding that gives us a f^2
Also is D synonomous with y in that D(f) is synonymous with y(x) where the (x) is ommitted for simplicity?
this is where you're making a mistake
yeah thats where im confused too
You did (operator 1)(operator2) f = [operator1(f)] * [operator2(f)]
operators do not distribute
operators act in sequence right to left
$\frac{d}{dx}\frac{d}{dx}f \neq \frac{df}{dx}\cdot \frac{df}{dx}$
riemann
idk what you want
I was expressing it as D(f)
that's already fine
Nathan
yea this is fine
so how would I factor that
you did here?
it's just this with f instead of y
but thats like me saying y(x) + x = (y+1)x
yes
but it doesnt like up with what I said
D(f) is not D*f just like y(x) is not yx
instead of 3, your example is just 1
$(f+g)(x)$ is very common notation for $f(x) + g(x)$
riemann
sure you can think of it that way
okay so then if I say
f = x^2
g = x
h = 17
(f+g+h)(x) = x^2+x+17
and
i = d^2/dx^2
j = d/dx
k = 1
(i+j+k)(y) = (D^2 + D + 1)(y) = d^2y/dx^2 + dy/dx + y
But then wouldnt that mean that (f+g+h)(x) is x^2+x+17x if (i+j+k)(y) = d^2y/dx^2 + dy/dx + y
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when is $\log(z^n)=n\log(z)$ false in the complex numbers?
UbuRoi
and can i always use that if i add 2kpi*i?
to solve equations like e^z=-1 for instance
by the definition of the principle branch of the complex logarithm, $\log(re^{i\theta})=\ln(r)+i\theta$, so $\log(r^ne^{ni\theta})=n\ln(r) + in\theta$
Bair
err wait
well i guess it's because z^n might not be in the domain of the complex log
hmm
if z=i and n=2 for example
but also, n theta might overcount the number of rotations
but like i CAN say in a equation like e^z=a then z=lna+2kpi*i, i can 'use' the property with the adjustment of adding the rotations
right?
ok good cause i solved a lot of questions doing that i'd be really sad if i couldn't do that would have to redo all of it
it didn't occur to me the thought until i saw this fake proof
ye, it's because e^ix is not injective
so the mistake here is on step two already yeah?
right, you would need to account for the rotation
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Need help
Is my proof for this limit correct?
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this is too small to read
Here you go the pdf
why can you just l'hopital here but not in the very beginning ?
No I am proving the limit
but also you don't need l'hopital to prove that limit
Think she has to use delta epsilon proof
delta epsilon rarely ever invokes l'hopital
So I just verify the limit holds as I did substitution
Because it’s undefined when I finished substitution
So I had to verify and I didn’t have scratch paper with me so I used LH there
doesn't mean you can use l'hopital
I used LH to verify if my substitution holds not to prove the limit
wat
u Goes to 1 is not defined for the fraction
I have to mention or its indeterminate
I have to show that it tends something then I prove the limit by incorporating delta - epsilon notation
I don’t see where lh is used
Didn’t u jsut do u sub
And then divide numerator and denom by u-1
yea but that doesn't mean you can use l'hopital
if you can use it at all, you can just use it in the very beginning
Proof is different than showing value
when you do delta epsilon you can't use l'hopital from what i know
Ye
you replace the x in the limit by a sequence
and the limit of that sequence should be the value for which the initial limit goes to
using l'hopital at all is supporting your proof
in this case 0
either you can use it in the beginning before the subsitution or you can't use it at all
Yes that’s true kinda…
Yes
You’re right 😭😭
I forgot that part
Then still this limit is actually only solvable by Taylor expansion
you don't need l'hopital to prove this limit is correct
Can I claim the limit to be 1/3 then prove it using delta epsilon?
I actually never used substitution for this kind of proof
i don't think you usually use substitution for proving limits
I couldn’t find articles so
you just say that any sequence for which the limit when k -> infinity = 0 in this case
Yes but k is defined
this is a proof that i did
much simpler than your example but it should be the same method
hopefully my handwriting isn't horrible
Why can’t I use substitution if I may ask? 😭
I think it can, I just need to isolate x at one point
Other wise that limit is probably only provable by Taylor in term of basic method
maybe
you probably are learning more advance proofs than i am
becasue i never had to prove anything like what you have
This limit I think is quite textbook styled
Can you guys verify another solution
Bar{number) is equivalent classes in the modulo
Please help me with this divisibility proof <@&286206848099549185>
This should hold right?
Another question
iono
Is the method rigorous?
try looking up this book: Mathematical Proofs- A Transition to Advanced Mathematics
it might have some answers to your questions
@sinful breach Has your question been resolved?
<@&286206848099549185>
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Guys is this method rigorous help me please ☺️☺️☺️
No please help me 😭😭
It’s actually quite some multiplicative groups and subgroups and some quotient rings
is this Algebra?
I basically rewrote my high school number theory using ring theory
Literally
It looks very fancy (knowledge points are fancy) but it was actually my youth ☺️☺️🤭🤭
damn taking Algebra rn but havent gone through ring theory yet
Ring theory, and majority is group theory though
This is effectively making a high schooler’s homework equivalent to fancy stuff however it is still just basic modular arithmetics
What are you trying to prove?
And it’s fun to see people’s reactions on those notions
10 | n
😂
n is that huge number
I used cyclicity of multiplicative subgroup to prove it
Involving Lagrange theorem
Subgroup
Basic property of rings
Blablabla
,w 3^(14102019)+3 mod 10
Basically for modular arithmetics
Some basic group is actually quite advantageous for these problems especially if you draw Cayley s table
But high schooler group theory 🫠🫠
how would you figure -3
All calculations are based on equivalent classes instead of basic arithmetics
Because of the quotient ring and some basic property of equivalent classes
Basically it is isomorphic to bar(10-3)
Is it very strict to call it isomorphic since it’s really structured at this point
But this is still basic modular arithmetics
Ok I would have explained it more in the proof, at least for me it's not obvious
Yes, this is a very fancy revisit of that definition which is in modular arithmetics
Being decorated within algebraic structure 🤭🤭
One of my mischievous game 🤭
In essence these elements are subgroups so they are cyclic
In a order (cardinality of the group) raised to phi(10)
By Lagrange theorem the order of subgroup must divide the order of the multiplicative group
Though I must say using cyclicity makes things a lot easier
yup
Though notation can be a bit too rigorous
But in high school especially EEA, misuse dan be common especially elementary number theory is quite abstract for high schooler
after that you defined the prime remainder class of Z_10 if I understand correctly?
Yes I think I just took subgroup <3> and analyse its cyclicity
See if it matches 7 for the cycle wise
It doesn’t have to be prime remainder though
Relative prime to 10 otherwise it’s an ideal (principle ideal) instead of a subgroup
You can’t have a principle ideal in multiplicative group
&\mathbb{Z}/10\mathbb{Z} \cong \mathbb{Z}/5\mathbb{Z} \oplus \mathbb{Z}/2\mathbb{Z}$
why won’t my latex work
I actually can make it even more trivialized but it would be even more abstract I can perform ring decomposition: \[ \mathbb{Z}_{10}\mathbb{Z} \conj \mathbb{Z}_{5}\mathbb{Z} \oplus \mathbb{Z}_{2}\mathbb{Z} \] by taking it into direct sum I can just do a lcm to verify it.
lmao
&\mathbb{Z}/10\mathbb{Z} \cong \mathbb{Z}/5\mathbb{Z} \oplus \mathbb{Z}/2\mathbb{Z}$
I got it wrong the directly of dividing/
&\mathbb{Z}/10\mathbb{Z} \cong \mathbb{Z}/5\mathbb{Z} \oplus \mathbb{Z}/2\mathbb{Z}$
Or not
imma behonest after the Let part I dont get the idea nor what you tried to do lol prob due to my lack of algebra knowledge
$ $ not & $
It’s CRT
Omg I am probably day dreaming 😭😭
$\mathbb{Z}/10\mathbb{Z} \cong \mathbb{Z}/5\mathbb{Z} \oplus \mathbb{Z}/2\mathbb{Z}$
meow
Exactly this one
what does that even mean 
Cong means isomorphic
ye
It means that LHS is isomorphic to the right hand side
With smaller n
In which case simpler
What does Z/10Z denote
ok
Fancier way of saying modulo though
hm
Though using decomposition is a thing probably not allowed for basic number theory
are you talking to yourself
No?
a crazy scientist
Well I study those stuff 😭😭
me too, teach me :)
Like undergraduate mathematics
I am only good with real analysis set theory and number theory
you probably have a doctor degree
I am yet to be a master student
i see you like 🥧
shit that's like 10 = 5 + 2 but multiplied
so is e^pi i think a classic question is to compare which larger
then you're rational?
Is not only a classic question it’s a open question
Ignore me I am just daydreaming
Can’t type properly
see talking to yourself hahahaha
Direct sum is only called a sum though
That’s why 2+5 = 10
It means to sum the structure together
If you study math then you know at the second year of math journey
well i do and i do know nothign
If you take algebraic structure then you know probably first off
tbh i dont like that algebra stuff as opposed to analysis or calculus
is that good or bad
how bad
But I am very good with analysis on the other hand 😢
try harder next time
Which is why I felt weird
i see you are prob a theorist
Like failed cal for once while excelled in real analysis
Functional analysis I got a 3.7
Hahaha
Contradiction
what is the stuff you do in real analysis
ok i think i had that
And some axiomatic proofs
I am particularly good with axiomatic proofs they are fun
And delta epsilon too
i sometimes struggle delta epsilon
Ap is like high school right?😭
If you are good with axiomatic proof delta and epsilon would be much easier I think
Tbh real Analysis once you mastered delta epsilon notation everything will be fine I think
what are axiomatic proofs
Like using only most elementary axioms to proof higher stuff
achso
Assume no other theorem than axiom
this reminds me more of linear algebra then
I think real analysis actually relies on axioms a lot
But it makes it less abstract so introduced lemma’s
you are a math student right?
Like lower upper bound which you can construct using axiom of completeness instead of using definition derived lemma
How so?
A bit
a cs student
Let me show you something the. You guess
a little crazy too
[
\begin{tikzcd}[column sep=5em, row sep=5em]
% Diagram F: J -> C
J_1 \arrow[r, "f"] \arrow[d, "g"'] & J_2 \arrow[d, "h"] \
J_3 \arrow[r, "k"'] & J_4
% Limit Cone
& L \arrow[ul, "\pi_1"'] \arrow[dl, "\pi_2"] \arrow[dd, "u"', dashed] & \\
K \arrow[ur, "\phi_1"] \arrow[dr, "\phi_2"'] & & \\
& M \arrow[r, "m"'] & N
\end{tikzcd}
]
]
meow
Compile Error! Click the
reaction for more information.
(You may edit your message to recompile.)
real
\begin{tikzcd}[column sep=5em, row sep=5em]
J_1 \arrow[r, "f"] \arrow[d, "g"'] & J_2 \arrow[d, "h"] \\
J_3 \arrow[r, "k"'] & J_4
& L \arrow[ul, "\pi_1"'] \arrow[dl, "\pi_2"] \arrow[dd, "u"', dashed] & \\
K \arrow[ur, "\phi_1"] \arrow[dr, "\phi_2"'] & & \\
& M \arrow[r, "m"'] & N
\end{tikzcd}
\end{document}
meow
Compile Error! Click the
reaction for more information.
(You may edit your message to recompile.)
\begin{tikzcd}[column sep=5em, row sep=5em]
J_1 \arrow[r, "f"] \arrow[d, "g"'] & J_2 \arrow[d, "h"] \\
J_3 \arrow[r, "k"'] & J_4
& L \arrow[ul, "\pi_1"'] \arrow[dl, "\pi_2"] \arrow[dd, "u"', dashed] & \\
K \arrow[ur, "\phi_1"] \arrow[dr, "\phi_2"'] & & \\
& M \arrow[r, "m"'] & N
\end{tikzcd}
meow
Compile Error! Click the
reaction for more information.
(You may edit your message to recompile.)
if that's LaTeX it doesnt work always with TeXit
\begin{tikzcd}[column sep=4em, row sep=4em]
% Original Diagram F: J -> C
J_1 \arrow[r, "f"] \arrow[d, "g"'] & J_2 \arrow[d, "h"] \
J_3 \arrow[r, "k"'] & J_4
% Limit Object L
\arrow[dashed, bend left=30, from=J_1, to=J_2]
% Additional Cone (K, φ_j)
K \arrow[ur, "\phi_1"] \arrow[dr, "\phi_2"'] \arrow[dd, "u"', dashed] & & \\
& L \arrow[ul, "\pi_1"'] \arrow[dl, "\pi_2"] & \\
M \arrow[r, "m"'] & N &
\end{tikzcd}
meow
Compile Error! Click the
reaction for more information.
(You may edit your message to recompile.)
Just a simple commutative diagram
begin{tikzcd}[column sep=4em, row sep=4em]
J_1 \arrow[r, "f"] \arrow[d, "g"'] & J_2 \arrow[d, "h"] \
J_3 \arrow[r, "k"'] & J_4
\arrow[dashed, bend left=30, from=J_1, to=J_2]
K \arrow[ur, "\phi_1"] \arrow[dr, "\phi_2"'] \arrow[dd, "u"', dashed] & & \
& L \arrow[ul, "\pi_1"'] \arrow[dl, "\pi_2"] & \
M \arrow[r, "m"'] & N &
\end{tikzcd
begin{tikzcd}[column sep=4em, row sep=4em]
J_1 \arrow[r, "f"] \arrow[d, "g"'] & J_2 \arrow[d, "h"] \
J_3 \arrow[r, "k"'] & J_4
\arrow[dashed, bend left=30, from=J_1, to=J_2]
K \arrow[ur, "\phi_1"] \arrow[dr, "\phi_2"'] \arrow[dd, "u"', dashed] & & \
& L \arrow[ul, "\pi_1"'] \arrow[dl, "\pi_2"] & \
M \arrow[r, "m"'] & N &
\end{tikzcd}
\begin{tikzcd}[column sep=4em, row sep=4em]
J_1 \arrow[r, "f"] \arrow[d, "g"'] & J_2 \arrow[d, "h"] \
J_3 \arrow[r, "k"'] & J_4
\arrow[dashed, bend left=30, from=J_1, to=J_2]
K \arrow[ur, "\phi_1"] \arrow[dr, "\phi_2"'] \arrow[dd, "u"', dashed] & & \
& L \arrow[ul, "\pi_1"'] \arrow[dl, "\pi_2"] & \
M \arrow[r, "m"'] & N &
\end{tikzcd}
meow
Compile Error! Click the
reaction for more information.
(You may edit your message to recompile.)
i never understood these diagrams
well i guess i will never enter the room
I hate theories
It’s actually fun
You may guess now
I actually don’t play chess
see i am very awful at taking a guess
I’m studying theoretical mathematics
bruh
The kind of math that has nothing to do with any application
Applied mathematics?
yes
Then after real analysis it’ll much easier
we are supposed to hate each other 🔥
Like for me after real analysis gig goes directly to hell
Yes
Like impurity of mathematics
haha
Impurifies the noble art of mathematics
you started it 
Tbh I’m thinking taking stochastic process though
Make a bit more money before going to spend rest of my life in academia
Calculus? That thing is too dense in computation
Like integration is almost like too complex
why would you spend your life in academia, you wanna become a professor?
Like I remember one I failed is like the surface area of ellipsoid derivation
Which was like bizarre and I failed
exactly what i love
we are the very opposite
Calculus I was expecting one for final which my professor didn’t give us one
Otherwise I would be passing it without resitting
Delta-epsilon proof
that's all you know and need
I can manipulate delta epsilon too a point to direct bound delta by using different inequalities though sometimes
For first year indeed
Delta epsilon is like my strongest weapon
this isn't uncommon fwiw
it saves lifes from what i hear 
what math might that be? 
these days, it seems hard to find any math that has zero application whatsoever 
Like how the fuck can one do 3 iterations on such shit thing x^lnx/(lnx)^x
I am saying theoretical math is nothing without application. The same vice versa, applied math is nothing without some theory behind it.
No direct application
I think calling either "nothing" without the other is too harsh
the world is harsh
x^lnx/(lnx)^x I still can’t do 3 iterations of derivative on this thing
unless you wanna become a professor, try to make some money just knowing pure maths
For pure math it’s actually easier I would say
but vice versa, if you only know to calculate stuff, and apply formulas, you are not much more than an AI
so both go hand in hand
Easy contracted job for faculty especially some rare abstract field these days
People went god digging
Even I wanted to do a bit of stochastic process study
in what year are you currently meow
For that sake
cake
😂
I think there is a distinction to be made between the fields and the practioners 
the fields of study would likely be okay if a single individual didn't go down one or the other
The thing I hate the most during calculus is that one
this is indeed a modular cycle study the stuff only to teach it to others, that will also teach it to others and that cycle never stops
any given person would be in a tougher spot
Lhopital Taylor’s and fancy integral
wdym
hm? why? 
you must hate diff geo then
Yes
as Matplotlib said, becoming a professor as someone who studies math is your typical golden route

HAHA
mhm
i loved diff geo
I hate hearing PDEs
ODEs
yea
ODEs actually saved my life for calculus
i dont think that will work
Otherwise I failed it twice
i got lost when it came to tensors and einstein summation 

but i passed reasonably well
I see
i understood like 70 % maybe but the very deep theory was too much for me to bear
i can try 😄
70% is probably better than what I'm doing rn 
$otimes$
meow
,, \otimes
higher!
$\otimes$
meow
Tensor product
but you are like studying math and me applied math so that might make a difference e.g. be more difficult for you
Remember the property of tensor product? For its determinant
meh, someone who knows DG is someone who knows DG
I don't care if they do applied math or not
I don't discriminate
I'm not a fan of the holier than thou attitude that a few people exhibit sometimes
I remember I had a la exam which had a very weird matrix like no one solved it but I did, find diagonalizer for \begin{bmatrix}
0 & A \
A & 0
\end{bmatrix}
Can you actually guess how did I find the diagonalizer in just one or two steps 🤭🤭🤭🤭
Though I dare not write the process on exam paper bcs I used something in LA 2
So I basically said set the unknowns to be blablabla haha
The things I covered tho were like mathematical basics with introduction to vector analysis and tensor calculation, space curves, areas/surfaces iirc and "special surfaces"
icic, so focused on curves and surfaces
yes
Master?
Btw any of you figured out how to find diagonalizer for that one in a simple way
