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#help-10
obtuse pebbleBOT
brisk matrix
crank, please ignore
obtuse pebbleBOT
Available help channel!
worthy pivot
can anyone tell me how to solve this?
obtuse pebbleBOT
can anyone tell me how to solve this?
worthy pivot
<@&286206848099549185>
reef grotto
can anyone tell me how to solve this?
what have you tried
worthy pivot
worthy pivot
yes because
reef grotto
the notation makes it ambiguous
worthy pivot
it wouldn't be true
reef grotto
but it's probably
worthy pivot
for any x
it can be 2.5
and give another value for example
since here they didn't tell us if x is included in N
reef grotto
$\sum^{E(\frac{1}{|x|})}_{i=1}i$
warm shaleBOT
asa_gao
reef grotto
this is what makes the most sense
worthy pivot
still, i don't know what shall we do with this
reef grotto
$\frac{1}{2} E(\frac{1}{|x|})(E(\frac{1}{|x|})+1)$
warm shaleBOT
asa_gao
worthy pivot
$\lim_0 x^2\frac{1}{2} E(\frac{1}{|x|})(E(\frac{1}{|x|})+1)$
warm shaleBOT
Alitoo
reef grotto
$\lim_{x\to 0} \frac{x^2 \cdot E(\frac{1}{|x|})(E(\frac{1}{|x|})+1)}{2}$
warm shaleBOT
asa_gao
reef grotto
there
worthy pivot
if we suppose that E(X) = X in infinity then it will give use the result 1/2 as in the corretion sheet
but idk if that's a thing
reef grotto
idk if you can just assume E(1/|x|)^2 behaves like 1/x^2
or you have to use the squeeze theorem
reef grotto
if we suppose that E(X) = X in infinity then it will give use the result 1/2 as ...
to be safe i'd use the théorème des gendarmes
reef grotto
E(1/|x|)
worthy pivot
oh in the types of test that im going to pass you just need the answer lol
no demonstration
nothing
obtuse pebbleBOT
obtuse pebbleBOT
Available help channel!
dreamy field
Let (A7, ⊗7)=({1, 2, 3, 4, 5, 6}, ⊗7) is a group. It has two sub groups X and Y. X={1, 3, 6}, Y={2, 3, 5}. What is the order of union of subgroups?
This is a very basic question, won't it's answer be 6?
obtuse pebbleBOT
Let (A7, ⊗7)=({1, 2, 3, 4, 5, 6}, ⊗7) is a group. It has two sub groups X and Y....
wise talon
Let (A7, ⊗7)=({1, 2, 3, 4, 5, 6}, ⊗7) is a group. It has two sub groups X and Y....
are they asking for the minimum subgroup generated by the union of the subgroups?
because if you take that union it's not actually a subgroup
dreamy field
are they asking for the minimum subgroup generated by the union of the subgroups...
no it is asking for the number of elements in the union of its subgroup and now i understood it because 3 was in both the subgroups and i counted it twice instead of 1 so the answer is 5
.close
obtuse pebbleBOT
wise talon
cool
obtuse pebbleBOT
Available help channel!
next ravine
obtuse pebbleBOT
next ravine
Which is the correct answer ?
Thanks in advance!!
obtuse pebbleBOT
Which is the correct answer ?
Thanks in advance!!
As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.
spare orchid
What have you tried
obtuse pebbleBOT
@next ravine Has your question been resolved?
obtuse pebbleBOT
obtuse pebbleBOT
Available help channel!
hollow river
obtuse pebbleBOT
hollow river
Is this right
candid yarrow
you can check by differentiating the result you got
obtuse pebbleBOT
obtuse pebbleBOT
Available help channel!
vestal hull
direct image
obtuse pebbleBOT
direct image
vestal hull
haven't solved it yet
obtuse pebbleBOT
@vestal hull Has your question been resolved?
obtuse pebbleBOT
@vestal hull Has your question been resolved?
zenith raft
this looks like it assumes f is invertible (which it should not)
also you should use words to introduce the variables you are talking about
like, the very first thing you write could be let $y\in f(P)-f(Q)$
warm shaleBOT
slayyla
zenith raft
the “let” goes a long way in communicating the proof
but it does not make sense to introduce the thing in f(P) - f(Q) as f(x)
there is no x there yet
vestal hull
yep, I was just doing an sketch to write the proof properly afterwards
zenith raft
this looks like it assumes f is invertible (which it should not)
i didn’t emphasize it but this was the more important issue
i don’t wanna comment more on that unless you wanna rewrite what you have with more detail tho
vestal hull
I did it using this definition
y∈f_(A) → x∈A
Cause using the cuantificators looked complicated
zenith raft
bc it’s not totally clear what you were saying
vestal hull
i don’t wanna comment more on that unless you wanna rewrite what you have with m...
yes, I want
I don't understand how I should do it at all
zenith raft
I did it using this definition
```
y∈f_(A) → x∈A
```
Cause using the cuantificat...
this doesn’t make sense… what’s the definition as you learned it?
vestal hull
y∈f_(A) → (∃x∈A)(f(x) = y) (direct image of a set)
zenith raft
I did it using this definition
```
y∈f_(A) → x∈A
```
Cause using the cuantificat...
you probably mean x to be the one with the property that f(x) = y? but there could be more than one (and one of them could be outside A)
zenith raft
y∈f_(A) → (∃x∈A)(f(x) = y) (direct image of a set)
typo here, f(x) = y
vestal hull
you probably mean x to be the one with the property that f(x) = y? but there cou...
I don't get you
zenith raft
I did it using this definition
```
y∈f_(A) → x∈A
```
Cause using the cuantificat...
what is x here?
vestal hull
an element in the domain A
zenith raft
well it looks like you’re claiming that’s a property it has, but what is it?
lone dirge
Hello
I'm joining @zenith raft in the request to clearly repeat what you've got so far, including the question please
vestal hull
Let f: A→B a function such that P, Q ⊂ A. Prove that f_(P) - f_(Q) ⊂ f_(P-Q)
This is an "sketch" I did
f(x)∈f_(P)-f_(Q)
f(x)∈f_(P) ∧ f(x)∉f_(Q)
x∈P ∧ x∉Q
x∈(P-Q)
f(x)∈f_(P-Q)
Where $f(P) = {f(x) : x \in P}$?
warm shaleBOT
cain0196
agile thorn
f(x) being in P does not mean that x is in P
vestal hull
Oh
vestal hull
f(x) being in P does not mean that x is in P
in f_(P)*
zenith raft
f(x) being in P does not mean that x is in P
there’s no claim of f(x) in P anywhere
agile thorn
there’s no claim of f(x) in P anywhere
but he wrote f(x)∈f_(P) ∧ f(x)∉f_(Q) => x∈P ∧ x∉Q
lone dirge
```
f(x)∈f_(P)-f_(Q)
f(x)∈f_(P) ∧ f(x)∉f_(Q)
x∈P ∧ x∉Q
x∈(P-Q)
f(x)∈f_(P-Q)
```
yes, f(P) not P
But that's true
@lag is right
And that's exactly what @zenith raft said before
About f being invertible
So, your claim is false @vestal hull .
agile thorn
however you can just take x to be in P from the very start
zenith raft
f(x) being in P does not mean that x is in P
umm i think lag still has a typo here
vestal hull
So, your claim is false <@803752118014967839> .
Ahh
That's why I thought it was rare
zenith raft
then yes that’s like what i was trying to say but there’s still an issue never even introducing an x
agile thorn
i think it's better to start with a y in f(P)-f(Q) rather than f(x)
vestal hull
however you can just take x to be in P from the very start
thats what lag is trying to say here, right?
vestal hull
i think it's better to start with a y in f(P)-f(Q) rather than f(x)
yes
agile thorn
i think it's better to start with a y in f(P)-f(Q) rather than f(x)
then introduce x
y=f(x) where x is in P
now you just gotta prove that x is not in Q
agile thorn
well after introducing the y in f(P)-f(Q), yes
vestal hull
```
f(x)∈f_(P)-f_(Q)
f(x)∈f_(P) ∧ f(x)∉f_(Q)
x∈P ∧ x∉Q
x∈(P-Q)
f(x)∈f_(P-Q)
```
supposing I already introduced y∈f(P)-f(Q) and x∈P, is the process that I did here wrong?
vestal hull
can I just use y instead without making it clear that f(x) = y?
agile thorn
don't think so
vestal hull
Thanks
obtuse pebbleBOT
@vestal hull Has your question been resolved?
subtle whale
LINEAR ALGEBRA / MATRIX
obtuse pebbleBOT
LINEAR ALGEBRA / MATRIX
subtle whale
BOOK SOLUTION ^
MY SOLUTION ^
I am not done yet with mine, but if I were to continue, my RREF would look different. I thought matrices in RREF are supposed to look the same
I took different steps than the book, but I do not think I have done anything illegal, so what went wrong?
If I continued, based on the rules of RREF, my final form would look like this, which is completely different than the book's... any help please?
nocturne minnow
You messed up the math for R2 in your second step
If you did -3 * R1, that's -3 * (1 -3 4 -3 | 2) = (-3 9 -12 9 | -6) add that to R2 which is (3 -7 8 -5 | 8) and -12 + 8 is not equal to positive 4, like you wrote
obtuse pebbleBOT
@subtle whale Has your question been resolved?
nocturne minnow
Yes
subtle whale
If you did -3 * R1, that's -3 * (1 -3 4 -3 | 2) = (-3 9 -12 9 | -6) add that to ...
Oh R2 is supposed to be
[0 2 -4 14 2] then right?
nocturne minnow
No
You're adding (-3 9 -12 9 | -6) and (3 -7 8 -5 | 8) together
So why are you getting 14?
What's 9 + -5?
nocturne minnow
Yes
subtle whale
Oh okay thanks a lot
nocturne minnow
Because of your math error there, it carried through and the rest was wrong
vocal pond
obtuse pebbleBOT
tardy epoch
Can someone give me a full walk through on number 1 please
not really how this server works
vocal pond
I have the answer which is
tardy epoch
just do some of your own work and show
vocal pond
Like I have the steps I just don’t understand what my teacher did to get that answer
Like that very first step? What is going on
tardy epoch
which step do you get lost first
do you know partial derivatives
and have you taken multivariable calculus?
vocal pond
I’ve only taken Calc 2
That’s the highest Calc I’ve taken
What is that backwards 6 thing?
See where it says E=-6V/6y
Why are those 6s backwards
vocal pond
Bruh
obtuse pebbleBOT
@vocal pond Has your question been resolved?
obtuse pebbleBOT
obtuse pebbleBOT
Available help channel!
tranquil wren
obtuse pebbleBOT
normal oracle
What have you tried?
tranquil wren
I tried taking square root of left side and trying to manipulate the numerator, but no success. I tried factoring out sqrt(2)/2 from left side, but also couldn't get any progress. I tried a little to start with the right side instead, but also couldn't get anywhere
normal oracle
Get rid off denominator first.
tranquil wren
ok
tranquil wren
I tried factoring out sqrt(2)/4, eventually produced (sqrt(3) + 1)/sqrt(8), but then went in a circle and came to the original expression
normal oracle
First, get rid off denominator, after that, square both sides.
And show what you get after squaring.
tranquil wren
ok
wait, if I squared both sides, wouldn't that be assuming that which I'm trying to prove? (That left side equals right side)
normal oracle
That was your first statement.
Your previous statement is proving already, but you want to compare now, not to prove.
tranquil wren
ok
normal oracle
So, after squaring do you know how to compare them?
tranquil wren
honestly no, I'm not sure 😅
normal oracle
write your progress after squaring
tranquil wren
I can continue and get 2sqrt(3) = 2sqrt(3), but I don't know how it helps me write a proof
tranquil wren
normal oracle
yes but you wrote in a way it's assuming I know what you're talking about
you can leave like this
but you already compared both and realized they are the same, which was what you wanted to
clear condor
in the first row you multiplied by 4 but you wrote 4/4 :/
normal oracle
lol, that's true, i didn't even check that one cause I assumed he did correctly
because the rest was correct
tranquil wren
oops yeah that's wrong
tranquil wren
but I'm still confused, this didn't prove what i wanted to show
normal oracle
whatever, u understand what i mean
yes it did
you compared 2 things and became to the conclusion they are the same
that's a proof
other thing is, you didn't get to the second expression from the first without the help of the comparisson
but that doesn't make any sense because
there are infinit fractions that are equivalent
also, it's better you use the word equivalent and not same or equal
tranquil wren
But in order to multiply both sides by a number, aren't I assuming the two quantities are equal? That's exactly what I want to prove
normal oracle
proofs sometimes begin with
let's assume this and that are equal
after realizing your skilling maths to those things, you can come to the conclusion it's false or true
assumption is a tool you need in math
candid yarrow
When you write a proof, you're supposed to start with one side of the equation and transform it into the other side
tranquil wren
I have not yet succeeded at figuring out the algebra on this one yet
normal oracle
you can get this step only with the LHS
after that, you factor
sqrroot, and divide by 4
now you get your RHS
tranquil wren
I'm trying now to start with only the left hand side and transform it
normal oracle
You can also do that
clear condor
do everything backwards?
normal oracle
the easiest way is to square and squareroot
it's 3 steps
the equal red is incorrect
i'm assuming the future square-root
my painting skills are not the best, but i think u know which one is inside the sqroot
obtuse pebbleBOT
@tranquil wren Has your question been resolved?
tranquil wren
.close
violet hull
Can someone please help me solve this question?
There is a total of 15 large and small plates. 6 apples are placed on the large plates and 4 apples are the small plates. If there are 80 apples in total, find the number of large plates?
A. 5
B. 7
C. 10
D. 12
obtuse pebbleBOT
Can someone please help me solve this question?
obtuse pebbleBOT
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'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above
step 2
timid silo
show me your work
violet hull
6L + 4S = 80
timid silo
you've a system of equations
violet hull
could you explain further?
timid silo
L + S = 15
6L + 4S = 80
violet hull
yes
timid silo
You can solve this with elimination or substitution
violet hull
so that goes on to (15 - S) + (15 - L) = 80?
hmm
so uh 4(15 -L) + L = 80
4(15) - 2L = 80
violet hull
I would recommend doing S = 15 - L, though.
am I doing it right?
hold on
I mucked up
violet hull
L + S = 15
6L + 4S = 80
so uh we can substitute
L = 15 - S
and S = 15 - L
We can replace 6L + 4S = 80 with 6(15 - S) + 4S = 80?
90 - 6S + 4S = 80
hmm
<@&286206848099549185>
nvm
I found it
M is 5
6(15 - m) + 4m = 80
90 - 6m + 4m = 80
90 - 2m = 80
subtract 90 from both sides
-2m = -10
divide both sides by -2
n = 5
.close
restive pilot
for ring theory and modules: Does Z/1Z even exist?
obtuse pebbleBOT
for ring theory and modules: Does Z/1Z even exist?
brisk matrix
you can think of {0} as that ring
restive pilot
you can think of {0} as that ring
what would Z/0Z be then?
cause regarding to my lecture that exists as well
brisk matrix
Z
brisk matrix
what is 0Z
restive pilot
oh right I take no elements out
restive pilot
what is 0Z
0
brisk matrix
then yeah it’s just Z / {0} = Z
restive pilot
then yeah it’s just Z / {0} = Z
is the 0 then actually taken out or not?
brisk matrix
in the case of Z/nZ
restive pilot
just know what the result looks like, for postive integers it logical kidna at least
brisk matrix
yeah it’s just the integers mod n
if you know the definition of quotient rings it’s just that
i don’t have too much visual intuition here either, just try to understand the notion of things being “equal” if their difference is in the ring being modded out
obtuse pebbleBOT
Available help channel!
lethal ridge
obtuse pebbleBOT
obsidian isle
what you'll first want to do is start with the bigger numbers and work your way down, in terms of the prime factorizations
slender sundial
<@709655283953041408>
sorry my msg got sent late
lethal ridge
Nah all g gl
obsidian isle
so you could start with:
10 = 2x5, so cross 2 and 5 off your list, then
9 = 3x3, so cross one 3 off your list, then...
lethal ridge
Sorry im not quite sure I understand
obsidian isle
can you write down the prime factorization of every number from 1 through 10? shouldn't take too long
runic void
1 isn’t a prime tho
runic void
We typically don’t consider 1 as being a prime because it will take away the uniqueness in fundamental theorem of arithmetic
lethal ridge
Oh shoot mb
obsidian isle
1 is not a prime either, don't list 1
lethal ridge
Ok
obsidian isle
also 8 is not 2x4
lethal ridge
Yeah im changing tn
obsidian isle
we're looking for PRIME factorizations
runic void
Write them as product of powers of prime
obsidian isle
doc seems like he can handle this, gl
obsidian isle
alr
lethal ridge
This good now?
lethal ridge
Omg istg my brain is dead
obsidian isle
Lmao!
lethal ridge
2^3
obsidian isle
yes
so now how many unique prime factors would you need in the answer?
(this is the same question as: how many prime numbers are less than or equal to 10?)
lethal ridge
So like 2,3,5,7? Sorry the phrasing is kinda throwing me off
Like how many
Or like which ones
Im so confused sorry
lethal ridge
Ok
obsidian isle
read up on it then return
lethal ridge
Ok
obsidian isle
if you're unaware of it then you probably don't even really understand what a prime number is
obsidian isle
every number has a unique factorization as a product of primes
lethal ridge
Yes
obsidian isle
yes. so suppose we take the product of all numbers from 1 - 10. let's call this number x.
then the prime factors of x should be exactly every prime number less than or equal to 10
make sense?
lethal ridge
Ok so all numbers multiplied equals x and the prime numbers of x are the prime numbers 1,2,3,5,7?
obsidian isle
so we can write:
$x = 2^a \ 3^b \ 5^c \ 7^d$
where a,b,c,d are yet to be determined
lethal ridge
Ah I c
warm shaleBOT
FriendTimes
obsidian isle
does it make sense how i got this?
lethal ridge
Yes
lethal ridge
Because we do not know the prime factors of said number we just know the range of numbers?
obsidian isle
not quite
lethal ridge
No ok
obsidian isle
consider that just because a number is large, it doesn't necessarily mean it has a lot of prime factors
can you give me a REALLY large number that has only one prime factor?
lethal ridge
Shoot mb
lethal ridge
God im awful at this um
obsidian isle
yeah. read up on FTA
lethal ridge
Yeah ok
obsidian isle
yes
lethal ridge
Ok
obsidian isle
now can you tell me why we need a,b,c,d?
(hint: what is the difference between the number 3 and the number 27? both have 3 as their only prime factor)
lethal ridge
Dude at this point idk 😭 im sorry I haven’t done this stuff in agea
obsidian isle
what's the prime factorization of 3?
obsidian isle
what's the prime factorization of 27?
yes
so the answer lies in the fact that just because we know the prime factors of a number, doesn't mean we know HOW MANY COPIES of each prime factor are needed
lethal ridge
Ok
obsidian isle
that's why we need a,b,c,d
lethal ridge
Yeah ok
obsidian isle
great.
so we have:
10 = 2x5
9 = 3x3
8 = 2x2x2
7 = 7
6 = 2x3
5 = 5
4 = 2x2
3 = 3
2 = 2
lethal ridge
Yeah?
obsidian isle
now all you need to do is take the LCM of these numbers
obsidian isle
and to do that, we just take the largest power of each prime factor present in any individual number
for instance, 8 = 2^3, and 8 contains the largest power of 2. so we know that the power of 2 in the prime factorization of x will be 3, so a = 3
so now we have:
$x = 2^3 \ 3^b \ 5^c \ 7^d$
warm shaleBOT
FriendTimes
obsidian isle
sure
lethal ridge
Ok so 2^3 3^2 5 7
lethal ridge
Now do I solve?
obsidian isle
2^3 = ?
lethal ridge
8
obsidian isle
3^2 = ?
lethal ridge
9
obsidian isle
8 x 9 = ?
lethal ridge
360?
lethal ridge
2520
obsidian isle
there you go
obtuse pebbleBOT
Available help channel!
thorny forge
obtuse pebbleBOT
royal basin
no, because then you subject yourself to the question of "why is (k-1)! always divisible by r! (k-r)! ?"
and in effect you've proved nothing
royal basin
that last one
you have not shown why those fractions are integers
also bad notation and bad phrasing and bad variable naming
royal basin
what series
thorny forge
well it ends with 3, 2, 1
royal basin
we talked about this exact problem yesterday
thorny forge
i know
royal basin
i almost walked you through the entire solution save a few steps
AND YOU CHOSE TO THROW IT ALL AWAY!
royal basin
what you've written is unsalvageable bullshit.
thorny forge
well thats crude
royal basin
the key idea, to reiterate from yesterday, is this:
if you multiply together several numbers each of which is strictly less than a prime number p, then the product won't be divisible by p.
there is no real way around making use of that fact eventually.
thorny forge
doesnt my thing kinda show that
with the
(k k-1)
its strictly less than k which is to say p
since all the other terms are smaller than (k k-1)
royal basin
no your thing shows absolutely nothing
no all the other terms arent smaller than C(k, k-1)
thorny forge
.close
obtuse pebbleBOT
polar fossil
you ever have a tangled mess of cables, like christmas lights? gotta untangle it one step at a time
rich plume
Try to write complete LHS as a power of x
Now use the exponent rule
(x^a)*(x^b)=x^(a+b)
oh yes
also enclose 7-k in brackets
Yes
now use the exp rule
x is just x^1
polar fossil
sure, that'll work
rich plume
Yes
polar fossil
perf
rich plume
Yes
polar fossil
should be, at least in the real numbers
we didn't square anything or take a square root
royal basin
always count on WA to overshare...
but yeah 14/3 is correct
your book is wrong
no
again WA just likes oversharing and didn't give the fraction form at the start fsr
polar fossil
i think yours is right, $(\sqrt2)^{14/3} = (2^{1/2})^{14/3} = 2^{7/3} = \sqrt[3]{2^7} = 2^2\sqrt[3]{2}$
rocky goblet
wolframalpha will just give you loads of stuff about whatever input you give it
urban patrol
maybe they meant 7/3 as an intermediate answer?
warm shaleBOT
Hayley
urban patrol
like not the actual answer of the problem
polar fossil
your method was sound and your work was good
urban patrol
but like a step you need to get to the answer
but idk
just trying to figure out how they got 7/3
sure
obtuse pebbleBOT
alpine raven
rich plume
That's the thing.
You need a quadratic to get 2 values of x
Think of values of m which would result in a quadratic
obtuse pebbleBOT
@fierce horizon Has your question been resolved?
obtuse pebbleBOT
@fierce horizon Has your question been resolved?
obtuse pebbleBOT
obtuse pebbleBOT
Available help channel!
thorny forge
How many different "words" can be formed using the letters GOOGLE?
obtuse pebbleBOT
How many different "words" can be formed using the letters GOOGLE?
runic void
as there are 6 letters, any possible re-arrangement will too have the same
for first letter, i have choice of all 6 letters
second choice, is 5 letters
third choice is 4 letters and so on
using multiplication theorem of counting, there are $65432*1$ or 6! ways
candid yarrow
using multiplication theorem of counting, there are $6*5*4*3*2*1$ or 6! ways
it is less than that
whole dock
Cause the two g's and two o's
candid yarrow
!nosols
obtuse pebbleBOT
As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.
@thorny forge Has your question been resolved?
candid yarrow
No, how did you get that
candid yarrow
how did you get that
thorny forge
how did you get that
6 letters in google, and 2 repeated letters twice
candid yarrow
Yes
candid yarrow
sorry i meant 6!/(2! * 2!)
this is right
thorny forge
what if it says
How many different five letter "words" can be formed using the letters GOOGLE?
is the correct answer 6p5/(2!x2!)
candid yarrow
how did you get that
thorny forge
the same way
except only 5 letters
instead of 6
so we pick a 5 letter word from 6 where the order matters also known as 6p5
we overcount by a factor of 2! 2!
so we divide by that
candid yarrow
it is possible to check your answer by counting a different way
thorny forge
wdym
candid yarrow
count how many five letter words you get from GOOGL, GOOGE, GOOLE, and so on
and add them all up
candid yarrow
no, there are only 6 combinations
candid yarrow
googl: 5!/2!
googe: 5!/2!
and so on
lone dirge
Let's call the positions on the circle 1,2,...,8
Let's say the first jester gets position j
So the second is near him if he gets position j+1,j-1 (mod 8)
thorny forge
only mod i know is modulus argument
lone dirge
yes... forget it
So the answer is 5/7
Can you show your work?
I'll tell you where you get it wrong
lone dirge
Try explaining your thought process
thorny forge
so
circle@theorem
if there are 8 people in a circle
there are 7! ways of arranging them
lone dirge
True
thorny forge
and then i divided by repeats
the J thing is the arrangements with two jesters next to teach other
each
lone dirge
How many are there?
thorny forge
two
lone dirge
Arrangments with the jesters together
thorny forge
grouping them gives us 7 elements in a circle
lone dirge
Yes
thorny forge
so that’s how i get that
dividing by the repeats
and multiplying by 2 for arrangements
ohh
i don’t multiply by two
wait do we assume them indistinguishable
thorny forge
so is that the mistake
thorny forge
dividing by the repeats
then the arrangements of the jesters not being next to each other is the total-them being next to each other
so i do that
then you put those arrangements ober total arrangements for the answer
lone dirge
So check if you get the right answer now
lone dirge
YW
thorny forge
idk why i divided that was so weirrd
i should also have divided by jesters
if i divided by q and k
lone dirge
yes and not multiply by 2
thorny forge
no thats correct
spark verge
Hi, I'm back with another tiling question:
The game Tetris is played with five different shapes — the five shapes that can be obtained by piecing together four unit squares. We also allow these pieces to be “flipped over.”
(a) Is it possible to perfectly cover a 4 × 5 chessboard using each of these shapes exactly once? Prove that it is impossible, or show by example that it is possible.
(b) Is it possible to perfectly cover an 8 × 5 chessboard using each of these shapes exactly twice? Prove that it is impossible, or show by example that it is possible.
I have solved (a) by noticing that a 4 × 5 chessboard has the same amount of white and black squares and only the T piece never covers the same amount of black and white squares.
This fact doesn't show that (b) is impossible, since you can include two complementary T pieces, which take up the same amount black and white squares. I can't find an example for (b) which suggests that it might impossible, or I'm just bad at Tetris.
I'd appreciate a someone gave me a hint or would be interested in solving this with me.
obtuse pebbleBOT
Hi, I'm back with another tiling question:
> The game Tetris is played with fiv...
candid yarrow
what did you get for the corners
lone dirge
YW
spark verge
.close
obtuse pebbleBOT
https://www.tumblr.com/mathlyblog/721922809649119232?source=share
timid silo
Hey guys. I have problem solving this question
candid yarrow
Hey guys. I have problem solving this question
!show
obtuse pebbleBOT
Hey guys. I have problem solving this question
Show your work, and if possible, explain where you are stuck.
candid yarrow
I posted the link
Show your work, and if possible, explain where you are stuck.
wooden cipher
Please show it here, we are always suspicious of links
timid silo
Please show it here, we are always suspicious of links
I can't upload it on discord it fails for some reason. i posted the tumblr link
I'm not gonna hack anyone LOL
candid yarrow
Hey guys. I have problem solving this question
Show your work, and if possible, explain where you are stuck.
candid yarrow
I can't upload it on discord it fails for some reason. i posted the tumblr link
I can see the picture, what do you need help with? where are you stuck?
timid silo
I can see the picture, what do you need help with? where are you stuck?
I can't solve it
I have no idea how to slove it
candid yarrow
Consider the denominator
timid silo
Consider the denominator
I don't undrestand how I am sopposed to find the vertical asymptote. Can you solve it for me?
candid yarrow
I don't undrestand how I am sopposed to find the vertical asymptote. Can you sol...
!nosols
obtuse pebbleBOT
I don't undrestand how I am sopposed to find the vertical asymptote. Can you sol...
As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.
timid silo
As a helper, please do not give out answers that could be copied as a homework s...
Its not a homework it is summer break ffs.
candid yarrow
I don't undrestand how I am sopposed to find the vertical asymptote. Can you sol...
The vertical asymptote will be at the roots of the denominator
so you can use the intermediate value theorem to find an interval that contains a root of the denominator
then that will be a vertical asymptote
candid yarrow
.close
obtuse pebbleBOT
obtuse pebbleBOT
Available help channel!
silent meadow
Hi,
My problem is I have n 3D rectangular of any size and i want to place them in the most optimal position in a 3D space.
For exemple I have 3 same sized rectangular of 6 by 8 by 5. (we don't care about the unit here). We have a lot of possibility to place them but whet is the most optimal possibility.
I don't really know how to describe the most optimal. I was thinking about multiple factor like the area + standard deviation of dimension (like 4 by 4 is better than 2 by 8).
The desired result is just the final lenght, width, height, i don't really care about the position.
obtuse pebbleBOT
Hi,
My problem is I have n 3D rectangular of any size and i want to place them ...
rotund lark
Hello
silent meadow
Hi !
polar fossil
are you trying to optimize for length + width + height, or total surface area, or what?
daring rock
What do you mean by optimal here? As long as you pack the boxes into a rectangular prism, there won't be any wasted space
silent meadow
To start i'm trying to optimize for length + width + height to get the smaller standard deviation
And in a second time, we have a space of 200 and the problem is to know if every box of any size can fit in it
(I'm trying to code all of that)
obtuse pebbleBOT
@silent meadow Has your question been resolved?
delicate drum
guys I've taken in school a theorem for limits that says the limit of (x^m -c^m)/(x^n-c^n) when x approaches c is equal to (m/n) * c^(m-n)
obtuse pebbleBOT
guys I've taken in school a theorem for limits that says the limit of (x^m -c^m)...
delicate drum
sry I don't know how to use the bots
so this theorem I don't see it in textbooks
though I think it's very important
Does anyone know what is the proof of this theorem?
faint juniper
Can anyone explain what does Laplace transform mean and what does it actually do?
delicate drum
Can anyone explain what does Laplace transform mean and what does it actually do...
bro go to a free channel
no one will help you here
faint juniper
I am new how to use this
next reef
I am new how to use this
Read #❓how-to-get-help
obtuse pebbleBOT
delicate drum
.close
obtuse pebbleBOT
Available help channel!
sweet sparrow
How do you factor a quadric if you have the solutions
obtuse pebbleBOT
How do you factor a quadric if you have the solutions
kind hawk
(x-x1)(x-x2)
fiery fox
obtuse pebbleBOT
fiery fox
<@&286206848099549185>
timid silo
!15m
obtuse pebbleBOT
Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.
@fiery fox Has your question been resolved?
obtuse pebbleBOT
Available help channel!
obtuse pebbleBOT
royal basin
are you allowed comparison?
are you perhaps allowed stirling's approximation formula 
though given the ban on both root and ratio there's not much hope for that
placid smelt
are you allowed comparison?
Yes but after thinking about it, i'm not sure how to find the sequence of inequalities to end up with a simple series I can show converges.
placid smelt
are you perhaps allowed stirling's approximation formula <:thonk:406575732563902...
I can not use stirilings, don't know what that is...
royal basin
ok well first off sin(pi/2 (2n+1)) = (-1)^n just to Acknowledge That™️
so if we maybe wanna go for absolute convergence to see if we get it
we can ignore that as it becomes 1 upon taking absolute values
placid smelt
ok well first off sin(pi/2 (2n+1)) = (-1)^n just to Acknowledge That™️
How so?
Is this because sin((π/2)(2n + 1)) = sin(πn + π/2) = cos(πn) = (-1)^n. ?
warm shaleBOT
Ann
royal basin
$\frac{n!}{n^n} = \frac{1}{n} \cdot \frac{2}{n} \cdot \frac{3}{n} \cdots \frac{n-1}{n} \cdot \frac{n}{n}$
warm shaleBOT
Ann
royal basin
do you follow?
placid smelt
Yes and that is <= (1/n) ⋅ (2/n) ⋅ 1 ⋅ 1 ⋅1 ⋅ 1 = 2/n^2
royal basin
great see
don't use the letter x for multiplication
but you have just come up with a working comparison
placid smelt
And we know 2/n^2 converges by the p-test as (p = 2 > 1)
royal basin
yeah
placid smelt
but how does the sin((π/2)(2n + 1)) come into play?
royal basin
ok well first off sin(pi/2 (2n+1)) = (-1)^n just to Acknowledge That™️
so if we maybe wanna go for absolute convergence to see if we get it
we can ignore that [ (-1)^n ] as it becomes 1 upon taking absolute values
placid smelt
cuz i'm pretty sure I can't use absolute convergence
royal basin
you cannot refer to the concept thereof at all?
placid smelt
I can only use the integral test, comparison test, alternating series and geometric series
royal basin
so you've been expressly told that everything not listed here is FORBIDDEN under some grave penalty
yes?
royal basin
nyeh
this is an alternating series thanks to that (-1)^n
so you could prove n!/n^n -> 0
and be golden
royal basin
so you could prove n!/n^n -> 0
and for that youve already got something to squeeze it with
warm shaleBOT
Ann
royal basin
spose you also actually need to show it's monotone, but god knows it's impossible to prove that without really skirting the boundary of this toolset restriction you've got placed on you (ratio test banned, and computing the ratio of adjacent terms really doesnt look all that rule-abiding!)
placid smelt
wait... is this true:
(n!/n^n) sin(pi/2 (2n+1)) <= (n!/n^n)?
placid smelt
true technically
How do we know this is true?
royal basin
(-1)^n ≤ 1.
also i'd prefer if you broke free from your hesitance to acknowledge that sin(pi/2 (2n+1)) is (-1)^n ...
placid smelt
hmm ye
If i'm using the alternating series test
I would have my series as:
n!/n^n (-1)^n
and then my a_n would be n!/n^n
and I have to show that the limit as n -> inf of n!/n^n is 0?
but how would I show that the limit as n -> inf of n!/n^n is 0?
hybrid gull
but how would I show that the limit as n -> inf of n!/n^n is 0?
Idk how rigorous you have to be, but you can rewrite the numerator as:
n(n-1)(n-2) ... 2 * 1
Denom is n * n * n ... * n * n
As n -> infinity, you get 1 * (something slightly less than 1) * (something slightly less than 1) * .... (something just above 0) * (something just above 0) * (something just above 0)
royal basin
and I have to show that the limit as n -> inf of n!/n^n is 0?
yay you understood what i was saying all along
placid smelt
> and I have to show that the limit as n -> inf of n!/n^n is 0?
yay you understo...
Do u know how I can show that limit
royal basin
you already know that n!/n^n ≤ 2/n^2
like, you're not using this inequality for a comparison test, but we cannot really blind ourselves to it, can we?
hell, even n!/n^n ≤ 1/n will do alright
placid smelt
you already know that n!/n^n ≤ 2/n^2
but how can I use this when finding the limit?
royal basin
squeeze theorem
placid smelt
but squeeze theorem would require me to have something to the left of the inequality too which I don't know
royal basin
0
obtuse pebbleBOT
@placid smelt Has your question been resolved?
royal basin
what's unclear?
tardy epoch
but squeeze theorem would require me to have something to the left of the inequa...
Usual first thing to try is the negative of the upper bounding sequence
royal basin
or, yknow, you could acknowledge that n!/n^n is positive.
placid smelt
what's unclear?
oh nothing, forgot to close sorry. ty for the help!
royal basin
why react with a ❌ then lmfao
placid smelt
.close
obtuse pebbleBOT
placid smelt
why react with a ❌ then lmfao
misclick
obtuse pebbleBOT
Available help channel!
crimson vapor
If I have a 9x9 grid, where each 'pixel' of the grid can have 2 values (1, 0) then that grid has 2^81 possible combinations by my understanding.
I want to build a subset of that where only exactly 25 pixels within that grid have a value of 1, and every pixel must be connected to every other pixel (no disconnected or disjointed pixels or groups of pixels, and they only connect via their 4 sides, not diagonally)
Is there a meaningful way to define this mathematically or programmatically, to either generate this, check a randomly generated field of 81 to see if it fits this criteria, or even calculate the total number of possibilities which fit in this subset?
I tried searching for ways to mathematically define a "group" vs "disconnected" or a way to count the number of pixels in a group, because then I just screen for groups of 25 and total value of 25, but that would be slow.
Where would I even begin to look to solve this problem?
One thought I had is, possibly if you pick a random point on the grid, change its value to 1, then randomly select one of the 4 directions and change its value to 1... etc till the total fields value is 25 then that could get you close to defining it programatically, but how do you use this to generate EVERY possible combination? How do you refine it to deterministically resolve issues like getting trapped, overstepping its self, bumping into walls, etc. It's an interesting start that could yield results but I don't think there is a way forwards with it that is 'complete'
Thank you.
obtuse pebbleBOT
If I have a 9x9 grid, where each 'pixel' of the grid can have 2 values (1, 0) th...
thick fog
It would be easier to iteratively remove elements until you have a valid configuration than to iteratively construct it without accounting for things like a dead end path
crimson vapor
like, pick a random edge pixel, remove it, repeat until the total value is 25?
thick fog
yes
crimson vapor
Hmm, okay, how do you define an "outermost" pixel? This would certainly make many types of formations unsafe to generate (what if an edge pixel wraps around and forms a loop, cutting off several '1' pixels?) but it's closer than what I have
thick fog
You don't. You just remove a random one and check that you still have a connected component of at least 25 pixels
If you remove a bad pixel you can white list that pixel and never select it again
crimson vapor
okay how do you mathematically define a "connected component"s size
obtuse pebbleBOT
@crimson vapor Has your question been resolved?
crimson vapor
<@&286206848099549185>
obtuse pebbleBOT
@crimson vapor Has your question been resolved?
crimson vapor
what do I do if it's been over an hour since I posted the question and no one responded to the helpers tag
lone dirge
What exactly is your goal
Property testing?
Pick a problem
You have mentioned many problems there
Say, given a grid, does it have the property...
How fast do you want to do this?
Would you like to try to think about the combinatiorial question first?
How many of these are there?
Just choose one question 🙂
crimson vapor
the goal is to deterministically/pseudorandomly generate clusters of 25 connected squares on a grid with no unconnected blocks, and no more or less squares. it doesn't need to be stupendously fast, at absolute maximum, given a seed for random number generators (which comes to the same result every time from a given seed) it should take 2-3 minutes to generate the layout programmatically, though less is obviously better.
the "how many are there" question is definitely an aside. the primary question is how
crimson vapor
hmmmm - like an agent which steps in a random direction, has a counter of 25, and whenever it changes a field on the grid from 0 to 1 it loses a number until the counter is 0 - that would make it so walking over its self would still guarantee a valid result
lone dirge
You choose a valid neighbor, from the zeros around. If you can't start over
crimson vapor
it wouldnt even need to be a valid neighbor
it can walk over its self, if it tries to walk off the grid it just skips that number
lone dirge
Yes
crimson vapor
basically generate a large enough series of numbers ranging from 1 to 4 and follow that string untill the counter is 0
okay I can work with that
lone dirge
1 to 4 meaning <- -> down up?
yeah you could do this... you don't have to generate these though
simply randomly decide whether to go up down left right on demand
If it takes too long to generate a pattern just give up and restart the random walk
Have this sort of condition -> if you tried say, 500 steps, and the counter is still > 0. Restart.
Play with these numbers see how long does it usually takes to create such pattern. I assume it should be very fast
crimson vapor
I'll try something out now thanks for this! Do you have any idea how I could determine the total number of possible layouts that exist adhering to those rules?
lone dirge
I'd try to start small and see how it behaves
2,3,4,5,6,7 pixels
This may be helpful https://en.wikipedia.org/wiki/Polyomino
To think about an inductive solution
crimson vapor
i appreciate it!
lone dirge
(:
crimson vapor
.close
shadow lava
obtuse pebbleBOT
shadow lava
wooden cipher
d/dx(x/2)=?
shadow lava
i'm not understanding d/dx (1/2 * e^x)
wooden cipher
thats the same thing as 1/2(d/dx(x))
shadow lava
d/dx [1/2 * x] = 1/2 * x^0 = 1/2 * 1 = 1/2
hmmm
it's power rule and chain rule combined?
wooden cipher
Recall d/dx(c×f(x))=c×d/dx(f(x))
vapid hamlet
it's just linearity of differentiation
warm canopy
like Garlic just wrote, you can pull constants outside the derivative if they're timesing a function
warm canopy
because thats two different functions you're differentiating
shadow lava
the first function is 2 to the power
the second function is 2 times x?
warm canopy
that is what it says yes
shadow lava
alright
oh right and we always do that
but when it's 2^x
the second function is just 1 anyways
so we can skip that step
warm canopy
in the chain rule?
shadow lava
yeah
2^x is the first function
^x is the second function
find derivatives of each, multiply together
warm canopy
yeah you are always secretly doing the chain rule in any derivative but a lot of the time the derivative is just 1
shadow lava
(2^x)(ln(2))(1)
warm canopy
yes
shadow lava
OK
this is equally as valid, ya?
i just put the * 2 at the end, instead of +1 in the exponent
warm canopy
sure no one will kill you for not doing that, but not as nice a form imo
wooden axle
sorry for the bother but can someone check my answer I'm so lost and have no idea what I'm doing
obtuse pebbleBOT
sorry for the bother but can someone check my answer I'm so lost and have no ide...
wooden axle
I started by calculating the volume of the square then the prism and adding them together
I'm not sure if I have to subtract anything like how it's done with surface area
tardy helm
That is correct
wooden axle
omg thank you so much 😭 I thought there was meant to be more steps in this
surface area version is so much more complicated
tardy helm
Nope, your clean and organized writing is nice and very helpful when checking. NJ!