#help-33
1 messages · Page 37 of 1
some help with this question please, i can see OABC is a rhombus but im not sure where to go from there
@thin frost Has your question been resolved?
@thin frost Has your question been resolved?
@thin frost Has your question been resolved?
OABC is a rhombus therefore all its sides are equal
draw a line OC
what can you say about the two triangles OBC and OAC
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I'm getting undefined for the tangent
I differentiated and got $\frac{8x-3y}{3x+2y}$
$3(2)+2(-3)=0$
,w implicit derivative 4x^2 - 3xy - y^2 = 25
checks out
and you got that y' is undefined at this point?
you can either go through the same song and dance but to find dx/dy instead of dy/dx,
or you could acknowledge that you just have a vertical tangent at this point.
Yeah..0 is in the denominator
How does he know it's vertical??
who's he
?
so the tangent is x=some real number?
@still temple is he
_____ does not have a pronoun role, so i would refrain from assuming he/him pronouns for them
i dont care about pronouns for myself
sure is, and you know exactly what that real number is.
then get an "any pronouns" role so there is no question of that
oh ok
Oops sry
by finding dx/dy and then plugging in values?
oh....
it's just x=2
can u explain what this means though?
this is what the answer key says
It means that on a Cartesian plane the relationship between the x and y of all points the tangent passes thru is given by that equation
Isn’t the equation x=2
Because it’s tangent is vertical
Equation for a line must involve both x and y
Ok
Desmos go brrrrr
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Sup
How can i approach this question?
is it given that x,y are integers or what condition?
integers, right?
Yes
Done
Now i can subsitute it in any of those 2 eq right?
you can just check the integer values that
(x+1)(y+1)=5 is valid
all those 4 pairs of (x,y)
and plug them back to the original equation to see if they are valid
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Isn't this slide incorrect?
It should be "strongly connected"
Here is the accompanying graph:
the above graph is strongly connected, not weakly connected
Or am I misunderstanding something here?
Complete wild guess (and I almost certainly won’t remember any more to help extra), but does strongly connected imply weakly connected?
I'm not sure what you mean here
they are two different states, I believe
weakly connected would be where some vertices only have an in-degree or out-degree, but not both
an euler circuit is a circuit that visits each edge exactly once, yes?
given that a circuit comes back where it starts, the existence of one does imply strong connectedness and not weak...
so it looks like the slide is in error
The same error seems to be repeated here as well.
How can there be so many typos that seem to be significant
ah, no, for an euler path you don't need strong connectedness
wait
let me show the accompanying example
this graph is strongly connected, not weakly
this is why I feel this is another typo
do you mean to claim "this graph is strongly connected and fails to be weakly connected"?
or it seems to be based on the example given
yes, a graph can not be both strongly and weakly, afaik
do you have a definition of weak connectedness on hand? we could check your graph against it.
or as far as i have been taught so far
i think either you misheard your teacher or they chose to misinform you.
as far as i know, strong connectedness implies weak connectedness.
But this is a reason why I feel a graph can't be both.
Here is one of the self test questions are teacher gave us during lecture:
as you can see, the options are strongly OR weakly connected
not both
Not the way I read the wording “…and if not, whether…”
^
well, that's because there is a third implied option
which is "not connected at all"
strongly connected means both, in this sense.
any strongly connected graph is also weakly connected.
so really, it's "Strongly connected", "Not strongly connected, but weakly connected", and "Not connected"
so, if this is true, what does this mean for my original question?
this one?
I'm not sure I understand how this negates the potential typo
this screenshot has a typo. "weakly" should be "strongly" here.
because even if we accept your interpretation, then why would the slide read "weakly" instead of "strongly"?
one is listed instead of the other because there is a difference intended to be shown
so, this slide would also have a typo:
no, i cannot say whether that screenshot has a typo. i know for a fact that there exists a graph that is weakly and only weakly connected, and which has an euler path. therefore the "weakly" there should not be changed to "strongly".
i do not know how else to communicate my point to you...
Could you share such a graph?
the example here is clearly incorrect, but maybe the definition isn't
a -> b -> c -> d
this graph is just a path
its sole euler path is itself
it is weakly connected but not strongly connected
in what way?
do you think there are not two vertices with unequal in- and out-degrees?
or that there are exactly two of them, but the differences between their in- and out-degrees are not what they should be?
you mean a doesn't have any incoming edges!
that's different from it not having an indegree at all!
the number zero exists!
ok, that's true
a: in: 0 out: 1
b: in: 1 out: 1
c: in: 1 out: 1
d: in: 1 out: 0
ok, you're right
it does match
then, maybe the example is just incorrect
in what way?
the graph is strongly connected
I know you feel this implies weakly connected
but the slides says "weakly"
specifically
not implied
i do not "feel" that way
it is an easy theorem that any strongly connected graph is also weakly connected
being strongly connected does not mean you fail to be weakly connected too
for instance, I'm sure I will get a 0 for a problem that requires an answer of "strongly", but I write "weakly"
this logic will not hold, I don't think
do you mean problems of the kind you showed?
yes
i.e. "determine if this graph is strongly connected; if it fails to be strongly connected, determine whether it is weakly connected"?
this isn't the wording of the problems above
then indeed, writing "weakly connected" would mean that you determined the graph to be weakly and not strongly connected, and that is what would contradict said graph being strongly connected.
"Determine whether this graph is strongly connected, and if not, whether it is weakly connected."
yes, this
are you or are you not now going to argue that, because my wording isn't the same-to-the-letter as that, i have committed misinterpretation thereof?
no, you are likely very correct here
but the issue is answering to my professor's standards
if strongly implies weakly, then, an answer of 'weakly' should be correct
but i will get that marked wrong every time, I assume
yes, because analyzing a strongly connected graph and following the instructions provided cannot result in an answer of "weak"
yes, so the slide's example is incorrect
not the description, but the example
I'm trying to keep repeating that to be clear here
def answer_assignment(G):
if is_strongly_connected(G):
print "Strong"
else:
if is_weakly_connected(G):
print "Weak"
else:
print "Not connected"
FOR THE ASSIGNMENT you are supposed to follow this pseudocode i just wrote here
do you agree or disagree with this
i agree
okay
in that case
let me attempt for the third time to clear up the apparent confusion
on what
that will print "Strong"
yes, but you didn't let me finish.
so, the slide is incorrect
not to cut you, but why would there be any follow up if we follow the algorithm?
"answer_assignment(G) prints Weak" is NOT equivalent to "G is weakly-connected".
that's all I wanted to inquire
that's the follow-up.
oh.
hmmm
our algorithm prints Weak for graphs that are WEAKLY CONNECTED WHILE FAILING TO BE STRONGLY CONNECTED, and no others.
but the definition does NOT call for a graph specifically like that.
it calls for a weakly connected graph, which ALSO INCLUDES STRONGLY CONNECTED GRAPHS.
this is confusing to me so let me follow up with this question
why is "weakly" specifically used here instead of "strongly"?
here as in where? the euler path slide?
so the euler path slide, yes.
if what you are saying is indeed the case, then there must be a reason why the professor chose one term over the other
yeah, of course there's a reason.
requiring that the graph be STRONGLY connected will exclude graphs that aren't strongly connected yet have an euler path.
the slide says "a graph has an euler path if and only if such-and-such conditions."
but "strongly" implies "weakly", not vice versa
which is why I don't understand this statement:
all strongly-connected graphs are weakly-connected, therefore "weakly-connected graphs" includes strongly-connected ones.
maybe you can share a weakly connected graph that is also strongly connected?
okay, sure...
i can't depict it in plain text easily, but take 4 vertices a, b, c, d with edges ab, bc, cd and da
this graph is both weakly connected and strongly connected
does this or does this not satisfy you
how is this strongly connected
both vertices do not have an in-degree and out-degree greater than 0
what "both" vertices? there aren't two of them, there are four.
oh, ok
then, that isn't weakly connected
maybe, we have different ideas of what 'weakly' means
do you agree that this is weakly?
yeah and you have the wrong one
you think that we are only allowed to call a graph weakly connected if it isnt also strongly connected
and that isnt the case
is this "weakly" to you?
AND YOU STILL REFUSE TO BUDGE
YES THIS GRAPH IS WEAKLY CONNECTED I NEVER ARGUED AGAINST THAT
why must you get so infuriated when someone doesn't understand your position?
it's not that serious
i've made my position as clear as i possibly could've, several times in a row.
and if it is then, perhaps return to this later when you've had a break?
Are these the defns you are using:
- 2 vertices are (weakly) connected if there exists a path between them
- 2 vertices are strongly connected if there exists a path in both directions between them
I feel either this description is incorrect:
OR
the accompanying example is incorrect:
or both, i guess
Without delving into the graph theory
Do you understand this point
i think so
ok
in which case what you've said here doesn't make sense
Prove all strongly connected graphs are also weakly connected
ah, ok
ahh!
I totally see your point now
by "weakly", my professor is just saying that there needs to be a path between the vertices
not necessarily in both directions, just that there is a path
then, my professor was never incorrect in any of the slides
even in my initial query, the slide is correct
"weakly" connected does mean ONLY weakly connected
will look later if someone else didnt
it just means the minimum requirement
great to have this cleared up
now, i don't need to email them
ty
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Hi, why is it true please ?
If you think about integral as the area under a graph, it might help you out.
Note that for every t in the interval 1/(k + 1) <= 1/t <= 1/k
@sick mesa Has your question been resolved?
I dont see how it help 😭the area is bigger than theses points
Yes but for the integral ?
Now integrate these from k to k+1
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If X is uncountable, equipped with the cocountable topology, how do I show that
You'll probably have more luck in #point-set-topology btw
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Yes
if a divides b, then lcm(a, b) = b
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can we discuss induction
its like recursion right
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Hi, when completing the square and getting to the square root process. I know that you just remove the ^2 but what happens to the 4?
4 is also a square number
Does it just become 2
yes
Alright thanks
np
Here’s the full thing, do you checking if I’ve done anything wrong?
Oh wait
There should be a +- sign
but I need to get the x by itself right
I can’t leave it as just 2x
you multiplied the whole thing by 4 but you only have 4x^2
Oh I thought I take the leading coefficient or something during cts
yes but that affects what's inside
Oh so I should be distributing it inside the square root first, and then removing the square?
I don't know what you mean
try and copmlete the square ont his again
actually it might help to just start from 4x^2 + 7x - 1
This
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Where should I go from here?
did you try and draw the line y=mx out and see how it cuts the shape?
Well how should I draw it if I don’t know the slope
O wait
It’s gonna be
Like this
yea, something similar, you dont really have to be exact, just to get the picture it's okay
Ok and now I just do the integral
now that it is divided into 2 parts
Thanks
yea
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So here you'll see the question is already solved, but can someone walk me through how to solve it?
@still temple Has your question been resolved?
There are several ways to prove that
Do you know what eigenvalues and eigenvectors are?
No. We've never discussed it in class or studied it
Ok
In that case, the first step is to find sequences in the form aₙ = xⁿ which satisfy the first equation, but not necessarily the two other ones
So you have x^n = x^(n-1) + 6 x^(n-2), right?
not ^ but _.
But yeah
But
We are actually looking for solutions in which a_n equals x^n, for some x ∈ ℝ
Then this is a polynomial equation. What are its solutions?
I don't know
Well
Either x = 0, which is a solution, or x ≠ 0, in wich case we can divide the equation by x^(n-2)
How do you get those as solutions?
What are you refering to with "those"? The sequences or the number x?
Where does x = 0 and x ≠ 0 come from?
That is a trick that can always be used
A number can either equal 0 or not equal 0
So if x ≠ 0 then why do we divide the equation by x^(n-2)?
To obtain a quadratic equation
Which we can solve
We had a polynomial equation of degree n. To simplify it, we can divide by x^(n-2) to obtain an equation of degree 2, but if we divide by x then x should not equal 0
So if we divide by x^(n-2), what do we obtain?
It would be something like this, right?
That is still wrong
Applying this rule, what would be each of the numbers a, b and c in the left side?
2?
Yes
Therefore, what should be changed here?
x^2 = x^1 + 6?
Exactly
that was embarassing
No
Everyone makes mistakes like that one in math
It is not an embarrassing thing, it is indeed very common
So we have x^2 = x^1 + 6, now what?
How did you reach that solution?
This is not correct, what steps did you make?
I know it's not correct. I'm redoing it
Ok
so this right?
No
I think you wrote twice something inside the square root
There are two 4, but there sould only be one
Be careful with the - signs
This isn't correct?
And consider this
So this and we now know x is 3 and -2
Here
^
x = -2
You forgot the -
fixed
So we obtained two sequences
a_n = 3^n and b_n = (-2)^n
What do we know about these sequences?
Do we need b_n?
I wrote b_n to use a different letter
We are not going to choose any of them yet
Oh
They are sequences we obtained after solving a different problem from what we originally wanted
We wanted that a_n = a_(n-1) + 6 a_(n-2), that a_0 = 3, and that a_1 = 6
But a_n and b_n satisfy the first one, right?
Do you understand why?
Sorry but I need you to refresh me, Where does this come in?
Remember what I said here
So all of this was step 1?
Yes
Ah
But step 2 is shorter
I see
Now we have a_n and b_n
And a_n = a_(n-1) + 6 a_(n-2)
And b_n = b_(n-1) + 6 b_(n-2)
Do you understand why?
Remember that a_n = 3^n and b_n = (-2)^n
So we found sequences in the form of aₙ = xⁿ?
Yes
And since -2 and 3 are solutions to the equation x^n = x^(n-1) + 6 x^(n-2)
We can substitute x^n with a_n, x^(n-1) with a_(n-1) et cetera
And then we have this
Do you understand the logic?
So what about 6a_n-2?
You can also substitute x^(n-2) with a_(n-2)
I will write it in a more clear way
You know that -2 and 3 hold the equation
x^n = x^(n-1) + 6 x^(n-2)
Therefore, the following identities are true:
3^n = 3^(n-1) + 6*3^(n-2)
(-2)^n = (-2)^(n-1) + 6*(-2)^(n-2)
Furthermore, we defined a_n = 3^n and b_n = (-2)^n. Because of that, the previous lines can be rewritten as
a_n = a_(n-1) + 6*a_(n-2)
b_n = b_(n-1) + 6*b_(n-2)
Do you understand this @still temple ?
Yes
So what is step 2?
But do you understand why what we have done so far makes sense and works?
Or are you just writing down all what I'm saying, even though you don't understand it?
I'm confused
Why are you confused?
So my professor's solution to the question was a_n = (3/5)((-2)^n) + (12/5)((3)^n)
Yes, and that is the only correct solution to the problem
So you are confused because we obtained something different
Yeah
Well
As I said before, what we have made so far is to solve an auxiliary problem, a more simple one. It is actually a different problem, so of course the solution is different
But we need it to obtain the solution of the original problem
What I want is that you understand why this is true, regardless of what we are going to use it for
I don't.
I'm so sorry if I'm wasting your time
Don't worry
First, what does a_n = a_(n-1) + 6 a_(n-2) mean?
What is a_n?
3^n
Yeah, but how does that relate to my initial question?
Because it is important to understand what you are doing
You should not just memorize how to solve every kind of problem every time you see it for the first time
Because that is just too difficult
Instead, understand math in general is easier in the long term, because you have to memorize less things
If you write down how to solve this and memorize it, you may make mistakes when you solve it again, but if you understand why the solution works you don't have to memorize it
Alright. Sorry but give me a moment
This being said, I'm gonna review on my textbook. I'll understand if you leave
Ok
Thanks for your time
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Just started a discrete math course, I have a question that I can’t get my head off of.
From this post, how does ExE!y differ from E!xEy?
Any help would be greatly appreciated.
there exists an x and there exists only one y
vs
there exists one x and there exists a y
yeah
How about E!xE!y?
yeah thats one the solutions if you scroll down
yea
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Radioactive radium has a half-life of approximately 1599 years. What percent of a given amount remains after 100 years?
Let the output of this function be in percentage
Then C would be 100
As at t=0, it would be at 100% and then will decay afterwards
So the function becomes 100e^(kt)
Why is C 100?
This is why
Do you know the correct answer?
Yeah it’s 95.76%
Yes so, its given that half life is 1599 yrs, so it decays to 50% in 1599 years. So, at t=1599, its output must be 50. Substituting the values we get...
100e^(1599k) = 50
=>e^(1599k)=0.5
=>k=-ln(2)/1599 = 0.000433
Are you getting it till this?
Ya sorry k=-0.000433
All good
So, now putting k back in equation, we get
100e^(-0.000433t)
Now we need to find the decay percent at t=100 years, so substituting t=100 in it, and you'll get the answer 95.76%
Got it?
And the y value is the half life percentage at t= amount of years
And I’m assuming half life is always 50%?
Yes
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can anyone help me with this?
its urgent
There are multiple ways you can solve this
But the best (for me) is using the equation
Thank you

Bruh
$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
Ann
Backward slash
Sorry forgot
given that this is a mulitple choice question
I mean in general
it might be less headache to simply expand all the answer options and see which matches
i dont think i can answer that
When it comes to solving complex polynomials this is literally the only way but in this case there are faster ways
you could look at the coefficient and constant
yea and that will make +7.x and -4.x which is d
you could check this by looking at the constants, 7*-4 is -28 so it checks out
Or you could do trial and error on each of the answers (try multiplying the numbers together to see if it matches the c coefficient and summing up together to see if they match the b coefficient
yea as you said there is a lot of ways to do this
x^2 + 7x - 4x - 28 = x(x + 7) - 4(x + 7) = (x + 7)(x - 4)
hold my ms paint
Lol
where do you get the 4 from
He re-writes 3x as 7x-4x
did it
but its positive in the equation
wdym
I wrote 3x as 7x - 4x, because 7 * (-4) = -28 and 7 - 4 = 3
i kind of get it
yea just try the factors of -28 constant and sum them
because of the 3 coeficcient in 3x, 4,-7 does not work because 4+(-7) would make -3
thats what i get
so it would be -3x instead of +3x
ohhh
yep
so -4*7?
i just switch them right
because 7 is higher than 4
@wheat sedge
yea
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How do I start?
@placid oak Has your question been resolved?
@placid oak Has your question been resolved?
@placid oak Has your question been resolved?
Can you cancel anything?
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differentiate w.r.t x
use quotient rule
i'm just learning derivatives and i'm not sure about if i'm supposed to leave the answer without simplifying it or simplify it cause the question just asks to differentiate w.r.t. x
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Is there a function where the derivative is undefined everywhere, by any chance?
and by "everywhere" i mean for all points of the function's domain
(-1)^x
On the reals plane
how about dirichlet's function
In mathematics, the Dirichlet function is the indicator function 1Q or
1
Q
{\displaystyle \mathbf {1} _{\mathbb {Q} }}
of the set of rational numbers Q, i.e. 1Q(x) = 1 if x is a rational number and ...
The among us function
but fractals can be continuous right? or am i missing smtg
yeah but u have more than 1 value at every x
continuous doesnt mean differentiable
ic, so u mean some fractals?
because some fractals can be conti iirc
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@marsh citrus
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Let W={(x,y,z)∣x=2y+z} be a subspace of R³. Let β={(2,1,0),(1,0,1)} be a basis of W.
Let T : W→R be a linear transformation such that T(2,1,0)=(1,0) and T(1,0,1)=(0,1).
How do I prove that T is an Isomorphism?
It maps basis to basis so it must be an isomorphism
T(x,y,z)=(y,z) or T(x,y,z)=(x-y-z,x-2y)
If you want to show a bit in detail, you would wanna prove 1-1 and onto
But notice the dimensions are same so proving either 1-1 OR onto is enough

How do I know which transformation is T?
That would be a bit hard to find, a better option is to consider inverse transformation T' : R2 -> W
then the transformation is easy to find
How do I prove T is one-one without knowing the transformation?
well, there's a theorem which says that T is 1-1 if and only if T maps linearly independent subsets to linearly independent subsets
but anyway, since you're given the explicit transformation that should be used
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[\frac{d}{dx}\frac{x^2 + x}{x^2 -x}]
[(x^2 - x)(2x+1)-((2x-1)(x^2+x))]
[\frac{-4x^2 + 2x^2}{(x^2 -x)^2}]
[\frac{-2x^2}{(x^2 -x)^2}]
my bad, use the first one
They might be equivalent but as per my calculations I got the 1st one
If T is the first one, then T(1,0,0)=(0,0) and T(0,0,0)=(0,0). So they are not one-one?
nah cause (1,0,0) doesn't lie in W
anyway get a different channel, I didn't notice you closed the channel
I think you answer is correct
It's just that wolfram simplified it by cancelling x
,w diff (x+1)/(x-1)
basically same as you got above
you might as well cancel x in the first place cause x won't be 0
@ionic owl
@ionic owl Has your question been resolved?
Alright thanks
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hi is this
what's the question?
simplify
I think you should be able to cancel some stuff before performing the fraction substraction
yep
Hmm, looks correct to me
as in the domain restrictions?
yesss
yep
thanks, can I ask something else?
sure
same question
the one at the right is my answer, is it correct as well?
yes, it is correct. I got the same.
thanks a lot, last question, am I just going to plug d^t and d^0 then solve for 1/f?
1/x^2-4x - 12 + 1/4x+8
like that
weird question but I think it does mean that only
alr, that's what I got, right?
yep, got the same
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Hi there can someone tell me how to deal with sine double angle identity
Like how is 2sin(x/2)cos(x/2)=sin(x)
x/2 is half of x
$\sin(\text{this}) = 2\sin(\frac{\text{this}}{2})\cos(\frac{\text{this}}{2})$
ℝamonov
Ohhhhh i see, thank you so much!!!
@gritty wren Has your question been resolved?
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how much do you need to get 9664 to 10058 help pls im stupid
So u want to do 10058 - 9664?
10058-9664
What’s troubling you with that
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Btw
@quaint hill there's an easier
Solution for the problem I had
U remember it right?
$\lim_{x\to\infty}x^0 = 1$
RulzerFly.
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if $p$ is prime and $p²+8$ is prime, Prove that $p^{3} + 8p + 2$ is prime too. Can someone gives me a hint?
phoestaclies
consider mod 3
I think I don't get the idea ?
when in doubt, try examples
go through the first few primes p, check if p^2+8 is prime and if yes check if p^3+8p+2 is prime
yes I did that with 3 and 7
p^2 + 8 is not prime when p is 7
if 2 examples is not enough then maybe do 4 or 5
euh yes sorry mb
tell that to grothendieck
To whom?
ok I will try to test other, and see what I get ty 🙂
Btw I noticed that p^2 + 8 is most of the times divisible by 3
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way to spoil the fun
What/Who is grothendieck?
famous mathematician who once when asked to name a prime said 57
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Actually I once picked 57 as prime for some reason as well 
Closed due to the original message being deleted
it does look quite primelike
as general advice btw, if any claim says "this formula will always be prime", then either the claim is false or the hypothesis only hold for finitely many cases
in 99% of cases anyway
ok thanks
Formulas for the nth prime number actually exist! One was cleverly engineered in 1964 by C. P. Willans. But is it useful?
References:
Herbert Wilf, What is an answer?, The American Mathematical Monthly 89 (1982) 289–292.
https://doi.org/10.1080/00029890.1982.11995435
C. P. Willans, On formulae for the nth prime number, The M...
That is what is meant in the remaining 1%
yes i saw that
everyone shitpost until bot queue's this job up
abusing wilsons theorem to give integers or not I guess with the cosine there
Every time that happens I just edit my last message to add what I want to say lol
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its already closed
bot's just slow
Stupid bot

10 minute timelimit before bot can move channel again
I wasn't thinking about the consequences of my question
how dare you
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does anyone know how'd you maximize a function of this type?
try to use the formula for geometric series ..then you get a simpler function on N
the formula (1-r^N+1)/(1-r)?
if so, good point. Thanks for help
yes but i think you have to add a -1 because the series doesn t strat from 0
yeah
alr how do I close this?
.close
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thanks
np
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could someone give me pointers for this
Please don't occupy multiple help channels.
@kindred spear Has your question been resolved?
<@&286206848099549185>
higher probability and stochastic processes
@kindred spear Has your question been resolved?
can u tell more
@kindred spear Has your question been resolved?
what's a geometric series
I think here you just need to show that Q(A) satisfies all of the axioms of a probability measure
which honestly I constantly forget, but it should be something like Q(emptyset) = 0, Q(the whole space) = 1, countable additivity, and nonnegativity
countable additivity is where you might need to use monotone convergence
<@&268886789983436800> could you check this person's msg history along with their deleted msgs
Seems to just be trolling
right I was thinking along those lines
I'll try it out
so for the case of Q(empty set) = 0, could we consider a decreasing sequence of events where the limit would be the empty set
right
im guessing for a set A in the collection of events H, the measure Q outputs a number [0,1]
do you think you could help with my understanding of Q (the numerator regarding that 1_A)
you don't know that this is a measure
you need to prove that it is a measure
do you know what 1_A means
no
why didn't you just look it up
i was trying to but i couldnt find smth explaining it
how do you expect to do a problem if you don't even know what it's saying
it's never defined in your textbook?
what's it called
it's called an indicator function
so what have you been wasting your time on
if you don't even know what the problem is saying
no it is not
do you know what a sample space is
let's say A is an event
do you know what an event is
do you know the definition of event
bruh
go review
sure
an event is a measurable subset of Omega
it is a set of outcomes
"measurable" means it belongs to H
no it is not
you need to review the definitions of sample space, probability measure, random variable
H is the set of all measurable sets
do you know how expectation is defined
do you know what a Lebesgue integral is
do you know what a random variable is
you kinda need to know the definitions
to do something that depends on them
after you review everything, to check your understanding, come back with an explanation of why E[1_A] = P(A)
@kindred spear Has your question been resolved?
you need to show it for general random variables
using the definition of expectation given by the Lebesgue integral
it's basically the same thing, because 1_A is a discrete random variable
but you need to understand what the Lebesgue integral is and where the formula for discrete RVs come from

