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1:00 PM
Indeed, that's one reason why the ignore button is no use. There exists messages that nobody can ignore
 
Let $x = t$, $y = t^2$ and $z = t^3$ where $t$ is a parameter in space. How do you eliminate $t$ from it?
 
even if your consciousness fools you that you have ignored it, later on as you recall memories and stuff, or reflecting your behaviour, you will realise those messages have affect you without notice
 
I can sure write it simply as $y = x^2$ which would be just one equation to represent it. But that is not sufficient.
 
in The h Bar, 6 mins ago, by heather
no, i'm not going to just solve a problem where you haven't really put any effort in to make it understandable to others.
in The h Bar, 7 mins ago, by heather
i've scrolled through the transcript and you've pinged a half-dozen people to "just solve the problem" for you.
@skullpatrol Now, do you know why?
 
So basically my question is establishing both ways equality between parametrised equation and cartesian coordinates curve, how do I ensure they are indeed equal.
 
1:10 PM
@gxyd Make sure every point on one is a point on the other.
 
Yes, that is what I want.
But I can't do that manually for every point.
 
You are right. Use some kind of reasoning then.
 
How do I prove that solution is $y^2 = xz$ and $xy = z$, -- (eq. 1) but not just $y^2 = xz$ -- (eq. 2). Since the second one also satisfies the required condition.
Or may be not exactly a rigorous proof, but a result for me would work.
Something like $n_{p} = no. of parameters$ and $n_e = no. of independent variables$, then number of equations is equal to $n_{p} - n_{e}$, or something like that.
But I also know that $y^2 = xz$ and $xy = z$ can simply be written down as $(y^2 - xz) (xy - z) = 0$.
So basically some formula for irreducible number of equations would be a better 'term'.
 
1:37 PM
Hi, $$\text{Why gcd}(2^{2a+1}+1,2^{2b+1}+1)=2^{2\text{gcd}(a,b)+1}+1 ?$$
a,b integer
a mistake
$$\text{why gcd}(2^{2a+1}+1,2^{2b+1}+1)=2^{\text{gcd}(2a+1,2b+1)}+1 ? $$
 
1:57 PM
@ACuriousMind Ser why did you delete my answer? I checked the homework policy and my answer didn't violate it! This is second or third time my answer got deleted!
Oops it was on Physics site
 
@samjoe 1. Physics business should go to the h bar. 2. Your answer was deleted because "If someone posts an answer to a homework-type question that gives away a complete or near-complete solution, in most cases it will be temporarily deleted." and your answer simply gave the numerical answer to the question, which is what we consider "complete".
 
it might as well be incorrect
 
Acuriousmind finally appeared into the maths chat
I have been waiting for this moment for a long time (reason I have no idea)
This is going to be interesting
 
2:17 PM
To celebrate this is an enigma:

A farmer has a rectangular ground of 100 m by 50 m, he wants to plant olive trees, in sufficiently spaced ways (to avoid exhaustion by the roots) at least 10 meters from each other.

How much can one hope to put at the most, effectively?
..............................................................Is very difficult, and I don't know an answer
0
Q: The farmer and the olive trees

DattierA farmer has a rectangular ground of 100 m by 50 m, he wants to plant olive trees, in sufficiently spaced ways (to avoid exhaustion by the roots) at least 10 meters from each other. How much can one hope to put at the most, effectively?

Bye
 
3:05 PM
I don't know why I am puzzling over this so much, but I need to find, for the normal distribution, the $\sigma$ which yields $P(-0.98 < x < 0.98) = 0.98$, when $\mu = 0$
 
@logical123 express that statement in terms of $\sigma$.
(and of course $z$ the standard score)
 
oh derp, I see now, thanks
lol 'that was easy'
$z_{-} = \frac{-0.98}{\sigma}$, $z_{+} = \frac{0.98}{\sigma}$
then z+ - z- = 0.98, solve for sigma
wait.. no, z doesn't directly represent the area under the curve
 
but you do know that it is symmetric
Express that in the form of $P(z<\frac{a}{\sigma})=b$ and find $a$ and $b$
 
$P(z<\frac{0.98}{\sigma}) = 0.49$ ?
 
not really
but you're close
 
3:19 PM
$P(z>\frac{0.98}{\sigma}) = 0.01$ ?
lol
 
right, and convert that back into the $<$ form?
 
Isn't the first one the ' $<$ ' form? I haven't had my coffee yet today :P
 
@logical123 but it's wrong
 
I believe you, is the a wrong or b wrong in the first one
or disastrously, both?
 
3:24 PM
$P(z<\frac{0.98}{\sigma}) = 0.99$
?
or $P(z<\frac{0.98}{\sigma}) = 0.98$ but because of symmetry im thinking =0.99
 
it's 0.99
 
so then using $\frac{1}{\sqrt{2\pi}} e^{\frac{-z^2}{2}} = \phi(z)$, just solve for sigma?
 
3:44 PM
@logical123 just use the table
 
Ok lol, I'll have to find one online real quick
 
when you're doing problems like this you pretty much have to use a table
 
@Dattier $\begin{array}{cl} &\gcd(2^m+1,2^n+1) \\=& \gcd(2^m+1-2^{n-m}(2^n+1),2^n+1) \\=& \gcd(2^{n-m}+1,2^n+1) \end{array}$
and $\gcd(2^a+1,2^0+1)=2$ when $a$ is odd
 
though part of the work there is putting the problem into a form where the table will suffice.
 
Are you talking about greatest common divisor @LeakyNun?
 
3:52 PM
@Abcd no, just me
what brings you here?
 
@LeakyNun Okay, I thought everyone is.
@LeakyNun Just came here for a few minutes... any problem?
 
@Abcd no problem
 
4:36 PM
when is $\binom N k \equiv 0\pmod 2$ for which k?
 
@RE60K what is the question?
 
I am trying to solve some computational problem and it relates to serpenski's triangle and for each row mathworld.wolfram.com/SierpinskiSieve.html says that is when the above is true
 
If I can find those k directly rather than trying all N values then it would be faster
@LeakyNun exactly, my point
 
@RE60K does the picture help?
 
4:39 PM
yes it does.. but how would an algorithm compute that, when ${\rm AND}({\rm NOT}(N),k)$ is zero
as said on that page
but trying for all values is $O(n)$, I am looking for something probably $O(\log n)$
 
how do you generate $\le n$ values in $O(\log n)$?
 
It is possible because one of my collegue solved this problem on an online Judge
 
@RE60K what is the full problem?
 
Hey everyone, a question in the definition of derivative with limits ex: f(x)-f(x0)/x-x0=f'(x0), can we do something with lim (x-x0)/(f(x)-f(x0) ?
 
@valer that's just 1/f'(x0)
 
4:44 PM
The contest is still going on, but it is equivalent to finding Q rows of sierpinski triangle of size atmost next power of 2 for N where $0\le N,Q \le 2\times10^5$ hence I think $Q\log(n)$
 
@LeakyNun thanks
 
@RE60K why didn't you give me the full problem ...
the full problem doesn't require generation of $n$ values!
 
well how should I tell you without making public the relation between the problem and this because the contest closes on 11 Sept.
 
@RE60K then why are you asking it at all?
 
I am asking not the problem but a question related to binomial questions, sorry if my behaviour offended you! :)
 
4:47 PM
@RE60K I hate with passion people asking questions with information hidden
re near power of 2: just read it out from the picture I gave
 
Don't worry :)
 
Hi, Demonark
 
Yo Leaky!
 
L m a o
I think it was fair, even if the other person takes it as being harsh that sorta thing is just nope
 
5:03 PM
@Daminark an eye for an eye
I take his comment as harsh
2
Q: Constructing the reals from the integers

José Carlos SantosA map $f\colon\mathbb{Z}\longrightarrow\mathbb Z$ is called a quasi-homomorphism if the set$$\{f(m+n)-f(m)-f(n)\,|\,m,n\in\mathbb{Z}\}$$is bounded. Let $R$ be the set of these functions. Let's consider the binary relation $\sim$ in $R$ defined by$$f_1\sim f_2\iff\{f_1(n)-f_2(n)\,|\,n\in\mathbb{Z}...

The best thing I came across today.
 
Oh boy here we go
 
@Daminark can one equivalence class have multiple elements there?
 
I think $f(n) = n+1$ and $f(n) = n+2$ are both quasihomomorphisms and are in the same equivalence class
In fact I think any translate of a quasihomomorphism is one as well, and in the same class?
 
can someone explain to me how to buidl the naturals using equivlance classes and unions?
 
5:19 PM
@Faust but they're classically built with sets or Peano's axioms
@Daminark oh
 
i have seens peano axioms
i i have seen 1 and the addition operation
nfi about equivlance classes and unions as a construction though
what if your 2 equivlance classes shared an element?
as in 5+4 =7
 
@Faust ??
 
nonsense i tell you!
 
I don't get you.
 
@Daminark Muffet is hard to beat
 
5:22 PM
for example {a,b,c,d,e} union {1,2,a,b} when u union them u dont get a set of 9 elements
it has seven
 
She's one of the tough bosses for sure
 
Use strong induction to prove that every positive integer can be expressed as the sum
and/or difference of distinct powers of 3
why do i need strong induction?
 
@Daminark The telegram just came
 
cant i just show it using distinct number excluding $3^0$
 
@Faust I've never seen such a construction of $\mathbb{N}$
 
5:23 PM
@Faust when did we say that we're defining natural numbers that way?
 
I mean I guess you can define them via cardinalities
 
@Daminark well you can define addition of ordinals in a manner similar to what Faust suggested, but using $\sqcup$ disjoint unions instead of $\cup$ unions
 
he said you can construct the natural numbers using equivlance classes and ther eunions
 
But then union won't correspond to addition unless the sets are disjoint
 
@Faust who said that?
1 min ago, by Faust
why do i need strong induction?
 
5:24 PM
the guy who teachs my real analysis course
 
well because it's hard to prove it using "weak" induction?
@Faust then ask him what he meant
 
it bothers me and hes obv not working till monday
 
because they're usually not constructed that way
 
ic
 
@Faust then disregard it for now
 
5:25 PM
i have also seen construction of using powersets of the empty set
 
@Faust Von Neumann's universe
but we don't usually use that
 
but never heard of whatever he was saying
 
@Daminark mettaton leave me alone
 
@Faust that one's real doe
 
@Shog9 ok. I'll leave then.
 
5:29 PM
Use strong induction to prove that every positive integer can be expressed as the sum
and/or difference of distinct powers of 3.

take a base case n=1 is simply $3^0$ then n=2 = 3 -3^0 how many base cases do i need?
long time no see turtle wave
 
@Faust who is turtle wave?
 
wait thats from DBZ Turtle... warrior?
 
are you drunk or something? I see no turtles here.
 
O.o
how do i know what u look like i cant see you through my monitor
 
what are you talking about?!?
 
5:34 PM
I always assumed you were a giant snapping turtle with four hands thats why u type so quickly
 
dude
are you trying to irritate me?
No, it's cause I do programming.
and im on here all the time
and I think a mile a minute sometimes.
 
well you do kind of snap at people sometimes
 
no
other people snap at me
you made me think of this
 
lmao
 
@Faust I better leave though
Someone will get upset again like before.
@Faust Keep going till you start noticing that the previous cases imply the latter cases in the general case. Probably have to use modular arithmetic for that or something.
 
5:44 PM
evens are a product of 2 numbers odds are a rpoduct of 3
but thtas not strong induction is it thats just weak induction twice
 
What do you mean?
Strong induction is just proving that all previous cases imply some case
 
even = 3^a +-3^b
 
that doesn't mean you actually have to use all those cases
if it ends up being that you only need the previous case that doesn't make it invalid
you still proved all previous cases imply the next
 
k i dont know how to do it then lol
 
me neither
tbh, I thought that was an unsolved conjecture
one of Fermat's theorems
is this your homework?
 
5:47 PM
yeah
 
wait
oh my god it's so trivial
well for the case of 3n
n can be expressed as 3^a +- 3^b right?
so 3n is also of that form
 
no
 
?
no to what?
 
pick 5
 
what about it?
 
5:48 PM
its odd need 3 numbers to make it\
 
"Use strong induction to prove that every positive integer can be expressed as the sum and/or difference of distinct powers of 3. "
that means two powers, right?
 
any number
distinct
 
well regardless
 
need 2 for even numbers
 
a series of them added together and subtracted
multiply them by 3
 
5:49 PM
3 for odd not multiples of 3
 
did you not just see what I said?
 
grr
not mulptiples
3^a are the only dds that can be done in less than 3 numbers
 
i know that
 
take the example 15
 
listen tio me
would you listen to me and stop grring at me. I'm trying to help.
you have m powers of 3.
added and or subtracted in some combination to make n
3n is therefore also a sum and difference of m distinct powers of 3
cause you can distribute the three through all the terms
 
5:52 PM
i was grring at my inability to convey my meaning not you
 
Anyone can help me on some questions on Gambler's Ruin?
 
neit
 
6:33 PM
0
Q: Mean time of hitting a boundary state in Gambler's Ruin problem

Little Rookiei am currently studying Stochastic Processes without Measure Theory. I have a question on Gambler's Ruin problem. Suppose each round of game are independent bernoulli trials $X_k$ for $k\geq 0$, whereby the player wins $\$1$ with probability $p$ and lose $\$1$ with probability $1-p$. $P(X_{k+1}...

 
6:55 PM
hi @Daminark
 
7:07 PM
2
Q: Dimension of $U+V$

Sachchidanand PrasadLet $U$ and $V$ be the null spaces of $ A=\begin{bmatrix} 1&1&0&0\\0& 0&1&1 \end{bmatrix} $ and $ B=\begin{bmatrix} 1&2&3&2\\0&1&2&1 \end{bmatrix} $. Then what will be the dimension of $U+V.$ I calculated the null space of $U$ and $V$ as follows: \begin{align*} U=\{x\in \mathbb{R}^4: Ax=0 \}\im...

I don't get it
my method is more straight-forward and does not require finding $U$ and $V$
why is the other answer getting more upvotes
2
 
well, the OP already computed U,V
so having already computed U,V the first answer is able to proceed just by inspection
so the context is a bit different than it would be had the OP not included that
 
OP listed his steps and requested a shorter method
presumably from scratch
@Semiclassical
 
7:29 PM
Can someone please help me out with tensor products when available?
 
7:42 PM
@BalarkaSen Are you here now ?
 
Yes.
 
Did you see my question ?
 
No.
 
What are some interesting application of proving a polynomial to be irreducible over the integers ?
I find proving irreducibility pretty boring.
 
Hell if I know.
 
7:44 PM
@BalarkaSen What do you mean ?
 
@AlexKChen it's a way to find finite fields.
 
Can you please elaborate @LeakyNun
 
To find a field with order $p^n$, find a polynomial $q$ irreducible in $\Bbb Z_p[x]$, and $\Bbb Z_p[x]/\langle q\rangle$ gives you the desired field.
 
thats not irreducibility over Z though
 
Oops sorry, but I don't know finite fields :\ Can you provide more elementary examples (Eg proving x^2 +- p to be irreducible over Q(zeta_p, i) implies QR) ?
QR means quadratic reciprocity.
 
7:49 PM
thats not irreducibility over Z either
 
lol that's definitely less elementary than what Leaky said
 
@BalarkaSen lol I don't understand that either.
Honestly, I don't think I know much about polynomials, but I don't feel motivated to learn about it either. That's why I'm asking this.
 
Hi, if some user is interested in partial differential equations, then he/she can read the last issue of the Notices of the AMS Volume 64, Number 8, Two Optimization Problems in Thermal Insulation by Bucur, Buttazzo, and Nitsch. Good weekend all users.
 
Okay, lemme ask more concrete examples.
What's the point of proving cyclotomic polynomials to be irreducible ?
?
 
I just realized that you could construct a contrived proof of row rank = column rank by symbol pushing with the rank-nullity theorem.
Just a trivia
 
7:55 PM
What about my question about cyclotomic poly ?
 
I don't know the answer.
 
@BalarkaSen how?
 
Maybe someone else does
 
Who ?
 
I don't know.
 
7:56 PM
(Also BTW isn't it midnight in India now Balarka ?)
 
@AlexKChen the author of this document, for example
(what, they managed to prove that every finite division ring is a field using the fact that cyclotomic polynomials are irreducible)
(this is freaking unheard of)
 
Does a two-sided arrow between objects ensure there exists a unique isomorphism between objects?
Or just an isomorphism.
 
@gian no, you need a sound proof to ensure that there exists any homomorphism at all between objects
 
@LeakyNun Rank-nullity says for an mxn matrix, dim im A + dim ker A = n, yeah? Now im A is literally the column space of the matrix; that means dim C(A) + dim ker A = n. Alternatively you know dim R(A) + dim ker A = n (because R(A), the rowspace, is the orthocomplement of ker A).
Equate it out
 
@BalarkaSen :O
the premises are true
 
7:59 PM
@LeakyNun, but what if I'm proving that two different tensor products of the same modules $A$ and $B$ are uniquely isomorphic?
 
Cute, isn't it
 
the result is amazing
and I can't poke any holes.
@gian what do you mean by uniquely isomorphic?
 
That the isomorphism between them is unique.
 
what do you mean that two tensor products are isomorphic?
I should probably step out before you realize that I have zero knowledge about tensors
 
Oh
 
8:02 PM
@LeakyNun Hello ! :D
 
@gian What you do is start with two things $A$, $B$ that satisfy some universal property. It's not true that there is only one isomorphism $A \to B$. What is true is that the universal property will end up providing you with a map $A \to B$, which you will check (using the universal property of $B$) has an inverse.
 
@KasmirKhaan hi
 
How to make a group table ?
On texstudio?
 
When one says that blah is unique up to unique isomorphism, it means that the universal property provides the inverse - it actually specifies the isomorphism.
 
its gonna be a relativly big one
one for S_3 and other from GL_2(Z/2)
 
8:03 PM
\[\begin{array}{c|c|c|c}
\cdot&e&a&b\\\hline
e&e&a&b\\\hline
a&a&b&e\\\hline
b&b&e&a
\end{array}\]
here's a group table for $\Bbb Z_3$
 
Thanks man you are the best :D
 
Right so when finding an isomorphism between tensor products, I need to treat two cases in which one of the tensor products is indeed a tensor product and the other is an arbitrary module.
 
Yo
 
ster
 
8:05 PM
I don't know what that means.
 
Shall the world, then, be ruled by oysters? No, no, horrible
 
Ohh I see now. It's unique isomorphism because the mapping property deals with a unique map that makes the diagram commute.
 
@LeakyNun can you make it for 6 elements please? or tell me how from the 3 elements one I can do it?
 
@KasmirKhaan just imitate.
 
Okay hope nothing wierd accures =p
 
8:06 PM
@BalarkaSen Nikolai Ivanovich Lobachevsky
 
so many refefefereneces
 
@MikeMiller, must one of the modules be left and the other be right in a tensor product?
I've seen texts that do and don't make this point.
 
depends on whether the ground ring you're working over is commutative
 
Oh gothca, thanks.
 
there are other cases, but the general rule is that if the ground ring is commutative, and $M$ is a left $R$-module, $N$ a right $R$-module, you can form the group $M \otimes_R N$; but you no longer have any $R$-actions
 
8:10 PM
But "left" and "right" can be disregarded if the ring is commutative.
 
on the other hand, when you work over a commutative ring, left modules and right modules are the same concept; define the right action by $ma = am$. this only works as a right action because $ab = ba$
precisely
 
Got it.
 
8:31 PM
> I have heard (from two sources) that at the University of Chicago a senior faculty member was temporarily banned from teaching undergraduate courses. The reason is that during a first semester undergraduate linear algebra course he did everything over the Quaternions.
 
What does that mean? @LeakyNun
quaternions?
 
@KasmirKhaan $\Bbb R^4$
with $i^2=j^2=k^2=ijk=-1$
 
Why would he banned from for doing that?
more advanced than needed?
 
@KasmirKhaan because you're supposed to teach linear algebra in just $\Bbb R$ or $\Bbb C$...
 
Oh =p
 
8:34 PM
esp. when it is a first semester undergraduate course
 
They should have just told him that not bann him
he is over entuisiastic
:D
 
lol you're missing the point
 
grrrr
 
@Daminark University of Chicago, eh?
Makes sense!
 
That sounds false
UChicago has never had semesters
And it only had a linear algebra class for a year
I know Nori taught difftop using sheaves though
And a junior faculty member this past year taught field extensions during linear algebra but barely talked about determinants
Ofc there's Peter May
 
 
2 hours later…
11:00 PM
taco
daminark
u there?
 
Yo
 
can i ask u graph theory question?
 
Sure! Though whether I'll be able to answer it or not... we shall see
 
If a tree is balanced how do i show that each of its partite sets contains a leaf? i mean i know if its balanced then half the edges belong to each set which imply by it being bi partite that each set has the same number of vertices
but id like to use a Hamiltonian path to prove my point but we havent learn that yet
 
What's a balanced tree?
Is it binary and symmetric?
 
11:15 PM
balanced just means you can split it into two subsets with equal number of vertices or edges
like K3,3 is balanced
k4,4 etc
like |W|=|V| where every vertex is in one or the other
 
Okay
 
can have higher balanced partites but all trees are bi partite
 
That's true
 
like i did really well in my first graphy theory course n i cant even read this dam assinment
and the intro to graph theory course isnt even a pre req
 
11:54 PM
0
Q: Question regarding Gambler's Ruin

Little RookieConsider a gambling process $(X_n)_{n∈\mathbb{N}}$ on the state space $S = {0, 1, . . . , N}$, with probability $p$, resp. $q$, of moving up, resp. down, at each time step. For $x = 0, 1, . . . , N$, let $τ_x$ denote the first hitting time, $τ_x := \inf\{n ≥ 0 : X_n = x\}$ Let $p_x := P(τ_{x+1} <...

 

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