12:09 AM
Working with the normed vector space $(\mathbb R^n,\mathbb R, ||\cdot||)$, how can one justify that $||x||_1\geq ||x||_2$? In particular, I cannot justify this step $\sqrt{ \sum_{i=1}^{n} | x_i |^{2} } \leq \sum_{i=1}^{n} \sqrt{ | x_ i |^{2} }$.

verify $n=2$ by calculation and then use induction

@johnny09 you can deduce this from the triangle inequality for $\| \cdot\|_2$. Let $v_i=(0,\dots,x_i,\dots,0) \in \Bbb R^n$ such that it's $0$ everywhere except at the $i$-th place. Then we get $\sum_{i=1}^nv_i = (x_1,\dots,x_n)=x$ and we get from repeatedly applying the triangle inequality $\|x\|_2 =\| \sum_{i=1}^n v_i \| \leq \sum_{i=1}^n \|v_i\|_2 = \sum_{i=1}^n \sqrt{|x_i|^2}=\|x\|_1$

12:30 AM
@Thorgott thanks for the hint. I think i got it although I used the fact that for nonnegative numbers $a,b$, we have $\sqrt{a}+\sqrt{b}\geq\sqrt{a+b}$
@MatheinBoulomenos not sure I understand. we have $x\in\mathbb R^n$ so how can we define $v_i$?

$x=(x_1, \dots, x_n)$
and then define $v_i$ as $0$ everywhere except at the $i$-th place where it is $x_i$
sorry, forgot to introduce the notation for $x_i$

2 hours later…
2:06 AM
290

Suppose we have data set $(X_i,Y_i)$ with $n$ points. We want to perform a linear regression, but first we sort the $X_i$ values and the $Y_i$ values independently of each other, forming data set $(X_i,Y_j)$. Is there any meaningful interpretation of the regression on the new data set? Does this ...

I have no words

2 hours later…
4:04 AM
Let $G$ be a group and let $\mathbf{L}(G)$ denote the subgroup lattice of $G$. Consider the set $\mathcal{S}:=\{\mathbf{L}(H):H~\text{is a subgroup of}~G \}$ and let, $$\mathscr{M}:=\{H:\mathbf{L}(H)~\text{is a maximal element of}~\mathcal{S}\}$$Is it true that $G\cong \displaystyle\bigoplus_{H\in \mathscr{M}} H$?

If $L$ is a finite separable field extension of the field $K$, then $Spec(L)\to Spec(K)$ is etale. Does this actually classify all etale morphisms $Y\to Spec(K)$ to a field? Or are there high dimensional etale covers etc?
I guess I expect there to be lots of other etale morphisms, like many smooth schemes such as $\Bbb A^n_K$, but perhaps someone knows one extra condition that makes the separable field extensions the entire class

4:56 AM
@user616128 etale = smooth + relative dimension 0

@loch So $Y$ has to have dimension $0$???

yes

How the hell have I not seen that definition :'(
How do I see that to be equivalent to smooth and unramified?

The unramified contains relative dim 0 somehow

Why is A^1_K not etale over Spec(K)?

@user616128 Intuitively, smooth means your morphism is surjective on tangent spaces and unramified means is injective on tangent spaces

But if I take unramified as locally of finite type, which seems to be satisfied

@user616128 this map is not unramified

and giving K\to O_{A^1,x}/m_x as a separable field extension
Then that seems satisfied too
Why not, with the definition on wiki for unramified en.wikipedia.org/wiki/Glossary_of_algebraic_geometry#U

5:04 AM
@RockDock interesting question

@RockDock You can show this by taking a standard cubic cover
and showing that multiplying by k^{-1} sends cubic cover to cubic cover

@Hansie Thanks .I am trying that question for long time but unsuccessful

And then you can see that the cubic covers have the same volume

@user616128 Never heard about cubic cover .Can we solve it using basic definitions .I am beginner in measure theory

By cubic cover, I mean you are computing the exterior measure, where in one dimension, a cubic cover is just a cover by closed intervals
But your statement about dilation holds in R^d with cubic covers, and the same argument described above

5:09 AM
@user616128 @RockDock said he want's to solve it using basic definitions and i think it can be solved using basic definitions only .
5

Bhowmick, Hansie, RockDock, are you all the same person
/ all working on the same homework assignment?
Just noticing the instant starring, and same typing style
What more basic definition is there than the exterior measure here??

@user616128 try taking $x$ to be the generic point. then you don't get a finite extension (and note that this is precisely because $\dim \mathbb{A}^1 = 1$!)

Crap! Good point
For some dumb reason, I've just been checking closed points
No wonder i'm super confused

@user616128 actually if you check closed points you'll also see a problem - the ideal $\mathfrak{n}$ is generated by $f^{\#} (\mathfrak{m})$. In this case $\mathfrak{m} = 0$, so $\mathfrak{n} = 0$ (hence is not maximal)
this is closer to the issue of induced map on tangent vectors not being injective

5:26 AM
pathetic
@user616128 they are the same person

@LeakyNun Well they got what they wanted I guess
Since they self upvoted their question, and they got an answer
Oh, I just realised there are 4 of them lol
@loch Apparently I didn't read this properly at all... Thanks (for some reason I took it to be the max ideal of O_Y,y and it contraction, rather than max ideal of O_X,x and its extension - oh well I better sleep before I make more mistakes)

A couple of weeks ago I was thinking that a person starring their own messages should be disallowed, but I hadn't considered people making multiple accounts.

@user616128 i didn't read it properly at first glance too - which is why i wasn't really happy with the generic point answer because it didn't give the intuition of injectivity on tangent vectors

@Rithaniel You already can't star your own message right? Unless you're a moderator
2

Ah, I thought you could.
Well, that's just me being dumb, then.
I recall there was a situation where a person starred, like, 8 sentences in a row.

5:53 AM
@LeakyNun One of our friends asked this question and rest of us were making fun of him here .Also it is textbook problem not homework or assignment

right
why don't you tell neraj to talk to us
if you still remember that account's password, that is

what you mean ?
@LeakyNun

@loch how do you say Cholesky?

@LeakyNun ko-les-key?
i'm not sure
maybe cho-les-key
idk

6:19 AM
So, I recently read a claim that the only multiplication rule which makes $\mathbb{Q}/\mathbb{Z}$ into a ring is zero multiplication. The proof accompanying this claim relied on the assertion that for $a,b,c,d\in\mathbb{Z}$ we have that $\frac{a}{b}\cdot\frac{c}{d}=a\cdot \frac{c}{bd}=0\cdot\frac{c}{bd}=0$, but something about this isn't sitting well with me.
I'm not opposed to the notion of $\mathbb{Q}/\mathbb{Z}$ only being a valid ring with zero multiplication, but the proof relies on multiplication rules derived from the rationals. It doesn't disprove the existence of multiplicative rules developed independently from rational multiplication, does it?

@Rithaniel multiplication by integer is unique once you have addition

@user616128 is kE^c and (kE)^c same thing ? definition is kE = {x : x/k ∈ E} . E^c means complement of E .
I guess both are same

And this multiplication is inherited by $\mathbb{Q}$ and therefore $\mathbb{Q}/\mathbb{Z}$ would only have two possible multiplicative rules? (namely zero multiplication and a multiplicative rule derived from the original, unique integer multiplication, which in this case is also zero multiplication)
Or do you mean in the abstract sense? Ie. if $a\in R$ where $R$ is a ring, $ka$ is unique (where $k\in\mathbb{Z}$).

6:38 AM
right
more precisely, if $(R,+,\times_1)$ and $(R,+,\times_2)$ are ring structures, then $k \times_1 a = k \times_2 a$ for all $k \in \Bbb Z$ and $a \in R$

Okay, then any ring modulo the integers has only zero multiplication? If so, that's kind of wild.

I'm not sure

Well, in order for $R/\mathbb{Z}$ to even be valid, $\mathbb{Z}$ must be an ideal of $R$. That's a necessary statement.

no, we're taking quotient as groups
$\Bbb Z$ is certainly not an ideal in $\Bbb Q$

Ah, right. Then for it to be valid, $\mathbb{Z}$ must be a normal subgroup of $R$ (also containing the commutator subgroup, since we're thinking about turning this into a ring).
Could this be extrapolated to a normal subgroup isomorphic to $\mathbb{Z}$, perhaps?

6:52 AM
rings are abelian groups
every subgroup is normal

Indeed, but since we're taking quotient as groups, we could just consider the case when $R$ is a group.
However, I don't know if this is going anywhere.
(and I probably need to get some sleep, at that)

7:15 AM
@MatheinBoulomenos if a degree $n$ polynomial has $n+1$ zeroes then it is $0$

2 hours later…
9:10 AM
hi chat

hi
we know that a function f is measurable if and only if inverse image of every open set is measurable .Is it true for closed set also ? , i.e if f is measurable then inverse image of every closed set is also measurable

yeah if i remember correctly
a little more generally, a function is measurable iff the inverse image of a measurable set is measurable
thats the definition
closed sets are measurable..

Yes, you can check measurability on any family of sets that generates the Borel $\sigma$-algebra

well, not all closed sets i suppose

All closed sets are measurable
2
All Borel sets are measurable and the Lebesgue $\sigma$-algebra is way bigger than the Borel one

9:25 AM
yeah i take that back

so @AlessandroCodenotti so we can say if f is measurable then inverse image of every closed set is also measurable

Yes
I'm assuming you're talking about measurable functions $\Bbb R\to\Bbb R$, with the Borel $\sigma$-algebras

yes thanks .

Similarly another sometimes useful criterion is that $f$ is measurable iff $f^{-1}((-\infty,a])$ is measurable for all $a$
You can also use $(-\infty,a)$, any family of sets that generates the Borel $\sigma$-algebra as I was saying above

yeah i got what you're saying

9:35 AM
Why does $k {n \choose k}=n {n-1 \choose k-1}$, not from the formula ?

what does not from the formula mean
its exactly from the formula

If we choose k elements from n, then chose one among that k, the number of choices would amount to the RHS
But if choose one among the n elements, "isolating" it, then choose k-1 among the rest, we would end up with k elements, one of which is "marked" by our first choice..
I want to see how other people put it

if f^2 is measurable then f is measurable ? If not what is the counter example without using borel set . Of course other way around (i.e f is measurable then f^2 is measurable is trivial) is easy .
there is counter example using borel so i am assuming in general it is not true

9:54 AM
8

Help finding a combinatorial proof of $k {n \choose k } = n {n - 1 \choose k -1}$ I have expanded it this far: $$\frac{k \cdot n!}{k!(n-k)!} = \frac{n \cdot (n-1)!}{(k-1)!(n-k)!}$$ but then I am stuck from where to go from there.

"Combinatorial interpretation"... way too fancy

so you are looking "from the formula"?
the identity you have on there is true

No, I was just complaining about the term, I should've used it instead.

but you expanded the combinatorial notation
"the formula"
why?
if thats not the proof you were after

Oh no, if you check the post you'll find that the first answer is the other proof
That's just the question

i think with vitali set we can prove that f^2 is measurable then f need not to be
yeah i got it

1 hour later…
11:07 AM
Here on P12-19, theorem is given that says that if $f$ continuous inside a connected open domain D, then $f$ has antiderivative iff $\int_Cf \,dz=0$ for closed curve $C$ lying in D. Then, in slides P20-22, $\int_C z^n\,dz=0$ is noted for $n\ne1$. My question is for negative n's, we can't apply the given theorem, right? It is zero for some other reason, right?
Oh! Even $z^n$ for $n>0$ is not continuous, right? So, how do we apply that theorem?

11:29 AM
Let $(X,\Sigma,\mu)$ be a measure space and $f : X \to \Bbb R$ measurable and $g : X \to \Bbb R$ integrable such that $|f(x)| \le g(x)$ a.s., then is it true that $f$ is also integrable?
@Silent it's continuous as long as you exclude 0 from the domain

@LeakyNun Thank you. Is $z^n$ continuous on entire $\Bbb C$ for $n\ge0$? And $z^n$ continuous except $0$ for $n\le2$?

11:46 AM
@LeakyNun Isn't that dominated convergence with a constant sequence of functions?

12:03 PM

Only of the dominating function

so it must have been a lemma...
unnamed
che faro

is this really true lol

I'm not following

2 hours later…
1:40 PM
is measure theory if liminf = limsup = infinity at a point , then can we say that sequence converges at that point or limit has to be finite ?

Let $M_{2\times2}(\Bbb R)$ be the vector space of all $2\times 2$ matrices over $\Bbb R$ and let $W_1=\{\begin{pmatrix}x&y\\ 0&x\end{pmatrix}:x,y\in \Bbb R\}$ and $W_2=\{\begin{pmatrix}x&y\\ z&0\end{pmatrix}:x,y,z\in \Bbb R\}$. What is dimension of $W_1\cap W_2$?

I think its 2, the sol i have says its 1.
Am i correct?

2:09 PM
No, You are wron@silent
Suppose $A\in W_1$, then diagonal elements must be same.
also note that$a_{21}=0$
If this $A\in W_2,$ then $a_{22}=0$
So, $x=0$ the diagonal elemnts must be zero
similarly, If $A\in W_2$ unless $z=0$ and $x=0$ this is not qualify to be a member of $W_1$.
So $W_1\cap W_2=\{\begin{pmatrix}0&y\\ 0&0\end{pmatrix}:y\in \mathbb R\}$

2:24 PM
Will x^(1/x), where $x \in \Bbb{+Q}$, always output an irrational number except for x=1?

2:54 PM
If $f(X) \in \Bbb{Q}[X]$ is irreducible and $\alpha$ is some root of $f$ in a field extension of $\Bbb{Q}$, is it true that $|\Bbb{Q}[\alpha] : \Bbb{Q}| = \deg f$?
I don't think so...I think we can only say that the degree of the field extension is at most $\deg f$, right? Because there could be a smaller polynomial in $\Bbb{Q}[X]$ for which $\alpha$ is a root...right?

@N.Maneesh That was eye opening! Thank you so much
@N.Maneesh, Let $T:\Bbb R^n\to\Bbb R^m$, $m>n$, then $\dim (T(\Bbb R^n))\ge n$. Is this statement correct?
I think its wrong by rank-nullity but sol says otherwise!

@Silent Yes I also think so, By rank nullity theorem

thank u

3:09 PM
if you get the solution of CSIR questions. Can you please send to that group?@Silent
I will also do the same

3:26 PM
"which is the categorical version of what is known as Cramer’s rule in undergraduate courses in linear algebra" :o

hi chat

3:38 PM
Is $x^5 - 2$ the minimal polynomial for $2^{1/5}$ over $\Bbb{Q}$?

Will x^(1/x), where $x \in \Bbb{+R}$, always output an irrational number except for x=1?

It's continuous and non-constant, so no

In mathematics, Volterra's function, named for Vito Volterra, is a real-valued function V defined on the real line R with the following curious combination of properties: V is differentiable everywhere The derivative V ′ is bounded everywhere The derivative is not Riemann-integrable. == Definition and construction == The function is defined by making use of the Smith–Volterra–Cantor set and "copies" of the function defined by f ( x ) = x 2 sin ⁡ ( ...
wtf? first time I've heard of that function ...

3:57 PM
Anyone here?
can someone please check this out?

https://math.stackexchange.com/questions/3195987/slightly-alternative-proof-to-the-converse-part-of-cauchys-general-principle

4:58 PM
@user193319 yes

5:12 PM
@LeakyNun How does one show this?

just show that $x^5-2$ is irreducible over $\Bbb Q$

Oh, by Eisenstein's criteria?
So, if there were a smaller degree polynomial, it would have to divide $x^5 - 2$; but since this poly. is irreducible, this is a contradiction.

How do I prove $\frac{1}{2nx+1}$ is not uniformly convergent in (0,1)
?
I am not able to apply weirstrauss M-test

@Mathphile If you restrict to rational numbers $x>1$, yes you will get only irrational numbers.

since it is strictly decreasing function

5:23 PM
@N.Maneesh Well, what do you need to show, by definition?

Nope, any method

Recall the definition

@karl

Ok, then come up with the logical negation of the definition.

If x and y are complex numbers such that x+y and xy are algebraic number then how to prove that x and y are also algebraic

5:31 PM
Hi, could anyone help me understand the difference between the questions 10 and 11 in Sheldon Axler's LA Done Right?
Link to questions 10 and 11 http://linearalgebras.com/5b.html
To me they both seem exactly the same, I am sure I am missing something very subtle but I can't figure it out :/

Is it correct to call this piecewise sequence constant ? $f_n(x)=\begin{cases}\frac{i-1}{2^n}, f(x)\in X\\ n,f(x)\in Y \end{cases}$ ? Because in the end $\frac{i-1}{2^n}$ are numbers and is a number $n$ too.

@KarlKronenfeld could you prove this?

@Mathphile My work is convoluted, but based on the classic proof that the square root of 2 is irrational.
@GaurangTandon The forward implication in Question 11 looks specific to the complex numbers (rather than an arbitrary field F).

@KarlKronenfeld i can prove it for $x \in \Bbb{+Z}$
but $x \in \Bbb {+Q}$ is another story

@KarlKronenfeld yes but the result of q11 and q12 seems to be that it should only hold for complex fields, not any arbitrary field. however, q10 does seem to be holding for arbitrary fields, hence my confusion

5:43 PM
@GaurangTandon Yep, so examine forward implication

oh, so only the forward implication holds in q10, while in q11, the backward also holds
cool thanks @KarlKronenfeld

@Glator Do you know alg. nums are alg. closed?

For what values of $p>2$ is $p^p-1$ prime?
are there a finite number of these values?

@KarlKronenfeld No
i have not read that concept

5:59 PM
@user178403 No. Constant sequence $\neq$ sequence of constant functions (the latter terminology applies)
@Glator Ah, basically if you had that, you could set up a polynomial with the given algebraic numbers as coefficients and see what the roots turn out to be.
@Mathphile No, it is never prime for integers $p>2$. Hint: $p$ must be even for $p^p-1$ to be prime.

@Glator Since $x+y$ and $xy$ are both algebraic, $z^2 - (x+y)z + xy$ is a polynomial in $z$ with algebraic coefficients.
But that is just $(z-x)(z-y)$!

That's why I wanted the algebraic numbers to be algebraically closed. :)

6:25 PM
> YouTube have made it clear that Fractal Art is not welcome on the platform
Turns youtube into a dedekind finite set to be peeled like onions

What's an easy example of a discriminant using a 2 by 2 determinant and it's algebraic number field?

@user76284 How to prove that root of non zero polynomial with algebraic coefficients is algebraic ?
Because what you said is true only if above is true
Or is their any other method ?

7:26 PM

7:43 PM
Any idea on what I did wrong for the straight line segment of this contour integral?
@TedShifrin just so you see this when you come around
@Semiclassical! how’s your complex analysis ;)

8:25 PM
1

Find the the splitting field of $X^5-2$ over $\mathbb{Q}$ and find it's degree. My approach: The roots of $X^5-2$ are $\{\sqrt[5]{2},\sqrt[5]{2}\omega,\sqrt[5]{2}\omega^2, \sqrt[5]{2}\omega^3, \sqrt[5]{2}\omega^4\}$ where $\omega=e^{2\pi i/5}$. It's quite easy to show that splitting field of $X... Why is$|\Bbb{Q}(2^{1/5},e^{\frac{2 \pi i}{5}}) : \Bbb{Q}|$at most$20$? Because$\Bbb Q(2^{1/5})$and$\Bbb Q(e^{2\pi i / 5})$have degree$5$and$4$over$\Bbb Q$respectively So, the degree is always the least common multiple? How does one prove that? The solution I am working through says the following: "The degree of$\Bbb{Q}[\xi, \alpha]$is$20$, since it must be divisible by$[\Bbb{Q}[\xi] : \Bbb{Q}]$and$[\Bbb{Q}[\alpha] : \Bbb{Q}] = 5$. @ÍgjøgnumMeg But I don't follow this logic. We also need to show that the degree of$\Bbb{Q}[\xi, \alpha]$is at most$20$; but I don't see this... tower law What's that? I know that$|\Bbb{Q}[\xi,\alpha] : \Bbb{Q}| \le (\deg (x^5 - 2))! = 5!$; but this isn't helpful. 8:56 PM Is a surface of the form z = c, where c is constant, considered to be closed? Is it compact?$z=c$represents a plane, right? If so, it won't be compact, because$\Bbb{R}^2$isn't compact. yes hmm, then how am I supposed to find the flux of a vector vield$xi +yj +zk/(x^2+y^2+z^2)$through a plane Not sure...I don't remember Calc 3. ahh, I need to use the dirac delta function @user193319 is it closed though? Yes, because you can write the plane$z=c$as the preimage of a singleton under a continuous function. @topologicalmagician The function being$f(x,y,z) = z$. 9:08 PM my topology isn't really good. Can you direct me to a good reference? So,$f : \Bbb{R}^3 \to \Bbb{R}$given by$f(x,y,z) = z$is continuous, right? And continuity means that, for any closed set$C$, the preimage$f^{-1}(C)$is closed (alternatively, we could have replaced "closed" with "open"). The set$\{c\}$is closed in$\Bbb{R}$. As a set, what is$f^{-1}(\{c\})$? 9:12 PM$\mathbb{R}$? No,$f^{-1}(\{c\})$will be a subset of$\Bbb{R}^3$. To get you started,$f^{-1}(\{c\}) = \{(x,y,z) \in \Bbb{R}^3 \mid f(x,y,z) \in \{c\} \}$(if necessary, look up the definition of 'preimage'). 9:44 PM Hold on, so for a surface z=c>0, does the surface contain the origin in its interior? Why does equating real parts and imaginary parts give different values for the integral? Hi, I am strugling with the proof of the following theorem: If S is saturated multiplicative closed set in a ring R than S is the union of prime ideals. My problem is in the direct implication, the second inculsion, i.e. @topologicalmagician If I understand you correctly, the answer is no. This is because the interior of some set$S$is a subset of$S$. Since the plane$z=c > 0$does not contain the origin, its interior cannot contain it. Then it is contained in its exterior, right? @user193319? No, because the interior of the plane is contained in the plane. In fact, I believe the interior of a plane in$\Bbb{R}^3$is empty. 10:05 PM @user193319 @topologicalmagician You're using "interior" in different ways. In the calculus setting, we do say that the origin is in the interior of the sphere$x^2+y^2+z^2=1$. This is not the topological use of the word. We're saying that the sphere bounds a ball and the origin is inside that ball. Hi :) @TedShifrin Hey Ted, so does that mean that the origin is in the interior of a plane z=c >0? No, a plane doesn't bound a region ... hmmm, i'm trying to compute the net flux of the inverse square vector field over a plane z = c > 0, but i'm stuck You have to do an explicit calculation. It doesn't follow from the Divergence Theorem. 10:07 PM can't I use the dirac delta distribution? Parametrize by polar coordinates and do the integral. and then apply the div theorem? No. You can only apply the Divergence Theorem when the surface bounds a (finite) region. @TedShifrin that's what I tried, but i'm not sure what the bounds of my integral should be It's an improper integral. You're integrating over the entire plane. In polar coordinates, you'll have$0\le r<\infty$. 10:09 PM and angle would be 0 to 2pi? Yup. okay. But what about the fact that the inverse square field is undefined at the origin, shouldn't I take that into account? and so the divergence at the origin is undefined No. You're integrating over a plane that doesn't go through the origin. I don't care about divergence. Problem: Let$\Gamma$be a tree, and$m \ge 2$. If$\Gamma$is an$m$-tree (all vertices have valence of$m$), then$\Gamma$is infinite. Proof: Suppose that$\Gamma$is finite and yet every vertex has valence$m$. Let$|V(\Gamma)| = n$. Then$|E(\Gamma)| = mn$. But since$\Gamma$is finite, we also know$|V(\Gamma)| = |E(\Gamma)| + 1$or$n = mn + 1$, which is a contradiction... Does this seem right? You're doing the surface integral directly. No Divergence Theorem (for the THIRD time). 10:11 PM I understand thank you soo much I don't believe your count on edges, @user193319. You're welcome, @topologicalmagician. just out of curiosity, may you please just clarify what the difference between interior in the calculus setting and the topological setting is? @TedShifrin Oh, wait. Switch everything. Let$|E(\Gamma)| = n$. Then$|V(\Gamma)| = mn$. Is that better? Nope...that's not right either. You should just say "inside" the surface in the calculus setting. The point is that the surface bounds a region and the point is an interior point (in the topological sense) of the region. Maybe it is...I've never been good at counting... 10:13 PM You're still missing something crucial, @user193319. Ahh, I see. Also, what if my region was z=c<0, I should still get the same answer as if z=c>0, right? Yes, by symmetry. @TedShifrin I really can't thank you enough. I wish you were the lecturer in all of my modules :P LOL, all I can do is offer you my YouTube videos, @topologicalmagician. @TedShifrin I was watching them, but I stopped because I've been stressing out the material that i'm being taught im gonna watch them again soon after im done, and I also bought your book statistics is killing me slowly lol 10:18 PM Statistics isn't as scary as math :P Statistics with programming is whats killing me Knowing both programming and some statistics is a smart idea in this day and age. True, I agree but in stats its not as rigorous as my other math modules Well, no one says every math course should be like a Rudin analysis course. @TedShifrin I can't seem to figure it out. I drew a graph with$n := |V(\Gamma)| = 4$and$m= 3$. I got$6$edges, but I don't know how to relate this to$m$and$n$. 10:21 PM yeah, but I guess I enjoy Rudin like courses, I guess that's why im going into pure mathematics lol My main observation, @user193319, is that each edge joins$2$vertices. Oh, so$|E(\Gamma)| = 2n$? Slow down and think. Also: Can there be more than one edge joining a given pair of vertices? @TedShifrin How did you use to study for mathematics? To answer your question, I don't think so because$\Gamma$is a tree, and what you describe sounds like a cycle. 10:23 PM I don't remember ancient history :P OK, @user193319, so that's good. Hmm...not for me...it seems to complicate the counting process... For an improper integral, I can still change the bounds of the integral, if I do a u-substitution, right? So each additional edge adds an additional vertex. @topologicalmagician: You mean to do the$r$integral by doing a substitution? Yes. yeah okay, perfect So, I start with a vertex and connect an edge to it; this introduces another vertex in the graph...Okay I buy that. 10:28 PM So each vertex after the first must yield$m-1$new vertices. Okay, I see that. Does this help me with determining$|E(\Gamma)|$? Sure. My complaint earlier was that you weren't taking into account that each edge accounts for two vertices, not one. So you have to count carefully. I'm still kind of confused. If$|V(\Gamma)| = n$, what does$|E(\Gamma)|$equal? Ok, so to make sure I understand things correctly, I should only consider the divergence theorem when my surface bounds a finite region, right? Baffling only because I don't understand why he didn't tell them to go #$#%# themselves but still it is easter look for the algorithm Gauss proposed it's classic en.wikipedia.org/wiki/Computus

10:46 PM
Guys, if I have multiple linear model what would be preffered, a response variable that depends on 2 or 3 variables?

I suppose it depends on what you want. Though, all other things being equal, I'd want 2 dependent variables.

What would happen to the n dependent variable model if I added a n+1 variable?
it could affect the variance of the model, right?
i
i'm not sure why though
I suppose its because that it causes more scattering
but I don't know if that's true

Well, it depends on how the new variable interacts with existing ones, but something we might be able to say is that it increases the solution space. (Ah, is this statistics? I glazed over statistics when I took it)

@TedShifrin Is $|E(\Gamma)| = n(m-1)$?
Really, I'm asking anyone. $n$ is the number of vertices and $m$ is the valence at each vertex. Given this setup, is it true that the number of edges is $n(m-1)$?

by valence you mean degree?

10:57 PM
Yeah, I think so. The number of edges containing the vertex.
My formula isn't right though...I just tested it against an example.

what type of graph is it?

Here's the problem I'm working on: Problem: Let $\Gamma$ be a tree, and $m \ge 2$. If $\Gamma$ is an $m$-tree (all vertices have valence of $m$), then $\Gamma$ is infinite.
I'm doing the proof by contradiction.
And somehow I am suppose to contradict the formula $|V(\Gamma)| = |E(\Gamma)| +1$.

not sure, i don't really know anything about trees, sorry

Well, in a tree you can't have cycles, right?

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Let $\Gamma$ be a tree, and $m \ge 2$. If $\Gamma$ is an $m$-tree (all vertices have valence of $m$), then $\Gamma$ is infinite. My idea is to prove it by contradiction. Suppose that $\Gamma$ is finite yet every vertex has a valence of $m$. I suspect that I should be able to contradict the f...

11:09 PM
how can we check that the jacobian $\partial f / \partial x= \begin{bmatrix} -1+x_2 & x_1 \\ -x_2 & 1-x_1 \end{bmatrix}$ is not uniformly bounded on $\mathbb R^2$?
do we take the norm of the jacobian and try to find an upper bound with a constant?

Well, suppose you have an $n-$tree with at least $2$ vertices. These vertices must be connected by a path and they must also each have at least one more additional path connecting them to vertices not yet accounted for in our description of the tree (nor can they be connected to the same vertice, as that would create a cycle). The vertices these paths connect to must similarly be connected to yet more vertices, etc.
So, question: Suppose you wanted to show something is a field, but don't want to go through and prove that each of the field axioms hold. What would be a few equivalent statements to "$X$ is a field?"

11:43 PM
@Rithaniel for the group part maybe show left inverse, left identity and associativity
and closure

Well, yeah, I imagine showing the underlying group to be an abelian group will need to check off the group axioms.

but its not a trivial result that if a set satisfies associativity, left identity, left inverse and closure then its a group

Indeed, but if you show that the operation is commutative you're golden.