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1:42 AM
Is delu/delx always just the reciprocal of delx/delu?
@LeakyNun can you help?
 
2:22 AM
Wondering if someone could take a look on my question math.stackexchange.com/questions/3355740/…
 
 
2 hours later…
4:35 AM
Does bisection method of finding roots converge linearly?
 
 
2 hours later…
6:27 AM
Hello everyone
By chance, is anyone a little versed in dimensional analysis/basic math modeling?
 
 
1 hour later…
7:42 AM
My prof defined radius of convergence of power series $\sum_{n=0}^{\infty}c_nx^n$ (where $c_n\in\Bbb R$) as $R=\sup\{r\ge0:(c_nr^n )\text{ is bounded}\}$. I have not seen this kind of definition before, can someone motivate. In particular, it seems very different from ${\displaystyle R=\sup \left\{|x|\ \left|\ \sum _{n=0}^{\infty }c_{n}x^{n}\ {\text{ converges }}\right.\right\}}$
 
 
2 hours later…
9:33 AM
Last night dream explore a finitary version of the limit operator
The following equality is frequently shown:
$$\text{an} s_n = \lim s_n$$
$\text{an}: \Bbb{N}^{<\omega}\to \Bbb{R} $ is a limit like operator that maps finite sequences to some real number
For example the an of $[1,1,3,5,7,7,3,3]$ is $5$
One can in theory formalise this easily by defining a topology on the set of natural numbers such that sequences sharing the same tail e.g. ... 5,5,5,5,... belong to the same equivalence class and then map their limits to another point not equal to its tail value
Such topology is expected to be quite non hausedoff as the neighbourhood of 5 has to contain 3
For a strict version, the very definition of limits has to be modified so that eventually constant sequences cannot converge to the constant value. The result may not be a topology
 
 
2 hours later…
11:23 AM
if $c_nr^n$ is bounded for $r \geq 0$, then for $0<r'<r$ we have that $\sum_{n=0}^\infty |c_n r'^n| =\sum_{n=0}^\infty |c_n r^n| \left(\frac{r'}{r}\right)^n \leq \sup_{n \in \Bbb N}(c_nr^n)\cdot \sum_{n=0}^\infty\left(\frac{r'}{r}\right)^n<\infty$. So $\sum_{n=0}^\infty c_n r'^n$ converges absolutely
On the other hand, if $\sum_{n =0}^\infty c_n x^n$ converges, then $\lim_{n \to \infty} c_n x^n =0$, so $c_nx^n$ is bounded and hence also $c_n|x|^n$ is bounded.
These two arguments give you two inequalities between the two definitions.
passing from $r$ to $r'$ doesn't change the sup, as we can still let $r' \to r$
 
 
3 hours later…
2:43 PM
I'm trying to show that $\Bbb{R}P^n$ is a smooth manifold, and I am wondering whether it reduces to showing that $\Bbb{R}P^2$. My intuition is that $\Bbb{R}P^n$ is a CW complex, formed by inductively gluing cells onto $\Bbb{R}p^2$ of increasing dimension, and $n$ cell is itself a smooth manifold Intuitively it seems that gluing smooth manifolds together should yield a smooth manifold...
On the other hand, gluing $[0,1]$ to $[0,1]$ at $0$ at some angle doesn't seem like it would be a smooth manifold...
 
How can the function $F(\mathbf{u})(t)=\mathbf{u}^{(n)}(t)+a_1\mathbf{u}^{(n-1)}(t)+...+a_n\mathbf{u}(t)$, where $\mathbf{u}\in U=C^n(\mathbf{R})$ (i.e. the space of all $n$ times differentiable continuous functions on $\mathbf{R}$) be a linear transformation (from $U$) to $V=C(\mathbf{R})$? In $F$, isn't the term $\mathbf{u}^{(n)}(t)$ not eliminable?
 
@user193319 If you glue together 3 half disks on the diameter at 120° from each other you have a CW complex which isn't even a topological manifold
For $\Bbb R\Bbb P^n$ there are easy explicit charts you can think about
 
 
2 hours later…
4:42 PM
How can I split the mod here. (4a^3-a^2)/3 mod 10^9+7
a is nearly of order 10^16
 
@schn I don’t really see what $u^{(n)}(t)$ being eliminable has to do with $F$ being a linear transformation. It’s not a bijection, certainly.
@prog_SAHIL is that mod 10^(9+7) or is it 7+10^9
 
@Semiclassical Yes, right.
 
(Projecting points onto a line is not a bijection, but it is nevertheless a linear transformation)
 
taking the second-derivative $u^{(2)}$ is a linear transformation
not to be confused with taking the square $u^2$
or applying the function twice $u(u(t))$
 
What is a basis for $C^n(\mathbf{R})$?
 
4:51 PM
I don't think I know of an explicit basis
not even for $C^0(\Bbb R$)
every vector space has a basis, but it often cannot be explicitly given
 
Yeah, cause the dimension n+1 should imply a basis of n+1 vectors, correct?
 
5:08 PM
@Semiclassical (10^9) + 7
 
of course not @schn
I don't know what you mean by n+1, but the n in $C^n(\Bbb R)$ is not the dimension
 
Hi. I had a dumb doubt
How do u find euler's theorem for z=log(x^2+y^2/(√x+√y)
For I take my function to be e^z and find partial derivative for it
Is this right?
 
5:34 PM
Euler’s theorem on homogenous functions? @Aladdin
(Better not be, since that function isn’t homogenous)
 
Yes
Why isn't e^z homogenous? ... I mean e^z=(x^2+y^2/(√x+√y) which I can simplify to f(y/x) ?
 
@MatheinBoulomenos can we characterize all additive functions from f.g. A-mod to N?
 
6:07 PM
@Semiclassical can you tell what's wrong with my way
 
Suppose I quadruple the values of x and y. Then x^2 and y^2 each increase by a factor of sixteen. But sqrt(x)+sqrt(y) doubles, so y^2/(sqrt(x)+sqrt(y)) will only change by a factor of 8
Hence x^2 + y^2/(sqrt(x)+sqrt(y)) does not increase by some overall factor.
Try plugging in numbers if you’re not convinced, eg x=y=1, x=y=2, x=y=4...
If it were homogenous, the factor of increase of z = f(x,y) from the first to the second should be the same as from the second to third.
Since it isn’t, this is not a homogenous function.
 
6:35 PM
(removed)
 
6:59 PM
@TedShifrin, I have a question
1
Q: Prove that $d(i,j)$ is a metric on $\{1,..,m\}$

Subhasis BiswasLet $w = \{w(i,j)\}_{1 \leq i,j \leq m}$ be an $m \times m$ symmetric matrix with non-negative real entries such that $w(i,j)=0$ if and only if $i=j$. Show that $$d(i,j)=\min\left\{\sum_{j=0}^{k-1}w(i_j,i_{j+1}) \mid k \geq 1, i_0=i, i_k=j,i_j \in \{1,2,...,m\}\right\}$$ is a metric on $\{1,...,m...

 
7:12 PM
Today I derived the fact that $\text{Aut}(\bigoplus\limits_{1\leq i\leq n}\mathbb{Z}/2\mathbb{Z})\cong S_{2^n-1}$
Now my brain hurts.
 
 
1 hour later…
8:30 PM
removes skull
 
9:02 PM
@TedShifrin I'm not a doctor but I'd advise against it
 
9:18 PM
demonic @Alessandro: I wasn't referring to mine. I was referring to our resident skullpatrol.
I just had to comment on a post that everything looked like total nonsense that has no mathematical meaning. I'm such a mean a__h_le.
oops, asterisks can't work.
hi @Erico, if in fact you're here.
 
9:37 PM
i am in fact not here
 
Nor am I.
 
@Rithaniel but that's wrong
 
10:11 PM
@AlessandroCodenotti you know how $\left(\dbinom n r\right) = \dbinom{n+r-1}{r-1}$ right
 
10:22 PM
I'm trying to prove the uniqueness a vector $\mathbf{r}$ such that its scalar projection on $\mathbf{u}$ is a real number $a$ and its scalar projection on $v$ is a real number $b.$ I'm working in $\mathbf{R^2}.$ I feel close - I've considered orientations and magnitudes but I'm not quite there yet. Anyone have any hints/suggestions for which direction to go?
 
10:38 PM
$\mathbf{u}$ and $\mathbf{v}$ are non-zero and not parallel to each other
 
infinite tetration of x converges for $e^{-e} \lt x \le e^{1/e} $
does infinite pentation converge in any range too?
 
@LeakyNun Is it not? The direct product of $n$ copies of the cyclic group of order 2. It's an abelian group with exponent 2, so any inner automorphism as well as any automorphism that maps any subgroup to itself is trivial. So any nontrivial automorphism must permute the subgroups of the group in some way. All subgroups are cyclic of order 2, and there exists one for each non-identity element.
There are $2^n-1$ such elements, so you get that the automorphism group is isomorphic to the symmetric group of $2^n-1$ elements.
 
11:15 PM
@LeakyNun I guess, why?
 
11:31 PM
@Rithaniel Surely $(\Bbb Z/2\Bbb Z\times\Bbb Z/2\Bbb Z)\times\Bbb Z/2\Bbb Z$ has a subgroup of order $4$?
The automorphism group of $(\Bbb Z/p\Bbb Z)^n$ is $\mathrm{GL}_n(\Bbb F_p)$, think in terms of invertible matrices to figure out how many elements it has
 
11:59 PM
apropos of some computations I'm doing: spherical trigonometry is one of those topics where, for all the neat stuff that's in it, the nagging question for me is "couldn't I do this just as easily with vectors and linear algebra?"
 

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