Spectral decomposition, version 1: For a symmetric $A \in M_n(\mathbb R)$, there exists an orthogonal $T \in O_n(\mathbb R)$ with $T^{\operatorname{tr}} \cdot A \cdot T = I_n$.
Spectral decomposition, version 2: Let $(V, \beta)$ be an Euclidean space and $\varphi \in \operatorname{End}(V)$ self-adjoint. Then there exists an orthonormal basis of $V$ (w.r.t. $\beta$) that consists of eigenvectors of $\varphi$.
I want to match each component from version 1 to version 2.
'symmetric $A$' corresponds to '$\varphi \in \operatorname{End}(V)$ self-adjoint', right?
$M_B(\varphi^*) = G^{-1} \cdot M_B(\varphi)^{\operatorname{tr}} \cdot G = M_B(\varphi)^{\operatorname{tr}}$ and since $\varphi = \varphi^*$, we get that $M_B(\varphi)$ is symmetric
From what I understand the Ihara zeta function, $\zeta_{\Gamma}(u)$, is obtained from a graph (or multigraph $\Gamma$) irrespective of an embedding. So, what changes if $\Gamma$ is embedded in $\Bbb R^3$ as a topological multigraph (i.e. edges are smooth curves)? How would this affect a meromorph...
modular: no, but it is not apparent to me even on a formal level that the purported definition on en.wikipedia.org/wiki/Ihara_zeta_function that zeta_Gamma(u) would depend in any way on an embedding of Gamma. the formula depends only on adjacency relationships in Gamma
so as a first stab at an answer to "what changes if Gamma is embedded in ..." i would guess "nothing"
i don't know. again, without looking into it, i don't see how "With smooth edges we can enrich gamma with a line bundle V gamma" defines a bundle, or even gestures toward some class of equivalent bundles under some unspecified notion of equivalence. in general, the same base can admit multiple line bundles, even if you allow yourself to regard different bundles as "the same" up to various natural notions of equivalence.
so how does "with smooth edges we can enrich" single one out, if Gamma is just an arbitrary graph?
offhand i'm not sure that i know very much, if anything, about bundles over base spaces that might not be manifolds, which embeddings of graphs in R^3 generally won't be
so i just don't know. it seems like an underspecified problem, or i am just outside of whatever audience can engage with this problem
Having a difficult time ATM dk why something seems different with me. Many of those interested in math have entirely different approach and values than I do
It is claimed in my book that it is easy to verify that $$A=\mathbb R\times\cdots\times\mathbb R\times (-\infty,a]\times\mathbb R\times\cdots\times\mathbb R$$with $a\in\mathbb R$ is a generating set of the Borel $\sigma$-algebra $\mathcal B(\mathbb R^n)$. I know $(-\infty,a]$ generates $\mathcal B(\mathbb R)$, but how can I verify that sets of the form $A$ generate $\mathcal B(\mathbb R^n)$?
words such as "second countable" or "Lindelof" come to mind
for example, if $U\subseteq \mathbb{R}^n$ and $U = \bigcup_{i\in I} U_i$ where $U_i$ are open, then we have a countable subfamily $I_0\subseteq I$ such that $U = \bigcup_{i\in I_0} U_i$
but you can justify those unions are countable in whichever way you like, of course
my way would be by saying that every subspace of $\mathbb{R}^n$ is a separable metric space, and so it's Lindelof
not saying this is most efficient, but it's what "jumps at me"
@Jakobian sorry to ask again, but I still struggle with putting the pieces together. I don't see the connection between what you wrote here and the set $A$. Do you think you could tell me once again?
Lindelof to me means every open cover of a second-countable metric space $X$ has a countable subcover.
@Jakobian Ok, so do you think my book means the same? I.e. that sets of the form $A$, with $(-\infty,a]$ not only in one fixed position (but in the first, second, etc.) generate the Borel sigma algebra?
I'm not really sure how to write down such a collection of sets efficiently :)
probably would have been much better to write $$A=\mathbb R\times\cdots\times\mathbb R\times U_i\times\mathbb R\times\cdots\times\mathbb R$$with $U_i=(-\infty,a]$ for $1\leq i\leq n$ and $a\in\mathbb R$.
@Jakobian by the way, how come we can replace $(-\infty,a]$ with $(-\infty,a)$?
@Jakobian for $n=2$, we have $A=\mathbb R\times (-\infty,a]$. And you are hinting at taking the union over $j\in\mathbb N$ of $\mathbb R\times (-\infty,a-1/j]$ to obtain $\mathbb R\times (-\infty,a)$, is that correct?
finite intersections of them form a basis for the topology of $\mathbb{R}^n$, and every open set of $\mathbb{R}^n$ is a countable union of such sets
@Jakobian I'm still stuck on a little detail. Here's what I've understood so far. We have $\sigma(\mathcal C)$ where $\mathcal C$ is the collection of all sets of the form $A$. You've shown that we can replace $(-\infty,a]$ with any open set of $\mathbb R$ in the definition of $A$. So since $\sigma(\mathcal C)$ is a $\sigma$-algebra, it contains the subbase, but it doesn't contain the open sets, right?
Those might be arbitrary unions of the subbase elements, not countable unions. Grateful if you could clarify :)
If $U$, $V$ are orthogonal matrices and $S$ diagonal and $u_i$, $v_i$ are the columns, then why is $$U \cdot S \cdot V^{\operatorname{tr}} = \sum_{i = 1}^n u_i \cdot s_{ii} \cdot v_i^{\operatorname{tr}}?$$
@Jakobian yes, I agree with that finite intersections of the subbase elements form a base, and those are in $\sigma(\mathcal C)$, but some open sets might be arbitrary unions of the base elements, no?
for example, if $U\subseteq \mathbb{R}^n$ and $U = \bigcup_{i\in I} U_i$ where $U_i$ are open, then we have a countable subfamily $I_0\subseteq I$ such that $U = \bigcup_{i\in I_0} U_i$
it feels odd to me because I literally told you the answer before, so I'm not sure if its an issue of not being able to connect two things with each other, or just really bad memory
we even had discussion that justified why this is true
proved that when $X$ has order topology, the order topology on $Y \subseteq X$ for a convex subset $Y$ is the same as the subspace topology of $Y$. Going via basis took a lot of cases, like 10+ cases
Let $\varphi: \mathbb R^n \to \mathbb R^n$ with $y \mapsto T \cdot y + u$ where $T$ is an orthogonal matrix and $u$ a vector. Then for all $x \in \mathbb R^n$, we have $x = \varphi(y)$ for some $y$. Why is this? My book just claims it without any proof.
We know that $||\varphi(x) - \varphi(y)|| = ||x - y||$ and $||T \cdot x|| = ||x||$.
ilike: something like that will work. it is still using finite dimensionality in that the orthogonality condition is that T^t T = I, while the verification of varphi(y) = x, when y is given by that formula for the putative phi^{-1} applied to x, uses the operator identity T T^t = I, and while AB = I does imply BA = I when A, B are operators on R^n, this would not necessarily follow for operators A, B on more general spaces.
so there is no escaping jakobian's "dimension issues" even if one offers a symbolic formula for the inverse mapping, they are baked into that formula
Hi everyone. What techniques can I use to analyze the behavior of the series $\sum_{n=1}^{\infty} e^{n + \frac{1}{3^n}} - e^n$ using the asymptotic comparison test?
i would be inclined to express the nth term as e^n (e^(1/3^n) - 1) and to recall that if f is differentiable at 0 and a_n is a nonzero sequence that goes to 0 then (f(a_n) - f(0))/a_n could be expected to go to f'(0)
and i can do this despite having no idea what the asymptotic comparison test is
i'm guessing some form of what my college calculus book (stewart calculus early transcendentals) would have called the limit comparison test
but i don't need to have that guess validated to have those thoughts
(Much of that ire is related to the fact that I am at my sister's house right now, and without paper and pencil to work things through... I have spent a lot of time with math.stackexchange.com/q/5017458, but I really need paper-and-pencil to make real progress.)
Yeah, that too. I am to the point where I think that brute force one a computer might be the right way to go.
I thought I had a nice inductive argument based on the number of turns remaining, but it turns out that it also depends on how many more red balls there are than blue balls (huh huh huh... you said "blue balls").
I can't keep it all straight without paper and pencil. :(
i would encourage students of this kind of thing to be inclined to see little or no difference between "using asymptotics" and something like stewart's limit comparison test. i think of them as the same thing in different languages
not one as a route to the other
but it is certainly a valuable insight for students to be able to see "~" on wikipedia and translate it into limit-ese
You can solve this as a Markov decision problem (stochastic dynamic programming) as follows. Let $D = \{\text{blue}, \text{red}\}$ be the set of two possible draws. Given $n$, let the states be
$$\{(k,b,r,d): k \in \{0,\dots,n\}, b \in \{1,\dots,n+1\}, r \in \{1,\dots,n-k+2-b\}, d \in D\},$$
wh...
You can solve this as a Markov decision problem (stochastic dynamic programming) as follows. Let $D = \{\text{blue}, \text{red}\}$ be the set of two possible draws. Given $n$, let the states be
$$\{(k,b,r,d): k \in \{0,\dots,n\}, b \in \{1,\dots,n+1\}, r \in \{1,\dots,n-k+2-b\}, d \in D\},$$
wh...
