TITLE: What is the difference between these two sets of symbols?
QUESTION [1 upvotes]: I am considering two sets of symbols. Are these two sets different or the same? Why?
\begin{align}(1)\qquad&\lim_{n\to\infty}\sum_{j=1}^n\\
(2)\qquad&\sum_{j=1}^\infty\end{align}

REPLY [1 votes]: The series (2) in the form
  $\sum_{j=1}^{\infty}a_n$
  admits two different interpretations.

It can be seen as sequence  of partial sums
  \begin{align*}
\sum_{j=1}^{\infty}a_j:=\left(a_1,a_1+a_2,a_1+a_2+a_3,\ldots\right)=\left(\sum_{j=1}^na_j\right)_{n\geq 1}
\end{align*}
and it can be seen as limit of the sequence of partial sums
  \begin{align*}
\sum_{j=1}^{\infty}a_j:=\lim_{n\to\infty}\sum_{j=1}^na_j
\end{align*}

In the second case the symbols (1) and (2) have the same meaning whereby (2) is defined by (1).

Note: You can find this line of argumentation e.g. in K. Knopps Theory and application of infinite series.


TITLE: Discontinuous point of a distribution function on ${\Bbb R}^n$
QUESTION [2 upvotes]: Can  the points of discontinuity of a distribution function on $\mathbb R^n$, ($n\gt1$)  be uncountable？ What about $\left(\{-\infty\}\cup\Bbb R\cup\{+\infty\}\right)^n$?

REPLY [1 votes]: I'm not sure what your second sentence means, but here's a hint on your first question: a point of discontinuity of a distribution function has a jump size $\delta > 0$. Create a sequence $\{\epsilon_{_k}\} \searrow 0$. Consider how many points of discontinuity there can be that have jump size between $\varepsilon_{_k}$ and $\varepsilon_{_{k+1}}$ for each $k$, and then consider summing up this number as $k \rightarrow \infty$. How big can this sum be?


TITLE: If $\lambda \not= 0$ and $\lambda x = 0$, then $x = 0$
QUESTION [1 upvotes]: I am trying to use the definition of vector spaces to prove that, if $\lambda \not= 0$ and $\lambda x = 0$, then $x = 0$.
One proof I have seen begins as follows:

$\lambda 0 = 0$ for each $\lambda$ since $\lambda 0 = \lambda(0, + 0) = \lambda 0 + \lambda 0$.

It then proceeds as follows:

Therefore, since $\lambda \not= 0$,
$$\begin{align} \lambda x = 0 &\Rightarrow \lambda^{-1} (\lambda x) = \lambda^{-1} 0 \\ &\Rightarrow (\lambda^{-1} \lambda) x = 0 \\ &\Rightarrow 1x = 0 \\ &\Rightarrow x = 0 \end{align}$$

It is this part that I am confused about:

$\lambda 0 = 0$ for each $\lambda$ since $\lambda 0 = \lambda(0, + 0) = \lambda 0 + \lambda 0$.

It is not clear to me what purpose this has in the proof, nor is it clear to me that it is even valid (that is, it is not clear to me that it is a valid claim, given the definition of a vector space). This part seems out-of-place to me. I would greatly appreciate it if people would please take the time to explain this.

REPLY [1 votes]: Read the main proof again:

$$
\begin{align}
\lambda x=0
&\Rightarrow\lambda^{-1}(\lambda x)=\color{red}{\lambda^{-1}0}\\
&\Rightarrow(\lambda^{-1}\lambda)x=\color{red}{0}\\ &\Rightarrow\cdots\end{align}
$$

One needs to justify that $\lambda^{-1}0=0$. So, it suffices to prove that $\lambda0=0$ for every scalar $\lambda$. I think this is the purpose of the leading line of the proof.


TITLE: Why do we only consider Euclidean simplicial complex?
QUESTION [1 upvotes]: I'm reading J.Dieudonné - The history of algebraic and differential topology
Here is what written in page 4-5

From this beginning, the evolution of homology went throught a series of steps, the description of which constitutes Part I of this book. We summarize them below, not necessarily in chronological order
$\textbf{I}$ Poincaré had already proved that a subdivision of every cell into smaller cells gives the same homology for the subdivided triangulations. This allows one to only consider euclidean simplicial complexes in which every cell is a rectilinear simplex contained in some $\mathbb{R}^N$ with large $N$.
$\textbf{II} \cdots$

I don't get what exactly the line $\textbf{I}$ means. Why does the fact that subdivision does not affect homology lets us to only consider euclidean simplicial complexes? What does cell mean here?
So say, we define homology using $n$-gons with fixed orientations. (There is a natural way to define this as we do to define the usual simplicial homology) Does $\textbf{I}$ mean that this homology and the usual simplicial homology induced by simplices coincide, and that Poincaré proved this?

REPLY [1 votes]: Right, exactly.  On page 4, Dieudonné refers to "a triangulation $T$ consisting of cells homeomorphic to convex polyhedra".  So Poincaré originally considered cell structures where the cells were arbitrary convex polyhedra, not just simplices.  But you can subdivide such a cell structure to get one using only simplices, and Poincaré showed that this does not change the homology.


TITLE: Fourier Transform: If $H(\alpha)=e^{-t\alpha^{2}}F(\alpha)$, what is $h(x)$?
QUESTION [3 upvotes]: I am trying to solve the following problem from my textbook:

If $f(x) \in L^{2}(-\infty, \infty)$ has Fourier transform
  $F(\alpha)$, then what is the function of $x$ whose transform is the
  function $H(\alpha)=e^{-t\alpha ^{2}}F(\alpha)$? ($t$ is a positive real constant).

The answer given in the back of the book is $\displaystyle h(x) =  \left[4 \pi t\right]^{-1/2}\int_{-\infty}^{\infty}\exp \left[\frac{-(x-y)^{2}}{4t} \right]f(y)dy$, but I have no idea how they got that.
I attempted the problem by writing that $\displaystyle h(x) = e^{-t\alpha^{2}}\frac{1}{2\pi}\int_{-\infty}^{\infty}f(x)e^{-ix\alpha}dx = \frac{1}{2\pi}\int_{-\infty}^{\infty}f(x) e^{-t\alpha^{2}-ix\alpha} dx$, but from that point on, I don't know what to do.
Is this possibly a situation where a convolution is involved? Or is it some kind of a translation? I am really lost, and am trying to teach these things to myself. Could somebody please help me understand what's going on here, and how to get the solution the book has?
Thank you.

REPLY [2 votes]: Note that we can write the Fourier Transform of the convolution, $\mathscr{F}\{f*g\}=\mathscr{F}\left(\int_{-\infty}^\infty f(x')g(x-x')\,dx'\right)$, between two functions $f$ and $g$ as
$$\begin{align}
\mathscr{F}\{f*g\}(\alpha)&=\int_{-\infty}^\infty \left(\int_{-\infty}^\infty f(x')g(x-x')\,dx'\right)\,e^{i\alpha x}\,dx\\\\
&=\int_{-\infty}^{\infty}f(x')\left(\int_{-\infty}^\infty g(x-x')e^{i\alpha x}\,dx\right)\,dx'\\\\&=\int_{-\infty}^{\infty}f(x')\left(\int_{-\infty}^\infty g(x'')e^{i\alpha (x'+x'')}\,dx''\right)\,dx'\\\\
&=\left(\int_{-\infty}^{\infty}f(x')e^{i\alpha x'}\right)\left(\int_{-\infty}^{\infty}g(x'')e^{i\alpha x''}\,dx''\right)\\\\
&=\left(\mathscr{F}\{f\}(\alpha)\right)\left(\mathscr{F}\{g\}(\alpha)\right)\tag 1
\end{align}$$
Let $F(\alpha)=\mathscr{F}\{f\}(\alpha)$ and $e^{-t\alpha^2}=G(\alpha)=\mathscr{F}\{g\}(\alpha)$ in $(1)$.  Therefore, the inverse Fourier Transform $\mathscr{F}^{-1}\{F(\alpha)G(\alpha)\}$ of $F(\alpha)G(\alpha)$ is given by $f*g$.  Since $g$ is the inverse Fourier Transform of $e^{-t\alpha^2}$, then $g(x)=\frac{1}{\sqrt{4\pi t}}e^{-x^2/(4t)}$ and we have
$$(f*g)(x)=\frac{1}{\sqrt{4\pi t}}\int_{-\infty}^{\infty}f(x')e^{-(x-x')^2/(4t)}\,dx'$$
as was to be shown!


TITLE: Nil-Radical equals Jacobson Radical even though not every prime ideal is maximal?
QUESTION [15 upvotes]: Let's assume we have a commutative ring with identity. Can the Nil-Radical and the Jacobson Radical be equal in a non-trivial case (i.e. not every nonzero prime ideal in said ring is maximal)?
Are there any interesting examples of this case?

REPLY [4 votes]: Theorem 5.1 in T.Y. Lam's book "A First Course in Noncommutative Rings" states that every polynomial ring $R[T]$ over a commutative ring $R$ satisfies
$$rad \,\, R[T] = Nil(R[T]) = (Nil \,\,R)[T]$$


TITLE: $\zeta(2n)$ and Levy processes
QUESTION [2 upvotes]: I am missing some steps in the final derivation of a probabilistic computation of the even values of $\zeta$.  They show the Cauchy distribution is relate to a certian Levy process:
$$ |\mathbb{C}_1| \stackrel{\text{law}}{=} e^{\frac{\pi}{2} \hat{C}_1} \text{ with }\mathbb{E}\left[ e^{i\lambda \hat{C}_1} \right] = \frac{1}{\cosh \lambda} $$
 If $X_t$ and $Y_t$ be $\mathbb{R}$-valued Brownian motions, then we have a $\mathbb{C}$-valued Brownian motion.
$$ Z_t = X_t + i Y_t \text{ with }Z_0 = 1+0i$$
  If we switch to polar coordinates, there is a "time-change" such that it is still Brownian motion, e.g. by conformal invariance of Brownian motion, as discussed in these early chapters in SLE.

$\log R_t = \beta_{H_t} $ ($\log R_t$ is not a Brownian motion)
$\theta_t = \gamma_{H_t} $

Then thy considered the hitting time for the Y-axis: $T = \text{inf} \{ t: X_t = 0 \}$.  In the time-change parameterization:
$$ H_T := T^{\gamma, *}_{\pi/2} $$
The $Y$-corrdinate at the hitting time $T$ (or $H_T$) is distributed as the Cauchy distribution.
\begin{eqnarray}
 \log | \mathbb{C}_1 | &\stackrel{\text{law}}{=}& \beta_{T^{\gamma, *}_{\pi/2}} \\
\frac{2}{\pi} \, \log | \mathbb{C}_1 | & \stackrel{\text{law}}{=} &
\beta_{T^{\gamma, *}_{1}}
\end{eqnarray}
The log-radius and angle $\beta$ and $\gamma$ are independent Brownian motions, over time.  I was not able to varify this change of equations:
$$ \mathbb{E} \left[ e^{i\lambda \frac{2}{\pi} \log | \mathbb{C}_1|}  \right]  \stackrel{1}{=}
 \mathbb{E} \left[ e^{i\lambda \beta_{T^{\gamma,*}_{\pi/2}} }  \right]
 \stackrel{2}{=}
 \mathbb{E} \left[ e^{ - \frac{\lambda^2}{2}  T_1^{\gamma, *} }  \right]
\stackrel{3}{=} \frac{1}{\cosh \lambda}
 $$
I wish the authors would come up with different variable names for the hitting times.
Why 2 and 3 are correct? And can anyone explain the bigger picture of why these authors feel special values of zeta might be connected to these random processes?

Reference:

Euler's formulae for $\zeta(2n)$ and products of Cauchy variables Paul Bourgade, Takahiko Fujita,  Marc Yor

REPLY [4 votes]: The (2) is just the independence of the Brownian motion $ (\beta_t)_t $ with the ``clock'' $ (H_T)_T $. This is a classical result due to Lévy (and more generally, this is the Dambis-Dubins-Schwartz representation theorem that any continuous martingale is a changed-time Brownian motion, the BM being independent of the time change). You can find it the book by Revuz and Yor. You use the scaling of the BM and the fact that $ \beta_1 $ is a Gaussian to get this $ \frac{\lambda^2}{2} $.
The (3) comes from the first equality you gave. You should better write
$$ \frac{1}{\cosh \lambda} \stackrel{0}{=} \mathbb{E} \left[ e^{i\lambda \frac{2}{\pi} \log | \mathbb{C}_1|}  \right]  \stackrel{1}{=}
 \mathbb{E} \left[ e^{i\lambda \beta_{T^{\gamma,*}_{\pi/2}} }  \right]
 \stackrel{2}{=}
 \mathbb{E} \left[ e^{ - \frac{\lambda^2}{2}  T_1^{\gamma, *} }  \right]
 $$
and now, it is clear that (0) is the first equality.
I had a look at the article : the link with the special values of $ \zeta $ comes from an explicit expression for the density of $  \mathbb{C}_1 \mathbb{C}_2 $. The goal of the article is to provide a probabilistic proof of the identities (1) and (2) (in the article) using the planar BM. It appears that this is related with some stopping of the BM on a line after the computations given in the article. For the intuition, though, I can't really say anything.