TITLE: Integral Calc Proof
QUESTION [0 upvotes]: Let $f (p/q) = 1/q$ if the fraction $p/q$ is in lowest terms and $f (x) = 0$ for irrational $x$.
Prove that $f$ is continuous at $x$ if and only if $x$ is irrational or $x = 0$.

REPLY [1 votes]: Hint : just prove the limit of every point is all zero.


TITLE: probability problem
QUESTION [1 upvotes]: 3 machines (a1,a2,a3) produce parts
a1=45%
a2=35%
a3=20%
10% of a1 are defective
6% of a2 is defecyive
2% of a3 is defective
a)prob of a random part chosen to not be defective
b)part produced by a1
a) i thought of finding the defective first then subtracting it from 1 to get the not defective
so...defective= d
p(d|a1)=.1
p(d|a2)=.06
p(d|a3)=.02
p(d)= p(dUa1)+p(dUa2)+p(dUa3)=.1x.45+.06x.35+.02x.2
then 1-(above calculation) should give me the not damaged part right?
for b is it just .45? because it asks for the prob that it was produced by a1, so 45% is produced by a1 so 45/100?

REPLY [0 votes]: If you think (b) is too easy, it perhaps meant to ask what is the chance a defective part is produced by $a_1$?
If that's so, it is
$$
\frac{\mathbb{P}[\text{defective part} \cap \text{produced by} a_1]}
     {\mathbb{P}[\text{part is defective}}
 = \frac{0.1 \cdot 0.45}{0.1 \cdot 0.45 + 0.06 \cdot 0.35 + 0.02 \cdot 0.2}
 = \frac{9}{14} \approx 64.3 \%.
$$


TITLE: The current status of the Birch & Swinnerton-Dyer Conjecture
QUESTION [35 upvotes]: [Une traduction française suit la version anglaise.]
The question is only about elliptic curves $E$ over $\mathbb{Q}$ and concerns only the aspect
(order of vanishing of $L(E,s)$ at $s=1$)$\ =\ $(rank of $E(\mathbb{Q})$).
Let $r$ be the LHS and $d$ the RHS, so that (a special case of ) the Birch and Swinnerton-Dyer Conjecture is
BSD?. $r=d$.
By the end of the last millenium, we knew
Theorem (1977--2000). If $\ r=0,1$, then $d=r$ (and $\ \operatorname{Sha}(E)$ is finite).
Some years ago, I heard that there was some progress in proving $(r>0)\Longrightarrow (d>0)$ under the assumption of the finiteness of $\operatorname{Sha}(E)$.  What is the current status of the
Statement. Suppose that $\operatorname{Sha}(E)$ is finite.  If $r>1$, then $d>0$ ?

L'état actuel de la conjecture de Birch et Swinnerton-Dyer
On s'interesse uniquement aux courbes abéliennes $A$ sur $\mathbf{Q}$ et à
l'aspect
(ordre d'annulation de $L(A,s)$ en $s=1$)$\ =\ $(rang de $A(\mathbf{Q})$).
Désignons par $r$ le membre de gauche et par $d$ le membre de droite, de sorte
que la conjecture de Birch et Swinnerton-Dyer prédit (en particulier)
BSD? $r=d$.
Vers la fin du millénaire dernier, on avait démontré le
Théorème (1977--2000).  Si $r=0,1$, alors $d=r$ (et $\ \operatorname{Cha}(A)$ est fini).
Il y a quelques années, j'avais entendu dire qu'on a fait des progrès
concernant l'implication $(r>0)\Longrightarrow(d>0)$ sous l'hypothèse de la
finitude de Cha$(A)$. Quel est l'état actuel de l'
Énoncé. Supposons que $\operatorname{Cha}(A)$ est fini.  Si $r>1$, alors
$d>0$ ?

REPLY [4 votes]: Alice Silverberg has a "cheat sheet" on all things currently known related to the rank of an elliptic curve. The link is below.
http://math.uci.edu/~asilverb/connectionstalk.pdf


TITLE: square of primes above 5
QUESTION [1 upvotes]: Given that squares of all primes above 5 are either 1 (mod 30) or 19 (mod 30), is this just a curious coincidence, or is there some straight-forward explanation?
My research has not lead me to any explanation, but I do think there is more going on here than just a curious coincidence. Any tips or pointers for learning more about this would be greatly appreciated.
To be sure, not asking for a proof here.

REPLY [2 votes]: Any prime $>3$ can be written as $6n\pm1$
$$(6n\pm1)^2=36n^2\pm12n+1=24n^2+24\frac{n(n\pm1)}2+1\equiv1\pmod{24}\  \ \ \ (1)$$
Any prime $p>3$ can be written as $\equiv\pm1,\pm2\pmod5$
$$\implies p^2\equiv1,4\pmod 5\  \ \  \ (2)$$
Now apply CRT on $(1),(2)$


TITLE: Example of contour integration
QUESTION [1 upvotes]: Could someone help me evaluate the following integral with contour integration ?
$$\int_{0}^{2\pi}\frac{d\theta}{(a+b\cos\theta)^2}.$$
Constraints are: $a>b>0$.

REPLY [3 votes]: Using the parametrization $z = e^{i\theta}$, $0 \le \theta \le 2\pi$ for the unit circle, we can write the integral as the contour integral
$$\oint_{|z| = 1} \dfrac{1}{\left(a + b\frac{z + z^{-1}}{2}\right)^2} \frac{dz}{iz},$$
which can be rewritten
$$\oint_{|z| = 1} \dfrac{4z}{i(bz^2 + 2az + b)^2}\, dz,$$
or
$$\frac{4}{ib^2}\oint \frac{z}{(z^2 + \frac{2a}{b}z + 1)^2}\, dz.$$
The roots of $z^2 + 2(a/b)z + 1$ are
$$z_0 = \frac{-a-\sqrt{a^2 - b^2}}{b}\quad \text{and} \quad z_1 = \frac{-a + \sqrt{a^2 - b^2}}{b}.$$
The point $z_0$ lies outside the circle since
$$|z_0| = \frac{a + \sqrt{a^2 - b^2}}{b} > \frac{a}{b} > 1.$$
So $z_1$ lies inside the circle (shown by computation or by use of the fact $z_0z_1 = 1$ and $|z_0| > 1$). Hence, the function $z/(z - z_0)^2$ is analytic inside and on $|z| = 1$. Since $z^2 + 2(b/a)z + 1 = (z - z_0)(z - z_1)$, it follows from Cauchy's differentiation formula that
\begin{align}\frac{4}{ib^2}\oint_{|z| = 1} \frac{z}{(z + \frac{2b}{a}z + 1)^2}\, dz&= \frac{4}{ib^2} \cdot 2\pi i \frac{d}{dz}\bigg|_{z = z_1} \frac{z}{(z - z_0)^2}\\
&= \frac{8\pi}{b^2} \cdot \frac{-(z_1 + z_0)}{(z_1 - z_0)^3}\\
&= \frac{8\pi}{b^2} \cdot \dfrac{\frac{2a}{b}}{\frac{8(a^2 - b^2)\sqrt{a^2 - b^2}}{b^3}}\\
&= \frac{2\pi a}{(a^2 - b^2)\sqrt{a^2 - b^2}}.
\end{align}


TITLE: Evaluating $\int_{T}^{\infty} \exp\left[\beta - \phi t \right]t^{-\frac{3}{2}} \log ^{\kappa}(t) \mbox{d}t$
QUESTION [4 upvotes]: I am trying to evaluate the following integral
\begin{eqnarray}
I(\phi, \beta, \kappa) = \int_{T}^{\infty} \exp\left[\beta - \phi t \right]t^{-\frac{3}{2}} \log ^{\kappa}(t) \mbox{d}t
\end{eqnarray}
where $\phi$ and $\beta$ are positive constants and $\kappa > -4$.
Integration by parts for $I$
Let $u = t^{-\frac{3}{2}} \log ^{\kappa}(t)$ and $\frac{dv}{dt} = \exp\left[\beta - \phi t \right]$  then using integration by parts formula I got
\begin{eqnarray}
- t^{-\frac{3}{2}} \log ^{\kappa}(t) \phi^{-1} \exp\left[\beta - \phi t \right] \mid_T^{\infty} + \phi^{-1} \int_{T}^{\infty} \exp\left[\beta - \phi t \right]\frac{(2 \kappa -3 \log (t)) \log ^{\kappa -1}(t)}{2 t^{5/2}} \mbox{d}t
\end{eqnarray}
I do n't think this new integral is simpler.
Taylor Series expansion attempt for $I$
Using Taylor Series expansion for the exponential term I get
\begin{eqnarray}\nonumber
&=& \int_{T}^{\infty} \sum_{n=0}^{\infty} \frac{\left[\beta - \phi t \right]^n}{n!} t^{-\frac{3}{2}} \log ^{\kappa}(t) \mbox{d}t\\\nonumber
&=&   \int_{T}^{\infty} \sum_{n=0}^{\infty} \sum_{k=0}^{n} \frac{1}{n!} {{n}\choose{k}} \beta^{n - k} (- \phi t)^k  t^{-\frac{3}{2}} \log^{\kappa}(t) \mbox{d}t\\
&=&   \sum_{n=0}^{\infty} \sum_{k=0}^{n}  \frac{1}{n!} {{n}\choose{k}}  \beta^{n - k} (- \phi)^k \int_{T}^{\infty} t^{-\frac{3}{2} + k} \log^{\kappa}(t) \mbox{d}t
\end{eqnarray}
For $k \neq 0$ the last integral does not converge, but i think my original integral should converge, even if conditionally.
Any suggestions how to evaluate this integral?
Thank you

REPLY [3 votes]: Your integral $I\left( \phi ,\beta ,\kappa \right) $ can be evaluated for $\kappa =n\in N>0$:
$$I\left( \phi ,\beta ,n\right) =\exp \left( \beta \right) \int_{T}^{\infty
}\exp \left( -\phi \,t\right) t^{-\frac{3}{2}}\log \left( t\right) ^{n}dt$$
with the trick of [Jack D'Aurizio] (Integrating $\int_0^{\frac{\pi}{2}} x (\log\tan x)^{2n+1}\;dx$)
$$I\left( \phi ,\beta ,n\right) =\exp \left( \beta \right) \underset{\alpha
\rightarrow 0}{\lim }\frac{d^{n}}{d\alpha ^{n}}\int_{0}^{\infty }u^{\alpha -%
\frac{3}{2}}\exp \left( -\phi \,t\right) dt$$
Performing the integration:
$$I\left( \phi ,\beta ,n\right) =\exp \left( \beta \right) \sqrt{\phi }%
\underset{\alpha \rightarrow 0}{\lim }\frac{d^{2n+1}}{d\alpha ^{2n+1}}\phi
^{-\alpha }\Gamma \left( \alpha -\frac{1}{2},T~\phi \right)$$
the differentiation and take the limit as $\alpha \rightarrow 0$ leads to:
$$I\left( \phi ,\beta ,n\right) =\exp \left( \beta \right) \sqrt{\phi }%
\sum_{k=0}^{n}\binom{n}{k}\left( -1\right) ^{n-k}\left( \log [\phi ]\right)
^{n-k}\times $$
$$\times \Gamma \left( -\frac{1}{2},T~\phi \right) \left( \log [T~\phi
]\right) ^{k}+k!\sum_{m=1}^{k}\frac{\left( \log \left( T~\phi \right)
\right) ^{k-m}}{\left( k-m\right) !}G_{m+1,m+2}^{m+2,0}\left( T~\phi
\left\vert
\begin{array}{c}
1,1,...,1 \\
0,0,...,0,-\frac{1}{2}%
\end{array}%
\right. \right)$$
where the general Leibniz rule (https://en.wikipedia.org/wiki/Product_rule) and the Wolfram Function side is used.


TITLE: quotient of an object 2
QUESTION [0 upvotes]: Here I've asked what is a quotient of an object and the answer was that it is an equivalence class of epis.
But here on the first page they claim that regular quotient is an coequalizer of 2 morphisms. I do not follow why this coequalizer should be epi.

REPLY [3 votes]: It seems that your question is: If $c\colon B\to C$ is the coequalizer of the pair $f,g\colon A\to B$, why is $c$ an epimorphism?
To show that $c$ is an epimorphism, suppose $p,q\colon C\to D$ are two arrows such that $p\circ c = q\circ c$. We need to show that $p = q$.
Let $h = p\circ c = q\circ c\colon B\to D$. Since $c$ coequalizes $f$ and $g$, we have $c\circ f = c\circ g$, so $h\circ f = p\circ c \circ f = p\circ c \circ g = h\circ g$. Thus $h$ coequalizes $f$ and $g$. By the universal property of the coequalizer, there is a unique arrow $h'\colon C\to D$ such that $h'\circ c = h$. By uniqueness, $h' = p = q$.
An arrow is called a regular epimorphism if it is a coequalizer for some parallel pair of arrows. Dually, every equalizer is a monomorphism, and an arrow is called a regular monomorphism if it is an equalizer for some parallel pair of arrows.