TITLE: Sum of acceleration vectors
QUESTION [2 upvotes]: If a point mass has some accelerations $\mathbf{a_1} $ and $\mathbf{a_2} $, why is mathematically true that the "total" acceleration is $\mathbf{a}= \mathbf {a_1}+\mathbf {a_2}$?

REPLY [3 votes]: This is due to the superposition principle: when several forces act upon a body, the net force is the sum of the individual forces: $$\vec F_{net} = \sum \vec F_i $$
However, this is only true when the relation between the force and the acceleration is linear.
Let's take the gravitational force as an example: say you have three bodies and you have already calculated $\vec a_1$ and $\vec a_2$ - the accelerations felt from the third body due to the other two. Then the  force on the third one would be $$m \vec a =\vec F_1 + \vec F_2= m \vec a_1 + m \vec a_2 = m(\vec a_1 +  \vec a_2)=m\vec a_{1+2} = \vec F_{net} $$ since the force is linear in $\vec a$. Here $\vec a_{1+2}$ - the total acceleration - is really $\vec a_1 + \vec a_2$.
Counter-example:
If you  had an environment where the acceleration is proportional to the force squared then the superposition principle would not be true. Let's say that this quadratic relationship would be the case for the gravitational force, then the force on the third body would be (I'm just considering the x-component here):
$$\begin{align}
m a_x & = (F_{net})^2\\
&=( F_{1x} +  F_{2x})^2\\
&=(m  a_{1x} +  m a_{2x})^2\\
& = (m  a_{1x})^2+2m^2 a_{1x} a_{2x}+(m a_{2x} )^2\\
&=( F_{1x})^2+ (F_{2x})^2 + 2m^2 a_{1x} a_{2x}\\
\end{align}$$
The linearity is not given ($(a+b)^2\neq (a^2+b^2)$) and hence the superposition principle not valid. You can see this by looking at the $2m^2 a_{1x}...$ term: in principle the superposition principle just says that the sum of the forces has the same effect as the combination of the individual forces. Although here, the squared sum has the effect of the combined squared forces plus another term.
This in turn means that in this case, the total acceleration which you get on the right hand side is not just $\vec a_1 + \vec a_2$.

REPLY [1 votes]: While the other answer are all completely correct, I just want to write a more simplified answer.
It's much the same as distances. I you walk 1 meter North and 1 meter East, you can add the two distance vectors and get $\sqrt2$m North-East:
$$\vec{d}_1=1m[N]=(1,0),~~\vec d_2=1m[E]=(0,1)$$
$$\vec d=\vec d_1+\vec d_2=(1,1)=1m[N]+1m[E]=\sqrt2m[NE]$$
Adding acceleration vectors works the same way as adding distance vectors. You add the corresponding components (x with x, y with y, etc. whatever coordinates you are using) and the magnitude and direction will work itself out.


TITLE: Does there exist a function $f:\mathbb{R}→\mathbb{R}$ such that $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$; $f(1)=1$; and $f(\sqrt2)=3$?
QUESTION [3 upvotes]: Does there exist a function $f: \mathbb{R} \rightarrow \mathbb{R}$ such that $f(x+y)=f(x)+f(y)$ for all $x,y \in\mathbb{R}$; $f(1)=1$; and $f(\sqrt 2)=3$?
Intuitively, I believe the answer is no. However, I'm having a hard time proving this. My initial thought was a proof by contradiction. Somehow using  $f(1)=1$ and $f(\sqrt 2)=3$ to show that $f(x+y)\neq(x)+f(y)$ for some choice in $x$ and $y$. This didn't prove fruitful. My next thought was to use properties of functions satisfying the addition property of homomorphisms to prove that $f(\sqrt 2)\neq3$. For example, one such property is $f(nx)=nf(x)$ for all $n \in \mathbb{Z}, x \in \mathbb{R}$. With this, one could theoretically construct a proof by induction that $f(x)=x$ for all $x \in \mathbb{R}$. Once again, this didn't prove fruitful. After these failed attempts at showing such a function doesn't exist, I considered how to show that one does exist. For reference, this question was asked in the context of a Field Theory course. As such, the thought for how to construct a proof is to satisfy the addition property with some linear transformation and show that a correct selection of basis spans all of $\mathbb{R}$. This idea seems feasible, however I am not sure where to begin. Any advice on how to approach this problem is greatly appreciated.

REPLY [9 votes]: Yes, it does exist.
Since $\sqrt{2}$ is irrational, there is a Hamel basis $B$ of $\mathbb R$ over $\mathbb Q$ that contains $1$ and $\sqrt{2}$.  Any function $g: B \to \mathbb R$ extends uniquely to a $\mathbb Q$-linear (and thus additive) function $f: \mathbb R \to \mathbb R$.   So just take $g(1) = 1$, $g(\sqrt{2}) = 3$, and whatever you want on the rest of $B$.


TITLE: Question about proof using the chain rule
QUESTION [1 upvotes]: The following is a proof from my textbook:
If $u(x, t)$ is a solution of $(1)$, so is the dilated function $u(\sqrt{a}x, at)$,
for any $a > 0$. Prove this by the chain rule: Let $v(x, t) = u(\sqrt{a} x, at)$. Then $v_{t} = [\partial (at)/\partial t]u_{t} = au_{t}$ and $v_{x} =[\partial(\sqrt{a} x)/\partial x]u_{x} = \sqrt{a} u_{x}$ and $v_{xx} = \sqrt{a}\cdot \sqrt{a}u_{xx} = a u_{xx}$.
I was wondering how we get that $v_{t} = [\partial (at)/\partial t]u_{t} = au_{t}$, for example? Since writing out the right-hand side of $v_{t}$ we have $\dfrac{\partial (at)}{\partial t}\cdot \dfrac{\partial u}{\partial t}$, but $\partial u$ and $\partial t$ don't "cancel out."

REPLY [1 votes]: Hint: Write $v=u\circ g$ where $g(x,t)=(\sqrt{a} x, at)$. Then, $\frac{\partial v}{\partial t}$ can be computed by reading off from the product of the Jacobians $\mathcal J(u)\circ \mathcal J(g)=\mathcal J(v)=\begin{pmatrix}
\frac{\partial v}{\partial x} & \frac{\partial v}{\partial t}
\end{pmatrix}=\begin{pmatrix}
\frac{\partial u}{\partial x} & \frac{\partial u}{\partial t}
\end{pmatrix}\begin{pmatrix}
\frac{\partial g_1}{\partial x} & \frac{\partial g_1}{\partial t}\\
\frac{\partial g_2}{\partial x} & \frac{\partial g_2}{\partial t}
\end{pmatrix}$.


TITLE: Prove that series with non summable coefficients do not converge in Banach space
QUESTION [1 upvotes]: Let $X$ be a Banach space and $\{e_i\}_i$ be a sequence in $X$ such that $||e_i||=1$ and such that every $x \in X$ can be written as $ x = \sum_{i=1} ^\infty \lambda_i e_i$ with a sequence $ \lambda_i \in \mathbb{R}$ that satisfies $ \sum_{i = 1}^\infty |\lambda_i| < \infty $.
If we further assume that these representations by absolutely summable sequences of coefficients are unique, can we show that if $\sum_{i = 1}^\infty |\lambda_i| = \infty$, the corresponding sequence $x_n:=\sum_{i=1}^n\lambda_i e_i$ does not converge in $X$?
Note: the uniqueness condition is
$$\forall x\in X \colon |\{\lambda_i\in \mathbb{R}^\infty: (x=\sum\lambda_ie_i) \wedge( \sum |\lambda_i|<\infty )\}|=1$$ not $$\forall x\in X \colon |\{\lambda_i\in \mathbb{R}^\infty : x=\sum\lambda_ie_i \}|=1\wedge (x=\sum\lambda_ie_i\Rightarrow \sum |\lambda_i|<\infty )$$

REPLY [0 votes]: By the open mapping theorem, the obvious bijection $\ell^1\to X$ has a continuous inverse. Thus the sequence $x_n$ in question does not converge.


text:
\begin{definition}[Definition:Clockwise]
'''Clockwise''' is this direction:
:[[File:Clockwise.png|400px]]
\end{definition}

config: wiki


TITLE: quadratic reciprococity example
QUESTION [0 upvotes]: I am new to the number theory concept of quadratic reciprocity.  I worked this example.  My result is that the quadratic residues mod 5 are 0,1, and 4. Can someone tell me if I am correct?
Here is my reference
http://www2.math.ou.edu/~kmartin/nti/chap9.pdf
Here is my table

Thanks,
Matt

REPLY [0 votes]: Your table has a $X=6$ where it should say $X=4$ and $\mathbb Z_5^\times = \{1,2,3,4\}$ so the quadradic residues in $\mathbb Z_5^\times$ are $\{1,4\}$. The $0\equiv 5 \pmod5$ drops out because
$$\mathbb Z_n^\times := \{ 1\le k\le n-1 : \gcd(k,n) = 1\}$$
So for a prime $p$
$$\mathbb Z_p^\times = \{1\le k\le p-1\}$$
And quadratic residues only make sense in $\mathbb Z_n^\times$, but since your reference looks at $\mathbb Z_p$, you'd include $\{0\}$ by definition. ($0$ and $1$ are always quadratic residues because $0^2 = 0, 1^2 = 1$ in any $\mathbb Z_p$)
A few pages down he states the definition $\Box_p$ to exclude $0$ as a QR, if he had left it out altogether, he wouldn't need that distinction... Also note that $\mathbb Z_p = \mathbb Z / p\mathbb Z$ in your text.


TITLE: Linear independent
QUESTION [0 upvotes]: $V$ is a vector space over a field $\mathbb{R}$. Let $S,T$ be subspaces of $V$ with $S\cap T=\{\vec0\}$ and let $A$ be a set.
Suppose $f\in \text{ Fun}(A,S)$, $g\in\text{ Fun}(A,T)$, where $\text{Fun}(A,S)=\{f:A\to S \text{ is a function}\}$
Show that, regarded as vectors in $\text{Fun}(S,V)$, the vectors $f$ and $g$ are linearly independent.
First, suppose that $f,g$ are linearly dependent.
Then there exists not all zero $\alpha,\beta\in\mathbb{R}$ such that $\alpha f+\beta\ g=\vec{0}$. ($\vec{0}(x)=0$ for all $x\in A$)
So $\forall x\in A$, $(\alpha f+\beta g)(x)=\vec{0}(x)\Leftrightarrow \alpha f(x)+\beta g(x)=0$
WLOG let $\alpha \neq0$. Then $f(x)=\frac{-\beta}{\alpha}g(x)$
So $f=\frac{-\beta}{\alpha}g\in \text{ Fun}(A,T)$
I can't find contradiction. Could anyone help me?

REPLY [0 votes]: I think there are several things missing in your post that caused some misconception.
First, you most likely mean that $f,g \in \text{Fun}(A,V)$ and not $f,g \in \text{Fun}(S,V)$.
Second, your task should be to show that these functions are linearly indepedent under the simple assumption that both aren't null functions.
Then all of it makes sense. The proof then gives you $f \in \text{Fun}(A,T)$, but since you know that $f \in \text{Fun}(A,S)$ you can tell as $S \cap T=\{0\}$ that $f(x)=0$ for all $x$ (i.e. $f$ is a null function) which contradicts your assumption of $f$ not being a null function.
PS: It's also important that $A$ is a non-empty set.


TITLE: What is the physical reasoning behind the generation of equal but opposite force (Newton's third law)?
QUESTION [4 upvotes]: When I apply a force on a body, it will apply a force on me (Newton's third law). But where does this force on me (i.e. equal but opposite force) generate from? Does it generate from within the body, or from somewhere else?

REPLY [1 votes]: For example: I do a force on a broken car to move it. But instead of put my hand directly on it, I hold a spring, and there is another equal spring fixed on the car. So I do the force by contacting both springs.
No matter if the force is enough to move the car or not, the deflections of the springs are the same. So the forces are equal.
We could imagine a different outcome of the experiment without any logical contradiction I believe. So it is an experimental fact.


TITLE: What is the meaning of "on each bounded disc of the complex plane"?
QUESTION [0 upvotes]: My primary language for math is not English.
What is the meaning of "on each bounded disc of the complex plane"?
For example (from https://www.encyclopediaofmath.org/index.php/Uniformly-convergent_series):
"... is uniformly convergent on each bounded disc of the complex plane, ..."
More specifically, does "on each bounded disc of the complex plane" refer to the group $A=\{z:|z-z_0|\le r\}$ where $r\in \mathbb{R}$, $z_0\in \mathbb{c}$? (note that $A$ is what I would directly translate from my primary math language to "a closed circle").
Thanks.

REPLY [1 votes]: Every disc in the complex plane (open or closed) is a bounded set. A closed disc is a compact set.
Suppose that $R$ is a fixed positive number, and that $z_0$ is a fixed complex
number.

The circle of radius $R$ centred at $z_0$ is defined by
$C(z_0; R) := \{z \in \mathbb{C} : |z − z_0| = R\}$
The open disc of radius $R$ centred at $z_0$ is defined by
$D(z_0; R) := \{z \in \mathbb{C} : |z − z_0| < R\}$
The closed disc of radius $R$ centred at $z_0$ is defined by
$D(z_0; R) := \{z \in \mathbb{C} : |z − z_0| \leq R\} = D(z_0; R) ∪ C(z_0; R)$

And finally,

A subset $U$ of $\mathbb{C}$ is said to be bounded if there is some positive number $\Delta$ such that $U \subseteq D(0; ∆)$,
i.e., $|z| \leq \Delta$ for every $z \in U$. $U$ is unbounded if it is not bounded.


TITLE: showing group $G$ is cyclic if every number $m$ that divides $(G:1)$, there are atmost $m$ elements such that ...
QUESTION [0 upvotes]: The following problem is from J.S. Milne book

Let $G$ be a finite Abelian group. If $G$ has at most $m$ elements of order dividing $m$ for each divisor of $m$ of $(G:1)$, show that $G$ is cyclic.

I do not understand the notation $(G:1)$. Also I would highly appreciate any hints (need not solve the problem as it coincides with my assignment) for solving the problem.

REPLY [1 votes]: We are given $x^n=e$ has at most $n$ solutions for $n|o(G)$. Let $F(d)$ denote # of elements of order $d$ where $d|n$.
Claim : $F(d)\le \phi(d)$ $\forall d|o(G)$. Suppose there does not exist any element of order $d$ then $F(d)=0\le\phi(d)$. If there exists an element of order $d$ say $g$ Then $\{{g^i :i=1,2,...,d}\}$ are all the solutions of $x^d=e$. Hence we have there can exist at most $\phi(d)$ elements of order $d$. Now we have $$o(G)=\sum_{d|o(G)}F(d)$$ since every element has some order $d$. again we have $$\sum_{d|o(G)}\phi(d)=o(G)$$ This implies $F(d)=\phi(d)$ $\forall d|o(G)$ In particular $F(o(G))=\phi(o(G))$$\gt$$0$ and this proves $G$ is cyclic.


TITLE: Book recommendation on special functions
QUESTION [1 upvotes]: I am currently studying real analysis from rudin and really like the chapter on special functions. But Rudin does not give much knowledge about those topics. Reading the references I found book by Emil artin on The Gamma function and really liked it. Are there more books like Emil artin's gamma function on special functions.

REPLY [1 votes]: I used to use Special Functions and Orthogonal Polynomials for reference on special functions.


TITLE: Find the number of solutions to $ \lfloor x \rfloor + \lfloor 2x \rfloor + \lfloor 4x \rfloor + \ldots + \lfloor 32x \rfloor =12345$
QUESTION [3 upvotes]: Find the number of solutions of the equation
  $$\lfloor x \rfloor + \lfloor 2x \rfloor + \lfloor 4x \rfloor + \lfloor 8x \rfloor + \lfloor 16x \rfloor + \lfloor 32x \rfloor =12345,$$
  where $\lfloor\,\cdot\,\rfloor$ represents the floor function.

My work:
I use the fact that
$$\lfloor nx \rfloor =\sum_{k=0}^{n-1} \left\lfloor x +\frac kn \right\rfloor.$$
So the equation becomes
$$\lfloor x \rfloor +\sum_{k=0}^{1} \left\lfloor x +\frac k2 \right\rfloor +\sum_{k=0}^{3} \left\lfloor x +\frac k4 \right\rfloor +\sum_{k=0}^{7} \left\lfloor x +\frac k8 \right\rfloor   \\
\qquad {}+\sum_{k=0}^{15} \left\lfloor x +\frac k{16} \right\rfloor +\sum_{k=0}^{31} \left\lfloor x+\frac k{32} \right\rfloor   \\
= \lfloor x \rfloor + \left\lfloor x+\frac 12 \right\rfloor + \left\lfloor x+\frac 64 \right\rfloor + \left\lfloor x+\frac{28}{8} \right\rfloor   \\
\qquad {}+ \left\lfloor x+\frac{120}{16} \right\rfloor + \left\lfloor x+\frac{496}{32} \right\rfloor$$
What should I do next?

REPLY [3 votes]: Alt. solution: Write the fractional part of $x$ in binary:
$$
x = n + 0.b_1 b_2 b_3 \ldots = n + \sum_{i=1}^\infty \frac{b_i}{2^i},
$$
where $n \in \mathbb{Z}$ and $b_i \in \{0,1\}$.
Also give your function $\mathbb{R} \to \mathbb{R}$ a name:
$$
f(x) = \lfloor x \rfloor + \lfloor 2x \rfloor + \lfloor 4x \rfloor + \lfloor 8x \rfloor + \lfloor 16x \rfloor + \lfloor 32x \rfloor
$$
Then
\begin{align*}
f(x)
&= \lfloor x \rfloor + \lfloor 2x \rfloor + \lfloor 4x \rfloor + \lfloor 8x \rfloor + \lfloor 16x \rfloor + \lfloor 32x \rfloor \\
&= n + (2n + b_1) + (4n + 2b_1 + b_2) + \cdots + (32n + 16b_1 + 8b_2 + 4b_3 + 2b_4 + b_5) \\
&= 63n + 31b_1 + 15b_2 + 7b_3 + 3b_4 + b_5.
\end{align*}
We can choose $b_i \in \{0,1\}$ arbitrarily,
so it follows that the image of $f$ is exactly
the set of integers whose remainder is between $0$ and $31 + 15 + 7 + 3 + 1 = 57$, mod $63$.
In particular, $12345 \equiv 60 \pmod{63}$, so $12345$ is not in the image of $f$.


TITLE: Why is the sample space no longer important once a discrete random variable is introduced?
QUESTION [1 upvotes]: F.M. Dekking, C. Kraaikamp, H.P. Lopuhaa, L.E. Meester, "A Modern Introduction to Probability and Statistics", Page-43.

Once a discrete random variable X is introduced, the sample space Ω is
  no longer important. It suffices to list the possible values of X and
  their corresponding probabilities. This information is contained in
  the probability mass function of X.

Can you explain this?
Why is the sample space no longer important once a discrete random variable is introduced?

REPLY [1 votes]: Given a discrete random variable $\xi$ with values $x_1,\ldots,x_n$ and corresponding probabilities $\mathsf P\{\xi=x_i\}=p_i$ you can easily build a sample space for it, for example as $\Omega=\{x_1,\ldots,x_n\}$ and $\xi(x_i)=x_i$. So we can exhibit a sample space. Note however that a specific realization of the sample space is irrelevant to everything one wants to do with the random variable.
P. S. Strictly speaking, random variable as a function already contain domain in its definition (indeed, in set theory function defined as a special kind of a relation; relation is a tuple of domain $X$, codomain $Y$, and subset of $X\times Y$). But the fact your book mentioned is that for discrete random variable $\xi$ we can "forget" the original domain of $\xi$ and that we can easily build own domain for it. This is why often random variables introduced without explicit definition of its sample space.


TITLE: Finding the simplest form
QUESTION [1 upvotes]: How can we simplify the expression below:
For any real numbers $a$ and $b$,
$
\frac{\frac{\frac{a+b}{2}+ b}{3}+ b} {4}+ b \cdots $
This is same as asking what is  $$\frac{a+ b\sum_{i=1}^ni!}{n!}$$ as $n$ approaches to $+ \infty$?
Thanks.

REPLY [1 votes]: Let's rewrite your sum like this:
$\dfrac{a+ b\sum_{i=1}^ni!}{n!}=\dfrac{a+ b \left( 1!+2!+ \cdots + \left( n-2\right)! + \left( n-1\right)! +n! \right)}{n!}$
$=\dfrac{a}{n!}+b \left( \dfrac{1!}{n!} +\dfrac{2!}{n!} + \cdots \dfrac{ \left( n-2\right)!}{n!} + \dfrac{\left( n-1\right)!}{n!} + \dfrac{n!}{n!}  \right)$
$=\dfrac{a}{n!}+b \left( \dfrac{1}{n!} +\dfrac{2}{n!} + \cdots \dfrac{ 1 }{n \left( n-1 \right)} + \dfrac{1}{n} + 1  \right)$
Take the limit as $n$ approches $+ \infty$ and you get $b$.


TITLE: Volume of the symmetric difference between a parallelotope and its translated.
QUESTION [2 upvotes]: Let $A$ be a n-dimensional parallelotope and $v \in \mathbb{R}^n$  a vector. Is there a formula giving the volume of the symmetric difference $A \Delta (v+A)$?

REPLY [0 votes]: The overlap (if any) of the two versions of the parallelotope is again a parellelotope $B$, with the same face normals as $A$. The volume of the symmetric difference is just twice the difference of the volumes of $A$ and $B$. How to put that into a formula depends on how the parallelotope is given.


TITLE: Expectation of the minimum of n independent Exponential Random Variables
QUESTION [0 upvotes]: Let $X_1$, $X_2$, · · · , $X_n$ be n independent Exponential random variables with mean 1. Find an expression
for E(min($X_1$, $X_2$, · · · , $X_n$ )).
I have been studying random variables for a while but I don't know how to approach this one. Any help is appreciated!

REPLY [0 votes]: Hint
Let $Y=\min(X_1,...,X_n)$.
$$\mathbb P\{Y\geq y\}=\mathbb P\{X_1\geq y,...,X_n\geq y\}=\mathbb P\{X_1\geq y\}^n,$$
where the last equality come from independence. The density function is therefore given by $$f_Y(y)=-n\lambda e^{-\lambda x}(e^{-\lambda x}-1)^{n-1}\cdot \boldsymbol 1_{[0,\infty )}(x).$$
Can you continue from here ?

REPLY [0 votes]: $P(min(X_1,..,X_n)>t)=P(X_1>t)^{n}=(e^{-t})^{n}=e^{-nt}$. So the density function of $min(X_1,..,X_n)$  is $ne^{-nt}$ (for $t>0$). Can you find the expectation from this?


TITLE: Show that $\exists c\in (0,1)$ such that $e^{f'(c)}=f(c)+1.$
QUESTION [5 upvotes]: Question: Let $f:[0,1]\in\mathbb{R}$ be a continuous function satisfying $\int_0^1f(x)dx=0$ and $f(0)f(1)>0$. Assume further that $f'$ is continuous on $(0,1)$. Show that $\exists c\in (0,1)$ such that $$e^{f'(c)}=f(c)+1.$$

My solution: It is given that $f(0)f(1)>0\implies f(0)>0,f(1)>0$ or $f(0)<0, f(1)<0$. Let us assume WLOG, that $f(0)>0,f(1)>0$.
Now since $\int_0^1f(x)dx=0$ and $f(0)>0,f(1)>0$, thus we must have a point $a\in (0,1)$ such that $f(a)<0$. Now since $f$ is continuous on $[0,1]$, therefore by IVT we can conclude that $\exists c_1\in (0,a)$, such that $f(c_1)=0$ and $\exists c_2\in(a,1),$ such that $f(c_2)=0$. From this we can conclude, there exists at least two distinct roots of $f$ in $(0,1)$. Therefore we are sure to have a point $b,0<b<1$ such that $f(x)>0, \forall x\in (0,b)$ and $f(b)=0$. Also we are sure to have a point $a,0<a<e<1$, such that $f(x)<0, \forall x\in(a,e)$ and $f(e)=0$.
Now consider the interval $[0,b]$. We have $f(0)>0$ and $f(b)=0$. Thus by MVT, we can conclude that $\exists c_3\in (0,b)$, such that $$f'(c_3)=\frac{f(b)-f(0)}{b-0}=-\frac{f(0)}{b}<0.$$ Now since $c_3\in (0,b)\implies f(c_3)>0.$
Again consider the interval $[a,e]$. We have $f(a)<0$ and $f(e)=0$. Thus by MVT, we can conclude that $\exists c_4\in (a,e)$, such that $$f'(c_4)=\frac{f(e)-f(a)}{e-a}=-\frac{f(a)}{e-a}>0.$$ Now since $c_4\in (a,e)\implies f(c_4)<0.$
Now let $g(x):=e^{f'(x)}-f(x)-1, \forall x\in(0,1).$ Observe that $g$ is continuous on $(0,1)$.
We have $g(c_3)=e^{f'(c_3)}-f(c_3)-1.$
Now $f'(c_3)<0\implies e^{f'(c_3)}<1$ and $f(c_3)>0\implies f(c_3)+1>1\implies -f(c_3)-1<-1.$
Thus $e^{f'(c_3)}-f(c_3)-1<0\implies g(c_3)<0.$
A similar reasoning shows that $g(c_4)>0$.
Thus by IVT, we can conclude that $\exists c\in(c_3,c_4)\subseteq(0,1)$, such that $g(c)=0\implies e^{f'(c)}=f(c)+1.$
Does this solution works fine? And is there a more better solution?

REPLY [2 votes]: If such $c$ doesn't exist, either always $e^{f'}>f+1$ or always $e^{f'}<f+1$. Whenever $f=0$, the former means $f'>0$ always and the latter means $f'<0$ always. However $f$ must change sign at least twice, and this is impossible.


TITLE: Prove there no integers $x,y$ such that $x+y=100$ and $\gcd(x,y)=3$
QUESTION [0 upvotes]: I have not encountered a problem like this before.I was thinking that there are 2 methods to solve this.
1. Assume $x+y=100$ and then prove that $\gcd(x,100-x)\neq 3$.
2. Assume that $\gcd(x,y)=3$, and then prove that $x+y$ can never equal $100$ for any $x,y\in R$.
I would appreciate a little guidance on the initial setup. Thank you.

REPLY [4 votes]: You can use your second strategy. Indeed, if $3 = gcd(x,y)$ then $3$ divides both $x$ and $y$ and hence $3$ divides $x+y$.

REPLY [2 votes]: $\gcd(x,y) = 3 \implies 3 \mid x$ and $3 \mid y$, i.e., $x=3m$ and $x=3n$, where $m,n \in \mathbb{Z}$. This means we need $3m+3n = 100$, which is not possible.


TITLE: Which solvable ODE's correspond to algebraic curves?
QUESTION [3 upvotes]: This is a follow-up to my previous question: Are there periodic functions satisfying a quadratic differential equation?
Let $P(u,v)$ be a bivariate polynomial. Then its zero set $\{(u,v): P(u,v)=0\}$ is an algebraic curve. It might make it easier to assume in what follows that the curve has no singularities.
If we substitute $y$ and $y'$ into $P(u,v)$ to take the places of $u$ and $v$ respectively, then we get a possibly non-linear first-order ODE. I am curious about how many ODEs of this type have solutions.

1. How many/which first order ODE's formed by substituting into a bivariate polynomial have solutions?
2.  Does any solution to such an ODE parametrize the algebraic curve that characterizes the defining ODE?  EDIT: I realize now that the answer to this question is obviously yes, because it follows directly from the fact that $y$ and $y'$ are both functions of the same single variable (say $t$) and that they satisfy the equation $P(y,y')=0$, so if they exist they must parametrize the algebraic curve, as a result of their definition.

Some examples might help. For second-degree algebraic curves, we have that the equation of the unit circle corresponds to a first order ODE which has both sine and cosine as solutions, and that these two functions parametrize the unit circle as well: $$(y')^2 + y^2 -1=0  $$ Likewise, the hyperbolic sine and hyperbolic cosine parametrize the unit hyperbola, and both correspond to solutions of a first order ODE whose form is that of the unit hyperbola: $$(y')^2 - y^2  -1=0$$ Finally, smooth cubic curves in Weierstrass normal form can be parametrized by the Weierstrass $\wp$ functions, and the (complex) differential equations characterizing the Weierstrass $\wp$ functions have the form of a smooth cuvic curve in Weierstrass normal form: $$(\wp')^2 = 4[\wp^2] -g_2 \wp -g_3  $$ Then my question is essentially: how far does this go? For examples, if I were to choose the algebraic curve $$ v^4 - v^3 = u^5 +u^2 -7$$ would the first-order ODE $$(y')^4 - (y')^3 = y^5 + y^2 -7$$ have a solution? And would the solution to this ODE (if it existed) parametrize the original algebraic curve $v^4 - v^3 = u^5 + u^2 -7$? This example is completely arbitrary, but hopefully it makes clear to some extent the level of generality I am interested in.
Note: I am not sure how to properly tag this question.
This question is related but addresses second order ODE's -- however, I am not interested in second order ODE's, only first order: Algebraic Curves and Second Order Differential Equations
I think that this question is probably the most related, assuming that epicycloids can be defined by first-order ODEs (I don't know either way). Proof that Epicycloids are Algebraic Curves?

REPLY [1 votes]: What you're looking for is, probably, the following theorems: (Theorems A and B of "A Malmquist–Yosida type of theorem for the
second-order algebraic differential equations"  by Liangwen Liao)

Theorem (Malmquist): Let $R(z,f)$ be birational. If a differential equation of the form $$f'(z)=R(z,f)$$ admits a transcendental meromorphic solution, then the equation can be reduced into a Riccati differential equation $$f'(z)=a(z)+b(z)f(z)+c(z)f^2(z),$$ where $a(z), b(z)$ and $c(z)$ are rational functions.

That is, if your polynomial $P(y,y')$ has degree $1$ in $y'$, it must have degree at most $2$ in $y$.

Theorem (Steinmetz): Let $R(z,f)$ be birational. If the following differential equation $$f'(z)^n=R(z,f),$$ admits a transcendental meromorphic solution, then the differential equation can be reduced into $$f'(z)^n=\sum_{i=0}^{2n}a_i(z)f(z)^i,$$ where the $a_i(z)(i=0,1,...,n)$ are rational functions and at least one of then does not vanish.

So if your equation has degree $n$ in $y'$, then it must have degree at most $2n$ in $y$.
The general case with more then on power of $y'$ is adressed in the "Fusch conditions" that you can find in section $1$ of  "Meromorphic solutions of algebraic differentlal equations" by Eremenko.


TITLE: Taylor series of $\frac{1}{1-\exp(z-1)}$
QUESTION [2 upvotes]: As part ofa voluntary homework assignment, I want to find the Taylor series around 0 for the complex valued function
\begin{equation}
f(z):=\frac{1}{1-\exp(z-1)}
\end{equation}
My approach so far is to express $\exp(z-1)$ and $\frac{1}{1-z}$ as power series. Since $\exp(z)= \sum \frac{z^k}{k!}$ and $\frac{1}{1-z} = \sum{z^k}$, this leads me to
\begin{equation}
\frac{1}{1-\exp(z-1)} = \sum_{k=0}^{\infty} \sum_{m=0}^{\infty} \left( \frac{(z-1)^m}{m!} \right)^k = \sum_{k=0}^{\infty} \sum_{m=0}^{\infty} \left( \frac{(z-1)^{mk}}{(m!)^k} \right)
\end{equation}
To obtain a power series around 0, I use the binomial series:
\begin{equation}
\sum_{k=0}^{\infty} \sum_{m=0}^{\infty} \left( \frac{(z-1)^{mk}}{(m!)^k} \right) = \sum_{k=0}^{\infty} \sum_{m=0}^{\infty} \sum_{n=0}^{mk} \frac{1}{(m!)^k} \binom{mk}{n} z^n \cdot (-1)^{mk-n}
\end{equation}
However, I am not sure how to proceed at this point. Can the last expression be simplified?
Of course, one could find the solution by using derivatives, but I should find the coefficients of the Taylor series based on the geometric and exponential power series. Am I overlooking something?

REPLY [1 votes]: I think it is more convenient to start with
  \begin{align*}
f(z)=\frac{1}{1-e^{z-1}}&=\sum_{n=0}^\infty e^{(z-1)n}
\end{align*}
  and calculate the Taylor series via
  \begin{align*}
f(z)=\sum_{n=0}^\infty f^{(n)}(0)\frac{z^n}{n!}
\end{align*}
We obtain
  \begin{align*}
\color{blue}{\left.\left(\frac{d^k}{dz^k}f(z)\right)\right|_{z=0}}&=\left.\sum_{n=0}^\infty n^ke^{(z-1)n}\right|_{z=0}\tag{1}\\
&=\sum_{n=0}^\infty n^k\left(\frac{1}{e}\right)^n\tag{2}\\
&=\sum_{n=0}^\infty \sum_{j=0}^k{k\brace j}n^{\underline{j}}\left(\frac{1}{e}\right)^n\tag{3}\\
&=\sum_{j=0}^k{k\brace j}\left(\frac{1}{e}\right)^j\sum_{n=0}^\infty n^{\underline{j}}\left(\frac{1}{e}\right)^{n-j}\tag{4}\\
&=\sum_{j=0}^k{k\brace j}\left(\frac{1}{e}\right)^j\left.\left(\frac{d^j}{dz^j}\frac{1}{1-z}\right)\right|_{z=\frac{1}{e}}\tag{5}\\
&=\sum_{j=0}^k{k\brace j}\left(\frac{1}{e}\right)^j\frac{j!}{\left(1-\frac{1}{e}\right)^{j+1}}\tag{6}\\
&\,\,\color{blue}{=e\sum_{j=0}^{k}{k\brace j}\frac{j!}{\left(e-1\right)^{j+1}}}\\
\end{align*}

Comment:

In (1) we differentiate $k$ times.
In (2) we evaluate the expression at $z=0$.
In (3) we represent $n^k$  with the help of the Stirling numbers of the second kind as sum of falling factorials $n^{\underline{j}}=n(n-1)\cdots(n-j+1)$.
In (4)  we exchange the sums.
In (5) we use the geometric series expansion.
In (6) we differentiate $j$ times  and evaluate at $z=\frac{1}{e}$.

We conclude
  \begin{align*}
\color{blue}{\frac{1}{1-e^{z-1}}=e\sum_{n=0}^\infty \left(\sum_{j=0}^{n}{n\brace j}\frac{j!}{\left(e-1\right)^{j+1}}\right)\frac{z^n}{n!}}
\end{align*}


TITLE: An upper bound for the ratio of one random variable to the square root of the sum of squares of random variables
QUESTION [1 upvotes]: Let $X_1, X_2, X_3$ be positive, iid, non-degenerate random variables with finite variances.
I wish to show that
$$\mathbb{E}\left[\dfrac{X_1 + X_2}{\sqrt{X_1^2 + X_2^2 + X_3^2}} \right] < \dfrac{2}{\sqrt{3}}\text{.}$$
My efforts. We can start off by observing that
$$\mathbb{E}\left[\dfrac{X_1 + X_2}{\sqrt{X_1^2 + X_2^2 + X_3^2}} \right] = 2 \cdot \mathbb{E}\left[\dfrac{X_1}{\sqrt{X_1^2 + X_2^2 + X_3^2}} \right]$$
so it is sufficient to prove that
$$\mathbb{E}\left[\dfrac{X_1}{\sqrt{X_1^2 + X_2^2 + X_3^2}} \right] < \dfrac{1}{\sqrt{3}}\text{.}$$
I am not sure how to go about this step.
One could try to handwave replacing $X_2$ and $X_3$ with $X_1$, but I can't think of a way of theoretically justifying that (it's probably a bad idea).

REPLY [0 votes]: Consider the distribution of $(X_1, X_2, X_3)$. Because $X_1, X_2, X_3$ are iid, it follows that they are exchangeable, hence the distribution of $(X_1, X_2, X_3)$ is the same as $(X_2, X_1, X_3)$.
By linearity of expectation, we have that
$$\mathbb{E}\left[\dfrac{X_1 + X_2}{\sqrt{X_1^2 + X_2^2 + X_3^2}} \right] = \mathbb{E}\left[\dfrac{X_1}{\sqrt{X_1^2 + X_2^2 + X_3^2}} \right] + \mathbb{E}\left[\dfrac{X_2}{\sqrt{X_1^2 + X_2^2 + X_3^2}} \right]\text{.}$$
The two terms of the right-hand side of the above equation are equal by exchangeability, hence we know that
$$\mathbb{E}\left[\dfrac{X_1 + X_2}{\sqrt{X_1^2 + X_2^2 + X_3^2}} \right] = 2 \cdot \mathbb{E}\left[\dfrac{X_1}{\sqrt{X_1^2 + X_2^2 + X_3^2}} \right]$$
so it suffices to prove that
$$\mathbb{E}\left[\dfrac{X_1}{\sqrt{X_1^2 + X_2^2 + X_3^2}} \right] < \dfrac{1}{\sqrt{3}}\text{.}$$
By Cauchy-Schwarz, observing that $\dfrac{X_1}{\sqrt{X_1^2 + X_2^2 + X_3^2}} > 0$ with probability $1$, and therefore $\mathbb{E}\left[\dfrac{X_1}{\sqrt{X_1^2 + X_2^2 + X_3^2}} \right] > 0$, we have that
$$\mathbb{E}\left[\dfrac{X_1}{\sqrt{X_1^2 + X_2^2 + X_3^2}}\right] = \mathbb{E}\left[\dfrac{X_1}{\sqrt{X_1^2 + X_2^2 + X_3^2}} \cdot 1 \right] \leq \sqrt{\mathbb{E}\left[\dfrac{X_1^2}{X_1^2 + X_2^2 + X_3^2} \right]}\text{.}$$
Observe that
$$\mathbb{E}\left[\dfrac{X_1^2 + X_2^2 + X_3^2}{X_1^2 + X_2^2 + X_3^2} \right] = 1 = \sum_{i=1}^{3}\mathbb{E}\left[\dfrac{X_i^2}{X_1^2 + X_2^2 + X_3^2} \right]\text{.}$$
Exploiting exchangeability again, we have that
$$1 = \sum_{i=1}^{3}\mathbb{E}\left[\dfrac{X_i^2}{X_1^2 + X_2^2 + X_3^2} \right] = 3 \cdot \mathbb{E}\left[\dfrac{X_1^2}{X_1^2 + X_2^2 + X_3^2} \right]$$
hence
$$\mathbb{E}\left[\dfrac{X_1^2}{X_1^2 + X_2^2 + X_3^2} \right] = \dfrac{1}{3}$$
hence
$$\mathbb{E}\left[\dfrac{X_1}{\sqrt{X_1^2 + X_2^2 + X_3^2}} \right] \leq \dfrac{1}{\sqrt{3}}\text{.}$$
However, equality holds if and only if $\dfrac{X_1}{\sqrt{X_1^2 + X_2^2 + X_3^2}}$ is constant (i.e., proportional to $1$) almost surely, so we can drop the equality.


TITLE: Burau matrix of braid
QUESTION [4 upvotes]: What is the definition of a Burau matrix of a braid? Where can I find a definition?

REPLY [4 votes]: The other answer seems to be correct, but in the interest of providing a non-link-only answer, I'll give the definition here.
The Burau matrix of a braid is the matrix representing the braid in the Burau representation, so it's a matrix over $\mathbb Z [t, t^{-1}]$, i.e. the entries of the matrix are Laurent polynomials in the variable $t$ with integer coefficients. The Burau representation is a representation (i.e. group homomorphism) $\psi_n: B_n \rightarrow \operatorname{GL}_n(\mathbb Z [t, t^{-1}])$. As with any group homomorphism, it's sufficient to define it on a generating set and check that the resulting map is well-defined.
We define in block form as
$$\psi_n : \sigma_i \mapsto  \left( \begin{array}{ccc}
I_{i-1} & 0 & 0 & 0 \\
0 & 1-t & t & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 0 & I_{n-i-1} \end{array} \right) ,$$
where $I_m$ is the $m \times m$ identity matrix, and $i$ ranges from $1$ to $n-1$. For this to be a valid homomorphism, we need to check that these matrices are invetible and that the braid relations are preserved, i.e. that $\psi_n(\sigma_i) \psi_n(\sigma_j) = \psi_n(\sigma_j) \psi_n(\sigma_i)$ for $|i-j| \ge 2$, and that $\psi_n(\sigma_i) \psi_n(\sigma_{i+1}) \psi_n(\sigma_i) = \psi_n(\sigma_{i+1}) \psi_n(\sigma_i) \psi_n(\sigma_{i+1})$ for $i=1, \ldots, n-2$. I'll exhibit the inverse matrix here, but leave it up to you to check that the computations work out.
$$\psi_n(\sigma_i)^{-1} = \left( \begin{array}{ccc}
I_{i-1} & 0 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & t^{-1} & 1-t^{-1} & 0 \\
0 & 0 & 0 & I_{n-i-1} \end{array} \right)$$
With that out of the way, for any braid $\gamma$, to compute its Burau matrix, just write $\gamma = \sigma_{i_1}^{e_1} \cdots \sigma_{i_m}^{e_m}$ as a word in the generators, and then you have $\psi_n(\gamma) = \psi_n (\sigma_{i_1})^{e_1} \cdots \psi_n (\sigma_{i_m})^{e_m}$, which is now easy to compute.
For some intuition, the case $t=1$ is good to look at. Setting $t=1$, we get a permutation matrix. This is the permutation matrix corresponding to the underlying permutation of the braid, i.e. the composite homomorphism $B_n \rightarrow S_n \rightarrow \operatorname{GL}_n(\mathbb Z)$ where $S_n$ is the symmetric group on $n$ letters. The Burau representation is thus 1-parameter deformation of the permutation representation.
The above is, of course, only one of several conventions, but it's the one I usually see chosen in the literature when one needs to do these calculations explicitly.


text:
\section{Nonlocal elasticity model}
\label{sec:nlmodel}

We consider a version of the Eringen's nonlocal elasticity model given
in~\cite{polizzotto01} (see also~\cite{eringen} and the references therein).
Let \(\O\subset\R^n\), \(n=1, 2\), or \(3\) be an open bounded domain with
Lipschitz boundary \(\Gamma = \partial \Omega\).
Let further \(\R_+ \ni d \mapsto \tilde{A}(d)\) be a function describing the
non-local interaction between the points in the model at a distance
\(d\ge 0\) from each other.
It will be convenient to evenly extend \(\tilde{A}\) onto the whole real line,
that is, we put \(\tilde{A}(d) = \tilde{A}(-d)\), \(\forall d\le 0\).
We define the kernel \(A: \R^n\times\R^n \to \R\) as
\(A(x,x') = \tilde{A}(|x-x'|)\).
We will assume that \(A \in L^1(\Omega\times\Omega)\), and that it is
a \emph{strictly} positive definite kernel:
\begin{equation}
  \label{pd}
  \int_{\O} \int_{\O} A(x,x') \phi(x) \phi(x') \,\mathrm{d}x'\,\mathrm{d}x > 0,
  \quad \forall \phi \in C^\infty_c(\Omega) \setminus \{0\},
\end{equation}
where the inequality above is known as Mercer's condition.
Often \(\tilde{A}\) is taken to be a non-negative, smooth function with
small compact support -- which results in \(A\) being a typical convolution
kernel for mollifying (see~\cite{brezis}).

The rest of this section should be understood as a preliminary informal
discussion where we do  not pay attention to the smoothness or integrability
requirements, which individual functions and the function spaces they are
contained in should satisfy.
The precise details will be added later on.

The nonlocal elasticity model that we consider can be stated as follows:
find the displacements \(u : \O \to \Rn\), the local strains
\(\varepsilon: \O \to \Sn \),
and the nonlocal stresses \(\sigma :\O \to \Sn\), where \(\Sn\) is
the set of \(n\times n\) symmetric matrices, such that
\begin{equation}\label{ne}
\left\{\begin{alignedat}{-1}
  -\diver(\sigma)&=f, && \text{in \(\O\)},\\
  \varepsilon &=\frac{1}{2}[\nabla u+(\nabla u)^\mathrm{T}], && \text{in \(\O\)},\\
  \sigma &=  \int_{\O} A(x,x')C\varepsilon(x')\,\mathrm{d}x', && \text{in \(\O\)},\\
\sigma \cdot \hat{n}&=g, && \text{on \(\Gamma_N\)},\\
u&=\bar{u}, &&\text{on \(\Gamma_D\)},\\
\end{alignedat}\right.
\end{equation}
where \(\Gamma_D,\Gamma_N \subset \Gamma\), \(\Gamma_D\cap\Gamma_N=\emptyset\),
\(\overline{\Gamma_D\cup\Gamma_N} = \Gamma\) are the Dirichlet and Neumann parts
of the boundary, respectively; \(\hat{n}\) is the outwards facing unit normal
for \(\O\) on \(\Gamma\); and \(C\) is the fourth-order stiffness tensor
with the usual symmetries.\footnote{
Unless it is explicitly stated otherwise, we assume that the stiffness tensor
\(C\) is constant in \(\O\).}
In~\eqref{ne}, the equations from top to bottom are the equilibrium, kinematic
compatibility, and non-local constitutive equations,
and traction (Neumann) and displacement (Dirichlet) boundary conditions,
respectively.
The  boundary conditions are in turn defined by the traction forces
\(g : \Gamma_N \to \R^n\) and the prescribed displacements
\(\bar{u} : \Gamma_D \to \R^n\).
The equilibrium equations are written with respect to the applied external
volumetric forces \(f : \O \to \R^n\).

As in the case of local elasticity, we assume that the
stiffness tensor is bounded and positive definite, that is, that there exist
constants \(\overline{C}\ge \underline{C} > 0\) such that
\begin{equation}
  \label{eq:localcoercivity}
\underline{C} \epsilon: \epsilon \le
C\epsilon: \epsilon \le
\overline{C} \epsilon: \epsilon,\quad
\text{for any \(\epsilon\in \Sn\)},
\end{equation}
where \(:\) stands for the Frobenius inner product in \(\Rnn\), that is
for \(\alpha=(\alpha_{km})_{1\le k,m\le n}\) and
\(\beta=(\beta_{km})_{1\le k,m\le n}\) we put
\(\alpha:\beta=\sum_{k,m=1}^n \alpha_{km}\beta_{km}\).

The weak formulation of~\eqref{ne} is obtained in the usual manner.
Namely we substitute the kinematics and the constitutive equations into the
equilibrium equation, multiply the latter with a test function
\( v \in V = \{\, \tilde{v} : \O \to \Rn \mid \tilde{v} = 0
\text{\ on \(\Gamma_D\)} \,\}\) and integrate by parts.
As a result we obtain the problem of finding
\( u\in u_0 + V\), where \( u_0 : \O \to \Rn \) is some fixed
function satisfying the Dirichlet boundary conditions \(u_0=\bar{u}\) on
\(\Gamma_D\), and such that
\begin{equation}\label{wf}
a(u,v)=\ell(v), \quad \forall v\in V,
\end{equation}
where
\begin{equation}\label{eq:al}
  \begin{aligned}
    a(u,v)&=\int_\O\int_{ \O}
    A(x,x') C \varepsilon_u(x): \varepsilon_v(x')\,\mathrm{d}x'\,\mathrm{d}x,
    \quad \text{and}\\
    \ell(v)&=\int_\O f(x)\cdot v(x)\,\mathrm{d}x
        +\int_{\Gamma_N} g(x)\cdot v(x)\,\mathrm{d}x,
  \end{aligned}
\end{equation}
and finally
\(\varepsilon_u = [\nabla u+(\nabla u)^\mathrm{T}]/2\) and
\(\varepsilon_v = [\nabla v+(\nabla v)^\mathrm{T}]/2\).

config: arxiv


TITLE: prove that $|\{f|f:A\rightarrow \{0,1\}\}|=|P(A)|$
QUESTION [1 upvotes]: Let A be a set ,
prove that $|\{f|f:A\rightarrow \{0,1\}\}|=|P(A)|$
I tried to prove that $g:|\{f|f:A\rightarrow \{0,1\}\}|\rightarrow |P(A)|$ is one to one and onto
but I didn't find the right function and I don't know how to find cardinality of functions
thoughts?

REPLY [0 votes]: Hint: define $g: \{f|f:A\rightarrow \{0,1\}\} \rightarrow P(A)$ by
$$g(f) = f^{-1}(\{1\})$$
where the RHS is the inverse image of $\{1\}$, namely all the elements in $A$ that are mapped to $1$ (you could equally well take the inverse image of $\{0\}$, the choice is arbitrary). Now show that $g$ is one-to-one and onto.


TITLE: Theorem or Conjecture?
QUESTION [1 upvotes]: I understand the definitions "theorem" and "conjecture" in mathematics, but I wasn't sure for physics. I mean, if it's proved mathematically, it's a theorem, otherwise it's a conjecture. But for physics (i.e. when the idea is meant to correspond to a physical reality - and - can only said to 'true' if tested experimentally), how does one make a distinction?

REPLY [3 votes]: In physics a theory is a mathematical model that provides a description of the real world in some limited circumstances. That is, it is a computational framework that allows us to feed in initial conditions and calculate what will happen.
All theories are approximate because they are based on assumptions that apply only within certain bounds. For example Newton's Theory of Gravitation gives a description of the Solar System that is pretty accurate (precession of Mercury apart). However we know that it fails at high speeds and energy densities and needs to be replaced by Einstein's Theory of general relativity. In turn we expect GR to fail at short distances and need to be replaced by some theory of quantum gravity.
As far as I know theorem means much the same in physics and maths, though the physics community is more pragmatic in the proof of theorems. I'm not sure that the word conjecture is used in any other than a casual way. In a sense all our theories are conjectures since they're not based on any rigorous mathematical proof.


TITLE: $\mathbb{C}\left(z\right)\left[\left[w\right]\right]\cap\mathbb{C}\left(w\right)\left[\left[z\right]\right]\overset{?}{=}\mathbb{C}\left(z,w\right)$
QUESTION [0 upvotes]: Consider the formal power series: $$f\left(z,w\right)=\sum_{m=0}^{\infty}\sum_{n=0}^{\infty}c_{m,n}z^{m}w^{n}$$
for some complex coefficients $c_{m,n}$. Defining:
$$\alpha_{m}\left(w\right)\overset{\textrm{def}}{=}\sum_{n=0}^{\infty}c_{m,n}w^{n}$$
$$\beta_{n}\left(z\right)\overset{\textrm{def}}{=}\sum_{m=0}^{\infty}c_{m,n}z^{m}$$
we can write: $$f\left(z,w\right)=\sum_{m=0}^{\infty}\sum_{n=0}^{\infty}c_{m,n}z^{m}w^{n}=\sum_{m=0}^{\infty}\alpha_{m}\left(w\right)z^{m}=\sum_{n=0}^{\infty}\beta_{n}\left(z\right)w^{n}$$
As such:
(1) If all the $\alpha_{m}$s and $\beta_{n}$s are rational functions of $w$ and $z$, respectively, is $f$ itself then necessarily a rational function of $z$ and $w$?
(2) (Generalization to power series in $d$ indeterminates) Working with a formal power series over $\mathbb{C}$ in $d$ indeterminates $z_{1},\ldots,z_{d}$:
$$f\left(z_{1},\ldots,z_{d}\right)=\sum_{n_{1}=0}^{\infty}\cdots\sum_{n_{d}=0}^{\infty}c_{n_{1},\ldots,n_{d}}z_{1}^{n_{1}}\cdots z_{d}^{n_{d}}$$
suppose that for every $j\in\left\{ 1,\ldots,d\right\}$, the series:
$$\sum_{n_{1}=0}^{\infty}\cdots\sum_{n_{j-1}=0}^{\infty}\sum_{n_{j+1}=0}^{\infty}\cdots\sum_{n_{d}=0}^{\infty}c_{n_{1},\ldots,n_{j-1},n,n_{j+1}\ldots n_{d}}z_{1}^{n_{1}}\cdots z_{j-1}^{n_{j-1}}z_{j+1}^{n_{j+1}}\cdots z_{d}^{n_{d}}$$
is a rational function of $ z_{1},\ldots,z_{j-1},z_{j+1}\ldots,z_{d}$ for all $n\geq0$. Is that sufficient to force $f$ to be a rational function of  $z_{1},\ldots,z_{d}$?

REPLY [1 votes]: Look at $e^{zw}=\sum_{n=0}^{\infty}\frac{(zw)^n}{n!}$. Each coefficient, as a $z$ series and as a $w$ series, is rational, but $e^{zw}$ isn't.


TITLE: Is $\pi_1(\widetilde{X/G})$ always finite if $\pi_1(X)$ is finite?
QUESTION [3 upvotes]: Let $X$ be a smooth complex manifold with finite fundamental group. Suppose that a finite group $G$ acts on $X$ and let $\widetilde{X/G}$ be a resolution of singularities. Is $\pi_1(\widetilde{X/G})$ always finite?
I think this is true but don't know a way to prove or a reference.

REPLY [5 votes]: The answer is yes when $X$ is simply connected. This can be proven as follows.
$\underline{\textrm{Step 1.}}$ The fundamental group  $\pi_1(X/G)$ is finite. More precisely, $\pi_1(X/G)= G/N$, where $N$ is the smallest  normal subgroup generated by those elements in $G$ which have fixed points on $X$.
For a proof, see [M. A. Armstrong, Calculating the fundamental group of an orbit space, Proceedings of the American Mathematical Society 84 (1982)].
$\underline{\textrm{Step 2.}}$ Since $X/G$ has only quotient singularities, one has $\pi_1(X/G)=\pi_1(\widetilde{X/G})$. This is proven in Theorem 7.8 of [J. Kollar, Shafarevich maps and plurigenera of algebraic varieties, Inventiones  Mathematicae 113 (1993)]. See also this MathOverflow question:
Comparing fundamental groups of a complex orbifolds and their resolutions.
The fact that $X$ is simply connected is only used in Step 1. Maybe this step can be refined in order to make it working also in the more general case where $X$ has finite fundamental group.


TITLE: A finite summation involving $2013$
QUESTION [11 upvotes]: Can you help me compute the summation below?
$$1+\frac{1}{2}+\cdots+\frac{1}{2013}+\frac{1}{1\cdot2}+\frac{1}{1\cdot3}+\cdots+\frac{1}{2012\cdot 2013}+\cdots+\frac{1}{1\cdot2\cdots2013}$$

REPLY [5 votes]: Here's how it works for $n=3$.
Find a common denominator:

$$\begin{align}
1 + S &= 1 + \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{1\cdot2} + \frac{1}{1\cdot3} + \frac{1}{2\cdot3} + \frac{1}{1\cdot2\cdot3}\\
&=\frac{1\cdot2\cdot3 + 2\cdot3 + 1\cdot3 + 1\cdot2 + 3 + 2 + 1 + 1}{1\cdot2\cdot3}\\
&= \frac{(1+1)(1+2)(1+3)}{1\cdot2\cdot3}\\
&= \frac{2\cdot3\cdot4}{1\cdot2\cdot3}\\
&= 4
\end{align}$$
  So, $S = 3$.

Generally, $(1 + S)n! = (x + 1)(x + 2) \cdots (x + n)\big|_{x=1} = (n + 1)!$, which gives $S = n$.


text:
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TITLE: Proving differentiability on every interval equals proving differentiability on all of $\mathbb{R}$?
QUESTION [1 upvotes]: The question actually came to me from series of functions. Suppose $f_n(x)$ is a sequence of continuous differentiable functions, that
$$\sum_{n=1}^\infty f_n'(x)$$
converges uniformly on any closed subinterval $[a,b]\subseteq \mathbb{R}$, and that $\sum_{n=1}^\infty f_n(x)$ converges pointwise for any $x$.
Then necessarily on any subinterval the sum of $f_n(x)$ converges to a function $f(x)$ such that  the sum of $f_n'(x)$ converges to $f'(x)$, and it seems somewhat intuitive to go from this to saying that the sum of $f_n(x)$ converges to a differentiable limit function on all of $\mathbb{R}$ as differentiation is a pointwise property.
But I'm still not sure about it.. I wasn't really able to prove it myself, and I'm not sure that the function I get for one interval will even match the function I get on another.
So is the claim true? Or is there an obvious counter-example?

REPLY [1 votes]: What you have to show is that for any given $x \in \mathbb{R}$ the sequence converges to a differentiable function the derivative of which is given by the series involving the derivatives. But since any $x$ is contained in a suitably chosen compact interval (e.g. $[-|x|-1, |x| + 1$) where your assumption applies, this is, of course, true.
You cannot, of course, expect uniform convergence (of the series of derivatives) on all of of $\mathbb{R}$


TITLE: Find $I:=\lim\limits_{R\to \infty}\int\limits_{-R}^R \frac{x \sin(3x)}{x ^2+4}dx$ using residues
QUESTION [3 upvotes]: Find $I:=\lim\limits_{R\to \infty}\int\limits_{-R}^R \frac{x \sin(3x)}{x ^2+4}dx$ using residues. Let $f(z)= \frac{z \sin(3z)}{z ^2+4}$. First define two contours:
$$\Gamma_1: z=t \text{ where } -R\leq t \leq R$$
$$\Gamma_2: z=Re^{i\theta} \text{ where } 0\leq \theta \leq \pi$$
And $\Gamma_3=\Gamma_1+\Gamma_2$. Basically $\Gamma_3$ is the closed half circle in the upper half of the complex plane and $\Gamma_1$ runs along the real axis from $-R$ to $R$. Now we have:
$$I=\lim\limits_{R\to \infty}\int\limits_{\Gamma_1}f(z)dz=\lim\limits_{R\to \infty}\oint_{\Gamma_3}f(z)dz-\lim\limits_{R\to \infty}\int_{\Gamma_2}f(z)dz$$
We can compute $\lim\limits_{R\to \infty}\oint_{\Gamma_3}f(z)dz=\lim\limits_{R\to \infty}\oint_{\Gamma_3}\frac{z\sin(3z)}{(z+2i)(z-2i)}dz$ fairly easily after recognizing that it has simple poles at $z_0=\pm 2i$ and only $z_0=2i$ is enclosed in $\Gamma_3$. By the residue theorem:
$$\lim\limits_{R\to \infty}\oint_{\Gamma_3}f(z)dz=2\pi i \left(\mathop{Res} _{z=2i}(f)\right)=2\pi i \lim\limits_{z\to 2i}(z-2i)f(z)=2\pi i \frac{2  i\sin(6 i)}{2 i +2i}=\pi i \sin(6i)$$
Now the only obstacle to finding the value of our original integral is finding $\lim\limits_{R\to \infty}\int_{\Gamma_2}f(z)dz$. If $f(z)$ were such that the degree of the polynomial in the denominator was at least 2 higher then the polynomial in the numerator (ignoring the $\sin$) we could easily show that the integral vanishes. I still think this last integral should vanish but find it hard to show why. Any help would be appreciated. Also any suggestions on different methods of solving for $I$ are welcome. Thanks!

REPLY [1 votes]: A not unrelated method of attack is to write the integral as
$$\frac{1}{2 i} \int_{-R}^R dx \frac{x}{x^2+4} \left (e^{i 3 x} - e^{-i 3 x} \right ) $$
Now work each piece separately.   For the $e^{i 3 x}$ piece, you use a contour in the upper half plane; for the $e^{-I 3 x}$ piece, use a contour in the lower half plane.