TITLE: The inverse of the function
QUESTION [0 upvotes]: How to find the inverse of \begin{equation}
f(x)=\frac{sinx}{cos^{2}x}
\end{equation}

REPLY [0 votes]: Your function is not one-to-one, so it does not have an inverse unless you restrict it to an appropriate interval.
In this case $(-\pi/2, \pi/2)$ could be one such interval.
$f(x) = y$ is equivalent to $g(\sin(x)) = y$, where $g(s) = s/(1-s^2)$.  So
$$\sin(x) = \frac{-1 \pm \sqrt{1+4y^2}}{2y}$$
If $y$ is real, $\sqrt{1+4y^2} > |2y|$, so only the $+$ term has a chance to give you a result in $[-1,1]$.  Thus the inverse to the restriction of your function to $(-\pi/2, \pi/2)$ is
$$ f^{-1}(y) = \arcsin \left( \frac{-1+\sqrt{1+4y^2}}{2y}\right) $$
defined on all of $\mathbb R$.
EDIT: As fleablood pointed out, the formula is undefined for $y=0$.  However, it's easy to see that $f^{-1}(0) = 0$.


TITLE: Can mass increase be measured by current technology?
QUESTION [3 upvotes]: One Kg of matter heated by 100° C is said to increase its mass by a few nanograms, can that increase be actually measured? Has it ever been verified?
If not, what is the indirect evidence of the mass increase because of hogher temp?

REPLY [2 votes]: Here's a recent paper in Metrologia which reports a Planck constant measurement using a Kibble balance (known more commonly as a "watt balance") with a relative uncertainty of about $2\times10^{-8}$.
That uncertainty is consistent with the state of the art described in this description of a much simpler LEGO-based Kibble balance.
Now, if you were to heat a kilogram of water from 0°C to 100°C, you'd have to add about 400 kilojoules of thermal energy. As you say, that energy has a relativistic mass equivalent of
$$
m_\text{equiv} = \frac E{c^2} = \frac{\rm4\times10^5\,J}{\rm9\times10^{16}\,m^2/s^2}
\approx \rm\frac 12\times10^{-11}\,kg
\approx \rm 4\,nanogram
$$
So no: the state of the art for high-precision mass measurements would need to improve its precision by about four orders of magnitude to see the effect you have in mind.


TITLE: What does the backslash denote in probability theory
QUESTION [1 upvotes]: I believe this draws from set theory but my question is:
What does the backslash denote in probability theory?
For example: what does a \ b denote?
In an application of a probability problem you could be given an arbitrary probability like:
P ( a \ b) = .50 or 50%
What would this effectively mean? How could you use this information then to determine the probabilities for two independent events P(A) and P(B)?

REPLY [3 votes]: The backslash denotes the "relative complement" or "set difference". The set $A \setminus B$ consists of all those elements that are in $A$ but not in $B$. In a probability $P(A \setminus B)$, it means "the probability that event $A$ occurs but event $B$ does not occur". If $A, B$ are independent, then
$$
P(A \setminus B) = P(A \cap B^c) = P(A)P(B^c) = P(A)(1- P(B)),
$$
but this is not enough to determine $P(A)$ or $P(B)$ by itself.


TITLE: Solution to $xe^{e^x}$
QUESTION [6 upvotes]: The problem $xe^{e^x}=e$ came up another day and I wondered if it were solvable.
My attempt was the following substitution,$$x=W(u)$$$$W(u)e^{e^{W(u)}}=e$$Where I used a Lambert W identity to get $$W(u)e^{\frac u{W(u)}}=e$$And attempted to solve.
I got this far:$$-\frac1{W(u)}e^{-\frac u{W(u)}}=-\frac1e$$However, I couldn't continue.  Its so darn close... but alas, its not quite there.
So I've come to you guys for help, well knowing that most of you will say "no solution" of closed form, but that's ok.  I will even accept answers that attempt to continue or start from the beginning to try out a different path for the solution.
I also considered the following: If you could get into a form where adding/subtracting/multiplying/dividing/whatever will cancel a part of the equation with a previous form, then that'll be great e.g.$$f(u)W(u)=g(u)$$$$xf(u)=W(u)$$Divide the two and you get $$\frac{W(u)}x=\frac{g(u)}{W(u)}$$$$W^2(u)=xg(u)$$Noting that you can switch around with substitutions as long as you are consistent with your substitutions.
I also realize that in the beginning, I could have used the substitution $x=-W(x)$ to flip the resulting fraction to put the $W$ on top and repeat the Lambert identity process, only to create$$-W(u)[\frac{u}{W(u)}]^{1/u}=e$$ Which is still not solvable!

REPLY [0 votes]: Is it solvable means two different things :
A solution exists and it can be calculated numerically, so in this sense it is solvable.
In terms of special functions I would say that if any solution exists it is always possible to express it as some expression involving special functions.  It is a question of how much effort and what special function you want to employ.  The question is whether that effort is useful to you or not.


TITLE: Explicit upper and lower bounds on the Prime counting function and Riemann Hypothesis
QUESTION [0 upvotes]: I have been researching equivalent statements on the Riemann Hypothesis. Schoenfeld proved that Riemann Hypothesis is equivalent to the bound $$|\pi(x) - {\rm Li}(x)| < \frac{1}{8\pi}\sqrt{x}\log x$$
for $x\geq 2657$. Excepting Schoenfeld's explicit bound, I have not been able to find explicit upper and lower bounds on the Prime counting function such that they would imply that Riemann Hypothesis is true.
In particular, I am wondering if they can exist $A$ and $B$ such that $$\frac {x}{\log(x)}+A<\pi(x)<\frac {x}{\log(x)}+B$$ and such that they would imply that Riemann Hypothesis is true.
Can these bounds exist? Are they known?
Thanks in advance!

REPLY [3 votes]: I suspect you want $A$ and $B$ to be constants, since you don't indicate any dependence in them on $x$. So you appear to be asking if RH might be equivalent to $\pi(x) = x/\log x + O(1)$ with explicit constants in the role of the $O(1)$ bound.
If you look at the graph at the bottom of the page https://primes.utm.edu/howmany.html, where $\pi(x)$ is in red and $x/\log x$ is in green, what it suggests there is true: $\pi(x) - x/\log x \to \infty$ as $x \to \infty$.  Specifically, the oldest form of the prime number theorem with error term
$$
\pi(x) = {\rm Li}(x) + O(xe^{-c\sqrt{\log x}})
$$
implies
$$
\pi(x) = {\rm Li}(x) + O_k\left(\frac{x}{(\log x)^k}\right)
$$
for all $k > 0$, and integration by parts twice tells us
$$
{\rm Li}(x) = \frac{x}{\log x} + \frac{x}{(\log x)^2} +
O\left(\frac{x}{(\log x)^3}\right),
$$
and together these estimates imply
$$
\pi(x) = \frac{x}{\log x} + \frac{x}{(\log x)^2} +
O\left(\frac{x}{(\log x)^3}\right),
$$
so $\pi(x) - x/\log x \sim x/(\log x)^2$.  Thus $\pi(x) - x/\log x \to \infty$ as $x \to \infty$.
In contrast to that, Littlewood proved that the difference $\pi(x) - {\rm Li}(x)$ changes sign infinitely often.
While the Riemann hypothesis is equivalent to $\pi(x) = {\rm Li}(x) + O(\sqrt{x}\log x)$, there is no estimate of the form $\pi(x) = x/\log x + O(x^{1-\delta})$ for some $\delta$ in $(0,1)$: using the last displayed estimate on $\pi(x)$ above, if $\pi(x) = x/\log x + O(x^{1-\delta})$ then
$$
\frac{x}{(\log x)^2} = O\left(\frac{x}{(\log x)^3}\right) + O(x^{1-\delta}) = O\left(\frac{x}{(\log x)^3}\right)
$$
for large $x$, which is false.
The reality is that the use of $x/\log x$ as an approximation for $\pi(x)$ works in the asymptotic estimate in the prime number theorem but not in non-asymptotic estimates that you will find in equivalent reformulations of the Riemann hypothesis. We usually write PNT in terms of $x/\log x$ because it has the attraction of being a simple expression in comparison to ${\rm Li}(x)$, but you have to give up on working with $x/\log x$ as an approximation to $\pi(x)$ when you want to look at more sophisticated things like RH.


TITLE: Function field of $\mathbb{P}^1$
QUESTION [0 upvotes]: What is the function field of $\mathbb{P}^1$ when $\mathbb{P}^1$ is viewed as a projective curve? From what I read, I think it should be $k(x)$, but I don't understand how.

REPLY [5 votes]: The only regular functions on $\mathbb P^1$ are constants, so you have to pass to nontrivial open subsets to get anything interesting. But once you have deleted a single point, you are now looking at $\mathbb A^1$ or a subset thereof, and the ring of regular functions on $\mathbb A^1 \setminus \left\{ a_1,a_2,...,a_n \right\}$ is $k\left[x,\frac{1}{x-a_1}, \frac{1}{x-a_2},...,\frac{1}{x-a_n}\right]$. Now as you vary over all possible finite subsets you get a directed system of rings starting with $k \subset k[x]$ (and the other possible ways to include $k$ into a one variable polynomial ring by deleting a point from $\mathbb P^1$). Taking the colimit over this system you obtain $k(x)$, essentially as a consequence of the universal property of localization.


TITLE: Representation of finite group and Classifying finite simple group
QUESTION [3 upvotes]: My question is quite clear as the title of it. I had studied theory of representation of finite group and I also had known something about the program of classifying finite simple group, and in general, the problems of two subjects is not far different, but in fact, I have not seen any mathematical paragraph that tells us about the connection(of these two subject).
But, I think the two subject both give us the structure of finite group. Could any MSer point it out for me?
Thanks.

REPLY [6 votes]: Representation theory of finite groups over the complex numbers yields information about groups which can be be determined by group characters. The most important characters in this context are the irreducible characters, and complete information about these can be encoded within the character table, which is a $k \times k$ array, where $k$ is the number of conjugacy classes of the group in question, say $G.$ It can be determined by inspection from the character table whether there are any normal subgroups of $G$ other than the trivial group and $G$ itself, ie whether $G$ is simple or not. However, it is sometimes possible to use remarkably small amounts of character-theoretic information to prove the non-simplicity of $G.$ There is a long tradition of using the embedding of certain subgroups of $G$ to produce enough character-theoretic information to show that $G$ is not simple. Two early examples of such theorems were Burnside's $p^a q^b$-theorem and Frobenius theorem about subgroups of $G$ which have trivial intersection with all their distinct conjugates. This tradition continued until the 1960s and beyond, and was give fresh impetus by the work of Richard Brauer, who showed how to use modular methods to further refine character-theoretic information. These methods were particularly effective when dealing with simple groups of low $2$-rank, that is without elementary Abelian $2$-subgroups of order $8$ or more. In particular, character theory played a prominent role in the proof by Feit and Thompson of the solvability of finite groups of odd order, and in the classification of finite simple groups with dihedral or semi-dihedral Sylow $2$-subgroups (as well as proving that no finite simple group has generalized quaternion Sylow $2$-subgroups). Another high point of the use of character theory was Glauberman's $Z^{*}$-theorem, which proved that if $G$ is a finite non-Abelian simple group, and $t$ is an involution of $G,$ then $t$ commutes with one of its other conjugates, which was used as a starting point for many group-theoretic analyses in the classification programme. It is probably fair to say that in the later stages of the classification programme, ordinary character theory played less of a role, with the honorable exception of the $Z^{*}$-theorem. On the other hand, most of the infinite families of finite simple groups are best described in terms of natural representations over finite fields.


TITLE: How do I form a group of permutations and determine the multiplication table for the group?
QUESTION [0 upvotes]: Let $S=\{1,2,3,4,5,6,7\}$ and let
$\sigma_0=(1)(2)(3)(4)(5)(6)(7)$
$\sigma_1=(1,2,3)(4)(5)(6)(7)$
$\sigma_2=(1,3,2)(4)(5)(6)(7)$
$\sigma_3=(1)(2)(3)(4)(5,6)(7)$
$\sigma_4=(1,2,3)(4)(5,6)(7)$
$\sigma_5=(1,3,2)(4)(5,6)(7)$
Prove that $G=\{σ0,σ1,σ2,σ3,σ4,σ5\}$ form a group of permutations and determine the multiplication table for the group. Is the obtained group commutative?
I don't understand what exactly I have to do here, do I have to create a $6 \times 6$ table and multiply each permutation with each other? What exactly is a group of permutations and a multiplication table?
Is the final table on this page what I have to do?
http://groupprops.subwiki.org/wiki/Determination_of_multiplication_table_of_symmetric_group:S3

REPLY [1 votes]: For $1$-cylce it can be omitted.
$\sigma_0=1$(identitiy function)
$\sigma_1=(123),\sigma_2=(132),\sigma_3=(56),\sigma_4=(123)(56),\sigma_5=(132)(56)$
The law of composition defined for symmetric group is not multiplication but is composition of mapping.
For example, $\sigma_1\sigma_4=(123)(123)(56)$. Look from right to left, $1\rightarrow2\rightarrow3$, $3\rightarrow1\rightarrow 2$, $2\rightarrow 3\rightarrow1$, $5\rightarrow6$, $6\rightarrow5$.
So $\sigma_1\sigma_4=(132)(56)$. By using this method, you can create a multiplication table and prove that $G$ satisfies axioms for closure, identity and inverse. It is not necessary to prove the axiom for associativity because the composition of mapping is itself associative.
Another interpretation: I write $\alpha=(123),\beta=(56)$. Then $$G=\{1,\alpha,\alpha^{-1},\beta,\alpha\beta,\alpha^{-1}\beta\}$$
Since $\alpha\beta=\beta\alpha$(you can easily verify it or by using the fact that they are disjoint), $G$ is commutative. And in fact $G$ is cyclic group of order $6$.


TITLE: Euler-Lagrange equation
QUESTION [2 upvotes]: This is PDE Evans, 2nd edition: Chapter 8, Exercise 2:

Find $L=L(p,z,x)$ so that the PDE $$-\Delta u + D\phi \cdot Du = f \quad \text{in }U$$ is the Euler-Lagrange equation corresponding to the functional $I[w] := \int_U L(Dw,w,x) \, dx$.
(Hint: Look for a Lagrangian with an exponential term.)

First, one must note that $U \subset \mathbb{R}^n$ and $L(p,z,x)=L(Du,u,x)$, where $Du=\nabla u =\sum_{i=1}^n u_{x_i}$.
One should also note that the Euler-Lagrange equation associated with the energy functional $I[w]:=\int_U L(Dw(x),w(x),x) \, dx$ is
$$-\sum_{i=1}^n (L_{p_i}(Du,u,x))_{x_i}+L_z(Du,u,x) = 0 \quad \text{in }U.$$
Here is my work so far:
I set $(L_{p_i})_{x_i}:= (p_i)_{x_i}-\phi_{x_i} p_i$ and $L_z:=-f(x)$ so that the LHS of the Euler Lagrange PDE is
\begin{align}
0 &=-\sum_{i=1}^n (L_{p_i}(Du,u,x))_{x_i}+L_z(Du,u,x) \\
&=-\sum_{i=1}^n\left[(p_i)_{x_i} - \phi_{x_i} p_i \right] - f(x) \\
&= -\sum_{i=1}^n \left[u_{x_ ix_i} - \phi_{x_i} u_{x_i} \right] - f(x) \\
&= -\sum_{i=1}^n u_{x_ ix_i} + \sum_{i=1}^n  \phi_{x_i} u_{x_i} - f(x) \\
&= -\Delta u + D\phi \cdot Du - f(x).
\end{align}
This gives us the PDE $-\Delta u + D\phi \cdot Du = f$ as written in Exercise 2.
Now, how can I write $L(p,z,x)$? I was thinking $$L(p,z,x)=\frac 12|p|^2 - \qquad \qquad - zf(x),$$ so that, upon taking the derivative with respect to $p_i$ or with respect to $z$, we would obtain both the $L_{p_i}$ and $L_z$ that I have, respectively. However, I have trouble writing the middle term, which is why I left a blank space. The textbook author did say "look for a Lagrangian with an exponential term".

REPLY [3 votes]: Using the hint, we set
$$L(p, z, x) = e^{-\phi(x)}\left(\frac 12|p|^2 - zf(x)\right).$$
Then the functional becomes
$$I(u) = \int_\Omega e^{-\phi} \left(\frac 12 |\nabla u|^2 - uf(x)\right)\, dx.$$
Thus
$$\frac{\partial}{\partial t}\bigg|_{t=0} I[u_t] = \int_\Omega e^{-\phi} (\nabla u \cdot \nabla v + vf) dx,$$
where $v = \partial_t u |_{t=0}$. Then by divergence theorem, it's
$$\int_\Omega v\big(-\nabla(e^{-\phi} \nabla u) -e^{-\phi} f\big)\, dx $$
which is
$$\int_\Omega e^{-\phi} v\big(\nabla \phi \cdot \nabla u - \Delta u  -f \big)\, dx $$
which means that
$$\nabla \phi \cdot \nabla u - \Delta u - f = 0$$
and this is the same as your equation.
Remark How I come up with this expression: First of all, as you know, I thought something like $D\phi \cdot Du$ would suffices. It's not true though, as if we write down $I'[v]$, we would get something like
$$ D\phi \cdot Dv$$
and there is no way to isolate $v$ using integration by part and still have $D\phi$. The above term is also a hint that $L$ should contains only $\phi$, but not $\nabla \phi$ (or the Euler-Lagrange equation would contains second derivatives of $\phi$).
So I tried $L = \frac 12 \phi |p|^2 - z f(x)$, then we have
$$I'[v] = \int_\Omega \left( \phi \nabla u \cdot \nabla v - v f\right) \,dx = \int_\Omega v\left( -\nabla(\phi \nabla u) - f\right) \,dx.$$
This is almost correct as the EL equation is
$$-\phi \Delta u - D\phi \cdot Du = f. $$
Then I realize that to deal with that $\phi$ in front of $\Delta u$, I should do $e^{\phi} |p|^2$ because I would get a term
$$D e^\phi \cdot Du = e^\phi D\phi \cdot Du$$
and this extra $e^\phi$ can be grouped to that $e^\phi$ in front of $\Delta u$.


TITLE: Matrix inequality: Eigenvalues
QUESTION [2 upvotes]: A is a $n \times n$ non symmetric matrix. Can $\sigma_1(A)$ - the greatest singular value of A -  be upper bounded by some function of the eigenvalues of A?

REPLY [3 votes]: You can not have such a boundary for a generic matrix basing only on the its eigenvalues. For example, $A=\begin{pmatrix}0&1\\0&0\end{pmatrix}$. Both eigenvalues are zeros, yet $\sigma_1(A)=1$. Take now $B=\begin{pmatrix}0&0\\0&0\end{pmatrix} $. Both eigenvalues of $B$ are also zeros, yet $\sigma_1(B)=0$.
You can, however, write $$\sigma_1(A)=\sup_{x\ne 0}\frac{\|Ax\|^2}{ \|x\|^2} = \|A\|^2\ge \max_j(|\lambda_j|^2),$$where $\lambda_j$ are the eigenvalues of $A$.
Edit to answer the comment
You are essentially asking to estimate the norm of the matrix by a function of its spectral radius, which is impossible without any additional info on the matrix (its eigenvectors and generalised eigenvectors, for example).
You can rewrite the matrix $A=PJP^{-1}$ where $J$ is its Jordan norm. You can write the norm of $J$ in terms of eigenvalues of $A$ and the structure of eigenvectors and generalised eigenvectors, then say that $\sqrt{\sigma_1(A)}=\|A\|\le cond(P)\|J\|$. However, the conditionment number $cond(P)=\|P\|\|P^{-1}\|$ is, in general case, very large; this, coupled with quite unstable behaviour of structure of Jordan form and of the matrix $P$ itself (with respect to coefficients of $A$), the above estimation on $\|A\|$ has little to no viable applications. On top of that, in real life we don't really have a reliable method of finding eigenvalues for non-symmetric matrices.