TITLE: Can the Borromean rings be unlinked if we allow each component to pass through itself?
QUESTION [9 upvotes]: Every knot and link can be untied if you allow it to pass through itself. (This is why the unknotting number is always finite.) This changes if you only allow each component of the link to pass through itself, prohibiting them from passing through any other component. The Whitehead link can be untied that way: the central crossing only involves one component, and flipping it unties the link. The Hopf link, on the other hand, cannot. This is because it has linking number $\pm1$ (depending on the relative orientations of the two components), and the linking number doesn't change when a component passes through itself.
Can the Borromean rings be unlinked if we allow each component to pass through itself? The above argument with the linking number doesn't work anymore, since the linking number between any two components is $0$.

REPLY [4 votes]: The Borromean rings are not link homotopic to the three component unlink, that is, one cannot transform the Borromean rings into the three component unlink by a deformation that allows components to pass through themselves, but does not allow one component to pass through another. One can use Milnor invariants to show this. Milnor's original work was part of his PhD thesis. I will use a formulation that can be found in this paper of Mellor and Melvin.
For a three component link, a complete invariant of link homotopy is given by the three pairwise linking numbers $p$, $q$, and $r$, and another residue class $\mu$ modulo the greatest common divisor of $p$, $q$, and $r$. The Borromean rings have $p=q=r=0$, but $\mu=\pm 1$ (depending on how one orients the link). The unlink has $p=q=r=\mu=0$, and so the Borromean rings are not link homotopic to the three component unlink.
The residue class $\mu$ is defined as follows. Let $X$, $Y$, and $Z$ be longitudes of the three components of the link, and let $F_X$, $F_Y$, and $F_Z$ be the corresponding Seifert surfaces of the components of $L$. Define $w_X$ to be a word in $y$ and $z$ as follows. Start at some point on $X$ and traverse the longitude once. Each time one passes through the Seifert surface $F_Y$, append either $y$ or $y^{-1}$ to $w_X$ depending on the orientation of the intersection. Similarly, append $z$ or $z^{-1}$ each time one passes through the Seifert surface $F_Z$. Define $m_{X}$ to be a signed count of the following patterns in $w_X$:

($y\dots z$): An instance of $y$ followed eventually by $z$ increases $m_X$ by one.
($y^{-1}\dots z^{-1}$): An instance of $y^{-1}$ followed eventually by $z^{-1}$  increases $m_X$ by one.
($y \dots z^{-1}$): An instance of $y$ followed eventually by $z^{-1}$  decreases $m_X$ by one.
($y^{-1}\dots z$): An instance of $y^{-1}$ followed eventually by $z$  decreases $m_X$ by one.

Similarly define $m_Y$ and $m_Z$. Let $t$ be the signed count of the number of triple points among the three Seifert surface $F_X$, $F_Y$, and $F_Z$. Then
$$\mu = m_X + m_Y + m_Z - t \mod \gcd(p,q,r).$$

The picture of the Borromean rings above is taken from the Mellor and Melvin paper. Let $F_X=F_i$, $F_Y=F_j$, and $F_Z=F_k$. Then $m_X=1$, $m_Y=m_Z=0$, and there are no triple points so $t=0$. Hence for the Borromean rings with the depicted orientation, $\mu=1$. Thus the Borromean rings are not link homotopic to the three component unlink.


TITLE: Could you explain the expansion of $(1+\frac{dx}{x})^{-2}$?
QUESTION [2 upvotes]: Could you explain the expansion of $(1+\frac{dx}{x})^{-2}$?
Source: calculus made easy by S. Thompson.
I have looked up the formula for binomial theorem with negative exponents but it is confusing. The expansion stated in the text is:
$$\left[1-\frac{2\,dx}{x}+\frac{2(2+1)}{1\cdot2}\left(\frac{dx}{x}\right)^2 - \text{etc.}\right] $$
Please explain at a high school level.

REPLY [2 votes]: Consider the expression $$A=(1+a)^{-2}=\frac 1{(1+a)^2}=\frac 1 {1+2a+a^2}$$ and, hoping you already know it, perform the long division. Limiting to first terms, you will arrive to $$A=1-2 a+3 a^2-4 a^3+5 a^4+\cdots$$ Now, replace in this last expression $a$ by $\frac{dx}x$; this will lead to $$A=1 - 2\left(\frac{dx}{x}\right) +3\left(\frac{dx}{x}\right)^2 - 4\left(\frac{dx}{x}\right)^3 + 5\left(\frac{dx}{x}\right)^4+\cdots$$
Another way, would be to consider  $$B=(1+a)^{-1}=\frac 1{1+a}$$ and perform the long division again. This will give $$B=1-a+a^2-a^3+a^4-a^5+\cdots$$ But $$\frac{dB}{da}=-\frac 1{(1+a)^2}=-A$$ which makes $$-A=\frac{d}{da}\Big(1-a+a^2-a^3+a^4-a^5+\cdots \Big)=-1+2 a-3 a^2+4 a^3-5 a^4+\cdots$$ Multiplying both sides by $-1$ leads to the previous result.


TITLE: Difference between "Lower sum" and "Riemann lower sum"
QUESTION [2 upvotes]: I am not getting the difference between "lower sum" and "Reimann lower sum" . Lemme explain through an example. (i know i am explaining it wrong, but i need to do it so u guys can correct me where i am wrong as i am not getting it).
Let f(x) be defined as :
f(x) =
$ \left\{
  \begin{array}{l l}
    \mid x \mid & \quad \text{if x $\neq$0}\\\\\\\
    1 & \quad \text {if x=0 }
  \end{array} \right.\]$
now, if we do lower Riemann sum of this function on the interval [-1,1] subdivided into 2 equal intervals,then we will have the partition as {0} such that :
-1 < 0 < 1.
so,we will do the Riemann lower sum as :
f(-1) (1) + f(1) (1) = 2
it can also be taken as :
f(0) (1) + f(0) (1) =2
or
f(-1) (1) + f(0) (1) =2
or
f(0) (1) + f(1) (1) =2 (as the infimum of f(x) in all these cases will be 1 for f(1),f(-1),f(0), so they can be written either way)
But then i came across the statement "The lower sum with respect to this partition, on the other hand, is truly zero.Therefore this particular lower sum can not be a Riemann sum." Why is it "0"???
having that said,i also saw that the supremum of f(x) between these 2 subintervals is also "1" . Thus the infimum and supremum is the same for the function. Is that possible??? I guess its not possible.
My question is whats the difference between a "lower sum" and a "Riemann lower sum" and why a continuous bounded function only has Riemann upper and lower sums??? [i know the fact that a continuous bounded function will have a max and min in all the subintervals it is divided into, but how to relate this fact with it being Riemann upper and lower sums???). I know its an easy one, if any of u guys has time to explain it plz do...thanks.

REPLY [1 votes]: There's no difference between a "lower sum" and a "lower Riemann sum."  There is a difference between a "left (Riemann) sum" and a "lower (Riemann) sum."  In particular, the left Riemann sum uses the leftmost point of each interval, while the lower Riemann sum uses the point on each interval at which the function is at a minimum.
(Of course for an increasing function the two are the same.)


TITLE: Positive integer solutions to $a^x-b^y=2$
QUESTION [2 upvotes]: Does anybody know the all positive integer solution of
$$
a^x-b^y=2
$$
under the condition
$$
x \geq 2, y\geq2
$$
I didn't find any solution without
$$
x=5,y=3,a=2,b=3
$$

REPLY [2 votes]: See Catalan's conjecture (Mihăilescu's theorem) for $a^x-b^y=1$, where it's also written about the general $a^x-b^y=n$.
https://oeis.org/A076427 says there's exactly one solution when $a^x,b^y<10^{18}$.
The solution is $(a,b,x,y)=(3,5,3,2)$.


TITLE: Find Tangent Points for Common Tangent to Two Curves Described with Parametric Equations
QUESTION [0 upvotes]: Consider one ellipse and one circle given by the parametric equations:
For the ellipse
$$ x = x_0 + a\cdot \cos(t_1)$$
$$ y = y_0 + b\cdot \sin(t_1)$$
and for the circle
$$ x = x_1 + \rho\cdot\cos(t_2) $$
$$ y = y_1 + \rho\cdot\sin(t_2) $$
How do I find the value of $t_1$ and $t_2$ for the two tangent points of their common tangent at the up side of the curves? See this image.

REPLY [0 votes]: We need the normals to the curves. They are vectors radiating away from the curves (ellipse or circle) such that they are always at a right angle to the curve. If normals to the curves are parallel, then your tangent line will intersect each norm (one on the ellipse and one on the circle) at a right angle. We know that the desired tangent line will intersect norms that are equal on the two curves.
Take the derivative of each curve to get the tangent. It will be the tangent at all points. Circle: $$ \left[\begin{array}{c}x\\y\end{array}\right]=\left(\begin{array}{c}x_{1}\\y_{1}\end{array}\right)+\rho\left(\begin{array}{c}cos(u)\\sin(u)\end{array}\right)$$
Tangent: $$\left[\begin{array}{c}x'\\y'\end{array}\right]=\left(\begin{array}{c}-\rho\cdot sin(u)\\\rho\cdot cos(u)\end{array}\right)$$ To get the norm, swap vector positions with the tangent elements and then change the sign of one of them.  It would be nice to be consistent about this. I have set the vectors to go away from the curves.
$$norm=\left(\begin{array}{c}\rho\cdot cos(u)\\\rho\cdot sin(u)\end{array}\right)$$
Similarly, the norm of the ellipse will be $$norm=\left(\begin{array}{c}b\cdot cos(t)\\a\cdot sin(t)\end{array}\right)$$ I have changed to using $t$ and $u$ for the parameters instead of $t_1$ and $t_2$. FYI to create a slope from a direction vector, use $y/x$.
Now that we have some preliminaries out of the way, here is the general idea for solution. My unknowns are point $C(x,y)$ and $D(x,y)$ and parameter $u$ and parameter $t$.  Therefore there are 6 unknowns. Take a look at the picture to understand where things are.
Equation 1: The y part of the parameterized circle. $$D_y=P_{1y}+\rho\cdot sin(u)\qquad\text{eq:01}$$
Equation 2: The x part of the parameterized circle. $$D_x=P_{1x}+\rho\cdot cos(u)\qquad\text{eq:02}$$
Equation 3: The y part of the parameterized ellipse. $$C_y=P_{y}+b\cdot sin(t)\qquad\text{eq:03}$$
Equation 4: The x part of the parameterized ellipse. $$C_x=P_{x}+a\cdot cos(t)\qquad\text{eq:04}$$
Equaton 5: The slope of the two vectors in the picture have to be equal. Call $m$ the slope. $$m=\frac{a\cdot sin(t)}{b\cdot cos(t)}=\frac{\rho\cdot sin(u)}{\rho\cdot cos(u)}\qquad\text{eq:05}$$
Now if there is a tricky part, then this is it. We will write the equation of a line (in point:slope form) that has slope $-1/m$ and goes through point $C$.  Then we will write the line equation that also has slope $-1/m$ and goes through point $D$. They are the same line, so we will equate them and lose the x and y to get the 6th equation.
$$y-C_y=-\frac{1}{m}\cdot (x-C_x)\qquad\text{1st line}$$ $$y-D_y=-\frac{1}{m}\cdot (x-D_x)\qquad\text{2nd line}$$
$$-\frac{1}{m}(x-C_x)-y+C_y=-\frac{1}{m}(x-D_x)-y+D_y\qquad\text{equated lines}$$ Equation 6: After some minor algebra: $$\frac{C_x}{m}+C_y=\frac{D_x}{m}+D_y\qquad\text{eq:06}$$
I am not going to solve these symbolically! You can if you have to.
Let $a=0.2\;b=0.5\;\rho=0.5\;\;P=(1,1.75)\;P_1=(2,2)$ That is shown in the picture. There are actually 4 tangent lines, 2 on each side of the circle and ellipse and 2 that cross over one another on the interior.
Just for the ones in the picture, point $C=(0.98,2.25)$ and $D=(1.87,2.48)$ $t=1.68$ and $u=1.83$


TITLE: Is it possible that everyone is a pseudo-winner in a tournament with 25 people?
QUESTION [2 upvotes]: Is it possible, that everyone is a pseudo-winner in a tournament with 25 people?
(pseudo-winner means that either he won against everyone, or if he lost against someone, then he beated someone else, who beated the one who he lost to).
In the "language" of graph theory: Is it possible in the directed K(25) (all edges drawn with 25 vertices), that from all vertices, every vertices can be reached with a maximum of 2-long route.
I thought about this, and my answer is no. I tried to do the proving, starting with K(3), which is a triangle. Triangle works, but I don't know how to continue this until 25 vertices.
Maybe any easier solutions?
Thanks!

REPLY [0 votes]: The answer is a well-known graph generalizing a classical 3-vertex construction:

To see that RPS-25 satisfies the conditions, it's enough to check that each shape beats exactly $12$ others. As a result, if shape $A$ loses to shape $B$, then among the $12$ shapes that shape $A$ beats, at most $11$ can lose to shape $B$: so there must be a shape $C$ that beats $B$ and loses to $A$.
(And so it's easy to find other examples, for any odd $n$: just take any regular tournament.)


TITLE: If a, b, c are three linearly independent vectors show that the vectors a × b, b × c, c × a are also linearly independent.
QUESTION [1 upvotes]: If $a$, $b$, $c$ are three linearly independent vectors show that the vectors $a \times b$, $b \times c$, $c \times a$ are also linearly independent.

I have tried proving with three linearly independent vectors and working out the determinants of the cross products, but I cannot seem to get them to equal 0.
Any help would be appreciated, thank you.

REPLY [1 votes]: Hint It's enough (in fact, equivalent) to show that the triple product
$$\newcommand{\bfa}{{\bf a}}\newcommand{\bfb}{{\bf b}}\newcommand{\bfc}{{\bf c}} \newcommand{\bfx}{{\bf x}}\newcommand{\bfy}{{\bf y}}\newcommand{\bfz}{{\bf z}}[(\bfa \times \bfb) \times (\bfb \times \bfc)] \cdot (\bfc \times \bfa)$$ of $\bfa \times \bfb, \bfb \times \bfc, \bfc \times \bfa$ is nonzero.
We can rewrite this quantity using the occasionally useful cross product identity
$$
(\bfx \times \bfy) \times (\bfx \times \bfz) = (\bfx \cdot(\bfy \times \bfz)) \bfx .
$$

Substituting gives that the triple product is $$(\bfa \cdot (\bfb \times \bfc))^2,$$ but this is just the square of the triple product of the original vectors $\bfa, \bfb, \bfc$; since the set of these vectors is linearly independent, this latter triple product is nonzero.


TITLE: Question about the principles behind AC generators
QUESTION [0 upvotes]: I've learnt that when the normal drawn from the surface of a coil (the area vector) is in the same direction as the magnetic field, i.e. when the field is perpendicular to the coil, then there are a maximum number of flux lines passing through the coil - which means as soon as the coil is brought into the field/the field is "switched on" emf is induced in the coil as per Faraday's law.
However, in the AC generator, when a coil is placed parallel to the field (with its area vector normal to the field) maximum emf is induced, whereas when it is placed perpendicular there is no emf induced. How is this?
Also, if I were to push a coil into a magnetic field parallel to the field (making an angle of 0), would the emf induced be maximum here? More than if the coil were perpendicular?
As you can see I'm more than a little confused here, a little clarification would be great.

REPLY [0 votes]: For the following discussion, assume that a rectantular coil of wire is rotating in a magnetic field, with the axis of rotation being down the length of the rectangle (e.g., the longer dimension of the rectangle).  Further assume that the axis of rotation is perpendicular to the magnetic field lines, and that all of the coil except the ends is "bathed" in the magnetic field.  For these circumstances, maximum magnetic flux is passing through the rotating coil when the plane of the coil is perpendicular to the magnetic field lines, so it would seem that maximum emf is produced at this point.  However, as non-intuitive as it seems, this is when the induced emf in the coil drops to zero, because the wires that make up both sides of the coil are moving parallel to the magnetic field lines when the coil is in this position.  As such, the charge carriers in the wires do not feel a magnetic force on them at this instant in time.  As the coil continues to turn in the magnetic field, it soon gets to the point where the plane of the coil is parallel to the magnetic field lines, and the wire on each side of the coil is cutting through the magnetic field lines at a 90 degree angle and at the maximum rate, which is causing the maximum rate of change of magnetic flux in the coil, leading to the maximum emf at that point in the rotational cycle.  Thus, maximum emf is generated when the wires of the coil are cutting through the magnetic field lines at the maximum rate, which occurs when the coil has minimum magnetic flux "flowing" through it.
Regarding pushing a coil into a magnetic field, the maximum emf is induced in the coil whenever you are changing the magnetic flux in the coil at the maximum rate.    This will occur when the coil is oriented such that the plane of the coil is perpendicular to the magnetic field, which seems "opposite" to the comments above.  However, keep in mind that the two different motions of the coil (e.g., one case is rotating the coil in a magnetic field while the other case is manually pushing a coil into a magnetic field) lead to two different orientations that produce maximum emf.  In both cases, the maximum emf is produced when the coil is experiencing the maximum rate of change of magnetic flux, which is not necessarily the point where the maximum magnetic flux through the coil occurs.


text:
An element in a free group is \emph{primitive} if it is an element of
some basis, or free generating set.
Failure of primitivity can be quantified:
define the \emph{imprimitivity rank} of an element to be the minimal
rank of a subgroup containing it as an imprimitive element, if such a
subgroup exists, or infinite otherwise.
An element has imprimitivity rank 0 if and only if it is trivial, 1 if and
only if it is a proper power, and $\infty$ if and
only if it is a primitive element.
In these cases the quotient of the free group by the subgroup
normally generated by the element is a hyperbolic
group, either a free group, in the first and third cases, or a
one-relator group with torsion, which is hyperbolic by
the B.\ B.\ Newman spelling theorem \cite{MR0222152}, in the second case.
Nonelementary, torsion-free two-generator one-relator groups have
relators of imprimitivity rank 2.
There are many nonhyperbolic
groups of this form, such as $\mathbb{Z}^2=\langle a,b\mid
aba^{-1}b^{-1}\rangle$, the Baumslag-Solitar
groups $BS(m,n)=\langle a,b\mid
ab^ma^{-1}b^{-n}\rangle$, and the groups considered by Gardam and
Woodhouse \cite{MR3876736}.
Louder and Wilton \cite[Theorem~1.4]{LouWil18a} show that
two-generated subgroups of a higher imprimitivity rank one-relator
group are free.
Thus, they are of Type $F$ and have no
Baumslag-Solitar subgroups.
It is a long-standing open question whether such groups must be hyperbolic.
Louder and Wilton conjecture \cite[Conjecture~1.6]{LouWil18a} a positive answer for
one-relator groups.

We offer experimental support for their conjecture.
Fix a basis for a free group, so that a group element can be uniquely represented as a freely reduced word, a product of basis elements and their
inverses, of a well-defined length.
 \begin{theorem}\label{rank3}
    Let $w$ be a word in $\F_r$ of length $L$ and imprimivity
    rank not equal to 2. Then $\F_r/\llangle w\rrangle$ is hyperbolic
    if $r\leq 4$ and $L\leq 17$.
  \end{theorem}

  These results are achieved computationally, by a combination of efficient enumeration
  of representatives and brute force\footnote{We ran 12 x 4 core
    Intel Core i5-4670S @ 3.10GHz for two months.}.
  Details are in \fullref{sec:experiment}.

 We also observe that a well-known result about hyperbolicity of
 one-relator groups is consistent with the conjecture. In these
 results $|w|_a$ denotes the total number of occurrences of $a$ and
 $a^{-1}$ in $w$, where $w$ is freely reduced and $a$ is an element of the chosen basis.
 \begin{proposition}\label{prop:ISirank}
    The nonhyperbolicity criteria of Ivanov and Schupp \cite[Theorems~3~\&~4]{MR1401522}
    imply imprimitivity rank 2.
  \end{proposition}
  The proposition is proven in \fullref{sec:IS}.
As a consequence, we have:
  \begin{corollary}\label{IS}
    Let $w$ be a cyclically reduced word in $\F_r$ of imprimitivity rank not equal to 2
    such that $0<|w|_a<4$ for some basis element $a$. Then
    $\F_r/\llangle w\rrangle$ is hyperbolic.
  \end{corollary}
  \begin{corollary}\label{ISbound}
    Let $w$ be cyclically reduced word in $\F_r$ of length less than $4r$ and
    imprimitivity rank not equal to 2 such that every generator or its
    inverse occurs in $w$.
    Then $\F_r/\llangle w\rrangle$ is hyperbolic.
  \end{corollary}

Combining these results with our experimental results, we have:
  \begin{corollary}
    Let $w$ be a word in $\F_r$ of length at most 17 and imprimitivity
    rank not equal to 2. Then $\F_r/\llangle w\rrangle$ is hyperbolic.
  \end{corollary}
  \begin{proof}
    $\F_r/\llangle w\rrangle$ is hyperbolic when the imprimitivity
    rank of $w$ is 0, 1, or $\infty$, so suppose it is finite and at least
    3.
    Up to replacing $w$ by an element in the same automorphic orbit,
    we may, without increasing the length of $w$, assume that it is
    cyclically reduced and that there is $s$ such that taking the
    first $s$ basis elements and the last $r-s$ basis elements gives a
    splitting  $\F_r=\F_s*\F_{r-s}$ where the $\F_s$
    factor is the smallest free factor containing $w$.
    Since $w$ is imprimitive in $\F_r$, it is imprimitive in $\F_s$,
    so $s$ is an upper bound on imprimitivity rank, which implies
    $s\geq 3$.
    Furthermore, since $\F_s$ is the smallest free factor containing
    $w$, all of the generators of $\F_s$ or their inverses occur in $w$.
    Since $\F_r/\llangle w\rrangle\cong (\F_s/\llangle w\rrangle) *
    \F_{r-s}$ is hyperbolic if and only if $\F_s/\llangle w\rrangle$
    is,
    we conclude by applying \fullref{rank3} or
    \fullref{ISbound} to $\F_s/\llangle w\rrangle$, according to
    whether  $s<5$ or $s\geq 5$, respectively.
  \end{proof}

  \subsection*{Additional conjectures}
  In checking the hyperbolicity conjecture, we enumerated the
  $\Aut(\F_4)$ orbits of cyclic subgroups of $\F_4$ that have a
  representative that can be generated by a word of at most length 16.
  We also computed imprimitivity ranks for these words.
  Armed with this data, we can test other questions involving
  imprimitivity rank.
  We check two additional conjectures and find that they are consistent with the
  data up to length 16. The first of these concerns uniqueness of the
  subgroup in the definition of imprimitivity rank, see
  \fullref{uniqueWsubgroups}.
  The second concerns the relationship between imprimitivity rank and
  stable commutator length, see \fullref{scl}.

  \subsection*{Acknowledgements}
We thank Henry Wilton for his comments on an earlier draft, in
particular for the suggestion to check Heuer's conjecture.

\endinput

config: arxiv


TITLE: Nonequivalent Colorings of the corners of a pentagon
QUESTION [0 upvotes]: It is known that there are eight nonequivalent colorings of the corners of a regular pentagon with the colors red and blue. Explicitly determine eight nonequivalent colorings.
Using Burnside's Lemma I found that there exist exactly eight nonequivalent colorings for a pentagon using 2 colors.
I know how to find different permutations of a certain coloring to get equivalent colorings but how do I find the nonequivalent colorings?

REPLY [1 votes]: A good way of finding all possibilities is to distinguish cases which are clearly different. In this example we can take the number of red sides as such a distinguishable property (as two colorings are always nonequivalent if they haven't the same number of red corners):
0 corners red: 1 possibility
1 corners red: 1 possibility (they are equivalent by rotation)
2 corners red: 2 possibilities (1 when the red dots are neighbours and 1 when they are not)
3 corners red: 2 possibilities (1 when the blue dots are neighbours and 1 when they are not)
4 corners red: 1 possibility
5 corners red: 1 possibility
Total: 8 possibilities
Note that it wasn't necessairely to do the work for 3, 4 and 5 red corners as we could have noticed that there should be 2, 1 and 1 possibility by symmetry (as m red corners and n blue corners should give the same number of possibilities as n red and m blue corners).


TITLE: Why does $A=UDU^H$ = $\lambda_1 u_1 u_1^H + \lambda_2 u_2 u_2^H$?
QUESTION [1 upvotes]: Just trying to figure out why this is true:
$$A=UDU^H \quad\Leftrightarrow\quad A= \lambda_1u_1u_1^H + \lambda_2u_2u_2^H$$
$U$ is a unitary matrix composed of the eigenvectors of hermitian matrix A.
$$Au_1=\lambda_1u_1$$
$$Au_2=\lambda_2u_2$$
So
$$Au_1u_1^H+Au_2u_2^H = \lambda_1u_1u_1^H + \lambda_2u_2u_2^H$$
So why is
$$Au_1u_1^H+Au_2u_2^H = A$$
????
EDIT:
Let $u_i=(a_i,b_i)^T$
$$u_i^Hu_i=\overline{a_i}a_i+\overline{b_i}b_i=1$$
$$u_iu_i^H+u_ju_j^H=\pmatrix{a_i\overline{a_i}&a_i\overline{b_i}\\b_i\overline{a_i}&b_i\overline{b_i}}+\pmatrix{a_j\overline{a_j}&a_j\overline{b_j}\\b_j\overline{a_j}&b_j\overline{b_j}}$$
So now how to arrive from the above statement to identity?

REPLY [1 votes]: The easiest way to see this is with block-matrix multiplication.
Let $u_1,u_2$ denote the columns of $U$.  We have
$$
UDU^H = \\
\pmatrix{u_1 & u_2} \pmatrix{\lambda_1\\&\lambda_2} \pmatrix{u_1^H\\u_2^H}
= \\
\pmatrix{u_1 & u_2} \pmatrix{\lambda_1u_1^H\\ \lambda_2u_2^H} = \\
u_1(\lambda_1 u_1^H) + u_2(\lambda_2 u_2^H) = \\
\lambda_1 u_1 u_1^H + \lambda_2 u_2 u_2^H
$$


TITLE: root of unity in quotient field
QUESTION [1 upvotes]: Let R be an commutative ring, in which $n \ge 2$ is invertible. Show that the modulo class of $y$ in the quotient ring $ R'=R[y]/\langle y^n-1\rangle $ is not an primitiv $n$th-root of unity in R'.
Hi,
this is a problem which bugs me for quite a while.
I don't know exactly where I shall begin. I know n is invertible which means there exists $n^{-1} \in R$ and n is invertible in R' too, and the modulo class of $y$ is $y~mod~y^n-1$. Can someone give me an hint how I can  proceed from here? Thank you.
Greets Bernd.

REPLY [0 votes]: I think you need $n$ to be even, and $R\neq0$ for this to be true.
As I indicated in the comments, I would first prove that the polynomials $y-1$ and $y^{n-1}+\cdots+1$ are relatively prime in $R[y]$, i.e., the ideal
$$
(y-1,y^{n-1}+\cdots+1)
$$
in $R[y]$ is the trivial ideal $(1)$. In order to show this, use transitivity of quotients:
$$
R[y]/(y-1,y^{n-1}+\cdots+1)=
(R[y]/(y-1))/(\bar y^{n-1}+\cdots+1)=
R/(1+\cdots+1)=R/n=0
$$
since $n$ is invertible in $R$, where $\bar y$ denotes the class of $y$ mod $y-1$, i.e., $\bar y=1$ in $R[y]/(y-1)=R$. Therefore
$$
(y-1,y^{n-1}+\cdots+1)=(1)
$$
in $R[y]$.
By the Chinese remainder theorem, the quotient morphisms
$$
R[y]\rightarrow R[y]/(y-1)=R
\quad\text{and}\quad
R[y]\rightarrow R[y]/(y^{n-1}+\cdots+1)
$$
induce an isomorphism of rings
$$
R'=R[y]/(y^n-1)\rightarrow R\times R[y]/(y^{n-1}+\cdots+1).
$$
This morphism send $y$ to the element $(1,\bar y)$, where $\bar y$ now denotes the class of $y$ modulo $y^{n-1}+\cdots+1$.
It is now clear that $y$ is not a primitive $n$-th root of unity in $R'$, i.e., that $y$ does not generate the subgroup of $(R')^\times$ of elements having order a divisor of $n$. Indeed, since $n$ is even $(-1,1)$ is a $n$-th root of unity but
$$
(-1,1)\neq (1,\bar y)^m
$$
for all integers $m$, otherwise $1=-1$ in $R$, hence $2=0$ in $R$, and $n=0$, and then $R=0$, which was excluded. This proves that $y$ is not a primitive $n$-root of unity in $R'$ if $n$ is even and $R\neq0$.
If $n$ is odd then the statement is not true.
Take $n=3$ and $R=\mathbf Z$. One gets $R'=\mathbf Z\times\mathbf Z[j]$, where $j$ is a primitive cubic root of unity, and the class of $y$ in $R'$ corresponds to the element $(1,j)$. The group of cubic roots of unity in $\mathbf Z\times\mathbf Z[j]$ is $\{(1,1),(1,j),(1,j^2)\}$, and $(1,j)$ is a generator. Hence, $y$ is a primitive cubic root of unity this case.


TITLE: Linear Algebra: Change of Basis Matrix and standard basis
QUESTION [0 upvotes]: Find the matrix of transformation $ \mathbb{R}^3 \to \mathbb{R}^4$ taking $v_1 = (1,2,3)^T$ to $w_1 = (1,0,0,2)^T$, $v_2 = (2,3,4)^T$ to $w_2 = (1,2,3,4)^T$ and $v_3 = (1,1,2)^T$ to $w_3 = (0,5,1,7)^T$ by first finding the change of basis matrix from the basis ${v_1,v_2,v_3}$ to the standard basis and then multiplying it by the matrix which takes the standard basis to $w_1, w_2$ and $ w_3,$ respectively. Explain why this gives you the right answer.

REPLY [1 votes]: We traditionally write vectors as column vectors. Since this takes up a lot more space on a piece of paper, we write them transposed. For example, if i write $[1,2,3]^T$ in this sentence, it takes one line, but $\begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}$ takes up a bit more than 3, which makes things hard to read and look ugly. But both denote the same vector, $\begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}$.


TITLE: Computing dilutions
QUESTION [1 upvotes]: I have a source bottle that is $32$ oz. at $10\%$ concentration.
Customer support said I need $1.5$ ounce per $10$ gallons for the final product?  What is this final product concentration?  How do I figure that out?
How will I dilute future batches of different final product size (1 gallon, 2 gallon, etc.) to match this final product concentration?

REPLY [0 votes]: Welcome to Mathematics Stack Exchange.
To figure out the volume ratio, convert gallons to ounces.
There are $128$ ounces in a gallon, so $1280$ in $10$.
Therefore, the final product concentration is $\dfrac {1.5}{1280}\times 10\%\approx0.012\%.$
If you need $ 1.5$ ounce for $10$ gallons, you need $0.15$ ounce per $1$ gallon.


TITLE: Quotient-Remainder Theorem Proving
QUESTION [1 upvotes]: This theorem is obviously correct. Now I try to prove it by well-ordering principle.
But I don't know where to start the proving....

REPLY [1 votes]: Suppose $\,S\subset \Bbb Z \,$ and $\,0<b\in\Bbb Z.\,$ If $\,S\,$ has a nonnegative integer $\,a,\,$  and $\,n\in S\,\Rightarrow\,n\!-\!b\in S\,$ then $\,S\,$ has least nonnegative element $\,\ell < b\,$ (otherwise $\,\ell\!-\!b\in S\,$ and $\,0 < \ell\! -\!b < \ell,\,$ contra minimality of $\,\ell).\,$  Therefore if $\ a\ge 0\,$ then $\,S = a-b\,\Bbb Z\,$ has least nonnegative element $< b.$


TITLE: $\sum_{n=1}^{\infty}\frac{1}{n^22^n}$ by integration or differentiation
QUESTION [12 upvotes]: There is an infinite sum given:
$$\sum_{n=1}^{\infty}\frac{1}{n^22^n}$$
It should be solved using integration, derivation or both.
I think using power series can help but I don't know how to finish the calculation. Any help will be appreciated!

REPLY [2 votes]: My answer to a duplicate question says:
$$\frac{1}{1-x}=\sum_{n=0}^{\infty} x^{n}$$
 Integrating from 0 to t we get
$$\int_{0}^{t}\frac{1}{(1-x)}dx=\sum_{n=0}^{\infty}\int_{0}^{t} x^{n}dx$$$$-\ln(1-t)=\sum_{n=1}^{\infty} \frac{t^{n}}{n}$$
Dividing by t and integrating
$$\int_{0}^{0.5}-\frac{\ln(1-t)}{t}dt=\sum_{n=1}^{\infty}\int_{0}^{0.5} \frac{t^{n-1}}{n}dt=\sum_{n=1}^{\infty} \frac{1}{n^{2}2^{n}}$$
This on calculating is $$ \dfrac{\pi^2}{12}-\dfrac{ln^2(2)}{2},$$


TITLE: Geometric imagination of differential forms
QUESTION [37 upvotes]: In order to explain to non-experts what a vector field is, one usually describes an assignment of an arrow to each point of space. And this works quite well also when moving to manifolds, where a generalized arrow will be a tangent vector.
My question is:
What are similar objects that can help with imagining differential forms?
Positive qualities for such an object would be, for example:

it helps justify change-of-coordinate formulas and formulas for the pullback via a function;
it is "easily drawable";
it helps understand more complicated differential-form-based concepts, e.g. connections, cohomology groups, etc.

REPLY [0 votes]: I'd like to comment on some of the ideas here:
You should ask yourself whether you want to make the (implicit) assumption that your manifold is already endowed with a volume form $\omega$. If the answer is "yes", then this gives you an isomorphism between $k$-vectors (antisymmetric tensors of degree $k$ of tangent vectors) and $n-k$-forms (where $n$ is the dimension of $M$) by inserting the $k$-vector into $\omega$. From this you get your interpretation: $n-k$-forms describe orientation and length/area/volume (choose depending on dimension) of a small $k$-dimensional cube. This isomorphism between $k$-vectors and $n-k$-forms essentially is the one that lets us describe the orientation of a flat surface in 3 dimensions by one arrow perpendicular to it, or by two non-parallel arrows inside of it.
But what if you do not want to make this assumption? Then you have to decide which one of forms and tangent vectors are the "arrow perpendicular to it" and which are the "arrow inside of it". I'd say that the tangent vectors are "arrows inside of it" and forms are "arrows perpendicular to it". For example, the differential of a smooth function are the arrows perpendicular to the levelsets of the function, while the tangent vectors of a curve are the arrows "along" the curve, which becomes "inside" the curve if we zoom in so that the curve becomes just a line.
Whatever version you choose, it might always be good to keep in mind that our intuition likes to implicitly apply the isomorphism between forms and vectors, so we have to be careful if this is not allowed.


TITLE: If heat is supplied to a system consisting of water and ice at 0 degrees Celsius, which will absorb heat first and what is the underlying physics?
QUESTION [0 upvotes]: I am supposed to answer this question in terms of the Zeroth and First Law of Thermodynamics. So, I wrote this:
“If heat is supplied to a system, the system will absorb the heat first. In this case, the system i.e. both water and ice will absorb the heat simultaneously. However, since both water and ice are at 0° C, according to the Zeroth Law of Thermodynamics, there is not going to be any change in temperature rather the heat absorbed by the system will be used as heat of fusion to transform ice to water. This further conserves the First Law of Thermodynamics as the heat absorbed is used to do work on the ice thus increasing its internal energy.”
However, I am still unsure of the accuracy of my explanation, and that is why I am here.

REPLY [0 votes]: It may be possible for both the water and ice to absorb the heat simultaneously if the system is heated reversibly. Consider the following thought experiment:
Let the ice and water mixture be in a perfectly insulated metal container for a long time so that we can be assured the ice and water are in thermal equilibrium with one another at 0$^o$C.
We now remove the thermal insulation from the metal container and immerse the container in a large thermal bath (constant temperature thermal reservoir) whose temperature is infinitesimally greater than 0$^o$C. This allows us to heat the ice water mixture reversibly. (If necessary, to insure uniform heating, we could include a small stirrer to circulate the water and ice, and include the stirrer work as part of the energy transfer to the mixture).
So since the thermal bath is in thermal equilibrium with the water and in thermal equilibrium with the ice, per the Zeroth law the water and ice should be in thermal equilibrium with one another.
Finally, per the first law and assuming all the ice doesn't melt, since there is no temperature increase the heat transfer from the thermal bath will equal the mass of ice in the mixture that melts times the latent heat of fusion.
Hope this helps.


TITLE: Functional equation involving three different functions: $ f ( x + y ) = g ( x ) + h ( y ) $
QUESTION [1 upvotes]: If $ f , g , h : \mathbb R \to \mathbb R $ all are continuous functions such that
$$ f ( x + y ) = g ( x ) + h ( y ) \text , \quad \forall x , y \in \mathbb R \text , $$
find $ f $, $ g $ and $ h $.

I have literally no idea where to begin. I mean, what can I even say or claim and prove? How should I go about this?
Also, a quick doubt I had was if a function is continuous over $\mathbb{R}$, can we safely say that it must be a linear function? If not, Why?

REPLY [1 votes]: Note that
$$g(x) + h(y) = f(x + y) = f(y + x) = g(y) + h(x) \implies h(x) - g(x) = h(y) - g(y)$$
for all $x, y \in \Bbb{R}$. That is, $h - g$ is a constant function, i.e. there exists some $k \in \Bbb{R}$ such hat $h(x) = g(x) + k$ for all $x \in \Bbb{R}$.
This gives us the equivalent functional equation
$$f(x + y) = g(x) + g(y) + k.$$
Note that, when $y = 0$, we simply see that
$$f(x) = g(x) + g(0) + k,$$
hence
$$g(x + y) + g(0) + k = g(x) + g(y) + k \implies g(x + y) + g(0) = g(x) + g(y).$$
Let $L(x) = g(x) - g(0)$. Then, the above equation simplifies to
$$L(x + y) = L(x) + L(y),$$
which is Cauchy's functional equation. Since $g$ is continuous, so is $L$, and hence $L$ is linear. On $\Bbb{R}$, this means $L(x) = ax$ for some $a \in \Bbb{R}$.
So, rebuilding, we have
\begin{align*}
g(x) &= ax + c \\
h(x) &= ax + c + k \\
f(x) &= ax + 2c + k,
\end{align*}
where $a, c, k \in \Bbb{R}$ are parameters. Checking this family of possible solutions, we get
$$f(x + y) = a(x + y) + 2c + k = ax + c + ay + c + k = g(x) + h(y),$$
verifying that all functions of the above form are indeed solutions, yielding a complete family of solutions.


TITLE: Find the range of $a$ for $a^2 - bc - 8a + 7 = 0$ and $b^2 + c^2 + bc - 6 a + b = 0$.
QUESTION [5 upvotes]: Let $a, b, c$ be real numbers such that
\begin{cases}
a^2 - bc - 8a + 7 = 0\\
b^2 + c^2 + bc - 6 a + b = 0\\
\end{cases}
Find the range of $a$.
By adding the two equations I have $$(a - 7)^2 + (b + \frac 12)^2 + c^2 = \frac {169}4,$$
thus $a \in [\frac 12, \frac {27}2]$. Then I found that $a$ cannot obtain the boundaries so $a \in (\frac 12, \frac {27}2)$. But this is only a rough restriction. Exactly how do I solve this?
Edit: When I got this question from my friend, it was a $b$, but after I saw the official solution, it should be a $6$ instead. So the official answer I posted is actually not the answer to the question here but the answer to the original question. I'm really sorry for this mistake.

REPLY [0 votes]: This is the "official" solution which I found later:
\begin{cases}
a^2 - bc - 8a + 7 = 0\\
b^2 + c^2 + bc - 6 a + \mathbf6= 0\\
\end{cases}
is equivalent to
\begin{cases}
bc=a^2 - 8a + 7\\
b+c=\pm(a-1)\\
\end{cases}
So $b,c$ are the roots of the equation
$$x^2\pm(a-1)x+(a^2-8a+7)=0$$
A necessary and sufficient condition that $b,c\in\mathbb{R}$ is
$$\Delta=(a-1)^2-4(a^2-8a+7)\ge0$$ Solving this gives the answer $a\in[1,9]$.


TITLE: Surface terms for path integrals in field theory?
QUESTION [9 upvotes]: My question relates to something that I´ve seen in many books and appears in all its glory here: Ryder, pg 198
My question is about eq. 6.74. Which I repeat below:
$$i \int {\cal D}\phi \frac{\delta \hat{Z} [\phi] }{\delta \phi} \exp \left(i\int J(x) \phi(x) dx\right)$$
$$= i\; \exp \left(i\int J(x) \phi(x)  dx\right) \hat{Z}[\phi] \Bigg|_{\phi\rightarrow\infty}$$
$$+ \int {\cal D}\phi J(x) \hat{Z}[\phi] \exp \left(i\int J(x) \phi(x)  dx\right).\tag{6.74}$$
$\phi$ is a scalar field, $J$ is a source, $x = x_{\mu}$ in 4D Minkowski space and $$\hat{Z}[\phi] = \frac{e^{iS}}{\int {\cal D}\phi\; e^{iS}}.$$
The author is clearly doing a integral by parts and the first term on the right hand side is a kind of surface term for the path integral. He then considers this term to be zero and the second one gives us:
$$i \int {\cal D}\phi \frac{\delta \hat{Z} [\phi] }{\delta \phi} \exp \left(i\int J(x) \phi(x) dx\right) = J(x) Z[J]$$
The trick thing here is that integral limits for $\int{\cal D}\phi$ are not very obvious (at least not to me). You are in fact summing up for all field configurations. So, there are actually two problems in my mind:

For what configuration of $\phi$ is the surface term calculated? (the author says it is $\phi \rightarrow \infty$)

Assuming the author is right about taking huge $\phi$: why is this term zero?

This applies to path integrals in general: can we do the usual trick of throwing out surface terms safely?

REPLY [1 votes]: Simple way to get the identity
IMHO the matter has nothing to do with the behavior of $\phi$ at infinity. It has to do with a simple property of the path integral, namely, that the path integral of a functional derivative is zero (I shall explain it in a second).
To explain how it is employed, recall that functional derivatives obey the product rule. Therefore we have $$\dfrac{\delta}{\delta \phi(z)}\bigg[\hat{Z}[\phi]\exp\left(i \int d^dx \ J(x)\phi(x)\right)\bigg]=\dfrac{\delta\hat{Z}[\phi]}{\delta \phi(z)}\exp\left(i\int d^dx \ J(x)\phi(x)\right)+\\+\hat{Z}[\phi] iJ(z)\exp\left(i\int d^dx \ J(x)\phi(x)\right)\tag{1}.$$
Multiplying by $i$ and functionally integrating against $\mathfrak{D}\phi$ and employing the property I have claimed the LHS functionally integrates to zero. Therefore we obtain the identity $$i \int \mathfrak{D}\phi \dfrac{\delta\hat{Z}[\phi]}{\delta \phi(z)}\exp\left(i\int d^dx J(x)\phi(x)\right)=\int \mathfrak{D}\phi \hat{Z}[\phi]J(z)\exp\left(i \int d^d x \ J(x)\phi(x)\right)\tag{2}.$$
This is the equation in the OP with the first term already discarded since it corresponds to the functional integral of the functional derivative which I have just said that vanishes. Moreover we have been more careful to denote the free variable as $z$ instead of $x$ to avoid confusion with the integration variable in the exponent.
At first sight this would seem to be a result of assuming some sort of Stokes theorem holds here allowing the conversion of functional integrals of functional derivatives into boundary terms in function space together with the assumption that such boundary terms vanish. This is what I think the notation $\phi\to \infty$ means here: a term in the "boundary of function space".
Particularly I don't buy this argument (and honestly have no idea how to describe the boundary of function space in useful terms). Therefore I have a much simpler one to offer to explain this identity.
Functional integral of a functional derivative is zero
There is a simple argument for this if we assume translation invariance of the path integral measure $\mathfrak{D}\phi$ (which Physicists do assume all the time anyway). In the end of the day the claim is that if $F[\phi]$ is an arbitrary functional of $\phi$ which is functionally differentiable then $$\int \mathfrak{D}\phi \dfrac{\delta F[\phi]}{\delta \phi(z)}=0.\tag{1}$$
To understand this consider one infintiesimal variation $\delta \phi$ of the field configuration $\phi$ and evaluate the change in $F$:
$$\delta F[\phi]:=F[\phi+\delta\phi]-F[\phi]\tag{2}.$$
Now functionally integrate (2). Assuming linearity of the functional integral you'll have two functional integrals. In the first we change varaibles defining $\phi+\delta \phi = \phi'$. Assuming translation invariance of $\mathfrak{D}\phi$ we have $\mathfrak{D}\phi'=\mathfrak{D}\phi$ and therefore we have $$\int \mathfrak{D}\phi \delta F[\phi]=\int \mathfrak{D}\phi F[\phi+\delta \phi]-\int\mathfrak{D}\phi F[\phi]=\int \mathfrak{D}\phi' F[\phi']-\int \mathfrak{D}\phi F[\phi] = 0.\tag{3}$$
Therefore the functional integral of $\delta F$ is always zero. Now expand $\delta F$ in a functional Taylor series: $$\int \mathfrak{D}\phi \delta F[\phi] = \int \mathfrak{D}\phi \left[\int d^dz \dfrac{\delta F[\phi]}{\delta \phi(z)}\delta \phi(z)+\dfrac{1}{2!}\int d^dz_1 d^dz_2\dfrac{\delta^2 F[\phi]}{\delta \phi(z_1)\delta \phi(z_2)}\delta \phi(z_1)\delta \phi(z_2)+\cdots\right]\tag{4}$$
this thing must be zero. Therefore we equate it to zero term by term. The first term gives $$\int \mathfrak{D}\phi\int d^dz \dfrac{\delta F[\phi]}{\delta \phi(z)}\delta\phi(z)=0.\tag{5}$$
Finally assuming that we can interchange the path integral and the spacetime integral yields the equation $$\int d^dz \left(\int \mathfrak{D}\phi \dfrac{\delta F[\phi]}{\delta \phi(z)}\right)\delta \phi(z) =0. \tag{6}$$
Since the variation $\delta \phi(z)$ is arbitrary, equation (6) tells us that in the distributional sense equation (1) must hold.
This derivation of (1) is formal as many manipulations in Physics with path integrals but notice that the central assumption is translation invariance of $\mathfrak{D}\phi$ which is a standard assumption in Physics anyway. In the end, no need to talk about any boundary of function space nor any functional Stokes' theorem (which IMHO would also be formal and even more formal than this approach).