TITLE: Multiplication group for $\mathbb Z_n$ modulo $n$
QUESTION [1 upvotes]: By definition:
Let $\mathbb Z^+_n = \{[0],[1],[2],\ldots,[n−1]\}$
$\mathbb Z^+_4 = \{[0],[1],[2],[3]\},$
but
how $\mathbb Z^*_{12}$ is $\{[1],[5],[7],[11]\}$ ?
how $\mathbb Z^*_{7}$ is $\{[1],[2],[3],[4],[5],[6]\}$ ?

REPLY [1 votes]: $\require{cancel}$
\begin{align}
\frac 1 {12} & = \frac 1 {12} & \text{No cancellation occurs here.} \\[6pt]
\frac 2 {12} & = \frac {\cancel2\times1}{\cancel2\times6} = \frac 1 6 \\[6pt]
\frac 3 {12} & = \frac {\cancel3\times 1}{\cancel3\times 4} = \frac 1 4 \\[6pt]
\frac 4 {12} & = \frac{\cancel4\times1}{\cancel4\times3} = \frac 13 \\[6pt]
\frac 5 {12} & = \frac 5 {12} & \text{No cancellation occurs here.} \\[6pt]
\frac 6 {12} & = \frac{\cancel6\times1}{\cancel6\times2} = \frac 1 2 \\[6pt]
\frac 7 {12} & = \frac 7 {12} & \text{No cancellation occurs here.} \\[6pt]
\frac 8 {12} & = \frac{\cancel4\times2}{\cancel4\times3} = \frac 2 3 \\[6pt]
\frac 9 {12} & = \frac{\cancel3\times3}{\cancel3\times4} = \frac 3 4 \\[6pt]
\frac{10}{12} & = \frac{\cancel2\times5}{\cancel2\times6} = \frac 5 6 \\[6pt]
\frac{11}{12} & = \frac{11}{12} & \text{No cancellation occurs here.} \\ {}
\end{align}


TITLE: Why a series-wired light bulbs consume less energy than a parallel-wired light bulbs?
QUESTION [0 upvotes]: Why a series-wired light bulbs  consume less energy than a parallel-wired light bulbs ?

REPLY [2 votes]: As a counterpoint to David Schwartz's answer which assumes a constant voltage source, here is what happens if you have a constant current source instead.
In the parallel case, the current splits so each of the bulbs gets a fraction of the current. The power is $I^2 R$. Let's assume two bulbs: then the current through each is $I/2$, so the power for each is $({I \over 2})^2 R = {I^2 \over4} R$ so the power for both is twice that: ${1\over2} I^2 R$.
In the series case, the current is $I$ through each resistor, so the total power is $2 I^2 R$ which is more than in the parallel case.
Admittedly, a constant voltage source is much more common, but it is important to state your assumptions clearly and avoid making unwarranted generalizations (e.g. "resistances in series consume less power" - well, no).


TITLE: solving an equation-differential
QUESTION [1 upvotes]: I was trying to solve this - $$ (y^4 -2x^3y)dx + (x^4 - 2xy^3)dy = 0\ $$
Pretty direct--homogeneous equation that can be solved with y=vx
It is also non-exact, with integrating factor $$\frac1{(y^4 -2x^3y)x + (x^4 - 2xy^3)y } $$
1/(Mx + Ny)
I wanted to know if there is a shorter way around the problem .

REPLY [1 votes]: Apart the trivial solutions $y(x)=0$ and $x(y)=0$, then let $y=xt$
$((xt)^4-2x^3(xt))dx+(x^4-2(xt)^3x)*(xdt+tdx)=0$
$(t^4-2t)dx+(1-2t^3)*(xdt+tdx)=0$
$(-t^4-t)dx+(1-2t^3)*xdt=0$
$\frac{1}{x}dx=\frac{1-2t^3}{t^4+t}dt$
and you integrate it.


TITLE: OPE of descendant fields
QUESTION [1 upvotes]: How do I calculate OPE of two descendant fields $\chi_1$ and $\chi_2$, if I know OPE of $\phi_1$ and $\chi_2$. The field $\chi_1$ is a descendant of the primary field $\phi_1$.

REPLY [1 votes]: In $d>2$, all descendents are derivatives so all you have to do is to take derivatives of the OPE of primaries.
In $d=2$, a generic descendent can always be described by a contour integrals of the OPE with the primary and several stress tensors. The OPE between two descendents then becomes the contour integral of the operator product of the two primaries with a bunch of stress-tensors.
For e.g. let $O_1$ and $O_2$ be two primaries and consider the descendent $L_{-3} O_2$. This is
$$
L_{-3} O_2(0) = \oint \frac{dz}{2\pi i} z^{-2} T_{zz} (z) O_2(0)
$$
Thus,
$$
O_1(z) L_{-3} O_2(0) = \oint_C \frac{dw}{2\pi i} w^{-2} O_1(z)   T_{ww} (w) O_2(0)
$$
The contour $C$ surrounds the origin and $z$ is outside the contour. Then, I can push the contour outside to include $z$ at the cost of adding the poles at $z=w$. The residue at this pole is given entirely by the $TO_1$ OPE which is fixed since $O_1$ is primary. Finally, we take $z \to 0$ and the singularities one gets is controlled by the $O_1 O_2$ OPE which, I assume, is known.

REPLY [1 votes]: It's pretty much trivial. I assume you're talking about global conformal symmetry i.e. $SL(2,C)$ for $d=2$ and $SO(d,2)$ in higher $d$. Say the OPE is given by
$$\phi_1(x)\phi_2(y) = \frac{C}{|x-y|^{a}} \left[ O(y) + b_1 (x-y)^\mu \frac{\partial}{\partial y^\mu} O(y) + b_2 (x-y)^2 \Box_y O(y) \ldots \right] + \text{other multiplets}
$$
for some other primary $O(y)$. Here $C$ is the OPE coefficient, and $b_1, b_2, \ldots$ are coefficients that are fixed by conformal symmetry. They depend on $d$ and the scaling dimensions and spins of $\phi_1$, $\phi_2$ and $O$.
Now if $\chi_1$ is a descendant of $\phi_1$, it can be written as
$$\chi_1^{\mu_1 \dotsm \mu_J}(x) = \partial^{\mu_1} \dotsm \partial^{\mu_J} \Box^n \phi_1(x)$$
and likewise for $\chi_2$. Schematically, there exist some different operators $D_{1,2}$ such that
$$\chi_1(x) = D_1(\partial_x) \phi_1(x)$$ and similarly for $\chi_2$. But then we get the OPE $\chi_1 \times \chi_2$ just by applying these operators:
$$\chi_1(x) \chi_2(y) = D_1(\partial_x) D_2(\partial_y) \left[ \phi_1(x) \phi_2(y) \right] = D_1(\partial_x) D_2(\partial_y) \left[ \frac{C}{|x-y|^{2a}} O(y) + \ldots \right].$$
End of story!


TITLE: Distribute different objects into identical bins
QUESTION [1 upvotes]: There are 6 identical pots and 26 different coins. In how many ways can we distribute these coins into pots where any pot can contain any number of coins.
For small numbers like 3 pots and 4 coins, it was easy to do because I was manually able to list the configurations possible and solve the sum like (4,0,0) , (3,1,0) etc.
How about this sum?

REPLY [1 votes]: The number can be found by adding  Stirling numbers of the second kind, each corresponding to a different number of non-empty bins.$$ S(26,1)+S(26,2)+S(26,3)+S(26,4)+S(26,5)+S(26,6).$$ The Stirling numbers can be calculated by a simple recurrence relation - found  on Wikipedia.
(Answer modified after discussion - see comments.)


TITLE: What is the physical explanation of why a stationary fermion is experiencing zero force from an external static magnetic field?
QUESTION [1 upvotes]: Since fermions like free electrons have an intrinsic spin magnetic dipole moment shouldn't they when subjected to an external static magnetic field experience a magnetic force even assuming they are stationary (stationary in a vacuum for example)?
I understand that because their angular momentum they may not be moved by an external magnet and resist any linear motion but saying that they will experience zero magnetic force $F_{(M)}=q(\mathbf{v} \times \mathbf{B})$ when they are stationary $v=0$, I find peculiar and had like to know the physical explanation for this if any?
Are there any cases, experiments or theory that prove otherwise?
Question Update 17 Feb 2022:
According to an answer here by @Hoody and other comments, a free unbound hypothetical stationary electron under a static homogeneous magnetic field would not experience a force that would force it to translational motion but nevertheless would experience a torque the would change its initial magnetic moment $\vec{m}$ orientation in space:
$\begin{array}{l}
\vec{F}=q \vec{V} \times \vec{B}=0 \\
\vec{\tau}=\vec{m} \times \vec{B} \neq 0
\end{array}$
However, this still does not answer the core of my question here which is, even so with this applied torque mentioned, why is then a free electron as suggested by known physics,  under this torque not forced to any translational motion (i.e. orbital or linear)?
Is this above  paragraph actually true or just never experimentally or analytically verified up to now?
1) What happens when I approach the pole of a permanent magnet (non-homogeneous field) close to a copper rod? Will the free electrons inside the copper rod initiate an orbital motion besides there known gyromagnetic motion under the influence of this torque applied by the external magnetic field, opposite to the orbital direction of the unpaired electrons inside the permanent magnet and therefore generate also the diamagnetic property of the copper material?
2) What about a free stationary electron inside a vacuum environment? Will it start to move towards a magnet even if it was initially stationary under the influence of a non-homogeneous magnetic field? Will the same behavior occur with a homogeneous magnetic field and if not why?
If this is the case for the free electrons in the copper rod or the vacuum is this then not an actual prove that electrons DO move translational (free electrons forced to orbital motion besides their gyromagnetic motion) under a static magnetic force even if they were initially stationary?
Is it possible and is this ever experimentally verified, given a very strong external magnetic field so that the arc trajectory of this orbital motion under the applied torque the magnetic moments of the free electrons are subjected with, to be so large that it will force a free electron practically in a linear motion?
Finally, does not the above mentioned, leads in the inescapable conclusion that magnetism is not an exclusive emergent (side effect) phenomenon of electron charge translational motion but actually originates from the intrinsic property of spin magnetic dipole moment of the electron similar as electric phenomenon originates from the intrinsic property of the charge of the electron?
Therefore the origin of the electromagnetism phenomenon in general  must exclusively be attributed to the whatever unknown intrinsic mechanics of the electron itself?
As an implication of the above mentioned and conclusions reached is that the classical electromagnetism, magnetic force equation of the Lorentz force:
$$F_{(M)}=q(\mathbf{v} \times \mathbf{B})$$ for initial $v=0$ should be amended with a torque factor.

REPLY [1 votes]: Most generally, the force $\vec F_m$ and torque $\vec \tau$ due to the magnetic field $\vec B$ acting on a particle with a charge $q$ and magnetic moment $\vec\mu$, moving with velocity $\vec v$ can be written as
\begin{align}
\vec F_m &= q \vec v \times \vec B + \vec\nabla (\vec B\cdot\vec\mu),
  &&&&(1)\\
\vec\tau &= \vec r \times \vec F_m + \vec\mu \times \vec B. &&&&(2)
\end{align}
The first term in (1) is the standard Lorenz term, while the latter is usually not mentioned as a fundamental law since classical electrodynamics did not assume point-like particles with spin and magnetic moment but rather modeled magnetic dipoles by an electric current loop, behaviour of which in $\vec B$ can be derived just from the first term (focres act on parts of a the electric circuit because there are moving charges inside).
In Eq. (2), the first term $\vec r \times \vec F$ is the usual textbook definition of a torque in the case of a force acting on a point particle - it is just a tool to understand how will the tracejtory of the particle be curved due to $\vec F_m$ (i.e., it tells us something about orbital motion).
Apparently, it depends on the choice of your coordinates.
In contrary (but analogously to the previous paragraph), the second term in Eq. (2), $\vec \mu \times \vec B$, is an entirely new piece of physical law for a point-like particle. It is not present usual chapters on classical mechanics since they do not assume that that particles have spin (it is hard to imagine a rotating point-like particle). Again, if you modeled magnetic dipoles by an electric-current loop, you can derive the second term in (2) from $\vec F_m = q \vec v \times \vec B$.

However, this still does not answer the core of my question here which is, even so with this applied torque mentioned, why is then a free electron as suggested by known physics, under this torque not forced to any translational motion (i.e. orbital or linear)?

Now, let us have a look at what forces and torques actually do: Changes in motion of the center-of-mass of a body are always determined solely by the force, according to the famous Newton's law $\vec F = m \vec a$. Denoting the body's momentum by $\vec p=m\vec v$, this law can be also reformulated as $$\vec F = \frac{\mathrm d\vec p }{\mathrm dt}. \qquad (3)$$
In contrary, torque causes changes in angular momentum $\vec J$ (i.e., in "rotation"):
$$ \vec \tau = \frac{\mathrm d \vec J}{\mathrm d t}. \qquad (4)$$
Generally, angular momentum of a moving particle with spin is a sum of two terms, $$\vec J = \vec L + \vec S$$ where $\vec L = \vec r \times \vec p = m\vec r \times \vec v$ is the usual orbital angular momentum while $\vec S$ is the particle's internal angular momentum, i.e., spin.
Now, if $\vec F_m = 0$, Eq. (3) tells us that there will be no acceleration. This is perfectly consistent with Eq. (4) which yield
$$
 \vec \tau = \frac{\mathrm d\vec J}{\mathrm dt}
= \frac{\mathrm d\vec L}{\mathrm dt} + \frac{\mathrm d\vec S}{\mathrm dt}
= \frac{\mathrm d(\vec r \times \vec p)}{\mathrm dt} + \frac{\mathrm d\vec S}{\mathrm dt}
=\vec v \times (m\vec v) + \vec r \times \vec F + \frac{\mathrm d\vec S}{\mathrm dt}
=0+0+\frac{\mathrm d\vec S}{\mathrm dt}.
$$
I.e., in the absence of a magnetic force, the magnetic torque acts solely on the spin of the particle. Using Eq. (2):
$$\vec \mu \times \vec B =\vec \tau =  \frac{\mathrm d\vec S}{\mathrm dt}. $$
Btw., notice that since the particle's magnetic moment  $\vec \mu$  is proportional to its spin $\vec S$, the torque is always perpendicular to $\vec S$ and hence may cause a change in the orientation of the spin (i.e. precession) but not of its size (i.e. electrons always have spin of 1/2 $\hbar$ - but let's not dig into the quantum world in this discussion).

Is this above paragraph actually true or just never experimentally or analytically verified up to now?

It is true as far as classical (non-quantum) electrodynamics is true. As noted by others in the comments, a theory cannot be verified but only disproved. Electrodynamics is the most precisely measured part of physics whatsoever.

What happens when I approach the pole of a permanent magnet (non-homogeneous field) close to a copper rod? Will the free electrons inside the copper rod initiate an orbital motion besides there known gyromagnetic motion under the influence of this torque applied by the external magnetic field, opposite to the orbital direction of the unpaired electrons inside the permanent magnet and therefore generate also the diamagnetic property of the copper material?

Yes, diamagnetism is a property of all materials. But notice that the origin of the torque causing the orbital motion is not the magnetic moment of the electron but the first term, $\vec r \times \vec F_m.$

What about a free stationary electron inside a vacuum environment? Will it start to move towards a magnet even if it was initially stationary under the influence of a non-homogeneous magnetic field?

Yes.

Will the same behavior occur with a homogeneous magnetic field and if not why?

No. For a non-moving electron in a homogeneous magnetic field, both terms in Eq. (1) are zero. The electron might only undergo spin precession.

If this is the case for the free electrons in the copper rod or the vacuum is this then not an actual prove that electrons DO move translational (free electrons forced to orbital motion besides their gyromagnetic motion) under a static magnetic force even if they were initially stationary?

As explained above, this does not happen in vaccum. In the material, we must be careful about what we mean by stationary. In the quantum language, an electron in its atomic orbital is stationary; however, its classical analogue is that it is somehow circling around the nucleus and hence it undergoes an orbital motion eve initially before you apply the external magnetic field. Hence, the answer is NO.

Is it possible and is this ever experimentally verified, given a very strong external magnetic field so that the arc trajectory of this orbital motion under the applied torque the magnetic moments of the free electrons are subjected with, to be so large that it will force a free electron practically in a linear motion?

Are you asking wheather we can compensate the curving effect of the magnetic force $q\vec v \times \vec B$ by the interaction of the electron's magnetic dipole $\mu$? I think the answer is NO, since both effects are directly proportional to the same field $\vec B$.

Finally, does not the above mentioned, leads in the inescapable conclusion that magnetism is not an exclusive emergent (side effect) phenomenon of electron charge translational motion but actually originates from the intrinsic property of spin magnetic dipole moment of the electron similar as electric phenomenon originates from the intrinsic property of the charge of the electron?

Indeed. If you accept the semi-classical picture of a point-like electron with non-zero spin and magnetic moment, then there are two sources of magnetism (moving charges and particle's spin), which also both respond to an external magnetic field (first and second terms in Eqs. (1) and (2).)
Nevertheless, note that in relativistic quantum theories, both "sources of magnetism" naturally emerge from a single physical law (i.e. Lagrangian of quantum electrodynamics).

Therefore the origin of the electromagnetism phenomenon in general must exclusively be attributed to the whatever unknown intrinsic mechanics of the electron itself?

No, not all magnetic phenomena are caused by the magnetic dipoles of elementary particles - see the first terms in Eqs. (1) and (2). Note that pions are charged particles without spin or magnetic moment but their trajectories can still be bent by magnetic fields (and it is an experimental routine).


TITLE: Proof that $\bigcup X$ is in bijection with $X$.
QUESTION [3 upvotes]: Let $X$ be an infinite set such that for all $x \in X$, exists an injection $f_x: x \to X$. Proof that $\bigcup X$ is in bijection with $X$.
By definition, $y \in \bigcup X \iff \exists x \in X, \, y \in x$.
My attempt: By Cantor-Schroder-Bernstein theorem,  it's sufficient showing that exists an injection $\bigcup X \to X$ and an injection $X \to \bigcup X$.
If I take $f \doteq \bigcup_{x \in X} f_x$, we have a function $f: \bigcup X \to X$. However, this is not injective.
By the other side, I couldn't see any injection from $X$ in $\bigcup X$.
Help?

REPLY [2 votes]: The result you're trying to prove is not true.  For instance, let $X=\mathcal{P}(\mathbb{N})$, the set of all subsets of $\mathbb{N}$. Then $X$ is uncountable, but $\bigcup X=\mathbb{N}$ is countable.
It is true that there is an injection $\bigcup X\to X$, but as you observed, your construction does not work since your $f$ need not be injective.  As G. Sassatelli commented, another more minor problem is that your $f$ may not even be a function, since the domains of the functions $f_x$ may overlap.
Here's one thing you can do instead.  Let $<$ be a well-ordering on $X$.  For each $a\in\bigcup X$, let $g(a)$ be the $<$-least element of $X$ such that $a\in g(a)$.  Define $f:\bigcup X\to X\times X$ by $f(a)=(g(a),f_{g(a)}(x))$.  I will leave it to you to verify that this $f$ is an injection.  Composing $f$ with a bijection $X\times X\to X$ then gives an injection $\bigcup X\to X$.


TITLE: Indeterminate form limits question
QUESTION [1 upvotes]: $$\lim_{x\to 0}\frac{10^x - 2^x - 5 ^ x + 1 } {x\tan x} $$
This is an indeterminate limit.
I want help in solving this problem. Thanks in advance

REPLY [2 votes]: $$\frac{10^x - 2^x - 5 ^ x + 1 } {x\tan x}$$
$$=\frac{(5^x-1)(2^x-1)}{x\tan x}$$
$$=\frac{\dfrac{5^x-1}x\cdot\dfrac{2^x-1}x\cdot\cos x}{\dfrac{\sin x}x}$$
Now as $\displaystyle a^h=\left(e^{\ln a}\right)^h=e^{h\cdot\ln a}$,
$\displaystyle\lim_{h\to0}\frac{a^h-1}h=\ln a\cdot\lim_{h\to0}\frac{e^{h\ln a}-1}{h\ln a}=\ln a$


TITLE: How are the 2 triangles used to prove the centripetal acceleration similar?
QUESTION [0 upvotes]: I know that when we prove the centripetal acceleration we use similar triangles that are isosceles. How exactly, can we prove that they are similar, after all those 2 similar triangles could have different angles?

REPLY [1 votes]: If we call the two radius vectors $\vec r_1$ and $\vec r_2$, then $|\vec r_1| = |\vec r_2|=r$ and the angle between $\vec r_1$ and $\vec r_2$ is $\theta$. So the triangle formed by $\vec r_1$, $\vec r_2$ and their difference $\Delta \vec r$ is an isosceles triangle with apex angle $\theta$.
Since the velocity vector is always perpendicular to the radius vector, the angle between $\vec v_1$ and $\vec v_2$ will also be $\theta$ (this is why the two triangles have the same apex angle). And $|v_1|=|v_2|=v$. So the triangle formed by $\vec v_1$, $\vec v_2$ and their difference $\Delta \vec v$ is also an an isosceles triangle with apex angle $\theta$, and so it is similar to the radius triangle. So we have
$\displaystyle \frac {|\Delta \vec v|}{|\Delta \vec r|} = \frac {|v_1|}{|r_1|} = \frac v r$
But $|\Delta \vec r| \approx |\vec v |\Delta t = v \Delta t$ so
$\displaystyle \frac {|\Delta \vec v|}{v \Delta t} \approx \frac v r
\\ \displaystyle \Rightarrow \frac {|\Delta \vec v|}{\Delta t} \approx \frac {v^2} r
\\ \displaystyle \Rightarrow a = \lim_{\Delta t \rightarrow 0} \frac {|\Delta \vec v|}{\Delta t} = \frac {v^2} r$


TITLE: Closed form for number of integers in a rational interval
QUESTION [0 upvotes]: Inspired by this question (which is hardly relevant here) I'm curious to know whether there is a closed form for the number of integers in a half-open rational interval $(\tfrac{a}{b},\tfrac{c}{d}]$ where $a,b,c,d\in\Bbb{Z}$ and $b,d\neq0$. It is either $\lfloor\tfrac{c}{d}-\tfrac{a}{b}\rfloor$ or $\lfloor\tfrac{c}{d}-\tfrac{a}{b}\rfloor+1$, but is there a general closed form? That is, is there a well-behaved description of the function
$$f:\ \Bbb{Z}\times(\Bbb{Z}-\{0\})\times\Bbb{Z}\times(\Bbb{Z}-\{0\}) \longrightarrow\ \Bbb{N}: (a,b,c,d)\ \longmapsto\ \left|(\tfrac{a}{b},\tfrac{c}{d}]\cap\Bbb{Z}\right|,$$
preferably in terms of elementary functions, gcd's, lcm's, maxima, minima, and the likes.

REPLY [1 votes]: Sure. The number of integers in the interval $(x,y]$ (where $x\le y$) is $\lfloor y\rfloor -\lfloor x\rfloor$.


TITLE: A question on the metric approximation property
QUESTION [0 upvotes]: Let $X,Y$ be Banach spaces. Suppose that $X^{***}$ has the metric approximation property. Let $T:X^{**}\rightarrow Y$ be a finite-rank operator and let $\epsilon>0$. Is there a finite-rank operator $S:X\rightarrow X$ such that $\|S\|\leq 1+\epsilon$ and $\|T-TS^{**}\|<\epsilon$?
Furthermore, I want to know whether there are other conditions on $X$ to ensure that this question holds true.

REPLY [2 votes]: No. The operator $T$ can vanish on $X$ while any such $S^{**}$ necessarily has its range contained in $X$.


TITLE: Existence of solutions to y'' = f(x)y
QUESTION [1 upvotes]: I was reading some results earlier concerning properties of solutions to equations of the form
$$y'' = f \left(x \right) y$$
paired with initial conditions $y (x_0) = y_0$, $y' (x_0) = y_0'$, where $f$ is continuous. I was wondering -- is there an easy proof that solutions exist in general? My work in progress is to allow $z$ to solve the first-order nonlinear nonhomogeneous system
$$z' + z^2 = -f (x)$$
Then if we set
$$y = e^{\int z dx}$$
we obtain that
$$y'' = (z' + z^2)y = f (x) y$$
I'm not happy with this solution however -- there are definitely some pieces missing (the integral of $z$ existing being just one of them) and I don't know if this method can be fixed.
I'm aware that there are (complicated) existence and uniqueness results for more general systems, but I want to avoid this. I'm fine with using Picard-Lindelof, however. Any help is appreciated.

REPLY [1 votes]: You can use Picard-Lindelof. Let $u = y'$, Then $$u' = f(x) y$$ so the vector $v = \binom u y$ satisfies the first order equation $$v' = \begin{pmatrix} 0 & f(x) \\1&0 \end{pmatrix}v =  A(x)v$$ which locally has a unique solution due to Picard-Lindelof (in fact, if $f$ is bounded, there is a unique global solution).


TITLE: Determining if ___ is an ideal:
QUESTION [0 upvotes]: I am reading some notes on algebra, with polynomial rings and there is an exercise that asks to determine if 1-5 are ideals. I am not very familiar and am hoping that the examples will give a more concrete understanding of what an ideal is (I understand the definition). I would appreciate if someone would answer with a reason why as well (though it does not need to be a full proof).
(1) k, the field of coefficients;
(2) a subring $k[x_1,...,x_m] \subset R = k[x_1,...,x_n]$, where $0 < m < n$;
(3) polynomials with no constant term;
(4) $R\leq d$, polynomials of degree at most $d$;
(this one would not be a subring, say $S$, because $x^d \in S$ but $x^d*x^d \not \in S$)?
(5) homogeneous polynomials, i.e., polynomials with all terms of the same
degree.
(as with 4, this would not be closed under multiplication)?
Thank you!

REPLY [3 votes]: Considering that the list of justifications hasn't been fully populated, I will provide an answer so as to remove this from the unanswered queue.
(1) $k$ is not an ideal, since it is not closed under multiplication using elements from outside $k$. For instance $1x_1\not\in k$.
(2) This is basically the same as (1).
(3) This is actually an ideal!
(4) You are correct.
(5) You are correct, it is not closed under multiplication by outside elements. To be clear, we are multiplying by something like $x+1$. Also not closed under addition if the degrees vary.


TITLE: Positive semi-definiteness of the adjoint matrix
QUESTION [1 upvotes]: I am studying the conditions of positive semi-definiteness of a $(n+1)\times(n+1)$ symmetric matrix $\mathbf{M}$ built in the following way:
$$
\mathbf{M}=\begin{pmatrix}
\mathbf{A}   & \mathbf{b} \\
\mathbf{b}^T & c
\end{pmatrix}
$$
where $\mathbf{A}$ is a symmetrix $n\times n$ matrix, $\mathbf{b}$ is a $n$-dimensional column vector and $c$ is a real number.
The first $n$ leading principal minors of $\mathbf{M}$ are the leading principal minors of $\mathbf{A}$, so $\mathbf{A}$ should be positive semi-definite.
The last condition is $\det\mathbf{M}=|\mathbf{M}|\geq0$. By a simple calculation, I obtained
$$
|\mathbf{M}|=c|\mathbf{A}|-\mathbf{b}^T\mathbf{A}^*\mathbf{b}\geq0
$$
where $\mathbf{A}^*$ is the adjoint matrix of $\mathbf{A}$, i.e. the transpose of the matrix of cofactors.
This condition can be written
$$
c|\mathbf{A}|-\mathbf{b}^T\mathbf{A}^*\mathbf{b}=
\begin{cases}
|\mathbf{A}|\left(c-\mathbf{b}^T\mathbf{A}^{-1}\mathbf{b}\right), & \text{if }|\mathbf{A}|>0 \\
-\mathbf{b}^T\mathbf{A}^*\mathbf{b}, & \text{if }|\mathbf{A}|=0
\end{cases}
$$
So, when $|\mathbf{A}|>0$ the condition simply becomes
$$
c\geq\mathbf{b}^T\mathbf{A}^{-1}\mathbf{b}\geq0,
$$
given that $\mathbf{A}^{-1}$ is positive definite.
When $|\mathbf{A}|=0$ the condition becomes
$$
\mathbf{b}^T\mathbf{A}^*\mathbf{b}\leq0,
$$
so I am interested to know if $\mathbf{A}^*$ is positive semi-definite when $\mathbf{A}$ is positive semi-definite.
In the case $|\mathbf{A}|>0$, using spectral decomposition
$$
\mathbf{A}=\sum_{i=1}^n\lambda_i\mathbf{e}_i\otimes\mathbf{e}_i,
$$
where $\lambda_i$ are the eigenvalues and $\mathbf{e}_i$ the unit eigenvectors, so we have
$$
\mathbf{A}^*=|\mathbf{A}|\mathbf{A}^{-1}=\left(\prod_{k=1}^n{\lambda}_k\right)\sum_{i=1}^n\frac{1}{\lambda_i}\mathbf{e}_i\otimes\mathbf{e}_i = \sum_{i=1}^n\left(\prod_{k=1,k\neq i}^n{\lambda}_k\right)\mathbf{e}_i\otimes\mathbf{e}_i,
$$
so $\mathbf{A}^*$ is positive definite when $\mathbf{A}$ is, given that its eigenvalues are expressed as the product of eigenvalues of $\mathbf{A}$, excluded one in turn.
I suspect that this last expression represents $\mathbf{A}^*$ also when $|\mathbf{A}|=0$, probably by considering a positive semi-definite matrix with vanishing determinant as the limit of a positive definite matrix when one or more eigenvalues tends to zero.
So my questions:

are my calculation correct?
the last expression of $\mathbf{A}^*$ is valid also when $|\mathbf{A}|=0$?
how can this be proved?

REPLY [1 votes]: Yes, your equations are correct. Yes, the last expression you wrote is valid when $|A| = 0$. Note in particular that $\mathbf A^* = 0$ whenever the kernel of $\mathbf A$ has dimension at least $2$.
For a quick proof, we could simply note that both sides of the equation
$$
\mathbf{A}^* = \sum_{i=1}^n\left(\prod_{k=1,k\neq i}^n{\lambda}_k\right)\mathbf{e}_i\otimes\mathbf{e}_i
$$
are continuous functions of the entries of $\mathbf A$. If the equation holds for all strictly positive definite $\mathbf A$, then it must hold for positive semidefinite $\mathbf A$ "by continuity". In particular, if we define $\mathbf A_{\epsilon} = \mathbf A + \epsilon \mathbf I$ and $\lambda_{k}^{\epsilon}$ to be the $k$th eigenvalue of $\mathbf A_{\epsilon}$, then we can say that for a positive semidefinite $\mathbf A$ we have
$$
\mathbf{A}^* =
\lim_{\epsilon \to 0^+}\mathbf{A}_{\epsilon}^*
=
\lim_{\epsilon \to 0^+}\sum_{i=1}^n\left(\prod_{k=1,k\neq i}^n{\lambda}_k^{\epsilon}\right)\mathbf{e}_i\otimes\mathbf{e}_i
=
\sum_{i=1}^n\left(\prod_{k=1,k\neq i}^n{\lambda}_k\right)\mathbf{e}_i\otimes\mathbf{e}_i.
$$

For a direct proof: we note that $\dim\ker \mathbf A \geq 2$ implies that $\mathbf A^* = 0$, which is positive semidefinite. For the case where $\dim\ker \mathbf A = 1$,  we see that $\mathbf A$ is symmetric and $\mathbf A \mathbf A^* = 0$ implies that $\mathbf A^*$ has rank at most $1$, which means that $\mathbf A^*$ can be written in the form $\mathbf A^* = k \mathbf {xx}^T$ for some unit vector $\mathbf x$ and some $k \in \Bbb R$.  We note that $k$ satisfies $\operatorname{tr}(\mathbf A^*) = k$.
With that, it suffices to note that
$$
\operatorname{tr}(\mathbf A^*) = -\frac{d}{dt}|_{t = 0} \det(t\mathbf I - \mathbf A) = -\frac{d}{dt}|_{t = 0} (t - \lambda_1) \cdots (t - \lambda_n).
$$


TITLE: Groups of real numbers
QUESTION [1 upvotes]: Possible Duplicate:
Groups of real numbers

A group as we all know is a set A with a binary  operation ""  defined on it such that for every  a,b belongs to A ab belongs to A, the operation is assocaitive, there exists an identity element and for every a in A there exists a unique p in A such that a*p=identity element.
Can anyone provide me examples of groups where the operation defined on the set is not  addition + or multiplication x.

REPLY [1 votes]: I think it's important to note that the way you want to think about groups is not necessarily as number systems (a role better suited for rings), but as symmetries or in terms of actions.
To make this more mathematically precise, for any set $X$, you can form a group by looking at all one-to-one, onto maps from $X$ onto $X$, let's call this $\operatorname{Perm}(X)$. The binary operation is composition, the identity is the the map $x\mapsto x$, and the inverse is the compositional inverse. Lots of examples of groups can be found by considering subgroups of $\operatorname{Perm}(X)$, namely those that preserve additional "structure".
So for concrete examples, if $X=\{1,2,\dotsc,n\}$, then $\operatorname{Perm}(X)$ is the $n$-th symmetric group and is denoted $S_n$.
If $X=\mathbb{R}^n$, then we can look at the non-singular linear maps from $\mathbb{R}^n$ onto $\mathbb{R}^n$. This group is called $GL_n(\mathbb{R})$, and it's the subgroup of $\operatorname{Perm}(\mathbb{R}^n)$ that preserves the vector space structure of $\mathbb{R}^n$.
Neither of these (for $n>1$) are subgroups of $\mathbb{R}$ with addition or multiplication.


TITLE: Write universal quantifier in terms of existential quantifier
QUESTION [0 upvotes]: By definition I have learned that:
$ (\exists x)B \mbox{ stands for }(\lnot((\forall x)(\lnot B)))
$
But in a logical proof I need the identity:
$(\forall x) B \mbox{ stands for } (\lnot((\exists x)\lnot B))
$
Can I just derive this identity or need I prove it (with the axioms, modus ponens, generalization or the deduction theorem)?
In the last case, it is hard. Since it seems so trivial, I can't think of a clever way to prove it, other than claiming it is trivial, but in logic everything needs explanation..

REPLY [2 votes]: $$(\forall x) B = \neg(\neg((\forall x) \neg(\neg(B)))) = \neg((\exists x) \neg B))$$


TITLE: Holomorphy for functions with quaternion variables.
QUESTION [1 upvotes]: Why we cannot define the notion of holomorphy for functions with quaternion variables ?

REPLY [0 votes]: This is the wrong and outdated answer. Theory of quaternionic differentiability and quaternionic holomorphic functions can be built on principles fully similar (essentially adequate) to ones of complex holomorphic functions.
We refuse to consider only the left or only the right approach (regarding another as equivalent)  when defining a quaternionic derivative. The left and the right derivatives should be considered only together. It follows that **quaternionic  holomorphic functions are those quaternionic functions (in quternon space), whose the left and right derivatives become equal after the transition to 3D space see theory principles here. The quaternionic multiplication of so defined holomorphic functions behave as commutative. The function f(q)=q^2 is quaternionic holomorphic.


text:
\begin{document}
\maketitle

\begin{abstract}
A seminal result of Hajnal and Szemer\'edi states that if a graph $G$ with $n$ vertices has minimum degree $\delta(G) \ge (r-1)n/r$ for some integer $r \ge 2$, then $G$ contains a $K_r$-factor, assuming $r$ divides $n$. Extremal examples which show optimality of the bound on $\delta(G)$ are very structured and, in particular, contain large independent sets. In analogy to the Ramsey-T\'uran theory, Balogh, Molla, and Sharifzadeh initiated the study of how the absence of such large independent sets influences sufficient minimum degree. We show the following two related results:
\begin{itemize}
	\item For any $r > \ell \ge 2$, if $G$ is a graph satisfying $\delta(G) \ge \frac{r - \ell}{r - \ell + 1}n + \Omega(n)$ and $\alpha_\ell(G) =o(n)$, that is, a largest $K_\ell$-free induced subgraph has at most $o(n)$ vertices, then $G$ contains a $K_r$-factor. This is optimal for $\ell = r - 1$ and extends a result of Balogh, Molla, and Sharifzadeh who considered the case $r = 3$.

	\item If a graph $G$ satisfies $\delta(G) = \Omega(n)$ and $\alpha_r^*(G) =o(n)$, that is, every induced $K_r$-free $r$-partite subgraph of $G$ has at least one vertex class of size $o(n)$, then it contains a $K_r$-factor. A similar statement is proven for a general graph $H$.
\end{itemize}
\end{abstract}

\section{Introduction}

Given graphs $H$ and $G$, a collection of vertex-disjoint copies of $H$ in $G$ is called an \emph{$H$-tiling}. A \emph{perfect $H$-tiling} of $G$, or an \emph{$H$-factor} for short, is an $H$-tiling which covers all the vertices of $G$. Note that a perfect matching corresponds to a $K_2$-factor, thus the notion of an $H$-factor is a natural generalisation from edges to arbitrary graphs.

Determining sufficient conditions for the existence of an $H$-factor is one of the fundamental lines of research in Extremal Graph Theory. Textbook results of Hall and Tutte give a sufficient condition for the existence of a perfect matching (e.g.\ see \cite{diestel}). The first treatment of a more general case is by Corr\'adi and Hajnal \cite{corradi1963maximal} who showed that if a graph with $n$ vertices has minimum degree $2n/3$ then it contains a $K_3$-factor, assuming an obvious divisibility condition. This is easily seen to be the best possible bound: consider a complete tripartite graph with parts of size $n/3 - 1, n/3$ and $n/3 + 1$. The minimum degree of such a graph is $2n/3 - 1$ and as every triangle has to be traversing it does not contain a triangle-factor. Hajnal and Szemer\'edi \cite{hajnalszemeredi} showed that $\delta(G) \ge (r-1)n/r$ is sufficient for the existence of a $K_r$-factor and a similar example shows that this is optimal. Characterising the best possible bound on a minimum degree for an arbitrary graph $H$ has attracted significant attention (some of the milestones along this road include \cite{alon1996h,komlos2000tiling,komlos2001proof}) until it was finally settled by K\"uhn and Osthus \cite{kuhn2009minimum}.

Knowing that extremal examples for these results have very special structure, a natural follow-up question is how much the bound on a minimum degree can be weakened if we exclude graphs with such a structure? Here we consider exclusion of one particular feature of  extremal examples, namely the existence of a large independent set. This was first studied by Balogh, Molla, and Sharifzadeh \cite{balogh2016} who showed that if the independence number of $G$ is $o(n)$ and $\delta(G) \ge n/2 + \eps n$, for any constant $\eps > 0$, then $G$ contains a triangle-factor. This significantly weakens the bound $\delta(G) \ge 2n/3$ from the Corr\'adi-Hajnal theorem. By considering a graph $G$ consisting of two cliques of size $n/2 - 1$ and $n/2 + 1$, we see that this bound on minimum degree is the best possible.

More generally, let $\alpha_\ell(G)$ denote the size of a largest $K_\ell$-free induced subgraph of $G$ (the independence number of $G$ corresponds to $\alpha_2(G)$). Generalising the result of Balogh, Molla, and Sharifzadeh we ask the following question.

\begin{question} \label{question}
  Let $r \ge \ell \ge 2$ be integers and let $G$ be a graph with $n$ vertices and $\alpha_\ell(G) = o(n)$. What is the best possible minimum degree condition on $G$ that guarantees a $K_r$-factor?
\end{question}

There is a clear analogy between this question and the Ramsey-T\'uran theory: a central topic in the Ramsey-T\'uran theory is to determine $\mathbf{RT}_\ell(n, H, o(n))$, the smallest number of edges which guarantees that every graph graph $G$ witn $n$ vertices with $\alpha_\ell(G) = o(n)$ contains a copy of $H$. For more on the Ramsey-T\'uran theory see \cite{erdos70,erdHos1994turan,erdHos1983more,simonovits2001ramsey}. Here we replace a T\'uran-type conclusion with one along the lines of the Hajnal-Szemer\'edi theorem.

The case $r = 3$ and $\ell = 2$ of Question \ref{question} was answered in \cite{balogh2016}. Our first main result gives some progress for the general values of $r$ and $\ell$.

\begin{theorem} \label{thm:main1}
  Let $r > \ell \ge 2$ be integers. For any $\eps > 0$ there exists $\eps' > 0$ such that if $G$ is a graph with $n$ vertices where $r$ divides $n$, $\delta(G) \ge \frac{r - \ell}{r - \ell + 1}n + \eps n$ and $\alpha_\ell(G) \le \eps' n$ then $G$ contains a $K_r$-factor.
\end{theorem}

This shows that the minimum degree sufficient for the existence of a $K_{r - \ell + 1}$-factor is also sufficient for a $K_r$-factor if every small induced subgraph contains a copy of $K_\ell$. For $\ell = r - 1$ this becomes $\delta(G) \ge (1/2 + \eps)n$ which is easily seen to be optimal: consider a graph $G$ consisting of two disjoint complete graphs of order $n/2 - 1$ and $n/2 + 1$. Reiher and Schacht (see the appendix in \cite{balogh2016}) observed that the techniques from \cite{han2009perfect} can be used to show that $\delta(G) \ge (1/2 + \eps)n$ is sufficient for the existence of a $K_r$-factor if additionally every subset of $G$ of size $\eps' n$ induces quadratically many edges. Thus Theorem \ref{thm:main1} improves upon this by only requiring a much weaker condition $\alpha_{r-1}(G) \le \eps' n$. We discuss the optimality of Theorem \ref{thm:main1} in the case $2 \le \ell < r - 1$ in Section \ref{sec:remarks}.

Let us briefly discuss the excluded case $r = \ell$ in Theorem \ref{thm:main1}. The required minimum degree if $\alpha_r(G) = o(n)$ is clearly at most as large as if we would only know $\alpha_{r - 1}(G) = o(n)$. As the previous extremal example also shows that for $\ell = r$ we cannot go below $n/2$, we conclude that the minimum degree for $\ell = r$ is the same as for $\ell = r - 1$. Our second main result shows that if we instead require $\alpha^*_r(G) = o(n)$, where $\alpha^*_r(G)$ is defined as the smallest number such that every $r$-partite subgraph of $G$ induced by $r$ disjoint sets of size at least $\alpha_r^*(G)$ contains a copy of $K_r$, then one can take arbitrarily small degree. More generally, we prove such a statement for an arbitrary graph $H$.

\begin{definition}
  Let $H$ and $G$ be graphs. We define $\alpha_H^*(G)$ as the smallest number such that every $v(H)$-partite graph of $G$ induced by $v(H)$ disjoint subsets $V_1, \ldots, V_{v(H)} \subseteq V(G)$, with each $|V_i| \ge \alpha_H^*(G)$, contains a copy of $H$ with one vertex in each $V_i$.
\end{definition}

\begin{theorem} \label{thm:main2}
  Let $H$ be a fixed graph with $h$ vertices. For any $\eps > 0$ there exists $\eps' > 0$ such that if $G$ is a graph with $n$ vertices where $h$ divides $n$, $\delta(G) \ge \eps n$ and $\alpha_H^*(G) \le \eps' n$, then $G$ contains an $H$-factor.
\end{theorem}

Let us mention a connection between Theorem \ref{thm:main2} and a recent result of Balogh, Treglown, and Wagner \cite{balogh17perturbed} on $H$-factors in dense \emph{randomly perturbed graphs}. It was shown in \cite{balogh17perturbed} that if $p \ge C n^{-1/d(H)}$, where
$$
  d(H) = \max\left\{ \frac{e(H')}{v(H') - 1} \mid H' \subseteq H, v(H') \ge 2\right\},
$$
and $G$ is a graph with $n$ vertices and $\delta(G) \ge \eps n$, then with high probability $G \cup G(n,p)$ contains an $H$-factor. Here $G(n, p)$ denotes the binomial random graph with $n$ vertices and edge probability $p$. Theorem \ref{thm:main2} implies a stronger version of this result: with high probability $G' \sim G(n,p)$ has the property that for \emph{every} graph $G$ with $\delta(G) \ge \eps n$ we have that $G \cup G'$ contains an $H$-factor (the result from \cite{balogh17perturbed} is for one fixed $G$). This follows from the fact that with high probability $\alpha_H^*(G(n,p)) < \eps' n$ for arbitrarily small $\eps'$ (which requires sufficiently large $C = C(\eps')$ in $p$). Advantages over the proof from \cite{balogh17perturbed} is that the proof of Theorem \ref{thm:main2} is significantly shorter, simpler, and avoids the use of the Regularity Lemma.

\subsection{Proof strategy}

The proofs of Theorem \ref{thm:main1} and Theorem \ref{thm:main2} use the \emph{absorbing method}. This general method was pioneered by R\"odl, Ruci\'nski and Szemer\'edi \cite{rodl2009perfect}, though it can be traced further back to Erd\H{o}s, Gy\'arf\'as, and Pyber \cite{erdHos1991vertex} and Krivelevich \cite{krivelevich1997triangle}. In our case the construction of absorbers, the heart of the proof, is based on ideas of Montgomery \cite{montgomery2014embedding}. The first author has learned this construction from discussions with Asaf Ferber. These ideas were used in a similar context, for example, by Kwan \cite{kwan2016almost} to show that a typical Steiner triple system contains a perfect matching.

The main idea behind the absorbing method is, given a fixed graph $H$ on $h$ vertices, to find a subset $A \subseteq V(G)$, called an \emph{absorber}, such that for every $R \subseteq V(G) \setminus A$ such that $h$ divides $|R| + |A|$ and $|R| \le \xi n$, for some small $\xi > 0$, the induced subgraph $G[A \cup R]$ contains an $H$-factor. This reduces the problem to finding an $H$-tiling in $G \setminus A$ which covers all but at most $\xi n$ vertices, which is usually much simpler.

Widely used constructions of absorbers by R\"odl, Ruci\'nski and Szemer\'edi \cite{rodl2009perfect} and H{\`a}n, Person, and Schacht \cite{han2009perfect} rely on the property of $G$ that for every subset $S \subseteq V(G)$ of size $v(H)$ there are $\Omega(n^{v(H) t})$ \emph{$S$-absorbers} of size $v(H) t$ for some $t \in \mathbb{N}$, that is, subsets $A_S \subseteq V(G) \setminus S$ of size $|A_S| = v(H) t$ such that both $G[A_S]$ and $G[A_S \cup S]$ contain an $H$-factor. In many problems on finding $H$-factors such a property indeed holds (e.g.\ see \cite{han2017exact,lo2015f,treglown2012exact}). However, as pointed out in \cite{balogh2016}, the conditions on $G$ in the case $r = 3$ and $\ell = 2$ in Theorem \ref{thm:main1} are not strong enough to guarantee this property. Consequently, the absorber construction from \cite{han2009perfect,rodl2009perfect} fails. Moreover, this turns out to be the case for an arbitrary $r$ and $\ell$: Consider a graph $G$ obtained by taking an $r$-partite complete graph with vertex classes $V_1, \ldots, V_r$ and in each $V_i$ place a graph $\Gamma$  with $|V_i|$ vertices such that $\alpha_\ell(\Gamma) = o(n)$ and $\Delta(\Gamma) = o(n)$ (see the proof of Proposition \ref{claim:lower} for the existence of such graphs). Take an arbitrary independent set $S \subseteq V_1$ of size $r$ and consider some fixed $t \in \mathbb{N}$. Any $S$-absorber $A_S \subseteq V(G) \setminus S$ of size $|A_S| = rt$ which does not contain edges of $\Gamma$ needs to intersect each $V_i$ equally. However, any $K_r$-tiling of $A_S \cup S$ has to be traversing (that is, each copy of $K_r$ contains exactly one vertex from each $V_i$), which leaves at least $r$ vertices of $(A_S\cup S)\cap V_1$ unmatched. Therefore, $A_S$ needs to contain an edge from some $V_i$, which implies an upper bound of order $o(n^{rt})$ on the number of such sets. To summarise, as soon as $\delta(G) \le (r - 1)n/r$, we cannot use constructions from \cite{han2009perfect,rodl2009perfect}.

We show that a much weaker property, namely that for every $v(H)$-subset $S$ there exists a family of $\Theta(n)$ pairwise disjoint $S$-absorbers, suffices (see Lemma \ref{lemma:absorbing}).

\paragraph{Notation.} We follow the standard graph-theoretic notation (see \cite{diestel}). In particular, given a graph $G$ and a vertex $v \in G$, we denote with $N_G(v)$ the set of neighbours of $v$ in $G$. Given a set $V$ and an integer $k \in \NN$, we denote with $\binom{V}{k}$ the family of all $k$-subsets of $V$. Throughout the paper we assume that $n$ is sufficiently large and, for brevity, avoid the use of floors and ceilings.

\section{The Absorbing Lemma} \label{sec:proof_strategy}

\begin{definition}
  Let $H$ be a graph with $h$ vertices and let $G$ be a graph with $n$ vertices.
  \begin{itemize}
    \item We say that a subset $A \subseteq V(G)$ is \emph{$\xi$-absorbing} for some $\xi > 0$ if for every subset $R \subseteq V(G) \setminus A$ such that $h$ divides $|A| + |R|$ and $|R| \le \xi n$ the induced subgraph $G[A \cup R]$ contains an $H$-factor.

    \item Given a subset $S \subseteq V(G)$ of size $h$ and an integer $t \in \mathbb{N}$, we say that a subset $A_S \subseteq V(G) \setminus S$ is \emph{$(S, t)$-absorbing} if $|A_S| = h t$ and both $G[A_S]$ and $G[A_S \cup S]$ contain an $H$-factor.
  \end{itemize}
\end{definition}

The following lemma gives a sufficient condition for the existence of $\xi$-absorbers based on abundance of disjoint $(S, t)$-absorbers.

\begin{lemma} \label{lemma:absorbing}
    Let $H$ be a graph with $h$ vertices and let $\gamma > 0$ and $t \in \NN$ be constants. Suppose that $G$ is a graph with $n \ge n_0$ vertices such that for every $S \in \binom{V(G)}{h}$ there is a family of at least $\gamma n$ vertex-disjoint $(S, t)$-absorbers. Then $G$ contains an $\xi$-absorbing set of size at most $\gamma n$, for some $\xi = \xi(h, t, \gamma)$.
\end{lemma}

The proof of Lemma \ref{lemma:absorbing} is based on ideas of Montgomery \cite{montgomery2014embedding} and relies on the existence of `robust' sparse bipartite graphs given by the following lemma.

\begin{lemma} \label{lemma:blueprint}
	Let $\beta > 0$. Then for every $m \ge m_0$ there exists a bipartite graph $B_m$ with vertex classes $X_m \cup Y_m$ and $Z_m$ and maximum degree $40$, such that $|X_m|= m + \beta m$, $|Y_m| = 2m$ and $|Z_m| = 3m$, and for every subset $X'_m \subseteq X_m$ of size $|X'_m| = m$ the induced graph $B[X'_m \cup Y_m, Z_m]$ contains a perfect matching.
\end{lemma}

Lemma \ref{lemma:blueprint} is a corollary of \cite[Lemma 2.8]{montgomery2014embedding}. We are now ready to prove Lemma \ref{lemma:absorbing}.

\begin{proof}[Proof of Lemma \ref{lemma:absorbing}]
  From the assumption that for every $S \in \binom{V(G)}{h}$ there are $\gamma n$ disjoint $(S, t)$-absorbers we conclude that for every vertex $v \in V(G)$ there is a family of at least $\gamma n$ copies of $H$ which contain $v$ and are otherwise disjoint. Let us denote the family of sets of vertices of each such copy (without the vertex $v$) by $\mathcal{H}_v$.

  Choose a subset $X \subseteq V(G)$ by including each vertex of $G$ with probability $q = \gamma / (500 h t)$. The parameter $q$ is chosen such that the calculations work and for now it is enough to remember that $q$ is a sufficiently small constant. A simple application of Chernoff's inequality and a union bound show that with positive probability $|X| \le 2nq$ and for each vertex $v \in V(G)$ at least $q^{h - 1} |\mathcal{H}_v| / 2$ sets from $\mathcal{H}_v$ are contained in $X$. Therefore, there exists one such $X$ for which these properties hold. Let us denote the family of sets from $\mathcal{H}_v$ completely contained in $X$ with $\mathcal{H}_v'$.

  Set $\beta = q^{h-1} \gamma / 4$ and $m = |X| / (1 + \beta)$. Let $B_m$ be a graph given by Lemma \ref{lemma:blueprint}. Choose disjoint subsets $Y, Z \subseteq V(G) \setminus X$ of size $|Y| = 2m$ and $|Z| = 3m(h-1)$ and arbitrarily partition $Z$ into subsets $\calZ = \{Z_i\}_{i \in [3m]}$ of size $h-1$. Take any injective mapping $\phi_1 \colon X_m \cup Y_m \rightarrow X \cup Y$ such that $\phi_1(X_m) = X$, and any injective $\phi_2 \colon Z_m \rightarrow \calZ$. We claim that there exists a family $\{A_e\}_{e \in B_m}$ of pairwise disjoint $(ht)$-subsets  of $V(G) \setminus (X \cup Y \cup Z)$ such that for each $e = \{w_1, w_2\} \in B_m$, where $w_1 \in X_m \cup Y_m$ and $w_2 \in Z_m$, the set $A_e$ is $(\phi_1(w_1) \cup \phi_2(w_2), t)$-absorbing.

  Such a family can be chosen greedily. Suppose we have already found desired subsets for all the edges in some $E' \subset B_m$. These sets, together with $X \cup Y \cup Z$, occupy at most
  \begin{align*}
    |X| + |Y| + |Z| + h t |E'| &< 4m + 3m(h-1) + ht \cdot 40 |Z_m|
    \\ &\le 4 hm + 120 h t m \le 124 htm < 240 ht nq \le \gamma n / 2
  \end{align*}
  vertices in $G$. Choose arbitrary $e = \{w_1, w_2\} \in B_m \setminus E'$. As there are $\gamma n$ disjoint $(\phi_1(w_1) \cup \phi_2(w_2),t)$-absorbers, there are at least $\gamma n / 2$ ones which do not contain any of the previously used vertices. Pick any and proceed.

  We claim that
  $$
    A = X \cup Y \cup Z \cup \left( \bigcup_{e \in B_m} A_e \right)
  $$
  has the $\xi$-absorbing property for $\xi = \beta / (h-1)$. Consider some subset $R \subseteq V(G) \setminus A$ such that $|R| + |A| \in h \mathbb{Z}$ and $|R| \le \xi n$. As
  $$
    |\mathcal{H}_v'| \ge q^{ht} \gamma n / 2 \ge 2 \beta n = 2 \xi n (h-1) \ge 2|R| (h-1)
  $$
  we can greedily choose a subset $A_v \in \mathcal{H}_v'$ for each $v \in R$ such that all these sets are pairwise disjoint (recall each set in $\calH_v'$ is of size $h - 1$ and forms a copy of $H$ with $v$). This takes care of vertices from $R$ and uses exactly $|R|(h-1) \le \beta m$ vertices from $X$. If $|R|(h-1) < \beta m$ then $|A| + |R| \in h \mathbb{Z}$ implies $\beta m - |R|(h-1) \in h \mathbb{Z}$, thus we can cover the remaining vertices from $X$ with disjoint copies of $H$ such that there are exactly $m$ vertices remaining. Again, $|\mathcal{H}_v'| \ge 2 \beta n > 2 \beta m$ implies that such copies of $H$ can be found in a greedy manner.

  Let $X'$ denote the remaining vertices from $X$ and set $X_m' = \phi_1^{-1}(X')$. By Lemma \ref{lemma:blueprint} there exists a perfect matching $M$ in $B_m$ between $X_m' \cup Y_m$ and $Z_m$. For each edge $e = \{w_1, w_2\} \in M$ take an $H$-factor in $G[\phi_1(w_1) \cup \phi_2(w_2) \cup A_e]$ and for each $e \in B_m \setminus M$ take an $H$-factor in $G[A_e]$. All together, this gives an $H$-factor of $G[A \cup R]$.
\end{proof}

\section{Proofs of main results}

The proof of Theorem \ref{thm:main2} is somewhat easier thus we use it to demonstrate an application of Lemma \ref{lemma:absorbing}.

\begin{proof}[Proof of Theorem \ref{thm:main2}]
 Let $\gamma = \eps / 8h^2$ and $\xi = \xi(h, \gamma, t)$ be as given by Lemma \ref{lemma:absorbing}, and set $\eps' =  \min \{\xi / h, \gamma\}$.

	The proof consists of two parts: (i) show that $G$ contains an $\xi$-absorbing set $A \subseteq V(G)$, and (ii) cover all but at most $\xi n$ vertices in $V(G) \setminus A$ using vertex-disjoint copies of $H$. Note that part (ii) is almost trivial: greedily pick disjoint copies of $H$ in $V(G) \setminus A$ as long as possible. The resulting set is $H$-free thus it has to be smaller than $h \alpha_H^*(G) \le \xi n$.

  With Lemma \ref{lemma:absorbing} at hand, to prove part (i) it suffices to show that for an arbitrary
  $S \in \binom{V(G)}{h}$ there is a family of $\gamma n$ vertex-disjoint $(S, t)$-absorbers for some $t\in \NN$. In this case we take $t=h$. First, for each $w \in S$ choose a subset $N_w \subseteq N_G(w) \setminus S$ of size $\eps n / (2h)$ such that all these sets are pairwise disjoint. From $h \alpha_H^*(G) < \eps n / 4h$ we have that $G[N_w]$ contains a family of $\eps n / 4h^2$ disjoint copies of $H$. Let $V_w$ be a set containing one vertex from each such copy of $H$. Each copy of $H$ which traverses all $V_w$'s, that is, it contains a vertex from each $V_w$ for $w \in S$, forms an $(S, h)$-absorber with the corresponding copies of $H$ from $N_w$'s. Greedily pick such disjoint traversing copies of $H$ one after the other, again as long as possible. As long as we have at least $|V_w|/2 > \alpha_H^*(G)$ unused vertices in each $V_w$, that is we have found less than $|V_w|/2$ traversing copies of $H$ so far, the process continues. This way we construct a family of at least $|V_w| / 2 \ge \gamma n$ disjoint $(S, h)$-absorbers, which finishes the proof.
\end{proof}

The proof of Theorem \ref{thm:main1} follows along the same lines as the proof of Theorem \ref{thm:main2}, though it it slightly more technically involved. In particular, the covering part does not come for free, thus we handle it with with the following lemma.

\begin{lemma} \label{lemma:cover}
  Let $r > \ell \ge 2$ be integers. For any $\eps, \xi > 0$ there exists $\eps' > 0$ such that if $G$ is a graph with $n \ge n_0$ vertices, $\delta(G) \ge \frac{(r - \ell)n}{r - \ell + 1} + \eps n$ and $\alpha_\ell(G) \le \eps' n$, then for every $A \subseteq V(G)$ of size at most $\eps n / 2$, there exists a $K_r$-tiling in $G \setminus A$ which covers all but at most $\xi n$ vertices.
\end{lemma}

The proof of Lemma \ref{lemma:cover} is fairly standard, however as it involves Szemer\'edi's regularity lemma we postpone it until the end of the section. We now show how together with Lemma \ref{lemma:absorbing} it implies Theorem \ref{thm:main1}.

\begin{proof}[Proof of Theorem \ref{thm:main1}]
  To prove Theorem \ref{thm:main1}, it suffices to show that for every subset $S \in \binom{V(G)}{r}$ there exists a family of $\gamma n$ vertex-disjoint $(S, t)$-absorbers for some $t\in \NN$. In particular, we do that for some $\gamma < \eps/2$ and $t=r$. Then by Lemma \ref{lemma:absorbing} there exists a $\xi$-absorbing subset $A \subseteq V(G)$ of size at most $\eps n / 2$ and by Lemma \ref{lemma:cover} there exists a $K_r$-tiling  in $G \setminus A$ covering all but at most $\xi n$ vertices. All together this implies the existence of a $K_r$-factor in $G$.

  Consider some $S \in \binom{V(G)}{r}$. Partition randomly $V(G) \setminus S$ into $r+1$ sets denoted by $V_1, \ldots, V_{r+1}$. Each $V_i$ is of size $(n - r)/(r+1)$ and by Chernoff's inequality and union bound, with high probability every vertex has at least
  $$
    \left( \frac{r - \ell}{r - \ell + 1} + \eps/2 \right) \frac{n}{r+1}
  $$
  neighbours in each $V_i$. Therefore, there exists a partition $V_1, \ldots, V_{r+1}$ for which this holds.

  Let us enumerate the vertices in $S$ as $v_1, \ldots, v_r$. We show that for every $X_i \subseteq V_i$ of size at most $\eps n / 4(r+1)$ there exists a copy of $K_{r}$ in $V_{r+1}$, with vertices labelled $w_1, \ldots, w_r$, and a copy of $K_{r-1}$ in $N_G(v_i) \cap N_G(w_i) \cap (V_i \setminus X_i)$ for every $i \in \{1, \ldots, r\}$. Note that such copies of $K_{r-1}$ together with the copy of $K_r$ in $V_{r+1}$ form an $(S, r)$-absorber. This allows us to greedily form a family of $\eps n / 4(r+1)r$ disjoint $(S, r)$-absorbers, which finishes the proof.

  The previous claim follows almost trivially from the bound on the minimum degree and $\alpha_\ell(G) \le \eps' n$. First, note that each vertex has at least
  $$
     \left( \frac{r - \ell}{r - \ell + 1} + \eps/4 \right) \frac{n}{r+1}
  $$
  neighbours in each $V_i \setminus X_i$. As $|V_i| < n/(r+1)$ this implies that any set of $r - \ell + 1$ vertices has a common neighbourhood of size at least $\eps n / 4(r+1)$ in $V_i \setminus X_i$. This means we can start with an arbitrary vertex $w_1 \in V_{r+1} \setminus X_{r+1}$ and iteratively for $2 \le i \le r - \ell$ pick a vertex $w_i$ which is in the neighbourhood of $w_1, \ldots, w_{i-1}$. Such vertices form $K_{r - \ell}$ and as $\alpha_\ell(G) \le \eps n / 4(r+1)$ there exists a copy of $K_\ell$ in their common neighbourhood in $V_{r+1} \setminus X_{r+1}$. This gives us a copy of $K_r$ in $V_{r+1} \setminus X_{r+1}$. Now for each $i \in [r]$  repeat a similar argument in order to find a copy of $K_{r-1}$ in $N_G(v_i) \cap N_G(w_i) \cap (V_i \setminus X_i)$. We omit the details.
\end{proof}

\subsection{Proof of Lemma \ref{lemma:cover}}

The proof of Lemma \ref{lemma:cover} is based on a standard application of the regularity method (e.g. see \cite{komlos96}). We quote below some basic definitions and a statement of the Regularity Lemma.

\begin{definition}
  Given a graph $G$ and disjoint subsets $A, B \subseteq V(G)$, we say that the pair $(A, B)$ is \emph{$\mu$-regular} if for all $X\subseteq A$, $|X|\geq \mu |A|$ and $Y\subseteq B$, $|Y|\geq \mu |B|$ we have
  \[\left|d(X,Y)-d(A,B)\right|\leq \mu\]
  where $d(X,Y)=e(X,Y)/|X||Y|$ and $e(X, Y)$ is the number of edges in the bipartite subgraph of $G$ induced by $X$ and $Y$.
\end{definition}

The following version of the Regularity Lemma is known as the \emph{``degree form''} (see \cite[Theorem 1.10]{komlos96}).

\begin{lemma} \label{lem:rl}
  For every $\mu > 0$ there is an $N = N(\mu)$ such that if $G$ is a  graph with $n$ vertices and $d \in [0,1]$ is any real number, there exists a partition $V(G)=V_0\cup...\cup V_k$ and a spanning subgraph $G'\subseteq G$ with  the following properties:
  \begin{enumerate}[(a)]
    \item $k \leq N$;
    \item $|V_0| \leq \mu n$;
    \item $|V_i|=m$ for all $1\leq i\leq k$, where $m\leq \mu n$;
    \item $d_{G'}(v)>d_{G}(v)-(d+\mu)|G|$ for all $v\in V(G)$;
    \item all $V_i$ are independent sets in $G'$;
    \item all pairs $(V_i,V_j)$ are $\mu$-regular in $G'$ with density 0 or at least $d$.
  \end{enumerate}
\end{lemma}

We only give a sketch of the proof of Lemma \ref{lemma:cover} as it is a straightforward application of the regularity method.

\begin{proof}[Sketch of the proof of Lemma \ref{lemma:cover}]

Let $\mu \ll \epsilon,\xi$, that is $\mu$ is sufficiently smaller than $\epsilon$ and $\xi$, and $d=\epsilon/4$. Apply the Regularity Lemma (Lemma \ref{lem:rl}) on $G$ with $\mu$ and $d$ to obtain a partition $V_0,...,V_k$ of $V(G)$ and a spanning subgraph $G'\subseteq G$ with the properties (a)--(f) as stated. Let $R$ be the \emph{reduced graph} of this partition: $R$ has vertex set $\{1, \ldots, k\}$ and there is an edge between $i$ and $j$ if and only if the pair $(V_i, V_j)$ has density at least $d$ in $G'$. Property (d) implies that $R$ has minimum degree at least
$$
  \left(\frac{r - \ell}{r - \ell + 1} + \eps/2 \right) k,
$$
thus by the Hajnal-Szemer\'edi theorem it contains a $K_{r-\ell+1}$-tiling which covers all but at most $r - \ell$ vertices (in case $k$ is not divisible by $r - \ell + 1$). For the rest of the proof we ignore $V_0$ and $V_i$'s for $i \in [k]$ which corresponds to vertices not covered by such a tiling. This way we ignore at most $\mu n + (r - \ell) \mu n < \xi n / 2$ vertices.

Consider one of the copies of $K_{r - \ell + 1}$ in the obtained tiling in $R$. Without loss of generality we may assume that it corresponds to vertex classes $V_1, \ldots, V_{r - \ell + 1}$. We show that we can find a $K_r$-tiling in $G[V_1 \cup \ldots V_{r - \ell + 1}]$ which covers all but at most $\xi m / 2$ vertices in each $V_j$. Applying this to every copy of $K_{r - \ell + 1}$ from the tiling of $R$ we find a $K_r$-tiling of $G$ covering all but at most $\xi n$ vertices, as desired.

To show that there exists a $K_r$-tiling in $G[V_1 \cup \ldots V_{r - \ell + 1}]$ which covers all but at most $\xi m / 2$ vertices in each $V_j$, it suffices to show that for any $z \in [r - \ell + 1]$ and any choice of subsets $V_j' \subseteq V_j$ of size $|V_j'| \ge \xi m / 8$ for $j \in [r - \ell + 1]$, there exists a copy of $K_r$ in $G[V_1' \cup \ldots \cup V_{r - \ell + 1}']$ with exactly one vertex in each $[r - \ell + 1] \setminus z$ and $\ell$ vertices in $V_z'$. By repeatedly applying this $(1 - \xi/4) m / r$ times for each $z \in [r - \ell + 1]$, each time removing vertices from the obtained $K_r$, we obtain the desired $K_r$-tiling.

Consider some subsets $V_j' \subseteq V_j$ for $j \in [r - \ell + 1]$ such that $|V_j'| \ge \xi m / 8$. By the Slicing Lemma (see \cite[Fact 1.5]{komlos96}) each pair $(V_i', V_j')$ is $\mu'$-regular for $\mu' = 8\mu / \xi$. Without loss of generality we may assume $z = r - \ell + 1$. Our goal is to find a vertex $w_j \in V_j'$ for each $1 \le j \le r - \ell$ such that these vertices form $K_{r - \ell}$ and their common neighbourhood $N_z \subseteq V_z'$ in $V_z'$ is of size at least
$$
  |N_z| \ge (d/2)^{r - \ell} |V_z'| \ge (d/2)^{r - \ell} \xi m / 8 \ge (d/2)^{r - \ell} \xi n / 16N \ge \eps' n,
$$
for sufficiently small $\eps'$ (recall that $N$ is a constant). The proof of this can be found, for example, in a textbook by Diestel \cite[Theorem 7.5.2]{diestel}, thus we omit it. Finally, as $\alpha_\ell(G) \le \eps' n$ we can find a copy of $K_\ell$ in $G[N_z]$ which completes a desired copy of $K_r$.
\end{proof}

\section{Concluding remarks} \label{sec:remarks}

As far as we are aware, it is not known whether Theorem \ref{thm:main1} is optimal for $r - 1> \ell \ge 2$. In particular, a construction from \cite{balogh2016} shows that $\delta(G) \ge (r - 2)n / r$ is necessary for $\ell = 2$, while Theorem \ref{thm:main1} shows that $(r - 2)n/(r-1) + o(n)$ is sufficient. We now generalise their construction to arbitrary $r > \ell \ge 2$.

\begin{proposition}\label{claim:lower}
If $r > \ell \ge 2$ then there exists $n_0\in \NN$ such that for every $n \ge n_0$ there is a graph $G$ on $n$ vertices such that
$$
  \delta(G) \ge
  \begin{cases}
    \frac{r - \ell}{r}n, &\text{ if } \ell \le r/2 \\
    1/2, &\text{ otherwise},
  \end{cases}
$$
and $\alpha_\ell(G) = o(n)$ which does not contain a $K_r$-factor.
\end{proposition}
\begin{proof}
Let $\Gamma_\ell(n)$ be a $K_{\ell+1}$-free graph with $n \ge n_0$ vertices, $\alpha_\ell(G) = o(n)$ and maximum degree $o(n)$. Such a graph can be obtained, for example, by considering a random graph $G(n,p)$ with edge probability $p = n^{-2/(\ell+1)}$: By applying the FKG inequality we have
\begin{align*}
  \Pr[G(n,p) \text{ is } K_{\ell+1}\text{-free}] &\ge \prod_{S \in \binom{[n]}{\ell+1}} \Pr[G(n,p)[S] \neq K_{\ell+1}] \\
  &\ge \left(1 - p^{\binom{\ell+1}{2}}\right)^{\binom{n}{\ell+1}} \ge \exp \left( -2p^{\binom{\ell+1}{2}} n^{\ell+1} \right) = \exp (-2n).
\end{align*}
In the third inequality we used $1 - x > e^{-2x}$ for $0 < x < 1/2$. On the other hand, for $p \ge n^{-2/\ell + \eps}$ for some $\eps > 0$, we have that a fixed subset of $G(n,p)$ of size $n^{1 - \eps/2}$ contains $K_\ell$ with probability $1 - \exp(\Omega(n^{2 - \eps} p))$. A union bound over all subsets of such size shows that $\alpha_\ell(G(n,p)) \le n^{1 - \eps/2}$ with probability $1 - 2^n \exp(\Omega(n^{2-\eps} p))$. By choosing sufficiently small $\eps > 0$, with positive probability we have both $\alpha_\ell(G(n,p)) = o(n)$ and $G(n,p)$ is $K_{\ell+1}$-free.

Having graphs $\Gamma_\ell(n)$ at hand, we can finish the proof of the claim. As mentioned before, the construction we present is a straightforward generalisation of the construction from \cite{balogh2016}. Consider some $2 \le \ell \le r/2$ and let $r = x \ell + y$ for some $x, y \in \NN$ and $1 \le y \le \ell$. We create a graph $G$ by taking an $(x+1)$-partite complete graph with one set $V_1$ of size $y n/r - 1$, one set $V_2$ of size $\ell n/ r + 1$ and all other sets $V_3, \ldots, V_{x+1}$ of size $\ell n / r$, and within each set $V_i$  put the graph $\Gamma_\ell(|V_i|)$. Such a graph has minimum degree $y n / r - 1 + (x - 1)\ell n / r = (r - \ell) n / r - 1$. Because $V_i$ does not contain $K_{\ell+1}$, any $K_r$ in such a graph $G$ has to contain at least $y$ vertices from $V_1$ and cannot contain more than $\ell$ from any other set. Therefore, a $K_r$-tiling can have at most $\lfloor |V_1|/y \rfloor < n/r$ copies of $K_r$, which is not enough to cover all the vertices in $V_2$.
\end{proof}

To summarise, Theorem \ref{thm:main1} and Proposition \ref{claim:lower} give an upper and lower bound on Question \ref{question}. The difference between these bounds gets larger as $\ell$ goes from $2$ to $r/2$ and then decreases again as it goes further to $r - 1$. For $\ell = r - 1$ Theorem \ref{thm:main1} matches Proposition \ref{claim:lower}, thus resolving Question \ref{question} in this case. It would be very interesting to determine the correct minimum degree condition for all other cases. This adds to a list of open problems posed in \cite{balogh2016}.

Finally, it is worth mentioning that Balogh et al.\ \cite{balogh2018triangle} showed that if $\alpha_2(G) = o(n)$ and $\delta(G) \ge n/3 + o(n)$ then there exists a triangle-tiling covering all but at most $4$ vertices. Moreover, in the case when $3$ divides $n$ they characterise a `barrier' for the existence of a triangle-factor. In the same spirit, as a step towards answering Question \ref{question} it would be interesting to show that $\delta(G) \ge (r - 2)n/r + o(n)$ is sufficient for the existence of a $K_r$-tiling covering all but constantly many vertices for $r \ge 4$. The construction from Proposition \ref{claim:lower} can be modified to show that there exists a graph $G$ with $\alpha_2(G) = o(n)$ and $\delta(G) = (r-2)n/r - \eps n$ such that no $K_r$-tiling of $G$ covers more than $(1 - \eps)n$ vertices.

\paragraph{Acknowledgement.} The authors would like to thank Anita Liebenau for the valuable discussions.

\bibliographystyle{abbrv}
\bibliography{references}

\end{document}

config: arxiv


TITLE: Complex mod proof
QUESTION [0 upvotes]: Prove that if $|z| = 1, |w| = 1$ and $\overline{z}w \ne 1$ then $\bigg| \frac{z - w}{1 - \overline{z}w} \bigg| = 1$

So we have two cases, $|z| = 1$ and $|w| = 1$.
Case 1: |z| = 1
I'd guess we begin with:
$\bigg| \frac{z - w}{1 - \overline{z}w} \bigg| = \bigg| \frac{z - w}{1 - \overline{z}w} \cdot \frac{\overline{1 - \overline{z}w}}{\overline{1 - \overline{z}w}} \bigg| = \bigg|\frac{(z-w)(1 - z\overline{w})}{(1 - \overline{z}w)(1 - z\overline{w})}\bigg| = \frac{z - z^2\overline{w} + wz + |w|^2}{1 - z\overline{w} - \overline{z}w + |z|^2|w|^2}$
But this doesnt lead me anywhere.
Am I going in the right direction?

REPLY [3 votes]: Let $z = e^{\theta i}, w = e^{\beta i}\implies |z-w| = |e^{\theta i}-e^{\beta i}|= |e^{\theta i}||1-e^{-\theta i}e^{\beta i}|= |1-\bar{z}w|$. This implies the result.


TITLE: How to prove L'Hospital's rule using $\varepsilon$-$\delta$ method?
QUESTION [3 upvotes]: For $\varepsilon$-$\delta$ proofs, basically we need to find a $\delta$ such that $|F(x)-L|<\epsilon$ whenever, $0<|x-a|<\delta$ (for a small positive number $\epsilon$).
To prove L'Hospital's rule (for when numerator and denominator function both tend to $0$ as $x\rightarrow a^{+}$)  let us assume $F(x)=\frac{f(x)}{g(x)}$. Where, $\lim_{x\rightarrow a^{+}}f(x)=0$ and $\lim_{x\rightarrow a^{+}}g(x)=0$.
I claim that $L=\lim_{x \rightarrow a^{+}}\frac{f'(x)}{g'(x)}$. Now I need to prove this $L$ is indeed the limit.
$$\left|\frac{f(x)}{g(x)}-\lim_{x \rightarrow a^{+}}\frac{f'(x)}{g'(x)}\right|<\epsilon.$$
But after this I cannot understand how to find $\delta$ as a function of $\epsilon$, so that I can complete the proof. How should I proceed?

REPLY [2 votes]: You need to assume that $$\lim_{x\rightarrow a}\frac{f^\prime(x)}{g^\prime(x)}$$ exists. This means that for any $\varepsilon > 0 $ there is $\delta > 0 $ such that
$$\tag{$\ast$}\quad\quad\quad\quad\left| \frac{f^\prime(y)}{g^\prime(y)}- \lim_{x\rightarrow a}\frac{f^\prime(x)}{g^\prime(x)}\right|< \varepsilon$$
whenever $|y-a|< \delta$.
Now the Cauchy version of the mean value theorem states that, if $f$ and $g$ are differentiable on $[b,a]$, say, and $x\in [b,a]$, and if $g(x)-g(a)\neq 0$, then there exists $c\in (x, a)$ such that
$$\frac{f(x)-f(a)}{g(x)-g(a) }=\frac{f^\prime(c)}{g^\prime(c)}$$
I assume this to be known, the tricky part is to be able to choose $c$ for both the numerator and denominator simultaneously. This also is true only if $a\neq \infty$, so the following proof of l'Hopital only works under this additional assumption.
Now let's look at the claim. Choose $\varepsilon > 0$. You want to estimate
$$\left| \frac{f(y)}{g(y)}- \lim_{x\rightarrow a}\frac{f^\prime(x)}{g^\prime(x)}\right| < \varepsilon$$ for $y$ close to $a$. Choose $\delta >0 $ such that $(*) $ is true for your choice of $\varepsilon$.
 By the mean value theorem (Cauchy version) and the assumption $f(a)=g(a) =0$, the difference you are looking at is just the same as
$$\left| \frac{f^\prime(c)}{g^\prime(c)}- \lim_{x\rightarrow a}\frac{f^\prime(x)}{g^\prime(x)}\right|$$
for some $c\in (y,a)$ which is $<\varepsilon$ by $(*)$ if only $|y-a| < \delta$.


TITLE: How to find the remaining segment of this triangle without the Law of Cosines?
QUESTION [0 upvotes]: I am given a triangle $T$ with vertices $A, B,$ and $C$ and the the following info about $T:$

$\overline{AC} = 6$ Km
$\overline{BC} = 9$ Km
The angle formed by these two segments is $120^{\circ}$

I have to find $\overline{AB}.$ Now, I know one straight way to do this is using the Law of Cosines. My question is, is there any other way to solve this problem without the Law of Cosines?

REPLY [1 votes]: Consider the diagram below.

Draw altitude $\overline{AD}$ to the extension of side $\overline{BC}$ to form right triangle $ABD$.
Since $\angle ACD$ is supplementary to $\angle ACB$, its measure is $180^\circ - 120^\circ = 60^\circ$.
Thus,
$$|\overline{CD}| = 6~\text{km}\cos(60^\circ) = 3~\text{km}$$
Hence,
$$|\overline{BD}| = |\overline{BC}| + |\overline{CD}| = 6~\text{km} + 3~\text{km} = 9~\text{km}$$
and
$$|\overline{AD}| = 6~\text{km}\sin(60^\circ) = 3\sqrt{3}~\text{km}$$
To find $|\overline{AB}|$, use the Pythagorean Theorem on right triangle $ABD$.


TITLE: Why does $\frac{1}{4}x^{-3/4} = \frac{1}{4x^{3/4}} = \frac{1}{4\sqrt[4]{x^3}}$?
QUESTION [2 upvotes]: This is taken from Khan Academy, I don't understand how these equate:
$$\frac{1}{4}x^{-3/4}
= \frac{1}{4x^{3/4}}
= \frac{1}{4\sqrt[4]{x^3}}$$
How come the minus was remove from the original exponent?

REPLY [12 votes]: These are properties of exponentiation. In particular,
$$a^{-b} = \frac{1}{a^b}$$
combined with
$$a^{\frac{m}{n}} = \sqrt[\leftroot{-2}\uproot{2}n]{a^m}.$$
In your case,
$$\frac{1}{4}x^\frac{-3}{4} = \frac{1}{4x^\frac{3}{4}} = \frac{1}{4\sqrt[\leftroot{-2}\uproot{2}4]{x^3}}.$$


TITLE: Uniqueness in Matrix Multiplication
QUESTION [1 upvotes]: I'm sure there is an answer to this somewhere else, but I'm simply not sure how to find it or what to call it. I looked online, but couldn't find anything.
The question is as follows:
Let $A$ and $C$ be $n \times n$ matrices.
If $AB = C$, where $A$ and $C$ are fixed, is $B$ necessarily unique?
Obviously $B$ is $n \times n$, but I'm not sure whether it could vary. It seems like it should be unique but I can't think how I would prove it.
EDIT
I should mention neither A or C are the 0 matrix. If A was the 0 matrix obviously B could be anything. Note sure what happens when C = 0, but it doesn't apply to the cases I'm considering.

REPLY [4 votes]: Certainly there will be cases where, for some matrix $A$ and vector $b$, the equation $Ax = b$ has infinitely many solutions in vectors $x$.
Now let $C$ be the matrix where each column is $b$.

REPLY [4 votes]: If $A$ is invertible, then $B=A^{-1}C$, so $B$ is unique. Otherwise, $B$ can certainly fail to be unique.
For example, let $A=\left(\begin{array}{cc}0 & 1 \\ 0 & 0\end{array}\right)$ and let $C=\left(\begin{array}{cc}1 & 1 \\ 0 & 0\end{array}\right)$. Then $B$ can be any matrix of the form $B=\left(\begin{array}{cc}* & * \\ 1 & 1\end{array}\right)$, where the $*$ entries are free to take any value.


TITLE: Numerical methods for approximating roots of degree $n+1$ polynomials
QUESTION [0 upvotes]: I am wondering how exactly do you use numerical methods to approximate roots of any degree $n+1$ polynomials? Like how would you solve $x^{n+1}+2x^n+1=0$?

REPLY [0 votes]: As said in answer and comments, as soon as $n\geq 4$, you need numerical methods for solving the equation.
However, for shortcut calculations, you can make reasonable aproximations. Considering the root which is close to $-2$, you could make a Taylor expansion of the equation around this value and get
$$x_n\approx-2-(-2)^{-n}\tag 1$$ For a better approximation, you could  use Padé pproximants and the simplest one would give
$$x_n\approx-2-\frac{2}{n+(-1)^n\, 2^{n+1}} \tag 2$$
Checking for a few values of $n$
$$\left(
\begin{array}{cccc}
 n & \text{exact} & (1) & (2) \\
 4 &  -2.055967397 & -2.062500000 & -2.055555556 \\
 5 &  -1.965948237 & -1.968750000 & -1.966101695   \\
 6 &  -2.014942526 & -2.015625000 & -2.014925373 \\
 7 &  -1.991964197 & -1.992187500 & -1.991967871   \\
 8 & -2.003846663 & -2.003906250 & -2.003846154 \\
 9 & -1.998029470 & -1.998046875 & -1.998029557   \\
 10 &  -2.000971830& -2.000976563 & -2.000971817 \\
 11 &  -1.999510402 & -1.999511719 & -1.999510404   \\
 12 & -2.000243784 & -2.000244141 & -2.000243784
\end{array}
\right)$$
 Notice that both $(1)$ and $(2)$ provide rational numbers.
$$\left(
\begin{array}{ccc}
 n &  (1) & (2) \\
 4 & -\frac{33}{16} & -\frac{37}{18} \\
 5 & -\frac{63}{32} & -\frac{116}{59} \\
 6 & -\frac{129}{64} & -\frac{135}{67} \\
 7 & -\frac{255}{128} & -\frac{496}{249} \\
 8 & -\frac{513}{256} & -\frac{521}{260} \\
 9 & -\frac{1023}{512} & -\frac{2028}{1015} \\
 10 & -\frac{2049}{1024} & -\frac{2059}{1029} \\
 11 & -\frac{4095}{2048} & -\frac{8168}{4085} \\
 12 & -\frac{8193}{4096} & -\frac{8205}{4102}
\end{array}
\right)$$


TITLE: Finding approximation of largest eigenvalue
QUESTION [3 upvotes]: Given the following matrix, find an approximation of the largest eigenvalue.
$$ A =
    \begin{bmatrix}
    3 & 2 \\
    7 & 5 \\
    \end{bmatrix}
$$
And I was also given $$\vec x=
    \begin{bmatrix}
    1 \\
    0 \\
    \end{bmatrix}
$$
How my professor solves this is by calculating the slopes of $A\vec x = \vec b_1$, $A^2 \vec x = \vec b_2$, $A^3 \vec x = \vec b_3$ and so on until we get the slope of $\vec b_i$ converging to the same value. Then when we get the approximated $\vec b$, he plug into $A \vec b =  \lambda \vec b$, and the corresponding $\lambda$ is the largest eigenvalue.
Since slope is $\frac yx$ , it works fine for $2 x 2$ matrix. But how do I apply this method for a bigger matrix?

REPLY [2 votes]: What you mention, is a known numerical analysis method for the approximation of the largest (by absolute value) eigenvalue of a matrix.

Let $A\in \mathbb R^{n\times n}$ have $n$ linearly independent eigenvalues $\{ u \}_{i=1}^n$ as well as a unique eigenvalue $λ_1$ such that : $|λ_1| < |λ_2| \leq \dots \leq |λ_n|$ where $λ_1 \in \mathbb R$ and $u_1 \in \mathbb R^n$ : $Au_1=λ_1u_1.$
The method "Power Iteration" :
$$\begin{cases} x_k= Ax_{k-1} \to x_k = A^kx_0 \quad k=1,2,\dots \\ x_0 \end{cases}$$
Thorem : The method "Power Iteration" converges $\forall x_0$ (that is adequate for the problem given) and it is :
$$\lim_{k\to \infty} ε_k\frac{x_k}{||x_k||_2}=u_1$$
$$\lim_{k \to \infty} \frac{x_{k,i}}{x_{k-1,i}}=λ_1 \quad \forall \space i=1,2,\dots,n \quad \text{with} \quad u_{1,i} \neq 0 $$
where $\{ ε_k\}_{k=1}^\infty = \{ \pm1\}$ and $u_1$ eigenvector of $A$ with $||u_1||_2=1$.

I've given you a formal explanation of the method according to my old notes and my knowledge, for more, check here.

REPLY [1 votes]: You normalize the vector at each iteration by dividing by its length, and wait until the resulting sequence of unit vectors has gotten close enough to converging for your purpose. (This is called the power method. It's pretty much the worst iterative method for eigenvalues there is, but it is the theoretical basis for lots of better methods.)
This is similar to what you're doing when you measure the slope in the 2D case, except in that case you divide by $x$ instead of dividing by $\sqrt{x^2+y^2}$. The point is that all that matters is the direction, not the magnitude.

REPLY [1 votes]: Here's a hint: You want to determine when $\mathbf{b}_n$ is a near-scalar multiple of $\mathbf{b}_{n-1}$. In $\mathbb{R}^2$, (nonzero) vectors are scalar multiples of one another iff their slopes are equal. A possibly more useful definition is that two vectors $\mathbf{v}$ and $\mathbf{w}$ are scalar multiples of one another if and only if
$$\hat{\mathbf{v}} = \pm\hat{\mathbf{w}},$$
where
$$\hat{\mathbf{v}} = \frac{\mathbf{v}}{|\mathbf{v}|},$$
which extends more nicely to multiple dimensions.

REPLY [0 votes]: The method you are thinking of is called Power Iteration. More formally, take any vector $x_0\ne 0$. (In practice, it is good to choose $x_0$ randomly. Then define
$$
x_k := \frac{Ax_{k-1}}{\|Ax_{k-1}\|_2}, \quad k\in\{1,2,\ldots\}
$$
With high probability, $x_k$ will converge to an eigenvector of $A$ corresponding to the largest eigenvalue $\lambda_1$ of $A$ with
$$
\lambda_1 = \lim_{k\to\infty} x_k^T Ax_k.
$$
In practice, in computer arithmetic, this method is numerically stable. For a better method, see for example, this.


text:
\section{``Iff'' is not expressible in IFG logic}

Let \(\phi \himplies{J} \psi\)\index{$\phi \himplies{J} \psi$} be an abbreviation for \({\hneg\phi} \hor{J} \psi\), and let \(\phi \hiff{J} \psi\)\index{$\phi \hiff{J} \psi$} be an abbreviation for
\[
(\phi \himplies{J} \psi) \hand{J} (\psi \himplies{J} \phi).
\]

It will be useful to know when \(\A \modeltf \phi \himplies{J} \psi[V]\) and \(\A \modeltf \phi \hiff{J} \psi[V]\). It follows immediately from the definitions that for \(\phi \himplies{J} \psi\),
\begin{itemize}
	\item[(+)] \(\A \modelt \phi \himplies{J} \psi[V]\) if and only \(\A \modelf \phi[V_1]\) and \(\A \modelt \psi[V_2]\) for some \(V = V_1 \cup_J V_2\),
	\item[($-$)] \(\A \modelf \phi \himplies{J} \psi[W]\) if and only if \(\A \modelt \phi[W]\) and \(\A \modelf \psi[W]\).
\end{itemize}

Similarly for \(\phi \hiff{J} \psi\),
\begin{itemize}
	\item[(+)] \(\A \modelt \phi \hiff{J} \psi[V]\) if and only if \(\A \modelf \phi[V_1]\) and \(\A \modelt \psi[V_2]\) for some \(V = V_1 \cup_J V_2\), and \(\A \modelt \phi[V_3]\) and \(\A \modelf \psi[V_4]\) for some \(V = V_3 \cup_J V_4\),
	\item[($-$)] \(\A \modelf \phi \hiff{J} \psi[W]\) if and only if \(\A \modelt \phi[W_1]\), \(\A \modelf \psi[W_1]\), \(\A \modelf \phi[W_2]\), and \(\A \modelt \psi[W_2]\) for some \(W = W_1 \cup_J W_2\).
\end{itemize}

In particular, the semantics for \(\phi \himplies{N} \psi\) are
\begin{itemize}
	\item[(+)] \(\A \modelt \phi \himplies{N} \psi[V]\) if and only if \(\A \modelf \phi[V]\) or \(\A \modelt \psi[V]\),
	\item[($-$)] \(\A \modelf \phi \himplies{N} \psi[W]\) if and only if \(\A \modelt \phi[W]\) and \(\A \modelf \psi[W]\),
\end{itemize}

and for \(\phi \hiff{N} \psi\),
\begin{itemize}
	\item[(+)] \(\A \modelt \phi \hiff{N} \psi[V]\) if and only if \(\A \modelt \phi[V]\) and \(\A \modelt\psi[V]\), or \(\A \modelf \phi[V]\) and \(\A \modelf \psi[V]\),
	\item[($-$)] \(\A \modelf \phi \hiff{N} \psi[W]\) if and only if \(\A \modelt \phi[W]\) and
	\(\A \modelf \psi[W]\), or \(\A \modelf \phi[W]\) and \(\A \modelt \psi[W]\).
\end{itemize}

For example, \(\A \modelt \phi \hiff{N} \psi[V]\)
if and only if \(\A \modelt ({\hneg\phi} \hor{N} \psi) \hand{N} (\phi \hor{N} {\hneg\psi})[V]\)
if and only if \(\A \modelt ({\hneg\phi} \hor{N} \psi)[V]\) and \(\A \modelt (\phi \hor{N} {\hneg\psi})[V]\)
if and only if \(\A \modelf \phi[V]\) or \(\A \modelt \psi[V]\), and \(\A \modelt \phi[V]\) or \(\A \modelf \psi[V]\)

if and only if \(\A \modelf \phi[V]\) and \(\A \modelf \psi[V]\), or \(\A \modelt \psi[V]\) and \(\A \modelt \phi[V]\).

\begin{prop}
\label{hiff_0}
For any IFG$_N$-formulas $\phi$ and $\psi$, \(\A \modelt \phi \hiff{\emptyset} \psi\) if and only if \(\norm{\phi}_\A = \norm{\psi}_\A\) and both are perfect.
\end{prop}

\begin{proof}
Suppose \(\A \modelt (\phi \hiff{\emptyset} \psi)[\^NA]\). Then there exist \(V, V' \subseteq \^NA\) such that
\(\A \modelt \phi[V]\), \(\A \modelf \psi[\^NA\setminus V]\), \(\A \modelf \phi[V']\), and \(\A \modelt \psi[\^NA \setminus V']\). Thus \(V \cap V' = \emptyset\) and \((\^NA \setminus V) \cap (\^NA \setminus V') = \emptyset\). Therefore \(V' = \^NA \setminus V\), and \(\norm{\phi}_\A = \tuple{\powerset(V), \powerset(\^NA \setminus V)} = \norm{\psi}_\A\).
\end{proof}

\begin{prop}
\label{hiff_N}
For any IFG$_N$-formulas $\phi$ and $\psi$, \(\A \modelt \phi \hiff{N} \psi\) if and only if \(\norm{\phi}_\A = \norm{\psi}_\A \in \set{0,1}\).
\end{prop}

\begin{proof}
Suppose \(\A \modelt (\phi \hiff{N} \psi)[\^NA]\). Then \(\A \modelt \phi[\^NA]\) and \(\A \modelt\psi[\^NA]\), in which case \(\norm{\phi}_\A = \norm{\psi}_\A = 1 \), or \(\A \modelf \phi[\^NA]\) and \(\A \modelf \psi[\^NA]\), in which case \(\norm{\phi}_\A = \norm{\psi}_\A = 0\).
\end{proof}

\begin{defn}
An \emph{IFG$_N$-schema}\index{IFG$_N$-schema|mainidx} involving $k$ formula variables is an element of the smallest set $\Xi$ satisfying the following conditions.
\begin{enumerate}
	\item The formula variables \(\alpha_0, \ldots, \alpha_{k-1}\) belong to $\Xi$.
	\item For all \(i,j < N\), the formula \(v_i = v_j\) belongs to $\Xi$.
	\item If $\xi$ belongs to $\Xi$, then ${\hneg\xi}$ belongs to $\Xi$.
	\item If $\xi_1$ and $\xi_2$ belong to $\Xi$, and \(J \subseteq N\), then \(\xi_1 \hor{J} \xi_2\) belongs to $\Xi$.
	\item If $\xi$ belongs to $\Xi$, \(n < N\), and \(J \subseteq N\), then \(\hexists{v_n}{J}\xi\) belongs to $\Xi$.
\end{enumerate}
Note that the symbols \(\alpha_0, \ldots, \alpha_{k-1}\) are distinct from the usual variables $v_0$, \ldots, $v_{k-1}$.  If $\xi$ is an IFG$_N$-schema involving $k$ formula variables, and \(\phi_0, \ldots, \phi_{k-1}\) are IFG$_N$-formulas, then the IFG$_N$-formula \(\xi(\phi_0, \ldots, \phi_{k-1})\) is called an \emph{instance}\index{instance of a schema|mainidx} of $\xi$.
\end{defn}

\begin{defn}
Every IFG$_N$-schema $\xi$ has a corresponding term $T_\xi$ in the language of IFG$_N$-algebras. $T_\xi$ is defined recursively as follows:
\begin{enumerate}
	\item \(T_{\alpha_i} = X_i\),
	\item \(T_{v_i = v_j} = D_{ij}\),
	\item \(T_{\hneg\,\xi} = \n{(T_\xi)}\),
	\item \(T_{\xi_1 \hor{J} \xi_2} = T_{\xi_1} +_J T_{\xi_2}\),
	\item \(T_{\hexists{v_n}{J}\xi} = C_{n,J}(T_\xi)\).
\end{enumerate}
\end{defn}

\begin{lem}
\label{schema terms}
Let $\xi$ be an IFG$_N$-schema involving $k$ formula variables, and let $T_\xi$ be its corresponding term. Then for any IFG$_N$-formulas \(\phi_0, \ldots, \phi_{k-1}\) and any suitable structure $\A$,
\[
\norm{\xi(\phi_0, \ldots, \phi_{k-1})} = T_\xi^{\Cyls_{\mathrm{IFG}_{N}}(\A)}(\norm{\phi_0}, \ldots, \norm{\phi_{k-1}}).
\]
\end{lem}

\begin{proof}
If $\xi$ is a formula variable $\alpha_i$, then \(T_\xi = X_i\), so
\[
\norm{\xi(\phi_0, \ldots, \phi_{k-1})} = \norm{\phi_i} = T_\xi^{\Cyls_{\mathrm{IFG}_{N}}(\A)}(\norm{\phi_0}, \ldots, \norm{\phi_{k-1}}).
\]
If $\xi$ is \(v_i = v_j\), then \(T_\xi = D_{ij}\), so
\[
\norm{\xi(\phi_0, \ldots, \phi_{k-1})} = \norm{v_i = v_j} = D_{ij}^{\Cyls_{\mathrm{IFG}_{N}}(\A)} = T_\xi^{\Cyls_{\mathrm{IFG}_{N}}(\A)}(\norm{\phi_0}, \ldots, \norm{\phi_{k-1}}).
\]
Now assume that
{\allowdisplaybreaks
\begin{align*}
	\norm{\xi(\phi_0, \ldots, \phi_{k-1})}
		&= T_\xi^{\Cyls_{\mathrm{IFG}_{N}}(\A)}(\norm{\phi_0}, \ldots, \norm{\phi_{k-1}}), \\
	\norm{\xi_1(\phi_0, \ldots, \phi_{k-1})}
		&= T_{\xi_1}^{\Cyls_{\mathrm{IFG}_{N}}(\A)}(\norm{\phi_0}, \ldots, \norm{\phi_{k-1}}), \\
	\norm{\xi_2(\phi_0, \ldots, \phi_{k-1})}
	&= T_{\xi_2}^{\Cyls_{\mathrm{IFG}_{N}}(\A)}(\norm{\phi_0}, \ldots, \norm{\phi_{k-1}}).
\intertext{Then}
	\norm{\hneg\xi(\phi_0, \ldots, \phi_{k-1})}
		&= \n{\norm{\xi(\phi_0, \ldots, \phi_{k-1})}} \\
		&= \n{\left(T_\xi^{\Cyls_{\mathrm{IFG}_{N}}(\A)}(\norm{\phi_0}, \ldots, \norm{\phi_{k-1}})\right)} \\
		&= T_{\hneg\,\xi}^{\Cyls_{\mathrm{IFG}_{N}}(\A)}(\norm{\phi_0}, \ldots, \norm{\phi_{k-1}}), \\[10pt]

	\norm{\xi_1 \hor{J} \xi_2(\phi_0, \ldots, \phi_{k-1})}
		&= \norm{\xi_1(\phi_0, \ldots, \phi_{k-1})} +_J \norm{\xi_2(\phi_0, \ldots, \phi_{k-1})} \\
		&= T_{\xi_1}^{\Cyls_{\mathrm{IFG}_{N}}(\A)}(\norm{\phi_0}, \ldots, \norm{\phi_{k-1}}) \\
		&\qquad +_J T_{\xi_2}^{\Cyls_{\mathrm{IFG}_{N}}(\A)}(\norm{\phi_0}, \ldots, \norm{\phi_{k-1}}) \\
		&= T_{\xi_1 \hor{J} \xi_2}^{\Cyls_{\mathrm{IFG}_{N}}(\A)}(\norm{\phi_0}, \ldots, \norm{\phi_{k-1}}), \\[10pt]

	\norm{\hexists{v_n}{J}\xi(\phi_0, \ldots, \phi_{k-1})}
		&= C_{n,J}\norm{\xi(\phi_0, \ldots, \phi_{k-1})} \\
		&= C_{n,J}\left(T_\xi^{\Cyls_{\mathrm{IFG}_{N}}(\A)}(\norm{\phi_0}, \ldots, \norm{\phi_{k-1}})\right) \\
		&= T_{\hexists{v_n}{J}\xi}^{\Cyls_{\mathrm{IFG}_{N}}(\A)}(\norm{\phi_0}, \ldots, \norm{\phi_{k-1}}).
\end{align*}}
\end{proof}

\begin{prop}
\label{term operation implies hereditarily simple}
Any double-suited IFG-algebra that has a term operation $T(X,Y)$ such that
\(T(X,Y) = 1\) if and only if \(X = Y\) is hereditarily simple.
\end{prop}

\begin{proof}
Suppose $\C$ is a double-suited IFG-algebra that has such a term oper\-ation. Then for any \(X \not= Y\) we have \(\tuple{1, Z} = \tuple{T(X,X), T(X,Y)} \in \Cg(X,Y)\), where $Z$ is some element different than 1. Hence $\Cg(X,Y)$ is the total congruence. Thus $\C$ is simple.
Furthermore, the sentence
\[
\forall X \forall Y[ T(X,Y) = 1 \iff X = Y]
\]
is universal, and so must hold in every subalgebra of $\C$. Hence $\C$ is hereditarily simple.
\end{proof}

\begin{thm}
\label{no schema}
There is no IFG$_1$-schema $\xi$ involving two formula variables such that for every pair of IFG$_1$-formulas $\phi$ and $\psi$, and every suitable structure $\A$, we have
\[
\A \modelt \xi(\phi, \psi) \quad \text{iff} \quad \norm{\phi}_\A = \norm{\psi}_\A.
\]
\end{thm}

\begin{proof}
Suppose $\xi$ were such a schema. Then the corresponding term $T_\xi$ would have the property that for any $\A$ and any \(\norm{\phi}_\A, \norm{\psi}_\A \in \Cyls_{\mathrm{IFG}_{1}}(\A)\),
\begin{align*}
	T_\xi^{\Cyls_{\mathrm{IFG}_{1}}(\A)}(\norm{\phi}_\A, \norm{\psi}_\A) = 1 \quad
		&\text{iff} \quad \norm{\xi(\phi,\psi)}_\A = 1 \\
		&\text{iff} \quad \A \modelt \xi(\phi,\psi) \\
		&\text{iff} \quad \norm{\phi}_\A = \norm{\psi}_\A.
\end{align*}
Thus every \(\Cyls_{\mathrm{IFG}_{1}}(\A)\) would be hereditarily simple. However \(\Cyls_{\mathrm{IFG}_{1}}(\mathbf{3})\) is not hereditarily simple.
\end{proof}