TITLE: proving that a graph has only one minimum spanning tree if and only if G has only one maximum spanning tree
QUESTION [0 upvotes]: Is this claim true?
I thought about it and it seems true but for proving it i started with one direction by assuming that i have one minimum spanning tree and i want to show that from this i have also one maximum spanning tree.
Can i claim from having one minimum spanning tree the weight of every edge is different and then use this proof Show that there's a unique minimum spanning tree if all edges have different costs ?

REPLY [2 votes]: It's not true.
Let $G$ be $K_n$ and let $P$ be a Hamilonian path in $G=K_n$. [Say $i$ and $j$ are adjacent in $P$ iff $|i-j|=1$, then two vertices $i$ and $j$ are adjacent in $G \setminus E(P)$ iff $|i-j| \ge 2$. One can check that $G \setminus E(P)$ is connected for $n \ge 6$ and has many cycles.]
Give every edge in $P$ a weight of 100 and every edge in $G \setminus E(P)$ a weight of 1. Then $G$ has one maximum-weight spanning tree--namely $P$, but many minimum spanning trees; indeed $G \setminus E(P)$ is connected and has many cycles so any spanning tree of $G \setminus E(P)$ is a minimum spanning tree, and $G \setminus E(P)$ has more than one spanning tree.


TITLE: Definition of supremum clarification (Abbott - Understanding Analysis Lemma 1.3.8)
QUESTION [0 upvotes]: In Abbott, Understanding Analysis page 17 we have the following:
Lemma 1.3.8. Assume $s \in \mathbb R$ is an upper bound for a set $A \subseteq \mathbb R$. Then, $s = \sup A$ if and only if, for every choice of $\epsilon > 0$, there exists an element $a \in A$ satisfying $s-\epsilon < a$.
But if we take $A = [1, 2]$ then $s = \sup A =2$ and choose $\epsilon = 5 > 0$, say, then the lemma fails since $s-\epsilon = 2-5=-3 \notin A$?

REPLY [0 votes]: $s-\epsilon$ need not be in $A$. The Lemma says that the is an element of $A$ that is bigger than $s-\epsilon$, although $s$ is bigger than (or equal) to all elements of $A$.
In your example, where we take $\epsilon=5$ all the elements of $A$ are bigger than $s-\epsilon=2-5=-3$. In fact, for every choice of $\epsilon>0$, the element $2\in A$ is bigger than $s-\epsilon=2-\epsilon$.
Every number $s'$ greater than 2 cannot be the supremum, since we can take $\epsilon=\frac{s'-2}{2}$ so $s'-\epsilon=\frac{2+s'}{2}>2$, so for this choice of $\epsilon$, all the elements of $A$ are less than $s'-\epsilon$.


TITLE: Is the cesaro sum any less valid than the formal sum? (And the arbitrary nature of mathematics.)
QUESTION [0 upvotes]: The title basically says it all. I've just watched a video explaining (without any formal definitions) that the 'sum' of the natural numbers is equal to -1/12. Now, the more I read about mathematics the more it seems to be quite arbitrary - but then this isn't really a problem because from what I understand we are just playing with arbitrary rules.
The problem I have is that I think a lot of people feel naturally inclined to attach some real-world meaning to mathematics. I often see what seems to be sensationalist mathematics explained in layman's terms to provoke a feeling of 'mathematics broke reality!' or something like that.
I'm actually struggling to comprehend the idea of mathematics at the moment so my writing may not be entirely together but a representation of my current thoughts. I would appreciate some clarity - possibly to explain the notion of applying 'realness' to certain elements of mathematics over others.

REPLY [5 votes]: If you want to understand something about mathematics, the way to start is to forgot about this video as quickly as possible. It's not that this video is wrong - it's neither wrong or right, it's a bunch of confusing non-arguments.
One of the achievements of modern mathematics is to provide a consistent foundation for reasoning about infinite sums and infinite quantities. Those foundations require a bit of an effort to learn, but once you learn them, they enable to you formally deal with questions such as "what is the sum of $1+2+3+\ldots$". Those foundation then show that there are different ways to interpret such an expression, and depending on the interpretation you get different results. That shouldn't be surprising - after all, $1+2+3+\ldots$ isn't really a sum - it isn't something you can compute by doing arithmetic - but rather an abstract expression that you first have to assign some meaning to before it makes sense to speak about it's value.
The video, OTOH, ignores all that, and uses essentially arbitrary and ad-hoc arguments to argue that the sum "is obviously" $-\frac{1}{12}$. That's not mathematics, that's just ildly playing with numbers.


TITLE: Estimating age of the universe by Hubble's law?
QUESTION [5 upvotes]: Hubble's law states that $v=Hx$
Age of the universe is calculated by $T= \frac{x}{v} = \frac{1}{H}$
but the velocity is not constant; it changes with distance, so I think that this equation cannot be applied because simply the velocity is NOT uniform
$v$ should have been replaced with $\frac{\mathrm{d}x}{\mathrm{d}t}$, so it gives this differential equation $\frac{\mathrm{d}x}{\mathrm{d}t}=Hx$.
Nevertheless, when this differential equation is solved, and we substitute for initial condition ($x=0$ when $t=0$), it appears that there are no solutions, because it will be an exponential function, which can't take the value $0$ ever.
What is wrong with my understanding?

REPLY [5 votes]: You're actually pretty close to the correct method for estimating the age of the Universe, but $H$ is not constant with time, it is $H=H(t)$.
One of the many ways of writing the equation to solve is:
$$t(z) = \frac{1}{H_0}\int_z^\infty \frac{dz}{(1+z)E(z)}$$
Here $z$ is redshift; $z\rightarrow\infty$ at the Big Bang, and $z=0$ now, so if you integrate from $0$ to $\infty$ you get $t(0)$ , the age of the Universe.
$E(z)$ is a function describing the relative content of the Universe at different redshifts, one example of $E(z)$ is the one for the $\rm\Lambda CDM$ model:
$$E(z) = \sqrt{\Omega_{\Lambda,0}+(1-\Omega_0)(1+z)^2+\Omega_{m,0}(1+z)^3+\Omega_{r,0}(1+z)^4}$$
The $\Omega$ are density parameters, which you can read more about here or here. The $0$ subscripts indicate that these should be the values measured at $z=0$, so these are constants and you can write the age of the Universe in terms of them (or you could, if that integral had an analytic solution - I don't suggest trying to solve it by hand, unless you make some simplifying assumptions or approximations first!).
The reason that $t(0)\approx1/H_0$ works reasonably well is that the Universe has been expanding at very nearly the rate it is expanding at now for the majority of the time since the Big Bang. So approximating $v$ as constant at the present value actually gets you pretty close to the right answer.


TITLE: Check whether the set is connected or compact?
QUESTION [3 upvotes]: Suppose $ f\colon\mathbb{R}\to\mathbb{R}$ be a continuous function and $A$ be a subset of $\mathbb{R}$ defined as:
$$A = \bigl\{\, y\in \mathbb{R} :y= \lim_{n\to\infty} f(x_n)\text{, for  some  sequence  $x_n \rightarrow \infty$}\,\bigr\}.$$
Then the set $A$ is necessarily connected? compact? a singleton set?
Clearly, choosing $f(x)=\sin(x)$ implies $A=[-1,1]$. Hence $A$ need not be a singleton set. Intuitively it seems $A$ should be a connected set. But I am unable to produce a proof.

REPLY [3 votes]: Assume $a,b \in A$, say $f(a_n)\to a$, $f(b_n)\to b$ with $a_n,b_n\to\infty$. Wlog. $a>b$. Let $c\in(a,b)$. For almost all $n$, we have $f(a_n)>c>f(b_n)$. By the IVT, there exists $c_n$ between $a_n$ and $b_n$ with $f(c_n)=c$. Then $c_n\to\infty$, which shows $c\in A$. We conclude that $A$ is convex, hence connected.
$A$ need not be compact, consider $f(x)=x\sin x$.


TITLE: Is QC with Superpositioned Quantum Gates any different than normal Quantum Computation?
QUESTION [4 upvotes]: This might be more appropriate for theoretical CS stackexchange, but it feels sufficiently low level to be relevant here.
Consider the following thought experiment:
I have a Quantum FPGA, it is a Quantum Computer, whose gates themselves can be controlled programmatically.
For example: Suppose I can have a gate object G which can be in a superposition of being a 1 Qubit Pauli X or a 1 Qubit Hadamard gate. The gate object could then be superpositioned into:
$$a_0  \left| P_x \right>  + a_1  \left|  H \right>$$
So when I apply this gate to a single Qubit $Q = Q_0 \left| 0 \right>+ Q_1 \left| 1 \right>$
The resulting state is
$$ a_0 \left| P_x Q \right> + a_1 \left|H Q \right>$$
$$ a_0  \left|  \left( Q_1  \left| 0 \right> + Q_0 \left| y \right> \right)  \right> +a_1 \left| \left( \frac{Q_0 + Q_1}{\sqrt{2}} \left| 0 \right> + \frac{Q_0 - Q_1}{\sqrt{2}}\left| y \right> \right) \right>  $$
As far sampling the Qubit goes, this state looks identical to perhaps some other composition of concrete gates, but at a high level, it may be possible to determine for example the nature of $G$, in which case, the Qubit collapses to a smaller superposition.
So my question:
Is Quantum Computation with Concrete Gates Equivalent in its computational power to Quantum Computation with Superpositioned Gates?
Obviously they must be equivalent up to Polynomial time differences, simply because the first is capable of simulating any quantum system including the latter in polynomial time. But do we know for fact that the latter class, isn't say polynomially faster?
Something to note here (and switching to matrix notation).
The system is initialized with qubits $A,Q$ of the form:
$$Q =  \begin{bmatrix} Q_0 \\ Q_1 \end{bmatrix} , A =  \begin{bmatrix} A_0 \\ A_1 \end{bmatrix} $$
We have a gate G, that can act as either unitary operator $$ C =  \begin{bmatrix} C_{00} & C_{01} \\ C_{10} & C_{11} \end{bmatrix}, D =  \begin{bmatrix} D_{00} & D_{01} \\ D_{10} & D_{11} \end{bmatrix}   $$
Depending on $A$, and acts on $Q$. So the resulting output is
$$ \begin{bmatrix} A_0 (C_{00} Q_0 + C_{01} Q_1)+ A_1(D_{00}Q_0 + D_{01} Q_1) \\ A_0 ( C_{10} Q_0 + C_{11} Q_1) + A_1 (D_{01} Q_0 + D_{11} Q_1 ) \end{bmatrix} $$
This is a distinctly non-linear process. that cannot be represented using a unitary transform since the output Qubits have probabilities of the form $A_i Q_j$
So while this may be simulated by a Quantum Machine it definitely is doing something a bit different.

REPLY [4 votes]: If a superposed gate is equivalent to a choice of gates controlled by some pre-initialized ancilla qubits, then you can get the exact same effect with a normal gate. Just have the appropriately initialized ancilla be passed in, instead of hidden inside.
I don't think hiding the ancilla inside will give any polynomial benefit in gate count or other metrics. Unless you play games with counting the cost of passing in the ancilla for normal gates, but not counting the cost for superposed gates.

Also, be aware that a superposition of gates will act like a probability distribution of gates. At least, it will if the ancilla backing the superposition are not used for anything else.
Clearly you could measure the hidden backing ancilla at the end of the circuit without affecting the already-measured result. But measurement commutes with controls, and the only things on the gates' ancilla qubits are controls. So you can just slide those measurements all the way to the start of the circuit without affecting the expected outcome.
If the backing qubits were already measured at the start, you have a probability distribution of gates instead of a superposition of gates. Which somehow seems a lot less promising, but must be equivalent.


TITLE: How to find the CDF of the sum of independent uniformly distributed random variables?
QUESTION [3 upvotes]: $X,Y$ are independent random variables with uniform distribution on $[0,1]$, and let the random variable $Z=X+Y$.
The density of $Z$ is:
$$f_{X+Y}(z)=\int_0^z f_X(x)f_Y(z-x)dx$$
What is the formula for the probability $P(Z \leq m)$?
Any help would be appreciated.
Sorry, I have no attempt.

REPLY [2 votes]: $\textbf{hint}$
$$
F(a) = P(X\leq a)
$$
Where $F(X)$ is the cdf.
The following is how to compute the cdf from a given pdf $f(x)$
$$
F(a) = \int^{a}_{x_{min}}f(x)dx
$$
where the minimum of your support of the pdf is $x_{min}$ i.e for uniform random variable between 0 and 1, $x_{min}$ = 0.
Can you see how you can solve your problem?
it doesn't seem right generally and by that I mean
$f_Z(z) = \begin{cases}{
      z, \ \ \ \ 0\leq z < 1 \\
      2-z, \ \ \  \ 1\leq z\leq 2 \\
      0, \ \ \ \  \text{otherwise}}
\end{cases}
$
ps. can someone reformat this horredous expression for me :).
So can you take it from there?


TITLE: Why does $-b < a < b \iff |a| < b$?
QUESTION [0 upvotes]: I don't have any intuition for this because it's just something I memorized. I only understand that $|a| = a$ if $a$ is already positive (or $0$), and $|a| = -a$ if $a$ is negative since we want to turn it positive ($-(-a)$). So does that mean we will have a piecewise inequality also?
But how do we get absolute value from $-b < a < b$?

REPLY [0 votes]: First of all you should remark that: $-b < a < b$ implies that $b \geq 0$. Then you get:
$$-b < a \Rightarrow -a < b$$
Because of the usual rules of inequalities.
You also have the inequality:
$$a < b$$
Which implies:
$$ max(-a, a) < b$$
And
$$|a| = max(a, -a)$$
So you can conclude from here.
The converse follow the same path of reasonning backward


TITLE: System of ODEs on a graph
QUESTION [0 upvotes]: I have a directed graph consisting of nodes and edges. A system of ordinary differential equations describes how the value at each node changes with respect to time, as a function of the value of the surrounding nodes that are connected by edges.
After numerical integration I can see how the value at each node changes as a function of time. Is there a way for me to see how each edge affects the change at a single node (using the Jacobian perhaps)?
Example of graph
In this example, if we pick node A, I want to know the relative contributions of each of those vertices (flow coming from nodes B, C and D) at each moment in time.

REPLY [1 votes]: You can consider the partial derivative of $x_i(t)$ (the value at node $i$) with respect to a constant $a_{jk}$ of the system corresponding to edge $jk$. These will satisfy
differential equations as well, since
$$ \dfrac{d}{dt} \dfrac{\partial x_i}{\partial a_{jk}} = \dfrac{\partial}{\partial a_{jk}} \dfrac{d x_i}{dt} $$
You will thus end up with a larger system of differential equations involving the $x_i$ and $\partial x_i/\partial a_{jk}$, which can again be solved numerically.


TITLE: Diffeomorphism between a sphere and ellipsoid in $\mathbb R^3$.
QUESTION [3 upvotes]: Diffeomorphism between a sphere and ellipsoid in $\mathbb R^3$.

I've been looking for an diffeomorphism between a sphere in $\mathbb R^3$ and an ellipsoid of the form $$\{ (x,y,z) \in \mathbb R^3 \mid \frac {x^2} {a^2} + \frac {y^2} {b^2} + \frac {z^2} {c^2} = 1\}$$
I know from linear algebra such an ellipsoid can be represented as a matrix product $x^T A x = 1$, but how can this be used to get a diffeomorphism ?
Can such a diffeomorphism be found for a general ellipsoid ?

REPLY [6 votes]: Denote by $E$ your ellipsoid, and by $S^2$ the sphere $\{x \in \mathbf R^3 \mid x_1^2 + x_2^2 + x_3^2 = 1\}$. Define a map $f \colon E \to S^2$ by
$$ f(x) = \left(\frac{x_1}a, \frac{x_2}b, \frac{x_3}c \right) $$
Then $f$ is a diffeomorphism.


TITLE: An integration in PDE
QUESTION [1 upvotes]: $$\dfrac{y}{\pi}\int_{-\infty}^{+\infty}\dfrac{{\rm d}\xi}{(1+\xi^2)\left[(\xi-x)^2+y^2\right]}$$
I am stuck here. Any help would be appreciated!

REPLY [2 votes]: Outline of the solution:
Suppose $x$ and $y$ are real numbers and $(x,y)\ne(0,1)$ and denote the integral by $I$. Take the upper hemispherical contour in the complex plane. The part at infinity vanishes. WLOG $y>0$. Let the integrand be $f(z)$.
\begin{align}\frac{\pi}{y}2\pi iI &= \sum\text{Res}[f(\text{poles above the real axis})] \\
&= \frac{1}{2i((i-x)^2+y^2)}+\frac1{(1+(x+iy)^2)(2iy)}.
\end{align}


text:
\begin{document}

\pagestyle{empty}

\setcounter{page}{0}

\hspace{-1cm}

\begin{center}

{\LARGE{\sffamily  New path integral representation for Hubbard model: II. Spinless case}}\\[1cm]

{\large V.Kirchanov, V. Zharkov \footnote{vita@psu.ru,  Kirchanv@rambler.ru}\\[.484cm]
\textit{ Perm State Technical university, Komsomolcky Prospect, 29a, Perm, 614600, Russia\\[.242cm]
Natural Sciences Institute of Perm State university,\\
 Genkel st.4, Perm,614990, Russia. }}
\end{center}

\vspace{2.5em}

\begin{abstract}
The Hubbard model is used to study an electronic system.
In this paper we present the new path integral representation for Hubbard
model. We have constructed the new supercoherent state for spinless electrons which appears from a set
of eigenfunctions of atomic limit of strongly correlated systems. Exact
calculation of nonlinear representation of a supergroup has been carried. This group
defines the transformation of atomic base. The general formalism we
elaborate for Hubbard model is the one widely used in the gauge field theory of the nonlinear
representation of a superconformal group.
\end{abstract}

\newpage
\pagestyle{plain}

\section{Introduction}

\noindent

The Hubbard model was originally constructed to describe
a metal-insulator transition for spin-dependent fermions
in a simple way \cite{hubbard,fradkin,fulde}.

Today this  model is still remain  the main workspace for
investigation of the strong electron's correlation.There exist many approaches to
this model for describing many electrons system: the band limit approximation
for the weak interaction between electrons and the atomic limit for electrons with the
strong coulomb repulsion. We start the series of papers in which we are intending to elaborate the new
approach to the Hubbard model.

 We will present a new path integral approarch to the strong
interaction regime. The main ingredients for us is the usage of a supercoherent state
acting upon an atomic base and the work with effective functional for an electronic
system. We will develop  the procedure of geometric quantization for strongly
correlated systems of electrons. Let us make the brief sketch of the program
firstly developed in the series of papers of one of the  authors \cite{zharkov}. One of the main
distinction of this approach to the Hubbard model is that we treat the  local
supergroup of the local space-time  as the main object of our theory. This
supergroup generates the transformation of the space-time coordinates which assigns the arguments
for any function describing this system. This representation of a supergroup in the
superspinor space has to contain a Lorents or an  SO(4) generators  equal to the even
subalgebra of the Hubbard operators. Odd Hubbard operators produce some superextension of the
Lorents group into some supergroup. This dynamic supergroup is given by
local dynamic superfields  describing the local degrees of freedom of the
strongly correlated electrons system. We shall introduce the supercoherent
state depending on generalized angles  equal to the bosonic and fermionic
fields of the system. Parametrised by x,y,z,t space-time manifold which
determines the arguments of wave function defines us the spinor and superspinor
bundle. This superbundle is determined by the supercoherent state.

There are two kinds of the gauge fields in superspinor bundle: one sort of fields is
the composite fields equal to the quadratic combination of odd grassmann fields
and other sort are nonlinear fields determining the local coordinates frame of the
four-dimensional space-time. In general, the local superspinor bundle defines the nonlinear representation of a
superconformal group as a maximal group of 4-dimentional space-time of
interacting fermionic system.

As the first step of quantization of electronic system with strong coulomb
repulsion we perform the reformulation of Hubbard model into the atomic limit
formalism.  This approach is well known in  Mott-Hubbard insulators theory.
We want to point out that our approach includes all elements of
geometric quantisation \cite{Woodhouse}: for example possess  the algebra of 4-dimensional
rotations (Lorents) group, Cartan differential one-forms which give us the
lagrangian of system, the nonlinear representation of underlying supergroup as a
ground for the supercoherent state.

In the second paper of this series we will fulfill the exact calculation of the nonlinear
representation of spinless  supergroup generators of which appears in the  so called "tower of symmetry" in the Hubbard model \cite{zharkov}. This gives us the possibility to introduce the
supercoherent state for some deformed version of spinless algebra of the strongly correlated electron system and as a result to
obtain the effective action in a future  paper.

Let us give the brief description of the calculation method for the finding
representation of dynamic supergroup in strongly interacting models.
Our task is to find in this model such group structure which could help us
to describe the specificity of strong correlation. We take the following construction as a base:

1) we will collect space coordinates together with the time coordinate and will consider some curved
space-time as a base in which Lorents subalgebra of superconformal group act in the spinor base on dynamical fields.

2) full bases of electronic operators gives us the  supergroup which is
parametrised by  2 dynamical fermionic fields: this fields comprise the
real massless spinor.

3) the superspinor representation of a supergroup gives us the supercoherent state
described by the nonlinear function over odd grassmanian fields. This function
characterises the local properties of the strongly correlated system.

\section{Atomic description of Hubbard model}

We consider the Hubbard model:

\begin{equation}
H=-W\sum_{ij\sigma}\alpha_{\sigma,i}^{+}\alpha_{\sigma,j}+U\sum_{i,\sigma
}n_{\sigma,i}n_{-\sigma,i}+\mu \sum_{\sigma,i}n_{\sigma,i},
\end{equation}

here \ $\alpha_{\sigma,i}^{+}\alpha_{\sigma,j}$\  \ -electron creation and
annihilation operators. \ $n_{\sigma,i}$\ - electron \  \ density operator
 W, U,  \ -band width, one- site electron repulsion and chemical potential.

At first we represent the Hubbard model in atomic bases which determine the atomic limit.
This limit appear as a result of the following procedure. A zero approximation of the atomic limit is described by one-site repulsion term:
$$U\sum_{i,\sigma}n_{\sigma,i}n_{-\sigma,i}+\mu \sum_{\sigma,i}n_{\sigma
,i}.$$

This hamiltonian can be diagonalized by the following one-site atomic eigenfunction:

\begin{equation}
|0\succ;|+\succ=\alpha_{\uparrow}^{+}|0\succ;|-\succ=\alpha_{\downarrow}
^{+}|0\succ;|2\succ=\alpha_{\uparrow}^{+}\alpha_{\downarrow}^{+}|0\succ.
\end{equation}

 This bases gives us the fundamental representation of some supergroup
in the space of dimension $(2,2)$. Point out that the states   \ $|+\succ;|-\succ;$\  \ are fermions \ but
\ $|0\succ;$\  \ $|2\succ$\  \  \ are bosons in our construction. \ We insert
later odd grassmann fields and make this fact obvious ( ie states
\ $|+\succ;|-\succ;$\ will be depend on odd order of the grassmann fields but
\ states \  \ $|0\succ;$\  \ $|2\succ$\ -on even order of the grassmann fields). All
operators in this bases will be the matrix which are determined by commutation
and anticommutation relation giving some superalgebra. Full set of the Hubbard
operators have 16 operators part of which $$(X^{0+},X^{0-},X^{+0},X^{-0}
,X^{+2},X^{-2},X^{2+},X^{2-})$$ are the fermionic operators, but other part
$$(X^{+-},X^{-+},X^{++}-X^{--},X^{02},X^{20},X^{00}-X^{22})$$--are the bosonic
operators. $X^{ij}$ -Hubbard operators contain only one non-zero element equal
1 sitting on site $(i,j)$ in the matrix representation. Point out that this set of
operators gives some bases for some superalgebra.

 We have the following representation for creation-annihilating
operators in this bases:

\begin{equation}
\alpha_{\uparrow}^{+}=X^{+0}+X^{2-}\\
\alpha_{\downarrow}^{+}=X^{-0}+X^{2+}
\end{equation}

The Hubbard model in this representation has the form:

\begin{equation}
 H=U\sum_{i,p}X_{i}^{pp}-W\sum_{ij\alpha \beta}X_{i}^{-\alpha}X_{j}^{\beta}
\end{equation}

\section{ Supercoherent state for Hubbard model}

In constracting the supercoherent state we use the following interesting
observation in interpretation of the set of atomic operators and function for
on-site Hubbard repulsion. This observation can be formulated as the following statement: six even Hubbard
operators constitute the subalgebra isomorphic with algera of Lorentz group or algebra of four dimensional rotation group in spinor representation. Complete derivation of this statement will be obtain in subsequent paper.

 To characterise the state of the system by coherent
state we input some fields which depend on coordinates $x,y,z$ and the time $t.$
We have three component dynamic vector of the electricl field
$$
\mathbf{E}=(E^{+}(x,y,z,t),E^{-}(x,y,zt),E^{z}(x,y,z,t)),
$$
three component dynamical vector of the magnetic field
$$
\mathbf{h}=(h^{+}(x,y,z,t),h^{-}(x,y,z,t),h^{z}(x,y,z,t))
$$
and four component dynamical odd grassmann fields

$$\chi^{\ast}(x,y,z,t),\chi(x,y,z,t),$$  which are the fermionic fields giving the
components of maiorana spinor. All dynamical fields appear in supercoherent
state in the followng manner:

\begin{equation}
\label{cs}\mid G>=exp\left[
\begin{array}
[c]{cccc}
E_{z} &\chi & \chi & E^{+}\\
\chi^{*} & h_{z} & h^{+} &\chi\\
\chi^{*} & h^{-} & -h_{z} &\chi\\
E^{-} & \chi^{*} & \chi^{*} & -E_{z}
\end{array}
\right]  \mid0>,
\end{equation}

 Exponent here act in space of atomic eigenfunctions $(\mid
0>,\mid+>,\mid->,\mid2>)$, Function  $\mid0>$ is highest weight vector of
 representation space of supergroup which is given by exponent.

\section{ Evolution operator for electronic system}

 The transition amplitude of the evolution operator of the quantum systems is
given by the following expresion: $<Z_{f}|e^{-iH(t_{f}-t_{i})}|Z_{i}>$.
We want to obtain the expression for the effective functional using the states
$|Z>$. Time evolution of the system is given by the following operator:

$$
U(t,t_{0})=T_{ord}exp(-i\int_{t_{0}}^{t}H(\tau)d\tau);
$$

 if \ $t-t_{0}=\delta t$\  \ is small, ie $\delta t<<1$,\ then

$$
U(t_{0}+\delta t,t_{0})=1-i\int_{t_{0}}^{t_{0}+\delta t}H(\tau)d\tau.
$$

It is follow from this expression that the symbol for evolutionary
operator has the following form:

$$
U(Z,Z^{\ast}|t_{0}+\delta t,t_{0})=exp(-i\int_{t_{0}}^{t_{0}+\delta
t}H(Z,Z^{\ast}|\tau)d\tau).
$$

 We devide time interval $[t_{0},t]$ by the number N and obtain N small
intervals for finding the expression for symbol $U(Z,Z^{\ast}|t,t_{0})$. Consider the
matrix elements of evolution operator $exp(-iH(t_{f}-t_{i}))$ between the states
$<Z_{f}|$ and $|Z_{i}>$. Factorising operator $exp(-iH(t_{f}-t_{i}))$ by inserting the identity operator
$\int d\mu(Z)|Z><Z|=1$\ we obtain the following representation:

$$
<Z_{f}|exp(-iH(t_{f}-t_{i}))|Z_{i}>=\int \prod_{k=1}^{N}d\mu(Z_{k})<Z_{f}
|Z_{N}>
$$

$$
<Z_{N}|e^{-i\epsilon H}|Z_{N-1}>....<Z_{k-1}|e^{-i\epsilon H}|Z_{k}
>...<Z_{1}|e^{-i\epsilon H}|Z_{i}>,
$$

 here $\epsilon=\frac{t_{f}-t_{i}}{N}.$  In first order of $\epsilon$
\ we can transform this formular and place the symbol of operator H in the exponent

\begin{equation}
\frac{<Z_{k+1}|e^{-i\epsilon H}|Z_{k}>}{<Z_{k+1}|Z_{k}>}=\frac{<Z_{k+1}
|(1-i\epsilon H)|Z_{k}>}{<Z_{k+1}|Z_{k}>}=e^{-i\epsilon \frac{<Z_{k+1}
|H|Z_{k}>}{<Z_{k+1}|Z_{k}>}}+O(\epsilon^{2}).
\end{equation}

 As a result we obtain the representation

\begin{equation}
<Z_{f}|e^{-iH(t_{f}-t_{i})}|Z_{i}> = \lim_{N->\infty}\int \prod_{k=1}^{N} d
\mu(Z_{k}) <Z_{k+1}|Z_{k}> e^{-i\epsilon \sum_{k=1}^{N}\frac{<Z_{k+1}|
H|Z_{k}>}{<Z_{k+1}|Z_{k}>}},
\end{equation}

here $|Z_{0}>=|Z_{i}>;<Z_{N+1}|=<Z_{f}|.$  Let define a variation of the following type $|Z>:|\delta Z_{k+1}>=|Z_{k+1}
>-|Z_{k}>.$ We have:

$$
<Z_{f}|e^{-iH(t_{f}-t_{i})}|Z_{i}> = \lim_{N->\infty}\int \prod_{k=1}^{N} [d
\mu(Z_{k}) <Z_{k}|Z_{k}>]<Z_{f}|Z_{N}>
$$

$$
exp(\sum_{k=1}^{N}(Ln(1-\frac{<Z_{k}|\delta Z_{k}>}{<Z_{k}|Z_{k}>}
)-i\epsilon \frac{<Z_{k}| H|Z_{k-1}>}{<Z_{k}|Z_{k-1}>}).
$$

 in linear-slice approximation. We take the following expresion for the time derivative: Considering the first
order in $\epsilon$\ we can take the folllowing expresion for the time derivative:

$$
\frac{d|Z>}{dt}=\frac{|\delta Z>}{\epsilon}.
$$

In first order in $\epsilon$, we obtain the final path integral representation
of the evolutionary operator in coherent state formalizm

\begin{equation}
<Z_{f}|e^{-iH(t_{f}-t_{i})}|Z_{i}>=\int_{|Z(t_{i})>=|Z_{i}>}^{|Z(t_{f}
)>=|Z_{f}>}D(Z,Z^{\ast})e^{-iS[Z,Z^{\ast}]};
\end{equation}
$$
S[Z,Z^{\ast}]=\int_{t_{i}}^{t_{f}}dt\int_{V}dr(\frac{<Z(r,t)|i\frac
{\partial}{\partial t}-H|Z(r,t>}{<Z(r,t|Z(r,t>}
$$
$$
-i[Ln(<Z_{f}|Z(t_{f})>)-Ln(<Z_{i}|Z(t_{i})>)]).
$$
Measure of integration is given by the following expresion:

$$
D[Z,Z^{*}] = \prod_{t_{i}<t<t_{f}} \prod_{r} d\mu[Z(r,t)^{*}
,Z(r,t)]<Z(r,t)|Z(r,t)>.
$$

 This form of path intergal representation will be the starting point
of our consideration.

\section{ Nonlinear representation of supergroup in Hubbard model}

In construction of supercoherent state we have the followng supermatrix which
we must compute analyticaly:

$$
U=\exp \left(
\begin{array}
[c]{cccc}
E_{z} & \chi & \chi & E^{+}\\
\chi^{*} & h_{z} & h^{+} & \chi\\
\chi^{*} & h^{-} & -h_{z} & \chi\\
E^{-} & \chi^{*} & \chi^{*} & -E_{z}
\end{array}
\right)
$$

In this expression we have the fields of different statistics: for example, set of
the fields

$$(\chi^{*}(t,x,y,z),\chi(t,x,y,z))$$

are the odd grassman valued function of space-time coordinates and
describe the fermionic degree of freedom but fields

$$(E_{z}(t,x,y,z),E^{+}
(t,x,y,z),E^{-}(t,x,y,z),h_{z}(t,x,y,z),h^{+}(t,x,y,z),h^{-}(t,x,y,z)) $$

describe the electro-magnetic degree of freedom, equal to two three component
vectors of the space-time coordinates and are bosonic. In this paper we concentrate in calculating exact representation of exponent of the supermatrix in the coherent state. We take the dynamic electrical, magnetic and grassmann fields which depend on coordinates of 4-dimensional space-time manifold on definite the coordinates and omit  $(t,x,y,z)$\ coordinates in subsiquent formulas.

Our general strategy will be to expand the supermatrix to N-order in fields. Them
we can isolate and collect certain series and get some recurrent formular for general term in infinite series. Using this formula we can sum all terms to anytical compact representation. Analytical representation of the supermatrix elements will be the final point of our work. As a starting point we have the following exponential expression for the representation of super extension of the Lorentz group in the spinor representation.

Expanding this exponent in series we can obtain first and second order in
the parameters $ b$ and $ h $. We see that the polynomial series on the grassmann numbers can be classified in grassmann order n. All the supermatrix elements can be represented as a coefficients in grassmann polynomials of order n, where n=0,1,2,3,4.

\section{ Matrix series for the nonlinear representation of supergroup}

 First of all point out that the supermatrix \ in exponent have two
submatrix: one is the odd grassmann matrix and the other is the even submatrix
containing only the fields of type $E_{i}(t,x,y,z)$ \ and $h_{i}(t,x,y,z)$ \ type.
\ As a first step we make expansion of the exponent for the even matrix.
Expanding in series this exponent we can obtain first, second order and 3,4,5
order in bosonic fields $E_{i},i=1,2,3.$ and \ $h_{i},i=1,2,3.$ For example, the series for n=0, 1,2 has the following form:

$$\left(
\begin{array}
[c]{cccc}
1 & 0 & 0 & 0\\
0 & 1 & 0 & 0\\
0 & 0 & 1 & 0\\
0 & 0 & 0 & 1
\end{array}
\right)  +$$

$$\left(
\begin{array}
[c]{cccc}
E_{z} & 0 & 0 & E^{+}\\
0 & h_{z} & h^{+} & 0\\
0 & h^{-} & -h_{z} & 0\\
E^{-} & 0 & 0 & -E_{z}
\end{array}
\right)  +\frac{1}{2!}\left(
\begin{array}
[c]{cccc}
b^{2} & 0 & 0 & 0\\
0 & h^{2} & 0 & 0\\
0 & 0 & h^{2} & 0\\
0 & 0 & 0 & b^{2}
\end{array}
\right)  +$$

$$\frac{1}{3!}\left(
\begin{array}
[c]{cccc}
b^{2}E_{z} & 0 & 0 & b^{2}E^{+}\\
0 & h^{2}h_{z} & h^{2}h^{+} & 0\\
0 & h^{2}h^{-} & -h^{2}h_{z} & 0\\
b^{2}E^{-} & 0 & 0 & -b^{2}E_{z}
\end{array}
\right)  +\frac{1}{4!}\left(
\begin{array}
[c]{cccc}
b^{4} & 0 & 0 & 0\\
0 & h^{4} & 0 & 0\\
0 & 0 & h^{4} & 0\\
0 & 0 & 0 & b^{4}
\end{array}
\right)  +$$

$$  \frac{1}{5!}\left(
\begin{array}
[c]{cccc}
b^{4}E_{z} & 0 & 0 & b^{4}E^{+}\\
0 & h^{4}h_{z} & h^{4}h^{+} & 0\\
0 & h^{4}h^{-} & -h^{4}h_{z} & 0\\
b^{4}E^{-} & 0 & 0 & -b^{4}E_{z}
\end{array}
\right)  $$

 here we introduce the following abbriviation \ $b=\sqrt{E_{z}
^{2}+E^{+}E^{-}},h=\sqrt{h_{z}^{2}+h^{+}h^{-}}$

For the odd grassmann number we \ have following expansion series of exponent. We write here two terms for the grassmann fields $(\chi^{*},\chi)$ for obtaining coefficients in higher order in $E_{i}$ \  \ and
\ $h_{i}$ .

 $$ \left(\begin{array}
[c]{cccc}
0 & \chi & \chi & 0\\
\chi^{*} & 0 & 0 & \chi\\
\chi^{*} & 0 & 0 & \chi\\
0 & \chi^{*} & \chi^{*} & 0
\end{array}
\right)  +$$

$$\frac{1}{2!}\left(
\begin{array}
[c]{cccc}
0 & \chi^{*} E^{+}+\chi p1 & \chi^{*} E^{+}+\chi p3 & 0\\
\chi^{*}E_{z}+\chi ( E^{-}+ h^{+}+ h_{z}) & 0 & 0 & \chi^{*} E^{+}+\chi(- E_{z}+ h^{+}+ h_{z})\\
\chi^{*} E_{z}+\chi (E^{-}+ h^{-}- h_{z}) & 0 & 0 & \chi^{*} E^{+}+\chi(- E_{z}+ h^{-}- h_{z})\\
0 & \chi^{*}p2+\chi E^{-} & \chi^{*}p4+\chi E^{-} & 0
\end{array}
\right)  $$

 For even order of the grassmann variables we have following series for the
composite bosonic fields:

$$\left(
\begin{array}
[c]{cccc}
2 \chi \chi^{*} & 0 & 0 & 0\\
0 & 0 & 0 & 0\\
0 & 0 & 0 & 0\\
0 & 0 & 0 & 0
\end{array}
\right)$$
$$  +\frac{1}{3!}\left(
\begin{array}
[c]{cccc}
4 E_{z} \chi \chi^{*}+(h^{+}+h^{-})\chi \chi^{*} & 0 & 0 & 0\\
0 & -2 E_{z} \chi \chi^{*} & -2 E_{z} \chi \chi^{*} & 0\\
0 & -2 E_{z} \chi \chi^{*} & -2 E_{z} \chi \chi^{*} & 0\\
0 & 0 & 0 &-4 E_{z} \chi \chi^{*}+(h^{+}+h^{-})\chi \chi^{*}
\end{array}
\right)  +$$

$$\frac{1}{4!}\left(
\begin{array}
[c]{cccc}
2 mm \chi \chi^{*} & 0 & 0 & 4 E^{+}E_{z} \chi \chi^{*}\\
0 & -2 E_{z}(2 h_{z}+X) \chi \chi^{*} & -4 E_{z}h^{+}\chi \chi^{*} & 0\\
0 & -4 E_{z}h^{-}\chi \chi^{*} &  -2 E_{z}(-2 h_{z}+X) \chi \chi^{*} & 0\\
-4 E^{-}E_{z} \chi \chi^{8} & 0 & 0 & 2 mm \chi \chi^{*}
\end{array}
\right)  $$

 here $ m=e^2+h^2,X=h^{+}+h^{-}, pm=h_{z}+h^{-},pp=h_{z}+h^{+},mp=h^{+}-h_{z},mm=h^{-}-h_{z}$
and $ mm=m+2 E_{z}^2+E_{z} X$
 The expansion series for third order of the grassmann variable has the
following form of first two terms
$$\frac{m}{4!} *$$
$$\left(
\begin{array}
[c]{cccc}
0 & E^{+} \chi^{*} +p1 \chi & E^{+} \chi^{*} +p3 \chi & 0\\
E^{-} \chi +(E_{z}+pp)\chi^{*} & 0 & 0 & E^{+} \chi^{*} +(-E_{z}+pp)\chi\\
E^{-} \chi +(E_{z}+mm)\chi^{*} & 0 & 0 & E^{+} \chi^{*}+( -E_{z}+mm)\chi\\
0 & E^{-} \chi+p2 \chi^{*} & E^{-} \chi+p4 \chi^{*} & 0
\end{array}
\right)  $$
here $p1=E_{z}+pm, p2=-E_{z}+pm, p3=E_{z}+mp, p4=-E_{z}+mp $

 Expanding this series on 12 order in even and odd parameters we can
obtain the analytical representation for the supermatrix elements.

\section{ Analytical representation of supergroup}

 Having series representation for supergroup for high order in fields
we can obtain the general analytical form for the matrix elements. Our task is to
obtain exact dependence of the matrix element over component of fields:
$E_{z,}E^{+},E^{-};h_{z},h^{+},h^{-}$ but not dependence over $b$ and $h$.

For example the general form of $u_{11}$ \ supermatrix elements in orders
higher than 8 as a functions of the dynamic even and odd grassmanian fields
are given by following expressions

$$u_{11}=cosh(e)+E_{z} sinh(e)/e +$$
$$(2 f_3 +X f_2 +2 E_{z}(2 f2EE+X f1EE)+2 E_{z}^2(2 f1EE+X fEE))\chi \chi^{*}$$

here the coefficients $f_{i}, i=2,3,4$  are some series in $e$ and
$h$ variables:

$f2EE, f1EE,fEE; f2h$ have some infinite series over $e$ and $h$ variables.\

Another matrix element equal to

$$u_{21}=\chi^{*} (f_4 + (E_{z}+h^{+}+h_{z}) f_3 +  E_{z} (h^{+}+h_{z}) f_2) + \chi E^{-} (f_3+(h^{+}+h_{z})f_2)$$

and we have some function for the coefficients: $f_4,f_3,f_2$

 For $u_{31}$ we have

$$u_{31}=\chi^{*} (f_4 + (E_{z}+h^{-}- h_{z}) f_3 +  E_{z} (h^{-}- h_{z}) f_2) + \chi E^{-} (f_3+(h^{+}+h_{z})f_2)$$

 Last element of first column is

$$u_{41}=E^{-} sinh(e)/e +2 E_{z}E^{-}( 2 f1EE + X fEE)\chi \chi^{*}$$

 For coefficients in this expression we have: $ X=h^{+}+h^{-} $

 For second column we have

$$u_{12}=\chi (f_4 + (E_{z}+h^{-}+h_{z}) f_3 + E_{z} (h^{-}+h_{z}) f_2) + \chi^{*} E^{+} (f_3+(h^{-}+h_{z})f_2)$$

For $u_{22}$ we have:

$$u_{22}=cosh(h)+h_{z} sinh(h)/h + E_{z}( f2h +X h_{z} fhh +(2 h_{z}+X) f1hh)(-2\chi \chi^{*})$$

For $u_{32}$ we have:

$$u_{32}=h^{-} sinh(h)/h + E_{z}( f_2 + 2 h^{-} f1hh + h^{-}X fhh)(-2\chi \chi^{*})$$

For $u_{42}$ we have:

$$u_{42}=\chi^{*} (f_4 + (-E_{z}+h^{-}+h_{z}) f_3 -  E_{z} (h^{-}+h_{z}) f_2) + \chi E^{-} (f_3+(h^{-}+h_{z})f_2)$$

For $u_{13}$ we have:

$$u_{13}=\chi (f_4 + (E_{z}+h^{+}- h_{z}) f_3 + E_{z} (h^{+}- h_{z}) f_2) + \chi^{*} E^{+} (f_3+(h^{+}- h_{z})f_2)$$

For $u_{23}$ we have:

$$u_{23}=h^{+} sinh(h)/h + E_{z}( f_2 + 2 h^{+} f1hh + h^{+}X fhh)(-2\chi \chi^{*})$$

For $u_{33}$ we have:

$$u_{33}=cosh(h)- h_{z} sinh(h)/h + E_{z}( f2h - X h_{z} fhh +(- 2 h_{z}+X) f1hh)(-2\chi \chi^{*})$$

For $u_{43}$ we have:

$$u_{43}=\chi^{*} (f_4 + (-E_{z}+h^{+}- h_{z}) f_3 -  E_{z} (h^{+}- h_{z}) f_2) + \chi E^{-} (f_3+(h^{+}- h_{z})f_2)$$

For $u_{14}$ we have:

$$u_{14}=E^{+} sinh(e)/e +2 E_{z}E^{+}( 2 f1EE + X fEE)\chi \chi^{*}$$

For $u_{24}$ we have:

$$u_{24}=\chi (f_4 + (-E_{z}+h^{+}+h_{z}) f_3 -  E_{z} (h^{+}+h_{z}) f_2) - \chi^{*} E^{+} (f_3+(h^{+}+h_{z})f_2)$$

For $u_{34}$ we have:

$$u_{34}=\chi (f_4 + (-E_{z}+h^{-}- h_{z}) f_3 -  E_{z} (h^{-}- h_{z}) f_2) +$$
$$ \chi^{*} E^{+} (f_3+(h^{-}- h_{z})f_2)$$

For $u_{44}$ we have:

$$u_{44}=cosh(e)- E_{z} sinh(e)/e +$$
$$(2 f_3 +X f_2 - 2 E_{z}(2 f2EE+X f1EE)+2 E_{z}^2(2 f1EE+X fEE))\chi \chi^{*}$$

We see later that many series in our list are equivalent to each other. After such selection between similar ones we have only some different series.

\section{Analytical representation for series}

Collecting the terms in series expansion for $a$ and \ $b$
\ coefficients to 12 order  we obtain for example the following representation for:

$$ f_{2}=\frac{1}{3!}+\frac{b^{2}+h^{2}}{5!}+\frac
{b^{4}+h^{2}b^{2}+h^{4}}{7!}+\frac{b^{6}+h^{2}b^{4}+h^{4}b^{2}+h^{6}}
{9!}+\frac{b^{8}+h^{2}b^{6}+h^{4}b^{4}+h^{6}b^{2}+h^{8}}{11!}+
$$
$$
\frac
{b^{10}+h^{2}b^{8}+h^{4}b^{6}+h^{6}b^{4}+h^{8}b^{2}+h^{10}}{13!}+$$

$$\frac{b^{12}+h^{2}b^{10}+h^{4}b^{8}+h^{6}b^{6}+h^{8}b^{4}+h^{10}b^{2}+h^{12}
}{15!}+
$$
$$
\frac{b^{14}+h^{2}b^{12}+h^{4}b^{10}+h^{6}b^{8}+h^{8}b^{6}+h^{10}
b^{4}+h^{12}b^{2}+h^{14}}{17!}+.........$$

Let us show how to sum following infinite series for $b$ \ for
example. \ We have

$$1+\frac{b^{2}}{2!}+\frac{b^{4}}{4!}+\frac{b^{6}}{6!}+\frac{b^{8}}
{8!}+\frac{b^{10}}{10!}+......+E_{z}(1+\frac{b^{2}}{3!}+\frac{b^{4}}{5!}
+\frac{b^{6}}{7!}+\frac{b^{8}}{9!}+\frac{b^{10}}{11!}+\frac{b^{12}}
{13!}+........)$$

It is seen that the first series equal to $$cosh(b)=1+\frac{b^{2}
}{2!}+\frac{b^{4}}{4!}+\frac{b^{6}}{6!}+\frac{b^{8}}{8!}+\frac{b^{10}}
{10!}+......$$

and the second series equal to $$sinh(b)/b=1+\frac{b^{2}}{3!}+\frac{b^{4}}
{5!}+\frac{b^{6}}{7!}+\frac{b^{8}}{9!}+\frac{b^{10}}{11!}+\frac{b^{12}}
{13!}+........$$

For sum of two series we have the following expression  $\cosh
(b)+E_{z}\sinh(b)/b$

The main series for us is the following expansion:

$$f=\frac{1}{5!}+\frac{b^{2}+h^{2}}{7!}+\frac{b^{4}+h^{2}b^{2}+h^{4}}{9!}
+\frac{b^{6}+h^{2}b^{4}+h^{4}b^{2}+h^{6}}{11!}+\frac{b^{8}+h^{2}b^{6}
+h^{4}b^{4}+h^{6}b^{2}+h^{8}}{13!}+
$$
$$
\frac{b^{10}+h^{2}b^{8}+h^{4}b^{6}
+h^{6}b^{4}+h^{8}b^{2}+h^{10}}{15!}+$$

$$\frac{b^{12}+h^{2}b^{10}+h^{4}b^{8}+h^{6}b^{6}+h^{8}b^{4}+h^{10}b^{2}+h^{12}
}{17!}+
$$
$$
\frac{b^{14}+h^{2}b^{12}+h^{4}b^{10}+h^{6}b^{8}+h^{8}b^{6}+h^{10}
b^{4}+h^{12}b^{2}+h^{14}}{19!}+.........$$

 Let us show how the summation of this series can be performed.  We can make the summation of subpart of hole series:

$$\frac{b^{5}}{5!}+\frac{b^{7}}{7!}+\frac{b^{9}}{9!}+\frac{b^{11}}{11!}
+\frac{b^{13}}{13!}+...=\sinh(b)-b-\frac{b^{3}}{3!}$$

$$h^{2}(\frac{b^{7}}{7!}+\frac{b^{9}}{9!}+\frac{b^{11}}{11!}+\frac{b^{13}}
{13!}+....)=h^{2}(\sinh(b)-b-\frac{b^{3}}{3!}-\frac{b^{5}}{5!})$$

$$h^{4}(\frac{b^{9}}{9!}+\frac{b^{11}}{11!}+\frac{b^{13}}{13!}+....)=h^{4}
(\sinh(b)-b-\frac{b^{3}}{3!}-\frac{b^{5}}{5!}-\frac{b^{7}}{7!})$$

Having this series representation we can rewrite expression for f

$$\frac{1}{5!}+\frac{b^{2}+h^{2}}{7!}+\frac{b^{4}+h^{2}b^{2}+h^{4}}{9!}
+\frac{b^{6}+h^{2}b^{4}+h^{4}b^{2}+h^{6}}{11!}+\frac{b^{8}+h^{2}b^{6}
+h^{4}b^{4}+h^{6}b^{2}+h^{8}}{13!}+
$$
$$
\frac{b^{10}+h^{2}b^{8}+h^{4}b^{6}
+h^{6}b^{4}+h^{8}b^{2}+h^{10}}{15!}+$$

$$\frac{b^{12}+h^{2}b^{10}+h^{4}b^{8}+h^{6}b^{6}+h^{8}b^{4}+h^{10}b^{2}+h^{12}
}{17!}+....=$$

$$\frac{1}{b^{5}}(\sinh(b)-b-\frac{b^{3}}{3!})+\frac{1}{b^{7}}h^{2}
(\sinh(b)-b-\frac{b^{3}}{3!}-\frac{b^{5}}{5!})+\frac{1}{b^{9}}h^{4}
(\sinh(b)-b-\frac{b^{3}}{3!}-\frac{b^{5}}{5!}-\frac{b^{7}}{7!})=$$

$$\sinh(b)/b^{5}(1+\frac{h^{2}}{b^{2}}+\frac{h^{4}}{b^{4}}+...)+(-b-\frac
{b^{3}}{3!})/b^{5}(1+\frac{h^{2}}{b^{2}}+\frac{h^{4}}{b^{4}}+...)+$$

$$(-\frac{b^{5}}{5!})\frac{h^{2}}{b^{7}}(1+\frac{h^{2}}{b^{2}}+\frac{h^{4}
}{b^{4}}+.....)+(-\frac{b^{7}}{7!})\frac{h^{4}}{b^{9}}(1+\frac{h^{2}}{b^{2}
}+\frac{h^{4}}{b^{4}}+...)+...$$

It is seen that the series of the type \ $1+\frac{h^{2}}{b^{2}}+\frac{h^{4}
}{b^{4}}+.....$\  \  \ describe geometric series and gives the following result: \

$$1+\frac{h^{2}}{b^{2}}+\frac{h^{4}}{b^{4}}+.....=\frac{1}{1-\frac{h^{2}}
{b^{2}}}=\frac{b^{2}}{b^{2}-h^{2}}$$

If we insert this result we obtain the representation for:

$$\frac{1}{5!}+\frac{b^{2}+h^{2}}{7!}+\frac{b^{4}+h^{2}b^{2}+h^{4}}{9!}
+\frac{b^{6}+h^{2}b^{4}+h^{4}b^{2}+h^{6}}{11!}+\frac{b^{8}+h^{2}b^{6}
+h^{4}b^{4}+h^{6}b^{2}+h^{8}}{13!}+
$$
$$
\frac{b^{10}+h^{2}b^{8}+h^{4}b^{6}
+h^{6}b^{4}+h^{8}b^{2}+h^{10}}{15!}+....$$

$$=\frac{\frac{\sinh(b)}{b^{3}}}{b^{2}-h^{2}}+(-b-\frac{b^{3}}{3!})/b^{5}
\frac{b^{2}}{b^{2}-h^{2}}+(-\frac{b^{5}}{5!})\frac{h^{2}}{b^{7}}\frac{b^{2}
}{b^{2}-h^{2}}+(-\frac{b^{7}}{7!})\frac{h^{4}}{b^{9}}\frac{b^{2}}{b^{2}-h^{2}
}+...=$$

$$\frac{\frac{\sinh(b)}{b^{3}}}{b^{2}-h^{2}}-\frac{\frac{1}{b^{2}}}{b^{2}
-h^{2}}+(-\frac{1}{3!}-\frac{h^{2}}{5!}-\frac{h^{4}}{7!}-....)\frac{1}
{b^{2}-h^{2}}=
$$
$$
\frac{\frac{\sinh(b)}{b^{3}}}{b^{2}-h^{2}}-\frac{\frac{1}{b^{2}
}}{b^{2}-h^{2}}+(-\sinh(h)/h^{3}+\frac{1}{h^{2}})\frac{1}{b^{2}-h^{2}}=
$$
$$
\frac{\frac{\sinh(b)}{b^{3}}-\frac{\sinh(h)}{h^{3}}}{b^{2}-h^{2}}+\frac
{1}{b^{2}h^{2}}$$

Let us consider two series: one is \ $f$ and second is \ $f_{1}$. If we multiply $f$ \  \ by coefficient $a^{5}$ and make following substitusion
\ $b->ba,h->ha$ we obtain the following series

$$a^{5}\frac{1}{5!}+a^{7}\frac{b^{2}+h^{2}}{7!}+a^{9}\frac{b^{4}+h^{2}
b^{2}+h^{4}}{9!}+a^{11}\frac{b^{6}+h^{2}b^{4}+h^{4}b^{2}+h^{6}}{11!}+$$
$$a^{13}\frac{b^{8}+h^{2}b^{6}+h^{4}b^{4}+h^{6}b^{2}+h^{8}}{13!}+
$$
$$
a^{15}
\frac{b^{10}+h^{2}b^{8}+h^{4}b^{6}+h^{6}b^{4}+h^{8}b^{2}+h^{10}}{15!}+$$

$$a^{17}\frac{b^{12}+h^{2}b^{10}+h^{4}b^{8}+h^{6}b^{6}+h^{8}b^{4}+h^{10}
b^{2}+h^{12}}{17!}+.....$$

It is obvious that if we take derivative we can reduce factorial in our
series for example:

$$f_{1}=\frac{\partial (a^5 f)}{\partial a},\quad f_{2}=\frac{\partial f_{1}
}{\partial a},\quad f_{3}=\frac{\partial f_{2}}{\partial a},\quad f_{4}
=\frac{\partial f_{3}}{\partial a}$$

here we must insert $b->a b, h->a h$ and $a$ put to 1 after differentiation.

Taking derivatives we obtain the analytical expression for $f_{i}$ :

$$f_{1}=\frac{\frac{\cosh(b)}{b^{2}}-\frac{\cosh(h)}{h^{2}}}
{b^{2}-h^{2}};\quad f_{2}=\frac{\frac{\sinh(b)}{b}-\frac{\sinh(h)}{h}}
{b^{2}-h^{2}};\quad f_{3}=\frac{\cosh(b)-\cosh(h)}{b^{2}-h^{2}};
$$
$$
f_{4}=\frac{b\sinh(b)-h\sinh(h)}{b^{2}-h^{2}}$$

Series for $ f_{i}$ have the following forms:

$$f_{1}=\frac{1}{4!}+\frac{b^{2}+h^{2}}{6!}+\frac{b^{4}+h^{2}
b^{2}+h^{4}}{8!}+\frac{b^{6}+h^{2}b^{4}+h^{4}b^{2}+h^{6}}{10!}+\frac
{b^{8}+h^{2}b^{6}+h^{4}b^{4}+h^{6}b^{2}+h^{8}}{12!}+
$$
$$
\frac{b^{10}+h^{2}b^{8}+h^{4}b^{6}+h^{6}b^{4}+h^{8}b^{2}+h^{10}}{14!}+$$

$$\frac{b^{12}+h^{2}b^{10}+h^{4}b^{8}+h^{6}b^{6}+h^{8}b^{4}+h^{10}b^{2}+h^{12}
}{16!}+
$$
$$
\frac{b^{14}+h^{2}b^{12}+h^{4}b^{10}+h^{6}b^{8}+h^{8}b^{6}+h^{10}
b^{4}+h^{12}b^{2}+h^{14}}{17!}+....$$

$$\ f_{2}=\frac{1}{3!}+\frac{b^{2}+h^{2}}{5!}+\frac{b^{4}+h^{2}
b^{2}+h^{4}}{7!}+\frac{b^{6}+h^{2}b^{4}+h^{4}b^{2}+h^{6}}{9!}+\frac
{b^{8}+h^{2}b^{6}+h^{4}b^{4}+h^{6}b^{2}+h^{8}}{11!}+
$$
$$
\frac{b^{10}+h^{2}
b^{8}+h^{4}b^{6}+h^{6}b^{4}+h^{8}b^{2}+h^{10}}{13!}+$$

$$\frac{b^{12}+h^{2}b^{10}+h^{4}b^{8}+h^{6}b^{6}+h^{8}b^{4}+h^{10}b^{2}+h^{12}
}{15!}+
$$
$$
\frac{b^{14}+h^{2}b^{12}+h^{4}b^{10}+h^{6}b^{8}+h^{8}b^{6}+h^{10}
b^{4}+h^{12}b^{2}+h^{14}}{17!}+........$$

$$\ f_{3}=\frac{1}{2!}+\frac{b^{2}+h^{2}}{4!}+\frac{b^{4}+h^{2}b^{2}+h^{4}}
{6!}+\frac{b^{6}+h^{2}b^{4}+h^{4}b^{2}+h^{6}}{8!}+\frac{b^{8}+h^{2}b^{6}
+h^{4}b^{4}+h^{6}b^{2}+h^{8}}{10!}+
$$

$$
\frac{b^{10}+h^{2}b^{8}+h^{4}b^{6}
+h^{6}b^{4}+h^{8}b^{2}+h^{10}}{12!}+$$

$$\frac{b^{12}+h^{2}b^{10}+h^{4}b^{8}+h^{6}b^{6}+h^{8}b^{4}+h^{10}b^{2}+h^{12}
}{14!}+
$$
$$
\frac{b^{14}+h^{2}b^{12}+h^{4}b^{10}+h^{6}b^{8}+h^{8}b^{6}+h^{10}
b^{4}+h^{12}b^{2}+h^{14}}{16!}+........$$

$$\ f_{4}=1+\frac{b^{2}+h^{2}}{2!}+\frac{b^{4}+h^{2}b^{2}+h^{4}}
{4!}+\frac{b^{6}+h^{2}b^{4}+h^{4}b^{2}+h^{6}}{6!}+\frac{b^{8}+h^{2}b^{6}
+h^{4}b^{4}+h^{6}b^{2}+h^{8}}{8!}+
$$
$$
\frac{b^{10}+h^{2}b^{8}+h^{4}b^{6}
+h^{6}b^{4}+h^{8}b^{2}+h^{10}}{10!}+$$

$$\frac{b^{12}+h^{2}b^{10}+h^{4}b^{8}+h^{6}b^{6}+h^{8}b^{4}+h^{10}b^{2}+h^{12}
}{12!}+
$$
$$
\frac{b^{14}+h^{2}b^{12}+h^{4}b^{10}+h^{6}b^{8}+h^{8}b^{6}+h^{10}
b^{4}+h^{12}b^{2}+h^{14}}{14!}+........$$

 If we take the series for $f$ and multiply it by $b^{2}$and take
following derivative $\frac{\partial(b^{2}f)}{\partial(b^{2})}$ we obtain the
series for $fEE.$

  $$fEE=\frac{1}{5!}+\frac{2b^{2}+h^{2}}{7!}+\frac{3b^{4}
+2h^{2}b^{2}+h^{4}}{9!}+\frac{4b^{6}+3h^{2}b^{4}+2h^{4}b^{2}+h^{6}}{11!}+
$$
$$
\frac{5b^{8}+4h^{2}b^{6}+3h^{4}b^{4}+2h^{6}b^{2}+h^{8}}{13!}+
$$
$$
\frac
{6b^{10}+5h^{2}b^{8}+4h^{4}b^{6}+3h^{6}b^{4}+2h^{8}b^{2}+h^{10}}{15!}$$

 For series $fhh$  we must make multiplication of $f$ \ on $h^{2}$
and make derivative on $h^{2}:fhh=\frac{\partial(h^{2}f)}{\partial(h^{2})}$

$$fhh=\frac{1}{5!}+\frac{2h^{2}+b^{2}}{7!}+\frac{3h^{4}+2b^{2}
h^{2}+b^{4}}{9!}+\frac{4h^{6}+3b^{2}h^{4}+2b^{4}h^{2}+b^{6}}{11!}+
$$
$$
\frac{5h^{8}+4b^{2}h^{6}+3b^{4}h^{4}+2b^{6}h^{2}+b^{8}}{13!}+
$$
$$
\frac{6h^{10}+5b^{2}h^{8}+4b^{4}h^{6}+3b^{6}h^{4}+2b^{8}h^{2}+b^{10}}{15!}$$

$$f_{2}EE=\frac{1}{3!}+\frac{2b^{2}+h^{2}}{5!}+\frac{3b^{4}
+2h^{2}b^{2}+h^{4}}{7!}+\frac{4b^{6}+3h^{2}b^{4}+2h^{4}b^{2}+h^{6}}{9!}+
$$
$$
\frac{5b^{8}+4h^{2}b^{6}+3h^{4}b^{4}+2h^{6}b^{2}+h^{8}}{11!}+
$$
$$
\frac
{6b^{10}+5h^{2}b^{8}+4h^{4}b^{6}+3h^{6}b^{4}+2h^{8}b^{2}+h^{10}}{13!}$$

 To evaluate $f1EE,f2EE,f1hh$ we multiply $fEE$ by $a^{5}$
and make the following substitution  $b-> ba, h->ha.$ After calculation we fix $a=1.$

It is seen that expression for $f1EE,f2EE,f1hh$ are the following:
 $$f_{1}EE=(\frac{\partial fEE}{\partial a})_{a=1}$$

$$f1EE=\frac{1}{4!}+\frac{2b^{2}+h^{2}}{6!}+\frac{3b^{4}+2h^{2}
b^{2}+h^{4}}{8!}+\frac{4b^{6}+3h^{2}b^{4}+2h^{4}b^{2}+h^{6}}{10!}+
$$
$$
\frac
{5b^{8}+4h^{2}b^{6}+3h^{4}b^{4}+2h^{6}b^{2}+h^{8}}{12!}+
$$
$$
\frac{6b^{10}
+5h^{2}b^{8}+4h^{4}b^{6}+3h^{6}b^{4}+2h^{8}b^{2}+h^{10}}{14!}+........ $$

Repeating all operation we can obtain for \ $f_{2}EE=(\frac
{\partial^{2}fEE}{\partial^{2}a})_{a=1}$

$$f_{2}EE=\frac{1}{3!}+\frac{2b^{2}+h^{2}}{5!}+\frac{3b^{4}+2h^{2}
b^{2}+h^{4}}{7!}+\frac{4b^{6}+3h^{2}b^{4}+2h^{4}b^{2}+h^{6}}{9!}+
$$
$$
\frac{5b^{8}+4h^{2}b^{6}+3h^{4}b^{4}+2h^{6}b^{2}+h^{8}}{11!}+
$$
$$
\frac{6b^{10}
+5h^{2}b^{8}+4h^{4}b^{6}+3h^{6}b^{4}+2h^{8}b^{2}+h^{10}}{13!}$$

and for  $ f1hh=(\frac{\partial fhh}{\partial a})_{a=1}$  and for  $ f2hh=(\frac{\partial f1hh}{\partial a})_{a=1}$

$$f1hh=\frac{1}{4!}+\frac{2h^{2}+b^{2}}{6!}+\frac{3h^{4}+2b^{2}
h^{2}+b^{4}}{8!}+\frac{4h^{6}+3b^{2}h^{4}+2b^{4}h^{2}+b^{6}}{10!}+
$$
$$
\frac{5h^{8}+4b^{2}h^{6}+3b^{4}h^{4}+2b^{6}h^{2}+b^{8}}{12!}+
$$
$$
\frac{6h^{10}+5b^{2}h^{8}+4b^{4}h^{6}+3b^{6}h^{4}+2b^{8}h^{2}+b^{10}}{14!}+.......$$

$$f2h=\frac{1}{3!}+\frac{3h^{2}+b^{2}}{5!}+\frac{5h^{4}+3b^{2}
h^{2}+b^{4}}{7!}+\frac{7h^{6}+5b^{2}h^{4}+3b^{4}h^{2}+b^{6}}{9!}+
$$
$$
\frac{9h^{8}+7b^{2}h^{6}+5b^{4}h^{4}+3b^{6}h^{2}+b^{8}}{11!}+
$$
$$
\frac{11h^{10}+9b^{2}h^{8}+7b^{4}h^{6}+5b^{6}h^{4}+3b^{8}h^{2}+b^{10}}{13!}+.......$$

In analytical form we have for $ f2h= 2 f2hh-f_2 $

 Series for $DEDEf$ \ equal to derivative of $fEE$ on $b^{2}$. It is
seen if we compare series for $fEE$ and $DEDEf$: \  \  \ $DEDEf=\frac{\partial
fEE}{\partial(b^{2})}$

$$DEDEf=\frac{2}{7!}+\frac{6b^{2}+2h^{2}}{9!}+\frac{12b^{4}
+6h^{2}b^{2}+2h^{4}}{11!}+\frac{20b^{6}+12h^{2}b^{4}+6h^{4}b^{2}+2h^{6}}
{13!}+
$$
$$
\frac{30b^{8}+20h^{2}b^{6}+12h^{4}b^{4}+6h^{6}b^{2}+2h^{8}}
{15!}+........$$

We can obtain the following representation for $DEDEf1:$ \  \  \ $DEDEf1=\frac
{\partial fhh}{\partial(b^{2})}$

$$ DEDEf1=\frac{2}{6!}+\frac{6b^{2}+2h^{2}}{8!}+\frac{12b^{4}
+6h^{2}b^{2}+2h^{4}}{10!}+\frac{20b^{6}+12h^{2}b^{4}+6h^{4}b^{2}+2h^{6}}
{12!}+
$$
$$
\frac{30b^{8}+20h^{2}b^{6}+12h^{4}b^{4}+6h^{6}b^{2}+2h^{8}}
{14!}+........$$

Comparing series for $DEDhf2$ and series for $f_{2}$ we can obtain
$DEDhf2=\frac{\partial^{2}f_{2}}{\partial(b^{2})\partial(h^{2})}$

$$DEDhf2=\frac{1}{7!}+\frac{2b^{2}+2h^{2}}{9!}+\frac{3b^{4}
+4h^{2}b^{2}+3h^{4}}{11!}+\frac{4b^{6}+6h^{2}b^{4}+6h^{4}b^{2}+4h^{6}}{13!}+
$$
$$
\frac{5b^{8}+8h^{2}b^{6}+9h^{4}b^{4}+8h^{6}b^{2}+5h^{8}}{15!}+$$

$$\frac{6b^{10}+10h^{2}b^{8}+12h^{4}b^{6}+12h^{6}b^{4}+10h^{8}b^{2}+6h^{10}
}{17!}+.......$$

and for $DEDhf3$ we can obtain the following representation:
$DEDhf3=\frac{\partial^{2}f_{3}}{\partial(b^{2})\partial(h^{2})}$

$$
 DEDhf3=\frac{1}{6!}+\frac{2b^{2}+2h^{2}}{8!}+\frac{3b^{4}
+4h^{2}b^{2}+3h^{4}}{10!}+\frac{4b^{6}+6h^{2}b^{4}+6h^{4}b^{2}+4h^{6}}
{12!}+
$$
$$
\frac{5b^{8}+8h^{2}b^{6}+9h^{4}b^{4}+8h^{6}b^{2}+5h^{8}}{14!}+$$

$$
\frac{6b^{10}+10h^{2}b^{8}+12h^{4}b^{6}+12h^{6}b^{4}+10h^{8}b^{2}+6h^{10}}{16!}+.......$$

Summing our calculation let's write all analytical formula for the function in the matrix elements:

$$fEE=\frac{2\sinh(h)b^{3}+h\left(  b^{2}-h^{2}\right)  \cosh
(b)b+\left(  h^{3}-3b^{2}h\right)  \sinh(b)}{2b^{3}h\left(  b^{2}
-h^{2}\right)  ^{2}};
$$
$$
f1EE=\frac{-2b\cosh(b)+2b\cosh(h)+\left(b^{2}-h^{2}\right)  \sinh(b)}{2b\left(  b^{2}-h^{2}\right)  ^{2}};$$

$$f2EE=\frac{b\left(  b^{2}-h^{2}\right)  \cosh(b)-\left(  b^{2}
+h^{2}\right)  \sinh(b)+2bh\sinh(h)}{2b\left(  b^{2}-h^{2}\right)  ^{2}}$$

$$f2=\frac{h\sinh(b)-b\sinh(h)}{b^{3}h-bh^{3}}; \quad f3=\frac
{\cosh(b)-\cosh(h)}{b^{2}-h^{2}};
$$
$$
f4=\frac{b\sinh(b)-h\sinh
(h)}{b^{2}-h^{2}};\qquad f=\frac{\frac{\sinh(b)}{b^{3}}-\frac
{\sinh(h)}{h^{3}}}{b^{2}-h^{2}}$$

\section{ Conclusion}

 We have calculated the exact representation of the supergroup as well as the
supercoherent state in the Hubbard model. These constructions naturally appear in the
strongly correlated electronic systems in the case of introducing atomic bases in limit
of large on-site Hubbard repulsion. The dynamical supergroup which operates in a
local superbundle determined by any on-site eigenfunction gives us the wave function
in the form of a superspinor. This superspinor describes a local supercoordinates
frame in the curved supermanifold. The operator spinor part acting in tangent and
cotangent bandles of this supermanifold in supergroup can be reformulated in the
terms of the atomic Hubbard operators. Next step lies in the calculation of
effective functional of Hubbard model.

config: arxiv


TITLE: Length of path traced
QUESTION [1 upvotes]: Let $P$ is a variable point moves such that $OP = 5$ and $AP = 13$. If $A(6, 12,4)$, $O$ is origin and then we have to find the length of path traced by $P$.

My attempt

$$(x - 6)^2 + (y - 12)^2 + (z - 4)^2 = 13^2$$
$$\because x^2+y^2+z^2=25$$
$$\therefore12x + 24y + 8z- 196 = 25 - 169$$
$$\Rightarrow12x + 24y + 8z = 52 $$
$$\Rightarrow3x+6y+4z=13~~~~ $$
But now how to proceed.

REPLY [1 votes]: As you state in your attempt, point $P$ lies on the intersection of two spheres $S_A$ and $S_O$:
$$S_A: (x - 6)^2 + (y - 12)^2 + (z - 4)^2 = 13^2,$$
$$S_O:x^2+y^2+z^2=5^2.$$
That intersection of two spheres is a circle. The equation you obtained, $3x+6y+2z=13$, is simply the plane where that circle lies.
So now you can compute the distance from the origin to the plane and use the radius of $S_O$ and Pythagorean Theorem to compute the radius of the circle.
Finally, knowing that the circle center lies on the line between $O$ and $P$ and the distance from $O$, you can also compute its coordinates.


TITLE: Use Stokes Theorem to Prove Cauchy Integral Formula
QUESTION [3 upvotes]: In my undergrad complex analysis, our professor used a quick way to derive the Cauchy integral formula (without Goursat's theorem and all that) with the following setup: Let $U$ be an open set of $\mathbb{C}$. Consider a complex function $f:U\to\mathbb{C}$ as a zero form on $U$, i.e., $f\in \Lambda^0(U).$ Then analyticity of $f$ is equivalent to Cauchy-Riemann equation $\partial_{\bar{z}}f=0$, which is equivalent to the form $f(z)dz\in\Lambda^1(U)$ being closed:
$$d(fdz)=df\wedge dz=\partial_{\bar{z}}fd\overline{z}\wedge dz+\partial_{z}fdz\wedge dz=0.$$
Then one can prove the Cauchy integral formula as follows:

Suppose $\Omega\subset U$ is a relatively compact set with piecewise smooth boundary and $U$ is an
open subset of complex plane, and $f:U\to\mathbb{C}$ is holomorphic. Then for any $a\in \Omega$, we have $$f(a)=\frac{1}{2\pi I}\int_{\partial\Omega}\frac{f(z)}{z-a}dz.$$

Proof: Let $\Omega_\epsilon=\Omega-D_\epsilon(a).$ Then we consider the one form $f(z)/(z-a)\in\Lambda^1(\Omega_\epsilon).$ By Stoke's theorem ($\int_{\Sigma}d\omega=\int_{\partial \Sigma}\omega$) we have
$$
\int_{\partial\Omega_\epsilon}\frac{f(z)}{z-a}dz=\int_{\Omega_\epsilon}d\left(\frac{f(z)}{z-a}dz\right)=0
$$
since $f/(z-a)$ is holomorphic on $\Omega_\epsilon$. Now
$$
\int_{\partial\Omega_\epsilon}\frac{f(z)}{z-a}dz=\int_{\partial\Omega}\frac{f(z)}{z-a}dz-\int_{\partial D_\epsilon(a)}\frac{f(z)}{z-a}dz=0.
$$
Taking $\epsilon\to 0$ the second term goes to $2\pi if(a).$ This concludes the proof.
However, my graduate complex analysis professor told us that one cannot apply Stoke's theorem directly until one verifies that $f'$ is integrable. I never learned Stoke's theorem properly so I wonder:

What do we need to know prior to the proof above about $f,f'$ in order for the proof to be valid? (differentiability of $f'$ or something like that, or even smoothness of $f$?) What should we have proved before this point to make the above proof work?

Is there a way of using differential forms and Stokes theorem to prove the Cauchy integral formula, without going into the traditional Goursat's theorem for triangles and polygons and all that? Are there any recommended references on this topic?

Thanks in advance

REPLY [4 votes]: The usual proof of Stokes's theorem requires the differential form to be at least of class $C^1$ (i.e., differentiable everywhere with continuous first partial derivatives). There are versions that work with weaker hypotheses, but just assuming the existence of derivatives is definitely not enough.
Your proof of Cauchy's formula works fine as long as you assume that $f$ is of class $C^1$ and satisfies the Cauchy-Riemann equations. But the Cauchy-Goursat argument is needed if you only assume that $f'(z)$ exists for all $z\in\Omega$, because then Stokes’s theorem doesn't apply.


TITLE: Solve many linear equations of similar structure
QUESTION [5 upvotes]: Given

G: real and symmetric square matrix
v: real column vector

I need to solve n linear systems of the form
\begin{align} A = \begin{pmatrix} G & v \\\ v^T & 0 \end{pmatrix}\end{align}
\begin{align} Ax = b\end{align}
Where

n is large
G: real and symmetric square matrix, constant for all n systems
v: real column vector, changes for each system (Combination vector where at most 2 values are nonzero)
b: is zero column vector with exception of the last element

I want to know if there is a fast method to solve these many systems via exploiting this structure and suspect that there is a way to do this via eigenvalue decomposition of sums of hermitian matrices. However, I am unsure of how to combine the results.
I currently solve n systems via a hermitian solver which doesn't scale well.
For convenience, I provide the following equivalent python code
import numpy as np
import scipy.linalg as sp_linalg

np.set_printoptions(threshold=np.inf, linewidth=100000, precision=3, suppress=True)

N = 10 # Size of A-1

G = np.random.random(size=(N, N))
G += G.T
G *= 2

v = np.zeros((N, 1))
v[np.random.choice(N, 2)] = 1.0

A = np.block([[G, v], [v.T, 0.0]])
A_G = np.block([[G, np.zeros((N, 1))], [np.zeros((1, N+1))]])
A_v = np.block([[np.zeros((N, N)), v], [v.T, 0.0]])

b = np.concatenate((np.zeros((N, 1)), np.random.random((1,1))))

###

x = sp_linalg.solve(A, b, assume_a='sym') # General solution to compare against

###

# for eigenvalue decomposition
# lambda_G, Q_G = np.linalg.eigh(A_G)
# lambda_v, Q_v = np.linalg.eigh(A_v)

Thanks!
Solution:
I've taken the solution mentioned by eepperly16 and further generalized the problem. Now

G: NxN random symetric matrix constant for all n systems
v: NxM matrix of random variables

The big idea is since v is now a matrix, an inverse of $-v^\top G^{-1} v$ rather than doing a simple divide. These changes include...

$x_2 = -y_2 / (v^\top G^{-1}v)$ Becomes $x_2 = (v^\top G^{-1}v)^{-1} -y_2$
$x_1 = y_1 - x_2G^{-1}v$ Becomes $x_1 = y_1 - G^{-1}vx_2$

Since the result of this is always symmetric, that can be exploited with similar factorization. Note, however, that now the time complexity of the second stage expands proportionately to $O(M^2)$.
And finally the code with benchmark
import numpy as np
import scipy.linalg as sp_linalg
import timeit

np.random.seed(40)
np.set_printoptions(threshold=8, linewidth=1000, precision=3, suppress=True)

N = 100 # Size of square matrix G
M = 10 # Number of columns in v

# Setup problem and randomize
def setup_and_randomize():

    # Create random symmetric matrix G on range (-1.0, 1.0)
    G = 2.0 * np.random.random(size=(N, N)) - 1.0
    G += G.T
    G *= 0.5

    # Create random rectangular matrix v on range (-1.0, 1.0)
    v = 2.0 * np.random.random(size=(N, M)) - 1.0

    A = np.block([[G, v], [v.T, np.zeros((M, M))]])

    b_1 = np.zeros((N, 1))
    b_2 = np.ones((M, 1))
    b = np.concatenate((b_1, b_2), axis=0)

    return A, G, v, b, b_1, b_2

# General solution to compare against
def naive_method(A, b):
    return sp_linalg.solve(A, b, assume_a='sym')

# Generalised solution crated from eepperly16's solution Part 1
def answer_method_precompute(G, b_1, b_2):
    P, L, U = sp_linalg.lu(G, overwrite_a=True, check_finite=False)
    L_inv = sp_linalg.solve_triangular(L, np.eye(N), lower=True, trans='N', overwrite_b=True)
    U_inv = sp_linalg.solve_triangular(U, np.eye(N), lower=False, trans='N', overwrite_b=True)
    G_inv = U_inv @ L_inv @ P.T

    y_1 = G_inv @ b_1
    y_2 = b_2 - v.T @ y_1
    return y_1, y_2, G_inv

# Generalised solution crated from eepperly16's solution Part 2
def answer_method_main(v, y_1, y_2, G_inv):
    G_inv_dot_v = G_inv @ v

    # IF M >= 1 -----------------------------------------------------
    B = v.T @ G_inv_dot_v
    P, L, U = sp_linalg.lu(B, overwrite_a=True, check_finite=False)
    L_inv = sp_linalg.solve_triangular(L, np.eye(M), lower=True, trans='N', overwrite_b=True)
    U_inv = sp_linalg.solve_triangular(U, np.eye(M), lower=False, trans='N', overwrite_b=True)
    B_inv = U_inv @ L_inv @ P.T

    x_2 = B_inv @ -y_2
    x_1 = y_1 - G_inv_dot_v @ x_2

    # IF M == 1 -----------------------------------------------------
    # x_2 = -y_2 / (v.T @ G_inv_dot_v)
    # x_1 = y_1 - (x_2 * G_inv_dot_v)

    return np.concatenate((x_1, x_2), axis=0)

if __name__ == "__main__":

    # Verify Same Solution ------------------------------------------
    A, G, v, b, b_1, b_2 = setup_and_randomize()

    x_naive = naive_method(A, b)

    y_1, y_2, G_inv = answer_method_precompute(G, b_1, b_2)
    x_answer = answer_method_main(v, y_1, y_2, G_inv)

    print('Naive Solution:\t', x_naive.T)
    print('Final Solution:\t', x_answer.T)

    # Benchmark Performance ----------------------------------------------
    n_tests = 1000

    A, G, v, b, b_1, b_2 = setup_and_randomize()
    print('\nTimeit on naive_method', timeit.timeit('naive_method(A, b)', globals=globals(), number=n_tests))
    print('Timeit on answer_precompute', timeit.timeit('answer_method_precompute(G, b_1, b_2)', globals=globals(), number=n_tests))
    print('Timeit on answer_main', timeit.timeit('answer_method_main(v, y_1, y_2, G_inv)', globals=globals(), number=n_tests))

Which yields the following on my machine for 1000 iterations of N=100, M=10
Naive Solution:  [[ 0.33  -1.518  0.434 ... -0.394 -0.569  0.824]]
Final Solution:  [[ 0.33  -1.518  0.434 ... -0.394 -0.569  0.824]]

Timeit on naive_method 0.39002
Timeit on answer_precompute 0.46521499999999993
Timeit on answer_main 0.14545809999999992

Final Edit:
I understand that with scipy, there are better ways to compute the inverse that better tie into one of many BLAS style libraries. Below are 2 ways to compute the inverse of G that work better than the initial solution. Also, enabling more flags on the naive solver also makes that timing calculation fairer.
G_inv = sp_linalg.lu_solve(
            sp_linalg.lu_factor(G, overwrite_a=True, check_finite=False),
            np.eye(N), overwrite_b=True, check_finite=False)

L, D, perm = sp_linalg.ldl(G, overwrite_a=True, hermitian=True, check_finite=False)
    L_inv = sp_linalg.solve_triangular(L[perm, :], np.eye(N), lower=True, trans='N', overwrite_b=True, check_finite=False)[:, perm]
    G_inv = (L_inv.T / D.diagonal()) @ L_inv

REPLY [4 votes]: Notice that $A$ can be factored as
$$
A = \begin{bmatrix} G & v \\ v^\top & 0 \end{bmatrix} = \begin{bmatrix} G &0 \\ v^\top & 1 \end{bmatrix}\begin{bmatrix} I & G^{-1}v \\ 0 & -v^\top G^{-1} v\end{bmatrix}.
$$
Using this we can devise a scheme to solve $A$ for lots of different $G$'s. First, factorize $G$ using an $LU$ factorization (or a Cholesky factorization or $LDL^\top$ factorization or whatever). This requires time proportional to the cube of the size of $G$ ($O(n^3)$ operations), but once you have such a factorization you can compute $G^{-1}u$ in time proportional to the square of the size of $G$ ($O(n^2)$ operations). Now suppose you want to solve $Ax = b$. Write $x = \begin{bmatrix} x_1 \\ x_2 \end{bmatrix}$, where $x_2$ is the last entry of $x$. Write
$$
y = \begin{bmatrix} y_1 \\ y_2 \end{bmatrix} = \begin{bmatrix} I & G^{-1}v \\ 0 & -v^\top G^{-1} v\end{bmatrix}x.
$$
Then we have that
$$
Ax = \begin{bmatrix} G &0 \\ v^\top & 1 \end{bmatrix}\begin{bmatrix} y_1 \\ y_2 \end{bmatrix} = \begin{bmatrix} b_1 \\ b_2 \end{bmatrix}.
$$
Then we have that $Gy_1 = b_1$. Use your precomputed $LU$ factorization to solve $Gy_1 = b_1$ for $y_1$. Then we have that $v^\top y_1 + y_2 = b_2$ so $y_2 = b_2 - v^\top y_1$.
Next we compute $x$ from $y$. Write
$$
\begin{bmatrix} I & G^{-1}v \\ 0 & -v^\top G^{-1} v\end{bmatrix}\begin{bmatrix}x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} y_1 \\ y_2 \end{bmatrix}.
$$
Use your precomputed $LU$ factorization to compute $G^{-1}v$. Then we have that $(-v^\top G^{-1} v)x_2 = y_2$ so $x_2 = -y_2 / (v^\top G^{-1}v)$. We also have that $x_1 + x_2G^{-1}v = y_1$ so $x_1 = y_1 - x_2G^{-1}v$. We've now solved $Ax = b$ by using only two linear solves with $G$, which are much faster when we've precomputed the factorization of $G$.


TITLE: How to find inverse of this equation: $g(x)=e^{3x-1}+\sec{{\pi}x}$
QUESTION [0 upvotes]: Question:
How to find inverse of this equation: $g(x)=e^{3x-1}+\sec{{\pi}x}$
My thoughts on this
I know that If i was to normally find the inverse you would replace the Y with the X's and solve for y but I dont think that is how it is supposed to work in this problem?

REPLY [1 votes]: As one can quickly see from a graph, the function is not one-to-one, so it doesn't have an inverse function...


TITLE: Frequency response $H(e^{-j\omega})$ and calculating its magnitude
QUESTION [1 upvotes]: Determine the frequency response $H(e^{j\omega})$ of a system characterized by $h(n) = (0.9)^n u(n)$
using the definition
$$H(e^{j\omega})=\sum_{-\infty}^\infty h(n) e^{-j\omega n}$$
$H(e^{j\omega})=\sum_0^\infty (0.9)^n e^{-j\omega n}$  = $\sum_0^\infty (0.9 e^{-j\omega})^n$ = $$\frac{1}{1-0.9e^{-j\omega}}$$
I understand the above but my question comes when the book tries to get the magnitude of $H(e^{j\omega})$.
$$|H(e^{j\omega})| = \sqrt \frac{1}{(1-0.9 \cos\omega)^2+(0.9 \sin\omega)^2}$$
I know that the use of Euler's formula and the complex conjugate is involved. However, I need to see the math.

REPLY [1 votes]: Using that $\,e^{j \alpha}=\cos \alpha + j \sin \alpha\,$and $\,|z| = \sqrt{\operatorname{Re}^2(z)+\operatorname{Im}^2(z)}\,$:
$$
\begin{align}
\left|\frac{1}{1-0.9e^{-j\omega}}\right| &= \frac{1}{\;|1-0.9e^{-j\omega}|\;} \\
 &= \frac{1}{\;|1 - 0.9 \big( \cos(-\omega)+j \sin(-\omega)\big)|\;} \\
 &= \frac{1}{\;|\big(1 - 0.9 \cos(\omega)\big) - 0.9\,j \sin(\omega))|\;} \\
 &= \frac{1}{\sqrt{\big(1 - 0.9 \cos(\omega)\big)^2+\big(0.9 \sin(\omega)\big)^2}}
\end{align}
$$


TITLE: Notation of inverse trigonometric functions and exponentiation
QUESTION [1 upvotes]: Possible Duplicate:
$\arcsin$ written as $\sin^{-1}(x)$

I have worked a bit on trigonometry today, and something strikes me as inconsistent.
In the book, the notation for the inverse sine function is $\sin^{-1}$, but the same notation is also used in $\sin^2$ meaning $(\sin x)^2$.
Are there any alternate notations which avoid this ambiguity?

Some examples:
$\sin30^\circ = 0.5$
$\sin^2 30^\circ = (\sin 30^\circ)^2 = 0.25$
However, inverse sine does not work that way:
$\sin^{-1} 30^\circ \ne (\sin 30^\circ)^{-1}$
$\sin^{-1} 30^\circ = $ Error or complex number?
$(\sin 30^\circ)^{-1} = 2$
The potential confusion only gets worse if you use radians, as they are in the range [-1, 1] for [-57°, 57°].:
$\sin^{-1} 0.524 = 0.551$
$(\sin 0.524)^{-1} = 2$
And what if you want both at the same time? You are forced to use parentheses, thus breaking any consistency: $(\sin^{-1} x)^2$

REPLY [3 votes]: Carl Gauss also objected to this particular notational inconsistency, so you're in good company.  He proposed that $\sin^2(x)$ ought to mean $\sin(\sin(x))$, whereas $(\sin x)^2$ should be written in that way.
But the notation in the book you mention is standard.  I don't think there's any standard alternative.


TITLE: negative one times positive one is negative one
QUESTION [0 upvotes]: This is a question I was surprised that no one on this forum has asked(as far as my search went).
I gave a proof that negative times negative = positive.
But it relied on the assumption negative times positive = negative.
SO now I am asking myself how does one show this fact:

negative times positive = negative

I am guessing that maybe the axioms of the real numbers maybe able to help.
I am wondering if anyone has pondered this question.
A lot of people pondered the -1 x -1, but not this one.
Regards,
P

REPLY [2 votes]: $ (-1)1 = -1 $ because 1 is the multiplicative identity.
$$ x * -1 = x * ( 0 - 1) = (0x) - (1x) = \\ 0 - x = -x $$ by the distributive law.
Then you can generalize to $ x * -y $
$ x * -y = x * ( y * -1 ) = (x * y) * -1 = -(x * y) $

REPLY [1 votes]: For the equation $1\times-1$, we can use the following fact:
$$1\times A=A\quad\text{ for every }A$$
Substituting in $A=-1$, we get $1\times-1=-1$.
For the general case, consider $a\times-b$, where $a$ and $b$ are positive. We have:
\begin{align}
0&=a\times 0\\
&=a\times(b+-b)\\
&=(a\times b)+(a\times-b)
\end{align}
We know $a\times b$ is positive. Thus, $a\times-b$ must be negative, because otherwise, if it were positive, we would have $0={\rm positive}+\rm positive$, which is impossible.


TITLE: Probabilities to end up in certain state
QUESTION [0 upvotes]: I have 6 steps with a certain probability. Let say:

step > probability
0 > 0.5
1 > 0.1
2 > 0.1
3 > 0.1
4 > 0.1
5 > 0.05
6 > 0.05

This table presents a number of moves that I have to do in a next step, this means, if I get 3 I am moving 3 positions forward, if I get 5, I have to move 5 positions forward and with 0 I will quit the game. My maximum number of steps is 15 and I quit the game in two cases: or if I get 0, or if I reach the value 15.
Here is an example 1:

I get 4 > going to position 4
I get 2 > going to position 6
I get 6 > going to position 12
I get 1 > going to position 13
I get 2 or more > going to position 15
The game is finished!

Example 2:

I get 2 > going to position 2
I get 2 > going to position 4
I get 1 > going to position 5
I get 0 > staying at position 5 and the game is finished.

Can someone explain me, how to calculate the expected probability to be in all states between 0 and 15? I would like to have a table like:
$$\begin{array}{c|c|c|c|c|c|}
\text{state} & 0& 1 & 2 & 3 & 4& 5&... & 14 & 15 \\ \hline
\text{prob} &0.1& 0.01 &0.011&0.03&0.2&0.25&...& 0.0001 &0.00001& \\ \hline
\end{array}$$
Thanks for help!

REPLY [0 votes]: You'll construct a markov matrix that looks like this (assuming that if you cross 15, you win):

This markov matrix has two absorbing states, 0 and 15. Now, use the method detailed in section 7.3 here: https://ryu577.github.io/jekyll/update/2018/10/08/competitive_coin_tossing.html to get the probabilities of the state ending up in 0 or 15 (losing/winning; the two absorbing states).


TITLE: Counting positive integral solutions to an equation
QUESTION [1 upvotes]: How many positive integral solutions to $y_1+y_2+y_3+y_4+y_5+y_6=13$ are there?
I started with $y_1=8$ and distributed the rest to $y_2$ to $y_6$, but I feel like I will miss some of them by doing so.

REPLY [2 votes]: Consider $$y_1+y_2+y_3+y_4+y_5+y_6=13$$ where $1\leq y_i\leq8$ for each $i\in\{1,2,3,4,5,6\}$. Make the substitution $z_i=y_i-1$ for each $i$. This gives us $$z_1+z_2+z_3+z_4+z_5+z_6=7$$ where $0\leq z_i\leq 7$ for each $i$. The number of non-negative integral solutions to this equation corresponds to the number of positive integral solutions to the original equation. Thus there are $${7+6-1\choose 7}={12\choose 7}=792$$ positive integral solutions to $y_1+y_2+y_3+y_4+y_5+y_6=13$.