TITLE: Vector spaces isomorphic to its dual
QUESTION [0 upvotes]: Let $X$ be a finite-dimensional normed vector space. Is there an isometry $L:X\to X^*$? I know there is an isomorphism. I guess it's not true for all spaces, but I wonder on which conditions it is true. More precisely, I'm looking for conditions on a Banach $E$ to have: each finite-dimensional subspace of $E$ satisfies the property.

REPLY [2 votes]: Observe that $(\ell^n_p)^* =\ell^n_q$, where $1/p+1/q=1$. Therefore it isn't true that the dual is always isometric. It's true for $p=2$ and generally if the space is Hilbert. If $p=1$ or $p=\infty$ then it's also true but only if $n=2$ (balls are rectangles). It's sufficient that the balls in both spaces are similar (actually, more generally, affine equivalent, as Ilya Bogdanov observed) (recall that the unit ball determine the norm). It is related to the notion od dual polyhedra: if the ball is a polyhedron (of some dimension) then the ball in the dual space is a dual polyhedron.


text:
\begin{document}

\title{Negative dimensional approach to evaluating real integrals}
\author{Alfredo Takashi Suzuki}
\affiliation{Instituto de F\'{\i}sica Te\'orica,\\
Universidade Estadual Paulista, \\
Rua Pamplona, 145 \\
01405-900 Sao Paulo, SP - Brazil}
\date{\today}

\begin{abstract}
In solving the differential equation for a non damped harmonic oscillator
one meets, after subjecting the equation to a Fourier transformation, an
integration in the complex $\omega$ plane. In most cases such an integral is
evaluated by calculating residues together with some physical input such as
the principle of causality to define which pole residues are relevant to the
physical problem. For this kind of application, Cauchy's theorem or residue
theorem can be applied to evaluate certain real integrals. Here we present
an alternative approach based on the concept of negative dimensional
integration to treat such integrals and give an specific example on how this
is accomplished.

\vspace{5cm}
\end{abstract}

\pacs{02.10.Ud; 02.30.-f; 02.30.Cj}
\maketitle

Mathematics Subject Classification (2000): 28E05 Nonstandard Measure Theory

\section{Introduction}

In a textbook on complex variables we may find real definite integrals of
the type
\begin{equation}  \label{I0}
I_0=\int_{-\infty}^{+\infty} \! \frac{P(x)}{Q(x)}\,dx,
\end{equation}
where $P(x)$ and $Q(x)$ are polynomials in $x$ satisfying the following two
requisites: (a) the degree of $Q(x)$ must exceed the degree of $P(x)$ by at
least 2, to ensure convergence and vanishing of the integral over the
semicircle $C_{\mathrm{R}}$ in the limit of $\mathrm{R}$ extended to
infinity; and (b) $Q(x)$ is a polynomial with no real zeros. This second
requirement can be somehow relaxed by special modifications in order to make
the integral become defined in such a case. The standard result for such
integrals is
\begin{equation}  \label{I0_a}
I_0=\int_{-\infty}^{+\infty} \! \frac{P(x)}{Q(x)}\,dx = 2\pi\,i\sum_{+}
\mathrm{Res}
\end{equation}
where $\sum_{+}\mathrm{Res}$ is the sum of the residues of the integrand in
the complex upper half-plane.

This can be applied straightforwardly to the integral
\begin{equation}
I_{\mathrm{cos}} = \int_{-\infty}^{+\infty} \frac {\cos x}{x^2+a^2}\,dx =
\frac{\pi}{a}\,\mathrm{e}^{-a}.
\end{equation}
as the real part of the more general integral
\begin{equation}  \label{I_exp}
I_{\mathrm{exp}} = \int_{-\infty}^{+\infty} \frac {\mathrm{e^{ix}}}{x^2+a^2}
\,dx = \frac{\pi}{a}\,\mathrm{e}^{-a}.
\end{equation}

Now, it is a well-known result that all the positive dimensional polynomial
integrals of the form below vanishes:
\begin{equation}
I_{\mathrm{pol}} = \int_{-\infty}^{+\infty}(x^2)^n\,dx = 0, \qquad n\geq 0.
\end{equation}

However, when we allow an analytic continuation to negative dimensions, the
above integral has a non vanishing and nontrivial result \cite{ricotta}:
\begin{equation}
I_{\mathrm{pol}} = \int_{-\infty}^{+\infty}(x^2)^n\,d^\star x = (-1)^n\, n!\,
\sqrt{\pi}\,\delta_{n+\frac{1}{2},\,0}.
\end{equation}
where we have written the measure $d^\star x$ to remind ourselves that we
are in negative dimensional space.

In the next section we apply this concept of negative dimensional space to
evaluate (\ref{I_exp}).

\section{Negative Dimension Integration Method (NDIM)}

Since the integrand in (\ref{I_exp}) has an exponential function which is
not in the form a polynomial, first of all, we need to convert it in a
polynomial form to apply the negative dimensional technique. We can do this
by using the series expansion for the exponential function, namely,
\begin{equation}  \label{I_G}
I_{\mathrm{exp}} = \sum_{m=0}^{+\infty} \frac{i^m}{m!}\int_{-\infty}^{+
\infty}\,dx\,\frac{x^m}{x^2+a^2}
\end{equation}

Let us now define the Gaussian generating functional of the negative
dimensional integration \cite{labtesting} corresponding to the integral in (
\ref{I_G}), namely,
\begin{equation}  \label{generating}
\mathcal{G}=\int_{-\infty}^{+\infty}\!dx\;\mathrm{{e}^{-\alpha x}\,{e}
^{-\beta(x^2+a^2)}}
\end{equation}
which following the NDIM procedure we write as
\begin{equation}  \label{sum1}
\mathcal{G} = \sum_{r,s=0}^{+\infty} (-1)^{r+s}\frac{\alpha^r \beta^s}{r!\,s!
}\,\int_{-\infty}^{+\infty}\,d^\star x\,x^{r}(x^2+a^2)^s
\end{equation}

Observe now that for $s=-1$ the integral
\begin{equation}
I_0^{\star}(s) \equiv \int_{-\infty}^{+\infty}\,d^\star x\,x^{r}(x^2+a^2)^s
\end{equation}
is exactly the integral that appears in (\ref{I_G}).

This means that after $I_0^{\star}(s)$ is performed with positive values of $
s$ and negative dimension, we need to analytic continue it to negative
values of $s$ and positive dimension.

On the other hand, the Gaussian generating functional integral (\ref
{generating}) can be evaluated, yielding
\begin{equation}
\mathcal{G} = \sqrt {\frac{\pi}{\beta}}\,\mathrm{e}^{-\beta a^2+\frac {
\alpha^2}{4\beta}}
\end{equation}

Expanding the exponential function $\mathrm{e}^{-\beta a^2+\frac {\alpha^2}{
4\beta}}$ in power series, we get
\begin{equation}  \label{sum2}
\mathcal{G} = \pi^{1/2} \sum_{k,l=0}^{+\infty}\frac{(-1)^k(a^2)^k}{
4^l\;k!\,l!}\,\alpha^{2l}\, \beta^{k-l-1/2}
\end{equation}

Now, comparing (\ref{sum1}) and (\ref{sum2}) term by term, we have to have
the following constraints satisfied:
\begin{eqnarray}
2l & = & r  \nonumber \\
k-l-\frac {1}{2} & = & s,
\end{eqnarray}

Expressing the different factorials that appear in terms of gamma functions,
i.e., $p! = \Gamma(p+1)$ and using the Pochhammer's symbols for ratios of
gamma functions, i.e.
\begin{equation}
(p)_q \equiv \frac{\Gamma(p+q)}{\Gamma(p)}
\end{equation}
we get
\begin{equation}
I_0^{\star}(s) = \frac{(-\pi)^{1/2}(a)^{r+2s+1}}{2^r\;(r+1)_{-r/2}
\,(s+1)_{r/2+1/2}}
\end{equation}

We need now to analytic continue this result to allow for negative values of
$s$ and positive dimension. Note here that only the exponent $s$ needs to be
analytic continued to allow for negative values, since $r$ is always
positive \cite{tensorial}. This is accomplished by using the property of the
Pochhammer's symbols, namely,
\begin{equation}
(p)_q = \frac{(-1)^{-q}}{(1-p)_{-q}}
\end{equation}
for the $s$ bearing Pochhammer, so that
\begin{equation}
I_0^{\star, \mathrm{AC}}(s) = \frac{i^r}{2^r}\;\pi^{1/2}\;(a)^{r+2s+1}\frac{
(-s)_{-r/2-1/2}}{(r+1)_{-r/2}}
\end{equation}

This, for the particular value of $s=-1$ which is of interest for us, yields
\begin{equation}
I_0^{\star, \mathrm{AC}}(-1) = \frac{\pi}{a}(-a)^r
\end{equation}
which introduced into (\ref{I_G}) gives
\begin{equation}
I_0 = \sum_{m=0}^{+\infty} \frac{\pi}{a}\frac{(-a)^m}{m!} = \frac {\pi}{a}\;
\mathrm{e}^{-a}
\end{equation}
which agrees with the one obtained through the use of complex plane residue
theorem.

\section{Conclusions}

In this work we have considered a class of real integrals which usually is
performed with the help of Cauchy's residue theorem. We have presented here
an alternative method whereby this type of real integrals can also be
performed without difficulty.

config: arxiv


TITLE: Find if a form is symmetric or skew-symmetric
QUESTION [0 upvotes]: Consider the set of all n × n matrices in R. Given the defined function
Φ: $M$(n,n)× $M$(n,n) → R , which
Φ(A,B) = $tr$(A$^T$JB) ,
where J is a skew-symmetric n × n matrix , define if Φ is a bilinear form, and in case it is define if it is symmetric or skew-symmetric.
So far, I've already proved that Φ is a bilinear form. I know that a form is symmetric if:
Φ(A,B) = Φ(B,A)
and skew-symmetric if
Φ(A,B) = -Φ(B,A)
I tried to prove it by doing:
Φ(A,B) = $tr$(A$^T$JB)
Φ(B,A) = $tr$(B$^T$JA)
but I'm not sure how to proceed in order to prove if Φ is symmetric or skew-symmetric.

REPLY [1 votes]: Hint:
$$
\DeclareMathOperator{\tr}{tr}
\tr(A^TJB) = \tr([A^TJB]^T) =
\tr(B^TJ^TA) = -\tr(B^TJA)
$$


TITLE: Possibility of a number line that has variable density.
QUESTION [1 upvotes]: In my real analysis class, I have been informed that if you have a continuous line say $[0,1]$, and you do a mapping say $A\to B$ such that any element in $A$ is equal to $B^2$, where $A$ is every element in $[0,1]$  you would ultimately get a line that is denser toward the $0$ and less dense toward the $1$.
I have two question regarding this result,

Can we note that every element in $A$, say variable $a$  has both its square and square root already existing in the line $[0, 1]$, therefore the mapping is only an exchange of the position of points,

for example we know that if we have point $0.5$
both $0.5^2$ and $\sqrt{0.5}$ exist in $[0,1]$  therefore if we do the mapping, $0.5 \mapsto 0.25$ doesn't the other point $\sqrt{0.5}$ replaces $0.5$'s original position ?

How is it possible that you can have a denser part in a continuous line, does this not ultimately suggest that there will be a gap around the $1$'s side?

Any comment would be much appreciated

REPLY [1 votes]: All that is meant by the statement that the line is "more dense" towards 0 is this: say that you pick some small number $\epsilon>0$. Then the interval $[0,\sqrt{\epsilon}]$ all gets mapped (by $x\mapsto x^2$) in to the interval $[0,\epsilon]$; so, the preimage of $[0,\epsilon]$ under this map has total length $\sqrt{\epsilon}$.
On the other side, the numbers $[\sqrt{1-\epsilon},1]$ are mapped by $x\mapsto x^2$ to the interval $[1-\epsilon,1]$; so, in this case, the preimage of $[1-\epsilon,1]$ has length $1-\sqrt{1-\epsilon}$.
The interesting thing here: even though the intervals $[0,\epsilon]$ and $[1-\epsilon,1]$ have the same length... their preimages do not.  In fact, for every $\epsilon\in(0,1)$, it turns out that $\sqrt{\epsilon}$ (the "size" of the preimage of $[0,\epsilon]$) is larger than $1-\sqrt{1-\epsilon}$ (the "size" of the preimage of $[1-\epsilon,1]$).
So, all that they really meant was that different intervals of $[0,1]$ get more or less "spread out" by the mapping $x\mapsto x^2$.


TITLE: Why is Skolem normal form equisatisfiable while the second order form equivalent?
QUESTION [4 upvotes]: I asked in another
question when is it appropriate to de-Skolemize a statement. The answer, I'm not sure I'm satisfied with yet, relies on a second order logical equivelance, but Skolem normal form isn't logically equivalent. I've read first order Skolemization can be defined through a second order logic equivelance:
$\forall$x$\exists$y $\phi$(x,y) $\iff$ $\exists$F$\forall$x $\phi$(x,F(x))
From this we can omit the quantifiers, and treat all variables as implicitly universally quantified:
$\phi$(x,f(x))
I also understand that Skolem normal form is equisatisfiable:
$\forall$x$\exists$y $\phi$(x,y) $\vDash$ $\phi$(x,f(x))
[I apparently incorrectly understood it, see accepted answer]
But apparently this is not equivalent.
What does this mean? Can we freely move between Skolem normal form and the second order sentence even though they aren't equivalent? I only see equisatisfiability with Skolemization used to prove satisfiability or unsatisfiability, and I haven't seen it used to prove logical consequence from first order formulas except through unsatisfiability with searching for a contradiction.
Why is the second order form of a first order sentence that moves existential quantifiers out of scope of universal quantifiers logically equivalent, while Skolem normal form only equisatisfiable?

REPLY [8 votes]: It's perhaps best to think about this first in the context of a particular structure and a particular formula, where you can easily see what's going on.  Consider the structure of the natural numbers with the usual ordering relation $<$.  The sentence $\forall x\exists y\,(x<y)$ is true in this structure, because for every natural number $x$ there is a larger one $y$, for example $x+1$ (or $x+17$ or $(2x)!$ or lots of other choices).  Equivalently, the second-order sentence $\exists F\forall x\,(x<F(x))$ is true in the same structure, witnessed by functions like $F(x)=x+1$ (or $F(x)=x+17$ or $F(x)=(2x)!$, or lots of other functions $F$).
On the other hand, we cannot say (as you did) that $\forall x\exists y\,(x<y)\models\forall x\,(x<F(x))$.  To see this, notice first that this "logical consequence" relationship concerns structures for a language that has not only the relation symbol $<$ but also the function symbol $F$. Consider the structure of natural numbers with $<$ as before and some interpretation of the function symbol $F$. Then $\forall x\exists y\,(x<y)$ is true (regardless of the interpretation of $F$ because it doesn't mention $F$), but whether $\forall x\,(x<F(x))$ is true depends on how we interpret $F$.  It's true if we take $F(x)=x+1$ (or $F(x)=x+17$ or $F(x)=(2x)!$, or lots of other functions $F$), but it's false if we take $F(x)=0$ (or $F(x)=17$ or $F(x)=\lfloor x/2\rfloor$, or lots of other functions).  But $\models$ means that every structure satisfying what's on the left of $\models$ also satisfies what's on the right. So in our case, it would require that $\forall x\,(x<F(x))$ is true regardless of how we interpret $F$ as long as $\forall x\exists y\,(x<y)$ is true.  Since that doesn't hold when $F$ is "badly" interpreted ($0$ or $17$ or $\lfloor x/2\rfloor$), we don't have $\forall x\exists y\,(x<y)\models\forall x\,(x<F(x))$.  The most we can say is that, having satisfied $\forall x\exists y\,(x<y)$, we can, by suitably interpreting $F$, also satisfy $\forall x\,(x<F(x))$ --- and vice versa. That's what "equisatisfiable" means.


TITLE: Conditional probability that a rotated inner product is zero
QUESTION [9 upvotes]: Consider a two random vectors $v=v_1,\dots, v_n$ and $w=w_1,\dots, w_n$, each of $n$ elements, each of which is independently $\pm1$ with prob $1/2$ .  Let $n$ be even and let $X$ be the inner product of $v$ and $w$. We know:
$$P(X = 0) =  {n \choose n/2} \frac{1}{2^n}.$$
Now let $Y$ be the inner product of $v$ and the vector $z=w_2,\dots,w_n, w_1$.

What is $P(Y=0|X=0)$?

REPLY [7 votes]: First of all, for the inner product to be $0$, $n$ must be even.
We get that $u\cdot v=0$ exactly when the the element-wise product of the vectors, $v_kw_k$, has as many $+1$s as $-1$s. After a shift, the element-wise product changes every where that $w_kw_{k+1}=-1$ (with wrapping). Thus, for each $-1$ of $w_kw_{k+1}$ in the $+1$s of $v_kw_k$, there must be a $-1$ of $w_kw_{k+1}$ in the $-1$s of $v_kw_k$.
The number of transitions that leave the dot product unchanged is
$$
\sum_{k=0}^{n/2}\binom{n/2}{k}\binom{n/2}{k}=\binom{n}{n/2}\tag{1}
$$
The total number of transitions is
$$
\sum_{k=0}^{n/2}\binom{n}{2k}=2^{n-1}\tag{2}
$$
Note that $(2)$ only holds for $n\gt0$. For $n=0$, the sum is $1$, not $\frac12$.
Thus, for $n\gt0$, the probability that there is no change in the dot product is
$$
\binom{n}{n/2}2^{1-n}\tag{3}
$$
Which is exactly twice the probability of having a dot product of $0$ in the first place. For $n=0$, the probability is $1$.


TITLE: Let $a,b,c \in [0,1[$ such that $a+b+c=2$. Prove that $a^3+b^3+c^3+2abc\leq 2$
QUESTION [2 upvotes]: Let $a,b,c \in [0,1[$ such that $a+b+c=2$. Prove that
$$a^3+b^3+c^3+2abc\leq 2$$

My attempt:
put $a=x^{\frac{2}{3}}$,$b=y^{\frac{2}{3}}$,$c=z^{\frac{2}{3}}$
\begin{align*}
&x^{2}+y^{2}+z^{2}+2x^{\frac{2}{3}}y^{\frac{2}{3}}z^{\frac{2}{3}} \\
&\quad \leq x^2+y^2+z^2+\frac{2}{3}(x^{2}+y^{2}+z^{2})\\
&\quad =\frac{5}{3}(x^{2}+y^{2}+z^{2}),
\end{align*}
because $\frac{a+b+c}{3}\geq (abc)^{\frac{1}{3}}$. And
\begin{align*}
\frac{5}{3}(x^{2}+y^{2}+z^{2})
&= \frac{5}{3}(x^{\frac{1}{3}} \cdot x^{\frac{5}{3}} + y^{\frac{1}{3}} \cdot y^{\frac{5}{3}} + z^{\frac{1}{3}} \cdot z^{\frac{5}{3}}) \\
&\leq \frac{5}{3} \sqrt{x^{\frac{10}{3}} + y^{\frac{10}{3}} + z^{\frac{10}{3}}}\sqrt{x^{\frac{2}{3}} + y^{\frac{2}{3}} + z^{\frac{2}{3}}}\\
&= \frac{5\sqrt{2}}{3}\sqrt{x^{\frac{10}{3}} + y^{\frac{10}{3}} + z^{\frac{10}{3}}}
\end{align*}
Now I need to find a good frame to $\sqrt{x^{\frac{10}{3}}+y^{\frac{10}{3}}+z^{\frac{10}{3}}}$, or $\sqrt{a^5+b^5+c^5}$ for make
$$ \frac{5\sqrt{2}}{3}\sqrt{x^{\frac{10}{3}}+y^{\frac{10}{3}}+z^{\frac{10}{3}}}\leq 2. $$
I have two question:

Is my attempt correct? (I mean if my algebraic manipulation is true.)

Can you find a good frame to $\sqrt{x^{\frac{10}{3}}+y^{\frac{10}{3}}+z^{\frac{10}{3}}}$, or $\sqrt{a^5+b^5+c^5}$ for make
$$\frac{5\sqrt{2}}{3}\sqrt{x^{\frac{10}{3}}+y^{\frac{10}{3}}+z^{\frac{10}{3}}}\leq 2 \quad ?$$

If you have an other method, you can post it, but it will be nice if you can complete my attempt.

REPLY [3 votes]: New proof:
Using $0 \le (1 - a)(1 - b) = 1 - a - b + ab$, we have $ab \ge a + b - 1$.
We have
\begin{align*}
 &a^3+b^3+c^3 + 2abc\\
 \le\,& a^2 + b^2 + c^2 + 2abc\\
 =\,& (a + b)^2 - 2ab + c^2 + 2abc\\
 =\,& (a + b)^2 - 2ab(1 - c) + c^2\\
 \le\,& (a + b)^2 - 2(a + b - 1)(1 - c) + c^2 \\
 =\,& (2 - c)^2 - 2(1-c)(1-c) + c^2 \\
 =\,& (2 - c)^2 - (1-c)^2 + c^2 - (1-c)^2\\
 =\,& 1 \cdot (3-2c) + (2c-1)\cdot 1\\
 =\,& 2.
\end{align*}
We are done.

Old proof:
WLOG, assume that $c = \min(a,b,c)$.
Using $0 \le (1 - a)(1 - b) = 1 - a - b + ab$, we have $ab \ge a + b - 1$.
We have
\begin{align*}
 &a^3 + b^3 + c^3 + 2abc\\
 =\,& (a + b)^3 - 3ab(a + b) + c^3 + 2abc \\
 =\,& (a + b)^3 - ab(3a + 3b - 2c) + c^3\\
 \le\,& (a + b)^3 - (a + b - 1)(3a + 3b - 2c) + c^3\\
 =\,& c^2 - c + 2\\
 \le\,& 2.
\end{align*}
We are done.


TITLE: What's the probability of a number being chosen if it can be chosen twice?
QUESTION [0 upvotes]: Given a set of {1,2,3,...,n} numbers and a number can be chosen randomly and there is also the possibility of assigning the same number chosen by two different participants.
My thought process is, since 2 people can choose the same number then there is $2^n$ ways each number could be chosen.
Suppose I want to choose the number k, then the probability someone is assigned that number k is $\dfrac{2^{n-1}}{2^n}$?
Please let me know if I am incorrect. Thank you!

REPLY [1 votes]: If there are $m$ independent and equally probable selections from $n$ numbers, then the probability a particular number is not selected is $\left(1-\frac1n\right)^m$
so the probability a number is selected is $1-\left(1-\frac1n\right)^m$
For $m=2$ this is $\frac2n - \frac1{n^2}$


TITLE: How to show that basis of space of Dirac gamma-matrices is given by following matrices?
QUESTION [3 upvotes]: How to show that 16 matrices
$$
\mathbf E , \quad \gamma^{\mu}, \quad \gamma^{5} = \frac{i}{4}\gamma^{0}\gamma^{1}\gamma^{2}\gamma^{3}, \quad \eta^{\mu \nu} = -\frac{1}{4}\left(\gamma^{\mu}\gamma^{\nu} - \gamma^{\nu}\gamma^{\mu}\right), \quad \gamma^{5}\gamma^{\mu}  \qquad (.1)
$$
create the basis in space of Dirac matrices, by having
$$
[\gamma^{\mu}, \gamma^{\nu}]_{+} = 2g^{\mu \nu}\mathbf E ?
$$
I don't want to calculate each of matrices $(.1)$.

REPLY [1 votes]: So far I can see only 12 different nonzero matrices. Perhaps you missed $\gamma^5\gamma^\mu$.
EDIT: There is a proof of linear independence of the 16 matrices, e.g., in Bogoliubov and Shirkov's Introduction to theory of quantized fields. They prove (using trace invariance under cyclic permutations and anticommutation relations) that each of the matrices, except E, has zero trace, assume that a nontrivial linear combination of the matrices vanishes and show (multiplying this linear combination separately by each of the matrices) that this assumption is in contradiction with the trace of each matrix (except E) being zero.


text:
\subsection{Cocharacter groups}
\label{subsec:CocharacterGroups}

In the representation theory of groups the \emph{characters} play a
fundamental role. For Lie groups also the notion of an infinitesimal
character (or derivations) is useful. We are interested in a theory of
\emph{cocharacters}, i.e. the theory dual to that of characters. This
theory will play an important role in characterizing the kernel of the
forgetful morphism from the equivariant Picard group to the standard
Picard group. In this section we make use of the characteristic zero
condition on $R$, as we need to be able to divide through integers and
hence $\mathbb{Q} \subseteq R$ is crucial.

A nice investigation of character groups of Hopf algebras with trivial
action is given in \cite{bogfjellmo.dahmen.schmedig:2015a}, the
equivariant version, even for a noncommutative target algebra, was
introduced in \cite{jansen.waldmann:2006a}. Basically, one considers
the algebra $\Hom(H, \algebra{A})$ of linear morphisms from the Hopf
algebra $H$ to the algebra $\algebra{A}$ with product being the
convolution product and finds suitable subgroups of this algebra as
character groups.

We now go the other way round and pair a coalgebra $C$ with a Hopf
algebra $H$, that is we consider the set $\Hom(C, H)$ with convolution
product. Recall that for the set of linear maps from a coalgebra $C$
to an algebra $\algebra{A}$ one has the convolution product
\begin{equation}
    \label{eq:ConvolutionProductAgain}
    f \conv g = \mu \circ (f \tensor g) \circ \Delta
\end{equation}
for $f,g \in \Hom(C,\algebra{A})$. This convolution product endows
$\Hom(C, \algebra{A})$ with the structure of an algebra. For a Hopf
algebra $H$ one can apply this on both sides, since $H$ has a
coalgebra and an algebra structure.

The first part is remodelling the group of which the character groups
are subgroups. We have to replace the unit by the counit. We call a
morphism $\phi \in \Hom(C,H)$ \emph{counitary} if it satisfies
$\counit_H \circ \phi = \counit_C$. We set
\begin{equation}
    \label{eq:CounitarityMonoidCocharacters}
    \Gamma_0 = \{ \phi \in \Hom(C,H) \; | \; \phi \text{ is counitary
      and convolution invertible}\}
\end{equation}
This is completely dual to the situation in the definition of unitary
characters as the unitarity condition $\phi(1) = 1$ is actually
$\phi \circ \unit_H = \unit_{\algebra{A}}$ due to linearity. The
corresponding Lie algebra is
\begin{equation}
    \label{eq:CochcaracterLieAlgebra}
    \lie{c}_0 = \{\phi \in \Hom(C,H) \; | \; \counit_H \circ \phi = 0
      \}.
\end{equation}
Obviously $\Gamma_0$ forms a group with respect to the the convolution
product $\conv$ and $\lie{c}_0$ a Lie algebra with Lie bracket being
the commutator to the convolution product. The $\lie{c}_0$ injects
into $\Gamma_0$ via the convolution exponential
$\exp_\conv(\phi) = \sum_{k \geq 0} \frac{\phi^{\conv k}}{k!}$ if $H$
is a filtered Hopf algebra. Recall that a filtration of a Hopf algebra
$H$ is a filtration of the underlying $R$-module, i.e.\ one has
submodules
$H^0 \subseteq H^1 \subseteq \dots \subseteq H^n \subseteq \dots$ with
$\bigcup_{n \geq 0} H^n = H$, such that one has
$\mu(H^p \tensor H^q) \subseteq H^{p+q}$,
$\Delta(H^n) \subseteq \sum_{p+q = n} H^p \tensor H^q$, and
$S(H^n) \subseteq H^n$. As example of a filtered Hopf algebra consider
the following construction of the coradical filtration. Consider $H$
as $(H,H)$-bicomodule and let $H^0 = \mathsf{soc} \; H$ be the
coradical, i.e.\ the sum of all simple subcoalgebras of $H$. If the
coradical $H^0$ is a Hopf subalgebra then one can recursively define a
filtration of $H$ via
\begin{equation}
    \label{eq:CoradicalFiltration}
    H^n = \Delta^{-1}( H \tensor H^0 + H^{n-1} \tensor H)
\end{equation}
which is a Hopf algebra filtration, see \cite[Thm.\ II.2.2, Rem.\
1]{Manchon.2001}. The condition that $H^0$ is a Hopf subalgebra is for
example fulfilled for every pointed Hopf algebra, i.e.\ Hopf algebras
whose simple Hopf subalgebra are all one-dimensional. These include
all cocommutative Hopf algebras, e.g.\ the group algebras and
universal enveloping algebras of Lie algebras.

Let us denote by $\coacts$ the coaction of $H$ on the coalgebra $C$.
We adapt the extended Sweedler notation from
Remark~\ref{remark:ExtendingSweedlerNotationToCoaction}.
\begin{definition}[$H$-equivariant cocharacters]
    \label{definition:CocharacterGroupOfHopfAlgebra}
    An element $\phi \in \Hom(C,H)$ is called $H$-equivariant
    cocharacter if it is convolution invertible and satisfies the
    following axioms:
    \begin{definitionlist}
        \item \label{item:CocharacterCounitary}
        $\counit_H \circ \phi = \counit_C$,
        \item \label{item:CocharacterCoactionCondition}
        $\phi(c)^{(1)} \tensor \phi(c)^{(2)} = \phi(c_{(1)}^{(0)})
        \tensor c_{(1)}^{(1)} \phi(c_{(2)})$,
        \item \label{item:CocharacterComoduleCondition}
        $c_{(1)}^{(0)} \tensor c_{(1)}^{(1)} \phi(c_{(2)})
        = c_{(2)}^{(0)} \tensor \phi(c_{(1)}) c_{(2)}^{(1)}$,
    \end{definitionlist}
    for all $c \in C$. The set of all $H$-equivariant cocharacters is
    denoted by
    \begin{equation}
        \label{eq:CocharacterGroup}
        \Gamma(C \Coacts H)
        = \{ \phi \in \Hom(C,H) \; | \; \phi \text{ is } H
        \text{-equivariant cocharacter} \}.
    \end{equation}
\end{definition}
Note that the conditions can be phrased completely in the terms of
morphisms as follows: for \eqref{item:CocharacterCoactionCondition}
one has
\begin{equation}
    \label{eq:CocharacterCoactionConditionInMorphisms}
    \Delta_H \circ \phi
    = (\id \tensor \mu) \circ (\phi \tensor \id \tensor \phi) \circ
    (\coacts \tensor \id) \circ \Delta_C
\end{equation}
and for \eqref{item:CocharacterComoduleCondition} one has
\begin{equation}
    \label{eq:CocharacterComoduleConditionInMorphisms}
    (\id \tensor \mu) \circ (\coacts \tensor \phi) \circ \Delta_C
    = (\id \tensor \mu) \circ \tau^{(1 2)} \circ (\phi \tensor \coacts)
      \circ \Delta_C.
\end{equation}
These conditions arise by dualizing the conditions of
\cite[Definition~A.1]{jansen.waldmann:2006a}. We call condition
\refitem{item:CocharacterCounitary} counit condition, condition
\refitem{item:CocharacterCoactionCondition} the coaction condition and
condition \refitem{item:CocharacterComoduleCondition} the comodule
condition.
\begin{proposition}
    \label{proposition:CocharactersFormGroup}
    The set $\Gamma(C \Coacts H)$ forms a group. The inverse of $\phi
    \in \Gamma(C \Coacts H)$ is given by
    \begin{equation}
        \label{eq:CocharacterInverse}
        \phi^{\conv -1}(c) = S(\phi(c^{(0)})) c^{(1)}.
    \end{equation}
\end{proposition}
\begin{proof}
    Let $\phi, \psi \in \Gamma(C \Coacts H)$. We have to check that
    $\phi \conv \psi \in \Gamma(C \Coacts H)$. The counit condition is
    clear. For the coaction condition we have for $c \in C$
    \begin{align*}
        (\phi \conv \psi)(c)_{(1)} \tensor (\phi \conv \psi)(c)_{(2)}
        &= \phi(c_{(1)})_{(1)} \psi(c_{(2)})_{(1)} \tensor
           \phi(c_{(1)})_{(2)} \psi(c_{(2)})_{(2)} \\
        &= (\mu \tensor \mu) \circ \tau^{(2 3)} (\phi(c_{(1)})_{(1)}
           \tensor \phi(c_{(1)})_{(2)} \tensor \psi(c_{(2)})_{(1)}
           \tensor \psi(c_{(2)})_{(2)}) \\
        &\stackrel{(a)}{=} (\mu \tensor \mu) \circ \tau^{(2 3)}
           (\phi(c_{(1)}^{(0)}) \tensor c_{(1)}^{(1)} \phi(c_{(2)})
           \tensor \psi(c_{(3)}^{(0)}) \tensor c_{(3)}^{(1)}
           \psi(c_{(4)})) \\
        &= \phi(c_{(1)}^{(0)}) \psi(c_{(3)}^{(0)}) \tensor
           c_{(1)}^{(1)} \phi(c_{(2)}) c_{(3)}^{(1)} \psi(c_{(4)}) \\
        &\stackrel{(b)}{=} \phi(c_{(1)}^{(0)}) \psi(c_{(2)}^{(0)})
           \tensor c_{(1)}^{(1)} c_{(2)}^{(1)} \phi(c_{(3)})
           \psi(c_{(4)}) \\
        &\stackrel{(c)}{=} (\phi \conv \psi)(c_{(1)}^{(0)}) \tensor
        c_{(1)}^{(1)} (\phi \conv \psi)(c_{(2)})
    \end{align*}
    where in ($a$) we used the coaction condition
    \refitem{item:CocharacterCoactionCondition} for $\phi$ and $\psi$,
    in ($b$) the comodule condition
    \refitem{item:CocharacterComoduleCondition}, and in ($c$) we used
    \eqref{eq:SweedlerCoactionCoflatnessCompatibleProduct}. Checking
    the comodule condition is comparatively easy:
    \begin{align*}
        c_{(1)}^{(0)} \tensor c_{(1)}^{(1)} (\phi \conv \psi)(c_{(2)})
        &= c_{(1)}^{(0)} \tensor c_{(1)}^{(1)} \phi(c_{(2)})
           \psi(c_{(3)}) \\
        &\stackrel{\mathclap{\refitem{item:CocharacterComoduleCondition}}}{=}
           c_{(2)}^{(0)} \tensor \phi(c_{(1)}) c_{(2)}^{(1)}
           \psi(c_{(3)}) \\
        &\stackrel{\mathclap{\refitem{item:CocharacterComoduleCondition}}}{=}
            c_{(3)}^{(0)} \tensor \phi(c_{(1)}) \psi(c_{(2)})
            c_{(3)}^{(1)} \\
        &=  c_{(2)}^{(0)} \tensor (\phi \conv \psi)(c_{(1)})
            c_{(2)}^{(1)},
    \end{align*}
    for $c \in C$.

    The candidate for the inverse satisfies
    \begin{align*}
        \phi^{\conv -1} \conv \phi(c)
        &= S(\phi(c_{(1)}^{(0)})) c_{(1)}^{(1)} \phi(c_{(2)}) \\
        &\stackrel{(a)}{=} S(\phi(c)_{(1)}) \phi(c)_{(2)} \\
        &\stackrel{(b)}{=} e(\phi(c)) \\
        &\stackrel{(c)}{=} e(c),
    \end{align*}
    where in ($a$) we used the coaction condition of $\phi$, in ($b$)
    the fact that $S$ is the convolution inverse to the identity,
    applied via $S \conv \id(\phi(c))$, and in ($c$) we used the
    counitarity of $\phi$. Now we have to check that $\phi^{\conv -1}
    \in \Gamma(C \Coacts H)$. The counitarity of $\phi^{\conv -1}$ is
    clear by definition.
    For the comodule condition, we calculate
    \begin{align*}
        c_{(1)}^{(0)} \tensor c_{(1)}^{(1)} \phi^{\conv -1}(c_{(2)})
        &= c_{(1)}^{(0)} \tensor c_{(1)}^{(1)} S(\phi(c_{(2)}^{(0)}))
           c_{(2)}^{(1)} \\
        &\stackrel{(a)}{=} c_{(1)}^{(0)} \tensor c_{(1)}^{(1)}
           S(\phi(c_{(2)}^{(0)})) S(c_{(1)}^{(1)}) c^{(1)} \\
        &\stackrel{(b)}{=} c_{(1)}^{(0)} \tensor c_{(1)}^{(2)}
           S(c_{(1)}^{(1)} \phi(c_{(2)}^{(0)})) c^{(1)} \\
        &\stackrel{(c)}{=} c_{(2)}^{(0)} \tensor c_{(1)}^{(1)}
           S(\phi(c_{(1)}^{(0)}) c_{(2)}^{(1)}) c^{(1)} \\
        &= c_{(2)}^{(0)} \tensor c_{(1)}^{(1)} S(c_{(2)}^{(1)})
           S(\phi(c_{(1)}^{(0)})) c^{(1)} \\
        &= c_{(2)}^{(0)} \tensor S(\phi(c_{(1)}^{(0)})) c^{(1)} \\
        &\stackrel{(d)}{=} c_{(2)}^{(0)} \tensor
           S(\phi(c_{(1)}^{(0)})) c_{(1)}^{(1)} c_{(2)}^{(1)} \\
        &= c_{(2)}^{(0)} \tensor \phi^{\conv -1}(c_{(1)})
           c_{(2)}^{(1)}
    \end{align*}
    where in ($a$) we used the identity
    \eqref{eq:AntipodeAndCoaction}, in ($b$) we used the properties of
    the antipode, in ($c$) the comodule condition for $\phi$ and in
    ($d$) the compatibility of the coaction with product of the Hopf
    algebra.
    Finally we have to check the coaction condition. Here we have
    \begin{align*}
        \phi^{\conv -1}(c_{(1)}^{(0)}) \tensor c_{(1)}^{(1)}
        \phi^{\conv - 1}(c_{(2)})
        &= S(\phi(c_{(1)}^{(0)})) c_{(1)}^{(1)} \tensor c_{(1)}^{(2)}
           S(\phi(c_{(2)}^{(0)})) c_{(2)}^{(1)} \\
        &\stackrel{(a)}{=} S(\phi(c_{(1)}^{(0)})) c_{(1)}^{(1)}
           c_{(2)}^{(1)} \tensor S(c_{(1)}^{(1)}) c_{(1)}^{(2)}
           S(\phi(c_{(2)}^{(0)})) c_{(2)}^{(2)} \\
        &= S(\phi(c_{(1)}^{(0)})) c^{(1)} \tensor
           S(\phi(c_{(2)}^{(0)})) \counit(c_{(1)}) c_{(2)}^{(1)} \\
        &= S(\phi(c_{(1)}^{(0)})) c^{(1)} \tensor
           S(\phi(c_{(2)}^{(0)})) S(c_{(1)}^{(1)}) c_{(1)}^{(2)}
           c_{(2)}^{(2)} \\
        &= S(\phi(c_{(1)}^{(0)})) c^{(1)} \tensor S(c_{(1)}^{(1)}
           \phi(c_{(2)}^{(0)})) c^{(2)} \\
        &\stackrel{(b)}{=} (S(\phi(c^{(0)}))^{(1)} c^{(1)} \tensor
           (S(\phi(c^{(0)})))^{(2)} c^{(2)} \\
        &= \phi^{\conv -1}(c)^{(1)} \tensor \phi^{\conv -1}(c)^{(2)}
    \end{align*}
    where we used in ($a$) the fact that $\id \conv S = e$, whence
    $\counit \circ \Delta = (\id \tensor S) \circ \Delta$, in ($b$)
    the coaction condition for $\phi$ and that $S$ is a coalgebra
    homomorphism, and the properties of antipode and counit.
\end{proof}
\begin{example}
    \label{example:TrivialCocharacterOnGrouplike}
    Consider the a group-like coalgebra $R[S]$ for some set $S$ and a
    Hopf algebra $H$. Assume the trivial coaction $\coacts \colon R[S]
    \to R[S] \tensor H$ given by $\coacts s = s \tensor 1$. Then the
    cocharacter group $\Gamma(C \Coacts H)$ is given by those
    morphisms $\phi \in \Hom(C, H)$ which send group-like elements to
    group-like elements, as the coaction condition in this case reads
    \begin{equation}
        \label{eq:TrivialCocharacterOnGrouplike}
        \phi(s)^{(1)} \tensor \phi(s)^{(2)} = \phi(s) \tensor \phi(s)
    \end{equation}
    for all $s \in S$.
\end{example}

There is a corresponding infinitesimal notion of coderivations:
\begin{definition}[$H$-equivariant coderivations]
    \label{definition:EquivariantCoderivations}
    An element $\phi \in \Hom(C,H)$ is called $H$-equivariant
    coderivation if it is convolution invertible and satisfies the
    following:
    \begin{definitionlist}
        \item \label{item:EquivCoderUnit}
        $\counit_H \circ \phi = 0$
        \item \label{item:EquivCoderCoaction}
        $\phi(c)^{(1)} \tensor \phi(c)^{(2)}
        = e(c_{(1)}^{(0)}) \tensor c_{(1)}^{(1)} \phi(c_{(2)}) +
          \phi(c_{(1)}^{(0)}) \tensor c_{(1)}^{(1)} e(c_{(2)})$
        \item \label{item:EquivCoderComodule}
        $c_{(1)}^{(0)} \tensor c_{[1)}^{(1)} \phi(c_{(2)})
        = c_{(2)}^{(0)} \tensor \phi(c_{(1)}) c_{(2)}^{(1)}$,
    \end{definitionlist}
    for all $c \in C$. The set of all $H$-equivariant coderivations is
    denoted by $\lie{c}(C \Coacts H)$.
\end{definition}
\begin{proposition}
    \label{proposition:EquivCoderivationExpToCocharacters}
    If $H$ is filtered, the set $\lie{c}(C \Coacts H)$ injects into
    $\Gamma(C \Coacts H)$ via $\exp(\phi) = \sum_{k = 0}^\infty
    \frac{\phi^{\conv k}}{k!}$.
\end{proposition}
\begin{proof}
    The filtration of $H$ is needed to ensure convergence of the
    exponential map.

    We first have to show that
    \begin{equation*}
        \phi^{\conv n}(c)^{(1)} \tensor \phi^{\conv n}(c)^{(2)}
        = \sum_{k=0}^n \binom{n}{k} \phi^{\conv k}(c_{(1)}^{(0)})
        \tensor c_{(1)}^{(1)} \phi^{\conv(n-k)}(c_{(2)})
    \end{equation*}
    holds for all for $\phi \in \lie{c}(C \Coacts H)$ and all
    $c \in C$. This is done by induction, for $n=1$ it is just the
    infinitesimal coaction condition, and for $n \to n+1$ we have
    \begin{align*}
        \Delta_H \circ \phi^{\conv n+1}(c)
        &= \mu( \Delta_H(\phi(c_{(1)})) \tensor \Delta_H(\phi^{\conv
           n}(c_{(2)}))) \\
        &= \sum_{k=0}^n \binom{n}{k} (e(c_{(1)}^{(0)}) \tensor
           c_{(1)}^{(1)} \phi(c_{(2)}) + \phi(c_{(1)}^{(0)}) \tensor
           c_{(1)}^{(1)} e(c_{(2)})) (\phi^{\conv k}(c_{(3)}^{(0)})
           \tensor c_{(3)}^{(1)} \phi^{\conv n-k}(c_{(4)})) \\
        &= \sum_{k=0}^n \binom{n}{k} (\phi^{\conv k}(c_{(1)}^{(0)})
           \tensor c_{(1)}^{(1)} \phi^{\conv n+1-k}(c_{(2)}) +
           \phi^{\conv k+1}(c_{(1)}^{(0)}) \tensor c_{(1)}^{(1)}
           \phi^{\conv n-k}(c_{(2)})) \\
        &= \sum_{k=0}^n \binom{n}{k} \phi^{\conv k}(c_{(1)}^{(0)})
           \tensor c_{(1)}^{(1)} \phi^{\conv n+1-k}(c_{(2)}) +
           \sum_{k=1}^{n+1} \binom{n}{k-1} \phi^{\conv
             k}(c_{(1)}^{(0)}) \tensor c_{(1)}^{(1)} \phi^{\conv n -
             (k+1)}(c_{(2)}) \\
        &= c_{(1)}^{(1)} \phi^{\conv n+1}(c_{(2)}) + \sum_{k=1}^{n+1}
           \binom{n+1}{k} \phi^{\conv k}(c_{(1)}^{(0)}) \tensor
           c_{(1)}^{(1)} \phi^{\conv n+1-k}(c_{(2)}) + \phi^{\conv
             n+1}(c_{(1)}^{(0)}) \\
        &= \sum_{k=0}^{n+1} \binom{n+1}{k} \phi^{\conv
            k}(c_{(1)}^{(0)}) \tensor c_{(1)}^{(1)} \phi^{\conv
            n+1-k}(c_{(2)}).
    \end{align*}
    This now allows to show that the exponential of the infinitesimal
    coaction condition is indeed the coaction condition:
    \begin{align*}
        \exp(\phi)(c)^{(1)} \tensor \exp(\phi)(c)^{(2)}
        &= \sum_{n \geq 0} \frac{1}{n!} \phi^{\conv n}(c)^{(1)}
           \tensor \phi^{\conv n}(c)^{(2)} \\
        &= \sum_{n \geq 0} \frac{1}{n!} \sum_{k=0}^n \binom{n}{k}
           \phi^{\conv k}(c_{(1)}^{(0)}) \tensor c_{(1)}^{(1)}
           \phi^{\conv n-k}(c_{(2)}) \\
        &= \sum_{k=0}^\infty \sum_{n=k}^\infty \frac{1}{k! (n-k)!}
           \phi^{\conv k}(c_{(1)}^{(0)}) \tensor c_{(1)}^{(1)}
           \phi^{\conv n-k}(c_{(2)}) \\
        &= \sum_{k=0}^\infty \frac{1}{k!} \phi^{\conv k}(c_{(1)}^{(0)})
           \tensor \sum_{n=0}^\infty \frac{1}{n!} c_{(1)}^{(1)}
           \phi^{\conv n}(c_{(2)}) \\
        &= \exp(\phi)(c_{(1)}^{(0)}) \tensor c_{(1)}^{(1)}
           \exp(\phi)(c_{(2)}).
    \end{align*}
    The counit and comodule condition follow immediately, hence the
    injection is clear.
\end{proof}


TITLE: How do we know that $^1S$ is the ground state of the helium atom?
QUESTION [0 upvotes]: Let $\psi=a_1\phi(1s(2) \ ^1S)+a_2\phi(1s(1)2s(1) \ ^1S)+a_3\phi(2s(2) \ ^1S)
+... $ be a state of the helium atom. Applying variationally calculus we can found the energy  expectation value of this state is almost exact to the experimental value.
Is by this comparation that we know that the total angular momentum $J$ of the helium atom is zero or can we proof it theoretically alone, or experimentally alone  ?

REPLY [0 votes]: The answer to this question can be given with a much simpler explanation. Hund's rules state that:
1)For a given electron configuration, the term with maximum multiplicity has the lowest energy. The multiplicity is equal to $2S+1$ , where $S$ is the total spin angular momentum for all electrons. The multiplicity is also equal to the number of unpaired electrons plus one. Therefore, the term with lowest energy is also the term with maximum $S$,  and maximum number of unpaired electrons.
2)For a given multiplicity, the term with the largest value of the total orbital angular momentum quantum number $L$ ,  has the lowest energy.
3)For a given term, in an atom with outermost subshell half-filled or less, the level with the lowest value of the total angular momentum quantum number $J$ (for the operator $J=L+S$ lies lowest in energy. If the outermost shell is more than half-filled, the level with the highest value of $J$ is lowest in energy.
Since the total spin is 0 and the angular momentum is 0, you have the ground state that you have written.


TITLE: Let $X_n$ a sequence such that $x_n \leq 7$ and $x_n < x_{n+1}$
QUESTION [0 upvotes]: Let $X_n$ a sequence such that $x_n \leq 7$ and $x_n < x_{n+1}$ show that $x_n$ converges without using Theorem of the Convergence of Monotonic Sequences
the problem is I don't understand how the sequence is defined, any hint of what should I do to prove it?

REPLY [1 votes]: The proof can be written with the following outline. The questions in parentheses are questions that should be answered in order to make the proof complete.

Let $S = \{x_n : n \in \Bbb N\}$, and let $\alpha$ denote the supremum $\alpha = \sup(S)$ (why does the supremum exist/why is it finite?) We will show that $x_n$ converges to $\alpha$.
Let $\epsilon > 0$. Take $m \in \Bbb N$ to be such that $x_m > \alpha - \epsilon$ (how do we know that such an $m$ exists?). For any $n \geq m$, we see that $|x_n - \alpha| \leq |x_m - \alpha| < \epsilon$ (how?). Thus, $x_n$ converges to $\alpha$.


TITLE: Derivative of inverse trigonometric functions: $f(x)=\arcsin\left(2x\sqrt{1-x^2}\right)$
QUESTION [0 upvotes]: While trying to find the derivative of
$$f(x)=\arcsin\left(2x\sqrt{1-x^2}\right)$$
we arrive at two different answers by substituting $x=\sin t$ or $x=\cos t$. The aim is to simplify it as $\arcsin(\sin2t)$ and further simplify it as $2t$ and take its derivative. Proceeding in that manner, I arrived at two different answers for the derivative. I could understand that it was because the function $f$ is defined differently in different intervals, and as I have solved the problem without considering the intervals, I hit this answer. Also, I thought that since $\sin t$ and $\cos t$ will coincide only when $t=\dfrac{\pi}4$, $x=\pm\dfrac1{\sqrt{2}}$ will be the points where the definition changes. I got some more clarity when I looked at the graph of $f(x)$ but I could not figure out a way to systematically figure out what answer suits what interval.
MAJOR EDIT:  I made a mistake in the function itself that I had given. My question earlier read:
$$f(x)=\arcsin\left(2x\sqrt{x^2-1}\right)$$
But I wanted only for what I've now altered it as. My sincere apologies...
EDIT: My question is how to figure out the interval in which the substitution applies for such problems in general. I don't need the expression for the  derivative of  $f(x)$.
Note: I am new to this community, so please point out any deviation from the policy, if I have deviated.
Also, I couldn't post my working as I do not have $10$ reputation.

REPLY [0 votes]: $$ f(x) = arcsin(2x\sqrt{x^2-1}) $$
Replace x=cosh$\theta$
then f(x) = arcsin(sinh$2\theta$)
by differentiating f(x),
f '(x) = $\frac{2cosh(2\theta)}{\sqrt{1-sinh^2(2\theta)}\sqrt{x^2-1}}$
Replacing $ cosh(2\theta) = 2x^2-1 $
f '(x) = $\frac{4x^2-2}{\sqrt{1-4x^4+4x^2}\sqrt{x^2-1}}$