text:
\subsection{Construction of the $\ci$ differential operators}\label{cioperatordefinition}

First we construct the $\ci$ differential operators.  Later, we will define $p$-adic differential operators through a similar construction.

Let
\begin{eqnarray}\label{cinfinitysplitting}
H^1_{DR}(\ci) = \omci\oplus \splci
\end{eqnarray}
be the canonical splitting of the Hodge filtration corresponding to the holomorphic and anti-holomorphic one-forms.  Here, $\Split(C^\infty)$ is the sheaf $\overline{\uo(\ci)}$ of anti-holomorphic one forms.  Note\smargin{The discussion of $\nabla$ needs to be moved to the appropriate section, when I reorganize.} that for each derivation $D\in Der(\OM^{\ci}, \OM^{\ci})$,
\begin{align*}
\nabla(D)(\overline{\uo(\ci)})\subset\overline{\uo(\ci)}.
\end{align*}

Since
\begin{align*}
\dr(\ci)^{\pm}\subseteq\omci^{\pm}\oplus\splci,
\end{align*}
the splitting \eqref{cinfinitysplitting} induces projections
\begin{align*}
\dr(\ci)^{\pm}\rightarrow\omci^{\pm},
\end{align*}
which induces a projection

\begin{align}\label{proji}
\dr(\ci)^\rho\otimes T^\bullet(\dr^+(\ci)\otimes\dr^-(\ci))&\rightarrow \omci^\rho\otimes T^\bullet(\uo^+(\ci)\otimes \uo^-(\ci))\\
&\isomto \omci^\rho\otimes T^\bullet(\Omega(\ci)).\nonumber
\end{align}

As usual, we associate $\uo^\pm$ with its image in $(\dr)^{\pm}$ under the inclusion coming from hypercohomology
\begin{align}\label{uointodr}
\uo\hookrightarrow\dr.
\end{align}\smargin{I need to note somewhere that $i(a)\s$ preserves the splitting of $\dr$!  If it doesn't preserve the splitting, then there can be trouble.}
As in (\ref{roin}),  the inclusion \eqref{uointodr} and the Kodaira-Spencer isomorphism \eqref{ks2} induce inclusions\smargin{I need to add $\ci$'s in many parts of this section.}
\begin{align}\label{uor}
\mformsnost\otimes(\Omega_{\moduli/R_0}^{\otimes e})\hookrightarrow (\dr)^{\rmrp}\otimes (\dr^-\otimes\dr^+).
\end{align}

Restricting $D^\rho$ to the image of \eqref{uor}, we get a map

\begin{small}
\begin{align}\nonumber
(D^\rho)^d\rvert_{\mformsnost\otimes(\Omega_{\moduli/R_0}^{\otimes e})}: \mformsnost\otimes(\Omega_{\moduli/R_0}^{\otimes e})\rightarrow \dr(\ci)^\rmrp\otimes (\dr^+\otimes \dr^-)^{\otimes e+d}.
\end{align}
\end{small}

We define the $C^\infty$-differential operator $\ciop$ to be the map\smargin{I may need to later expand to work over an extension of $\IQ$, rather than $\IQ$ itself, and consider different embeddings into $\IC$.  Check whether I need to work over larger field now, and modify soon if necessary.}
\begin{align}\nonumber
\ciop: \mformsnost\otimes(\Omega(\ci)^{\otimes \bullet})\rightarrow  \mformsnost\otimes(\Omega(\ci))^{\otimes \bullet+d})\end{align}
that is the composition of maps in the following commutative diagram:

\begin{small}
\begin{align}\nonumber
\xymatrix{
\mformsnost(\ci)\otimes(\Omega(\ci)^{\otimes \bullet})\ar@{.>}[rrrdddd]_{\ciop}\ar[rrr]^{(D^\rho)^d}  &  & &\dr(\ci)^\rho\otimes (\dr^+\otimes \dr^-)^{\bullet+d}\ar[dd]^{\mod \splci}\\
& &&\\
 & &&\mformsnost(\ci)\otimes (\uo^+\otimes\uo^-)^{\otimes \bullet+d}\ar[dd]^{\cong}\\
 & &&\\
&& & \mformsnost(\ci)\otimes (\Omega(\ci))^{\bullet+d}
 }
\end{align}
\end{small}

\begin{rmk}\label{adhocrmk}
For the reader who worries that defining the differential operators on $\mformsnost$ (instead of $\EEVr$) is too {\it ad hoc} or non-canonical, we note that the differential operators can equivalently be defined as morphisms from $\EEVr$ to $\EE_{V, (\rho_+\otimes\rhost)\otimes(\rho_-\otimes\rhost)}$.  Indeed, the composition of maps

\begin{footnotesize}
\begin{align}\nonumber
\xymatrix{
\EEVr\ar@{.>}[rrrr]\ar[dd]_{(\lambda, v)\mapsto \lambda(v)}&&&& \EE_{(V_-\otimes \IC^n)\otimes(V_+\otimes\IC^n), (\rho_-\otimes\rhost)\otimes(\rho_+\otimes\rho_+)}\\
&&&&\\
\mformsnost\ar[rrr]_{\ciop}&&&\mformsnost\otimes\Omega\ar[r]_{KS}&\mformsst\ar[uu]_{v\mapsto (\lambda, \lambda^{-1}(v))}
 }
\end{align}
\end{footnotesize}

gives an equivalent expression of our differential operator as an operator from  $\EEVr$ to $\EE_{(V\otimes(R^n)\otimes(R^n)), (\rho_+\otimes\rhost)\otimes(\rho_-\otimes\rhost)}$.
\end{rmk}

Let $Z$ be a $GL_n(\IC)\times GL_n(\IC)$-stable quotient of
\begin{align*}
\mformsnost(\ci)\otimes (\Omega_{\A/\moduli}(\ci))^{\bullet+d},
\end{align*} and let $\phi_Z$ be the projection of $\mformsnost(\ci)\otimes (\Omega_{\A/\moduli}(\ci))^{\bullet+d}$ onto $Z$.  We define the differential operator $\ciopo^Z$ by
\begin{align*}
\phi_Z\circ \ciop.
\end{align*}

\subsection{Algebraicity theorem for $\ci$-differential operators}
The following algebraicity theorems (Theorems \ref{cialgthm} and \ref{cialgthm2}) are important for our intended applications.  The statement of Theorem \ref{cialgthm} and the idea of the proof are essentially the same as what is done in Section 2.4 of \cite{kaCM}; the new parts are our generalizations from Katz's special scalar-valued case to the arbitrary (often vector-valued) case and to the case of projection onto subrepresentations.

Throughout this section, fix an $\OK$-algebra $R$ with an inclusion\smargin{Add additional conditions on $R$ later.}
\begin{align}\nonumber
\ir: R\hookrightarrow\IC.
\end{align}

In the special case $R=\bar{\IQ}$, the statement of Theorem \ref{cialgthm} is essentially the same as Theorem 14.9 (2) of \cite{shar}.  However, the methods or the proof of Theorem \ref{cialgthm} are different from the proof in \cite{shar}; the proof we present is similar to what is done in Section 2.4 of \cite{kaCM}.

We associate each automorphic form $f$ in $\agauto_{\rho\otimes\tau^e}(R)$ with its image in $\agauto_{\rho\otimes\tau^e}(\IC)$ via the extension of scalars induced by $\ir$.  We also associate each automorphic form in $\agauto_{\rho\otimes\tau^e}(\IC)$ with the corresponding holomorphic section of $\mformsnost(\ci)\otimes\underline{\Omega}(\ci)^{\otimes e}$ on $\moduli(\ci)$.

Let $x=\underline{X}$ be an $R$-valued point of $\moduli_R$, and let $\lambda$ be an element of $\EE_{X/R}$.  Suppose there is a splitting over $R$
\begin{align}\nonumber
\splxr\oplus\omxr\isomto\drxr.
\end{align}
Then we have an inclusion
\begin{align}\label{incah}
\Split(X/R)\hookrightarrow H^1(X^{an}_{\IC}, \IC)
\end{align}
coming from the composition of maps
\begin{align}\nonumber
\xymatrix{
\splxr\ar@{^{(}->}[rrrr]^{\mbox{ext. of scalars via }\ir}\ar@{.>}[rrrrrdd]&&&& \splxr\otimes\IC\ar@{^{(}->}[r]^{\eqref{incah}\otimes \id}&\dr(X/R)\otimes\IC\ar[dd]^{\cong}\\
&&&&&\\
&&&&& H^1(X^{an}_{\IC}, \IC)
 }
\end{align}

We say that the the pair $(x, \splxr)$ satisfies the condition \pstrict   if the following holds:

\begin{align}
\mbox{The image of $\splxr$ under the inclusion } &\mbox{\eqref{incah} is the antiholomorphic subspace}\nonumber\\ H^{0,1}&\subset H^1({X^{an}}_{\IC}, \IC),\nonumber\\
 \mbox{ i.e. }\splxr\otimes\IC& = \splci(x_{\IC}).\tag{\strict}
\end{align}

Note that this condition is essentially the same as condition (2.4.2) in \cite{kaCM}.

For our intended applications, the only points that will interest us are certain ordinary CM points.  We shall see later that for each such ordinary CM point, there is indeed a splitting satisfying condition \pstrict.

Note that in general, we only know that the values of $\ciop(f)$ at points $(x, \lambda)$ lie in a $\IC$-vector space.  (This is because even if $f\in \agauto_{\rho}(R)$, we only know that $\ciop(f)$ is a $\ci$-function and nothing about where its values at arbitrary points lie.)  We see in theorem \ref{cialgthm}, however, that we can say much more about the values of $\ciop(f)$ at points satisfying \pstrict.

\begin{thm}\label{cialgthm}
Suppose that $(x, \splxr)$ is an $R$-valued point of $\moduli_R$ that satisfies condition \pstrict.  Let $f$ be an automorphic form in $\agauto_{\rho\otimes\tau^e}(R)$ with values in an $R$-module $V$.  Then
\begin{align}
(\ciop f)(x, \lambda)_{\IC} = \ir (\partial(\rho, e, x, \lambda, \splxr, d)f).\label{surpeq}
\end{align}
Therefore,
\begin{align}\nonumber
(\ciop f)(x, \lambda)_{\IC}\in V\otimes_R \left(R^n\otimes R^n\right)^{\otimes d}.
\end{align}
\end{thm}
The proof we provide is similar to Katz's proof of Theorem 2.4.5 in \cite{kaCM}.
\begin{proof}
As it was defined in Section \ref{agdefmaps}, $\ir (\partial(\rho, e, x, \lambda, \splxr, d)f)$ lies in
\begin{align*}
V\otimes_R \left(R^n\otimes R^n\right)^{\otimes d}.
\end{align*}  So to prove the theorem, it suffices to prove that Equation \eqref{surpeq} holds.

By the extension of scalars from $R$ to $\IC$ given by $\ir$, we associate the automorphic form $f$ with its image $f_{\IC}$ in $\agauto_{\rho\otimes\tau^e}(\IC)$, the $R$-valued point $x$ with $R$-basis $\lambda$ with $x$ with basis $\lambda_{\IC}$, and $\splxr$ with its image in $\splxr\otimes\IC$.  Then, we see that
\begin{align}\nonumber
\ir (\partial(\rho, e, x_{\IC}, \lambda_{\IC}, \splxr_{\IC}, d)f_{\IC}) = \ir(\partial(\rho, e, x, \lambda, \splxr, d)f)
\end{align}
is $V\otimes_R \left(R^n\otimes R^n\right)^{\otimes d}$-valued.  So it suffices to show that Equation \eqref{surpeq} holds in the case $R=\IC$, which we will now do.

Consider the following commutative diagram \eqref{algcomdiag}.  (Note that commutativity in \eqref{algcomdiag} follows from the hypothesis that $(x, \splxr)$ satisfies condition \pstrict.)\smargin{Make Diagram \eqref{algcomdiag} look nicer later.}

\begin{footnotesize}
\begin{align}\label{algcomdiag}
\xymatrix{
(\dr)^\rho\otimes(\dr^+\otimes\dr^-)^{\otimes e+d}\ar[ddr]^{\mbox{     take fiber at } x}\ar[dd]_{\mod\splci}&\\
&\\
\mformsnost(\ci)\otimes(\uo^+(\ci)\otimes\uo^-(\ci))^{\otimes e+d}\ar[ddr]_{\mbox{     take fiber at } x}&(\dr(X/R))^\rho\otimes((\dr^+\otimes\dr^-)(X/R))^{\otimes e+d}\ar[dd]^{\mod \splxr}\\
&\\
&((\uo^+\otimes\uo^-)(X/R))^\rho\otimes((\uo^-\otimes\uo^+)(X/R))^{\otimes e+d}\ar[dd]^{\cong \mbox{ from choice of } \lambda}\\
&\\
&(V\otimes_R\IC)\otimes_{\IC}(\IC^n\otimes \IC^n)^{\otimes d}
 }
\end{align}
\end{footnotesize}

As usual, we associate $f$ with its image in $(\dr)^\rho\otimes(H^+\otimes H^-)^{\otimes e}$.  Then \\$(\partial(\rho, e, x, \lambda, \splxr, d)f)(x)$  is obtained by applying $(D^\rho)^d$ to $f$ and composing with the maps along the right side  of the commutative diagram \eqref{algcomdiag}.  Similarly, $(\ciop f)(x)$ is obtained by applying $(D^\rho)^d$ to $f$ and composing with the maps along the left side of the commutative diagram \eqref{algcomdiag}.  Therefore, \eqref{surpeq} holds for all $x$ admitting a splitting satisfying \pstrict.\end{proof}

We also obtain the following generalization of Theorem \ref{cialgthm}:

\begin{thm}\label{cialgthm2}
Suppose that $(x, \splxr)$ is an $R$-valued point of $\moduli_R$ that satisfies condition \pstrict.  Let $f$ be an automorphic form in $\agauto_{\rho\otimes\tau^e}(R)$.  Let $Z$ be a $GL_n(R)\times GL_n(R)$-stable $R$-quotient of $\uo_R^\rho\otimes (\Omega_{\A/\moduli_R})^{\bullet+d}$, and let $\phi_Z$ be the projection of $\omci^\rho\otimes (\Omega_{\A/\moduli}(\ci))^{\bullet+d}$ onto $Z\otimes_R\IC$.  Let $Z_x$ be the fiber of $Z$ at $x$.  Then
\begin{small}
\begin{align*}
(\ciop^Z f)(\x,\lambda)_{\IC} = \ir (\partial(\rho, e, x, \lambda, \splxr, d)^{Z_X}f).
\end{align*}
\end{small}
Therefore,
\begin{align}\nonumber
(\ciop^Z f)(x,\lambda)_{\IC}
\end{align}
actually takes values in the $R$-module $\lambda(Z)$.
\end{thm}

\begin{proof}
The proof is similar to the proof of Theorem \ref{cialgthm}, except that we replace the commutative diagram \ref{algcomdiag} with the commutative diagram

\begin{footnotesize}
\begin{align}\label{algcomdiag2}
\xymatrix{
(\dr)^\rho\otimes(\dr^+\otimes \dr^-)^{\otimes e+d}\ar[ddr]^{\mbox{     take fiber at } x}\ar[dd]_{\mod\splci}&\\
&\\
(\mformsnost)(\ci)\otimes((H^+\otimes H^-)(\ci))^{\otimes e+d}\ar[dd]_{\phi_Z}&(\dr(X/R))^\rho\otimes((H^+\otimes H^-)(X/R))^{\otimes e+d}\ar[dd]^{\mod \splxr}\\
&\\
Z\ar[ddr]_{\mbox{     take fiber at } x}&(\mformsnost(X/R))^\rho\otimes((\uo^+\otimes\uo^-)(X/R))^{\otimes e+d}\ar[dd]^{\phi_{Z_X}}\\
&\\
&Z_X\ar[dd]^{\cong \mbox{ from choice of } \lambda}\\
&\\
&\IC^{rank(Z_X)}
 }
\end{align}
\end{footnotesize}

Now, the proof goes in the same way as the proof of Theorem \ref{cialgthm}.\end{proof}

Note that similarly to in Remark \ref{adhocrmk}, restriction to $Z$ yields a canonical map
\begin{align*}
\EE_{V, \rho}\rightarrow \EE_Z.
\end{align*}

\subsection{Some properties of the $\ci$-differential operators}
In this section, we give some fundamental properties of the $\ci$-differential operators $\ciop$.

We denote by $(\ciopo)^d$ the composition of $\ciop$ with itself $d$ times.
\begin{thm}\label{impprop}
The differential operators satisfy the following properties.
\begin{enumerate}
\item{\begin{align}\label{afciopo}
\ciop = (\ciopo)^d
\end{align}
for each positive integer $d$.  }
\item{
Associating the space of automorphic forms of weight $\rho\otimes \tau^f$ with $\mformsnost(\ci)\otimes\Omega(\ci)^{\otimes f}$ via the natural isomorphism induced by the Kodiara-Spencer isomorphism, we have that $\ciop$ is the same as $\partial(\rho\otimes\tau^f, \ci, d)$, i.e.
\begin{align}
\ciop&\rvert_{\mformsnost(\ci)\otimes\Omega_{\A/\moduli}(\ci)^{\otimes e}}\nonumber\\
&= \partial(\rho\otimes\tau^f, \ci, d)\vert_{\mformsnost(\ci)\otimes\Omega_{\A/\moduli}(\ci)^{\otimes e}} \label{rhotorhotci}
\end{align}

for all non-negative integers $f\leq e$.}

\item{
\begin{align}
\partial(\rho\otimes\tau^{d-1}, \ci, 1)\circ\cdots\circ\partial(\rho\otimes\tau, \ci, 1)\circ\ciopo = \ciop\label{likeshci}
\end{align}
for all positive integers $d\geq 2$.
}
\end{enumerate}
\end{thm}
\begin{proof}
Recall that $\splci$ is horizontal\smargin{Need to say holo. horiz. in introductory section.} with respect to $\nabla$, in the sense that
\begin{align}\nonumber
\nabla(\splci)\subseteq \splci\otimes\Omega_{\A/\moduli}(\ci).
\end{align}
So from the definition of $D^{\rho}$, we see that
\begin{align}\nonumber
D^{\rho}(\splci)\subseteq \splci,
\end{align}
and hence, for all positive integers $d$,
\begin{align}\nonumber
(D^{\rho})^d(\splci) \subseteq \splci.
\end{align}
So
\begin{align}\nonumber
(D^{\rho})^d\circ(\mbox{``$\mod\splci$"}) = (D^{\rho}\circ(\mbox{``$\mod\splci$"}))^d,
\end{align}
where $(D^{\rho}\circ(\mbox{``$\mod\splci$"}))^d$ denotes $D^{\rho}\circ(\mbox{``$\mod\splci$"})$ composed with itself $d$ times.
Therefore, it follows directly from the definition of $\ciop$ that
\begin{align}
\ciop = (\ciopo)^d.\nonumber
\end{align}  So \eqref{afciopo} holds.

Equation \eqref{rhotorhotci} follows directly from the definition of $\tau$, our earlier explicit description of the Kodaira-Spencer isomorphism, and the definition of the map $\ciop$ for any representation $\rho$.

Now we prove that \eqref{likeshci} holds.  Note that for any $\rho$, $\ciopo$ has image in $\omci^{\rho}\otimes\sum_{e = 1}^{\infty}\Omega^{\otimes e}(\ci)$.  So \eqref{likesh} is a corollary of \eqref{afciopo} and \eqref{rhotorhotci}.
\end{proof}

\subsection{Explicit formulas for the $\ci$-differential operators}\label{rho04301}
In this section, we present explicit formulas for the $\ci$-differential operators and compare them to the formulas in \cite{shar}.  The reader familiar with \cite{BSY}, which handles the symplectic case, can compare our formulas with \cite{BSY} to see that we have constructed essentially the same $\ci$-operators that \cite{BSY} defines combinatorially.  The form in which we have chosen to write our formulas should help the reader see the connection between our operators and the ones in \cite{BSY}.  In particular, the reader may wish to compare our formulas with Formula (2.1) of \cite{BSY}.

Recall that a holomorphic (resp. $\ci$) automorphic form of weight $(\rho, V)=(\rho_-\otimes\rho_+, V_-\otimes V_+)$ and level $\Gamma\subseteq GU(\eta_n)$ may be viewed as a holomorphic (resp. $\ci$) function
\begin{align}\nonumber
f: &\hn\rightarrow V =V_-\otimes V_+\nonumber
\end{align}
 (where $V_{\pm}$ are complex vector spaces) that satisfies
 \begin{align}\label{modcond}
 f((A\tau+B)(C\tau+D)^{-1}) = \rho(C\tau+D)f(\tau)
 \end{align}
for all
\begin{align}\nonumber
\gamma=  \left( \begin{array}{cc}
A&B\\
C&D
\end{array}\right)\in \Gamma.
\end{align}

Let $D_{\rho}$ be Shimura's differential operator discussed in the introduction (cf. section 12.1 of \cite{shar}).  In Proposition \ref{rho0430}, we will show that our $\ci$-differential operator $\ciop$ is the {\it same} as Shimura's $\ci$-differential operator $D^d_{\rho}$.

\begin{prop}\label{rho0430}
Let $f: \hn\rightarrow V = V_-\otimes V_+$ be a $\ci$-function.  Let $\lambda\in\EE$.  Then
\begin{align}\label{frho}
\ciopo\left(\lambda(f)\right)= \lambda(D_{\rho} f).
\end{align}
\end{prop}

(In Equation \eqref{frho}, $\lambda$ refers to the induced map from $V_{\pm}$ to $\uo_{\pm}^{\rho_{\pm}}$.)

\begin{proof}

Define
\begin{align*}
v^{\pm}_{\lambda_{\pm}} = \lambda^{-1}(du^{\pm}_{\lambda_{\pm}})
\end{align*}
for each tuple $\lambda_{\pm}$.
Writing $f$ in terms of the basis $v_{\lambda_{\pm}}$ for $V^{\pm}$, we have
\begin{align}\label{sumoverl}
f = \sum_{\lambda_-, \lambda_+} f_{\lambda_-,\lambda_+}\lambda(v^-_{\lambda_-}\otimes v^+_{\lambda_+}),
\end{align}
for some $\ci$ complex valued functions $f_{\lambda_-, \lambda_+}$.  The sum in Equation \eqref{sumoverl} is, as usual, over all tuples $\lambda_{\pm}$ so that $v_{\lambda_{\pm}}$ is in $V^{\pm}$.  Note that
\begin{align}
\ciopo(\lambda(f(z))) &= \ciopo\left(\sum_{\lambda_-,\lambda_+}f_{\lambda_-, \lambda_+}\lambda(v^-_{\lambda_-}\otimes v^+_{\lambda_+})\right)\label{regbar1}\\
&= \ciopo\left(\sum_{\lambda_-,\lambda_+}f_{\lambda_-, \lambda_+}(du^--d\bar{u}^-)_{\lambda_-}\otimes (du^+-d\bar{u}^+)_{\lambda_+}\right).\label{regbar2}
\end{align}
We get from \eqref{regbar1} to \eqref{regbar2} by recalling that $\ciopo(\bar{\uo})=0$, since $\bar{\uo}$ is holomorphically horizontal with respect to $\nabla$.

Applying Equations \eqref{vecdui1} and \eqref{vecdui2} to Equation \eqref{regbar2}, we see that
\begin{align*}
\ciopo&(\lambda(f(z)))\\
& = \ciopo\left( \rho(\Xi(z)) \left(\sum_{\lambda_-,\lambda_+}f_{\lambda_-, \lambda_+}(\beta+\bar{\alpha}\beta)_{\lambda_-}\otimes (\beta+\alpha\beta)_{\lambda_+}\right)\right).
\end{align*}

Now recall that the sections $\beta_i+\bar{\alpha}\beta_i$ and $\beta_i+\alpha\beta_i$ are horizontal for the Gauss-Manin connection.  So their image under $\ciopo$ is zero.

Therefore

\begin{small}
\begin{align*}
\ciopo(\lambda(f(z))) = D\left( \rho(\Xi(z)) \left(\sum_{\lambda_-,\lambda_+}f_{\lambda_-, \lambda_+}(\beta+\bar{\alpha}\beta)_{\lambda_-}\otimes (\beta+\alpha\beta)_{\lambda_+}\right)\right)\mod\splci,
\end{align*}
\end{small}

where $D$ is the map \eqref{defnD}.  Applying \eqref{vecdui1} and \eqref{vecdui2} again, we obtain

\begin{small}
\begin{align*}
\ciopo&(\lambda(f(z)))\\
&=\rho^{-1}(\Xi(z)) D\left( \rho\left(\Xi(z)\right) \left(\sum_{\lambda_-,\lambda_+}f_{\lambda_-, \lambda_+}(du^--d\bar{u}^-)_{\lambda_-}\otimes (du^+-d\bar{u}^+)_{\lambda_+}\right)\right)\mod\splci\\
&= \rho^{-1}(\Xi(z)) D\left( \rho\left(\Xi(z)\right) \left(\sum_{\lambda_-,\lambda_+}f_{\lambda_-, \lambda_+}(du^-)_{\lambda_-}\otimes (du^+)_{\lambda_+}\right)\right)\\
& = \rho^{-1}(\Xi(z)) D\left( \rho\left(\Xi(z)\right) \lambda(f)\right).
\end{align*}
\end{small}
\end{proof}

Let $Z$ be a $GL_n(\IC)\times GL_n(\IC)$-stable quotient of
\begin{align*}
\omci^\rho\otimes (\Omega_{\A/\moduli}(\ci))^{\bullet+d},
\end{align*}
 and let
\begin{align}\nonumber
\mathcal{Z} = H^0(\hn, Z).
\end{align}
Let $\phi_{\mathcal{Z}}$ denote projection onto $\mathcal{Z}$.

Recall that Shimura defines a differential operator $D_{\rho}^\mathcal{Z}$ (see e.g. \cite{shar}) by
\begin{align}\nonumber
D_{\rho}^\mathcal{Z} = \phi_{\mathcal{Z}}D^d.
\end{align}
So as a corollary of Proposition \ref{rho0430}, we obtain the following
\begin{cor}
\begin{align}\nonumber
D_{\rho}^\mathcal{Z} = \ciopo^Z.
\end{align}
\end{cor}

From the proof of the above propositions, we see that

\begin{small}
\begin{align}
\image(\ciop)\subset \mformsnost\otimes \sym^d(\Omega(\ci))\subset\mformsnost\otimes\Omega(\ci)^{\otimes d},\label{syminc0501}
\end{align}
\end{small}
where we associate $\sym^d(\Omega(\ci))$ with its image in $\Omega(\ci)^{\otimes d}$ as in \eqref{syminc}.  Note that we could have defined another $\ci$-differential operators with similar properties to those of $\ciop$, by replacing $D^{\rho}$ with $\tilde{D}^{\rho}$ in the construction of $\ciop$.  By \eqref{syminc0501}, we see that in fact, the differential operators constructed from $\tilde{D}^{\rho}$ are the {\it same} differential operators as $\ciop$.

We have seen above that our $\ci$-differential operators are the same as the $\ci$-differential operators Shimura defines in \cite{shar}.  For our later applications, as well as a reference for the reader who wishes to apply the differential operators, we now provide explicit formulas for the action of some of the differential operators.

\begin{lem}\label{formro}
Let
\begin{align*}
f = \sum_{1\leq \lambda_-, \lambda_+\leq n} f_{\lambda_-, \lambda_+}du^-_{\lambda_-}\otimes du^+_{\lambda_+}
\end{align*}
be an automorphic form of weight $\rhost\otimes \rhost$, viewed as an element of $\uo^-\otimes \uo^+$.  Then
\begin{align}\label{formurhost}
\ciopo(f) = \sum_{1\leq \lambda^+\leq n}M_f^{\lambda_+}\cdot \left( \begin{array}{cc}
du_1^-\\
\vdots\\
du_n^- \end{array}\right)\otimes du^+_{\lambda_+} + \sum_{1\leq \lambda^-\leq n} du_{\lambda_-}\otimes M_f^{\lambda_-}\cdot \left( \begin{array}{cc}
du_1^+\\
\vdots\\
du_n^+
\end{array}\right),
\end{align}
where $M_f^{\lambda_{\pm}}$ is the row vector (dependent only on $f$) defined by
\begin{align*}
\frac{1}{2}\left( df_{1, \lambda_{\pm}}, df_{2, \lambda_{\pm}}, \ldots, df_{n, \lambda_{\pm}} \right) + \left(f_{1, \lambda_{\pm}}, \ldots, f_{n, \lambda_{\pm}}\right)\cdot \left(dz^{\pm}\right)\lp \Xi^{\pm}\rp
\end{align*}
\end{lem}

\begin{proof}
Note that
\begin{align*}
\ciopo(f) = \sum_{1\leq \lambda_-, \lambda_+\leq n}\ciopo\lp f_{\lambda_-, \lambda_+}du^-_{\lambda_-}\otimes du^+_{\lambda_+}\rp.
\end{align*}
We have that
\begin{align*}
\ciopo\lp f_{\lambda_-, \lambda_+}du^-_{\lambda_-}\otimes du^+_{\lambda_+}\rp & = \frac{1}{2} df_{\lambda_-, \lambda_+} du_{\lambda_-}^-\otimes du_{\lambda_+}^++ du_{\lambda_-}^-\otimes \frac{1}{2}dfdu^+\\
+ f&\ciopo(du^-)\otimes du^+ + du^-\otimes f\ciopo(du^+)
\end{align*}
So now it suffices to recall $\ciopo(du_{\lambda_{\pm}})$; applying \eqref{vecdui1} and \eqref{vecdui2}, we obtain \eqref{formurhost}.

One can alternatively derive the formula from Proposition \ref{rho0430}.
\end{proof}

Similarly, we can provide formulae for $\ciopo$ for representations other than $\rhost$.  For now, we conclude with the formulas for $\rho = \sym^{m_-}\otimes \sym^{m_+}$ and $\rho = \det^{m_-}\otimes \det^{m_+}$.

Using the above discussion, we see that if
\begin{align}\label{detform}
f = \tilde {f} (\wedge^n_{du_i^-})^{m_-}\otimes (\wedge^n{du_i^+})^{m_+}
\end{align}
is an automorphic form of weight $\rho = \rho_-\otimes \rho_+$ with $\rho_{\pm} = \det^{m_{\pm}}$, then
\begin{align*}
\ciopo(f) = df + \lp m^+\cdot \det(\Xi^+)^{-1}\cdot (dz)^++ m^-\cdot\det(\Xi^-)^{-1}(dz)^- \rp f
\end{align*}

In the following lemma, $\lambda^{\pm}$ denotes as usual an $n$-tuple of positive integers $(\lambda^{\pm}_1, \ldots, \lambda^{\pm}_n)$.

We have chosen these particular forms (\eqref{symform12} and \eqref{detform}) in which to express the action of $\ciopo$ for the determinant and symmetric representations because it helps illustrate the connection with Formula (2.1) for the operators on Siegel modular forms defined in \cite{BSY}.

\begin{lem}\label{symfor}
Let $f=$ be an automorphic form of weight $\rho = \rho^-\otimes \rho^+$, with
\begin{align*}
\rho^{\pm} = \sym^{m_{\pm}}
\end{align*}
for some integers $m_{\pm}$.  Viewing $f$ as an element $\sum_{\lambda_-, \lambda_+}f_{\lambda_-, \lambda_+}(du^-)^{\lambda_-}\otimes (du^+)^{\lambda_+}$ of $\sym^{m_-}(\uo^-)\otimes \sym^{m_+}(\uo^+)$, we have the following equality
\begin{align}\label{symform12}
&\ciopo \lp \sum_{\lambda_-, \lambda_+}f_{\lambda_-, \lambda_+}\lp du^-\rp^{\lambda_-}\otimes\lp du^+\rp^{\lambda_+} \rp = \\
&\sum_{\lambda_+}M_f^{\lambda_+}\cdot\left( \begin{array}{cc}
du_1^-\\
\vdots\\
du_n^- \end{array}\right)\otimes\lp du^+\rp^{\lambda_+}+\sum_{\lambda_+}\lp du^-\rp^{\lambda_-}\otimes M_f^{\lambda_-}\cdot \left( \begin{array}{cc}
du_1^+\\
\vdots\\
du_n^+ \end{array}\right),
\end{align}
where $M^{\lambda_\pm}$ is the row vector (dependent only on $f$) defined by

\begin{small}
\begin{align}\nonumber
M_f^{\lambda_{\pm}} =
&\frac{1}{2}\left( \sum_{\lambda_{\mp}} df_{\lambda_{\mp}, \lambda{\pm}}du_1^{\lambda^{\mp}_1-1}\prod_{j\neq 1}du_j^{\lambda^{\mp}_j}, \ldots, \sum_{\lambda_{\pm}}df_{\lambda_{\mp}, \lambda_{\pm}}du_n^{\lambda^{\mp}_n-1}\prod_{j\neq n}du_j^{\lambda^{\mp}_j} \right)\\
&+ \left( \sum_{\lambda_{\mp}}\lambda^{\mp}_1f_{\lambda_-, \lambda_+}(du_1^{\mp})^{\lambda_1^{\mp}-1}\prod_{j\neq 1}du_j^{\lambda^{\mp}_j}, \ldots, \sum_{\lambda_{\mp}}\lambda^{\mp}_nf_{\lambda_-, \lambda_+}(du_1^{\mp})^{\lambda_n^{\mp}-1}\prod_{j\neq n}du_j^{\lambda^{\mp}_j} \right)\cdot(dz^{\pm})(\Xi^{\pm})^{-1}\nonumber
\end{align}
\end{small}
\end{lem}
One sees that in the case $m_- = m+ = 1$, Lemma \ref{symfor} specializes to Lemma \ref{formro}.
\begin{proof}
The lemma follows by expressing the action of $\ciopo$ as follows.
\begin{align}\nonumber
\ciopo&\left(\sum_{\lambda_-, \lambda}f_{\lambda_-, \lambda_+}(du^-)^{\lambda_-}\otimes(du^+)^{\lambda_+}\right) =\\
\sum_{\lambda_-, \lambda_+} d&f_{\lambda_-, \lambda_+}(du^-)^{\lambda_-}\otimes(du^+)^{\lambda_+} + \sum_{\lambda_-, \lambda_+}f_{\lambda_-, \lambda_+}\ciopo\left((du^-)^{\lambda_-}\otimes(du^+)^{\lambda_+}\right).\nonumber
\end{align}
Now,
\begin{align*}
\ciopo\left((du^-)^{\lambda_-}\otimes (du^+)^{\lambda_+}\right)\\
= \ciopo\left( (du^-)^{\lambda_-} \right)&\otimes (du^+)^{\lambda_+}+\left(du^-\right)^{\lambda_-}\otimes\ciopo\left( (du^+)^{\lambda_+} \right),
\end{align*}
and
\begin{align}
\ciopo\left((du^{\pm})^{\lambda_{\pm}}\right) &= \ciopo\left( {du_1^{\pm}}^{\lambda_1^{\pm}}\cdots {du_n^{\pm}}^{\lambda_n^{\pm}} \right)\nonumber\\
& = \sum_{i=1}^n \left(\ciopo\left( (du_i^{\pm})^{\lambda_i^{\pm}}\right) \right)\cdot\prod_{j\neq i} {du_j^{\pm}}^{\lambda_j^{\pm}}\nonumber.
\end{align}
Note that
\begin{align*}
\ciopo\left( {du_i^{\pm}}^{\lambda_i^{\pm}} \right) = \lambda_i^{\pm}\left(du_i^{\pm})^{\lambda_i^{\pm}-1}\cdot(\ciopo\left(du_i^{\pm}\right)\right).
\end{align*}
So applying \eqref{vecdui1} and \eqref{vecdui2} and putting the above equations together, we obtain Equation \eqref{symform12}.
\end{proof}

The above formulas for the action of $\ciopo$ allow one to write down formulas for the action of $\ciop$ for $d>1$.

One can also similarly work out formulas for other representations, but the main ones of interest to us for our intended applications are powers of the determinant and symmetric product, which we just finished describing.

\begin{comment}
\begin{prop}\label{symim}
For each representation $\rho$,
\begin{align}
\image\lp(\ciop)\rvert_{\omci^{\rho}}\rp\subset \omci^{\rho}\otimes \sym^d(\Omega(\ci))\stackrel{\hookrightarrow}{\eqref{syminc}} \omci^{\rho}\otimes (\Omega(\ci))^{\otimes d}.\label{insym}
\end{align}
\end{prop}

\begin{proof}
By \eqref{afciopo} of Theorem \ref{impprop}, it suffices to prove that
\begin{align}
\image\lp(\ciopo)\rvert_{\omci^{\rho}\otimes \sym(\Omega(\ci))}\rp\subset \omci^{\rho}\otimes \sym(\Omega(\ci)).\label{onesuf}
\end{align}  Since $\ciopo$ is linear, it suffices to prove \eqref{onesuf} for elements in $\omci\otimes \sym^{\otimes e}(\ci)\subset \omci^{\rho}\otimes \sym(\Omega(\ci))$ for each non-negative integer $e$.  Let $e$ be a nonnegative integer, and let $w\in \omci\otimes \Omega^{\otimes e}(\ci)$.  Locally, we can write $w$ as an $\mathcal{O}_{\moduli}$-linear sum of horizontal elements
\end{proof}

Thus, by Proposition \ref{symim}, the inclusion \eqref{syminc} allows us to view the maps
\begin{align*}
\ciop: \omci^{\rho}\otimes T^\bullet(\Omega(\ci))\rightarrow \omci^{\rho}\otimes T^\bullet(\Omega(\ci))
\end{align*}
as maps
\begin{align*}
\omci^\rho\otimes T^\bullet(\Omega(\ci))\rightarrow \omci^{\rho}\otimes \sym^\bullet(\Omega(\ci))
\end{align*}
\end{comment}

\setcounter{equation}{0}


TITLE: Question about p-value for one-sided hypothesis testing
QUESTION [1 upvotes]: Is it true that the p-value is symmetric?
Let's say for a null hypothesis $H_{0}$: p = 0.75. Given that the z-score test statistics equals 1.44. Using the R command, I found out that
pnorm(1.44, lower=F)
## [1] 0.0749337

I want to test $H_{\alpha}: p > 0.75$. I concluded that p-value is equal to 0.0749337 because by definition, assume H0 is true, then the p-value is the probability that the test statistic z takes a value (in support of $H_{\alpha}$) as or more extreme than the one we observed. For a two-sided hypothesis testing, $H_{0}: p \neq 0.75$, the p-value would be 2*0.0749337.
Now, my quesiton is: can I claim that the p-value for $H_{\alpha}: p < 0.75$ is also 0.0749337?
Thank you!

REPLY [1 votes]: If you are doing an approximate normal test of $H_0: p = 3/4$ vs $H_a: p \ne 3/4,$ rejecting when $|Z_{\text{obs}}| > 1.96,$ where
$$Z_{\text{obs}} = \frac{\hat p = 3/4}{\sqrt{\frac{(3/4)[1-(3/4)]}{n}}},$$
then the two-sided P-value is $P(|Z| > Z_{\text{obs}}),$ where $Z \sim \mathsf{Norm}(0,1).$
So if $Z_{\text{obs}} = 1.44,$ then the P-value is $P(Z \le -1.44) + P(Z\ge 1.44) = 2P(Z \ge 1.44) \approx 0.15.$
However, if you do an exact 2-sided binomial test, then the null distribution of
the number $X$ of successes is $\mathsf{Binom(n, 3/4)},$ which is not a
symmetrical distribution. Opinions vary how to find the P-value in that case. Some authors and software packages interpret "more extreme" than $X_{\text{obs}}$ to mean farther in absolute value from the mean $np_0 = n(3/4)$
than $X_{\text{obs}}$ in either direction. Then finding the P-value may involve
adding two somewhat different tail probabilities, and I suppose this would not match your idea of 'symmetry'. Some simply double the probability from the 'observed tail`.
Note: I can't illustrate the exact binomial test for your case, because you have
not given the values of $X_{\text{obs}}$ and $n.$ However, since you have R,
you can look at the help screen at ? binom.test where the example shows
the same null and alternative hypotheses as yours with $X = 682$ and $n = 925.$
The P-value for the exact binomial test is given as 0.3825. With the
approximate normal test, I obtain a P-value $\approx 2(.186) = 0.372.$


TITLE: Radio vs optical telescope imaging
QUESTION [4 upvotes]: As I understand, the visible light from an optical telescope is focused on a sensor which correlates light exposure to an electrical voltage, which is then converted to an image.
A single antenna radio telescope's signal is focused on some sort a sensor but this signal represents only a single 'pixel.' To generate a composite image of an object, a time intensive scan pattern needs to be carried out.
Why does this difference exist? Why does a radio telescope see a single pixel, while optical telescopes see an entire image? Are radio telescope sensors only a single pixel, or is a radio telescope itself, regardless of the sensor, only capable of seeing a single pixel?

REPLY [2 votes]: It's all because of the wavelength of light.
In most bands the radio telescope is about the size of the wavelength it's observing - so it can only see a single point at once anyway. It would be like having an optical telescope that was a tiny microscopic pinhole - there wouldn't be much point in having a megapixel camera behind it.  Radio telescopes that work with mm wavelength band now have multi-pixel detectors.
To get comparable resolution to an optical telescope with wavelengths that are a million times longer we would need a telescope a million times larger. So we link radio telescopes together - to make a single telescope. Each telescope measures a single part of the incoming signal and we re-create the detailed picture later in a computer.


TITLE: what is a complete undergraduate level textbook for algebra.
QUESTION [0 upvotes]: i need a fairly elementary text(s) that includes all the topics that an undergraduate needs to know before going for a course.
i have : "higher algebra",by hall & knight.
         "higher algebra",by barnard & child.

REPLY [0 votes]: Well with all due respect to that book, I think Hall and Knight's Higher Algebra is rather elementary with respect to other books.
I would definitely recommend that you try "Mathematical Circles(Russian Experience)". I think this book has lots of interesting question. Check the e-book before you buy the actual book because everyone has a different taste, you may not like what I do....


TITLE: Is this known alternating sum for Euler's constant?
QUESTION [5 upvotes]: This probably is known, but Wolfram Alpha doesn't recognize it
and couldn't find it in Mathworld (there is something close,
but using floor).
We have
$\lim_{s \to 1} (\zeta(s)-1/(s-1)) = \gamma$
Also
$F(s) = \zeta(s) = \frac{1}{1-2^{1-s}}\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^s} $.
According to Maple 13:
$$\lim_{s \to 1} (F(s)-1/(s-1)) = \sum _{n=1}^{\infty }-{\frac { \left( -1 \right) ^{n-1}\ln  \left( n \right) }{n}} \left( \ln  \left( 2 \right)  \right) ^{-1}+1/2\,\ln \left( 2 \right) = \gamma \qquad (1) $$

Is (1) known and/or trivial?

I believe all terms and partial sums except the first of the sum are transcendental.

Intuitive explanation how (1) could be hypothetically rational?

Reference request? Was this known to Euler?

Numerically (1) is correct to precision at least $500$ decimal digits.
Sage code:
nsu=1/2*mpmath.log(2)-mpmath.nsum(lambda n:  (-1)**(n-1)*mpmath.log(n)/n ,[1, mpmath.inf])/ mpmath.log(2);nsu

REPLY [11 votes]: Denote
$$
f(s)=\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^s}=(1-2^{1-s})\zeta(s).
$$
Expanding into series and using $\zeta(s)=\frac1{s-1}+\gamma+O((s-1))$ leads to
$$
f(s)=\log 2+(s-1) \left(\gamma  \log 2-\frac{\log
   ^22}{2}\right)+O\left((s-1)^2\right).
$$
Differentiating both sides gives
$$
f'(1)=
-\sum _{n=1}^{\infty }{\frac { \left( -1 \right) ^{n-1}\ln  \left( n \right) }{n}}=
\gamma  \log 2-\frac{\log
   ^22}{2}.
$$


TITLE: Singularity or not
QUESTION [0 upvotes]: A point in which a function is not defined, for instance for $f(x)=\frac{1}{x-1}$ let consider $x=1$, it makes sense to know if it is a point of singularity? Or not since the study of derivability makes sense only in the domain?

REPLY [0 votes]: Well, you need to define what you mean by "singularity". People may call a point outside of the domain of a function "singular". Sometimes, a point where the function is not differentiable is also called "singular".
In complex analysis, there are several kinds of "singularities". For instance, near $z=0$, the functions $f(z)=\frac1z$ and  $g(z)=\frac{\sin z}{z}$ behave quite differently.


TITLE: Calculate volume of sphere cap
QUESTION [3 upvotes]: I found few equations here: http://mathworld.wolfram.com/SphericalCap.html
Developing one ($V_{\text{cap}}=\frac{1}{3}\pi h^{2}(3R-h)$) and considering that $R$ in my case is equals to $1$ and $V_{\text{cap}}$ is $0.4$ I can't figure out $h$.
I am stuck here: $0.4 = h^2 (3-h)$
Tried also to use various equation solvers but they all return 0.
Thanks a lot!
P.S.: I hope this doesn't sound as silly as it feels.

REPLY [1 votes]: If I interpreted your question correctly then you simply do as Américo Tavares does and substitute in the values. Given that $V_{cap}=0.4$ and $R=1$ you get:
$$V_{cap}=\frac{1}{3}\pi h^2(3R-h)\Longrightarrow0.4=\frac{1}{3}\pi h^2(3-h)$$
Simplifying gives:
$$0=5\pi h^3-15h^2+6$$
However, unfortunately it seems as though the zero's of the cubic seem to have rather nasty closed forms which can be found here.
An alternate method of finding the volume of a cap can be used using a little bit of calculus.
Say we have a circle with radius $r$ centered at $r$. Then we get a circle with equation $(x-r)^2+y^2=r^2$.
Here is an example with radius 3 with center $(3,0)$

We can simplify the general equation and get $y^2=r^2-(x^2-2xr+r^2)=2xr-x^2$. Since the volume of revolution of a circle is a sphere, the function for its volume is $$\int\pi y^2dx=\int \pi(2xr-x^2)dx$$
Now, if you want to find the volume of a cap with height $x$ you get:
$$V_{cap}(x)=\int_0^x\pi(2tr-t^2)dt=\pi r x^2-\frac{\pi x^3}{3}$$
Edit: My "alternate method" is in fact the exact same method as that givien on the link you provided, just with a slightly different proof. I lieu of this, please consider only the first part of the answer.


TITLE: Characteristic Polynomial Computation
QUESTION [0 upvotes]: Let $A_{0}=k[x,y]$, $\mathfrak{m}=(x,y)$. Let $A=(A_{0})_{\mathfrak{m}}$. We wish to compute the characteristic polynomial, $\chi_{\mathfrak{q}}$, of the $\mathfrak{m}$-primary ideals (i) $(x,y)$, (ii) $(x,y^2)$, (iii) $(x,y)^2$ and finally check that the degrees of each polynomial are equal.
I am only really familiar with char. poly. in linear algebraic terms, i.e., Jordan canonical form computations. I read through Atiyah-Macdonald, but I did not find it terribly helpful in direct computation. How might I do this problem? Any help would be appreciated!

REPLY [1 votes]: Well, it seems that $\chi_{\mathfrak q}(n)=l(A/\mathfrak q^n)$ for $n\gg0$.
In general, it is not an easy task to compute these lengths! That's why the theory reduces the finding of the degree from the general case to the most simple one when $\mathfrak q=\mathfrak m$. (However I won't exclude the possibility to perform such computation in your particular cases.) In your case this is easy to find since you know that $\mathfrak m^n/\mathfrak m^{n+1}$ is a $A/\mathfrak m\cong k$-vector space having a basis with $n+1$ elements, so $\chi_{\mathfrak m}(n)=n(n+1)/2$ is a polynomial of degree $2$ for $n\gg 0$. (But, of course, you know this from Krull-Chevalley-Samuel theorem which says that the degree of the characteristic polynomial equals the Krull dimension of $A$.)


TITLE: Non strictly singular operators
QUESTION [5 upvotes]: Let $X$ be a separable Banach space and let $T:X\to X$ be a bounded operator that is not strictly singular. Can we always find an infinite dimensional, closed, and complemented subspace $Y$ of $X$ such that $T:Y\to T(Y)$ is an isomorphism?
I cannot show this is true, and I cannot think of a counterexample. The only idea of counterexample I had was looking at HI spaces, where every operator is of the form $\lambda I+S$, where $S$ is strictly singular. However, as far as I know, such operators (when $\lambda\neq 0$) are isomorphisms on a closed, finite-codimensional subspace, thus complemented.

REPLY [1 votes]: This was answered in the negative in this post at MathOverflow.
See the first answer there and Bill Johnson's comment to the second answer.


TITLE: What makes a nucleus unstable?
QUESTION [4 upvotes]: My question is simply that - what makes a nucleus unstable? What exactly causes a nucleus to start breaking apart in the first place? Is it the Coulomb force between the neighboring protons?
I'm just curious.

REPLY [8 votes]: It is instructive to look at chart  of isotopes ,

number of neutrons on the x axis and protons on the y. The stable (black) isotopes diverge from the diagonal, more neutrons are needed to neutralize the coulomb repulsion of the protons, for stability.
The main forces are the coulomb force (repulsive) and the strong force (attractive) , but the specific energy levels will depend on quantum numbers . If there are too many neutrons, there exists a probability for the neutron decay to express itself , if too many protons a positron emission will bring  the balance.
Instability means that there are lower energy levels in the collective nuclear potential of the specific number of protons and neutrons to which the system will stabilize. The decay time can take from centuries to nanoseconds depending on the probability of decay for the given system. A number of models have been developed to explain the specific behavior.


TITLE: Derivative of Integral with variable bounds
QUESTION [5 upvotes]: I was testing my calculus knowledge when I found an example final exam from UCIrvine:
http://www.math.uci.edu/sites/math.uci.edu/files/2B_final_samp1.pdf
Number 2.) a.) asks to evaluate:
$$ {d\over dx}\int_{sin(x)}^{x^2}t^3tan(t)dt $$
This is what I did:
$$ {d\over dx}\int_{sin(x)}^{x^2}t^3tan(t)dt $$
$$ {d\over dx}\left(\int_{sin(x)}^{a}t^3tan(t)dt+\int_{a}^{x^2}t^3tan(t)dt\right) $$
$$ {d\over dx}\left(-\int_{a}^{sin(x)}t^3tan(t)dt+\int_{a}^{x^2}t^3tan(t)dt\right) $$
$$ -{d\over dx}\int_{a}^{sin(x)}t^3tan(t)dt+{d\over dx}\int_{a}^{x^2}t^3tan(t)dt $$
$$ -sin^3(x)tan(sin(x))cos(x)+{x^2}^3tan(x^2)2x $$
$$ 2x^7tan(x^2)-sin^3(x)cos(x)tan(sin(x)) $$
Is this correct?

REPLY [14 votes]: Yes is correct, remember that $$\frac{d}{dx}\int_{g(x)}^{f(x)}h(t)\,dt=h(f(x))\cdot f'(x)-h(g(x))\cdot g'(x) $$
this is by the second theorem of calculus and by chain rule.


TITLE: Prove that a group of order 30 has at least three different normal subgroups
QUESTION [4 upvotes]: Prove that a group of order 30 has at least three different normal subgroups.
Prove:
$30=2\cdot3\cdot5$
There are $2$-Sylow, $3$-Sylow and $5$-Sylow subgroups. If $t_p$= number of $p$-Sylow-subgroups. Then $t_2$=$1$, $3$, $5$, $15$ and $t_3$=$1$, $10$ and $t_5$=$1$, 6.
Therefore I can claim that the group is not simple. So $t_3$=1 or $t_5$=1. Which I can prove. So now we know that the group has another non-trivial normal subgroup. But this is just one normal subgroup. How can I show that there are (at least) three different normal subgroups ?

REPLY [4 votes]: Answer with hints that must be checked and solved:
As already proved, a group $\;G\;$ of order $\;30\;$ must have either one unique subgroup of order five or one unique subgroup of order three.
Take now the unique such subgroup, say $\;N\;$ , and one of the (possibly several) groups of the other order, say $\;H\;$ . Since clearly $\;N\cap H=1\;$ , we get that $\;NH\;$ is a subgroup (why?) of order $\;15\;$ and thus $\;NH\lhd G\;$ (why?).
But a group of order $\;15\;$ is necessarily cyclic (why?), and since any subgroup of a normal cyclic subgroup of a big group is normal in the big group (why?), we get that both $\;N,H\lhd G\;$


TITLE: Solve $n < e^{6 \sqrt{n}}$
QUESTION [3 upvotes]: Find for which values of $n \in \mathbb{N}$ it holds that $$n < e^{6 \sqrt{n}}.$$

I tried to use the inequality $(1 + x) \leq e^x$, but from this, I can only find that the inequality holds for $n > 36$. But I need to get $n$ as small as possible.
I also tried the induction on $n$, but I stucked in the induction step. In particular, in showing that $e^{6\sqrt{n}} + 1 \leq e^{6\sqrt{n+1}}$.
I appreciate any help and suggestions.

REPLY [3 votes]: We have $e^{3x} > 3x \ge x$ for $x\ge 0$. Apply this to $x = \sqrt{n}$ to get $e^{3\sqrt{n}} > \sqrt{n}$. Square both sides and you have $e^{6\sqrt{n}} > n$ for all $n\ge 0$.

REPLY [0 votes]: The inequality holds for all positive $x$, so effectively that a proof seems overkill :-)
(Notice that $0<e^{6\sqrt0}$ and $1<\dfrac6{2\sqrt x}e^{6\sqrt x}$.)

Or using Taylor,
$$n<1+6\sqrt n+18n+\cdots$$