\documentclass{homework}
\course{Math 5522H}
\author{Alex Li}
\input{preamble}

\usepackage[symbol]{footmisc}
\renewcommand{\thefootnote}{\fnsymbol{footnote}}

\begin{document}
\maketitle

\begin{inspiration}
  Wenn ich nur erst die S\"atze habe! Die Beweise werde ich schon
    finden.\footnote{``If only I first had the theorems, then I'd find
        the proofs somehow.''  Riemann knew all too well the challenge of
            PODASIPs.} \byline{Bernhard Riemann}
            \end{inspiration}

            \section{Terminology}

            \begin{problem}
              When $\Real s > 0$, what is the Gamma function $\Gamma(s)$?
              \end{problem}
              \begin{solution}
              \[
              \Gamma(s): \C^+ \to \C = \int_0^\oo x^{s-1}e^{-x}dx
              \]
              \end{solution}
              \begin{problem}
                When $\Real s > 1$, what is the Riemann zeta function $\zeta(s)$?
                \end{problem}
                \begin{solution}
                \[
                \zeta(s) = \sum_{k=1}^\infty k^{-s}
                \]
                \end{solution}
                \begin{problem}
                  What is the von Mangoldt function $\Lambda(n)$?
                  \end{problem}
                  \begin{solution}
                  \[
                  \Lambda(n) = \begin{cases}
                  \log(p) & n = p^k \text{ for some prime $p$ and $k\in \N$}\\
                  0 & \text{else}
                  \end{cases}
                  \]
                  \end{solution}
                  \section{Numericals}

                  \begin{problem}
                    Compute $\Res(\Gamma,s)$ for $s \in \{ 0, -1, -2, \ldots \}$.
                    \end{problem}
                    \begin{solution}
                    The subsequent problem shows that for $f= \Gamma(s)\Gamma(1-s), k\in \Z, \Res(f, k)= (-1)^k$. At each of these poles, $\Gamma(1-s)$ is holomorphic, so
                    \[
                    \Res(f, s) = \Gamma(1-s)\Res(\Gamma, s) = (-s)!\Res(\Gamma, s).
                    \]
                    And therefore
                    \[
                    \frac{(-1)^s}{(-s)!} = \Res(\Gamma, s).
                    \]
                    \end{solution}
                    \begin{problem}Evaluate $\Res(f,s)$ for the
                      function $f(s) = \Gamma(s) \Gamma(1-s)$.
                      \end{problem}
                      \begin{solution}
                      Considering the definition of the analytic extension of $\Gamma$, one can observe that the poles are at every integer.
                      First let's compute the residue at $s=0$:
                      \[
                      \lim_{s\to 0} s \Gamma(s) \Gamma(1-s) = \lim_{s\to 0} \Gamma(s+1)\Gamma(1) = \Gamma(1)\Gamma(1) = 1
                      \]
                      Then note the following periodicity property of $f$:
                      \[
                      f(s) = \Gamma(1-s)\Gamma(s) = -s\Gamma(-s) \cdot \frac{\Gamma(s+1)}{s} = -f(s+1)
                      \]
                      So the residue at $s \in \Z$ is $(-1)^s$.
                      \end{solution}
                      \begin{problem} % be careful to get the sign right!
                        Use \ref{euler-reflection-formula} to compute $\Gamma(1/2)$.
                        \end{problem}
                        \begin{solution}
                        The referenced problem shows that $\Gamma(1-s)\Gamma(s) = \frac{\pi}{\sin(\pi s)}$. At $s=\frac{1}{2}$,
                        \begin{align*}
                        \Gamma(.5)^2&= \frac{\pi}{1}\\
                        \Gamma(.5)&= \pm \sqrt{\pi}
                        \end{align*}
                        Since $\Gamma(.5) =\int_0^\infty x^{-.5}e^{-x} dx > 0$, $\Gamma(.5) = \sqrt{\pi}$

                        \end{solution}
                        \section{Exploration}

                        \begin{problem}
                          In previous courses, you have seen that
                            $\Gamma(s+1) = s \cdot \Gamma(s)$.  Use this fact to explain how to
                              define a meromorphic function on $\C$ agreeing with $\Gamma(s)$ for
                                $\Real s > 0$.  This is \textbf{analytic continuation} which succeeds
                                  here by using a \textbf{functional equation}.
                                  \end{problem}
                                  \begin{solution}
                                  We can define an extension named $\hat\Gamma$ as follows
                                  \[
                                  \hat\Gamma(s) = \begin{cases}
                                  \Gamma(s) &\Real(s) > 0\\
                                  (s+1)\hat \Gamma(s+1) &\Real(s) \leq 0\\
                                  \end{cases}
                                  \]
                                  We show this is meromorphic by induction. Assume $\hat\Gamma$ is meromorphic with poles at the negative integers for $s>k$ for $k\in \N\cup \{0\}$. This is true if $k=0$ since $\Gamma$ is meromorphic.

                                  Now we will prove $\hat\Gamma$ is meromorphic if $s>k-1$. $\hat\Gamma$ is holomorphic when $\Real(s)$ is not a nonnegative integer because it is defined to be $(s+1)\hat\Gamma(s+1)$, the product of two functions holomorphic when $\Real(s)$ is not a nonnegative integer. Furthermore, on the line $s=k-1$, $(s+1)\hat\Gamma(s+1)= \hat\Gamma(s)$ by definition, so the functions agree where they are defined, and thus we can extend $\Gamma$ back using this formula, creating another pole at $s=k$. By induction, we can extend $\Gamma$ down the entire real line.

                                  \end{solution}
                                  \begin{problem}\label{start-of-reflection-formula}For $\lambda \in (0,1)$, evaluate
                                    \[
                                        \int_{0}^\infty \frac{u^{\lambda - 1}}{1 + u} \, du.
                                          \]
                                            \textit{Hint:} make the substitution $u = e^x$ and invoke \ref{integral-for-euler-reflection}.
                                            \end{problem}
                                            \begin{solution}

                                            \begin{align*}
                                            \int_0^\oo \frac{u^{\lambda - 1}\, du }{1 + u} &=
                                            \int_0^\oo \frac{u^{\lambda}\, du }{u(1 + u)}
                                            \end{align*}
                                            Let $u=e^x$ so that $du = e^x dx = u dx$ and thus $\frac{du}{u} = dx$.
                                            \begin{align*}
                                            \int_0^\oo \frac{u^{\lambda - 1}\, du }{1 + u} &= \int_{-\oo}^\oo \frac{e^{x\lambda}\, dx }{1 + e^x}
                                            \end{align*}
                                            In a previous problem, we evaluated this integral to be
                                            \[
                                            \frac{2\pi i e^{i\pi \lambda}}{1 - e^{2\pi i \lambda}} = \frac{\pi}{\sin(\pi \lambda)}
                                            \]
                                            \end{solution}

                                            \begin{problem}\label{integrate-gamma-one-minus-s}For $s \in (0,1)$ and positive $t \in \R$, show that
                                              \[
                                                  \Gamma(1 - s) = t \int_0^\infty (xt)^{-s} e^{-xt}\, dx.
                                                    \]
                                                    \end{problem}
                                                    \begin{solution}
                                                    \begin{align*}
                                                    \Gamma(s) &= \int_0^\infty u^{s-1} e^{-u}\, du\qquad\color{purple}\text{Definition of }\Gamma\\
                                                    \Gamma(1 - s) &= \int_0^\infty u^{-s} e^{-u}\, du
                                                    \end{align*}
                                                    Now take $tx = u$ so that $t\, dx = du$
                                                    \begin{align*}
                                                    \Gamma(1 - s) = t\int_0^\infty (tx)^{-s} e^{-tx}\, dx
                                                    \end{align*}

                                                    \end{solution}

                                                    \begin{problem}\label{euler-reflection-formula}Combine \ref{start-of-reflection-formula} and \ref{integrate-gamma-one-minus-s} to conclude that
                                                      \[
                                                          \Gamma(1-s) \, \Gamma(s) = \frac{\pi}{\sin \left( \pi s \right)}.
                                                            \]
                                                              This is \textbf{Euler's reflection formula}.
                                                              \end{problem}
                                                              \begin{solution}
                                                              Let's suppose that $0<s<1$.
                                                              Combining \ref{integrate-gamma-one-minus-s} with the definition of $\Gamma$,
                                                              \begin{align*}
                                                              \Gamma(1-s) \, \Gamma(s) &= \left(t\int_0^\oo (tx)^{-s} e^{-tx}\, dx\right)\left(\int_0^\oo y^{s-1}e^{-y} dy\right)\\
                                                              &= \int_0^\oo \int_0^\oo (tx)^{-s} e^{-tx} ty^{s-1}e^{-y} dy dx
                                                              \end{align*}
                                                              Let $t=y$
                                                              \begin{align*}
                                                              &= \int_0^\oo \int_0^\oo (yx)^{-s} e^{-yx} y^{s}e^{-y} dy dx\\
                                                              &= \int_0^\oo \int_0^\oo x^{-s} e^{-y(x+1)} dy dx\\
                                                              &= \int_0^\oo \left\{x^{-s} \frac{-e^{-y(x+1)}}{x+1}\right\}\bigg|_0^\oo dx\\
                                                              &= \int_0^\oo \frac{x^{-s}}{x+1} dx
                                                              \end{align*}
                                                              Now let $\lambda = \frac{1}{x}$, $d\lambda = -\frac{dx}{x^2}$ to see that
                                                              \begin{align*}
                                                              \int_0^\oo \frac{x^{-s}}{x+1} dx &=
                                                              \int_0^\oo \frac{x^{-s+2}}{x^2(x+1)} dx\\
                                                              &= \int_0^\oo \frac{\lambda^{s-2}}{\frac{1}{\lambda}+1} dx\\
                                                              &= \int_0^\oo \frac{\lambda^{s-1}}{1 + \lambda} dx
                                                              \end{align*}
                                                              By problem \ref{start-of-reflection-formula}, this is \[\frac{\pi}{\sin(\pi s)}\]

                                                              So the formulas are equal on the real line from $0$ to $1$. Noting that
                                                              \[
                                                              f(s) = \Gamma(1-s)\Gamma(s) = -s\Gamma(-s) \cdot \frac{\Gamma(s+1)}{s} = -f(s+1)
                                                              \]
                                                              and that
                                                              \[
                                                              \frac{\pi}{\sin(\pi s)} = -\frac{\pi}{\sin(\pi (s+1))} \], we can extend this to the whole real line.

                                                              Subtracting the two functions, we have a meromorphic function that is 0 at all nonintegral points on the real line. Taking the limit going towards an integral points, this implies that the difference between these functions is 0 at integral points on the real line, so the difference is in fact an entire function. Then the identity theorem implies that they must be the same.
                                                              \end{solution}
                                                              \begin{problem}
                                                                Show that the series
                                                                  \[\displaystyle\sum_{n=1}^\infty \displaystyle\frac {1}{n^{s}}\]
                                                                    converges to a holomorphic function when $\Real s > 1$.
                                                                    \end{problem}
                                                                    \begin{solution}
                                                                    \begin{align*}
                                                                    \abs{\sum_{n=1}^\infty \frac {1}{n^{s}}} &\leq \sum_{n=1}^\infty \frac {1}{n^{\Real(s)}} \\
                                                                    &\leq \int_1^\oo \frac{1}{n^{\Real(s)}}dn \leq -\frac{1}{\Real(-s)+1}n^{\Real(-s)+1}\bigg|_1^\oo \\
                                                                    &= \frac{1}{\Real(-s)+1}\left(1 - \oo^{\Real{(-s)+1)}}\right)
                                                                    \end{align*}
                                                                    Since $s>1$, this is finite, this converges. Since each term of the series is holomorphic, each partial sum is holomorphic, so the limit of the partial sums is also holomorphic.
                                                                    \end{solution}
                                                                    \begin{problem}\label{euler-product-formula}
                                                                    Recall that you defined
                                                                      \(\displaystyle\prod_{n=1}^\infty \left( 1 + a_n \right)\) in
                                                                        \ref{terminology-infinite-product}.  Show that when $\Real s > 1$, we
                                                                          have
                                                                            \[
                                                                                \frac{1}{\zeta(s)} = \displaystyle\prod_{n=1}^\infty \left( 1 - {p_n}^{-s} \right),
                                                                                \]
                                                                                where $p_n$ is the $n$th prime number.
                                                                                \end{problem}
                                                                                \begin{solution}
                                                                                Let
                                                                                \[
                                                                                P(s) = \prod_{p_n} 1 + p^{-s}_n + p^{-2s}_n + \dots = \prod_{p_n} \frac{1}{1-p^{-s}}
                                                                                \]
                                                                                We will show that $\zeta(s) = P(s)$. First, we show $\zeta(s) \leq P(s)$.
                                                                                \begin{align*}
                                                                                \sum_{n=1}^N n^{-s} \leq \prod_{p_n\leq N} 1 + p^{-s}_n + p^{-2s}_n + \dots
                                                                                \end{align*}
                                                                                because every term on the LHS appears in the RHS and all terms are positive, this inequality is true for all $N$, and taking the limit as $N\to\oo$, we deduce that $\zeta(s) \leq P(s)$.

                                                                                Next, we show that $\zeta(s)\geq P(s)$
                                                                                \begin{align*}
                                                                                \sum_{n=1}^{N!} n^{-s} \geq \prod_{p|N!}\sum_{p^k|N!} p^{-sk}
                                                                                \end{align*}
                                                                                Note that both the right and left sides are montonically increasing and bounded if we take the limit as $N\to\infty$. After rewriting the right hand side as
                                                                                \[
                                                                                \lim_{N\to\infty} \prod_{p|N!}\sum_{p^k|N!} p^{-sk} \geq  e^{\lim_{N\to\infty}\sum_{p|N!} \log(\sum_{p^k|N!} p^{-sk} )}
                                                                                \]
                                                                                since the total sum is bounded by $\log(\zeta(s))$ which converges when $Re(s)>1$, we can use Fubini's theorem to evaluate the inner sum first
                                                                                \[
                                                                                \zeta(s)\geq \lim_{N\to \infty} e^{\sum_{p|N!} \log(\frac{1}{1 - p^{-s}})} =  \lim_{N\to \infty} \sum_{p|N!} \frac{1}{1 - p^{-s}} = P(s)
                                                                                \]
                                                                                Thus $\zeta(s) = P(s)$. Inverting both sides gets the desired equality.
                                                                                \end{solution}
                                                                                \begin{problem}
                                                                                Use \ref{euler-product-formula} and \ref{nonzero-infinite-product}
                                                                                to show that if $\Real s > 1$ then $\zeta(s) \neq 0$.
                                                                                \end{problem}
                                                                                \begin{proof}
                                                                                At $\zeta(0)$ we expand the formula in $\ref{euler-product-formula}$ and see it suffices to show that
                                                                                \[
                                                                                \prod_{n=1}^\infty (1- p_n^{-s})
                                                                                \]
                                                                                is constant. Now,
                                                                                \[
                                                                                \sum_{n=1}^\oo p_n^{-s} \leq \sum_{n=1}^\oo \abs{n^{-s}} \leq \infty
                                                                                \]
                                                                                when the real part of $s$ is more than 1. By \ref{nonzero-infinite-product}, the product converges. Therefore $\zeta(s)\neq 0$.
                                                                                \end{proof}
                                                                                \begin{problem}
                                                                                Use \ref{euler-product-formula} to show that
                                                                                \[
                                                                                \Real \left( \frac{ \zeta'(a+bi) }{ \zeta(a+bi) } \right) = - \sum_{n=1}^\infty \frac{\Lambda(n) \, \cos \left( b \log n \right)}{n^{a}}.
                                                                                \]
                                                                                \end{problem}
                                                                                \begin{proof}
                                                                                Let's compute $\Real(\log\zeta(a+bi))$, taking the derivative will get us the LHS of the desired equation. Using \ref{euler-product-formula},
                                                                                \begin{align*}
                                                                                \Real(\log\zeta(a+bi))&= \Real(\log \prod_{n=1}^\infty \frac{1}{1-p_n^{-(a+bi)}})\\
                                                                                &= \Real(\sum_{n=1}^\infty \log \frac{1}{1-p_n^{-(a+bi)}})\\
                                                                                &= \Real(-\sum_{n=1}^\infty \log (1-p_n^{-(a+bi)}))\\
                                                                                &= \sum_{n=1}^\infty \sum_{m=1}^\oo \Real(\frac{p_n^{-m(a+bi)}}{m}) \quad \text{ power series of log}\\
                                                                                \end{align*}
                                                                                Since the double sum converges absolutely, we can change the order of summation. Note that the only coefficients of $n^{-s}$ which are nonzero correspond to prime powers, so this is
                                                                                \begin{align*}
                                                                                \Real(\log\zeta(a+bi)) &= \sum_{n=1}^\infty \phi(n)\Real(n^{-(a+bi)})
                                                                                \end{align*}
                                                                                Where
                                                                                \[
                                                                                \phi(n) = \begin{cases}
                                                                                \frac{1}{k} & n= p^k\\
                                                                                1 & n = 0
                                                                                \end{cases}
                                                                                \]
                                                                                We next take a derivative with respect to $a+bi$
                                                                                \[
                                                                                \Real(\frac{\zeta'(a+bi)}{\zeta(a+bi)}) = -\sum_{n=1}^\infty \phi(n)\log(n)\Real(n^{-a+bi})
                                                                                \]
                                                                                Next we can notice that
                                                                                \begin{align}\label{n_to_a+bi_expansion}
                                                                                \Real(n^{-(a+bi)}) = n^{-a}\Real(e^{-bi\log n}) = n^{-a}\cos(b\log n)
                                                                                \end{align}
                                                                                and also that when $\phi(n)\neq 0$ so $n$ is a prime power $n=p^k$,
                                                                                \[
                                                                                \phi(p^k)\log(p^k) = \frac{1}{k}k\log(p) = \log(p) = \Lambda(n)
                                                                                \]
                                                                                With this, we can simplify to the given expression:
                                                                                \begin{align*}
                                                                                \Real(\frac{\zeta'(a+bi)}{\zeta(a+bi)}) &= -\sum_{n=1}^\infty \Lambda(n)\frac{\cos(b\log n)}{n^{a}}
                                                                                \end{align*}
                                                                                \end{proof}
                                                                                \begin{problem}
                                                                                Use the trigonometric fact $3 + 4 \cos \theta + \cos (2\theta) \geq 0$ to show that
                                                                                \[
                                                                                \Real \left( 3 \frac{ \zeta'(a) }{ \zeta(a) } + 4 \frac{ \zeta'(a+bi) }{ \zeta(a+bi) } +  \frac{ \zeta'(a+2bi) }{ \zeta(a+2bi) } \right) \leq 0.
                                                                                \]
                                                                                Knowing $\zeta$ has a simple pole at $s = 1$ and assuming
                                                                                $\zeta(1+bi) = 0$ for $b \in \R$, define
                                                                                \[
                                                                                f(a) = \zeta(a)^3 \cdot \zeta(a+bi)^4 \cdot \zeta(a + 2bi)
                                                                                \]
                                                                                and study $f$ near $a = 1$ to uncover a contradiction.  You will
                                                                                have shown that $\zeta(s) \neq 0$ when $\Real s = 1$.
                                                                                \end{problem}
                                                                                \begin{solution}
                                                                                First we use equation \eqref{n_to_a+bi_expansion} to notice that
                                                                                \[
                                                                                \Real(\log\zeta(x+yi)) = \sum_{n=1}^\infty n^{-x}\cos(y\log n)
                                                                                \]
                                                                                Then we take derivatives to see that
                                                                                \begin{align*}
                                                                                \Real(\log\zeta'(x+yi)) = -\sum_{n=1}^\infty \log(n) n^{-x}\cos(y\log n)
                                                                                \end{align*}
                                                                                Now we can check the inequality by plugging this in three times
                                                                                \begin{align*}
                                                                                &\Real\left(3\log\left(\frac{\zeta'(a)}{\zeta(a)}\right) + 4\log\left(\frac{\zeta'(a+bi)}{\zeta(a+bi)}\right) + \log\left(\frac{\zeta'(a+2bi)}{\zeta(a+2bi)}\right)\right)\\
                                                                                &= -\sum_{n=1}^\infty \log(n) n^{-a}(3 + 4\cos(b\log n) + \cos(2b\log n))
                                                                                \end{align*}
                                                                                Using the trig identity with $\theta = b\log n$, each of these terms is positive so the RHS is negative.

                                                                                Next, we assume that $\zeta(1 + bi) = 0$, and consider
                                                                                \[
                                                                                f(a) = \zeta(a)^3\zeta(a+bi)^4 \zeta(a+2bi)
                                                                                \]
                                                                                Note that we have just shown that $\Real(\frac{f'(a)}{f(a)}) \leq 0.$

                                                                                Now, since $\zeta(1)$ is a pole of degree 1 $\zeta(1+bi)$ is a zero of degree 4, and $\zeta(1+2bi)$ is not a pole, $f(x)\to 0$ as $x\to 1$. Thus $\log(f(x))\to \infty$, and so the derivative of $\log(f(x))$ is positive, contradicting the fact that $\Real(\frac{f'(a)}{f(a)}) \leq 0.$ Thus $\zeta$ must not have any zeros on the line with real part equal to 1.
                                                                                \end{solution}
                                                                                \section{Prove or Disprove and Salvage if Possible}

                                                                                \begin{problem}\label{nonzero-infinite-product}
                                                                                For a sequence $a_n \neq 0$ of complex numbers, suppose
                                                                                $\displaystyle\sum_{n=1}^\infty |a_n|$ converges.  Then
                                                                                $\displaystyle\prod_{n=1}^\infty \left( 1 + a_n \right)$ converges
                                                                                to a nonzero quantity.
                                                                                \end{problem}
                                                                                \begin{solution}
                                                                                False. We should assume $a_n\neq -1$ instead of $a_n \neq 0.$

                                                                                Choose a branch of the log with a branch cut given by a line that intersects none of the $a_i$. This is always possible since the set of $a_i$ is countable, but the set of angles is uncountable. We will use this branch for the rest of the proof.
                                                                                \begin{lemma}\label{log_prod_convergence}
                                                                                If
                                                                                \[
                                                                                \sum_{n=1}^\infty \log(1 + a_n)
                                                                                \]
                                                                                converges to $S_1$, then
                                                                                \[
                                                                                \prod_{n=1}^\infty (1 + a_n)
                                                                                \]
                                                                                converges to $e^{S_1}$.

                                                                                \end{lemma}
                                                                                \begin{proof}
                                                                                If the sum converges to $S_1$, then $e^{S_1}$ is also finite, and
                                                                                \[
                                                                                \exp(\sum_{n=1}^\infty \log(1 + a_n)) = \prod_{n=1}^\infty (1 + a_n)
                                                                                \]
                                                                                \end{proof}
                                                                                Now suppose that $\sum_{n=1}^\infty \abs{a_n}$ converges, then there are only finitely many terms $a_n$ with $\abs{a_n}>\frac{1}{2}$, and for the rest,
                                                                                \[
                                                                                \abs{\log(1 + a_n)} = \abs{\sum_{k=1}^\infty \frac{(-1)^{k-1}a_n^k}{k}} = \abs{a_n + a_n\sum_{k=1}^\infty \frac{(-1)^{k-1}a_n^k}{k}}
                                                                                \]
                                                                                Since
                                                                                \[
                                                                                \abs{\sum_{k=1}^\infty \frac{(-1)^{k-1}a_n^k}{k}}\leq
                                                                                \abs{\sum_{k=1}^\infty a_n^k} = \abs{\frac{a_n}{1-a_n}} \leq 1,
                                                                                \]
                                                                                we can conclude that
                                                                                \[
                                                                                \abs{\log(1 + a_n)}\leq \abs{2a_n}
                                                                                \]
                                                                                And since $a_n$ converges, $\log(1 + a_n)$ does as well, so by lemma \ref{log_prod_convergence}, $\prod(1+a_n)$ converges to $e^{S_1}\neq 0$.

                                                                                \end{solution}

                                                                                \begin{problem} % missing pole at s = 1
                                                                                Define the \textbf{Dirichlet eta function} via
                                                                                \[
                                                                                \eta(s) := \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^s}.
                                                                                \]
                                                                                This series converges to a holomorphic function when $\Real s > 0$.
                                                                                \end{problem}
                                                                                \begin{solution}
                                                                                True. Let
                                                                                \begin{align*}
                                                                                \eta(s) &= \sum_{n=1}^\infty (2n)^{-s} - (2n + 1)^{-s} \\
                                                                                \end{align*}
                                                                                Note that for $\Real(s) > 0$, $f(n, s) = (2n)^{-s} - (2n + 1)^{-s}$ is positive and a decreasing function in $n$, since
                                                                                \[
                                                                                \pfrac{f}{n} = -s(f(n, s+1)) < 0
                                                                                \]
                                                                                Thus we see that
                                                                                \begin{align*}
                                                                                \abs{\eta(s)}    &\leq \abs{\int_1^\infty (2n)^{-s} - (2n + 1)^{-s} dn}\\
                                                                                &= \frac{1}{2}\abs{\int_2^\infty n^{-s} - (n + 1)^{-s} dn}\\
                                                                                &= \frac{1}{2}\abs{\frac{n^{-s}}{\log(n)} - \frac{(n + 1)^{-s}}{\log(n+1)} \bigg|_{n=2}^\infty}
                                                                                \end{align*}
                                                                                If $\Real(s)>0$, then the numerator will go to zero faster than the denominator, so the value at infinity is 0 and thus $\abs(\eta(s))$ is bounded, so the series converges. Also, since $\sum_{n=1}^N \frac{(-1)^n}{n^s}$ is holomorphic, the limit as $N\to \infty$ is also holomorphic.
                                                                                \end{solution}
                                                                                \begin{problem}
                                                                                The Riemann zeta and Dirichlet eta functions are related via
                                                                                \[
                                                                                \left( 1 - 2^{1-s} \right) \zeta(s) = \eta(s).
                                                                                \]
                                                                                \end{problem}
                                                                                \begin{solution}
                                                                                We need to be a little careful about the domain: $\zeta$ isn't even defined at $s=1$. The claim is true if we restrict to $\Re(s)>1$, since in this case all the infinite sums converge aboslutely and so we can rearrange terms and make things cancel nicely:
                                                                                \begin{align*}
                                                                                \left( 1 - 2^{1-s} \right)\zeta(s) &=    (\sum_{n=1}^\infty \frac{1}{n^s}) -  \sum_{n=1}^\infty \frac{2}{(2n)^s} \\
                                                                                &= \sum_{n=1}^\infty \frac{(-1)^n}{n^s} = \eta(s)
                                                                                \end{align*}
                                                                                \end{solution}
                                                                                \end{document}


\documentclass[gr-notes.tex]{subfiles}

\begin{document}

\setcounter{chapter}{4}

\chapter{Preface to Curvature}

\setcounter{section}{7}

\section{Exercises}

\textbf{1}

(a)
Repeat the argument leading to Equation 5.1, but this time assume that only a fraction $\epsilon < 1$ of the mass's kinetic energy is converted into a photon.

If only a fraction $\epsilon$ of the energy is converted into a photon, then it will start with an energy of $\epsilon (m + mgh + \order{\boldsymbol{v}^4}$, but once it reaches the top it should have an energy of $\epsilon m$, as it loses the component due to gravitational potential energy. Thus
%
\begin{displaymath}
  \frac{E'}{E} =
  \frac{\epsilon m}{\epsilon (m + mgh + \order{\boldsymbol{v}^4)}} =
  \frac{m}{m + mgh + \order{\boldsymbol{v}^4}} =
  1 - gh + \order{\boldsymbol{v}^4}
\end{displaymath}

(b)
Assume Equation 5.1 does not hold. Devise a perpetual motion device.

If we assume that the photon does not return to an energy $m$ once it reaches the top, but instead has an energy $m' > m$, then we could create the perpetual motion device shown in Figure \ref{fig:ch5-problem-1b}. A black box consumes the photon with energy $m'$, and splits it into a new object of mass $m$, and a photon of energy $m' - m$. The object repeats the action of the original falling mass, creating an infinite loop.

\begin{figure}[ht]
  \centering
  \includegraphics[width=0.5\textwidth]{img/ch5_problem_1b}
  \caption{Problem 1: Perpetual motion device.}
  \label{fig:ch5-problem-1b}
\end{figure}

\textbf{2}
Explain why a uniform gravitational field would not be able to create tides on Earth.

Tides depend on there being a gravitational field gradient. If the curvature closer to the source of the field (e.g. the Moon) is greater than it is further away, then the closer side will move towards the source more than the further side, thus creating tides. In the absense of such a gradient, there would be no difference in curvature between the two sides, and thus they would not stretch relative to each other.

\textbf{7}
Calculate the components of $\tensor{\Lambda}{^{\alpha'}_\beta}$ and $\tensor{\Lambda}{^\mu_{\nu'}}$ for transformations $(x, y) \leftrightarrow (r, \theta)$.
~
\begin{align*}
  \mqty( \Delta r \\ \Delta \theta ) &=
  \mqty( \pdv*{r}{x}      & \pdv*{r}{y} \\
         \pdv*{\theta}{x} & \pdv*{\theta}{y} )
  \mqty( \Delta x \\ \Delta y )
  &
  \mqty( \Delta x \\ \Delta y ) &=
  \mqty( \pdv*{x}{r} & \pdv*{x}{\theta} \\
         \pdv*{y}{r} & \pdv*{y}{\theta} )
  \mqty( \Delta r \\ \Delta \theta )
  \\ &=
  \mqty( x / \sqrt{x^2+y^2} & y / \sqrt{x^2 + y^2} \\
         -y / (x^2 + y^2) & x / (x^2 + y^2) )
  \mqty( \Delta x \\ \Delta y )
  &&=
  \mqty( \cos\theta & -r \sin\theta \\
         \sin\theta &  r \cos\theta )
  \mqty( \Delta r \\ \Delta \theta )
  \\ &=
  \mqty( \cos\theta & \sin\theta \\
         -(1/r) \sin\theta & (1/r) \cos\theta )
  \mqty( \Delta x \\ \Delta y )
  &&=
  \mqty( x / \sqrt{x^2 + y^2} & -y \\
         y / \sqrt{x^2 + y^2} &  x)
  \mqty( \Delta r \\ \Delta \theta )
\end{align*}
%
\begin{align*}
  \tensor{\Lambda}{^r_x} &= x / \sqrt{x^2 + y^2} = \cos\theta
  &
  \tensor{\Lambda}{^x_r} &= \cos\theta = x / \sqrt{x^2 + y^2}
  \\
  \tensor{\Lambda}{^r_y} &= y / \sqrt{x^2 + y^2} = \sin\theta
  &
  \tensor{\Lambda}{^y_r} &= \sin\theta = y / \sqrt{x^2 + y^2}
  \\
  \tensor{\Lambda}{^\theta_x} &= -y / (x^2 + y^2) = -(1/r) \sin\theta
  &
  \tensor{\Lambda}{^x_\theta} &= -r \sin\theta = -y
  \\
  \tensor{\Lambda}{^\theta_y} &= x / (x^2 + y^2) = (1/r) \cos\theta
  &
  \tensor{\Lambda}{^y_\theta} &= r \cos\theta = x
\end{align*}

\textbf{8}

(a)
$f \equiv x^2 + y^2 + 2xy$, $\vec{V} \underset{(x,y)}{\to} (x^2 + 3y, y^2 + 3x)$, $\vec{W} \underset{(r,\theta)}{\to} (1, 1)$. Express $f = f(r, \theta)$, and find the components of $\vec{V}$ and $\vec{W}$ in a polar basis, as functions of $r$ and $\theta$.

\begin{align*}
  f &= x^2 + y^2 + 2xy = (x + y)^2
  \\ &=
  (r \cos\theta + r \sin\theta)^2 =
  r^2 \sin^2\theta + r^2 \cos^2\theta + 2 r^2 \sin\theta \cos\theta
  \\ &=
  r^2 (1 + \sin(2 \theta))
  \\
%%%%%
  \vec{V} &\underset{(x,y)}{\to}
  \mqty( r^2 \cos^2\theta + 3 r \sin\theta \\
         r^2 \sin^2\theta + 3 r \cos\theta )
  \\
  \vec{V} &\underset{(r,\theta)}{\to}
  \mqty( \cos\theta & \sin\theta \\
         -(1/r) \sin\theta & (1/r) \cos\theta )
  \mqty( r^2 \cos^2\theta + 3 r \sin\theta \\
         r^2 \sin^2\theta + 3 r \cos\theta )
  \\ &\underset{(r,\theta)}{\to}
  \mqty(
    r^2 \cos^2\theta + 6 r \sin\theta \cos\theta + r^2 \sin^3\theta
    \\
    -r \cos^2\theta \sin\theta - 3 \sin^2\theta +
     r \sin^2\theta \cos\theta + 3 \cos^2\theta
  )
  \\ &\underset{(r,\theta)}{\to}
  \mqty(
    r^2 (\sin^3\theta + \cos^3\theta) + 6 r \sin\theta \cos\theta
    \\
    r \sin\theta \cos\theta (\sin\theta - \cos\theta) +
    3 (\cos^2\theta - \sin^2\theta)
  )
  \\ &\underset{(r,\theta)}{\to}
  \mqty(
    r^2 (\sin^3\theta + \cos^3\theta) + 3 r \sin(2\theta)
    \\
    (r/2) \sin(2\theta) (\sin\theta - \cos\theta) + 3 \cos(2\theta)
  )
  \\
%%%%%
  \vec{W} &\underset{(r,\theta)}{\to}
  \mqty( \cos\theta & \sin\theta \\
         -(1/r) \sin\theta & (1/r) \cos\theta )
  \mqty( 1 \\ 1 )
  \\ &\underset{(r,\theta)}{\to}
  \mqty(
    \cos\theta + \sin\theta
    \\
    (1/r) (\cos\theta - \sin\theta)
  )
\end{align*}

(b)
Express the components of $\tilde{\dd}f$ in $(x,y)$ and obtain them in $(r,\theta)$ by:

(i) using direct calculation in $(r, \theta)$:

\begin{displaymath}
  \tilde{\dd}f \underset{(r,\theta)}{\to}
  \qty( \pdv*{f}{r}, \pdv*{f}{\theta} ) =
  \qty( 2 r (1 + \sin(2 \theta)), 2 r^2 \cos(2 \theta) )
\end{displaymath}

(ii) transforming the components in $(x, y)$:

\begin{displaymath}
  \tilde{\dd}f \underset{(x,y)}{\to}
  \qty( \pdv*{f}{x}, \pdv*{f}{y} ) =
  \qty( 2 (x+y), 2 (x+y) ) =
  \qty( 2 r (\cos\theta + \sin\theta), 2 r (\cos\theta + \sin\theta))
\end{displaymath}

\begin{align*}
  \mqty( (\tilde{\dd}f)_r & (\tilde{\dd}f)_\theta ) &=
  \mqty( 1 & 1 )
  \mqty( \cos\theta & -r\sin\theta \\ \sin\theta & r\cos\theta )
  \qty[ 2 r (\cos\theta + \sin\theta) ]
  \\ &=
  \mqty( 2 r  (\cos^2\theta + \sin^2\theta + 2 \sin\theta\cos\theta) &
         2 r^2 (\cos^2\theta - \sin^2\theta) )
  \\ &=
  \mqty( 2 r (1 + \sin(2 \theta)) & 2 r^2 \cos(2 \theta) )
\end{align*}

(c)
Now find the $(r,\theta)$ components of the one-forms $\tilde{V}$ and $\tilde{W}$ associated with the vectors $\vec{V}$ and $\vec{W}$ by

(i)
using the metric tensor in $(r,\theta)$:

\begin{align*}
  V_r &=
  g_{r\alpha} V^\alpha =
  g_{rr} V^r + g_{r\theta} V^\theta
  \\ &=
  r^2 (\sin^3\theta + \cos^3\theta) + 3 r \sin(2\theta)
  \\
  V_\theta &=
  g_{\theta r} V^r + g_{\theta\theta} V^\theta =
  (1/2) r^3 \sin(2\theta) (\sin\theta - \cos\theta) +
  3 r^2 \cos(2\theta)
  \\
%%%%%
  W_r &=
  g_{r\alpha} W^\alpha =
  g_{rx} W^x + g_{ry} W^y
  \\ &=
  1 (\cos\theta + \sin\theta) +
  0 \qty[ (1/r) (\cos\theta - \sin\theta) ]
  \\ &=
  \cos\theta + \sin\theta
  \\
%%%%%
  W_\theta &=
  g_{\theta x} W^x + g_{\theta y} W^y =
  \\ &=
  0 (\cos\theta + \sin\theta) +
  r^2 \qty[ r (\cos\theta - \sin\theta) ]
  \\ &=
  r (\cos\theta - \sin\theta)
\end{align*}

(ii)
using the metric tensor in $(x,y)$ and then doing a coordinate transformation:

\begin{align*}
  V_x &= V^x; \quad V_y = V^y
  \\
  V_r &=
  \tensor{\Lambda}{^\alpha_r} V_\alpha =
  \tensor{\Lambda}{^x_r} V_x + \tensor{\Lambda}{^y_r} V_y
  \\ &=
  \cos\theta V_x + \sin\theta V_y
  \\ &=
  r^2 \cos^3\theta + (3/2) r \sin(2\theta) +
  r^2 \sin^3\theta + (3/2) r \sin(2\theta)
  \\ &=
  r^2 (\cos^3\theta + \sin^3\theta) + 3 r \sin(2\theta)
  \\
  V_\theta &=
  \tensor{\Lambda}{^\alpha_\theta} V_\alpha =
  \tensor{\Lambda}{^x_\theta} V_x + \tensor{\Lambda}{^y_\theta} V_y
  \\ &=
  (-r \sin\theta) V_x + (r \cos\theta) V_y
  \\ &=
  -r^3 \cos^2\theta \sin\theta - 3 r^2 \sin^2\theta +
   r^3 \sin^2\theta \cos\theta + 3 r^2 \cos^2\theta
  \\ &=
  r^3 \sin\theta \cos\theta (\sin\theta - \cos\theta) +
  3 r^2 (\cos^2\theta - \sin^2\theta)
  \\ &=
  (1/2) r^3 \sin(2\theta) (\sin\theta - \cos\theta) +
  3 r^2 \cos(2\theta)
  \\
%%%%%
  W_x &= W^x = W_y = W^y = 1
  \\
  W_r &=
  \tensor{\Lambda}{^\alpha_r} W_\alpha =
  \tensor{\Lambda}{^x_r} W_x + \tensor{\Lambda}{^y_r} W_y
  \\ &=
  \cos\theta + \sin\theta
  \\
  W_\theta &=
  \tensor{\Lambda}{^\alpha_\theta} W_\alpha =
  \tensor{\Lambda}{^x_\theta} W_x + \tensor{\Lambda}{^y_\theta} W_y
  \\ &=
  -r \sin\theta + r \cos\theta
  \\ &=
  r (\cos\theta - \sin\theta)
\end{align*}

\textbf{11}
Consider $V \underset{(x,y)}{\to} (x^2 + 3y, y^2 + 3x)$.

(a)
Find $\tensor{V}{^\alpha_{,\beta}}$ in Cartesian coordinates.
%
\begin{displaymath}
  \tensor{V}{^x_{,x}} = 2 x; \quad
  \tensor{V}{^y_{,y}} = 2 y; \quad
  \tensor{V}{^x_{,y}} = \tensor{V}{^y_{,x}} = 3.
\end{displaymath}

(b)
\begin{align*}
  \tensor{V}{^{\mu'}_{;\nu'}} &=
  \tensor{\Lambda}{^{\mu'}_\alpha}
  \tensor{\Lambda}{^\beta_{\nu'}}
  \tensor{V}{^\alpha_{,\beta}}
  \\
  \tensor{V}{^r_{;r}} &=
  \tensor{\Lambda}{^r_x} \tensor{\Lambda}{^x_r} \tensor{V}{^x_{,x}} +
  \tensor{\Lambda}{^r_y} \tensor{\Lambda}{^y_r} \tensor{V}{^y_{,y}} +
  \tensor{\Lambda}{^r_x} \tensor{\Lambda}{^y_r} \tensor{V}{^x_{,y}} +
  \tensor{\Lambda}{^r_y} \tensor{\Lambda}{^x_r} \tensor{V}{^y_{,x}}
  \\ &=
  (\cos^2\theta) (2r \cos\theta) +
  (\sin^2\theta) (2r \sin\theta) +
  (\sin\theta \cos\theta) (3) +
  (\sin\theta \cos\theta) (3)
  \\ &=
  2r (\cos^3\theta + \sin^3\theta) +
  3 \sin(2\theta)
  \\
  \tensor{V}{^\theta_{;\theta}} &=
  \tensor{\Lambda}{^\theta_x} \tensor{\Lambda}{^\theta_r} \tensor{V}{^x_{,x}} +
  \tensor{\Lambda}{^\theta_y} \tensor{\Lambda}{^\theta_r} \tensor{V}{^y_{,y}} +
  \tensor{\Lambda}{^\theta_x} \tensor{\Lambda}{^y_\theta} \tensor{V}{^x_{,y}} +
  \tensor{\Lambda}{^\theta_y} \tensor{\Lambda}{^\theta_r} \tensor{V}{^y_{,x}}
  \\ &=
  (\sin^2\theta) (2 r \cos\theta) + (\cos^2\theta) (2 r \sin\theta) +
  (-\sin\theta \cos\theta) (3) + (-\sin\theta \cos\theta) (3)
  \\ &=
  \sin(2\theta) [ r (\sin\theta + \cos\theta) - 3 ]
  \\
  \tensor{V}{^r_{;\theta}} &=
  \tensor{\Lambda}{^r_x} \tensor{\Lambda}{^x_\theta} \tensor{V}{^x_{,x}} +
  \tensor{\Lambda}{^r_y} \tensor{\Lambda}{^y_\theta} \tensor{V}{^y_{,y}} +
  \tensor{\Lambda}{^r_x} \tensor{\Lambda}{^y_\theta} \tensor{V}{^x_{,y}} +
  \tensor{\Lambda}{^r_y} \tensor{\Lambda}{^x_\theta} \tensor{V}{^y_{,x}}
  \\ &=
  (-r \sin\theta \cos\theta) (2 r \cos\theta) +
  (r \sin\theta \cos\theta) (2 r \sin\theta) +
  (r \cos^2\theta) (3) + (-r \sin^2\theta)
  \\ &=
  r^2 \sin(2\theta) (\sin\theta - \cos\theta) +
  3 r \cos(2 \theta)
  \\
  \tensor{V}{^\theta_{;r}} &=
  \tensor{\Lambda}{^\theta_x} \tensor{\Lambda}{^x_r} \tensor{V}{^x_{,x}} +
  \tensor{\Lambda}{^\theta_y} \tensor{\Lambda}{^y_r} \tensor{V}{^y_{,y}} +
  \tensor{\Lambda}{^\theta_x} \tensor{\Lambda}{^y_r} \tensor{V}{^x_{,y}} +
  \tensor{\Lambda}{^\theta_y} \tensor{\Lambda}{^x_r} \tensor{V}{^y_{,x}}
  \\ &=
  (-(1/r) \sin\theta \cos\theta) (2 r \cos\theta) +
  ( (1/r) \sin\theta \cos\theta) (2 r \sin\theta) +
  (-(1/r) \sin^2\theta) (3) +
  ( (1/r) \cos^2\theta) (3)
  \\ &=
  \sin(2\theta) (\sin\theta - \cos\theta) +
  \frac{3}{r} \cos(2 \theta)
\end{align*}

(c) compute $\tensor{V}{^{\mu'}_{;\nu'}}$ directly in polars using the Christoffel symbols.

Recall that we have $\tensor{\Gamma}{^\mu_{rr}} = \tensor{\Gamma}{^r_{r\theta}} = \tensor{\Gamma}{^\theta_{\theta\theta}} = 0$, $\tensor{\Gamma}{^\theta_{r\theta}} = 1/r$, and $\tensor{\Gamma}{^r_{\theta\theta}} = -r$.

\begin{align*}
  \tensor{V}{^{\mu'}_{;\nu'}} &=
  \tensor{V}{^{\mu'}_{,\nu'}} +
  \tensor{V}{^{\alpha'}} \tensor{\Gamma}{^{\mu'}_{\alpha'\nu'}}
  \\
  \tensor{V}{^r_{;r}} &=
  \tensor{V}{^r_{,r}} +
  \tensor{V}{^{\alpha'}} \tensor{\Gamma}{^r_{\alpha'r}}
  \\
  \tensor{V}{^r_{,r}} &=
  \pdv*{V^r}{r} =
  2 r (\sin^3\theta + \cos^3\theta) + 3 \sin(2\theta)
  \\
  \tensor{V}{^\alpha} \tensor{\Gamma}{^r_{\alpha r}} &=
  V^r     \tensor{\Gamma}{^r_{rr}} +
  V^\theta \tensor{\Gamma}{^r_{\theta r}} =
  0
  \\
  \tensor{V}{^r_{;r}} &=
  \tensor{V}{^r_{,r}} =
  2 r (\sin^3\theta + \cos^3\theta) + 3 \sin(2\theta)
  \\
  \tensor{V}{^\theta_{;\theta}} &=
  \tensor{V}{^\theta_{,\theta}} +
  \tensor{V}{^{\alpha'}} \tensor{\Gamma}{^\theta_{\alpha'\theta}}
  \\
  \tensor{V}{^\theta_{,\theta}} &=
  \pdv*{V^\theta}{\theta} =
  (r/2) \sin(2\theta) (\sin\theta + \cos\theta) +
  r \cos(2\theta) (\sin\theta - \cos\theta) -
  6 \sin(2\theta)
  \\
  \tensor{V}{^{\alpha'}} \tensor{\Gamma}{^\theta_{\alpha'\theta}} &=
  \tensor{V}{^r} \tensor{\Gamma}{^\theta_{r\theta}} +
  \tensor{V}{^\theta} \tensor{\Gamma}{^\theta_{\theta\theta}}
  \\ &=
  [r^2 (\sin^3\theta + \cos^3\theta) + 3r \sin(2\theta)] (1/r)
  \\ &=
  r (\sin^3\theta + \cos^3\theta) + 3 \sin(2\theta)
  \\
  \tensor{V}{^\theta_{;\theta}} &=
  \sin(2\theta) [r (\sin\theta + \cos\theta) - 3]
  \\
  \tensor{V}{^r_{;\theta}} &=
  \tensor{V}{^r_{,\theta}} +
  V^r \tensor{\Gamma}{^r_{r\theta}} +
  V^\theta \tensor{\Gamma}{^r_{\theta\theta}}
  \\ &=
  \tensor{V}{^r_{,\theta}} +
  V^\theta \tensor{\Gamma}{^r_{\theta\theta}} =
  \pdv*{V^r}{\theta} - r V^\theta
  \\ &=
  6 r \cos(2\theta) +
  (3/2) r^2 \sin(2\theta) (\sin\theta-\cos\theta) -
  ((1/2) r^2 \sin(2\theta) (\sin\theta - \cos\theta) + 3 r \cos(2\theta))
  \\ &=
  r^2 \sin(2\theta) (\sin\theta - \cos\theta) + 3 r \cos(2\theta)
  \\
  \tensor{V}{^\theta_{;r}} &=
  \tensor{V}{^\theta_{,r}} +
  V^r \tensor{\Gamma}{^\theta_{rr}} +
  V^\theta \tensor{\Gamma}{^\theta_{\theta r}} =
  \tensor{V}{^\theta_{,r}} +
  \frac{1}{r}
  V^\theta
  \\ &=
  (1/2) \sin(2\theta) (\sin\theta - \cos\theta) +
  (1/2) \sin(2\theta) (\sin\theta - \cos\theta) +
  (3/r) \cos(2\theta)
  \\ &=
  \sin(2\theta) (\sin\theta - \cos\theta) +
  (3/r) \cos(2\theta)
\end{align*}

(d)
Calculate the divergence using the results from part (a)
%
\begin{displaymath}
  \tensor{V}{^\alpha_{,\alpha}} =
  \tensor{V}{^x_{,x}} + \tensor{V}{^y_{,y}} =
  2 (x + y) =
  2 r (\sin\theta + \cos\theta)
\end{displaymath}

(e)
Calculate the divergence using the results from either part (b) or (c).
%
\begin{align*}
  \tensor{V}{^{\mu'}_{;\mu'}} &=
  \tensor{V}{^{r}_{;r}} +
  \tensor{V}{^{\theta}_{;\theta}}
  \\ &=
  2 r (\sin^3\theta + \cos^3\theta) +
  3 \sin(2\theta) +
  \sin(2\theta) [r (\sin\theta + \cos\theta) - 3]
  \\ &=
  2 r (\sin\theta + \cos\theta)
\end{align*}

(f)
Compute $\tensor{V}{^{\mu'}_{;\mu'}}$ using Equation 5.56.

\begin{displaymath}
  \tensor{V}{^{\mu'}_{;\mu'}} =
  \frac{1}{r} \pdv{r} (r V^r) + \pdv{\theta} (V^\theta) =
  2 r (\sin\theta + \cos\theta)
\end{displaymath}

\textbf{12}
%
\begin{displaymath}
  \tilde{p} \underset{(x,y)}{\to} (x^2 + 3y, y^2 + 3x).
\end{displaymath}

(a)
Find the components $p_{\alpha,\beta}$ in Cartesian coordinates.

Since $p_{\alpha,\beta} = \pdv*{p_\alpha}{x^\beta}$, it's simply $p_{x,x} = 2x$, $p_{y,y} = 2y$, and $p_{x,y} = p_{y,x} = 3$.

(b)
Find the components $p_{\mu';\nu'}$ in polar coordinates by using the transformation $\tensor{\Lambda}{^\alpha_{\mu'}} \tensor{\Lambda}{^\beta_{\nu'}} p_{\alpha,\beta}$.

\begin{align*}
  p_{r;r} &=
  \qty(\tensor{\Lambda}{^x_r})^2 p_{x,x} +
  \qty(\tensor{\Lambda}{^y_r})^2 p_{y,y} +
  2 \tensor{\Lambda}{^x_r} \tensor{\Lambda}{^y_r} p_{x,y}
  \\ &=
  (\cos^2\theta) (2 r \cos\theta) +
  (\sin^2\theta) (2 r \sin\theta) +
  2 (\sin\theta \cos\theta) (3)
  \\ &=
  2 r (\sin^3\theta + \cos^3\theta) + 3 \sin(2\theta)
  \\
  p_{\theta;\theta} &=
  \qty(\tensor{\Lambda}{^x_\theta})^2 p_{x,x} +
  \qty(\tensor{\Lambda}{^y_\theta})^2 p_{y,y} +
  2 \tensor{\Lambda}{^x_\theta} \tensor{\Lambda}{^y_\theta} p_{x,y}
  \\ &=
  (-r \sin\theta)^2 (2 r \cos\theta) +
  (r \cos\theta)^2 (2 r \sin\theta) +
  2 (3 (-r \sin\theta) (r \cos\theta))
  \\ &=
  r^2 \sin(2\theta) (r (\sin\theta + \cos\theta) - 3)
  \\
  p_{r;\theta} &=
  \tensor{\Lambda}{^x_r} \tensor{\Lambda}{^x_\theta} p_{x,x} +
  \tensor{\Lambda}{^y_r} \tensor{\Lambda}{^y_\theta} p_{y,y} +
  \tensor{\Lambda}{^x_r} \tensor{\Lambda}{^y_\theta} p_{x,y} +
  \tensor{\Lambda}{^y_r} \tensor{\Lambda}{^x_\theta} p_{y,x}
  \\ &=
  (-r \sin\theta \cos\theta) (2 r \cos\theta) +
  (r \sin\theta \cos\theta) (2 r \sin\theta) +
  3 (r \cos^2\theta - r \sin^2\theta)
  \\ &=
  r^2 \sin(2\theta) (\sin\theta - \cos\theta) + 3 r \cos(2\theta),
\end{align*}
and by the symmetry of $p_{\alpha,\beta}$ in Cartesian coordinates, $p_{\theta;r} = p_{r;\theta}$.

(c)
Now find $p_{\mu';\nu'}$ using the Christoffel symbols.

\begin{align*}
  p_{r;r} &=
  p_{r,r} -
  p_r \tensor{\Gamma}{^r_{rr}} -
  p_\theta \tensor{\Gamma}{^\theta_{rr}} =
  p_{r,r} = \pdv*{p_r}{r}
  \\ &=
  \pdv*{r} \qty[ r^2 (\cos^3\theta + \sin^3\theta) + 3 r \sin(2\theta) ] =
  2 r (sin^3\theta + \cos^3\theta) + 3 \sin(2\theta)
  \\
  p_{\theta;\theta} &=
  p_{\theta,\theta} -
  p_r \tensor{\Gamma}{^r_{\theta\theta}} -
  p_\theta \tensor{\Gamma}{^\theta_{\theta\theta}} =
  p_{\theta,\theta} + r p_r =
  \pdv*{p_\theta}{\theta}
  \\ &=
  \pdv*{\theta} \qty[
    (1/2) r^3 \sin(2\theta) (\sin\theta - \cos\theta) +
    3 r^2 \cos(2\theta)
  ] +
  r \qty[
    r^2 (\cos^3\theta + \sin^3\theta) + 3 r \sin(2\theta)
  ]
  \\ &=
  r^2 \sin(2\theta) \qty[ r (\sin\theta + \cos\theta) - 3 ]
  \\
  p_{r;\theta} &=
  p_{r,\theta} -
  p_r \tensor{\Gamma}{^r_{r\theta}} -
  p_\theta \tensor{\Gamma}{^\theta_{r\theta}} =
  \pdv*{p_r}{\theta} - (1/r) p_\theta
  \\ &=
  r^2 \sin(2\theta) (\sin\theta - \cos\theta) + 3 r \cos(2\theta)
  \\
  p_{\theta;r} &=
  p_{\theta,r} -
  p_r \tensor{\Gamma}{^r_{\theta r}} -
  p_\theta \tensor{\Gamma}{^\theta_{\theta r}} =
  \pdv*{p_\theta}{r} - (1/r) p_\theta
  \\ &=
  r^2 \sin(2\theta) (\sin\theta - \cos\theta) + 3 r \cos(2\theta)
\end{align*}

\textbf{13}
Show in polars that $g_{\mu'\alpha'} \tensor{V}{^{\alpha'}_{;\nu'}} = p_{\mu';\nu'}$.

\begin{align*}
  g_{r\alpha'} \tensor{V}{^{\alpha'}_{;r}} &=
  g_{rr} \tensor{V}{^{r}_{;r}} + g_{r\theta} \tensor{V}{^{\theta}_{;r}}
  \\ &=
  1 \tensor{V}{^r_{;r}} =
  p_{r;r}
  \\
  g_{\theta\alpha'} \tensor{V}{^{\alpha'}_{;\theta}} &=
  g_{\theta r} \tensor{V}{^r_{;\theta}} +
  g_{\theta\theta} \tensor{V}{^\theta_{;\theta}}
  \\ &=
  r^2 \tensor{V}{^\theta_{;\theta}} =
  p_{\theta;\theta}
  \\
  g_{r \alpha'} \tensor{V}{^{\alpha'}_{;\theta}} &=
  g_{rr} \tensor{V}{^r_{;\theta}} +
  g_{r\theta} \tensor{V}{^\theta_{;\theta}}
  \\ &=
  1 \tensor{V}{^r_{;\theta}} =
  p_{\theta;r}
  \\
  g_{\theta\alpha'} \tensor{V}{^{\alpha'}_{;r}} &=
  g_{\theta r} \tensor{V}{^r_{;r}} +
  g_{\theta\theta} \tensor{V}{^\theta_{;r}}
  \\ &=
  r^2 \tensor{V}{^\theta_{;r}} =
  p_{\theta;r}
\end{align*}

\textbf{14}
Compute $\grad_\beta A^{\mu\nu}$ for the tensor $\boldsymbol{A}$ with components:
%
\begin{align*}
  A^{rr} &= r^2, & A^{r\theta} &= r \sin\theta,
  \\
  A^{\theta\theta} &= \tan\theta, & A^{\theta r} &= r \cos\theta
\end{align*}
%
\begin{align*}
  \tensor{A}{^{rr}_{,r}} &= 2r &
  \tensor{A}{^{rr}_{,\theta}} &= 0
  \\
  \tensor{A}{^{\theta\theta}_{,r}} &= 0 &
  \tensor{A}{^{\theta\theta}_{,\theta}} &= \sec^2\theta
  \\
  \tensor{A}{^{r\theta}_{,r}} &= \sin\theta &
  \tensor{A}{^{r\theta}_{,\theta}} &= r \cos\theta
  \\
  \tensor{A}{^{\theta r}_{,r}} &= \cos\theta &
  \tensor{A}{^{\theta r}_{,\theta}} &= -r \sin\theta
\end{align*}
%
\begin{displaymath}
  \grad_\beta A^{\mu \nu} =
  \tensor{A}{^{\mu \nu}_{,\beta}} +
  A^{\alpha \nu} \tensor{\Gamma}{^\mu_{\alpha \beta}} +
  A^{\mu \alpha} \tensor{\Gamma}{^\nu_{\alpha \beta}}
\end{displaymath}
%
\begin{align*}
  \grad_r A^{rr} &=
  \tensor{A}{^{rr}_{,r}} +
  A^{\alpha r} \tensor{\Gamma}{^r_{\alpha r}} +
  A^{r\alpha} \tensor{\Gamma}{^r_{\alpha r}}
  \\ &=
  \tensor{A}{^{rr}_{,r}} +
  A^{r r} \tensor{\Gamma}{^r_{r r}} +
  A^{\theta r} \tensor{\Gamma}{^r_{\theta r}} +
  A^{r r} \tensor{\Gamma}{^r_{r r}} +
  A^{r \theta} \tensor{\Gamma}{^r_{\theta r}}
  \\ &=
  \tensor{A}{^{rr}_{,r}} =
  2 r
  \\
  \grad_\theta A^{r r} &=
  \tensor{A}{^{r r}_{,\theta}} +
  A^{\alpha r} \tensor{\Gamma}{^r_{\alpha \theta}} +
  A^{r \alpha} \tensor{\Gamma}{^r_{\alpha \theta}}
  \\ &=
  \tensor{A}{^{r r}_{,\theta}} +
  A^{r r} \tensor{\Gamma}{^r_{r \theta}} +
  A^{\theta r} \tensor{\Gamma}{^r_{\theta \theta}} +
  A^{r r} \tensor{\Gamma}{^r_{r \theta}} +
  A^{r \theta} \tensor{\Gamma}{^r_{\theta \theta}}
  \\ &=
  (A^{\theta r} + A^{r \theta}) \tensor{\Gamma}{^r_{\theta \theta}} =
  -r^2 (\sin\theta + \cos\theta)
  \\
  \grad_r A^{\theta \theta} &=
  \tensor{A}{^{\theta \theta}_{,r}} +
  A^{\alpha \theta} \tensor{\Gamma}{^\theta_{\alpha r}} +
  A^{\theta \alpha} \tensor{\Gamma}{^\theta_{\alpha r}}
  \\ &=
  \tensor{A}{^{\theta \theta}_{,r}} +
  A^{r \theta} \tensor{\Gamma}{^\theta_{r r}} +
  A^{\theta \theta} \tensor{\Gamma}{^\theta_{\theta r}} +
  A^{\theta r} \tensor{\Gamma}{^\theta_{r r}} +
  A^{\theta \theta} \tensor{\Gamma}{^\theta_{\theta r}}
  \\ &=
  2 A^{\theta \theta} \tensor{\Gamma}{^\theta_{\theta r}} =
  (2/r) \tan\theta
  \\
  \grad_\theta A^{\theta \theta} &=
  \tensor{A}{^{\theta \theta}_{,\theta}} +
  A^{\alpha \theta} \tensor{\Gamma}{^\theta_{\alpha \theta}} +
  A^{\theta \alpha} \tensor{\Gamma}{^\theta_{\alpha \theta}}
  \\ &=
  \tensor{A}{^{\theta \theta}_{,\theta}} +
  A^{r \theta} \tensor{\Gamma}{^\theta_{r \theta}} +
  A^{\theta \theta} \tensor{\Gamma}{^\theta_{\theta \theta}} +
  A^{\theta r} \tensor{\Gamma}{^\theta_{r \theta}} +
  A^{\theta \theta} \tensor{\Gamma}{^\theta_{\theta \theta}}
  \\ &=
  \tensor{A}{^{\theta \theta}_{,\theta}} +
  (A^{r \theta} + A^{\theta r}) \tensor{\Gamma}{^\theta_{r \theta}} =
  \sin\theta + \cos\theta + \sec^2\theta
  \\
  \grad_r A^{r \theta} &=
  \tensor{A}{^{r \theta}_{,r}} +
  A^{\alpha \theta} \tensor{\Gamma}{^r_{\alpha r}} +
  A^{r \alpha} \tensor{\Gamma}{^\theta_{\alpha r}}
  \\ &=
  \tensor{A}{^{r \theta}_{,r}} +
  A^{r \theta} \tensor{\Gamma}{^r_{r r}} +
  A^{\theta \theta} \tensor{\Gamma}{^r_{\theta r}} +
  A^{r r} \tensor{\Gamma}{^\theta_{r r}} +
  A^{r \theta} \tensor{\Gamma}{^\theta_{\theta r}}
  \\ &=
  \tensor{A}{^{r \theta}_{,r}} +
  A^{r \theta} \tensor{\Gamma}{^\theta_{\theta r}} =
  2 \sin\theta
  \\
  \grad_\theta A^{r \theta} &=
  \tensor{A}{^{r \theta}_{,\theta}} +
  A^{\alpha \theta} \tensor{\Gamma}{^r_{\alpha \theta}} +
  A^{r \alpha} \tensor{\Gamma}{^\theta_{\alpha \theta}}
  \\ &=
  \tensor{A}{^{r \theta}_{,\theta}} +
  A^{r \theta} \tensor{\Gamma}{^r_{r \theta}} +
  A^{\theta \theta} \tensor{\Gamma}{^r_{\theta \theta}} +
  A^{r r} \tensor{\Gamma}{^\theta_{r \theta}} +
  A^{r \theta} \tensor{\Gamma}{^\theta_{\theta \theta}}
  \\ &=
  \tensor{A}{^{r \theta}_{,\theta}} +
  A^{\theta \theta} \tensor{\Gamma}{^r_{\theta \theta}} +
  A^{r r} \tensor{\Gamma}{^\theta_{r \theta}} =
  r (1 + \cos\theta - \tan\theta)
  \\
  \grad_r A^{\theta r} &=
  \tensor{A}{^{\theta r}_{,r}} +
  A^{\alpha r} \tensor{\Gamma}{^\theta_{\alpha r}} +
  A^{\theta \alpha} \tensor{\Gamma}{^r_{\alpha r}}
  \\ &=
  \tensor{A}{^{\theta r}_{,r}} +
  A^{r r} \tensor{\Gamma}{^\theta_{r r}} +
  A^{\theta r} \tensor{\Gamma}{^\theta_{\theta r}} +
  A^{\theta r} \tensor{\Gamma}{^r_{r r}} +
  A^{\theta \theta} \tensor{\Gamma}{^r_{\theta r}}
  \\ &=
  \tensor{A}{^{\theta r}_{,r}} +
  A^{\theta r} \tensor{\Gamma}{^\theta_{\theta r}} =
  2 \cos\theta
  \\
  \grad_\theta A^{\theta r} &=
  \tensor{A}{^{\theta r}_{,\theta}} +
  A^{\alpha r} \tensor{\Gamma}{^\theta_{\alpha \theta}} +
  A^{\theta \alpha} \tensor{\Gamma}{^r_{\alpha \theta}}
  \\ &=
  \tensor{A}{^{\theta r}_{,\theta}} +
  A^{r r} \tensor{\Gamma}{^\theta_{r \theta}} +
  A^{\theta r} \tensor{\Gamma}{^\theta_{\theta \theta}} +
  A^{\theta r} \tensor{\Gamma}{^r_{r \theta}} +
  A^{\theta \theta} \tensor{\Gamma}{^r_{\theta \theta}}
  \\ &=
  \tensor{A}{^{\theta r}_{,\theta}} +
  A^{r r} \tensor{\Gamma}{^\theta_{r \theta}} +
  A^{\theta \theta} \tensor{\Gamma}{^r_{\theta \theta}} =
  -r \sin\theta
\end{align*}

\textbf{15}
Find the components of $\tensor{V}{^\alpha_{;\beta;\gamma}}$ for the vector $V^r = 1$, $V^\theta = 0$.

We start by finding the components of $\tensor{V}{^\alpha_{;\beta}}$.
%
\begin{displaymath}
  \tensor{V}{^\alpha_{;\beta}} =
  \tensor{V}{^\alpha_{,\beta}} +
  V^\mu \tensor{\Gamma}{^\alpha_{\mu\beta}}.
\end{displaymath}
%
By noting that $\tensor{V}{^\alpha_{,\beta}} = V^\theta = \tensor{\Gamma}{^r_{rr}} = \tensor{\Gamma}{^r_{r\theta}} = 0$, we can simplify this to
%
\begin{displaymath}
  \tensor{V}{^\alpha_{;\beta}} =
  V^r \tensor{\Gamma}{^\alpha_{r\beta}},
\end{displaymath}
%
which means
%
\begin{displaymath}
  \tensor{V}{^r_{;r}} = \tensor{V}{^r_{;\theta}} = \tensor{V}{^\theta_{;r}} = 0;
  \quad
  \tensor{V}{^\theta_{;\theta}} = \frac{1}{r}.
\end{displaymath}
%
Now we can say
%
\begin{displaymath}
  \tensor{V}{^\alpha_{;\beta;\mu}} =
  \grad_\mu \tensor{V}{^\alpha_{;\beta}} =
  \tensor{V}{^\alpha_{;\beta,\mu}} +
  \tensor{V}{^\gamma_{;\beta}} \tensor{\Gamma}{^\alpha_{\gamma\mu}} -
  \tensor{V}{^\alpha_{;\gamma}} \tensor{\Gamma}{^\gamma_{\beta\mu}}.
\end{displaymath}
%
Note that $\tensor{V}{^\theta_{;\theta}}$ is a function only of $r$, and so $\tensor{V}{^\theta_{;\theta,r}} = -1/r^2$, and all other partial derivatives are zero. We can also see by inspecting the components, that $\tensor{V}{^r_{;\mu;\nu}} = \tensor{V}{^\theta_{;\mu}} \tensor{\Gamma}{^r_{\theta\mu}}$, as all other components go to zero. Likewise, we can see that $\tensor{V}{^\theta_{;r;\mu}} = -\tensor{V}{^\theta_{;\theta}} \tensor{\Gamma}{^\theta_{r\mu}}$. It then becomes easy to find all the individual components. I summarize their values in Table \ref{tab:ch5-ex15}.

\begin{table}[b]
  \centering
  \begin{tabular}{ccc|c}
    $\alpha$ &  $\beta$ &    $\mu$ & $\tensor{V}{^\alpha_{;\beta;\mu}}$ \\
    \hline
    $\theta$ & $\theta$ & $\theta$ & $0$      \\
    $\theta$ & $\theta$ &      $r$ & $-1/r^2$ \\
    $\theta$ &      $r$ & $\theta$ & $-1/r^2$ \\
    $\theta$ &      $r$ &      $r$ & $0$      \\
         $r$ & $\theta$ & $\theta$ & $-1$     \\
         $r$ & $\theta$ &      $r$ & $0$      \\
         $r$ &      $r$ & $\theta$ & $0$      \\
         $r$ &      $r$ &      $r$ & $0$      \\
  \end{tabular}
  \caption{Components of the tensor in Exercise 15.}
  \label{tab:ch5-ex15}
\end{table}

\textbf{16}
Repeat the steps leading from Equation 5.74 to 5.75.

Recalling that $g_{\alpha\mu;\beta} = 0$, we can rewrite Equation 5.72 as
%
\begin{displaymath}
  g_{\alpha\beta,\mu} =
  \tensor{\Gamma}{^\nu_{\alpha\mu}} g_{\nu\beta} +
  \tensor{\Gamma}{^\nu_{\beta\mu}} g_{\alpha\nu}.
\end{displaymath}
%
Now if we switch the $\beta$ and $\mu$ indices, and then switch the $\alpha$ and $\beta$ indices, we get two more equations,
%
\begin{align*}
  g_{\alpha\mu,\beta} &=
  \tensor{\Gamma}{^\nu_{\alpha\beta}} g_{\nu\mu} +
  \tensor{\Gamma}{^\nu_{\mu\beta}} g_{\alpha\nu},
  \\*
  g_{\beta\mu,\alpha} &=
  \tensor{\Gamma}{^\nu_{\beta\alpha}} g_{\nu\mu} +
  \tensor{\Gamma}{^\nu_{\mu\alpha}} g_{\beta\nu}.
\end{align*}
%
Now we add the first two equations and subtract the third, getting
%
\begin{align*}
  g_{\alpha\beta,\mu} + g_{\alpha\mu,\beta} - g_{\beta\mu,\alpha} &=
  \tensor{\Gamma}{^\nu_{\alpha\mu}} g_{\nu\beta} +
  \tensor{\Gamma}{^\nu_{\beta\mu}} g_{\alpha\nu} +
  \tensor{\Gamma}{^\nu_{\alpha\beta}} g_{\nu\mu} +
  \tensor{\Gamma}{^\nu_{\mu\beta}} g_{\alpha\nu} -
  \tensor{\Gamma}{^\nu_{\beta\alpha}} g_{\nu\mu} -
  \tensor{\Gamma}{^\nu_{\mu\alpha}} g_{\beta\nu}
  \\ &=
  {\color{red}
  \tensor{\Gamma}{^\nu_{\alpha\mu}} g_{\beta\nu}} +
  {\color{green}
  \tensor{\Gamma}{^\nu_{\beta\mu}} g_{\alpha\nu}} +
  {\color{blue}
  \tensor{\Gamma}{^\nu_{\alpha\beta}} g_{\nu\mu}} +
  {\color{green}
  \tensor{\Gamma}{^\nu_{\beta\mu}} g_{\alpha\nu}} -
  {\color{red}
  \tensor{\Gamma}{^\nu_{\alpha\beta}} g_{\nu\mu}} -
  {\color{blue}
  \tensor{\Gamma}{^\nu_{\alpha\mu}} g_{\beta\nu}}
  \\ &=
  2 \tensor{\Gamma}{^\nu_{\beta\mu}} g_{\alpha\nu}.
\end{align*}
%
Recalling that $g^{\alpha\gamma} g_{\alpha\nu} = \tensor{g}{^\gamma_\nu} = \tensor{\delta}{^\gamma_\nu}$, we divide both sides by $2$ and multiply by $g^{\alpha\gamma}$, arriving at Equation 5.75:
%
\begin{align*}
  \frac{1}{2}
  g^{\alpha\gamma}
  (g_{\alpha\beta,\mu} + g_{\alpha\mu,\beta} - g_{\beta\mu,\alpha} &=
  \frac{2}{2}
  g^{\alpha\gamma} g_{\alpha\nu}
  \tensor{\Gamma}{^\nu_{\beta\mu}}
  \\ &=
  \tensor{\Gamma}{^\nu_{\beta\mu}}
\end{align*}

\textbf{17}
Show how $\tensor{V}{^\beta_{,\alpha}}$ and $V^\mu \tensor{\Gamma}{^\beta_{\upsilon\alpha}}$ transform under change of coordinates. Neither follows a tensor transformation law, but their \emph{sum} does.

\begin{align*}
  \tensor{V}{^{\alpha'}_{,\beta'}} &=
  \pdv{V^{\alpha'}}{x^{\beta'}} =
  \tensor{\Lambda}{^\beta_{\beta'}}
  \pdv{x^\beta} \qty[ \tensor{\Lambda}{^{\alpha'}_\alpha} V^\alpha ]
  \\ &=
  \tensor{\Gamma}{^\beta_{\beta'}} \qty[
    V^\alpha \pdv{x^\beta} \tensor{\Lambda}{^{\alpha'}_\alpha} +
    \tensor{\Lambda}{^{\alpha'}_\alpha} \pdv{x^\beta} V^\alpha
  ]
  \\ &=
  \tensor{\Lambda}{^\beta_{\beta'}} V^\alpha
  \tensor{\Lambda}{^{\alpha'}_{\alpha,\beta}} +
  \tensor{\Lambda}{^\beta_{\beta'}} \tensor{\Lambda}{^{\alpha'}_\alpha}
  \tensor{V}{^\alpha_{,\beta}}
  \\ &\neq
  \tensor{\Lambda}{^\beta_{\beta'}} \tensor{\Lambda}{^{\alpha'}_\alpha}
  \tensor{V}{^\alpha_{,\beta}}
  \\
  \pdv{\vec{e}_{\alpha'}}{x^{\beta'}} &=
  \tensor{\Lambda}{^\beta_{\beta'}}
  \pdv{x^\beta} \qty[ \tensor{\Lambda}{^\alpha_{\alpha'}} \vec{e}_\alpha ]
  \\ &=
  \tensor{\Lambda}{^\beta_{\beta'}} \qty[
    \tensor{\Lambda}{^\alpha_{\alpha'}} \pdv{x^\beta} \vec{e}_\alpha +
    \vec{e}_\alpha \pdv{x^\beta} \tensor{\Lambda}{^\alpha_{\alpha'}}
  ]
  \\ &=
  \tensor{\Lambda}{^\beta_{\beta'}}
  \tensor{\Lambda}{^\alpha_{\alpha'}}
  \tensor{\Gamma}{^\mu_{\alpha\beta}} \vec{e}_\mu +
  \tensor{\Lambda}{^\beta_{\beta'}}
  \tensor{\Lambda}{^\alpha_{\alpha',\beta}}
  \vec{e_\alpha}
  \\ &\neq
  \tensor{\Lambda}{^\beta_{\beta'}}
  \tensor{\Lambda}{^\alpha_{\alpha'}}
  \tensor{\Gamma}{^\mu_{\alpha\beta}} \vec{e}_\mu,
\end{align*}
%
so we have shown that $\pdv*{\vec{e}_{\alpha'}}{x^{\beta'}}$ is not a tensor,
and since $\tensor{V}{^\mu}$ is a tensor, and the product of a tensor and a non-tensor is also not a tensor, then $V^\mu\tensor{\Gamma}{^\beta_{\nu\alpha}}$ is not a tensor.

According to Carroll, the precise transformation is
%
\begin{displaymath}
  \tensor{\Gamma}{^{\nu'}_{\mu'\lambda'}} =
  \tensor{\Lambda}{^\mu_{\mu'}}
  \tensor{\Lambda}{^\lambda_{\lambda'}}
  \tensor{\Lambda}{^{\nu'}_\nu}
  \tensor{\Gamma}{^\nu_{\mu\lambda}} +
  \tensor{\Lambda}{^\mu_{\mu'}}
  \tensor{\Lambda}{^\lambda_{\lambda'}}
  \tensor{\Lambda}{^{\nu'}_{\mu\lambda}}.
\end{displaymath}
%
Now we add the two expressions, in order to show that it is a tensor equation
%
\begin{align*}
  \tensor{V}{^{\nu'}_{,\lambda'}} +
  V^{\mu'} \tensor{\Gamma}{^{\nu'}_{\mu'\lambda'}} &=
  \tensor{\Lambda}{^\lambda_{\lambda'}}
  V^\nu \tensor{\Lambda}{^{\nu'}_{\nu,\lambda}} +
  \tensor{\Lambda}{^\lambda_{\lambda'}}
  \tensor{\Lambda}{^{\nu'}_\nu} \tensor{V}{^\nu_{,\lambda}} +
  \tensor{\Lambda}{^\lambda_{\lambda'}}
  \tensor{\Lambda}{^{\nu'}_\nu} V^\mu
  \tensor{\Gamma}{^\nu_{\mu\lambda}} +
  \tensor{\Lambda}{^\lambda_{\lambda'}}
  V^\nu \tensor{\Lambda}{^{\nu'}_{\lambda,\mu}}
  \\ &=
  \tensor{\Lambda}{^\lambda_{\lambda'}}
  \tensor{\Lambda}{^{\nu'}_\nu}
  \qty(
    \tensor{V}{^\nu_{,\lambda}} +
    V^\mu \tensor{\Gamma}{^\nu_{\mu\lambda}}
  )
\end{align*}
%
So it does in fact transform like a tensor equation, meaning $\tensor{V}{^\nu_{;\lambda}}$ is a tensor!

\textbf{18}

Verify Equation 5.78:
%
\begin{displaymath}
  \left.
  \begin{array}{r}
    \vec{e}_{\hat\alpha} \cdot \vec{e}_{\hat\beta} \equiv
    g_{\hat\alpha \hat\beta} =
    \delta_{\hat\alpha \hat\beta}
    \\
    \tilde\omega^{\hat\alpha} \cdot \tilde\omega^{\hat\beta} \equiv
    g^{\hat\alpha \hat\beta} =
    \delta^{\hat\alpha \hat\beta}
  \end{array}
  \right\}
\end{displaymath}

For the basis \emph{vectors}, we have
%
\begin{align*}
  g_{\hat{r} \hat{r}} &=
  \vec{e}_{\hat{r}} \cdot \vec{e}_{\hat{r}} =
  \vec{e}_r \cdot \vec{e}_r =
  g_{rr} =
  1
  \\
  g_{\hat\theta \hat\theta} &=
  \vec{e}_{\hat\theta} \cdot \vec{e}_{\hat\theta} =
  \qty(\frac{1}{r} \vec{e}_\theta) \cdot \qty(\frac{1}{r} \vec{e}_\theta) =
  \frac{1}{r^2} (\vec{e}_\theta \cdot \vec{e}_\theta) =
  \frac{1}{r} g_{r\theta} =
  1
  \\
  g_{\hat{r} \hat\theta} &=
  \vec{e}_{\hat{r}} \cdot \vec{e}_{\hat\theta} =
  \vec{e}_r \cdot \qty(\frac{1}{r} \vec{e}_\theta) =
  \frac{1}{r} (\vec{e}_r \cdot \vec{e}_\theta) =
  \frac{1}{r} g_{r\theta} =
  0
  \\
  g_{\hat\theta \hat{r}} &=
  g_{\hat{r} \hat\theta} =
  0
\end{align*}
%
So it is indeed true that $g_{\hat\alpha \hat\beta} = \delta_{\hat\alpha \hat\beta}$.

Now for the basis \emph{one-forms}, we have
%
\begin{align*}
  g^{\hat{r} \hat{r}} &=
  \tilde\omega^{\hat{r}} \cdot \tilde\omega^{\hat{r}} =
  \tilde\dd{r} \cdot \tilde\dd{r} =
  g^{rr} =
  1
  \\
  g^{\hat\theta \hat\theta} &=
  \tilde\omega^{\hat\theta} \cdot \tilde\omega^{\hat\theta} =
  (r \tilde\dd\theta) \cdot (r \tilde\dd\theta) =
  r^2 (\tilde\dd\theta \cdot \tilde\dd\theta) =
  r^2 g^{\theta\theta} =
  r^2 (1/r^2) =
  1
  \\
  g^{\hat{r} \hat\theta} &=
  \tilde\omega^{\hat{r}} \cdot \tilde\omega^{\hat\theta} =
  \tilde\dd{r} \cdot (r \tilde\dd\theta) =
  r (\tilde\dd{r} \cdot \tilde\dd\theta) =
  r g^{r\theta} =
  0
  \\
  g^{\hat\theta \hat{r}} &=
  g^{\hat{r} \hat\theta} =
  0
\end{align*}
%
So it is indeed true that $g^{\hat\alpha \hat\beta} = \delta^{\hat\alpha \hat\beta}$.

\textbf{19}
Repeat the calculations going from Equations 5.81 to 5.84, with $\tilde\dd{r}$ and $\tilde\dd\theta$ as your bases. Show that they form a coordinate basis.

\begin{align*}
  \tilde\dd{r} &=
  \cos\theta \dd{x} + \sin\theta \dd{y} =
  \pdv{\xi}{x} \tilde\dd{x} + \pdv{\xi}{y} \tilde\dd{y}
  \\
  \pdv{\xi}{x} &=
  \cos\theta; \quad
  \pdv{\xi}{y} =
  \sin\theta
  \\
  \pdv{y} \pdv{\xi}{x} &=
  \pdv{x} \pdv{\xi}{y} \implies
  \pdv{y} (x/r) = \pdv{x} (y/r),
\end{align*}
%
which is true, so we have shown that at least $\tilde\dd{r}$ may be part of a coordinate basis.
%
\begin{align*}
  \tilde\dd\theta &=
  -\frac{1}{r} \sin\theta \tilde\dd{x} +
   \frac{1}{r} \cos\theta \tilde\dd{y} =
  \pdv{\eta}{x} \tilde\dd{x} +
  \pdv{\eta}{y} \tilde\dd{y}
  \\
  \pdv{\eta}{x} &=
  -\frac{1}{r} \sin\theta; \quad
  \pdv{\eta}{y} =
   \frac{1}{r} \cos\theta
  \\
  \pdv{y} \pdv{\eta}{x} &=
  \pdv{x} \pdv{\eta}{y} \implies
  \pdv{y} \qty[ -\frac{1}{r} \sin\theta ] =
  \pdv{x} \qty[  \frac{1}{r} \cos\theta ],
\end{align*}
%
which is also true, and thus we have shown that $\tilde\dd{r}$ and $\tilde\dd\theta$ form a coordinate basis.

\textbf{20}
For a non-coordinate basis $\{ \vec{e}_\mu \}$, let $\tensor{c}{^\alpha_{\mu\nu}} = \grad_{\vec{e}_\mu} \vec{e}_\nu - \grad_{\vec{e}_\nu} \vec{e}_\mu$. Use this in place of Equation 5.74 to derive a more general expression for Equation 5.75.

$\boldsymbol{c}$ is antisymmetric w.r.t. its bottom indices.
%
\begin{align*}
  \tensor{c}{^\alpha_{\mu\nu}} \vec{e}_\alpha +
  \tensor{c}{^\alpha_{\nu\mu}} \vec{e}_\alpha &=
  (\grad_{\vec{e}_\mu} \vec{e}_\nu - \grad_{\vec{e}_\nu} \vec{e}_\mu) +
  (\grad_{\vec{e}_\nu} \vec{e}_\mu - \grad_{\vec{e}_\mu} \vec{e}_\nu) =
  0
  \\ \implies
   \tensor{c}{^\alpha_{\mu\nu}} \vec{e}_\alpha &=
  -\tensor{c}{^\alpha_{\nu\mu}} \vec{e}_\alpha
  \\ \implies
   \tensor{c}{^\alpha_{\mu\nu}} &=
  -\tensor{c}{^\alpha_{\nu\mu}}
\end{align*}
%
Expanding the covariant derivatives in the original expression, we get
\begin{align*}
  \tensor{c}{^\alpha_{\mu\nu}} \vec{e}_\alpha &=
  \vec{e}_{\nu;\mu} - \vec{e}_{\mu;\nu}
  \\ &=
  (\vec{e}_{\nu,\mu} - \vec{e}_\alpha \tensor{\Gamma}{^\alpha_{\nu\mu}}) -
  (\vec{e}_{\mu,\nu} - \vec{e}_\alpha \tensor{\Gamma}{^\alpha_{\mu\nu}})
  \\ &=
  \vec{e}_\alpha
  (\tensor{\Gamma}{^\alpha_{\mu\nu}} - \tensor{\Gamma}{^\alpha_{\nu\mu}})
  \\
  \tensor{c}{^\alpha_{\mu\nu}} &=
  \tensor{\Gamma}{^\alpha_{\mu\nu}} - \tensor{\Gamma}{^\alpha_{\nu\mu}}
\end{align*}
%
Now we recall the result from Exercise 16, but without assuming symmetry of the Christoffel symbols
%
\begin{align*}
  g_{\alpha\beta,\mu} + g_{\alpha\mu,\beta} - g_{\beta\mu,\alpha} &=
  \tensor{\Gamma}{^\nu_{\alpha\mu}} g_{\nu\beta} +
  \tensor{\Gamma}{^\nu_{\beta\mu}} g_{\alpha\nu} +
  \tensor{\Gamma}{^\nu_{\alpha\beta}} g_{\nu\mu} +
  \tensor{\Gamma}{^\nu_{\mu\beta}} g_{\alpha\nu} -
  \tensor{\Gamma}{^\nu_{\beta\alpha}} g_{\nu\mu} -
  \tensor{\Gamma}{^\nu_{\mu\alpha}} g_{\beta\nu}
  \\ &=
  {\color{red} \tensor{\Gamma}{^\nu_{\alpha\mu}} g_{\beta\nu}} +
  {\color{green} \tensor{\Gamma}{^\nu_{\beta\mu}} g_{\alpha\nu}} +
  {\color{blue} \tensor{\Gamma}{^\nu_{\alpha\beta}} g_{\nu\mu}} +
  {\color{green} \tensor{\Gamma}{^\nu_{\mu\beta}} g_{\alpha\nu}} -
  {\color{blue} \tensor{\Gamma}{^\nu_{\beta\alpha}} g_{\nu\mu}} -
  {\color{red} \tensor{\Gamma}{^\nu_{\mu\alpha}} g_{\beta\nu}}
  \\ &=
  g_{\beta\nu}
  (\tensor{\Gamma}{^\nu_{\alpha\mu}} - \tensor{\Gamma}{^\nu_{\mu\alpha}}) +
  g_{\alpha\nu}
  (\tensor{\Gamma}{^\nu_{\beta\mu}} + \tensor{\Gamma}{^\nu_{\mu\beta}}) +
  g_{\nu\mu}
  (\tensor{\Gamma}{^\nu_{\alpha\beta}} - \tensor{\Gamma}{^\nu_{\beta\alpha}})
  \\ &=
  g_{\beta\nu} \tensor{c}{^\nu_{\alpha\mu}} +
  g_{\alpha\nu}
  (\tensor{\Gamma}{^\nu_{\beta\mu}} +
   \tensor{\Gamma}{^\nu_{\mu\beta}} +
   \tensor{\Gamma}{^\nu_{\beta\mu}} -
   \tensor{\Gamma}{^\nu_{\beta\mu}}) +
  g_{\nu\mu} \tensor{c}{^\nu_{\alpha\beta}}
  \\ &=
  g_{\beta\nu} \tensor{c}{^\nu_{\alpha\mu}} +
  g_{\nu\mu} \tensor{c}{^\nu_{\alpha\beta}} +
  g_{\alpha\nu}
  (2 \tensor{\Gamma}{^\nu_{\beta\mu}} + \tensor{c}{^\nu_{\mu\beta}})
  \\
  g^{\nu\mu} 2 g_{\alpha\nu} \tensor{\Gamma}{^\nu_{\beta\mu}} &=
  g^{\nu\mu} (
    g_{\alpha\beta,\mu} +
    g_{\alpha\mu,\beta} -
    g_{\beta\mu,\alpha} -
    c_{\beta\alpha\mu}  -
    c_{\mu\alpha\beta}  -
    c_{\alpha\mu\beta}
  )
  \\
  \tensor{\Gamma}{^\nu_{\beta\alpha}} &=
  \frac{1}{2}
  g^{\nu\mu} (
    g_{\alpha\beta,\mu} +
    g_{\alpha\mu,\beta} -
    g_{\beta\mu,\alpha} -
    c_{\beta\alpha\mu}  -
    c_{\mu\alpha\beta}  -
    c_{\alpha\mu\beta}
  )
\end{align*}

\textbf{21}
A uniformly accelerated observer has world line
%
\begin{displaymath}
  t(\lambda) = a \sinh\lambda,
  \quad
  x(\lambda) = a \cosh\lambda
\end{displaymath}

(a) Show that the spacelike line tangent to his world line (which is parameterized by $\lambda$) is orthogonal to the line parameterized by $a$.

The line tangent to his world line is
%
\begin{displaymath}
  \vec{V} \to
  \dv{\lambda} (t, x) =
  (a \cosh\lambda, a \sinh\lambda).
\end{displaymath}
%
The line parameterized by $a$ is
%
\begin{displaymath}
  \vec{W} \to
  \dv{a} (t, x) =
  (\sinh\lambda, \cosh\lambda)
\end{displaymath}
%
If they are orthogonal, then their dot product must be zero
%
\begin{displaymath}
  \vec{V} \cdot \vec{W} =
  -(a \cosh\lambda \sinh\lambda) + (a \sinh\lambda \cosh\lambda) =
  0,
\end{displaymath}
%
which it is.

(b) To prove that this defines a valid coordinate transform from $(\lambda,a)$ to $(t,x)$, we show that the determinant of the transformation matrix is non-zero.
%
\begin{align*}
  \det\mqty( \pdv*{t}{\lambda} & \pdv*{t}{a} \\
             \pdv*{x}{\lambda} & \pdv*{x}{a} ) &=
  \pdv{t}{\lambda} \pdv{x}{a} - \pdv{t}{a} \pdv{x}{\lambda}
  \\ &=
  a \cosh^2\lambda - a \sinh^2\lambda = a
  \\ &\neq 0,
\end{align*}
%
and so it is indeed a valid coordinate transform.

To plot the curves parameterized by $a$, we take
%
\begin{align*}
  -t^2 + x^2 &=
  a^2 (\cosh^2\lambda - \sinh^2\lambda)
  \\ &=
  a^2,
\end{align*}
%
which gives us a family of space-like hyperbola, depending on the chosen value of $a$.

To plot the curves parameterized by $\lambda$, we take
%
\begin{align*}
  x &= a \cosh\lambda \implies a = x / \cosh\lambda
  \\
  t &= a \sinh\lambda = x \sinh\lambda / \cosh\lambda = x \tanh\lambda,
\end{align*}
%
which gives us a family of space-like lines, depending on the chosen value of $\lambda$.

A plot of these curves is given in Figure \ref{fig:ch5-problem-21}, from which it is clear that only half of the $t$--$x$ plane is covered.

When $\abs{t} = \abs{x}$, then $a = 0$, since $-t^2 + x^2 = a^2$. We already found that the determinant of the coordinate transformation is $a$, so this would make the determinant $0$, making it singular.

\begin{figure}[ht]
  \centering
  \includegraphics[width=0.75\textwidth]{img/ch5_problem_21}
  \caption{Lines of constant $\lambda$ and $a$ in Problem 21.}
  \label{fig:ch5-problem-21}
\end{figure}

(c) Find the metric tensor and Christoffel symbols in $(\lambda,a)$ coordinates.

First we find the basis vectors:
%
\begin{align*}
  \vec{e}_\lambda &=
  a (\cosh\lambda \vec{e}_t + \sinh\lambda \vec{e}_x),
  \\
  \vec{e}_a &=
  \sinh\lambda \vec{e}_t + \cosh\lambda \vec{e}_x.
\end{align*}

Now we find the components of the metric tensor $\vb{g}$ as
%
\begin{align*}
  g_{\lambda\lambda} &=
  a^2
  ( \cosh\lambda \vec{e}_t +
    \sinh\lambda \vec{e}_x )^2
  \\ &=
  a^2
  ( \cosh^2\lambda \eta_{tt} +
    \sinh^2\lambda \eta_{xx} +
    2 \sinh\lambda \cosh\lambda \eta_{tx} )
  \\ &=
  a^2 (\sinh^2\lambda - \cosh^2\lambda)
  \\ &=
  -a^2
  \\
  g_{aa} &=
  (\sinh\lambda \vec{e}_t + \cosh\lambda \vec{e}_x)^2
  \\ &=
  \sinh^2\lambda \eta_{tt} +
  \cosh^2\lambda \eta_{xx} +
  2 \sinh\lambda \cosh\lambda \eta_{tx}
  \\ &=
  1
  \\
  g_{\lambda a} = g_{a \lambda} &=
  a (\cosh\lambda \vec{e}_t + \sinh\lambda \vec{e}_x)
    (\sinh\lambda \vec{e}_t + \cosh\lambda \vec{e}_x)
  \\ &=
  a
  ( \cosh\lambda \sinh\lambda (\eta_{tt} + \eta_{xx}) +
    2 \sinh\lambda \cosh\lambda \eta_{tx} )
  \\ &=
  0
  \\
  \vb{g} &\underset{(\lambda,a)}{\to}
  \mqty( -a^2 & 0 \\ 0 & 1 )
\end{align*}
%
Now for the Christoffel symbols, since we know this is a coordinate basis, we can use
%
\begin{displaymath}
  \tensor{\Gamma}{^\gamma_{\beta \mu}} =
  \frac{1}{2} g^{\alpha \gamma}
  (g_{\alpha \beta , \mu} + g_{\alpha \mu , \beta} - g_{\beta \mu , \alpha})
\end{displaymath}
%
\begin{align*}
  \tensor{\Gamma}{^\lambda_{\lambda \lambda}} &=
  \frac{1}{2} g^{\alpha \lambda}
  (g_{\alpha \lambda , \lambda} + g_{\alpha \lambda , \lambda} - g_{\lambda \lambda , \alpha}) =
  \frac{1}{2} g^{a\lambda} (-g_{\lambda\lambda,a})
  \\ &=
  0
  \\
  \tensor{\Gamma}{^a_{a a}} &=
  \frac{1}{2} g^{\alpha a}
  (g_{\alpha a , a} + g_{\alpha a , a} - g_{a a , \alpha})
  \\ &=
  0
  \\
  \tensor{\Gamma}{^\lambda_{\lambda a}} &=
  \frac{1}{2} g^{\alpha \lambda}
  (g_{\alpha \lambda , a} + g_{\alpha a , \lambda} - g_{\lambda a , \alpha}) =
  \frac{1}{2} g^{\lambda\lambda} g_{\lambda\lambda,a} =
  \frac{1}{2} (-a^{-2}) (-2a)
  \\ &=
  1/a
  \\
  \tensor{\Gamma}{^a_{\lambda a}} &=
  \frac{1}{2} g^{\alpha a}
  (g_{\alpha \lambda , a} + g_{\alpha a , \lambda} - g_{\lambda a , \alpha}) =
  \frac{1}{2} g^{\lambda a} g_{\lambda\lambda,a}
  \\ &=
  0
  \\
  \tensor{\Gamma}{^\lambda_{a a}} &=
  \frac{1}{2} g^{\alpha \lambda}
  (g_{\alpha a , a} + g_{\alpha a , a} - g_{a a , \alpha})
  \\ &=
  0
  \\
  \tensor{\Gamma}{^a_{\lambda \lambda}} &=
  \frac{1}{2} g^{\alpha a}
  (g_{\alpha \lambda , \lambda} + g_{\alpha \lambda , \lambda} - g_{\lambda \lambda , \alpha}) =
  \frac{1}{2} g^{aa} (-g_{\lambda\lambda,a}) =
  \frac{1}{2} \cdot 2 \cdot a
  \\ &=
  a
\end{align*}

\textbf{22}
%
\begin{align*}
  U^\alpha \grad_\alpha V^\beta = W^\beta &\implies
  U^\alpha \tensor{V}{^\gamma_{;\alpha}} = W^\gamma
  \\ &\implies
  g_{\alpha\beta} U^\alpha \tensor{V}{^\gamma_{;\alpha}} =
  g_{\gamma\beta} W^\gamma
  \\ &\implies
  U^\alpha V_{\beta;\alpha} = W_\beta
  \\ &\implies
  U^\alpha \grad_\alpha V_\beta = W_\beta
\end{align*}

\end{document}


\documentclass[a4paper, 11 pt, article, accentcolor=tud7b]{tudreport}

\usepackage[utf8]{inputenc}
\usepackage{amsmath}

\title{CNuVS Exercise 10}
\author{Nils Rollshausen, Daniel Drodt}
\subtitle{Nils Rollshausen, Daniel Drodt}

\begin{document}
	\maketitle
	\section{M/M/1 Queues}
	$\lambda = 20, \mu = \frac{1000ms}{40ms/r} = 25$
	\subsection*{a) System time}
	$T = \frac{1}{\mu - \lambda} = \frac{1}{5} = 0.2s$

	\subsection*{b) Average number of requests}
	$N = \lambda \cdot T = 20 \cdot 0.2 = 4$

	\subsection*{c) System utilization}
	$\rho = \frac{\lambda}{\mu} = \frac{20}{25} = 0.8 = 80\%$

	\subsection*{d) Average idle time in 15min}
	$t_{idle} = 15 \cdot 60 \cdot (1 - \rho) = 900 \cdot 0.2 = 180s$

	\subsection*{e) Probability of 3 requests}
	$p_{3} = (1 - \rho) \cdot \rho^3 = 0.2 \cdot 0.8^3 = 0.1024 = 10.24\%$

	\section{M/M/m Queues I}

	$\lambda_{1} = 0.1, \lambda_{2} = 0.2, \lambda_{3} = 0.3, \mu = \frac{1}{2.5} = 0.4$

	\subsection*{a) Utilization}
	$$\rho_{1} = \frac{0.1}{0.4} = 0.25$$
	$$\rho_{2} = \frac{0.2}{0.4} = 0.5 $$
	$$\rho_{3} = \frac{0.3}{0.4} = 0.75$$

	\subsection*{b) Idle probability}
	$$p_{0} = (1-\rho_{1}) \cdot (1-\rho_{2}) \cdot (1-\rho_{3}) = 0.75 \cdot 0.5 \cdot 0.25 = 0.09375 = 9.375\%$$

	\subsection*{c) System time}
	$$T_{1} = \frac{1}{0.4 - 0.1} = \frac{10}{3} \approx 3.3333m$$
	$$T_{2} = \frac{1}{0.4 - 0.2} = 5m$$
	$$T_{3} = \frac{1}{0.4 - 0.3} = 10m$$

	\subsection*{d) Number of jobs}
  $$N_{1} = 0.1 \cdot T_{1} = \frac{1}{3} \approx 0.3333$$
	$$N_{2} = 0.2 \cdot T_{2} = 0.2 \cdot 5 = 1$$
	$$N_{3} = 0.3 \cdot T_{3} = 0.3 \cdot 10 = 3$$

	\section{M/M/m Queues II}

	\subsection*{a) Arrival rate and utilization}
	$$\lambda = \lambda_{1} + \lambda_{2} + \lambda_{3} = 0.1 + 0.2 + 0.3 = 0.6$$
	$$\rho = \frac{\lambda}{m \cdot \mu} = \frac{0.6}{3 \cdot 0.4} = 0.5 = 50\%$$

	\subsection*{b) Probability of empty system}
	\begin{align*}
	  p_{0} &= \frac{1}{(\sum_{k=0}^{m-1} \frac{(m \cdot \rho)^{k}}{k!}) + \frac{(m \cdot \rho)^{m}}{m!} \cdot \frac{1}{1 - \rho}} \\
	        &= \frac{1}{(\sum_{k=0}^{2} \frac{(3 \cdot 0.5)^{k}}{k!}) + \frac{(3 \cdot 0.5)^{3}}{6} \cdot 2} \\
	        &= \frac{1}{1 + 1.5 + \frac{9}{8} + \frac{9}{8}} \\
	        &= \frac{1}{4.75} \approx 0.2105
	\end{align*}

	\subsection*{c) Average system time}
	$$\delta = p_{0} \cdot \frac{(m \cdot \rho)^{m}}{m! \cdot (1-\rho)} = p_{0} \cdot \frac{1.5^3}{3} = 1.125 \cdot p_{0} \approx 0.2368 $$
	$$ T = \frac{1}{\mu} \cdot (1 + \frac{\delta}{m \cdot (1-\rho)}) = 2.5 \cdot (1 + \frac{0.2368}{1.5}) \approx 2.8947m$$

	\subsection*{d) Average number of jobs}
	$$N = m \cdot \rho + \frac{\rho \cdot \delta}{1 - \rho} = 1.5 + \frac{0.5 \cdot 0.2368}{0.5} = 1.5 + 0.2368 \approx 1.7368$$

	\section{Queueing Theory}

	\subsection*{a) M/M/1 Queues}
	A M/M/1 queue describes a queue with a single serving unit, where both customer interarrival times and serving time are exponentially distributed. As such, they are useful models for most simple real-world queueing scenarios.

	\subsection*{b) The Markov Process}
	A Markov process is a discrete stochastic process in which the probabilities for the next state depend only on the current state. Additionally, a Markov process requires that the process is memoryless (i.e. the probability of a state change does not change with the time spent in that state), homogenous, and that process times are exponentially (or geometrically) distributed.

	\subsection*{c) Post Office}
	A post office with four cash desks is probably best modeled by a M/M/4 queue. Even if separate queues exist for each counter, customers most likely switch queues when any desk is idle and do not queue at every counter independently. Thus, the customers act like queueing in a single queue even if there are physically separate queues for each counter.

	\subsection*{d) Birth-Death processes}
	Birth-Death processes are a specialization of Markov processes in which in any state, the next following state can only be an adjacent state (e.g. $k \rightarrow k+1$ or $k \rightarrow k-1$).
	\end{document}


\section{Dynamic Programming}
\subsection{LIS $O(n\log{n})$}
\lstinputlisting{"./dynamicProgramming/lis.cc"}
\subsection{LCS $O(n\log{n})$}
\lstinputlisting{"./dynamicProgramming/lcs.cc"}
\subsection{Sparse-Table}
\lstinputlisting{"./dynamicProgramming/sparse-table.c"}
\subsection{Improved by quadrilateral inequality}
\lstinputlisting{"./dynamicProgramming/inequ.cc"}
\subsection{Improved by Slope}
\lstinputlisting{"./dynamicProgramming/slope.cc"}


\documentclass{article}
\usepackage[utf8]{inputenc}

\usepackage{amssymb,amsfonts,amsthm,amsmath}
\usepackage[margin=1in]{geometry}
\usepackage[usenames,dvipsnames,svgnames]{xcolor}
\usepackage{hyperref,url}
\hypersetup{
    colorlinks=true,
    citecolor=MidnightBlue,
    urlcolor=Bittersweet,
}
\newcommand{\<}{\langle}
\renewcommand{\>}{\rangle}
\newcommand{\iprod}[2]{\left\langle #1 , #2 \right\rangle}

% Theorems
\newtheorem{theorem}{Theorem}
\newtheorem{lemma}{Lemma}
\newtheorem{remark}{Remark}
\newtheorem{corollary}{Corollary}%[section]
\newtheorem{proposition}{Proposition}%[section]
\newtheorem{definition}{Definition}%[section] % number this the same as theorem and lemma

\title{Analysis Toolbox: Strong convexity and Lipschitz continuity of gradients}
\author{
Stephen Becker \\
Applied Math, U.\ Colorado Boulder
\texttt{stephen.becker@colorado.edu}
}
\date{Created Sep 27 2019, last edited \today}

\begin{document}

\maketitle

\subsection*{Lipschitz continuity of derivative and/or strong convexity of $f$}
The definition of Lipschitz continuity of $\nabla f$  (with constant $L$) is
\begin{equation} \label{eq:LL}
    \forall x, y\quad \| \nabla f(x) - \nabla f(y) \| \le L \|x-y\|
\end{equation}
and the definition of $f$ being $\mu$ strongly convex means that the function $x \mapsto f(x) - \frac{\mu}{2}\|x\|^2$ is convex. In the lines below, if $L$ or $\mu$ appears, then we are assuming the gradient is Lipschitz with constant $L$ or $f$ is strongly convex with constant $\mu$, respectively. Most references to Nesterov's book are to his first edition~\cite{Nesterov_2004}, not the recent 2018  edition~\cite{Nesterov_2018}.

These two inequalities are very helpful; see, e.g., Thm 2.1.5 and Thm 2.1.10 from \cite{Nesterov_2004}.
\begin{align}
    f(y) &\le f(x) + \iprod{\nabla f(x)}{ y - x } + \frac{L}{2}\|x-y\|^2 \label{eq:L} \\
    f(y) &\ge f(x) + \iprod{\nabla f(x)}{ y - x } + \frac{\mu}{2}\|x-y\|^2 \label{eq:mu}
\end{align}
If we drop convexity but keep Lipschitz continuity of the gradient, then the first equation is still true, but the second equation is not true with $\mu=0$, but it is true with $\mu = -L$.  This is often written as
$\left| f(y) - ( f(x) + \iprod{\nabla f(x)}{ y - x } )\right| \le \frac{L}{2}\|x-y\|^2$.

The main inequalities can be summarized by:
% Not sure how to number; see https://tex.stackexchange.com/a/103894 for a way that won't work if I want this format
\begin{equation} \label{eq:big}
\left.\begin{aligned}
L^{-1} \| \nabla f(x) - \nabla f(y) \|^2  \quad\text{\small\textcolor{red}{(a)}} \\
\mu \|x-y\|^2
\quad\text{\small\textcolor{red}{(b)}} \\
\frac{\mu L}{\mu + L} \|x-y\|^2  + \frac{1}{\mu+L}\| \nabla f(x) - \nabla f(y) \|^2
\quad\text{\small\textcolor{red}{(c)}}
\end{aligned} \right\rbrace
\le \< \nabla f(x) - \nabla f(y), x-y \>
\le
\left\lbrace\begin{aligned}
\text{\small\textcolor{red}{(d)}} \quad& L \|x-y\|^2  \\
\text{\small\textcolor{red}{(e)}} \quad& \mu^{-1}  \| \nabla f(x) - \nabla f(y) \|^2
\end{aligned} \right.
\end{equation}
The inequality {\small\textcolor{red}{(a)}} %left-most $\le$ above in the line for $L$
really follows from the co-coercivity of gradients; this result is actually surprisingly strong, since it makes implicit use of the Baillon-Haddad theorem. The result {\small\textcolor{red}{(e)}} for $\mu$ also requires $f$ be continuously differentiable.
The {\small\textcolor{red}{(c)}} inequality
assumes both strong convexity and Lipschitz continuity of the gradient; see \cite[Thm. 2.1.12]{Nesterov_2004} for a derivation.

\paragraph{Sub-optimality bounds} Added Sept 17 2019.  For unconstrained smooth optimization, if $x^\star$ is a minimizer, then $\nabla f(x^\star) = 0$. Note there are 3 equivalent definitions of optimality: $x$ is optimal if
\begin{equation}
    \|x - x^\star\|=0, \quad
    f(x) - f^\star = 0, \quad
    \|\nabla f(x)\| = 0
\end{equation}
and this would be ``iff'' if we assume the optimal solution is unique. Now, given a Lipschitz continuous derivative, we can bound
\begin{align}
 \|\nabla f(x) \| = \| \nabla f(x) - \nabla f(x^\star)\| &\le L \|x-x^\star\| \quad \text{by \eqref{eq:LL}}\\
 f(x) - f^\star &\le \frac{L}{2}\|x-x^\star\|^2
 \quad\text{by \eqref{eq:L}} \\
 \|\nabla f(x) \|^2 &\le 2L \left( f(x) - f^\star\right) \quad\text{by Eq. (9.14) in \cite{BoydVandenbergheBook}} \label{eq:33}
\end{align}
and given $\mu$ strong convexity, we can bound in the other direction:
\begin{align}
\|x-x^\star\|^2 &\le \frac{1}{\mu^2}\|\nabla f(x)\|^2
\quad \text{by \eqref{eq:big}  {\small\textcolor{red}{(b)}} and  {\small\textcolor{red}{(e)}} }
\\
\|x-x^\star\|^2 &\le \frac{2}{\mu}\left( f(x) - f^\star\right)
\quad \text{by \eqref{eq:mu}, with $x=x^\star$, $y=x$} \\
f(x) - f^\star &\le \frac{1}{2\mu}\|\nabla f(x)\|^2 \quad\text{by Eq. (9.9) in \cite{BoydVandenbergheBook}. This is PL}
\label{eq:44}
\end{align}
Given both $L$ and $\mu$, we can combine the bounds, and bound any one of the 3 error metrics in terms of another, i.e.,
%\begin{align*}
 $\|\nabla f(x) \|^2 \le \frac{2L^2}{\mu}\left( f(x) - f^\star\right)$ and $% \\
 f(x) - f^\star \le \frac{L}{2\mu^2}\|\nabla f(x)\|^2. $
%\end{align*}.
But these are not good bounds; the bounds in Eq~\eqref{eq:33} and \eqref{eq:44} are better. Note: \eqref{eq:44} is the Polyak-Lojasiewicz (PL) inequality, see \href{https://arxiv.org/abs/1608.04636}{Karimi, Nutini, Schmidt} for details.

\begin{thebibliography}{AZQRY16}

\bibitem[BC11]{bauschke2011convex}
H. H. Bauschke and P.~L. Combettes, \href{https://link.springer.com/book/10.1007/978-1-4419-9467-7}{\emph{Convex analysis and monotone
  operator theory in {H}ilbert spaces}, 1st edition}, {S}pringer, 2011.

\bibitem[BC17]{bauschke2017convex}
H. H. Bauschke and P.~L. Combettes, \href{https://link.springer.com/book/10.1007/978-3-319-48311-5}{\emph{Convex analysis and monotone
  operator theory in {H}ilbert spaces}, 2nd edition}, {S}pringer, 2017.

\bibitem[BV04]{BoydVandenbergheBook}
S.~Boyd and L.~Vandenberghe.
\newblock \href{http://www.stanford.edu/~boyd/cvxbook/}{{\em Convex Optimization}}.
\newblock Cambridge University Press, 2004.

\bibitem[Nes04]{Nesterov_2004}
Yu.~Nesterov.
\newblock \href{https://link.springer.com/book/10.1007/978-1-4419-8853-9}{{\em Introductory Lectures on Convex Optimization: A Basic Course}},
  volume~87 of {\em Applied Optimization}.
\newblock Kluwer, Boston, 2004.

\bibitem[Nes18]{Nesterov_2018}
Yu. Nesterov.
\newblock \href{https://link.springer.com/book/10.1007/978-3-319-91578-4}{{\em Lectures on Convex Optimization}}.
\newblock Springer International Publishing, 2018.

\end{thebibliography}

\end{document}