finish the pure questions from step ii 2017

Author: EasonSYC

tex/2017.tex

@@ -1,4 +1,4 @@
 \Year{\currfilebase}
 % \input{\currfiledir\currfilebase/1}
-% \input{\currfiledir\currfilebase/2}
+\input{\currfiledir\currfilebase/2}
 \input{\currfiledir\currfilebase/3}

tex/2017/2.tex

@@ -7,8 +7,8 @@
 \input{\currfiledir\currfilebase/6}
 \input{\currfiledir\currfilebase/7}
 \input{\currfiledir\currfilebase/8}
-\input{\currfiledir\currfilebase/9}
-\input{\currfiledir\currfilebase/10}
-\input{\currfiledir\currfilebase/11}
+% \input{\currfiledir\currfilebase/9}
+% \input{\currfiledir\currfilebase/10}
+% \input{\currfiledir\currfilebase/11}
 \input{\currfiledir\currfilebase/12}
 \input{\currfiledir\currfilebase/13}

tex/2017/2/1.tex

@@ -1,2 +1,98 @@
 \Question{\currfilebase}
-\WorkInProgress
+
+\begin{enumerate}
+    \item Using integration by parts, we notice that
+          \begin{align*}
+              (n + 1) I_n & = (n + 1) \int_{0}^{1} x^n \arctan x \Diff x                                                             \\
+                          & = \int_{0}^{1} \arctan x \Diff x^{n + 1}                                                                 \\
+                          & = \left[\arctan x \cdot x^{n + 1}\right]_{0}^{1} - \int_{0}^{1} x^{n + 1} \Diff \arctan x                \\
+                          & = \arctan 1 \cdot 1^{n + 1} - \arctan 0 \cdot 0^{n + 1} - \int_{0}^{1} \frac{x^{n + 1}}{1 + x^2} \Diff x \\
+                          & = \frac{\pi}{4} - \int_{0}^{1} \frac{x^{n + 1}}{1 + x^2} \Diff x.
+          \end{align*}
+
+          Set \(n = 0\), and we have
+          \begin{align*}
+              I_0 & = (0 + 1) I_0                                                         \\
+                  & = \frac{\pi}{4} - \int_{0}^{1} \frac{x}{1 + x^2} \Diff x              \\
+                  & = \frac{\pi}{4} - \frac{1}{2} \cdot \left[\ln(1 + x^2)\right]_{0}^{1} \\
+                  & = \frac{\pi}{4} - \frac{1}{2} \cdot \left[\ln 2 - \ln 1\right]        \\
+                  & = \frac{\pi}{4} - \frac{\ln 2}{2}.
+          \end{align*}
+
+    \item Using the result in the previous part,
+          \begin{align*}
+              (n + 3) I_{n + 2} + (n + 1) I_n & = \left(\frac{\pi}{4} - \int_{0}^{1} \frac{x^{n + 3}}{1 + x^2} \Diff x\right) + \left(\frac{\pi}{4} - \int_{0}^{1} \frac{x^{n + 1}}{1 + x^2} \Diff x\right) \\
+                                              & = \frac{\pi}{2} - \int_{0}^{1} \frac{x^{n + 1} + x^{n + 3}}{1 + x^2} \Diff x                                                                                \\
+                                              & = \frac{\pi}{2} - \int_{0}^{1} \frac{x^{n + 1} \left(1 + x^2\right)}{1 + x^2} \Diff x                                                                       \\
+                                              & = \frac{\pi}{2} - \int_{0}^{1} x^{n + 1} \Diff x                                                                                                            \\
+                                              & = \frac{\pi}{2} - \frac{1}{n + 2} \left[x^{n + 2}\right]_{0}^{1}                                                                                            \\
+                                              & = \frac{\pi}{2} - \frac{1}{n + 2}.
+          \end{align*}
+
+          Letting \(n = 0\), and we have
+          \[
+              3 I_2 + I_0 = \frac{\pi}{2} - \frac{1}{2}.
+          \]
+
+
+          Letting \(n = 2\), and we have
+          \[
+              5 I_4 + 3 I_2 = \frac{\pi}{2} - \frac{1}{4}.
+          \]
+
+          Subtracting the first one from the second one, and hence
+          \[
+              5 I_4 - I_0 = \frac{1}{4}.
+          \]
+
+          Hence,
+          \[
+              I_4 = \frac{1}{5} \cdot \left[\frac{1}{4} + \left(\frac{\pi}{4} - \frac{\ln 2}{2}\right)\right] = \frac{1}{20} + \frac{\pi}{20} - \frac{\ln 2}{10}.
+          \]
+
+    \item Let \(n = 1\), and the statement says
+          \begin{align*}
+              (4n + 1) I_{4n} & = 5 I_4                                                      \\
+                              & = A - \frac{1}{2} \sum_{r = 1}^{2 \cdot 1}(-1)^r \frac{1}{r} \\
+                              & = A - \frac{1}{2} \left(- \frac{1}{1} + \frac{1}{2}\right)   \\
+                              & = A + \frac{1}{4}.
+          \end{align*}
+
+          Comparing to the previous expression, we claim that
+          \[
+              A = \frac{\pi}{4} - \frac{\ln 2}{2}.
+          \]
+
+          This shows the base case for \(n = 1\). For the induction step, we first introduce a lemma. Since
+          \[
+              (n + 5) I_{n + 4} + (n + 3) I_{n + 2} = \frac{\pi}{2} - \frac{1}{n + 4}, (n + 3) I_{n + 2} + (n + 1) I_{n} = \frac{\pi}{2} - \frac{1}{n + 2},
+          \]
+          subtracting the second one from the first one will give us
+          \[
+              (n + 5) I_{n + 4} - (n + 1) I_{n} = \frac{1}{n + 2} - \frac{1}{n + 4}.
+          \]
+
+          Setting \(n = 4m\), we have
+          \begin{align*}
+              (4(m + 1) + 1) I_{4 (m + 1)} & = (4m + 1) I_{4m} + \frac{1}{4m + 2} - \frac{1}{4m + 4}                                                             \\
+                                           & = (4m + 1) I_{4m} - \frac{1}{2} \cdot \left(-\frac{1}{2m + 1} + \frac{1}{2m + 2}\right)                             \\
+                                           & = (4m + 1) I_{4m} - \frac{1}{2} \cdot \left[(-1)^{2m + 1} \frac{1}{2m + 1} + (-1)^{2m + 2} \frac{1}{2m + 2}\right].
+          \end{align*}
+
+          Now we show the inductive step. Assume the statement is true for some \(n = k \geq 1\), i.e.
+          \[
+              (4k + 1) I_{4k} = A - \frac{1}{2} \sum_{r = 1}^{2n} (-1)^r \frac{1}{r}.
+          \]
+
+          Using the identity above, we have
+          \begin{align*}
+              (4(k + 1) + 1) I_{4 (k + 1)} & = (4k + 1) I_{4k} - \frac{1}{2} \cdot \left[(-1)^{2k + 1} \frac{1}{2k + 1} + (-1)^{2k + 2} \frac{1}{2k + 2}\right]                                      \\
+                                           & = A - \frac{1}{2} \sum_{r = 1}^{2k} (-1)^r \frac{1}{r} - \frac{1}{2} \cdot \left[(-1)^{2k + 1} \frac{1}{2k + 1} + (-1)^{2k + 2} \frac{1}{2k + 2}\right] \\
+                                           & = A - \frac{1}{2} \sum_{r = 1}^{2(k + 1)} (-1)^r \frac{1}{r}.
+          \end{align*}
+
+          Hence, the original statement is true for \(n = 1\) (as shown when determining the value of \(A\)), and given the original statement holds for some \(n = k \geq 1\), it holds for \(n = k + 1\). By the principle of mathematical induction, this statement holds for all \(n \geq 1\), where
+          \[
+              A = \frac{\pi}{4} - \frac{\ln 2}{2}.
+          \]
+\end{enumerate}

tex/2017/2/2.tex

@@ -1,2 +1,65 @@
 \Question{\currfilebase}
-\WorkInProgress
+
+We have
+\begin{align*}
+    x_{n + 2} & = \frac{a x_{n + 1} - 1}{x_{n + 1} + b}                                       \\
+              & = \frac{a \cdot \frac{a x_n - 1}{x_n + b} - 1}{\frac{a x_n - 1}{x_n + b} + b} \\
+              & = \frac{a (a x_n - 1) - (x_n + b)}{(a x_n - 1) + b (x_n + b)}                 \\
+              & = \frac{(a^2 - 1) x_n - (a + b)}{(a + b) x_n + (b^2 - 1)}.
+\end{align*}
+
+\begin{enumerate}
+    \item If the sequence is periodic with period \(2\), then for all integers \(n \geq 0\), we have
+          \begin{align*}
+              x_{n + 2} = x_n & \iff x_n \left[(a + b) x_n + (b^2 - 1)\right] = (a^2 - 1) x_n - (a + b) \\
+                              & \iff (a + b) x_n^2 - (a + b) (a - b) x_n + (a + b) = 0                  \\
+                              & \iff (a + b) (x_n^2 - (a - b) x_n + 1) = 0.
+          \end{align*}
+
+          We also have
+          \begin{align*}
+              x_{n + 1} = x_n & \iff x_n (x_n + b) = a x_n - 1    \\
+                              & \iff x_n^2 - (a - b) x_b + 1 = 0,
+          \end{align*}
+          and this means that for some \(n = k \geq 0\), we must have \(x_n^2 - (a - b) x_n + 1 \neq 0\) (otherwise, the sequence will have period \(1\)).
+
+          Therefore, for such \(n = k\), we must have \(a + b = 0\) for the first condition to be true, and hence this is a necessary condition.
+
+    \item Using the formula between \(x_{n + 4}\) and \(x_n\), we have
+          \begin{align*}
+              x_{n + 4} & = \frac{(a^2 - 1) x_{n + 2} - (a + b)}{(a + b) x_{n + 2} + (b^2 - 1)}                                                                                                                                                         \\
+                        & = \frac{(a^2 - 1) \cdot \frac{(a^2 - 1) x_n - (a + b)}{(a + b) x_n + (b^2 - 1)} - (a + b)}{(a + b) \cdot \frac{(a^2 - 1) x_n - (a + b)}{(a + b) x_n + (b^2 - 1)} + (b^2 - 1)}                                                 \\
+                        & = \frac{(a^2 - 1) \cdot \left[(a^2 - 1) x_n - (a + b)\right] - (a + b) \cdot \left[(a + b) x_n + (b^2 - 1)\right]}{(a + b) \cdot \left[(a^2 - 1) x_n - (a + b)\right] + (b^2 - 1) \cdot \left[(a + b) x_n + (b^2 - 1)\right]} \\
+                        & = \frac{\left[(a^2 - 1)^2 - (a + b)^2\right] x_n - \left[(a^2 - 1)(a + b) + (a + b)(b^2 - 1)\right]}{(a + b)\left[(a^2 - 1) + (b^2 - 1)\right] x_n + \left[(b^2 - 1)^2 - (a + b)^2\right]}.
+          \end{align*}
+
+          If sequence has period \(4\), we have \(x_{n + 4} = x_n\) for all integers \(n \geq 0\), and the sequence does not have period \(1\), \(2\) or \(3\).
+
+          We notice
+          \begin{align*}
+              x_{n + 4} = x_n & \iff x_n \cdot \left[(a + b)\left[(a^2 - 1) + (b^2 - 1)\right] x_n + \left[(b^2 - 1)^2 - (a + b)^2\right]\right] \\
+                              & \phantom{\iff} = \left[(a^2 - 1)^2 - (a + b)^2\right] x_n - \left[(a^2 - 1)(a + b) + (a + b)(b^2 - 1)\right]     \\
+                              & \iff (a + b) (a^2 + b^2 - 2) \left(x_n^2 - (a - b)x_n + 1 \right) = 0.
+          \end{align*}
+
+          From the previous part, we know that for some \(n = k \geq 0\), we must have
+          \[
+              (a + b) \left(x_k^2 - (a - b)x_k + 1 \right) \neq 0,
+          \]
+          which means \(a + b \neq 0\) and \(x_k^2 - (a - b) x_k + 1 \neq 0\). Hence, we must have \(a^2 + b^2 - 2 = 0\).
+
+          On the other hand, if \(a^2 + b^2 - 2 = 0\), \(a + b \neq 0\) and \(x_k^2 - (a - b) x_k + 1 \neq 0\) for some \(n = k \geq 0\), we know that the sequence does not satisfy \(x_{n + 1} = x_n\), does not satisfy \(x_{n + 2} = x_n\), and it satisfies \(x_{n + 4} = x_n\).
+
+          If \(x_{n + 3} = x_n\), then we have \(x_{n + 3} = x_{n + 4}\) which contradicts with not satisfying \(x_{n + 1} = x_n\). Hence, the sequence does not satisfy \(x_{n + 3} = x_n\), and it must have period \(4\).
+
+          Therefore, the sequence has period \(4\), if and only if
+          \[
+              \left\{
+              \begin{aligned}
+                  a + b                  & \neq 0,                                \\
+                  a^2 + b^2 - 2          & = 0,                                   \\
+                  x_k^2 - (a - b)x_k + 1 & \neq 0 \text{ for some } n = k \geq 0.
+              \end{aligned}
+              \right.
+          \]
+\end{enumerate}

tex/2017/2/3-diag1.tex

@@ -0,0 +1,14 @@
+\begin{tikzpicture}
+    \draw[->] (-pi - 0.5, 0) -- (pi + 0.5, 0) node[right] {\(x\)};
+    \draw[->] (0, -pi - 0.5) -- (0, pi + 0.5) node[right] {\(y\)};
+    \node at (0, 0) [below right] {\(0\)};
+    \draw (-pi, -pi) -- (pi, pi) node [above right] {\(y = x\)};
+    \draw (0, pi) -- (pi, 0) node [above right] {\(y = \pi - x\)};
+    \draw (-pi, 0) -- (0, -pi) node [below left] {\(y = -\pi - x\)};
+    \draw[dashed] (0, pi) -- (pi, pi) -- (pi, 0);
+    \draw[dashed] (-pi, 0) -- (-pi, -pi) -- (0, -pi);
+    \node at (0, pi) [left] {\(\pi\)};
+    \node at (pi, 0) [below] {\(\pi\)};
+    \node at (0, -pi) [right] {\(-\pi\)};
+    \node at (-pi, 0) [above] {\(-\pi\)};
+\end{tikzpicture}

tex/2017/2/3-diag2.tex

@@ -0,0 +1,9 @@
+\begin{tikzpicture}[trig format=rad, scale=3]
+    \draw[->] (-0.5, 0) -- ({pi / 6 + 0.5}, 0) node[right] {\(x\)};
+    \draw[->] (0, -0.5) -- (0, {pi / 2 + 0.5}) node[right] {\(y\)};
+    \draw [smooth, samples=200, domain=0:pi/2] plot (\x, {asin(0.5*sin(\x))});
+    \draw[dashed] (0, pi / 6) -- (pi / 2, pi / 6) -- (pi / 2, 0);
+    \node at (0, 0) [below right] {\(O\)};
+    \node at (pi / 2, 0) [below] {\(\pi / 2\)};
+    \node at (0, pi / 6) [left] {\(\pi / 6\)};
+\end{tikzpicture}

tex/2017/2/3-diag3.tex

@@ -0,0 +1,26 @@
+\begin{tikzpicture}[trig format=rad]
+    \draw[->] (-pi - 0.5, 0) -- (pi + 0.5, 0) node[right] {\(x\)};
+    \draw[->] (0, -pi - 0.5) -- (0, pi + 0.5) node[right] {\(y\)};
+    \draw [smooth, samples=200, domain=-pi:pi] plot (\x, {asin(0.5*sin(\x))});
+    \draw [smooth, samples=200, domain=0:pi] plot (\x, {pi - asin(0.5*sin(\x))});
+    \draw [smooth, samples=200, domain=-pi:0] plot (\x, {-pi - asin(0.5*sin(\x))});
+    \draw[dashed] (0, pi / 6) -- (pi / 2, pi / 6) -- (pi / 2, 0);
+    \draw[dashed] (0, -pi / 6) -- (-pi / 2, -pi / 6) -- (-pi / 2, 0);
+    \draw[dashed] (0, pi) -- (pi, pi) -- (pi, 0);
+    \draw[dashed] (0, 5 * pi / 6) -- (pi / 2, 5 * pi / 6) -- (pi / 2, pi / 6);
+    \draw[dashed] (0, -pi) -- (-pi, -pi) -- (-pi, 0);
+    \draw[dashed] (0, -5 * pi / 6) -- (-pi / 2, -5 * pi / 6) -- (-pi / 2, -pi / 6);
+
+    \node at (0, 0) [below right] {\(O\)};
+    \node at (-pi, 0) [above] {\(-\pi\)};
+    \node at (-pi / 2, 0) [above] {\(-\pi / 2\)};
+    \node at (pi / 2, 0) [below] {\(\pi / 2\)};
+    \node at (pi, 0) [below] {\(\pi\)};
+
+    \node at (0, -pi) [right] {\(-\pi\)};
+    \node at (0, -5 * pi / 6) [right] {\(-5 \pi / 6\)};
+    \node at (0, -pi / 6) [right] {\(-\pi / 6\)};
+    \node at (0, pi / 6) [left] {\(\pi / 6\)};
+    \node at (0, 5 * pi / 6) [left] {\(5 \pi / 6\)};
+    \node at (0, pi) [left] {\(\pi\)};
+\end{tikzpicture}

tex/2017/2/3-diag4.tex

@@ -0,0 +1,23 @@
+\begin{tikzpicture}[trig format=rad]
+    \draw[->] (-pi - 0.5, 0) -- (pi + 0.5, 0) node[right] {\(x\)};
+    \draw[->] (0, -pi - 0.5) -- (0, pi + 0.5) node[right] {\(y\)};
+    \draw [smooth, samples=200, domain=-pi:pi] plot (\x, {acos(0.5*sin(\x))});
+    \draw [smooth, samples=200, domain=-pi:pi] plot (\x, {- acos(0.5*sin(\x))});
+
+    \draw [dashed] (-pi, pi / 2) -- (-pi, -pi / 2) -- (pi, -pi / 2) -- (pi, pi / 2) -- cycle;
+    \draw [dashed] (0, pi / 3) -- (pi / 2, pi / 3) -- (pi / 2, -pi / 3) -- (0, -pi / 3);
+    \draw [dashed] (0, 2 * pi / 3) -- (-pi / 2, 2 * pi / 3) -- (-pi / 2, -2 * pi / 3) -- (0, -2 * pi / 3);
+
+    \node at (0, 0) [below right] {\(O\)};
+    \node at (-pi, 0) [above right] {\(-\pi\)};
+    \node at (-pi / 2, 0) [above right] {\(-\pi / 2\)};
+    \node at (pi / 2, 0) [below left] {\(\pi / 2\)};
+    \node at (pi, 0) [below left] {\(\pi\)};
+
+    \node at (0, -2 * pi / 3) [right] {\(-2\pi / 3\)};
+    \node at (pi, -pi / 2) [right] {\(-\pi / 2\)};
+    \node at (0, -pi / 3) [left] {\(-\pi / 3\)};
+    \node at (0, pi / 3) [left] {\(\pi / 3\)};
+    \node at (pi, pi / 2) [right] {\(\pi / 2\)};
+    \node at (0, 2 * pi / 3) [right] {\(2\pi / 3\)};
+\end{tikzpicture}

tex/2017/2/3.tex

@@ -1,2 +1,72 @@
 \Question{\currfilebase}
-\WorkInProgress
+
+\begin{enumerate}
+    \item Since \(\sin y = \sin x\), we must have
+          \[
+              y = x + 2 k \pi
+          \]
+          where \(k \in \ZZ\), or
+          \[
+              y = (2k + 1) \pi - x
+          \]
+          where \(k \in \ZZ\).
+
+          For the first case, since \(x \in [-\pi, \pi]\) and \(y \in [-\pi, \pi]\), we must have simply \(x = y\).
+
+          For the second case, within this range, we can have \(y = \pi - x\), and \(y = - \pi - x\).
+
+          Hence, the sketch looks as follows.
+
+          \begin{center}
+              \input{\currfiledir 3-diag1}
+          \end{center}
+
+    \item Differentiating with respect to \(x\), we have
+          \[
+              \cos y \DiffFrac{y}{x} = \frac{1}{2} \cos x.
+          \]
+
+          Since \(\sin y = \frac{1}{2} \sin x\), \(\cos y = \pm \sqrt{1 - \sin^2 y} = \pm \sqrt{1 - \frac{1}{4} \sin^2 x}\). Since \(0 \leq y \leq \frac{1}{2} \pi\), \(\cos y > 0\), and hence \(\cos y = \frac{1}{2} \sqrt{4 - \sin^2 x}\). Hence,
+          \[
+              \DiffFrac{y}{x} = \frac{\frac{1}{2} \cos x}{\frac{1}{2} \sqrt{4 - \sin^2 x}} = \frac{\cos x}{\sqrt{4 - \sin^2 x}}.
+          \]
+
+          Differentiating this again gives us
+          \begin{align*}
+              \NdiffFrac{2}{y}{x} & = \frac{(- \sin x) \sqrt{4 - \sin^2 x} - \frac{1}{2} \cdot (-2 \sin x) \cdot \cos x \cdot \frac{1}{\sqrt{4 - \sin^2 x}} \cdot \cos x}{4 - \sin^2 x} \\
+                                  & = \frac{(- \sin x) (4 - \sin^2 x) + \sin x \cos^2 x}{(4 - \sin^2 x)^{\frac{3}{2}}}                                                                  \\
+                                  & = \frac{-4 \sin x + \sin^3 x + \sin x (1 - \sin^2 x)}{(4 - \sin^2 x)^{\frac{3}{2}}}                                                                 \\
+                                  & = -\frac{3 \sin x}{(4 - \sin^2 x)^{\frac{3}{2}}},
+          \end{align*}
+          as desired.
+
+          Within this range of \(x\) and \(y\), we have
+          \[
+              y = \arcsin \left(\frac{1}{2} \sin x\right),
+          \]
+          and hence this is a function, and each \(x\) corresponds to a unique \(y\).
+
+          At \(x = 0\),
+          \[
+              y = 0, y' = \frac{\cos 0}{\sqrt{4 - \sin^2 0}} = \frac{1}{2},
+          \]
+          and at \(x = \frac{\pi}{2}\),
+          \[
+              y = \frac{\pi}{6}, y' = -\frac{\cos \frac{\pi}{2}}{\sqrt{4 - \sin^2 \frac{\pi}{2}}} = 0.
+          \]
+
+          Since \(y'' = - \frac{3 \sin x}{(4 - \sin^2 x)^{\frac{3}{2}}} < 0\) for \(x \in \left[0, \frac{\pi}{2}\right]\), this function is concave, and hence the graph looks as follows.
+          \begin{center}
+              \input{\currfiledir 3-diag2.tex}
+          \end{center}
+
+          Hence, for \((x, y) \in [-\pi, \pi]^2\), the graph looks as follows, by symmetry.
+          \begin{center}
+              \input{\currfiledir 3-diag3.tex}
+          \end{center}
+
+    \item The graph is as follows.
+          \begin{center}
+              \input{\currfiledir 3-diag4.tex}
+          \end{center}
+\end{enumerate}