static-site-pdfs/TokyoEntrance2016/TokyoEntrance2016.tex

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\title{Tokyo 2016 Graduate School Entrance Exam}
\author{Patrick Stevens}
\date{16th March, 2017}

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\maketitle

\tiny \begin{center} \url{https://www.patrickstevens.co.uk/misc/TokyoEntrance2016/TokyoEntrance2016.pdf} \end{center}
\normalsize

\section{Question 2}
\subsection{Prove \texorpdfstring{that $S = 2\pi \int_{-1}^1 F(y, y') \ \mathrm{d}x$}{a certain integral form for S}}
The surface may be parametrised as $$S(x, \theta) = (x, y(x) \cos(\theta), y(x) \sin(\theta))$$
where $\theta \in [0, 2\pi)$ and $x \in [-1,1]$.

Hence $$\dfrac{\partial S}{\partial x} = (1, y'(x) \cos(\theta), y'(x) \sin(\theta))$$ and $$\dfrac{\partial S}{\partial \theta} = (0, -y(x) \sin(\theta), y(x) \cos(\theta))$$
so the surface element $$\mathrm{d}\Sigma = \left| \left(1, y'(x) \cos(\theta), y'(x) \sin(\theta) \right) \times (0, -y(x) \sin(\theta), y(x) \cos(\theta)) \right| \mathrm{d}x \mathrm{d}\theta$$
i.e. $$y \sqrt{1+(y')^2}$$

The integral is therefore $$\int_{0}^{2 \pi} \int_{-1}^1 y \sqrt{1+(y')^2} \mathrm{d}x \mathrm{d}\theta$$
as required.

\subsection{Prove the first integral of the Euler-Lagrange equation}
We know the Euler-Lagrange equation $$\dfrac{\partial F}{\partial y} = \dfrac{\mathrm{d}}{\mathrm{d}x} \dfrac{\partial F}{\partial y'}$$

Now, $$\frac{\mathrm{d}F}{\mathrm{d}{x}} = \dfrac{\partial F}{\partial y} \dfrac{\mathrm{d}y}{\mathrm{d}x} + \dfrac{\partial F}{\partial y'} \dfrac{\mathrm{d}y'}{\mathrm{d}x}$$
so substituting Euler-Lagrange into this:
$$\frac{\mathrm{d}F}{\mathrm{d}{x}} = \dfrac{\mathrm{d}}{\mathrm{d}x} \left(\dfrac{\partial F}{\partial y'}\right) \dfrac{\mathrm{d}y}{\mathrm{d}x} + \dfrac{\partial F}{\partial y'} \dfrac{\mathrm{d}y'}{\mathrm{d}x}$$

Notice the right-hand side is just what we get by applying the product rule: it is $$\dfrac{\mathrm{d}}{\mathrm{d}x} \left( \dfrac{\partial F}{\partial y'} \dfrac{\mathrm{d}y}{\mathrm{d}x} \right)$$

The result follows now by simply integrating both sides with respect to $x$.

\subsection{Solve the differential equation}

Just substitute $F(y,y') = y \sqrt{1+(y')^2}$:
$$y\sqrt{1+(y')^2} - y' \left[ \frac{1}{2} y (1+(y')^2)^{-1/2} \cdot 2 y'\right] = c$$
which can be simplified to $$y (1+(y')^2)^{-1/2} \left[ (1+(y')^2) -(y')^2\right] = c$$
i.e. $$y^2 - c^2 = (c y')^2$$

If $c=0$ then this is trivial: $y=0$. From now on, assume $c \not = 0$; then since $y$ is known to be positive, $c > 0$.

Invert: $$\frac{c^2}{y^2-c^2} = \left(\frac{dx}{dy}\right)^2$$
so $$\dfrac{dx}{dy} = \pm \frac{c}{\sqrt{y^2-c^2}}$$
which is a standard integral: $$x = \pm c \log(y+\sqrt{y^2-c^2}) + K$$

Also $y(-1) = 2 = y(1)$, so $$\{1,-1\} = \{c \log(2+\sqrt{4-c^2}) + K, -c \log(2+\sqrt{4-c^2}) + K\}$$
which means $K = 0$.

Then $$\exp\left(\pm \frac{x}{c}\right) = y+\sqrt{y^2-c^2}$$
Since $y(1) = 2$, we have $$\exp(\pm 1/c) = 2+\sqrt{4-c^2}$$ and in particular (since $c>0$) we have the $\pm$ on the left-hand side being positive; that is the expression $c$ is required to satisfy.

Rearrange: $$y = \frac{c+\exp(2 \frac{x}{c})}{2 \exp\left(\frac{x}{c}\right)}$$
which completes the question.

\section{Question 3}
\subsection{Part 1}
We must put one ball into each box. Then we are distributing $n-r$ balls freely among $r$ boxes, so the answer is $$\binom{n-r-1}{r-1}$$
(standard stars-and-bars result).

\subsection{Part 2}
Consider the $n$ black balls laid out in a line; we are interspersing the $m$ white balls among them.
Equivalently, we have $n+1$ boxes (represented by the gaps between black balls) and we are trying to put $m$ balls into them.
By stars-and-bars again, the answer is $\binom{n}{m-1}$.

\subsection{Part 3}
Condition on the colour of the first ball, and write $l$ for the length of the first run.
Then
$$P_{n,m}(r,s) = \frac{n}{n+m} \sum_{l=0}^{n} P_{n-l,m}(r-1, s) + \frac{m}{n+m} \sum_{l=0}^m P_{n, m-l}(r, s-1)$$
Also $P_{n,m}(0,s) = \chi[n=0] \chi[s=1]$ where $\chi$ is the indicator function, and $P_{n,m}(r,0) = \chi[m=0] \chi[r=1]$.

\subsection{Part 4}

\subsection{Part 5}
If $m \leq n$, then the sum is $$\sum_{l=0}^m \binom{n}{l} \binom{m}{m-l}$$ which is the $x^m$ coefficient of the left-hand side and hence of the right-hand side.

If $m > n$, then the sum is $$\sum_{l=0}^n \binom{n}{n-l} \binom{m}{l}$$ which is the $x^n$ coefficient of the left-hand side and hence of the right-hand side.

For the second equation: this follows by setting $n \mapsto n-1$ in the above.

\subsection{Part 6}

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