2020 MEXT Japanese Government Scholarship Qualifying Examinations

Compute $\sin(\sin(...\sin(x)))$

Problem Find the limit $\sin(\sin(...\sin(x)))$ for any real number $x$. Solution The problem can be precisely formulated as follows: Find the limit $L$ such that $$ L \equiv\lim_{n\rightarrow\infty}{f_n(x)}, $$ where $x\in\mathbb{R}$, $n\in\mathbb{N}$, and $$\begin{align} f_1(x) &= \sin(x)\\ f_n(x) &= \sin(f_{n-1}(x)). \end{align}$$ We observe that, if the limit $L$ exists, $$\begin{align} \lim_{n\rightarrow\infty} f_{n-1}(x) &= L\\ \lim_{n\rightarrow\infty} f_n(x) &= L, \end{align}$$ and thus, $$ L = \sin{L}, $$ which we show to be $L=0$ in the following. First, we show that the limit exists. We accept the Axiom of Continuity of the Real Line, which says that any bounded monotonic sequence converges. Thus, to show that $L$ exists, we need only show that (1) $f_n(x)$ is a bounded sequence of $n$, and that (2) $f_n(x)$ is a monotonic sequence of $n$. $f_n(x)$ is clearly bounded because $f_n(...

Find the limit of the sequence. $a_1=1, a_{n+1} = \frac{3a_n+4}{2a_n+3}$

We take the limits of sequences defined recursively by first showing the existence of the limits then, actually computing them. Existence is demonstrated by showing that the sequence is both (1) monotonic and (2) bounded. Problem Find the limit of the following sequence. $$\begin{align} $a_1&=1\\ a_{n+1} &= \frac{3a_n+4}{2a_n+3}$ \end{align}$$ Solution Monotonicity : We use mathematical induction to show that for all $n\in\mathbb{N}$, $a_{n+1} - a_n \geq 0$ and therefore, $\{a_n\}$ is monotonic increasing. For the base case at $n=1$, it is easy to see that $a_{n+1}-a_n = \frac{7}{5} - 1 = \frac{2}{5} >0.$ For the inductive case at $n=k$, we assume that $$\begin{align} a_{k+1} - a_{k} & = \frac{3a_k+4}{2a_k+3} - \frac{3a_{k-1}+4}{2a_{k-1}+3}\\ & = \frac{a_k - a_{k-1}}{(2a_k+3)(2a_{k-1}+3)} \geq 0. \end{align}$$ For $n=k...

2020 MEXT Japanese Government Scholarship Undergraduate Students Natural Sciences Qualifying Examination Mathematics (B): Problem 1(5)

This problem appears at the Qualifying Examinations for Applicants for Japanese Government (MEXT) Scholarships 2020 . There are two mathematics exams: one for biology-related natural sciences (Mathematics A), and another for physics- and engineering-related natural sciences (Mathematics B). This problem is from the 2020 Mathematics (B) questionnaire . The official answer key is here . Problem 1(5) The angle $\theta\left( 0\lt \theta \lt \frac{\pi}{2}\right)$ between the two lines $y=(2-\sqrt{3})x$ and $y=(\sqrt{3}-2)x$ on the $xy$-plane is $\fbox{ A }$. Solution For convenience, let us name the lines $l_1$ and $l_2$. $$\begin{align} l_1: y = (2-\sqrt{3})x\\ l_2: y = (\sqrt{3}-2)x \end{align}$$ Let us also call the angle that $l_1$ makes with the positive $x$-axis $\theta_1$, and the angle that $l_2$ makes with the positive $x$-axis $\theta_2$. The problem seeks to find the angle $\theta$ which is the smaller of $|\theta_1-\theta_2|$ and $\pi-|\t...

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