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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-|\theta_1-\t

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

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(4) The division of a polynomial function $f(x)$ by $(x-1)^2$ gives the remainder of $x+1$, and that by $x^2$ gives the remainder $2x+3$. Thus, the remainder of the division of $f(x)$ by $x^2(x-1)$ is $$ \fbox{ A }x^2 + \fbox{ B }x + \fbox{ C } . $$ Solution We need to find the remainder when $f(x)$ is divided by $x^2(x-1)$. Because $x^2(x-1)$ is of order $n=3$, the remainder will be of at most the order $n=2$, which means that it is of the form $Ax^2 + Bx + C$. The problem is now to find the coefficients $A,B$ and $C$ such that $$ f(x) = Q(x)\

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(3) Let $a$ and $b$ be real numbers with $b \geq 0$. When the equation $$ x^4 + ax^2 + b = 0 $$ has exactly two real solutions, the minimum value of $a + 2b$ is $\fbox{ A }$, and the maximum value of $\lceil a - b \rceil$ is $\fbox{ B }$. Here, $\lceil r \rceil$ denotes the smallest integer that is larger than or equal to the real number $r$. Solution The most important hint is the phrase " has exactly two real solutions ". This means that there are no complex solutions, and at least one or both of the real solutions have a multiplicity greater

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(2) Let $f(x) = 1 + \frac{1}{x-1}(x\neq 1)$. The solution of the equation $f\left(f\left(x\right)\right) = f\left(x\right)$ is $x = \fbox{A }, \fbox{B }$. Solution The right-hand side of the equation is $f(x)$, which is already given in the problem to be $$ f\left(x\right) = 1 + \frac{1}{x-1}. $$ The left-hand side of the equation is $$\begin{align} f\left(f\left(x\right)\right) &= 1 + \frac{1}{f(x) - 1}\\ &= 1 + \frac{1}{\left(1+\frac{1}{x-1}\right) - 1}\\ &= 1 + \frac{1}{\frac{1}{x-1}}\\ &= 1 + {x-1}\\ &= x. \end{align}

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(1) The largest one among natural numbers that are less than $$ \log_2{3}\cdot\log_3{4}\cdot\log_4{5}\cdots\cdots\log_{2019}{2020} $$ is $ \fbox { A } $. Solution The key idea is the following relationship that holds for positive numbers $p, q,$ and $b$. $$ \log_p{q} = \frac{\log_b{q}}{\log_b{p}}, $$ where $p\neq1$ and $b\neq1$. For convenience, let $$ L \equiv \log_2{3}\cdot\log_3{4}\cdot\log_4{5}\cdots\cdots\log_{2019}{2020}. $$ Selecting $b=e$, we obtain $$\begin{align} L &=\frac{\cancel{\log{3}}}{\log{2}}\cdot \frac{\cancel{\

We prove identities involving inverse trigonometric functions. Sum formulas from trigonometry are exploited. $\arctan{\frac{1}{2}}+\arctan{\frac{1}{3}} = \frac{\pi}{4}$ Suppose that $$ \arctan{\frac{1}{2}}+\arctan{\frac{1}{3}} \neq \frac{\pi}{4}. $$ Then, $$\begin{align} \tan{\left(\arctan{\frac{1}{2}}+\arctan{\frac{1}{3}}\right)} &\neq \tan{\frac{\pi}{4}}\\ \frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{2}\cdot\frac{1}{3}} &\neq 1 \\ \frac{5}{6-5} &\neq 1 \\ 1 &\neq 1, \end{align}$$ which is a contradiction. Thus, $$ \arctan{\frac{1}{2}}+\arctan{\frac{1}{3}} = \frac{\pi}{4}.□ $$ $\arcsin{\frac{16}{65}}+\arcsin{\frac{5}{13}} = \arcsin{\frac{3}{5}}$ Suppose that $$ \arcsin{\frac{16}{65}}+\arcsin{\frac{5}{13}} \neq \arcsin{\frac{3}{5}}. $$ Then, $$\begin{align} \sin{\left(\arcsin{\frac{16}{65}}+\arcsin{\fra

Equations involving inverse trigonometric functions are solved for the unknown. The following relation is assumed proven: $$ \theta = \arcsin{\frac{a}{c}} = \arccos{\frac{\sqrt{c^2-a^2}}{c}}, $$ which is clear from the figure below. $\arcsin{\frac{3}{5}}=\arctan{x}$ $$\begin{align} \arctan{x} &= \arcsin{\frac{3}{5}} = \arctan{\frac{3}{\sqrt{5^2-3^2}}} = \arctan{\frac{3}{4}}\\ &\Rightarrow x = \frac{3}{4}.□ \end{align}$$ $\arcsin{\frac{3}{5}}=2\arctan{x}$ $$\begin{align} 2\arctan{x} &= \arcsin{\frac{3}{5}} = \arctan{\frac{3}{\sqrt{5^2-3^2}}} = \arctan{\frac{3}{4}}\\ \frac{3}{4} &= \tan{\left(2\arctan{x}\right)} = \frac{2\tan{\left(\arctan{x}\right)}}{1-\tan^2{\left(\arctan{x}\right)}} \\ &= \frac{2x}{1-x^2}\\ &\Rightarrow 0 = 3x^2 +8x -3 = (3x-1)(3x+3) \\ &\Rightarrow x = \frac{1}{3} ( x = -1 \hbox{ is extraneous}).□ \end{align}$$ References

The following exercises involve the simplification of numerical expressions involving inverse trigonometric functions. We use the co-function relationships and reciprocal relationships among trigonometric functions. $\sin{\left(2\arccos{\frac{2}{5}}\right)}=?$ $$\begin{align} &=2\sin{\left(\arccos{\frac{2}{5}}\right)}\cos{\left(\arccos{\frac{2}{5}}\right)}% =2\cdot\frac{2}{5}\sin{\left(\arccos{\frac{2}{5}}\right)}\\% &=\frac{4}{5}\sin{\left(\arcsin{\frac{\sqrt{5^2-2^2}}{5}}\right)}\\ &=\frac{4}{5}\cdot\frac{\sqrt{21}}{5}\\ &=\frac{4}{25}\sqrt{21}.□ \end{align}$$ $\arcsin{a}+\arccos{a}=?$ We observe that because sine and cosine functions are co-functions of each other, for any $a$ in the domain of inverse cosine and inverse sine functions, $$ y \equiv \sin(\theta) = \cos{\left(\frac{\pi}{2} - \theta\right)} \Rightarrow \arccos{y} = \frac{\pi}{2} -\arcsin{y}. $$ Thus, the gi

In the following problems, we find the limits of sequences $\{a_n\}$ given the functional form of the sequences. We assume that some fundamental limits such as $$ \lim_{n\rightarrow\infty}\frac{1}{n} = 0 $$ and $$ \lim_{n\rightarrow\infty}\left(1+\frac{1}{n}\right)^n \equiv e $$ are known. $a_n = \left(1-\frac{1}{n}\right)^n $ $$\begin{align} a_n &= \left(\frac{n-1}{n}\right)^n% = \left(\frac{n}{n-1}\right)^{-n}\\ &=\left(\frac{n-1}{n-1}+\frac{1}{n-1}\right)^{-n}\\ &=\left(1+\frac{1}{n-1}\right)^{-n}\\ &=\left(1+\frac{1}{n-1}\right)^{-(n-1)}\left(1+\frac{1}{n-1}\right)^{-1}\\ &=\frac{\left(1+\frac{1}{n-1}\right)^{-1}}{\left(1+\frac{1}{n-1}\right)^{(n-1)}}\\ % \lim_{n\rightarrow\infty}a_n% &=\lim_{n\rightarrow\infty}\frac{\left(1+\frac{1}{n-1}\right)^{-1}}{\left(1+\frac{1}{n-1}\right)^{(n-1)}}\\ &=\frac{\lim_{n\rightarrow\infty}\left(1+\frac{1}{n-1}\right)^{-1}}{\lim_{{n-1}\rightarrow\infty}\left(1+\fra

Problem Find the value of $S$. $$ (★): S \equiv \lim_{k\rightarrow 0} \sum_{n=1}^{\infty} \frac{(-1)^{n!+k!+n+2}}{k!(2n+1)!} $$ Solution $$\begin{align} S &= \lim_{k\rightarrow 0} \sum_{n=1}^{\infty} \frac{(-1)^{n!} (-1)^{k!} (-1)^n \cancelto{1}{(-1)^2}}{k!(2n+1)!}\\ &= \lim_{k\rightarrow 0} \frac{(-1)^{k!}}{k!} \sum_{n=1}^{\infty} \frac{(-1)^{n!} (-1)^n}{(2n+1)!} \end{align}$$ We make the following observations. $n!$ is always even all natural numbers $n \geq 2$ because $$ n! = n\cdot(n-1)\cdot(n-2)\cdot\ldots\cdot 2 \cdot 1. $$ Therefore, $(-1)^{n!} = 1$ for any natural number $n\geq2$. The limit in (★) only makes sense if we define $k!$ and $-1$ as follows. $$\begin{aligned} k! &\equiv \Gamma(k+1)\\ -1 &\equiv e^{j\pi} \end{aligned}$$ Here, $\Gamma(k+1)$ is the gamma function and $j$ is the imaginary unit. We then get $$ \lim_

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(

Let the people decide what is good for them. On the one hand, this principle is at the core of democracy and manifests in the phrase "majority rule." On the other hand, the protection of the voice of the minority is another principle a functioning democracy must enforce. Democracy requires both. The fundamental assumption is that, after free and informed debate, the majority will be reasonable enough to judge ideas based on their merits (and not on their emotional relationship with the proponents). Because in a democracy, the Government's wisdom is an extension of the people's wisdom, wise decision-making is the duty of every citizen. The "goodness" of a decision or an idea does not necessarily depend on the number of its proponents but on the independent practical assessment of available information. The difficulty in balancing the interests of opposing sides is highlighted on occasions when the minority is unwilling to accept the decision of the

Two years into the pandemic, many continue to suffer from the disease. Even two or three vaccine doses later, our lifestyle remains hostage to the risk of a sudden outbreak that could cripple our liberties, as does the recent surge in cases of infections of the Omicron strain. Twenty-twenty-two brings no guarantee of complete relief and restoration of our freedoms stolen by COVID-19. But, in spite of (and sometimes, because of) the pandemic, I've built new genuine friendships, gained godchildren, and witnessed blissful weddings of my best friends. I feel hopeful that more good things are to come in 2022. I look forward to being free to travel again, to getting rid of the fear of the virus and getting used to living with it, and to beginning a new life chapter (after I finish writing those chapters I've been working on for the last three years). It feels awful not being able to celebrate with the people I love. But finding ways to make up despite the distance is