Shoulders, Giants, and Cats🐈

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

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(