等于$\sin{(\frac{1}{2}\arcsin{x})}$的两个根式

hbghlyj

Hall and Knight - Higher Algebra Volume 2 page 500:
105. Shew that the expansion of $\sqrt{\frac{1-\sqrt{1-x^2}}{2}}$ is$$\frac{x}{2}+\frac{1.3}{2.4} \cdot \frac{x^3}{6}+\frac{1.3 .5 .7}{2.4 .6 .8} \cdot \frac{x^5}{10}+\ldots $$
106. If $\alpha, \beta$ are roots of the equations
$$
x^2+p \cdot x+q=0, \quad x^{2 n}+p^n x^n+q^n=0,
$$
where $n$ is an even integer, shew that $\frac{\alpha}{\beta}, \frac{\beta}{\alpha}$ are roots of
$$
x^n+1+(x+1)^n=0 . \quad\small \text { [Pembroke College, Cambridge] }
$$
107. Find the difference between the squares of the infinite continued fractions$$a+\frac{b}{2 a+} \frac{b}{2 a+} \frac{b}{2a+}\ldots \text { and } c+\frac{d}{2 c+}\frac{d}{2c+} \frac{d}{2c+}\ldots \quad\small\text{[Christ's College, Cambridge.]}$$

hbghlyj

math.stackexchange.com/questions/2341167
\[\sqrt{\frac{1-\sqrt{1-x^2}}{2}}=\frac{\sqrt{1+x}-\sqrt{1-x}}2, \quad|x|\le1,\]
By Binomial Theorem
\begin{align*}(1+x)^{\frac{1}{2}}&=1+\frac{1}{2} x-\frac{1}{2} \cdot \frac{x^{2}}{4}+\frac{1.3}{2.4} \cdot \frac{x^{3}}{6}-\frac{1.3 .5}{2.4 .6} \cdot \frac{x^{4}}{8}+\frac{1.3 .5 .7}{2.4 .6 .8} \cdot \frac{x^{5}}{10}-\ldots,\quad|x|\le1\\
(1-x)^{\frac{1}{2}}&=1-\frac{1}{2} x-\frac{1}{2} \cdot \frac{x^{2}}{4}-\frac{1.3}{2.4} \cdot \frac{x^{3}}{6}-\frac{1.3 .5}{2.4 .6} \cdot \frac{x^{4}}{8}-\frac{1.3 .5 .7}{2.4 .6 .8} \cdot \frac{x^{5}}{10}-\ldots,\quad|x|\le1
\end{align*}Substracting
\[\frac{\sqrt{1+x}-\sqrt{1-x}}2=\frac{x}{2}+\frac{1.3}{2.4} \cdot \frac{x^3}{6}+\frac{1.3 .5 .7}{2.4 .6 .8} \cdot \frac{x^5}{10}+\ldots,\quad|x|\le1\]

hbghlyj

Denote the values of the continued fractions by $x$ and $y$; then
$$
x-a=\frac{b}{2 a+(x-a)} ; \text { whence } x^2=a^2+b \text {. }
$$
Similarly $y^2=c^2+d$; thus $x^2-y^2=a^2+b-c^2-d$.

hbghlyj

$x^2+p \cdot x+q=0\implies\cases{\alpha+\beta=-p\\\alpha \beta=q}$
Subtracting $\cases{a^{2 n}+p^n a^n+q=0\\\beta^{2 n}+p^n \beta^n+q=0}$
we get: $a^{2 n}-\beta^{2 n}+p^n\left(\alpha^n-\beta^n\right)=0$
Dividing by $(\alpha^n-\beta^n)$ we get: $\alpha^n+\beta^n+p^n=0$;
Plugging in $\alpha+\beta=-p$ we get: $a^n+\beta^n+(\alpha+\beta)^n=0$ (since $n$ is an even integer);
and therefore $x^n+1+(x+1)^n=0$, where $x=\frac{a}{\beta}$ or $\frac{\beta}{a}$.

hbghlyj

在107.中的连分数是否收敛? 如果是, 收敛到二次方程的哪个根?