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Prove the relation $\cos^{-1}x_0=\dfrac{\sqrt {1-x^2_0}}{x_1\cdot x_2\cdot x_3\cdots \text{ ad inf.}}$ where the successive quantities $x_r$ are connected by the relation $x_{r+1}=\sqrt{\frac{1}{2}(1+x_r)}$

My attempt:

$$x_1=\sqrt{\frac{1}{2}(1+x_0)}$$

$$x_2=\sqrt{\frac{1}{2}(1+x_1)}$$

$$x_3=\sqrt{\frac{1}{2}(1+x_2)}$$

$$x_4=\sqrt{\frac{1}{2}(1+x_3)}$$

Multiplying all these, we get

$$x_1\cdot x_2\cdot x_3\cdots\text{ ad inf.}=\sqrt{\frac{1}{2}(1+x_0)\frac{1}{2}(1+x_1)\frac{1}{2}(1+x_2)\frac{1}{2}(1+x_3)\cdots \text{ ad inf.}}$$

Putting in equation,

$$\cos^{-1}x_0=\frac{\sqrt {1-x^2_0}}{\sqrt{\frac{1}{2}(1+x_0)\frac{1}{2}(1+x_1)\frac{1}{2}(1+x_2)\frac{1}{2}(1+x_3)\cdots \text{ ad inf.}}}$$

$$\cos^{-1}x_0=\frac{\sqrt {1-x_0}}{\sqrt{\frac{1}{2}(1+x_1)\frac{1}{2}(1+x_2)\frac{1}{2}(1+x_3)\cdots \text{ ad inf}}}$$

but i could not solve further. Can someone guide me in this question?

Michael Hardy
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Vinod Kumar Punia
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  • I would try to find a geometrical representation of the sequence generated by $(x_r, x_{r+1})$ or something. I strongly suspect the infinite product comes from some iterated construction. – AlexR Jul 29 '15 at 16:30
  • Did this come from Euler? Not only does the math seem like something he might write, but the words "ad inf." are not frequently used in the present day. ${}\qquad{}$ – Michael Hardy Jul 29 '15 at 18:28

2 Answers2

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Let $x_0:=\cos y$. Want to prove that $y = RHS$. Also we know, that in case of convergence, $x_r \to 1$ because $\bar x = \sqrt{\frac12(1+\bar x)} \Rightarrow \bar x^2 - \frac12 \bar x - \frac12 = 0$ and $\bar x \ge 0$. The roots of the quadratic are $1$ and $-\frac12$.

We now need to show that $$y = \frac{\sqrt{1-\cos^2 y}}{\prod_{r=1}^\infty x_r}$$ By restricting $y$ to $[0, \pi]$ we can thus write $$y = \frac{\sin y}{\prod_{r=1}^\infty x_r}$$ Or equivalently $$\frac{\sin y}y = \prod_{r=1}^\infty x_r$$ This identity might be related to some other product identity for the cardinal sine. Specifically looking at $$\frac{\sin y}y = \prod_{k=1}^\infty \cos \left(\frac y{2^k}\right)$$ looks promising.

In fact we can prove that $$x_r = \cos\left(\frac y{2^k}\right)$$ by induction. Note that $$\frac12(1+\cos x) = \frac12 (1+\cos(\frac x2 + \frac x2)) = \frac12(1 + \cos^2 \frac x2 - \sin^2 \frac x2) = \frac12 (2\cos^2 \frac x2) = \cos^2 \frac x2$$ Thus $$x_1 = \sqrt{\frac12(1+\cos y)} = \cos \frac y2$$ Now by induction $$x_r = \sqrt{\frac12(1+\cos \frac y{2^{r-1}})} = \cos \frac y{2^r}$$

AlexR
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Let $\cos(\theta)=x_0$, so that $\sqrt{1-x_0^2}=\sin(\theta)$.

From known trigonometry, we have

$$x_{r+1}=\cos\left(\frac{\arccos(x_r)}2\right),$$ and by recurrence $$x_r=\cos\left(\frac{\arccos(x_0)}{2^r}\right)=\cos\left(\frac{\theta}{2^r}\right).$$

Hence the denominator is the infinite product

$$\prod_1^\infty\cos\left(\frac\theta{2^r}\right)=\text{sinc}(\theta),$$ and the ratio is $\theta$.