4

I am looking for alternate proof for Viete's infinite product of nested radicals. (Reference - Wikipedia)

Basically we need to find $\lim_{n\to \infty}\prod_{k=1}^{n} T_k$ where $$T_{k+1} = \sqrt{\left(\frac{T_k + 1}{2}\right)}$$ and $T_1 = \sqrt{\frac{1}{2}}$. Series looks like

$$\sqrt{\frac{1}{2}} \cdot \sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}}} \cdot \sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}}}}...$$

Miss gave a solution treating $\cos(\theta) = \frac{1}{\sqrt2}$ that is $\theta = 45^\circ$. The series result is given easily using the identity $\cos(\theta) + 1 = 2 \cos^2(\theta/2)$ and using $\sin(2\theta) = 2\sin(\theta)\cos(\theta)$. The final result is $\frac{\sin(2\theta)}{2\theta} = \frac{2}{\pi}$.

I look for alternate ways to get to this! I am open to calculus methods.

Later
  • 722
  • 2
  • 8
  • 24
King Tut
  • 3,074
  • The final result Not that it matters in this case, but I think that should rather be $,\frac{\sin(2 \theta)}{2\theta},$. You can telescope algebraically by using that $,2 T_{k+1}^2 = T_k + 1 \implies T_{k+2} = \frac{1}{2}\sqrt{\frac{T_k-1}{T_{k+1}-1}},$, but that still leaves $,\lim_{n \to \infty} 2^n(T_n-1),$ to determine, which isn't obvious offhand. – dxiv Feb 02 '18 at 20:46
  • Yes obviously it is, I simply wrote sin 90 as 1 without mention – King Tut Feb 03 '18 at 02:41
  • 1
    I think almost all of the answers are going to come down to some form of the half-angle identity. This is, incidentally, the classic Viète product formula. (Note that Viète's own proof comes down to this as well; he computes areas of $(2^n)$-gons inscribed in a circle as $n\to\infty$.) – Steven Stadnicki Feb 06 '18 at 18:12
  • @Steven - Thank you for valuable comment, I think you are correct. – King Tut Feb 08 '18 at 12:36
  • 2
    See this https://math.stackexchange.com/questions/1066363/probabilistic-proof-of-prod-limits-i-1-infty-cos-left-frac-t2i-right – Guy Fsone Feb 08 '18 at 14:01
  • What about pure geometric derivation (without using trig identitites at all)? That's how Viete did it (see the first paragraph of Wikipedia page, section Derivation). Edit: Just read Steven's comment, I'll need to check if Viete used trigonometry or not, though I think it wasn't developed enought at the time?) – Yuriy S Feb 13 '18 at 09:10
  • Yes yuriy I'm too confused how viete did it, doesn't look like he used trigonometric formula although Wikipedia may seem to imply that from its derivation tab. – King Tut Feb 13 '18 at 10:19
  • @Guy Actually that link I have found useful, for proving the end equality. Thank you! – King Tut Feb 13 '18 at 14:43
  • @KingTut it is actually similar technics that I used in my answer – Guy Fsone Feb 13 '18 at 14:56
  • @GuyFsone No. I am not referring to xpaul's answer. – King Tut Feb 13 '18 at 15:11

1 Answers1

1

Answer: we have $$\lim_{n\rightarrow\infty}\prod_{k=1}^{n} T_k =\lim_{n\rightarrow\infty}\prod_{k=1}^{n} \cos\left(\frac{\pi}{2^{k+1}}\right) =\lim_{n\rightarrow\infty}\frac{\sin (\pi/2)}{2^{n}\sin\left(\frac{\sqrt{2}}{2^{n}}\right)} = \color{blue}{\frac{\sin (\pi/2)}{\pi/2}}$$

  1. First check that for all $k$ we have $0\le T_k\le 1$ this is obvious by induction since $$0\le T_1= \frac{\sqrt{2}}{2}\le 1$$
  2. Hence there exists $a_k \in [0,\frac{\pi}{2} ]$ such that $$T_k = \cos a_k, $$
  3. Easily, $T_1 =\frac{\sqrt{2}}{2}\implies a_1 =\frac{\pi}{4}$ and $$\cos (a_{k+1})= T_{k+1} = \sqrt{\left(\frac{T_k + 1}{2}\right)}= \sqrt{\left(\frac{\cos a_k + 1}{2}\right)} = \cos\left(\frac{ a_k }{2}\right)$$
  4. Since $x\mapsto \cos x$ realize a bijection in $[0,\frac{\pi}{2} ],$ It turn out that $a_k$ is geometric sequence with ratio $1/2$ that is we have $$\color{red}{a_{k+1} = \frac{ a_k }{2}\implies a_k = \frac{a_1}{2^{k-1}}=\frac{\pi}{2^{k+1}}}$$

  5. Therefore, we have

    $$\color{red}{T_{k} = \cos\left(\frac{\pi}{2^{k+1}}\right)}$$

    6.By double angle formula we have $$\sin\left(x\right)=2\sin\left(\frac{x}{2}\right)\cos\left(\frac{x}{2}\right)=4\sin\left(\frac{x}{4}\right)\cos\left(\frac{x}{4}\right)\cos\left(\frac{x}{2}\right)\\=\dots=2^{n}\sin\left(\frac{x}{2^{n}}\right)\prod_{k\leq n}\cos\left(\frac{x}{2^{k}}\right)$$ now remains to note that $$\lim_{n\rightarrow\infty}2^{n}\sin\left(\frac{x}{2^{n}}\right)= \lim_{n\rightarrow\infty} x\frac{\sin\left(\frac{x}{2^{n}}\right)}{\frac{x}{2^{n}}}=\lim_{h\rightarrow 0} x\frac{\sin h}{h}= x .$$

  6. Thus $$\lim_{n\rightarrow\infty}\prod_{k=1}^{n} T_k =\lim_{n\rightarrow\infty}\prod_{k=1}^{n} \cos\left(\frac{\pi}{2^{k+1}}\right) =\lim_{n\rightarrow\infty}\frac{\sin (\pi/2)}{2^{n}\sin\left(\frac{\pi}{2^{n}}\right)} = \color{blue}{\frac{\sin (\pi/2)}{\pi/2}}$$

Guy Fsone
  • 23,903