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Inspired by Vieta's Formula for $\pi$, $$\pi=2\cdot\frac{2}{\sqrt{2}}\cdot\frac{2}{\sqrt{2+\sqrt{2}}}\cdot\frac{2}{\sqrt{2+\sqrt{2+\sqrt{2}}}}\cdots$$ I became interested in a more generalized case for Vieta's Formula.

For $m=\sqrt{a+\sqrt{a+\sqrt{a+\sqrt{a+\cdots}}}}$, what is the closed form of $$m\cdot\frac{m}{\sqrt{a}}\cdot\frac{m}{\sqrt{a+\sqrt{a}}}\cdot\frac{m}{\sqrt{a+\sqrt{a+\sqrt{a}}}}\cdots$$?

To find the value of $m$, we can solve the following equation $$m=\sqrt{a+m}$$, which gives $$m=\frac{1+\sqrt{1+4a}}{2}$$

One example is that $$3=\sqrt{6+\sqrt{6+\sqrt{6+\sqrt{6+\cdots}}}}$$ And I want to find the exact value of $$3\cdot\frac{3}{\sqrt{6}}\cdot\frac{3}{\sqrt{6+\sqrt{6}}}\cdot\frac{3}{\sqrt{6+\sqrt{6+\sqrt{6}}}}\cdots\approx3.815$$ but to no avail. Could anyone provide any insight on the possible closed form for this infinite product?

Larry
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    @heepo: Huh??? That number is zero. – David G. Stork Dec 14 '19 at 01:47
  • Why not show insight into the problem and divide by $3$ throughout? – David G. Stork Dec 14 '19 at 01:49
  • @DavidG.Stork I see three favorites and one upvote, Also, more related: the meat of this question is probably concerned with finding $P_n = \sqrt{a+\sqrt{a\dots}}$ where $a$ appears $n$ times. I doubt there's an explicit formula for this, but a recursive one might give you something. – Descartes Before the Horse Dec 14 '19 at 01:49
  • I see $0$ favorites. – David G. Stork Dec 14 '19 at 01:50
  • Somewhat unrelated, but I am surprised at how many users favorite questions without upvoting. – Descartes Before the Horse Dec 14 '19 at 01:50
  • Re: number of favorites -- it doesn't update without you reloading the page. – PrincessEev Dec 14 '19 at 01:51
  • The limit of the $n$-th term in that product is $\dfrac{m}{\sqrt{a+1/4}+1/2}$. So if $m > \sqrt{a+1/4}+1/2$, the product grows without bound, and if $m < \sqrt{a+1/4}+1/2$, the product tends to $0$. So only the case where $m = \sqrt{a+1/4}+1/2$ remains. – JimmyK4542 Dec 14 '19 at 01:56
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    One of the reasons why the first product has a simple closed form is that $2x^2-1$ is a Chebyshev polynomial. The more general case of quadratic recurrence with closed-form solutions is discussed here, which seems to imply that it's very hard to find a closed form of the second product. –  Dec 14 '19 at 02:09
  • @heepo I use the Favourite function as an easy way to bookmark a question I'm interested in seeing answered. If I don't think it's an especially good question for whatever reason I won't upvote it. – Deepak Dec 14 '19 at 02:27
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    An interesting sub-question: just for the hell of it, define $$\Pi(a)=\prod_{n\ge0}\frac{1+\sqrt{1+4a}}{2q_n(a)}$$ where $q_{n+1}(a)=\sqrt{a+q_n(a)}$ and $q_0(a):=1$. Of course we see that $\Pi(2)=\pi$, but what is $\Pi(1)$? I take interest in this particular case, because it would involve everyone's favorite number, the golden ratio $\varphi=\frac{1+\sqrt5}{2}$. Perhaps the $\Pi(1)$ product, being a product of powers of $\varphi$, can be rephrased in terms of Fibonacci numbers. – clathratus Dec 14 '19 at 07:32

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Just a comment, as the comment section is really getting long, you may really want to know the connection why it is $2$ in the radical that gives this amazing result. It is due to it's connection to trigonometric functions. More specifically, a quick research into this can give that a result, (link) $$\frac{\sin x}{x}=\cos \frac{x}{2}\cdot\cos \frac{x}{4}\cdot\cos \frac{x}{8}...$$ And also the fact that $$\cos\frac{x}{2^n}=\frac{1}{2}\sqrt{2+\sqrt{2+\sqrt{2+...(n-1) times}}}$$ or more specifically speaking; $$A=\frac{\sin A}{\cos\frac{A}{2}\cdot \cos\frac{A}{4}\cdot \cos\frac{A}{8}...}$$ So a generalized version of this would be doubtful. And also one can obviously make a conjecture that that your stated problem will result in a transendental number which couldn't be expressed in terms of closed form algebraic numbers or functions.

A very good paper, if you're really interested in square roots will be this, by Dixon J. Jones

user712576
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