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If a function is defined as sum of radicals in another radicals. Then how to differentiate this function $$\sqrt{\cos x+\sqrt{\cos x+\sqrt{\cos x+\dots}}}\quad?$$

Robert Z
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Zonnie
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  • is it finite or infinite? – IntegrateThis Dec 08 '16 at 14:43
  • If it is finite then it is just a very messy application of the chain rule, if it is infinite then I have no idea. That is the more interesting case. – IntegrateThis Dec 08 '16 at 14:44
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    Assuming that your pattern is actually infinite, your function satisfies the equation $f(x) = \sqrt{\cos(x) + f(x)}$. You can use this information to solve for $f(x)$. Once you have arrived at a simpler representation, you can then differentiate it easily. – Tom Dec 08 '16 at 14:46
  • Alternatively, if it's possible to find a recurrence for the derivative of the $n$th iteration of the representation, you can take the limit as $n \to \infty$. – MathematicsStudent1122 Dec 08 '16 at 14:52
  • $g(x) = \sqrt{\cos x+\sqrt{\cos x+\sqrt{\cos x+\dots}}}\quad$ is the limit - if it exists - of the sequence of functions $g_0(x) = 1$, $g_{n+1}(x) = \sqrt{\cos(x)+g_n(x)}$. For the derivative, it might be complicated and it might be not defined. But if the sequence of derivatives $g_0'(x) = -\sin(x), g_{n+1}'(x) = \frac{-\sin(x)+g_n'(x)}{2\sqrt{\cos(x)+g_n(x)}}$ converges to some function $h(x)$ and if $g_n(0)$ converges, then $g'(x) = h(x)$ and $g_n(x)$ converges to $g(x)$. – reuns Dec 08 '16 at 15:03

2 Answers2

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Putting all convergence problems aside, let: $$f(x)=\sqrt{\cos(x)+\sqrt{\cos(x)+\sqrt{\cos(x)\cdots}}}$$ so that $$f(x)=\sqrt{\cos(x)+f(x)}$$ By the chain rule $$f'(x)=\dfrac{f'(x)-\sin(x)}{2\sqrt{\cos(x)+f(x)}}=\dfrac{f'(x)-\sin(x)}{2f(x)}$$ so that $$f'(x)\big(2f(x)-1\big)=-\sin(x)$$ that is $$f'(x)=\dfrac{\sin(x)}{1-2f(x)}$$ provided $f(x)\not=\dfrac{1}{2}$. We also see that $$f(x)=\dfrac{1}{2}\Rightarrow\dfrac{1}{2}=\sqrt{\cos(x)+1/2}\Rightarrow\cos(x)=-\dfrac{1}{4}$$ but $f(x)$ is not defined for such $x$.

Olivier Moschetta
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  • And what about "all convergence problems"? – Arnaldo Dec 08 '16 at 14:52
  • Setting $a=\cos(x)$, this boils down to whether the sequence $u_n$ defined by $u_0=\sqrt{a}$ and $u_{n+1}=\sqrt{a+u_n}, n\geq 1$ is convergent. Clearly $a\geq 0$ is a necessary condition. For such $a$ one can show by induction that $u_n$ is increasing and that $0\leq u_n\leq\sqrt{a}+1$ see here: http://math.stackexchange.com/questions/333050/how-do-i-prove-the-sequence-x-n1-sqrt-alpha-x-n-x-0-sqrt-alpha-alph – Olivier Moschetta Dec 08 '16 at 15:01
  • That's helping. Thank you @Olivier and all.. – Zonnie Dec 08 '16 at 16:22
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So $f^2(x) - \cos(x) = f(x)$. Differentiating, $2 f(x) f'(x) + \sin(x) = f'(x)$ and then $f'(x) = \frac{\sin(x)}{1 - 2 f(x)}$. Which suggests the requirement that "$f(x) \neq 1/2$". You can solve the original quadratic for $f$ if you want to remove the $f$ from the right-hand side of the derivative.

I make no claim that this $f$ or its derivative exists.

Logan Clark
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Eric Towers
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