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I was solving some differentiation problems when I found the function

$$g(x)=\sqrt{x+\sqrt{x+\sqrt{x}}}.$$

So I thought: If I define the function $f:\mathbb{R_{x>0}}\to \mathbb{R}$ as

$$f(x)=\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+...}}}}$$

What kind of informations have I about the function $f$? Is it continous or differentiable, in some "sense"? If yes, is it correct to say by implicit differentiation that

$$f'(x)=\frac{1}{2f(x)-1}?$$

Thanks so much.

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  • $\begingroup$ you haven't, technically, defined $f(x)$ without defining that infinite expression. But no, that's not the derivative. Even doing a possibly illegitimate chain rule, you'd get a numerator on the right side of $1+f'(x)$ not $1$. That turns out to be correct. - you can solve for $f'$. $\endgroup$ Aug 31, 2015 at 3:52
  • $\begingroup$ Thanks @ThomasAndrews. For the derivative of $f$, if $f>0$, so $(f)^2=x+f$ then by implicit differetiation, $2ff'=1+f'$. So, we conclude $f'=\frac{1}{2f-1},$ or not? $\endgroup$
    – Irddo
    Aug 31, 2015 at 5:02
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    $\begingroup$ This question is certainly not dumb. +1 for the thinking. If you read the answer, you will learn from your mistake also. $\endgroup$
    – Shailesh
    Aug 31, 2015 at 5:04
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    $\begingroup$ Now that you've added the $-1$ to the answer, sure. It wasn't there when I commented, @Irddo $\endgroup$ Aug 31, 2015 at 9:58

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If $$f(x)=\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+...}}}}$$ then assuming this function actually converges for a particular $x$ (in fact, as it stands, your domain may not be properly restricted for convergence, making your "function" not necessarily well defined.) Then $$f(x)=\sqrt{x+f(x)}$$ $$[f(x)]^2-f(x)=x$$ Implictly, $$2f'(x)f(x)-f'(x)=1$$ $$f'(x)=\frac{1}{2f(x)-1}$$ This entirely assumes your function is, indeed, a function. You can actually work this out. By the quadratic formula, we have $$f(x)=\frac{1\pm\sqrt{1+4x}}{2}$$ So clearly, at least, $x\in[-\frac{1}{4},\infty)$. Your chosen domain, $(0,\infty)$, then looks okay. Now to deal with that $\pm$ sign. Your function, in its original form is non-negative, so we can only take the negative branch for $$\frac{1-\sqrt{1+4x}}{2}\geq0$$ $$\sqrt{1+4x}\leq 1$$ $$1+4x\leq 1$$ $$x\leq 0$$ Which is wonderful, since by our restriction on the domain, we don't need to consider the negative branch. Therefore the function is well defined, and is: $$f(x)=\frac{1+\sqrt{1+4x}}{2},x>0$$ Which is continuous and differentiable on its domain.

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    $\begingroup$ It would be useful for @Irddo to take the derivative of that closed form for $f(x)$ to see that the result is correct. $2f(x)-1=2\sqrt{1+4x}$... $\endgroup$ Aug 31, 2015 at 14:36

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