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Given a function $$y=\sqrt{\log(x)+\sqrt{\log(x)+\ldots}}$$ then what is the value of $\frac{dy}{dx}$.

My approach was to assume the inner radical as $y$ and then by equation manipulation we get $y’(2y-1)=\frac{1}{x}$. I am getting the value of $y’$ but how can I prove this is rigorously (because this is an infinite series I have to use real analysis to make it rigorous) .

I am unable to show whether $y$ is convergent or divergent. How to solve this using epsilon delta definition of limits?

Sebastiano
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Sillyasker
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  • Any function which is greater than the natural log function but is a convergent function? It might help me to figure out the missing part. – Sillyasker Oct 25 '24 at 17:41
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    I would rather suggest using sequences, i.e., $y_1 = \log x, \ y_n = \sqrt{\log x + \sqrt{y_{n-1}}}.$ – Sean Roberson Oct 25 '24 at 17:42
  • But how can I use this for the convergence proof? @SeanRoberson – Sillyasker Oct 25 '24 at 17:43
  • Prove that $y_n$ converges to some limit. You may use properties instead of the $\epsilon - N$ definition (maybe monotone and bounded...?). – Sean Roberson Oct 25 '24 at 17:45
  • @SeanRoberson I think $y_1=\sqrt{\log x}$ and $y_n = \sqrt{\log x +{y_{n-1}}}$ might be more natural. You will want $x \ge 1$. – Henry Oct 25 '24 at 18:02

1 Answers1

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We observe that $\log(x)$ must be non-negative, and $$\iff y^2 = \log(x) + y \iff y^2 -y -\log(x) = 0 \tag{1}$$

The quadratic equation $(1)$ in function of $y$ has $\Delta = 1+4\log(x)$, the only non-negative solution is $$y = \frac{1+\sqrt{1+4\log(x)}}{2}$$

From what you prove, we deduce:

$$\frac{dy}{dx} = y' = \frac{1}{x(2y-1)} = \frac{1}{x\sqrt{1+4\log(x)}} $$

NN2
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    What you did is absolutely fine but I wanted something which has some use of real analysis and limits ideas . Can you provide me that kind of solution @NN2? – Sillyasker Oct 25 '24 at 18:46
  • @Sillyasker I think you want to prove that $y$ exists. in order words, the sequence $y_n$ (defined in the comment section of the question by Henry) converges? I found this and perhaps it is what you are seeking? – NN2 Oct 25 '24 at 19:09