I have a foolish question any way(maybe out of topic) does $\sqrt{x+\sqrt{x+\sqrt{x+\cdots}}}$ always converge? for example when we try to find the value of $\sqrt{2+\sqrt{2+\sqrt{2+\cdots}}}$ we assume that to be equal to some to real $x$ but it could also be $\infty$?~Aurn0b~ wrote: ↑Fri Apr 23, 2021 1:50 amLet $y=\sqrt{x+\sqrt{x+\sqrt{x+\cdots}}}$Mehrab4226 wrote: ↑Thu Apr 22, 2021 10:25 pmThis is actually a self made problem, just for fun.
Find $\frac{d}{dx}\sqrt{x+\sqrt{x+\sqrt{x+\cdots}}}$
Note:
$\Rightarrow y=\sqrt{x+y} \Rightarrow y^2=x+y$
$\Rightarrow\frac{dy^2}{dx}=\frac{dx}{dx}+\frac{dy}{dx}$
$\Rightarrow\frac{dy^2}{dy}\cdot \frac{dy}{dx}=1+\frac{dy}{dx}$
$\Rightarrow2y\cdot \frac{dy}{dx}-\frac{dy}{dx}=1$
$\Rightarrow \frac{dy}{dx}=\frac{1}{2y-1}=\frac{1}{2\sqrt{x+\sqrt{x+\sqrt{x+\cdots}}}-1}$
This could've also been solved by obtaining $\frac{dx}{dy}$ and using the fact that $(\frac{dx}{dy})^{-1}=\frac{dy}{dx}$
I dont have any good derivative problem, so anyone can post the new problem.
Differentiation Marathon!
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Hmm..Hammer...Treat everything as nail
Re: Problem: 6
define $y_n=\sqrt{x+y_{n-1}}$, where $y_1=\sqrt{x}$. first prove that y is increasing. Then prove that y has an upper-bound.. Both can be proved by induction.Asif Hossain wrote: ↑Tue Apr 27, 2021 2:39 pmI have a foolish question any way(maybe out of topic) does $\sqrt{x+\sqrt{x+\sqrt{x+\cdots}}}$ always converge? for example when we try to find the value of $\sqrt{2+\sqrt{2+\sqrt{2+\cdots}}}$ we assume that to be equal to some to real $x$ but it could also be $\infty$?~Aurn0b~ wrote: ↑Fri Apr 23, 2021 1:50 amLet $y=\sqrt{x+\sqrt{x+\sqrt{x+\cdots}}}$Mehrab4226 wrote: ↑Thu Apr 22, 2021 10:25 pmThis is actually a self made problem, just for fun.
Find $\frac{d}{dx}\sqrt{x+\sqrt{x+\sqrt{x+\cdots}}}$
Note:
$\Rightarrow y=\sqrt{x+y} \Rightarrow y^2=x+y$
$\Rightarrow\frac{dy^2}{dx}=\frac{dx}{dx}+\frac{dy}{dx}$
$\Rightarrow\frac{dy^2}{dy}\cdot \frac{dy}{dx}=1+\frac{dy}{dx}$
$\Rightarrow2y\cdot \frac{dy}{dx}-\frac{dy}{dx}=1$
$\Rightarrow \frac{dy}{dx}=\frac{1}{2y-1}=\frac{1}{2\sqrt{x+\sqrt{x+\sqrt{x+\cdots}}}-1}$
This could've also been solved by obtaining $\frac{dx}{dy}$ and using the fact that $(\frac{dx}{dy})^{-1}=\frac{dy}{dx}$
I dont have any good derivative problem, so anyone can post the new problem.