Differentiation Marathon!

For college and university level advanced Mathematics
Asif Hossain
Posts:194
Joined:Sat Jan 02, 2021 9:28 pm
Re: Problem: 6

Unread post by Asif Hossain » Tue Apr 27, 2021 2:39 pm

~Aurn0b~ wrote:
Fri Apr 23, 2021 1:50 am
Mehrab4226 wrote:
Thu Apr 22, 2021 10:25 pm
This is actually a self made problem, just for fun.
Find $\frac{d}{dx}\sqrt{x+\sqrt{x+\sqrt{x+\cdots}}}$
Note:
There might be 2 derivatives. I don't know 2 derivatives are a thing or not. But just for the fun, we can do it, right?
Let $y=\sqrt{x+\sqrt{x+\sqrt{x+\cdots}}}$
$\Rightarrow y=\sqrt{x+y} \Rightarrow y^2=x+y$
$\Rightarrow\frac{dy^2}{dx}=\frac{dx}{dx}+\frac{dy}{dx}$
$\Rightarrow\frac{dy^2}{dy}\cdot \frac{dy}{dx}=1+\frac{dy}{dx}$
$\Rightarrow2y\cdot \frac{dy}{dx}-\frac{dy}{dx}=1$
$\Rightarrow \frac{dy}{dx}=\frac{1}{2y-1}=\frac{1}{2\sqrt{x+\sqrt{x+\sqrt{x+\cdots}}}-1}$

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.
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 $\infty$? :?:
Hmm..Hammer...Treat everything as nail

~Aurn0b~
Posts:46
Joined:Thu Dec 03, 2020 8:30 pm

Re: Problem: 6

Unread post by ~Aurn0b~ » Wed Apr 28, 2021 1:46 pm

Asif Hossain wrote:
Tue Apr 27, 2021 2:39 pm
~Aurn0b~ wrote:
Fri Apr 23, 2021 1:50 am
Mehrab4226 wrote:
Thu Apr 22, 2021 10:25 pm
This is actually a self made problem, just for fun.
Find $\frac{d}{dx}\sqrt{x+\sqrt{x+\sqrt{x+\cdots}}}$
Note:
There might be 2 derivatives. I don't know 2 derivatives are a thing or not. But just for the fun, we can do it, right?
Let $y=\sqrt{x+\sqrt{x+\sqrt{x+\cdots}}}$
$\Rightarrow y=\sqrt{x+y} \Rightarrow y^2=x+y$
$\Rightarrow\frac{dy^2}{dx}=\frac{dx}{dx}+\frac{dy}{dx}$
$\Rightarrow\frac{dy^2}{dy}\cdot \frac{dy}{dx}=1+\frac{dy}{dx}$
$\Rightarrow2y\cdot \frac{dy}{dx}-\frac{dy}{dx}=1$
$\Rightarrow \frac{dy}{dx}=\frac{1}{2y-1}=\frac{1}{2\sqrt{x+\sqrt{x+\sqrt{x+\cdots}}}-1}$

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.
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 $\infty$? :?:
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.

Post Reply