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        riddles >> medium >> solve  sqrt(x+1999·sqrt(x+1999·sqrt(...))) = x
        
(Message started by: tony123 on Nov 10^th, 2007, 1:43am)

Title: solve sqrt(x+1999·sqrt(x+1999·sqrt(...))) = x
Post by tony123 on Nov 10^th, 2007, 1:43am

http://www.komal.hu/verseny/1999-11/b3317.gif

Title: Re: solve
Post by towr on Nov 10^th, 2007, 9:23am

hmm.. not interested enough in your own problems to notice it's illegible?

Title: V(x+1999V(x+1999V(x+1999V(x+1999V(2000x)))))=x
Post by JohanC on Nov 10^th, 2007, 2:27pm

on 11/10/07 at 09:23:16, towr wrote:

hmm.. not interested enough in your own problems to notice it's illegible?

Looks like a problem with the transparent color in the gif format.
So, why not save it as png (which saves 300 bytes)?
(Waiting for somebody to change the thread's title ...)

Title: Re: solve sqrt(x+1999·sqrt(x+1999·sqrt(...))
Post by Grimbal on Nov 10^th, 2007, 3:39pm

Obvious solutions are 0 and 2000.

And considering that the whole sqrt(...) expression is concave, there can be no more than 2 real solutions.

The other solutions can be found among the zeros of a polynomial of degree 30, which I am too lazy to compute...

Title: Re: solve sqrt(x+1999·sqrt(x+1999·sqrt(...))
Post by anhkind on Nov 18^th, 2007, 2:35am

sqrt(x+1999·sqrt(x+1999·sqrt(...))) = x
<=> sqrt(x+1999*x)=x
<=> 2000x=x^2
<=> x=0 V x=2000
Solved.

Title: V(x+1999V(x+1999V(x+1999V(x+1999V(2000x)))))=x
Post by FiBsTeR on Nov 18^th, 2007, 8:45am

on 11/18/07 at 02:35:33, anhkind wrote:

sqrt(x+1999·sqrt(x+1999·sqrt(...))) = x

The original equation was closed; you can't find all the solutions from this.

Aside: is it just a coincidence that the above equation yields solutions that are also solutions to the original equation?

Title: Re: V(x+1999V(x+1999V(x+1999V(x+1999V(2000x)))))=x
Post by towr on Nov 18^th, 2007, 9:10am

on 11/18/07 at 08:45:06, FiBsTeR wrote:

Aside: is it just a coincidence that the above equation yields solutions that are also solutions to the original equation?

No.
If you start with sqrt(x+1999*x) and substitute the second x with x=sqrt(x+1999*x), x=0 or x=2000; then x=0 and x=2000 will remain solutions.