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Topic: solve sqrt(x+1999·sqrt(x+1999·sqrt(...))) = x (Read 613 times) |
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tony123
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solve sqrt(x+1999·sqrt(x+1999·sqrt(...))) = x
« on: Nov 10th, 2007, 1:43am » |
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| « Last Edit: Nov 10th, 2007, 3:35pm by Grimbal » |
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towr
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Re: solve
« Reply #1 on: Nov 10th, 2007, 9:23am » |
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hmm.. not interested enough in your own problems to notice it's illegible?
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JohanC
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V(x+1999V(x+1999V(x+1999V(x+1999V(2000x)))))=x  Sqrt1.png
« Reply #2 on: Nov 10th, 2007, 2:27pm » |
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on Nov 10th, 2007, 9:23am, towr wrote:| hmm.. not interested enough in your own problems to notice it's illegible? |
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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 ...)
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Grimbal
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Re: solve sqrt(x+1999·sqrt(x+1999·sqrt(...))
« Reply #3 on: Nov 10th, 2007, 3:39pm » |
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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...
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| « Last Edit: Nov 10th, 2007, 3:49pm by Grimbal » |
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anhkind
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Re: solve sqrt(x+1999·sqrt(x+1999·sqrt(...))
« Reply #4 on: Nov 18th, 2007, 2:35am » |
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sqrt(x+1999·sqrt(x+1999·sqrt(...))) = x
<=> sqrt(x+1999*x)=x
<=> 2000x=x^2
<=> x=0 V x=2000
Solved.
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FiBsTeR
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V(x+1999V(x+1999V(x+1999V(x+1999V(2000x)))))=x
« Reply #5 on: Nov 18th, 2007, 8:45am » |
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on Nov 18th, 2007, 2:35am, anhkind wrote:| sqrt(x+1999·sqrt(x+1999·sqrt(...))) = x |
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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?
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| « Last Edit: Nov 18th, 2007, 8:46am by FiBsTeR » |
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towr
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Re: V(x+1999V(x+1999V(x+1999V(x+1999V(2000x)))))=x
« Reply #6 on: Nov 18th, 2007, 9:10am » |
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on Nov 18th, 2007, 8:45am, FiBsTeR wrote:| Aside: is it just a coincidence that the above equation yields solutions that are also solutions to the original equation? |
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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.
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