Hii !
Can You Help Me For This Equation ,
Solve in IR the Equation :
$\displaystyle x=a-\sqrt{a-\sqrt{x}}$
following the values of parameter a ( $\displaystyle a \in \mathbb{R^+}$ )
Salut !
$\displaystyle x=a-\sqrt{a-\sqrt{x}}$
$\displaystyle \sqrt{x}=\sqrt{a-\sqrt{a-\sqrt{x}}}$
$\displaystyle a-\sqrt{x}=a-\sqrt{a-\sqrt{a-\sqrt{x}}}$
$\displaystyle \sqrt{a-\sqrt{x}}=\sqrt{a-\sqrt{a-\sqrt{a-\sqrt{x}}}}$
$\displaystyle a-\sqrt{a-\sqrt{x}}=a-\sqrt{a-\sqrt{a-\sqrt{a-\sqrt{x}}}}$
$\displaystyle x=a-\sqrt{a-\sqrt{a-\sqrt{a-\sqrt{x}}}}$
and so on ...
Therefore taking the series defined by $\displaystyle u_0 = a$ and $\displaystyle u_{n+1}=f(u_n)$ where $\displaystyle f(x) = a-\sqrt{x}$
x can be seen as the limit (if it exists) of $\displaystyle \left(u_n\right)_n$
If it converges, it is towards l such as $\displaystyle l = f(l)$ which can be solved by a quadratic
At the end $\displaystyle x = \frac{2a+1\pm \sqrt{4a+1}}{2}$ ... if I am not mistaken
It seems to be "minus" but I let you see why !!
All what I have written is to be checked ...