Rukevwe wrote: Find the real number satisfying $x=\sqrt{1+\sqrt{1+\sqrt{1+x}}}$. Is answer is two real values of x?

Pi-rate_91 wrote: Rukevwe wrote: Find the real number satisfying $x=\sqrt{1+\sqrt{1+\sqrt{1+x}}}$. Is answer is 2? Unfortunately, no.

Rukevwe wrote: Find the real number satisfying $x=\sqrt{1+\sqrt{1+\sqrt{1+x}}}$. My method $x=\sqrt{1+\sqrt{1+\sqrt{1+x}}}$ Substituting value of $x$ in R.H.S from L.H.S $x=\sqrt{1+\sqrt{1+\sqrt{1+...}}}$ $x=\sqrt{1+x}$ $x^2=x+1$ $x^2-x-1=0$ As $D>0$ , two real values of x. $x=\frac{1\pm\sqrt{5}}{2}$

Rukevwe wrote: Pi-rate_91 wrote: Rukevwe wrote: Find the real number satisfying $x=\sqrt{1+\sqrt{1+\sqrt{1+x}}}$. Is answer is 2? Unfortunately, no. Oops, I meant 2 real values of $x$

Of course the argument in #5 is not very formal using $\dots$ (without talking about convergence of anything). Here is a very simple and formal way: If $\sqrt{1+x}>x$, then \[x=\sqrt{1+\sqrt{1+\sqrt{1+x}}}>\sqrt{1+\sqrt{1+x}}>\sqrt{1+x}>x,\]contradiction! The same argument shows that $\sqrt{1+x}<x$ leads to a contradiction. Hence we must have $\sqrt{1+x}=x$ (and this is clearly also sufficient) and now we just solve the quadratic equation.

Tintarn wrote: Of course the argument in #5 is not very formal using $\dots$ (without talking about convergence of anything). Here is a very simple and formal way: If $\sqrt{1+x}>x$, then \[x=\sqrt{1+\sqrt{1+\sqrt{1+x}}}>\sqrt{1+\sqrt{1+x}}>\sqrt{1+x}>x,\]contradiction! The same argument shows that $\sqrt{1+x}<x$ leads to a contradiction. Hence we must have $\sqrt{1+x}=x$ (and this is clearly also sufficient) and now we just solve the quadratic equation. Hmm...nice one

contain quadratic equation so it's not really bad