davidbeckham
Quote:
as the expression under the square root extends infinitely then expression in brackets would equal to xx itself
What is the rationale behind this?
Hey,
davidbeckham,
This confused me too and is the reason I'm on this thread.
I realized... because the expression under the square root extends infinitely, the expression under the first square root \(\sqrt{here ...}\) is still \(\sqrt{6+{\sqrt{6 + ...}}}\) . . . infinitely.
The expression under the second square root: \(\sqrt{6+{\sqrt{(here...)}}}\) is still \(\sqrt{6+{\sqrt{6 + ...}}}\) . . . infinitely. And so on and so forth.
We know that x = \(\sqrt{6+{\sqrt{6 + ...}}}\) infinitely.
So, really, you can plug x in for the "infinite" portion under the first, second, third, fourth (you get the picture) square root.
In other words:
x = \(\sqrt{6 + x}\)
x also = \(\sqrt{6+{\sqrt{6 + x}}}\) ... etc.
... Unless, I'm completely wrong.
Bunuel, is my understanding correct?