Icebreaker
[tex]f(x)=\sqrt{2x}[/tex]
[tex]x(f)=\sqrt{2f}[/tex]
Does this express:
[tex]\sqrt{2\sqrt{2\sqrt{...}}}[/tex]
[tex]x(f)=\sqrt{2f}[/tex]
Does this express:
[tex]\sqrt{2\sqrt{2\sqrt{...}}}[/tex]
Refresh yourself on how I defined the function. I took some care with that.dextercioby said:It's not true.The value DOES matter
[tex] \sqrt{2\sqrt{2\sqrt{2\sqrt{...}}}}=2 [/tex]
[tex] \sqrt{3\sqrt{3\sqrt{3\sqrt{...}}}}=3 [/tex]
Generally
[tex] \sqrt{k\sqrt{k\sqrt{k\sqrt{...}}}}=k ,k\geq 0[/tex]
Daniel.
The 'x' was to prove a point about how the limit is independent of the initial choice for x. And to illustrate that with the right choice of x (in this case, 2), you get immediate convergence to the same limit.dextercioby said:Okay,got it.You defined a sequence of functions.I don't see the relevance of "x",though.
[tex] f_{n}\left(k\right)=:\substack{\underbrace{\sqrt{k\sqrt{k\sqrt{...\sqrt{k}}}}}\\ \mbox{n times}} [/tex]
I thought that was your function for k=2.
Daniel.