Function question with expression

[Icebreaker]

$$f(x)=\sqrt{2x}$$

$$x(f)=\sqrt{2f}$$

Does this express:

$$\sqrt{2\sqrt{2\sqrt{...}}}$$

 

[HallsofIvy]

It doesn't express anything because you are using the same symbol, "x", to represent two different things. If $$f(x)= \sqrt{2x}$$, the x(f) could only mean the inverse function which is $$x(f) = \frac{f^2}{2}$$

IF you had said $$f(x)= \sqrt{2x}$$ and
y(f)= $$\sqrt{2}$$, then you could say
y(x)= $$\sqrt{2\sqrt{2x}}$$.

 

[James R]

If you have

$$f(x) = \sqrt{x}$$

then

$$f(f(x)) = \sqrt{\sqrt{x}}$$,

and

$$f(f(f(x))) = \sqrt{\sqrt{\sqrt{x}}}$$

etc.

Considering the function f as an operator, we can write the last expression above as, for example:

$$f^3 x = \sqrt{\sqrt{\sqrt{x}}}$$

and, in general,

$$f^n = \sqrt{\sqrt{...}}$$

where there are n square root signs.

 

[neurocomp2003]

perhaps he implied x(f) in the first equation

 

[Icebreaker]

Wait, is it possible to have "x" in "f(x)" or "f" in "x(f)" to be the function instead of the variable? So the infinite fraction described in the first post can be written the same way?

 

[Curious3141]

Consider the function

$$f(x) = \sqrt{2x}$$ defined on the non-negative reals.

Then,

$$f_2(x) = \sqrt{2{\sqrt{2x}}$$

(The reason I'm using the subscript rather than the exponent notation will become clear soon).

Then the nested square root thing can be defined by the recursion

$$f_1(x) = \sqrt{2x}$$

$$f_{n+1}^2(x) = 2f_n(x)$$ ---(eqn 1)

where the exponent of 2 on the LHS signifies squaring.

The infinitely nested square root thing can be represented by

$$\sqrt{2\sqrt{2\sqrt{...}}} = \lim_{n \rightarrow \infty} f_n(x)$$

At the limit, $$f_{n+1}(x) = f_n(x)$$

so using the recursion in eqn 1,

$$f_n^2(x) = 2f_n(x)$$

$$f_n(x)[f_n(x) - 2] = 0$$

giving a trivial solution of $$f_n(x) = 0$$ for $$x = 0$$

and a nontrivial solution $$f_n(x) = 2$$ for $$x > 0$$

So $$\lim_{n \rightarrow \infty} f_n(x) = \sqrt{2\sqrt{2\sqrt{...}}} = 2$$ for $$x > 0$$

For interest's sake, note that the actual value that you set for $x$ doesn't matter (as long as it's positive). The limit always converges to 2. The choice of value for $x$ only decides from which direction the nested functions converge to that limit. For $x < 2$, it's from the left, and for $x > 2$, it's from the right. For $x = 2$, convergence is immediate.

 

[Icebreaker]

Interesting. Thanks everyone.

 

[dextercioby]

It's not true.The value DOES matter

$$\sqrt{2\sqrt{2\sqrt{2\sqrt{...}}}}=2$$

$$\sqrt{3\sqrt{3\sqrt{3\sqrt{...}}}}=3$$

Generally

$$\sqrt{k\sqrt{k\sqrt{k\sqrt{...}}}}=k ,k\geq 0$$

Daniel.

 

[Curious3141]

It's not true.The value DOES matter

$$\sqrt{2\sqrt{2\sqrt{2\sqrt{...}}}}=2$$

$$\sqrt{3\sqrt{3\sqrt{3\sqrt{...}}}}=3$$

Generally

$$\sqrt{k\sqrt{k\sqrt{k\sqrt{...}}}}=k ,k\geq 0$$

Daniel.

Refresh yourself on how I defined the function. I took some care with that.

 

[dextercioby]

Okay,got it.You defined a sequence of functions.I don't see the relevance of "x",though.

$$f_{n}\left(k\right)=:\substack{\underbrace{\sqrt{k\sqrt{k\sqrt{...\sqrt{k}}}}}\\ \mbox{n times}}$$

I thought that was your function for k=2.

Daniel.

 

[Curious3141]

Okay,got it.You defined a sequence of functions.I don't see the relevance of "x",though.

$$f_{n}\left(k\right)=:\substack{\underbrace{\sqrt{k\sqrt{k\sqrt{...\sqrt{k}}}}}\\ \mbox{n times}}$$

I thought that was your function for k=2.

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.

In your definition, I could say $$f(x) = \sqrt{kx}$$ and the limit is k. Immediate convergence occurs when x = k.

It's just a minor point I wanted to illustrate.