Question

What is the graphic of `f(x) = sqrt(x+sqrt(x+sqrt(x+sqrt(x+...))))` for `x ge 0`?

What is the graphic of `f(x) = sqrt(x+sqrt(x+sqrt(x+sqrt(x+...))))` for `x ge 0`?

Answer 1

This is the continued-surd model for the equation of part of a parabola, in the first quadrant. Not in the graph, the vertex is at #(-1/4, 1.2) and the focus is at (0, 1/2).

Explanation:

As of now, `y = f(x)>=0`. Then `y =+sqrt(x+y), x>=0`.. Rationalizing,

`y^2=x+y.`. Remodeling,

`(y-1/2)^2=(x+1/4)`.

The graph is the part of a parabola that has vertex at `(-1/4, 1/2)`

and latus rectum 4a = 1.. The focus is at `(0, 1/2)`.

As `x and y >= 0`, the graph is the part of the parabola in the 1st

quadrant, wherein `y>1`..

I think it is better to restrict x as > 0, to avoid (0, 1) of the parabola.

Unlike parabola y, our y is single-valued, with `f(x) in (1, oo)`.

`f(4) = (1 + sqrt17)/2 = 2.56` nearly. See this plot, in the graph.

graph{(x+y-y^2)((x-4)^2+(y-2.56)^2-.001)=0[0.1 5 1 5] }

I make it for another g in continued-surd `y = sqrt(g(x)+y)` .

Let g(x) = ln x. Then `y = sqrt(ln x + sqrt(ln x + sqrt(ln x +...)))`.

Here, `x >= e^(-0.25) = 0.7788...`.Observe that y is single valued for

`x >=1`. See the plot is (1, 1).
graph{((ln x+y)^0.5-y)((x-1)^2+(y-1)^2-.01)=0[0..779 1 0.1 1] }