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#?

1 Answer
May 20, 2016

Answer:

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] }