What is x equal to in this equation

Question

1=sqrt(x+sqrt(x+sqrt(x+sqrt(x+...

Using a substition, it seems you can get 1=sqrt(x+1), but that leads to x=0, which seems wrong when you plug it back into the original equation.

Answer 1

You are correct, Alexander.  There is no solution.

When x = 0, sqrt(x + sqrt(x + sqrt(x + ...) = 0. 

When x is greater than 0, the value of sqrt(x + sqrt(x + ...) can be found by using sqrt(x + y) = y, or x + y = y^2, or y^2 - y - x = 0, or y = 1/2 +-  1/2 sqrt(1 + 4x); we throw out the negative solution and get  1/2 +  1/2 sqrt(1 + 4x)  Note that the limit of this as x  tends to 0 is 1.  However, we never reach this limit.

Answer 2

One way to address this problem is to ask:

 

What is the lower limit, as x approaches zero from above, of

    sqrt(x + sqrt(x + sqrt(x + sqrt(x + ...)    ))

 

 I do not know how to rigorously evaluate this limit.

 

However, some experimentation with a graphing calculator suggests that this limit is  =  1

 

This means the answer to the original question is  x = 0. 

 

This is the answer that the questioner posed.  It seems to be correct.