if y is the above, then y=sqrt(xy), and hence y=x
You should be able to solve this. What's the end result?
Carter Gray
Camden Bailey
You got it right.
If you look closely, the same expression can be rewritten as pic related.
Notice that the exponents form a geometric progression, and their sum will eventually reach 1.
Landon Gomez
I think it's lim a->inf x^(1/2^a) = 1
Jonathan Sanders
Call the whole equation y. y=sqrt(xsqrt(xsqrt(x.....)
Since y is the infinite equation you can rewrite it as y=(xy)^1/2. Solve for x.
Michael Edwards
If Y is the value of that term, then you have Y = sqrt(xY). Thus Y^2 = xY. Which has two solutions: either Y = 0, or Y = x.
Caleb Johnson
this is true.
between x=0 and x=1 it's too small.
x = 0.60, y = 1.6e-13
x = 0.93, y = 0.0152
x = 1.00, y = 1.00
when x > 1, y becomes very fast infinity.
x = 1.06 , y = 28
x = 1.25, y = 387542
x = 1.28, y = 1500000
x = 2.00, y = 2.2e+17
Nolan Turner
shut up
the correct root is x, it's going to converge to it. see
Nolan Powell
Going further, you find the infinite series (excuse my phoneposting)
sum((1/(2^n)),1,inf) for the exponents
Which can be represented as a gemetric series with starting point 1/2 and ratio 1/2. Using the geometric series formula
((first term - first missing term)/(1 - ratio))
you see that the exponents converge to one
Robert Johnson
>the correct root is x
ONE of the correct solutions is x. If you start at zero, it will stay there, which makes 0 another solution. This is a fixpoint problem, there can be more than one solution.
David Taylor
the only correct solution is x. if x = 0, this means that x, that is, 0, is the only correct solution.