You should be able to solve this. What's the end result?

if y is the above, then y=sqrt(xy), and hence y=x

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.

I think it's lim a->inf x^(1/2^a) = 1

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.

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.

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

shut up

the correct root is x, it's going to converge to it. see

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

>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.

the only correct solution is x. if x = 0, this means that x, that is, 0, is the only correct solution.

Shouldn't it approach 0 or am I retarded

y=√(xy)
y^2=xy
y=x

sqrt(xy) is between x and y. If you iterate you get closer and closer to x (if you iterate x, that is)

y=sqrt(x sqrt(x sqrt(x..
y^2=x sqrt(x sqrt(x sqrt(x....
y^2/x=sqrt(x sqrt(x sqrt(x...
y^2/x=y
y=x

Whatever the value of x, that will be the value of y.

Looks like you forgot to read your SICP today.
The correct way to code it is pic related and it turns out that (f x) is almost x for all x.

it approaches x as you add more root x

This gives me a math boner

>didn't see the ellipsis and thought it was [math]x^{\frac{15}{16}}[/math]

Fuck I'm such a brainlet

Why does your accumulator start at 1 though? Different values yield different results.

technically true
not an answer tho

Did you mean for those to be added? Because we aren't. It's the product.

It just equal's x. It's Pooinloo's infinite radical centered at x.