Iterative square root like square root recursion limit

[bo198214]

Surely you guys know the recursion limit formula (I use recursion to avoid the term iteration which is already occupied) for the square root of \( b \):

\( x_{n+1} = \frac{1}{2}\left(x_n+\frac{b}{x_n}\right) \)
\( \lim_{n\to \infty} x_n = \sqrt{b} \)

Now - also inspired the Newton formula analogy - I was asking myself whether we cant do a similar thing for the iterative square root \( w \) of \( f \), i.e. \( w\circ w =f \).

An analogon could be:
\( w_0 = f \)
\( w_{n+1} = \frac{1}{2}\left(w_n + {w_n}^{-1}\circ f\right) \), \( w = \lim_{n\to\infty} w_n \).

The numerical verification will take some more time for me and is currently too slow. Do you think that the above construction converge?

For matrices the method works to compute matrix square root. But matrix multiplication is both side distributive while function composition is only distributive from the right.

Because this restricted distributivity there may also be variants like
\( w_{n+1} = \frac{1}{2}\left(w_n + f\circ {w_n}^{-1}\right) \) or
\( w_{n+1} = \frac{1}{2}{w_n}^{-1}\left( w_n \circ w_n + f\right) \)

will they converge to a differnt iterative square root?
Tell me what you think!

[Gottfried]

Now - also inspired the Newton formula analogy - I was asking myself whether we cant do a similar thing for the iterative square root \( w \) of \( f \), i.e. \( w\circ w =f \).

Nice idea. I've used f(x) = 2x + x^2 , then

f05 = Ser(x)
f05 = 1/2*(f05 + f (serreverse(f05))) \\ recurse

to arrive at the same result like when using the matrix-root.

:-)

[update] it works also for the dxp(x) = exp(x)-1 function, matching the matrix-method.
For exp(x) Pari/GP cannot give a seriesinversion, so I can't check this [/update]
Gottfried