Quote:
Originally Posted by SDK I'm not sure what you mean by this. By "calculate" do you mean obtain a finite decimal representation? If so then obviously this is not possible for irrational numbers. But this is not a problem which is unsolved. The field of numerical analysis deals with this problem.
I think the criticism here is that you should not use the = sign or claim things are equal if they are a numerical approximation. This is what I mean when I ask you to be more clear about what you are actually doing. You haven't said you are computing a numerical approximation anywhere and you haven't said how you are computing it nor have you shown that it satisfies any accuracy bound.
I don't know what you are trying to say here. I agree that we $\cos^{-1}$ has a Taylor expansion away from branch cuts but you have not explained what this Taylor expansion has to do with your claims. The only formulas you have provided are the following:
\[ \sqrt{x} = \frac{1}{10^n \cos^{-1}\left(\frac{x}{x + 0.5\times 10^{-2n}} \right)} \qquad
\sqrt{x} = 10^n x \cos^{-1}\left(\frac{x}{x + 0.5\times 10^{-2n}} \right) \]
and both of these formulas are just plain false. If I assume you mean to say these are approximations then I still have no reason to believe them since you haven't provided any evidence, analysis, or other information.
This is just not true. When I say "Newton's method", I'm referring to algorithms which include both the Newton iteration and the choice of initial conditions. There are methods for identifying when convergence is failing as well as using this information to choose better initial conditions.
In fact, Hirsch and Smale provided a deterministic algorithm for finding roots using Newton's method which is proved to converge to a root for any finite dimensional Banach space. This was around 1979. A few years ago, a deterministic algorithm for finding roots of polynomials was published which is also polynomial time. Obviously, this means computing square roots (or any other root) is guaranteed to converge.
Nobody is discarding it, but nobody can understand what you are trying to say either. |
Both, impossible to calculate and represent infinite number digits or infinity
You can use the terminology you want my formulas are correct UP TO 2n decimals controlled by the pattern 0.5 10^(-2n)
The proof is the PATTERN in my work
you calculate X=x/(x+0.5 10^(-2n)) and you replace X in "acos" series to calculate it, then you replace "acos" in the formulas to calculate the square root of x
Obviously the formulas do not calculate decimals > 2n I used my own terminology "controlled precision output" you call it what you want.
There are certainly many developments and sophisticated algorithm for Newton's methods to guarantee the convergence this just proof it was not perfect, consult
https://en.wikipedia.org/wiki/Newton%27s_method
you choose the method you want, I am just offering an option for scientist
So if you do not agree or do not understand or you are not comfortable with my work simply ignore it.