January 5th, 2019, 05:07 PM  #11  
Senior Member Joined: Sep 2016 From: USA Posts: 642 Thanks: 406 Math Focus: Dynamical systems, analytic function theory, numerics  Quote:
If you mean that you are approximating square roots numerically $\cos^{1}$ somehow, then you should be very clear about how you are going about this. What are you implementing? How is it being computed? Are you iterating some function? If so, what function? For example, if you are calling an implementation of $\cos^{1}$ which you didn't write then any numerical scheme will almost certainly be circular. This is because any fast implementation for $\cos^{1}$ will compute using Newton's method (with some special tricks). Since $\frac{d}{dx} \cos^{1}(x) = \frac{1}{\sqrt{1x^2}}$, you would be implicitly relying on an existing square root implementation in your own square root implementation.  
January 5th, 2019, 09:45 PM  #12  
Newbie Joined: Dec 2018 From: syria Posts: 10 Thanks: 0  Quote:
https://en.wikipedia.org/wiki/Invers...tric_functions But this is an exact formula not an approximation you can calculate it through series expansion, you do not have to calculate each term from the beginning you can use accumulation I am not proficient in computer science to suggest a program  
January 5th, 2019, 10:51 PM  #13  
Newbie Joined: Dec 2018 From: syria Posts: 10 Thanks: 0  Quote:
My formulas are exact formulas not approximation, there is no iteration, they are based on the discovery of a strong pattern using Pythagorean theorem , which allow me to scientifically say that it is a mathematical law, the limited use of correct decimals is a calculation method adopted in all scientific and engineering fields for calculations, as I explained in my work: Since the angle is very small we can consider tan(a)=a=√x/( x 10^n)= 1/(√x 10^n) for an EXACT limited accuracy and correct decimals 2n that you choose initially and you can expand the accuracy as you want by simply increasing n May be this is a new innovative approach that I have developed, as a technical engineer, that need your contentment and the scientific community. which mean if you choose in advance n=6 your calculation are correct up to 2n= 12 correct decimals, and obviously you should then discard all the decimals after that, this is controlled by the pattern of the hypotenuse b= x10^n+0.5 10^(n) as shown in table (1) in my work If you can calculate acos using newton method with some tricks and you choose to return to iterations it will be an acceptable option if it is correct, you should be careful about circular feature in my formulas which is just another feature of the formulas see paragraph (3.3) it gives different results, Anyway we know how to calculate exactly acos using series expansion Thanks!  
January 6th, 2019, 07:12 AM  #14 
Senior Member Joined: May 2016 From: USA Posts: 1,310 Thanks: 551 
Your formula is not exact. It has been known for two and a half millennia that there is no exact numerical representation of $\sqrt{2}$. What you may mean is that your approximation is correct up to n decimal places. It is silly to say that an approximation that is correct only to n decimal places is exact. Moreover, two approximations that are both correct to n decimal places are the same so yours is not better. What sdk has said implies that your method of approximation will not necessarily be faster computationally than other methods. You should address his point. In any case, the square root function is already implemented in most decent calculators so you have not suggested why a new method of approximation is useful. I get your point that it may sometimes be advantageous to avoid having to make an initial guess, but for functions with multiple zeroes, we may want to make an initial guess. My recollection is that Newton's method will work for finding a square root of a real number from any initial guess, but bad guesses converge slowly. (I studied numerical methods decades ago and so may have an incorrect recollection.) 
January 7th, 2019, 12:52 AM  #15  
Newbie Joined: Dec 2018 From: syria Posts: 10 Thanks: 0  Quote:
So it is realistic and useful to restrict and control your calculation formulas to your need of the correct decimals, this is the useful idea about my work ! My formulas may not be faster as tested by different computer scientist, because of "acos" series calculations, but sure calculation 100% Newton's method with bad guess may never converge, My work is just an option, If you do not like it discard it, Thank you for your comments !  
January 7th, 2019, 01:19 AM  #16 
Senior Member Joined: Oct 2009 Posts: 863 Thanks: 328  
January 7th, 2019, 01:26 AM  #17 
Senior Member Joined: Oct 2009 Posts: 863 Thanks: 328  What do you mean "sure calculation 100%". Do you imply you always get the right number of digits your proposed earlier? You don't, since that depends on the acos term being completely precisely calculated. So you'll still need to do an iteration and guesswork and whatever to make sure the acos computes the right number of decimals.

January 7th, 2019, 07:10 AM  #18  
Senior Member Joined: Sep 2016 From: USA Posts: 642 Thanks: 406 Math Focus: Dynamical systems, analytic function theory, numerics  Quote:
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. Quote:
\[ \sqrt{x} = \frac{1}{10^n \cos^{1}\left(\frac{x}{x + 0.5\times 10^{2n}} \right)} \qquad \sqrt{x} = 10^n x \cos^{1}\left(\frac{x}{x + 0.5\times 10^{2n}} \right) \] 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.  
January 7th, 2019, 10:18 PM  #19  
Newbie Joined: Dec 2018 From: syria Posts: 10 Thanks: 0  Quote:
https://en.wikipedia.org/wiki/Newton%27s_method and read the paragraph : "Failure of the method to converge to the root"  
January 7th, 2019, 10:34 PM  #20  
Newbie Joined: Dec 2018 From: syria Posts: 10 Thanks: 0  Quote:
using approximation is just an option see "computation" at: https://en.wikipedia.org/wiki/Trigonometric_functions  

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