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mones

square root formulas

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3 hours ago, mones said:
23 hours ago, studiot said:

Thank you for your answer.

Why do you think using radians would make any difference?
It doesn't on mine and is slightly slower.

Both using DMS or rads is much slower than using the log / antilog formula, which is effectively instantaneous.

However

I cannot enter 123456.0000005 into my caclulator.
Not many  calculators have that many digits.

Mine displays 10.
I 'm not sure if it calculates to further guard digits, but I think so.

If I use 123456/123456.0005 then I get 11110.42436, not the right answer.

I am not trying to dismiss your formula, or discredit it. Who knows, there may one day be a use for it.  Knowledge should not be disgarded because it is currently unwanted.

I am taking you up on your request to examine its use in real situations and have offered you the results of my simple trials.

 

So if I was sitting on top of some mountain in Arabia (as I have been) and wanted a square root (which I have) but have not lugged a heavy computer up there (which I didn't) my thoughts are.

1) Modern (small) calculators have a square root button.

2) This is faster than trigonometric solutions.

3) If you, like me, did not have a calculator with even a square root button, let alone trig and log functions, you would have to rely on human ingenuity (as I did).

4) My companion took a whole afternoon to extract a single root, using a version of Newton's method on his calculator.

5) My digit bracketing method takes a few minutes per root extraction and automatically confirms the answer at the end of the extraction. Checking is important in real life.

 

the radians choice is imposed by the calculator you can choose degrees or grads it is the same,

Since it is 0.510^(-2n) you have to enter an even decimals number of "0"(2,4,6,8,10...) it is better to just add "0" decimals than to calculate 10^(-2n) and consuming time, it is the same result. in your example you entered 3 "0" which is incorrect.

I do not have facilities to benchmark a compared solution about square root calculation with my formula so I can not give a scientific reply, but you can calculate my formula with hand and trigonometric function by hands also using series expansion

the speed of calculation depend also on the quality of program and techniques and how it is performed

My formula is the result of hard work and high quality research with deep analysis resolving a very tough problem since the Babylonian, the speed of calculation is just an aspect of the formula that I wish to be the fastest formula, but I am not a computer scientist to work on this issue.

 

I am sorry you choose not to address my points, particularly about answering the question of demonstrating you have the right answer whn you have put numbers into your formula.

Checking is a vital part of any real world process.

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5 minutes ago, studiot said:

 

I am sorry you choose not to address my points, particularly about answering the question of demonstrating you have the right answer whn you have put numbers into your formula.

Checking is a vital part of any real world process.

I welcome your comments and attempt to discover more aspects, I never choose to ignore any question about my formula whatever is, ask me a specific question about my formula to answer, not general issues, I am not proficient in all science field, in spite of that I will try my best and will give an answer if I can may be it is useful.

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1 hour ago, studiot said:

If I use 123456/123456.0005 then I get 11110.42436, not the right answer

I think don't you mean 'even decimals', but even zeros?
Both 123456 and 123456.0005 have an even number of digits.

4 hours ago, mones said:

Since it is 0.510^(-2n) you have to enter an even decimals number of "0"(2,4,6,8,10...) it is better to just add "0" decimals than to calculate 10^(-2n) and consuming time, it is the same result. in your example you entered 3 "0" which is incorrect.

The best I can do on my calculator is

tan (arcsin(123456/123456.005)) /10 = 351.3620004 which only gives me 5 digit accuracy compared with  351.3630601

 

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2 hours ago, studiot said:

I think don't you mean 'even decimals', but even zeros?
Both 123456 and 123456.0005 have an even number of digits.

The best I can do on my calculator is

tan (arcsin(123456/123456.005)) /10 = 351.3620004 which only gives me 5 digit accuracy compared with  351.3630601

 

Correct, I mean only even zeros"0" decimals,  and add number 5 decimal at the end  which is conform with 0.510^(-2n) 

this is true because your calculator has limited digits, the accuracy of the formula is 2n decimals which mean in your example 5 digits correct with n=1 and 2 correct decimals and.

 

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1 hour ago, mones said:

Correct, I mean only even zeros"0" decimals,  and add number 5 decimal at the end  which is conform with 0.510^(-2n) 

this is true because your calculator has limited digits, the accuracy of the formula is 2n decimals which mean in your example 5 digits correct with n=1 and 2 correct decimals and.

 

 

You are carefully avoiding the question

How can any user trust your method or ascertain if the result answer is correct?

 

You method has clearly failed on my calculator compared to the proper method or using a different formula, both of which give the same ansswer, correct to the display capacity of the calculator.
 

Thus the error cannot be inherent in the calculator cpaacity, but must be inherent in your method.

Edited by studiot

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One more question for you to ponder.

 

Have you done any work finding out what sort of precision users of your formula might want?

For instance consider a surveyor measuring distance to 1mm (quite good actuallly)

She will probably make most measurements in the range 0 to 999m, (6 digits) more rarely on the range 1000 to 9999 m (7 digits)

She will also mostly measure angles to 20 seconds of arc, more rarely to a single second.

 

Have you considered the implications of error propagation in your formula when inserting measurements of this or other precisions?

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14 hours ago, studiot said:

 

You are carefully avoiding the question

How can any user trust your method or ascertain if the result answer is correct?

 

You method has clearly failed on my calculator compared to the proper method or using a different formula, both of which give the same ansswer, correct to the display capacity of the calculator.
 

Thus the error cannot be inherent in the calculator cpaacity, but must be inherent in your method.

My formula is an INITIALLY CONTROLLED DECIMALS precision output formula, THIS IS HOW IT WORKS, the problem is not in the calculator capacity.

It is a very trusted calculation because you know in advance the correct decimals that you are calculating.

example for the calculated  351.3630601

my formula control ONLY the correct DECIMALS (3630601) precision output, but it return ALWAYS a correct number digits (351)

if you need 7 correct decimals and you choose n=1 you get 2n=2 correct decimals (36) you  obviously never get 7 correct decimals, which is not wrong, it is a voluntary limited precision , the "error" is in your choice of n.

if you can not enter the choice n=4 to get 2n=8 correct decimals in your calculator, it means my formula is not adequate for the calculator so use another method.

You  must choose FIRST just the correct DECIMALS  (not all number digits) precision output which is equal to 2n by simply choosing n according to your need

Most of the scientific calculation users want to use  15 correct decimals as a maximum,   so you can choose n=8 to get 2n=16 correct decimals that will satisfy any need for less than 16 decimals.   you can increase the precision of correct decimals as you want by simply increasing n.

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1 hour ago, mones said:

My formula is an INITIALLY CONTROLLED DECIMALS precision output formula, THIS IS HOW IT WORKS, the problem is not in the calculator capacity.

It is a very trusted calculation because you know in advance the correct decimals that you are calculating.

example for the calculated  351.3630601

my formula control ONLY the correct DECIMALS (3630601) precision output, but it return ALWAYS a correct number digits (351)

if you need 7 correct decimals and you choose n=1 you get 2n=2 correct decimals (36) you  obviously never get 7 correct decimals, which is not wrong, it is a voluntary limited precision , the "error" is in your choice of n.

if you can not enter the choice n=4 to get 2n=8 correct decimals in your calculator, it means my formula is not adequate for the calculator so use another method.

You  must choose FIRST just the correct DECIMALS  (not all number digits) precision output which is equal to 2n by simply choosing n according to your need

Most of the scientific calculation users want to use  15 correct decimals as a maximum,   so you can choose n=8 to get 2n=16 correct decimals that will satisfy any need for less than 16 decimals.   you can increase the precision of correct decimals as you want by simply increasing n.

 

Look carefully at the number of views of this thread, listed in the forum list.

378     

78 of these since this time yesterday.

 

Why do you think other people have stopped responding to you, even though they have been onsite and looked?

I think it is because you don't respond to other people's concerns.
You either dismiss them out of hand or just ignore them.

 

How many schools, colleges universities, industries   can or need to calculate to 15 digits?

How many manufacturers of calculators make 15 digit ones?
How many town centre shops have one in stock?

So yes, repeatedly stating you have a method that can calculate square roots to some arbitrarily large precision doesn't prove it.
Nor does a large number of examples.
And I have seen no mathematics to prove this.
There is a large body of mathematics concerned with precision and accuracy in calculation.
It should be possible to show mathematically these confidence levels so that any potential user of your system can demonstrate that his root extraction is sound and reliable for his purposes. This is especially important for the few that need to calculate to such levels.

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19 hours ago, studiot said:

 

Look carefully at the number of views of this thread, listed in the forum list.

378     

78 of these since this time yesterday.

 

Why do you think other people have stopped responding to you, even though they have been onsite and looked?

I think it is because you don't respond to other people's concerns.
You either dismiss them out of hand or just ignore them.

 

How many schools, colleges universities, industries   can or need to calculate to 15 digits?

How many manufacturers of calculators make 15 digit ones?
How many town centre shops have one in stock?

So yes, repeatedly stating you have a method that can calculate square roots to some arbitrarily large precision doesn't prove it.
Nor does a large number of examples.
And I have seen no mathematics to prove this.
There is a large body of mathematics concerned with precision and accuracy in calculation.
It should be possible to show mathematically these confidence levels so that any potential user of your system can demonstrate that his root extraction is sound and reliable for his purposes. This is especially important for the few that need to calculate to such levels.

You definitely must calculate the formula using high precision program such as (https://www.wolframalpha.com/) to ensure the accuracy of the trigonometric function, because the numerical accuracy of the function "asin" near −π/2 and π/2 is ill-conditioned, or you can not get accurate result using commercial program or calculator with relatively low accuracy.

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4 hours ago, mones said:

You definitely must calculate the formula using high precision program such as (https://www.wolframalpha.com/) to ensure the accuracy of the trigonometric function, because the numerical accuracy of the function "asin" near −π/2 and π/2 is ill-conditioned, or you can not get accurate result using commercial program or calculator with relatively low accuracy.

 

Indeed so, finally a response to what I said in my first reply.

Quote

studiot

2)

Your methods claim a single step, but appear to require an inexhaustible supply of more advanced functions such as trigonometric ones to achieve this.
How many steps does the computer or calculator execute to obtain these, and what would do without them?

What about the accuracy obtainable for these?
For instance the tangent changes very rapidly indeed near its singularities.

 

You have put in a lot of good effort in creating this paper and I have tried to engage in constructive discussion but I find it very disappointing that you seem like a marathon runner who, have covered the first 26 miles has decided to sit on a milestone and admire the scenery rather than complete the course in not supporting your work.
So I will leave you and your thread there.

 

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12 minutes ago, studiot said:

 

Indeed so, finally a response to what I said in my first reply.

 

You have put in a lot of good effort in creating this paper and I have tried to engage in constructive discussion but I find it very disappointing that you seem like a marathon runner who, have covered the first 26 miles has decided to sit on a milestone and admire the scenery rather than complete the course in not supporting your work.
So I will leave you and your thread there.

 

I am engaged to support my work, I experienced the same problem when I was working on my formula, and I wanted to dismiss all the work,  the low accuracy program is lost of time for my formula for n>3 accuracy it may return a mess of numbers, until I discovered that the problem is in the trigonometric function accuracy, I clearly and widely explained that in my paper ( see Paragraph 4. Discussion), this is not because of my work or formula.

May be the development of more efficient method to calculate the trigonometric functions will resolve this problem,

I am convinced that my formula has many aspects that will not be developed in just few days, it is a wonderful formula anyway, that still fascinating me. 

 

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I have deduced a simpler formula with new aspect for the square root using only one function, and according to the same pattern, the square root is also proportional to the angle:

√x = 1/(10n acos( x / (x + 0.5. 10-2n )))

You can calculate the function  acos using this series expansion:

blob.png.3becfa029cf4fe49a4d5e883d88e2718.png

with

X=( x / (x + 0.5. 10-2n )

Using one function should be faster than the previous formula, Just  calculate it correctly and precisely to get the correct √x result with 2n correct decimals.

You can see my updates at :

https://www.researchgate.net/profile/Mones_Jaafar

I hope this will be useful.

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5 hours ago, mones said:

I hope this will be useful.

I hope to win the lottery.

I think my odds are roughly the same as yours.

I have never bought a ticket.

The best you can hope is that people think your formula is interesting.

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Well, the Babylonians stumbled on a very simple algorithm. I suspect they reasoned this wise

a^2=S

a=S/a

a/2=S/2a and therefore

a=a/2+S/2a

I can't imagine a simpler algorithm.

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