Forums - Open Redstone Engineers

I'm making a Java library that finds the square root of a number, yes there already is a java library for that but I don't care. I found the formula for finding the square root of a number, and it's N^0.50. But Java doesn't know what the operator ^ means, so I have to use addition, subtraction, multiplication, and division to substitute ^. But the problem is, I'm not a mathematician, how do I substitute ^? I NEED THE HELP PEOPLE!

TL;DR. N^0.50 = sqrt(N). Need substitute for "^"

If you're calculating square roots only, fractional exponentiation might not be what you're looking for. I would look into the Babylonian method first, which is represented by the recursive equation:



... where x is the number you're taking the square root of, and:



You can even have default to 1 if you really want, you will get to an accurate calculation in the end. The Babylonian method zones in on a correct square root very quickly ( will be much closer than ), so you can just calculate a few (10? play around with accuracy!) iterations of the Babylonian sequence and give that as a square root. You may also have to round at the end if you don't want or something, but that's all up to you.

Good luck!

Also, as a side note, if you insist on calculating arbitrary and you have exponential (and natural log), you can use the formula for an accurate exponent.

Another side note, just because I'm far too maths-happy for my own good, you could also have accuracy as an optional argument which would be a real feature that common sqrt functions don't have. So for instance, you could have something like

Just a thought.

Links:

http://en.wikipedia.org/wiki/Methods_of_...ian_method

http://stackoverflow.com/questions/94341...nentiation

http://en.wikipedia.org/wiki/Methods_of_...#Example_3 (This one is a last resort! Don't ruin it for yourself unless you're really stuck :3)

(03-11-2014, 06:06 PM)Iceglade Wrote: [ -> ]If you're calculating square roots only, fractional exponentiation might not be what you're looking for. I would look into the Babylonian method first, which is represented by the recursive equation:



... where x is the number you're taking the square root of, and:



You can even have default to 1 if you really want, you will get to an accurate calculation in the end. The Babylonian method zones in on a correct square root very quickly ( will be much closer than ), so you can just calculate a few (10? play around with accuracy!) iterations of the Babylonian sequence and give that as a square root. You may also have to round at the end if you don't want or something, but that's all up to you.

Good luck!

Also, as a side note, if you insist on calculating arbitrary and you have exponential (and natural log), you can use the formula for an accurate exponent.

Another side note, just because I'm far too maths-happy for my own good, you could also have accuracy as an optional argument which would be a real feature that common sqrt functions don't have. So for instance, you could have something like

Code:

public static double sqrt(int x, int accuracy) {
//calculations and stuff, in a loop based on accuracy
}

public static double sqrt(int x) {
return sqrt(x, //some default accuracy here)
}

Just a thought.

Links:

http://en.wikipedia.org/wiki/Methods_of_...ian_method

http://stackoverflow.com/questions/94341...nentiation

http://en.wikipedia.org/wiki/Methods_of_...#Example_3 (This one is a last resort! Don't ruin it for yourself unless you're really stuck :3)

Accuracy, well I don't exactly want to use the accuracy method because that puts strain on a computer and it can't always be 100% accurate. I still like the N^0.50 method, but I can still make alternate method based on one of your suggestions.

Eh, just use the internal sqrt(x) function. For exponentiation you can use pow. sqrt(x) = pow(x, 0.5).

(03-12-2014, 07:58 PM)xdot Wrote: [ -> ]Eh, just use the internal sqrt(x) function. For exponentiation you can use pow. sqrt(x) = pow(x, 0.5).

Yeah, I think I'll use math.pow. I made a few posts on reddit and they all suggested I use that functino.

(03-11-2014, 06:06 PM)Iceglade Wrote: [ -> ]Play with accuracy.

Did dat.


First one is using accuracy to the 100,000th degree. The second one is the actual value of pi. I gotta say, if you put a number big enough, it can really calculator good.

(03-17-2014, 08:46 PM)gelloe Wrote: [ -> ]

(03-11-2014, 06:06 PM)Iceglade Wrote: [ -> ]Play with accuracy.

Did dat.

[url=http://i.imgur.com/4mVBifz.png"

[/url]

First one is using accuracy to the 100,000th degree. The second one is the actual value of pi. I gotta say, if you put a number big enough, it can really calculator good.

:3 fun

(03-17-2014, 08:48 PM)Iceglade Wrote: [ -> ]

(03-17-2014, 08:46 PM)gelloe Wrote: [ -> ]

(03-11-2014, 06:06 PM)Iceglade Wrote: [ -> ]Play with accuracy.

Did dat.

[url=http://i.imgur.com/4mVBifz.png"

[/url]

First one is using accuracy to the 100,000th degree. The second one is the actual value of pi. I gotta say, if you put a number big enough, it can really calculator good.

:3 fun

Jesus Christ! You replied to me before I could finish correcting my post *-*

Hehe, well it was just luck. I happened to get home from school at that exact moment and didn't see the time on the post

umm for calculating square roots its something like sqrt(a)=10^(logbase10 of a)/(2)

Idk if this is what your looking for lmao.