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Java Forum / General / September 2006java trigonometric solver
Hi all, I'm trying to implement in Java a solver for this kind of trigonometric system... k * sin( a1 * f ) = b1 k * sin( a2 * f ) = b2 my job is to find "f" whenever is possible. Can you please help in finding an algorithm or a java implementation... thanks in advance GG > Hi all, > I'm trying to implement in Java a solver for this kind of [quoted text clipped - 4 lines] > > my job is to find "f" whenever is possible. It's always possible. Here's the solution: use an x-f substitution on this example: http://users.secsme.org.au/~prbarnes/wikka/uploads/MrTrickFiles/find-x.PNG yeah of course I was meaning a value for "f" not really "f" itself!!! GG > yeah of course > I was meaning a value for "f" not really "f" itself!!! Oh! Well, in that case, you know, of course, that life's complex, don't you? It has both real and imaginary components. So, do you mind telling us moe about your k, a1, a2, f, b1 and b2? Are they really ;-) nasty? Or 'merely' complex? Functions even? What kind of functions? Whatever they may be, however, I don't think that the solution is Java dependent. k is calculated in order to force -1 <= ai/k <= 1 for every i -1 <= bi/k <= 1 for every i ai and bi are integers (I mean they are stored in java integers) in general ai != aj for every i != j and the same condition should be verified for bi ... so they seems not to be nasty nor complex... The solution I'm looking for is a java one, if eventually someone already implemented it, otherwise I have to implement it in java ... thus I need some help. In fact I read something about sin(x) = alpha on condition -1 <= a <= 1 that is absolutely my case. it seems the solution to be: x = alpha + 2 k pi x = pi - alpha + 2 k pi but since I have x = ai * f and alpha = bi / k I get: f = (bi / k + 2 k pi) / ai f = pi - (bi / k + 2 k pi) / ai that it is not exacly what I was expecting... so I am a little bit stuck!!! please help!!! thanks in advance for your help GG > k is calculated in order to force > -1 <= ai/k <= 1 for every i You mean for both of them. ;-) But really, sin(x) \in [-1, 1]? Wow! ;-) > -1 <= bi/k <= 1 for every i > > ai and bi are integers (I mean they are stored in java integers) > in general ai != aj for every i != j and > the same condition should be verified for bi ... > so they seems not to be nasty nor complex... Hey, Gunitus, how much do you know about trigonometry? Or math for that matter? Specifically... forget the solution for a moment: do you understand the problem (t.i. "question")? > The solution I'm looking for is a java one, if eventually someone > already implemented it, > otherwise I have to implement it in java ... thus I need some help. > > In fact I read something about sin(x) = alpha Trying to explain those <khm/> unknowns, you introduce new ones. That's helpful! > on condition -1 <= a <= 1 > that is absolutely my case. [quoted text clipped - 10 lines] > f = (bi / k + 2 k pi) / ai > f = pi - (bi / k + 2 k pi) / ai Here you go! All you have to do now is translate these two lines into Java. :-) > that it is not exacly what I was expecting... so I am a little bit > stuck!!! What _were_ you expecting? Just curious... I was expecting a single function capable of giving me all values of "f" and not two distinct functions... GG Well anyway I don't really understand the problem... and I'm not very well suited in math ... the fact is that I should deliver this function as a small part of a bigger project I'm not allowed to know about?! weird thing I know but the fact is that I only know the input, the output and the CPU time I have to do the trick... oooh yes and I have to do the whole thing in a nice java function... thanks you very much anyway you really helped me. GG > Well anyway I don't really understand the problem... > and I'm not very well suited in math ... [quoted text clipped - 9 lines] > thanks you very much anyway > you really helped me. I can't help you if I don't know (exactly) what the problem is. As you obviously don't know either, I suggest you ask your tutor. And of course, there's the proverbial Usenet reluctance of doing someone elses homework. In order to get help, the least you should do is show some effort of trying to solve the problem yourself. Now, as you found out in another post, your f = f(ai, bj, k). A few additional conditions (ai and bi being integers and so on) puts a few additional restrictions on your solution, but that's about it. Oh, and, "translating" that into Java is the least of your problems. > Hi all, > I'm trying to implement in Java a solver for this kind of [quoted text clipped - 8 lines] > thanks in advance > GG Maybe you can ask on comp.constraints for ideas about suited constraint solvers, algorithms, ... They will now better whether suited implementations exists (maybe even in Java, but I doubt it), or how to implement it yourself in Java. I would first look for existing solvers and let them do the hard work, even if this is outside our beloved Java. Cheers, Peter > Hi all, > I'm trying to implement in Java a solver for this kind of [quoted text clipped - 8 lines] > thanks in advance > GG The only really Java-specific information that seems relevant to the problem is that Math.asin(x) returns an angle, in radians, whose sine is x. See the API documentation for details. Patricia Thanks you very much... it seems that an approach could be: fi =Math.asin(bi/A)/ai fi =pi - Math.asin(bi/A)/ai at this point I have just to solve the system and find the only possible values for "fi"... well I hope it will work I'll keep you post it, just in case you are interested in any positive result. GG > Thanks you very much... it seems that an approach could be: > [quoted text clipped - 6 lines] > well I hope it will work I'll keep you post it, > just in case you are interested in any positive result. It seems to me that there is an infinite number of values for fi, if I understand the problem correctly (and your explanation so far has not been very clear).  SignatureDale King > Thanks you very much... it seems that an approach could be: > [quoted text clipped - 7 lines] > just in case you are interested in any positive result. > GG Don't forget that any pair of angles separated by 2*pi radians have the same sine. Patricia Free MagazinesGet these publications absolutely FREE for up to 12 months. There are no hidden fees and no obligation. Simply choose a title, complete the application form and submit it. Read more ...
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