Java - resolution of f (x) = 0

[Caden Fernandes]

Hi
I am a student who needs your help in programming, the resolution of f (x) = 0 in Java or c. I started thinking but I've been lost I do not know where to start, If I am wrong forgiving me and share your opinion about proposed. Any help will be highly appreciated. Thanks in advance.

[Zecho]

Hello
Doing research in the course of numerical analysis, this problem is already solved long ago by a series of algorithms, did you have to choose one that suits you and implement it. This is very easy my by knowledge. You have to just deal with numbers and nothing else, remaining is simple work. Hope you will do it. Just give it a try and if you have problems post back.

[MindSpace]

Hi
First you must know is that the equation is of first or second degree or that it is polynomial of degree n. This can simplify the problem, if you specify this when you have to solve the parameters of the solution (eg if 2 degree calculate delta, b and b'). I first recommend you to know the type of the equation.

[Katty]

Hello
It is already a well-defined grammar for your equations, then it must be possible to evaluate the expression f (a) for any value a. To resolve the equation you have several methods of numerical calculation (dychotomie, Newton etc..). For more information on the numerical calculation methods you can visit the wikipedia, and search for java numeric methods.

[opaper]

Hi
It is mostly about a. In the general case (a function quite some), there is quite simply not producing algorithm. If we start to conditions such as continuity, then we arrive at something (dichotomy, typically), and was more info, it produces a more efficient algorithm. I think you should go with it. I know at first it is hard to program, but after a time period and a lot of practice you can solve it easily.

[Katty]

Hi
You should know already that whatever is the input.
1 - kind of an equation: 2x ^ 2-3xcos (x) = 0. ( '^' Operator is power)
2 - an equation of 2nd degree type equation with coefficient 1, 2, 3.
The calculation methods are generally iterative part of an initial interval [a0, b0] the best thing to have
f (a0) * f (b0) <0 and f (x) continuous => existence of at least one root in the interval [a0, b0] (theory of intermediate values).
To (1) is a problem more complicated because you must extract the coefficient and make a calculation expression for each value of x i.e
f (0) = 2 * 0 -3 * 0 * cos (0) = 0.
In developing that you can implement a solution algorithm.