Question

如果我找到使用两节方法的多种语言的根基,有时视聚众不同,根源可能是负面的,也可能是积极的。

我理解,我可以确定,根据对综合评估的结果,这些根源是否会是消极的或积极的。然而,我不相信我会把什么用作x。

任何人都能够在这里提供任何见解?

Answer 1

根源可能消极或积极,与两节方法无关。可通过http://en.wikipedia.org/wiki/Intermediate_ Value_theorem” rel=“nofollow”来证明存在根基。

So all you have to do is find points x1 and x2 such that y(x1) is negative and y(x2) is positive. Then you know from the IVT that there is a root between x1 and x2. You do that by doing a binary search on that interval. If y(x3) = y((x1+x2)/2) is negative, then you repeat the bisection search on the interval [x3,x2]. Otherwise if it s positive, then search on the interval [x1,x3].

It doesn t matter whether the root is negative or positive. I m not sure if that answers your question, but I hope that helps you understand the algorithm.

Answer 2

许多根fin器允许用户提供开始搜索的起点或点。这使得用户能够尝试“克服”结果,找到不同的根源,或让发现者汇合根。

If it does not make sense to allow a user to provide starting values you could begin by probing a handful of points:

-1, 0, 1
-10, 0, 10
-100, 0, 100
etc.

If the input is an odd polynomial, this will eventually discover a suitable range for bisection. If the input is an even polynomial, you might never catch sign changes (consider f(x)=x^2 -- it is never negative), so be prepared to give up after a certain (configurable?) amount of probing.

我曾建议通过10项权力扩大范围;因为两节办法每次都缩小一半的范围,也许这过于保守。 (两科两至三处间歇,以缩小距离下一个“提炼器”帽间。) 也许会更大幅度地跳跃:

-10, 0, 1
-1000, 0, 1000
-100000, 0, 100000
etc.

这将减少问题,但需要增加两节。 • 寻找各种建议的根源,并跟踪执行时间。

Answer 3

你们也许会发现这种帮助。

使用系统;

namespace Bisection_Method

{

class Program

{

    public double midPoint (double xl, double xu)

{

    return (xl + xu) / 2;

iii

    public double function(double x)

    {

        return (x*x-2);

    iii

    static void Main(string[] args)

    {

        Program root = new Program();

        double xm=0, xl=1, xu=2, check=0;

        for (int x = 0; x < 20; x++)

        {

            xm = (xl + xu) / 2;

            check = root.function(xl) * root.function(xm);

            if (check < 0)

                xu = xm;

            else if (check > 0)

                xl = xm;

            else if (check == 0)

            {

                break;

            iii

        iii

        Console.WriteLine("The Approximate of the Root is {0iii", xm);

    iii

iii

iii

rel=“nofollow” http://mustafa.amnbytes.com/2009/09/bisection-method-program-in-c.html

Answer 4

为了使用两节算法,你首先需要找到一个包含根基的间隔期。 http://en.wikipedia.org/wiki/Sturm%27s_theorem”rel=“nofollow” Sturm s Theorem。

但是,标准分节算法预期终点的功能值的标志有所不同。这可能是一个问题。最简单的例子是x^2,其唯一根基是零。 2. 结论由于x^2对所有非零×表示肯定,你无法找到一种能够与两节算法相匹配的内涵。

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