Question

我正在开发一个网络软件，用JSXGraph绘制一个带有2个变量的交互式多项式方程，我需要方程的导数为0的每个点都有一个可拖动的点，当我拖动一个点时，只有在曲线的拖动点之前1点和1点之间的间隔合在一起有没有人知道我该怎么做？

Answer 1

抱歉延误。我想，你需要的是 Hermite插值。我最近在Google网上论坛（https://groups.google.com/forum/#!topic/jsxgraph/OHls2NTOmUA）举了一个例子。 Hermite插值尚未集成在JSXGraph中，但似乎需要它，所以我们可以考虑它。

以下是代码：

JXG.Math.Numerics.hermitePolynomial = function (p, derivatives) {
        var w = [],
            Lii = [],

            /** @ignore */
            fct = function (x, suspendedUpdate) {
                var i, k, xi, s,
                    L, Hi, Hi_hat,
                    len = p.length,
                    num = 0,
                    denom = 0;

                if (!suspendedUpdate) {
                    // Compute denominator
                    for (i = 0; i < len; i++) {
                        w[i] = 1.0;
                        Lii[i] = 0.0;
                        xi = p[i].X();

                        for (k = 0; k < len; k++) {
                            if (k !== i) {
                                w[i] *= (xi - p[k].X());
                                Lii[i] += 1 / (xi - p[k].X());
                            }
                        }

                        w[i] = 1 / w[i];
                    }
                }

                s = 0.0;
                for (i = 0; i < len; i++) {
                    xi = p[i].X();
                    L = w[i];
                    for (k = 0; k < len; k++) {
                        if (k !== i) {
                            L *= (x - p[k].X());
                        }
                    }

                    Hi = (1.0 - 2 * (x - xi) * Lii[i]) * L * L;
                    Hi_hat = (x -xi) * L * L;
                    s += p[i].Y() * Hi + derivatives[i] * Hi_hat;
                }
                return s;
            };

        return fct;
    };

var board = JXG.JSXGraph.initBoard("jxgbox", {
            boundingbox: [-5, 6, 6, -5],
            axis: true
});

var p = [];
// Define the points
p[0] = board.create('point', [-1,2], {size:4});
p[1] = board.create('point', [1,-2], {size:4});
p[2] = board.create('point', [3,-1], {size:4});

// The derivates at this points are all zero.
var derivatives = [0, 0, 0];

var f = JXG.Math.Numerics.hermitePolynomial(p, derivatives);
var graph = board.create('functiongraph', [f, -10, 10], {strokeWidth:2});

现场演示位于https://jsfiddle.net/vcL7aepo/278/

祝福，阿尔弗雷德

如何在方程图中设置JSXGraph中可拖动点的初始位置？

1 个答案: