Javascript 更改函数时JSXgraph不移动点

Javascript 更改函数时JSXgraph不移动点,javascript,jsxgraph,Javascript,Jsxgraph,我用JSXgraph向学生展示LIMSUP和liminfs。这个版本在这里起作用- 代码如下: <div style="width: 500px; height: 40px;"> <p style="display: inline;">N:</p> // Create a slider for the values of N <input id="NSlide

我用JSXgraph向学生展示LIMSUP和liminfs。这个版本在这里起作用-

代码如下:

<div style="width: 500px; height: 40px;">
        <p style="display: inline;">N:</p>
       // Create a slider for the values of N
        <input id="NSlider" type="range" min="1" max="19" value="1" step="1" style="width: 50%;">
        <p style="display: inline;"><span id="NOut"></span></p>
        <script>
            // Get the value of N from the slider
            var NSlider = document.getElementById("NSlider");
            var NOutput = document.getElementById("NOut");
            NOutput.innerHTML = NSlider.value; // Output value for student to see
        </script>
    </div>
    <!-- Create board with points -->
    <div id="box" class="jxgbox" style="width:500px; height:500px;">
        <script type="text/javascript">
            // Create board
            var board = JXG.JSXGraph.initBoard('box', {
                boundingbox: [-1, 2, 21, -0.2],
                axis: true,
                grid: true
            });
            // Generate points in the sequence and graph points of the sequence
            var i;
            var s = [null];
            for (i = 1; i <= 20; i++) {
                s.push(1 + Math.pow(-1, i) / i);
                board.create('point', [i, s[i]], {
                    color: 'yellow',
                    fixed: true,
                    withLabel: false
                });
            }
            // Genereate liminfs and limsups
            var infs = [null],
                sups = [null];
            for (i = 1; i <= 20; i++) {
                infs.push(Math.min.apply(null, s.slice(i + 1)));
                sups.push(Math.max.apply(null, s.slice(i + 1)));
            }
            // Graph liminf and limsup points
            var liminf = board.create('point', [
            function() { return NSlider.value; },
            function() { return infs[NSlider.value]; }], {
                color: 'blue',
                withLabel: false
            });
            var limsup = board.create('point', [
            function() { return NSlider.value; },
            function() { return sups[NSlider.value]; }], {
                color: 'orange',
                withLabel: false
            });
     
            board.update()
        </script>
    </div>
        <script>
            // Set board to update when the N slider is updated
            NSlider.oninput = function () {
            NOutput.innerHTML = this.value; // Output value for student to see
                liminf.moveTo([NSlider.value, infs[NSlider.value]]);
                limsup.moveTo([NSlider.value, sups[NSlider.value]]);
            }
        </script>

N:

//为N的值创建一个滑块

//从滑块获取N的值 var NSlider=document.getElementById(“NSlider”); var NOutput=document.getElementById(“NOut”); NOutput.innerHTML=NSlider.value;//输出值供学生查看 //创建板 var board=JXG.JSXGraph.initBoard('box'{ 边界框:[-1,2,21,-0.2], 安讯士:没错, 网格:对 }); //在序列中生成点并绘制序列的点 var i; var s=[null];
对于(i=1;i
u
v
在您的绘图中是什么意思?我们如何计算它?在网站上的文本中,u\N是数组infs的元素,v\N是数组sups的元素DOM上有蓝点,但一些脚本正在设置
显示:无;可见性:隐藏;
样式。您必须取消错误。我在电路板上的点上添加了
visible:true
,但它没有解决问题。知道当我更改填充s的函数时为什么会这样做吗?我已经确定了问题。如果sups或infs中的值为负值,moveTo函数不会显示该点。