JS 一元四次方程计算器

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml"><head><meta http-equiv="Content-Type" content="text/html; charset=UTF-8">
<title></title>
<body>

<div align="center">例如输入:a=3, b=6, c=-123, d=-126 e=1080</div>
<form name="form1" action="">
	ax<sup>4</sup> + bx<sup>3</sup> + cx<sup>2</sup> + dx + e = 0
	<table border="0" align="center" colspan="4" cellspacing="0" cellpadding="0">
		<tbody>
			<tr>
				<td><input type="text" name="aIn" size="5" placeholder="a">x<sup>4</sup> +</td>
				<td><input type="text" name="bIn" size="5" placeholder="b">x³ +</td>
				<td><input type="text" name="cIn" size="5" placeholder="c">x² +</td>
				<td><input type="text" name="dIn" size="5" placeholder="d">x +</td>
				<td><input type="text" name="eIn" size="5" placeholder="e">e</td>
			</tr>
		</tbody>
	</table>
	
	<center>
		<input type="button" name="button1" value="计算" onclick="quad4solve(this.form)">
	</center>
	
	<table border="0" class="innerc" cellspacing="0" cellpadding="3">
		<tbody>
		<tr align="center">
			<td class="inner" colspan="4"><b>结果:</b></td>
		</tr>
		<tr>
			<td>x<sub>1</sub>: 
				<input type="text" name="x1Re" size="25"> + &nbsp;
				<input type="text" name="x1Im" size="25"> i 
			</td>
		</tr>
		<tr>
			<td>x<sub>2</sub>: 
				<input type="text" name="x2Re" size="25"> + &nbsp;
				<input type="text" name="x2Im" size="25"> i 
			</td>
		</tr>
		<tr>
			<td>x<sub>3</sub>: 
				<input type="text" name="x3Re" size="25"> + &nbsp;
				<input type="text" name="x3Im" size="25"> i 
			</td>
		</tr>
		<tr>
			<td>x<sub>4</sub>: 
				<input type="text" name="x4Re" size="25"> + &nbsp;
				<input type="text" name="x4Im" size="25"> i 
			</td>
		</tr>
		</tbody>
	</table>
</form>

<script type="text/javascript">
function quad4solve(dataForm)
{
	var a = parseFloat(dataForm.aIn.value);
	var b = parseFloat(dataForm.bIn.value);
	var c = parseFloat(dataForm.cIn.value);
	var d = parseFloat(dataForm.dIn.value);
	var e = parseFloat(dataForm.eIn.value);
	if (a == 0)
	{
		alert("The coefficient of the power four of x is 0. Please use the utility for a third degree quadratic.");
		return;
	} 
	if (e == 0)
	{
		alert("One root is 0. Now divide through by x and use the utility for a third degree quadratic to solve the resulting equation for the other three roots.");
		return;
	} 
	if (a != 1) 
	{
		b /= a;
		c /= a;
		d /= a;
		e /= a;
	}

	//立方解算器系数
	var cb, cc, cd;  
	var discrim, q, r, RRe, RIm, DRe, DIm, dum1, ERe, EIm, s, t, term1, r13, sqR, y1, z1Re, z1Im, z2Re;
	cb = -c;
	cc = -4.0*e + d*b;
	cd = -(b*b*e + d*d) + 4.0*c*e;
	if (cd == 0)  
	{
		alert("cd = 0.");
	}
	q = (3.0*cc - (cb*cb))/9.0;
	r = -(27.0*cd) + cb*(9.0*cc - 2.0*(cb*cb));
	r /= 54.0;
	discrim = q*q*q + r*r;
	term1 = (cb/3.0);
	if (discrim > 0) 
	{ 
		// 1 实数, 2 复数
		s = r + Math.sqrt(discrim);
		s = ((s < 0) ? -Math.pow(-s, (1.0/3.0)) : Math.pow(s, (1.0/3.0)));
		t = r - Math.sqrt(discrim);
		t = ((t < 0) ? -Math.pow(-t, (1.0/3.0)) : Math.pow(t, (1.0/3.0)));
		y1 = -term1 + s + t;
		} 
		else 
		{
			if (discrim == 0) 
			{
				r13 = ((r < 0) ? -Math.pow(-r,(1.0/3.0)) : Math.pow(r,(1.0/3.0)));
				y1 = -term1 + 2.0*r13;
			} 
			else 
			{                             	
				q = -q;
				dum1 = q*q*q;
				dum1 = Math.acos(r/Math.sqrt(dum1));
				r13 = 2.0*Math.sqrt(q);
				y1 = -term1 + r13*Math.cos(dum1/3.0);
			}
		} 
		// 确定了y1 分解立方的实根
		term1 = b/4.0;
		sqR = -c + term1*b + y1;  
		RRe = RIm = DRe = DIm = ERe = EIm = z1Re = z1Im = z2Re = 0;
		if (sqR >= 0) 
		{
			if (sqR == 0) 
			{
				dum1 = -(4.0*e) + y1*y1;
				if (dum1 < 0){
					z1Im = 2.0*Math.sqrt(-dum1);
				} else {
	  				z1Re = 2.0*Math.sqrt(dum1);
					z2Re = -z1Re;
				}
			} 
			else 
			{                       
			RRe = Math.sqrt(sqR);
			z1Re = -(8.0*d + b*b*b)/4.0 + b*c;
			z1Re /= RRe;
			z2Re = -z1Re;
		 } 
	} 
	else 
	{                           
		RIm = Math.sqrt(-sqR);
		z1Im = -(8.0*d + b*b*b)/4.0 + b*c;
		z1Im /= RIm;
		z1Im = -z1Im;
	} 
	z1Re += -(2.0*c + sqR) + 3.0*b*term1;
	z2Re += -(2.0*c + sqR) + 3.0*b*term1;

	//At this point, z1 and z2 should be the terms under the square root for D and E
	if (z1Im == 0)
	{               // Both z1 and z2 real
		if (z1Re >= 0)
		{
			DRe = Math.sqrt(z1Re);
		}
		else
		{
			DIm = Math.sqrt(-z1Re);
		}
		if (z2Re >= 0)
		{
			ERe = Math.sqrt(z2Re);
		}
		else
		{
			EIm = Math.sqrt(-z2Re);
		}
	}
	else 
	{                      
		r = Math.sqrt(z1Re*z1Re + z1Im*z1Im); 
		r = Math.sqrt(r);
		dum1 = Math.atan2(z1Im, z1Re); 
		dum1 /= 2; //Divide this angle by 2
		ERe = DRe = r*Math.cos(dum1); 
		DIm = r*Math.sin(dum1);
		EIm = -DIm;
	} 
	dataForm.x1Re.value = -term1 + (RRe + DRe)/2;
	dataForm.x1Im.value = (RIm + DIm)/2;
	dataForm.x2Re.value = -(term1 + DRe/2) + RRe/2;
	dataForm.x2Im.value = (-DIm + RIm)/2;
	dataForm.x3Re.value = -(term1 + RRe/2) + ERe/2;
	dataForm.x3Im.value = (-RIm + EIm)/2;
	dataForm.x4Re.value = -(term1 + (RRe + ERe)/2);
	dataForm.x4Im.value = -(RIm + EIm)/2;
	return;
}
</script>
</body>
</html>

1,886 thoughts on “JS 一元四次方程计算器”

  1. Hey are using WordPress for your site platform? I’m
    new to the blog world but I’m trying to get started and set up my own. Do you
    need any coding expertise to make your own blog? Any help would
    be really appreciated!

  2. Unquestionably imagine that that you stated.
    Your favorite justification appeared to be at the
    internet the easiest factor to have in mind of. I say to you,
    I definitely get annoyed whilst folks consider worries that
    they just don’t understand about. You managed
    to hit the nail upon the highest and outlined out the whole
    thing without having side-effects , other people can take a signal.

    Will probably be again to get more. Thanks

  3. Nice blog! Is your theme custom made or did you download it from
    somewhere? A theme like yours with a few simple adjustements would really make my blog shine.
    Please let me know where you got your theme. With thanks

发表回复

您的电子邮箱地址不会被公开。 必填项已用*标注

Back To Top

鄂ICP备17008157号-1