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. I’m impressed, I must say. Rarely do I come across a blog that’s equally educative and amusing, and let me tell you, you have
    hit the nail on the head. The problem is something that too few folks
    are speaking intelligently about. I’m very happy that I came across
    this during my search for something regarding this.

  2. Fantastic website you have here but I was wanting to know if you knew
    of any discussion boards that cover the same topics discussed in this article?
    I’d really like to be a part of community where I can get advice from other experienced people that share the same interest.
    If you have any recommendations, please let me know. 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