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. Hi there! Quick question that’s entirely off topic. Do you know how to make your site mobile friendly?
    My web site looks weird when viewing from my iphone. I’m trying to find a theme or plugin that might be able to resolve this issue.
    If you have any recommendations, please share.

    Thank you!

  2. I do accept as true with all the ideas you have introduced to your post.
    They’re very convincing and can definitely work.
    Nonetheless, the posts are too quick for starters.

    Could you please extend them a bit from next time? Thanks for the post.

  3. Greetings I am so grateful I found your weblog, I really
    found you by accident, while I was browsing on Yahoo for something else, Anyways I am
    here now and would just like to say cheers for a remarkable post and a all round entertaining blog
    (I also love the theme/design), I don’t have time to look over it all at the minute but I have book-marked it and also
    included your RSS feeds, so when I have time I will be back to read more, Please do keep up
    the excellent job.

  4. An outstanding share! I’ve just forwarded this onto a co-worker who was doing a little homework on this.

    And he actually bought me lunch due to the fact that I stumbled upon it for him…
    lol. So allow me to reword this…. Thank YOU for the meal!!
    But yeah, thanx for spending time to talk about this subject here on your internet site.

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