请教高手：用matlab计算反函数值

Question

a=0.206,b=0.1156,c=0.1608,d=0.5176;
(a*x)^2.5+(b*x)^3.33+(c*x)^3.33+(d*x)^3.33=0.0014;
求x
resolve用过了，没求出来。请求高手现身。

Answer 1

可以画出(a*x)^2.5+(b*x)^3.33+(c*x)^3.33+(d*x)^3.33-0.0014的图像，看他与x轴的交点坐标
把交点放大，直到交点左右两边的值相同时，就是比较准确的结果
syms x;
ezplot('(0.206*x)^2.5+(0.1156*x)^3.33+(0.1608*x)^3.33+(0.5176*x)^3.33-0.0014',[0,0.5]);
hold on;
plot([0,1000],[0,0]);
我算出来是0.2329

Answer 2

function eq=chzh(x) % 保存为 chzh.m 文件
a=0.206;b=0.1156;c=0.1608;d=0.5176;
eq=(a*x)^2.5+(b*x)^3.33+(c*x)^3.33+(d*x)^3.33-0.0014;
----------------------------------------------------------------------
运行以下：

ezplot('chzh',[0 1])
hold on
plot([0 1],[0 0],'r')
x=fzero('chzh',0.25) % 设初值 0.25

Answer 3

>MuPAD
%将a,b,c,d代人(a*x)^2.5+(b*x)^3.33+(c*x)^3.33+(d*x)^3.33=0.0014中，整理后为
%0.01926052190*x^2.5+0.1146164492*x^3.33 = 0.0014
> solve(0.1926052190e-1*x^2.5+0.1146164492*x^3.33 = 0.14e-2);

0.2329335024, -0.1171253771+0.2334213668*I, -0.1171253771-0.2334213668*I