Example 4 Solve the equation $y^{4}-6 y^{3}+10 y^{2}-4 y=0$
The original equation can be transformed into $\left(y^{2}-3 y+2\right)^{2}-3\left(y^{2}-3 y+2\right)+2-y=0$. Let $f(y)=$ $y^{2}-3 y+2$, then the two fixed points of $f(y)$ are $2+\sqrt{2}$ and $2-\sqrt{2}$. Thus, $2+\sqrt{2}, 2-\sqrt{2}$ are also two fixed points of $f[f(y)]$, and the other two fixed points of $f[f(y)]$ are obtained from $1+[1+(y-2+\sqrt{2})][1+(y-2-\sqrt{2})]=0$, i.e., $y^{2}-2 y=0$, which yields 0 and 2. Therefore, the four roots of the equation are $2+\sqrt{2}$, $2-\sqrt{2}$, 0, and 2.

This example shows: For a univariate algebraic equation $g(x)=0$, it can be converted into the equivalent equation $f(x)-x=$ $g(x)=0$, and by finding the fixed points of $f(x)$, the roots of $g(x)=0$ can be obtained.

The Newton's method for finding approximate roots of higher-order univariate algebraic equations, irrational equations, and transcendental equations is based on the above principle.

When using Newton's method to find the approximate root of $g(x)=0$, a number $x_{0}$ is first appropriately chosen, and then $x_{1}=x_{0}-$ $\frac{g\left(x_{0}\right)}{g^{\prime}\left(x_{0}\right)}, x_{2}=x_{1}-\frac{g\left(x_{1}\right)}{g^{\prime}\left(x_{1}\right)}, x_{3}=x_{2}-\frac{g\left(x_{2}\right)}{g^{\prime}\left(x_{2}\right)}, \cdots$ is calculated to obtain a sequence $\left\{x_{i}\right\}, i=$ $1,2, \cdots$ (where, $g^{\prime}(x)$ is the derivative of $g(x)$). Depending on the actual needs, a certain $x_{i}$ can be obtained. For example, to solve the equation $\mathrm{e}^{-\frac{x}{4}}(2-x)-1=0$, it is known from the graph that $g(x)=\mathrm{e}^{-\frac{x}{4}}(2-x)-1$ has a zero point in $(0,1)$. Using Newton's method to find the approximate value of this zero point, we can take $x_{0}=3$, and have
$$
x_{1}=x_{0}-\frac{g\left(x_{0}\right)}{g^{\prime}\left(x_{0}\right)}=-1.155999, x_{2}=x_{1}-\frac{g\left(x_{1}\right)}{g^{\prime}\left(x_{1}\right)}=0.189433, x_{3}=x_{2}-\frac{g\left(x_{2}\right)}{g^{\prime}\left(x_{2}\right)}=
$$
0.714043.

Depending on the actual needs, a more precise value $x_{i}$ can be obtained, satisfying $f\left(x_{i}\right)-x_{i}=g\left(x_{i}\right) \approx 0$. Therefore, the Newton's method for finding approximate roots of the equation $g(x)=0$ using the fixed points of $f(x)$ brings great convenience to computer operations (solving equations).