Example 2. Approximate the larger negative root of the equation $x^{3}-12 x-8=0$ using the tangent method with an accuracy of five decimal places after the decimal point.
Solution. Graphically separating the roots of the given equation, we conclude that the equation has three real roots; the larger negative root belongs to the interval $[-1,0]$. We specify an interval of lesser length, in which the root is located; this is the interval $[-0.7; -0.65]$.

Since

$$
f^{\prime \prime}(x)=6 x, f^{\prime \prime}(-0.65)0$, i.e., condition (12.4) is satisfied, we take $x_{0}=-0.65$ as the initial approximation. Using formula (12.3), we compute successive approximations, recording them in Table 12.2, from which it is clear that the desired root is $x=-0.694593$.

Table 12.2

| $\mathrm{n}$ | $x_{n}$ | $f\left(x_{n}\right)$ | $\frac{1}{f^{\prime}\left(x_{n}\right)}$ | $\frac{f\left(x_{n}\right)}{f^{\prime}\left(x_{n}\right)}$ | $x_{n+1}$ |
| :---: | :---: | :---: | :---: | :---: | :---: |
| 0 | $-0.65$ | $-0.474625$ | $-0.093175$ | $-0.044223$ | $-0.694223$ |
| 1 | $-0.694223$ | $-0.003902$ | $-0.094759$ | $-0.000370$ | $-0.694593$ |
| 2 | $-0.694593$ | $-0.000003$ | $-0.095711$ | $-0.0000003$ | $-0.694593$ |

## Problems

Find the real roots of the equations using the method of tangents:

1. $x^{3}+x-1=0$.
2. $x^{3}-x+1=0$.
3. $x^{3}+x-3=0$.
4. $x^{3}-x+3=0$.
5. $x^{3}+x+1=0$.
6. $x^{3}+2 x-4=0$.
7. $x^{3}-3 x+3=0$.
8. $x^{3}+8 x-6=0$.
9. $x^{4}+x+1=0$.
10. $x^{4}+5 x-3=0$.
11. $x^{5}+4 x-2=0$.
12. $x^{5}-6 x+4=0$.

## Answers

![](https://cdn.mathpix.com/cropped/2024_05_22_7ef2114a321cf223ba99g-374.jpg?height=45&width=1020&top_left_y=1120&top_left_x=81)
6. 1.179509 . 7. $-2.103803$. 8. 0.706011 . 9. 0.724492 . 10. $-1.875719 ; 0.577721$. 11. 0.492738 . $12 .-1.700040 ; 0.693378 ; 1.310235$.

## § 12.3. Method of Iterations

If an approximate value $x_{0}$ of the root of equation (12.1) is obtained by some means, then the refinement of the root can be achieved by the method of successive approximations or the method of iterations. For this, equation (12.1) is represented in the form

$$
x=\varphi(x)
$$

which can always be done, and in many ways, for example,

$$
x=x+c f(x)
$$

where $c$ is an arbitrary constant.

Let the number $x_{1}$ be the result of substituting $x_{0}$ into the right-hand side of equation (12.6): $x_{1}=\varphi\left(x_{0}\right)$; further, $x_{2}=\varphi\left(x_{1}\right), x_{3}=\varphi\left(x_{2}\right), \ldots$,

$$
x_{n}=\varphi\left(x_{n-1}\right)
$$

The process of successively computing the numbers $x_{n}(n=1,2,3, \ldots)$ by formula (12.8) is called the method of successive approximations or the method of iterations.

The iterative process converges $\left(\lim _{n \rightarrow \infty} x_{n}=\xi\right)$ if, on the interval $[a, b]$ containing the root $\xi$ and its successive approximations, the condition

$$
\max _{a \leq x \leq b}\left|\varphi^{\prime}(x)\right| \leq q<1
$$

is satisfied.

Remark. As $x_{0}$, any value from the interval containing the root can be taken; such an interval can be made sufficiently small.