Example 1. Construct a table of differences of various orders for the following values of $x$ and $f(x)$:

\[
\begin{gathered}
x_{0}=-1, x_{1}=-2, x_{2}=1, x_{3}=2, x_{4}=3 \\
y_{0}=0, y_{1}=7, y_{2}=30, y_{3}=-16, y_{4}=-45
\end{gathered}
\]
Solution. Using formulas (15.4), we find the first differences:

$$
\begin{gathered}
\Delta y_{0}=y_{1}-y_{0}=7-0=7 ; \Delta y_{1}=y_{2}-y_{1}=30-7=23 \\
\Delta y_{2}=y_{3}-y_{2}=-16-30=-46 ; \Delta y_{3}=y_{4}-y_{3}=-45-(-16)=-29
\end{gathered}
$$

In accordance with formulas (15.5), we calculate the second-order differences:

$$
\begin{gathered}
\Delta^{2} y_{0}=\Delta y_{1}-\Delta y_{0}=23-7=16 ; \Delta^{2} y_{1}=\Delta y_{2}-\Delta y_{1}=-46-23=-69 \\
\Delta^{2} y_{2}=\Delta y_{3}-\Delta y_{2}=-29-(-46)=17
\end{gathered}
$$

Using the definitions and corresponding formulas, we obtain the third-order differences:

$$
\begin{aligned}
& \Delta^{3} y_{0}=\Delta^{2} y_{1}-\Delta^{2} y_{0}=-69-16=-85 \\
& \Delta^{3} y_{1}=\Delta^{2} y_{2}-\Delta^{2} y_{1}=17-(-69)=86
\end{aligned}
$$

and the fourth-order difference $\Delta^{4} y_{0}=\Delta^{3} y_{1}-\Delta^{3} y_{0}=86-(-85)=171$.

The obtained differences can be represented in the form of Table 15.2.

Table 15.2

| $x$ | $y$ | $\Delta y$ | $\Delta^{2} y$ | $\Delta^{3} y$ | $\Delta^{4} y$ |
| :---: | :---: | :---: | :---: | :---: | :---: |
| -1 | 0 | 7 |  |  |  |
| -2 | 7 | 23 | 16 | -85 | 171 |
| 1 | 30 | -46 | -69 | 86 |  |
| 2 | -16 | -29 | 17 |  |  |
| 3 | -45 |  |  |  |  |
| $\Sigma$ |  | -45 | -36 | 1 |  |
| $S$ | -45 | -36 | 1 |  |  |

Remark. The last two rows serve for checking the calculations: in the $\sum$ row, the numbers are equal to the sums of the numbers in the corresponding columns, in the $S$ row - the differences between the last and first numbers in the corresponding columns. The coincidence of these numbers (along the diagonal) means that the calculations are correct.