1469. Calculate $\sqrt{1.004}$ with an accuracy of 0.0001.
Solution. We will use formula (4) §4:

$$
\sqrt{1.004}=\sqrt{1+0.004}=(1+0.004)^{\frac{1}{2}}
$$

Here $m=\frac{1}{2}, x=0.004$

$$
\begin{aligned}
\sqrt{1.004}=1 & +\frac{0.004}{2}+\frac{\frac{1}{2}\left(\frac{1}{2}-1\right)}{2!}(0.004)^{2}+\ldots= \\
& =1+0.002-\frac{(0.004)^{2}}{2!4}+\ldots
\end{aligned}
$$

Notice that the signs of the obtained series, starting from the second, alternate.

Limiting ourselves to the sum of the first two terms, we get

$$
\sqrt{1.004} \approx 1.002
$$

Estimate the absolute error

$$
\Delta<\frac{(0.004)^{2}}{2 \mid 4}=0.000002
$$

Therefore, the value of the root is found with the required accuracy. 1470. Calculate $\sqrt[3]{30}$ with an accuracy of 0.001.

Solution. $\sqrt[3]{30}=3 \sqrt[3]{\frac{30}{27}}=3 \sqrt[3]{1+\frac{3}{27}}=3\left(1+\frac{1}{9}\right)^{\frac{1}{3}}$. Here

$$
m=\frac{1}{3}, x=\frac{1}{9}
$$

$$
\begin{aligned}
\sqrt[3]{30} & =3\left[1+\frac{1}{3} \cdot \frac{1}{9}+\frac{1}{3}\left(\frac{1}{3}-1\right)\left(\frac{1}{9}\right)^{2} \frac{1}{2!}+\right. \\
+ & \left.\frac{1}{3}\left(\frac{1}{3}-1\right)\left(\frac{1}{3}-2\right)\left(\frac{1}{9}\right)^{3} \frac{1}{3!}+\ldots\right]= \\
& =3\left(1+\frac{1}{27}-\frac{2}{2!9^{3}}+\frac{2 \cdot 5}{3!3^{3} 9^{3}}-\ldots\right)
\end{aligned}
$$

We obtained an alternating series (after the first term), so if we discard all terms starting from the fourth, the absolute error

$$
\Delta<\frac{3 \cdot 2 \cdot 5}{313^{3} 9^{3}}=\frac{5}{3^{3} \cdot 9^{3}}=\frac{5}{19683}<0.001
$$

from here

$$
\sqrt[3]{30} \approx 3+\frac{3}{27}-\frac{6}{2!9^{3}} \approx 3+0.1111-0.0041=3.1070
$$

(calculation was performed with one extra digit).

Thus, $\sqrt[3]{30} \approx 3.107$ with an accuracy of 0.001.