Let's calculate the value of the square of the expression

$$
1+\frac{x}{2}-\frac{x^{2}}{8}+\frac{x^{3}}{16}-\frac{5 x^{4}}{128}
$$

to six decimal places for the cases $x= \pm 0.1, \pm 0.05, \pm 0.01$.
The desired values can be easily calculated from the polynomial form of the given expression's square. Indeed, from this, the $x^{2}$, $x^{3}$, and $x^{4}$ terms are missing:

$$
K^{2}=1+x-\frac{7}{128} x^{5}+\frac{7}{512} x^{6}-\frac{5}{1024} x^{7}+\frac{25}{16384} x^{8}
$$

Furthermore, the absolute value of the $x^{5}$ term is smaller than the value of the sixth decimal place for each prescribed $x$ value, since even for the largest absolute value of $x = \pm 0.1$,

$$
\left|\frac{7}{128} \cdot x^{5}\right| = \left|\frac{7}{140} \cdot x^{5}\right| = \left|0.05 \cdot x^{5}\right| = 0.0000005
$$

and therefore, in the rounding of the sought values to six decimal places, it must still be considered, on the other hand, that the sum of the smaller terms can also modify the rounded value if they do not individually modify it.

Consider the 3rd to 6th terms of $K^{2}$ first for the prescribed negative $x$ values, in which case all four terms are positive. For $x = -0.1$, calculating the first two significant digits, the $-7x^{5}/128$ term is less than 0.00000054. The next term, as we have seen, is one-fortieth of this, and the subsequent ones are even smaller, so the sum of the next three terms is less than three-fortieths of the third term, which is less than 0.00000005, and therefore the sum of the 3rd to 6th terms falls between 0.0000005 and 0.0000006. Taking into account the first two terms,

$$
K^{2}=1-0.1+0.0000005 \ldots, \text { rounded to } 0.900001
$$

The value of $x = -0.05$ is half of the previous $x$, so the third term is $1/2^{5} = 1/32$ of the previous one, and is therefore less than if a 2 were written in the 8th decimal place, the next three terms are each less than one-eightieth of the third term, so the sum of the four terms does not reach half the value of the 6th decimal place, so in this case $K^{2} = 1 - 0.05 = 0.95 = 0.950000$. (The last form ends with four significant zeros.)

Similarly, for $x = -0.01$, $K^{2} = 0.990000$.

For the prescribed positive $x$ values, the terms of $K^{2}$ from the second onward are of alternating signs. Moreover, as we have seen, they are each smaller than the one preceding them. Therefore, the sum of the 5th and 6th terms is of the same sign as the 5th term, and in absolute value smaller than it; hence adding this sum to the 4th term does not change its sign, just as adding the 5th term alone would not change it, so the sum of the 4th to 6th terms is of the same sign as the 4th term, and in absolute value smaller than it.

For $x = +0.1$, the third term of $K^{2}$ is less than $-0.00000053$, the sum of the 4th to 6th terms is positive, and (rounding the one-fortieth part) less than 0.00000002, so the sum of the 3rd to 6th terms falls between $-0.0000005$ and $-0.0000006$, and thus must be rounded to $-0.000001$ (since the first two terms of $K^{2}$ are a round number up to the 6th decimal place), $K^{2} = 1.099999$.

For $x = +0.05$ and $x = +0.01$, the absolute value of the third term is less than half the value of the 6th decimal place, and even more so for the absolute value of the sum of the 3rd to 6th terms, so in these cases the value of $K^{2}$ is 1.050000, or 1.010000.

Sándor Horváth (Budapest, István I. Secondary School)

Remark. Reversing our results, it is a plausible conjecture that the numbers $1.099999 \approx 1.1$, $0.900001 \approx 0.9$, as well as the numbers $1.05$, $0.95$, $1.01$, and $0.99$ are approximate values of the square root of the number $1+x$ when $x$ is replaced by $\pm 0.1$, $\pm 0.05$, and $\pm 0.01$ in $K$, generally, that $K$ gives an approximate value of the square root of the number $1+x$. The conjecture is correct under certain conditions, but determining these conditions would lead us far afield.