Compute sqrt(1.121) to three decimal places

Compute $\sqrt{1.121}$ to three decimal places.

Step 1: Binomial series with $x=0.121$

\[ \sqrt{1+x} = 1+\frac x2-\frac{x^2}8+\frac{x^3}{16}-\frac{5x^4}{128}+\cdots \tag{1} \]

Step 2: Compute terms through $x^3$

The fifth term, $-\tfrac{5x^4}{128}\approx-0.000008$, is well below the $5\times10^{-4}$ needed for three-decimal rounding, so terms through $x^3$ suffice: \[1 \;+\; \frac{0.121}2=0.060500 \;-\; \frac{(0.121)^2}8=-0.001830 \;+\; \frac{(0.121)^3}{16}=0.000111. \tag{2}\]

Step 3: Sum and round

\[ \sqrt{1.121} \approx 1+0.060500-0.001830+0.000111 = 1.058781 \approx 1.059. \tag{3} \]

Step 4: Verification

\[1.059^2=1.121481,\ 1.058^2=1.119364\text{: }1.121\text{ lies between, closer to }1.059\ \checkmark\]

\[\sqrt{1.121}=1.0587729\ldots\text{, matching the series sum }1.058781\text{ to within }8\times10^{-6}\ \checkmark\]