Compute: 1° sqrt5 with accuracy 10^-4; 2° log_1011

Compute: 1° $\sqrt5$ with accuracy $10^{-4}$; 2° $\log_{10}11$ with accuracy $10^{-5}$; 3° $\sin1^\circ$ with accuracy $10^{-8}$.

Step 1: Part 1°: $\sqrt5$ with accuracy $10^{-4}$

Write $\sqrt5=2\sqrt{1+\tfrac14}=2(1+u)^{1/2}$, $u=\tfrac14$. Since $|u|\lt 1$, the binomial series is alternating with decreasing terms, so the truncation error is bounded by the first neglected term. Summing through $k=5$ (the $k=6$ term is $\approx-10^{-5}$, already below $10^{-4}$):

\[ \sqrt5 \approx 2.000000+0.250000-0.015625+0.001953-0.000305+0.000053 = 2.236076. \tag{1} \]

\[\sqrt5 \approx 2.2361, \qquad |{\rm error}| \lt 10^{-4}\]

Step 2: Part 2°: $\log_{10}11$ with accuracy $10^{-5}$

Write $\log_{10}11=1+\dfrac{\ln1.1}{\ln10}$, and expand $\ln(1+x)$ at $x=0.1$; summing through $k=5$ (the $k=6$ term contributes $\approx7\times10^{-8}$ to the quotient, negligible):

\[ \ln1.1 \approx 0.095310, \qquad \frac{\ln1.1}{\ln10} \approx \frac{0.095310}{2.302585} \approx 0.041393. \tag{2} \]

\[\log_{10}11 \approx 1.04139, \qquad |{\rm error}| \lt 10^{-5}\]

Step 3: Part 3°: $\sin1^\circ$ with accuracy $10^{-8}$

Converting $1^\circ=\pi/180\approx0.017453293$ rad and using $\sin x=x-\tfrac{x^3}6+\cdots$; the cubic term is $\approx0.000000886$, and the next (quintic) term is $\approx1.3\times10^{-11}\ll10^{-8}$, so two terms suffice:

\[ \sin1^\circ \approx 0.017453293-0.000000886 = 0.017452407. \tag{3} \]

\[\sin1^\circ \approx 0.017452406, \qquad |{\rm error}| \lt 10^{-8}\]

Step 4: Verification

\[\sqrt5=2.2360679\ldots,\ |2.236076-\sqrt5|\approx8.0\times10^{-6}\lt 10^{-4}\ \checkmark\]

\[\log_{10}11=1.0413926852\ldots,\ \text{matching }1.04139\text{ to the required }10^{-5}\ \checkmark\]

\[\sin1^\circ=0.0174524064\ldots,\ |0.017452406-\sin1^\circ|\approx1.3\times10^{-11}\lt 10^{-8}\ \checkmark\]