Using Lagrange's theorem, compute arctan 1.1 approximately

Using Lagrange's theorem, compute $\arctan 1.1$ approximately.

Step 1: Set up

The nearest argument with a known value is $1$, where $\arctan1 = \dfrac{\pi}{4}$. Apply Lagrange's theorem to $f(x) = \arctan x$ on $[1,\,1.1]$, with $f'(x) = \dfrac{1}{1+x^{2}}$.

Step 2: The exact identity

\[ \arctan1.1-\frac{\pi}{4} = \frac{0.1}{1+\xi^{2}}, \qquad \xi\in\bigl(1,\,1.1\bigr). \tag{1} \]

Step 3: Bounds first

Since $1+\xi^{2}$ runs between $2$ and $1+1.21 = 2.21$, \[\frac{0.1}{2.21} \lt \arctan1.1-\frac{\pi}{4} \lt \frac{0.1}{2}, \qquad 0.04525 \lt \arctan1.1-\frac{\pi}{4} \lt 0.05,\]

\[ 0.83065 \lt \arctan1.1 \lt 0.83540. \]

The value is already pinned to two decimal places.

Step 4: A single estimate

Taking the midpoint $\xi \approx 1.05$ of the narrow interval: \[\arctan1.1 \approx \frac{\pi}{4}+\frac{0.1}{1+1.1025} = 0.785398+\frac{0.1}{2.1025} = 0.785398+0.047562 = 0.832960.\]

Step 5: Verification

\[\text{Exact: } \arctan1.1 = 0.8329813, \text{ inside the bounds} \;\;\checkmark\]

\[\text{The midpoint estimate } 0.832960 \text{ is low by } 2\cdot10^{-5}, \text{ four correct significant figures} \;\;\checkmark\]

\[\text{The cruder choice } \xi = 1 \text{ gives } \frac{\pi}{4}+0.05 = 0.8354, \text{ high by } 2.4\cdot10^{-3} \;\;\checkmark\]