Approximate f(x)=sqrt(1+x) by its degree-2 Maclaurin

Approximate $f(x)=\sqrt{1+x}$ by its degree-2 Maclaurin polynomial and estimate the error for $x\in[0,1]$.

Step 1: Derivatives at $x=0$

\[ f(0)=1,\quad f'(x)=\tfrac12(1+x)^{-1/2}\Rightarrow f'(0)=\tfrac12,\quad f''(x)=-\tfrac14(1+x)^{-3/2}\Rightarrow f''(0)=-\tfrac14,\quad f'''(x)=\tfrac38(1+x)^{-5/2}. \tag{1} \]

Step 2: Degree-2 polynomial and Lagrange remainder

\[ \sqrt{1+x} = 1+\frac x2-\frac{x^2}8+R_2(x), \qquad R_2(x)=\frac{f'''(\xi)}{3!}x^3=\frac{x^3}{16(1+\xi)^{5/2}}, \tag{2} \] with $0\lt \xi\lt x$.

Step 3: Bound on $[0,1]$

For $x\in[0,1]$: $x^3\le1$, and since $\xi>0$, $(1+\xi)^{5/2}\gt 1$ so $(1+\xi)^{-5/2}\lt 1$. Hence

\[ |R_2(x)| = \frac{x^3}{16(1+\xi)^{5/2}} \lt \frac{x^3}{16} \le \frac1{16}. \tag{3} \]

Step 4: Verification

\[x=1\text{: }P_2(1)=1+0.5-0.125=1.375,\ \sqrt2\approx1.41421,\ |{\rm error}|\approx0.0392\lt 0.0625\ \checkmark\]