Approximate f(x)=sqrt x by its degree-3 Taylor polynomial

Approximate $f(x)=\sqrt x$ by its degree-3 Taylor polynomial about $x=1$ and estimate the error for $x\in[\tfrac9{10},\tfrac{11}{10}]$.

Step 1: Derivatives at $x=1$

\[ f(1)=1,\quad f'(1)=\frac12,\quad f''(1)=-\frac14,\quad f'''(1)=\frac38,\quad f^{(4)}(x)=-\frac{15}{16}x^{-7/2}. \tag{1} \]

Step 2: Degree-3 Taylor polynomial

\[P_3(x) = 1+\frac12(x-1)-\frac18(x-1)^2+\frac1{16}(x-1)^3\]

Step 3: Lagrange remainder

\[ R_3(x) = \frac{f^{(4)}(\xi)}{4!}(x-1)^4 = -\frac{15}{384\,\xi^{7/2}}(x-1)^4, \qquad \xi\text{ between }1\text{ and }x. \tag{2} \]

Step 4: Bound on $[\tfrac9{10},\tfrac{11}{10}]$

The factor $\xi^{-7/2}$ is largest at the smallest $\xi=\tfrac9{10}$: $(10/9)^{7/2}\approx1.446$, so $|f^{(4)}(\xi)|\le\tfrac{15}{16}(1.446)\approx1.355$. With $|x-1|\le\tfrac1{10}$:

\[ |R_3(x)| \le \frac{1.355}{24}\left(\frac1{10}\right)^4 \lt 6\times10^{-6}. \tag{3} \]

Step 5: Verification

\[x=0.9\text{: }P_3(0.9)\approx0.9486875,\ \sqrt{0.9}\approx0.9486833,\ |{\rm error}|\approx4.2\times10^{-6}\lt 6\times10^{-6}\ \checkmark\]