Approximate f(x)=x²ln²x by its degree-4 Taylor polynomial

Approximate $f(x)=x^2\ln^2x$ by its degree-4 Taylor polynomial about $x=1$ and estimate the error for $x\in[\tfrac9{10},\tfrac{11}{10}]$.

Step 1: Derivatives at $x=1$

Successive differentiation (product then quotient rule) gives: \[f(1)=0,\ f'(1)=0,\ f''(x)=2\ln^2x+6\ln x+2\Rightarrow f''(1)=2, \tag{1}\]

\[ f'''(x)=\frac{4\ln x+6}{x}\Rightarrow f'''(1)=6, \qquad f^{(4)}(x)=\frac{-4\ln x-2}{x^2}\Rightarrow f^{(4)}(1)=-2, \qquad f^{(5)}(x)=\frac{8\ln x}{x^3}. \tag{2} \]

Step 2: Degree-4 Taylor polynomial

Since $f(1)=f'(1)=0$: $P_4(x) = (x-1)^2+(x-1)^3-\frac{(x-1)^4}{12}$.

Step 3: Lagrange remainder

\[ R_4(x) = \frac{8\ln\xi}{120\,\xi^3}(x-1)^5, \qquad \xi\text{ between }1\text{ and }x. \tag{3} \]

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

Evaluating $|f^{(5)}(\xi)|=8|\ln\xi|/\xi^3$ at the two endpoints: at $\xi=0.9$, $\approx1.16$; at $\xi=1.1$, $\approx0.57$. The maximum over the interval is near $\xi=0.9$, so $|f^{(5)}(\xi)|\lt 1.2$, giving

\[ |R_4(x)| \lt \frac{1.2}{120}\left(\frac1{10}\right)^5 = 10^{-7}. \tag{4} \]

Step 5: Verification

\[x=1.1\text{: }P_4(1.1)\approx0.0109917,\ (1.1)^2\ln^2(1.1)\approx0.0109917,\ |{\rm error}|\approx1.0\times10^{-8}\lt 10^{-7}\ \checkmark\]