Review : Taylor Series — Question 1

Question 1 – Solution

Strategy. Divide derivatives by factorials, then distinguish series convergence from finite-polynomial accuracy.

Step 1: Normalize the coefficients. For $n\ge 1$, $f^{(n)}(x)=(-1)^{n-1}(n-1)!/x^n$. Thus $$\boxed{T_4(2+h)=\ln 2+\frac h2-\frac{h^2}{8}+\frac{h^3}{24}-\frac{h^4}{64}.}$$ The student omitted division by $n!$; the first error is the $h^2$ coefficient, $-1/4$ instead of $-1/8$.

Step 2: Establish the represented function. Integrating the geometric series for $1/(2+h)$ from $0$ to $h$ gives $\ln(2+h)=\ln 2+\sum_{n\ge 1}(-1)^{n-1}h^n/(n2^n)$ for $|h|<2$. Uniform convergence on the integration segment justifies this step. The term ratio tends to $|h|/2$, so the radius is $2$; $h=-2$ gives the divergent negative harmonic series and $h=2$ gives the alternating harmonic series. The convergence interval is $\boxed{(0,4]}$ in $x$.

Step 3: Certify a uniform error. Since $f^{(5)}(x)=24/x^5$, on $[1.8,2.2]$, $$|f(x)-T_4(x)|\le\frac{24}{1.8^5}\frac{0.2^5}{120} =\boxed{\frac 1{5\cdot 9^5}<3.39\times 10^{-6}}.$$

Step 4: Determine the error sign. For $x\ne 2$, $f(x)-T_4(x)=h^5/(5\xi^5)$ for some positive $\xi$ between $2$ and $x$. Thus $T_4$ is above $f$ on $[1.8,2)$ and below it on $(2,2.2]$; equality holds at $2$. The plot magnifies this signed error.