1. **Problem statement:** (a) Find the fifth Taylor polynomial for $f(x) = \ln(x+1)$ centered at $x=0$. (b) Use this polynomial to approximate $\ln\left(\frac{2}{3}\right)$. (c) Estimate the error using the Taylor remainder formula. 2. **Find derivatives of $f(x) = \ln(x+1)$ at $x=0$: ** $$f(x) = \ln(x+1)$$ $$f'(x) = \frac{1}{x+1}$$ $$f''(x) = -\frac{1}{(x+1)^2}$$ $$f^{(3)}(x) = \frac{2}{(x+1)^3}$$ $$f^{(4)}(x) = -\frac{6}{(x+1)^4}$$ $$f^{(5)}(x) = \frac{24}{(x+1)^5}$$ Evaluate at $x=0$: $$f(0) = \ln(1) = 0$$ $$f'(0) = 1$$ $$f''(0) = -1$$ $$f^{(3)}(0) = 2$$ $$f^{(4)}(0) = -6$$ $$f^{(5)}(0) = 24$$ 3. **Construct fifth Taylor polynomial $P_5(x)$ at $x=0$: ** $$P_5(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f^{(3)}(0)}{3!}x^3 + \frac{f^{(4)}(0)}{4!}x^4 + \frac{f^{(5)}(0)}{5!}x^5$$ Substitute values: $$P_5(x) = 0 + 1 \cdot x + \frac{-1}{2}x^2 + \frac{2}{6}x^3 + \frac{-6}{24}x^4 + \frac{24}{120}x^5$$ Simplify coefficients: $$P_5(x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \frac{x^5}{5}$$ 4. **Approximate $\ln\left(\frac{2}{3}\right)$ using $P_5$: ** Note: $\ln\left(\frac{2}{3}\right) = \ln\left(1 - \frac{1}{3}\right)$ so use $x = -\frac{1}{3}$. Calculate: $$P_5\left(-\frac{1}{3}\right) = -\frac{1}{3} - \frac{\left(-\frac{1}{3}\right)^2}{2} + \frac{\left(-\frac{1}{3}\right)^3}{3} - \frac{\left(-\frac{1}{3}\right)^4}{4} + \frac{\left(-\frac{1}{3}\right)^5}{5}$$ Calculate powers step by step: $$\left(-\frac{1}{3}\right)^2 = \frac{1}{9}$$ $$\left(-\frac{1}{3}\right)^3 = -\frac{1}{27}$$ $$\left(-\frac{1}{3}\right)^4 = \frac{1}{81}$$ $$\left(-\frac{1}{3}\right)^5 = -\frac{1}{243}$$ Evaluate terms: $$= -\frac{1}{3} - \frac{1/9}{2} - \frac{1/27}{3} - \frac{1/81}{4} - \frac{1/243}{5}$$ but carefully note alternating signs: $$P_5\left(-\frac{1}{3}\right) = -\frac{1}{3} - \frac{1}{18} - \frac{1}{81} - \frac{1}{324} - \frac{1}{1215}$$ Sum all negative terms: $$= -\left(\frac{1}{3} + \frac{1}{18} + \frac{1}{81} + \frac{1}{324} + \frac{1}{1215}\right) \approx -0.40554$$ 5. **Estimate error via Lagrange remainder formula:** The error after fifth term for $f(x) = \ln(x+1)$ is: $$R_5(x) = \frac{f^{(6)}(c)}{6!} x^6$$ for some $c$ between 0 and $x$. Compute 6th derivative: $$f^{(6)}(x) = -\frac{120}{(x+1)^6}$$ Since $x=-1/3$, $c \in [-1/3,0]$, so $x+1$ between $2/3$ and $1$. Max absolute value of $f^{(6)}(c)$ occurs at smallest $c$: $$|f^{(6)}(-\frac{1}{3})| = \frac{120}{(2/3)^6} = 120 \times \left(\frac{3}{2}\right)^6 = 120 \times 11.39 = 1366.8$$ Calculate error bound: $$|R_5(x)| \leq \frac{1366.8}{720} \times \left| -\frac{1}{3} \right|^6 = 1.8983 \times \frac{1}{729} \approx 0.0026$$ **Final answers:** (a) $P_5(x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \frac{x^5}{5}$ (b) $\ln\left(\frac{2}{3}\right) \approx -0.40554$ (c) Error estimate $|R_5(-\frac{1}{3})| \leq 0.0026$