1. **Problem statement:** Find the Taylor polynomial of degree 3, $T_3(x)$, for $f(x) = \ln(1 + 2x)$ centered at $c=1$, and find an expression for the remainder $R_3(x)$. 2. **Formula and rules:** The Taylor polynomial of degree $n$ centered at $c$ is given by $$ T_n(x) = \sum_{k=0}^n \frac{f^{(k)}(c)}{k!}(x-c)^k $$ where $f^{(k)}(c)$ is the $k$-th derivative of $f$ evaluated at $c$. The remainder term in Lagrange form is $$ R_n(x) = \frac{f^{(n+1)}(z)}{(n+1)!}(x-c)^{n+1} $$ for some $z$ between $x$ and $c$. 3. **Calculate derivatives:** - $f(x) = \ln(1+2x)$ - $f'(x) = \frac{2}{1+2x}$ - $f''(x) = -\frac{4}{(1+2x)^2}$ - $f^{(3)}(x) = \frac{32}{(1+2x)^3}$ - $f^{(4)}(x) = -\frac{384}{(1+2x)^4}$ 4. **Evaluate derivatives at $c=1$:** - $f(1) = \ln(3)$ - $f'(1) = \frac{2}{3}$ - $f''(1) = -\frac{4}{9}$ - $f^{(3)}(1) = \frac{32}{27}$ 5. **Write $T_3(x)$:** $$ T_3(x) = f(1) + f'(1)(x-1) + \frac{f''(1)}{2!}(x-1)^2 + \frac{f^{(3)}(1)}{3!}(x-1)^3 $$ Substitute values: $$ T_3(x) = \ln(3) + \frac{2}{3}(x-1) - \frac{4}{9} \cdot \frac{(x-1)^2}{2} + \frac{32}{27} \cdot \frac{(x-1)^3}{6} $$ Simplify coefficients: $$ T_3(x) = \ln(3) + \frac{2}{3}(x-1) - \frac{2}{9}(x-1)^2 + \frac{16}{81}(x-1)^3 $$ 6. **Expression for remainder $R_3(x)$:** $$ R_3(x) = \frac{f^{(4)}(z)}{4!}(x-1)^4 = \frac{-384}{(1+2z)^4} \cdot \frac{(x-1)^4}{24} = -\frac{16}{(1+2z)^4}(x-1)^4 $$ where $z$ is between $x$ and $1$. 7. **Estimate error using Taylor's inequality on $[0.5,1.5]$:** - On $[0.5,1.5]$, $z$ lies between $0.5$ and $1.5$. - Minimum of $|1+2z|$ in this interval is at $z=0.5$: $1+2(0.5)=2$. - So maximum of $|f^{(4)}(z)| = \frac{384}{(1+2z)^4} \leq \frac{384}{2^4} = \frac{384}{16} = 24$. - Then $$ |R_3(x)| \leq \frac{24}{4!} |x-1|^4 = \frac{24}{24} |x-1|^4 = |x-1|^4 $$ - Since $|x-1| \leq 0.5$ on $[0.5,1.5]$, error bound is $$ |R_3(x)| \leq (0.5)^4 = 0.0625 $$ 8. **Maclaurin series for $\sqrt{27+x}$ and radius of convergence:** - Rewrite as $\sqrt{27+x} = \sqrt{27} \sqrt{1 + \frac{x}{27}} = 3\sqrt{3} \left(1 + \frac{x}{27}\right)^{1/2}$. - Use binomial series for $(1+u)^{1/2}$: $$ (1+u)^{1/2} = \sum_{n=0}^\infty \binom{1/2}{n} u^n $$ where $u = \frac{x}{27}$. - Radius of convergence is $|u| < 1 \Rightarrow |x| < 27$. **Final answers:** $$ T_3(x) = \ln(3) + \frac{2}{3}(x-1) - \frac{2}{9}(x-1)^2 + \frac{16}{81}(x-1)^3 $$ $$ R_3(x) = -\frac{16}{(1+2z)^4}(x-1)^4, \quad z \in (x,1) $$ Error bound on $[0.5,1.5]$: $$ |R_3(x)| \leq 0.0625 $$ Maclaurin series for $\sqrt{27+x}$ has radius of convergence: $$ R = 27 $$