1. Determine the Taylor polynomial $T_{2}(x)$ at $x=3$ .
$f(x)=\frac{x}{x+2}$
The first thing we want to do is work out the first two derivatives, $f'(x)=\frac{(x+2)-x}{(x+2)^{2}}=\frac{2}{(x+2)^{2}}$ and $f''(x)=\frac{-2\cdot2(x+2)}{(x+2)^{4}}=\frac{-4}{(x+2)^{3}}$
Using the definition of the Taylor polynomial $T_{n}(x)=\sum_{i=0}^{n}\frac{f^{(i)}(a)}{i!}(x-a)^{i}$ , we see that
$$T_{2}(x)=\sum_{i=0}^{2}\frac{f^{(i)}(a)}{i!}(x-a)^{i}\\T_{2}(x)=\frac{f(3)}{0!}(x-3)^{0}+\frac{f'(3)}{1!}(x-3)^{1}+\frac{f''(3)}{2!}(x-3)^{2}\\T_{2}(x)=f(3)+(x-3)f'(3)+\frac{1}{2}(x-3)^{2}f''(3)\\T_{2}(x)=\frac{3}{5}+\frac{2}{25}(x-3)-\frac{2}{125}(x-3)^{2}\\T_{2}(x)=\frac{3}{5}+\frac{2}{25}x-\frac{6}{25}-\frac{2}{125}x^{2}+\frac{12}{125}x-\frac{18}{125}\\T_{2}(x)=\frac{27}{125}+\frac{22}{125}x^{2}-\frac{2}{125}x^{2}$$.
We can verify this using desmos.
2. Show that $T_{2}(x)\approx F(x)$ with an error $<0.1$ on the interval $[2,4]$
Recall that the error term of a Taylor polynomial is given by $R_{n}(x)=\frac{\text{max}(f^{(n+1)}(a))}{(n+1)!}(x-a)^{n+1}$. So, we first find the third derivative $f'''(x)=\frac{4\cdot3(x+2)^{2}}{(x+2)^{6}}=\frac{12}{(x+2)^{4}}$. For $x>-2$ this is a strictly decreasing function, so its maximum on the interval $[2,4]$ is its value at $x=2$. Thus, $R_{2}=|\frac{\frac{12}{(2+2)^{4}}}{3!}(2-3)^{3}|=\frac{12}{6\cdot4^{4}}=\frac{1}{128}<0.1$.
5. Determine the interval of the convergence of the power series $\sum^{\infty}_{n=1}\frac{(-1)^{n}}{n\cdot3^{n}}(x-3)^{n}$
We know this converges when the ratio test is less than 1:
$$\lim_{n\to\infty}\left|\frac{\frac{(-1)^{n+1}}{(n+1)\cdot3^{n+1}}(x-3)^{n+1}}{\frac{(-1)^{n}}{n\cdot3^{n}}(x-3)^{n}}\right|<1\\\lim_{n\to\infty}\left|\frac{\frac{1}{(n+1)\cdot3^{n+1}}(x-3)\cdot3^{n+1}(n+1)n}{\frac{1}{n\cdot3^{n}}\cdot3^{n+1}(n+1)n}\right|<1\\\lim_{n\to\infty}\left|\frac{(x-3)n}{3(n+1)}\right|<1\\\lim_{n\to\infty}\left|\frac{(x-3)}{3}\right|<1\\-1<\frac{x}{3}-1<1\\0<x<6$$
6. Using your answer to pt. A, determine the interval of the convergence of the power series $\sum_{n=1}^{\infty}\frac{(-1)^{n}}{3^{n}}(x-3)^{n-1}$
As in the previous section, we use the ratio test:
$$\lim_{n\to\infty}\left|\frac{\frac{(-1)^{n+1}}{3^{n+1}}(x-3)^{n}}{\frac{(-1)^{n}}{3^{n}}(x-3)^{n-1}}\right|<1\\\lim_{n\to\infty}\left|\frac{\frac{1}{3^{n+1}}(x-3)\cdot3^{n+1}}{\frac{1}{3^{n}}\cdot3^{n+1}}\right|<1\\\lim_{n\to\infty}\left|\frac{x-3}{3}\right|<1\\-1<\frac{x}{3}-1<1\\0<x<6$$
Diophantus
