<s>[INST] Use Taylor’s Theorem to approximate $\sqrt{e}$ to within an error of magnitude at most $10^{-6}$ Could you please let me know if the following considered as a correct answer to this question.
Taylor Series
$f(x) = \sum_{k=0}^{n} \frac{f^{(k)}x_{0}}{k!}(x - x_{0})^{k}+ \frac{f^{(n+1)}(c)}{(n+1)!}(x - x_{0})^{(n+1)}$.
Take $f(x) = e^x$ and $x_{0} = 0$.
$f^{(0)}(x) = e^x \Rightarrow f^{(0)}(0) = 1$
$f^{(1)}(x) = e^x \Rightarrow f^{(1)}(0) = 1$
$f(x) = e^{x} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + .. +  \frac{e ^{c} x^{(n+1)}}{(n+1)!}$
Now, we take $x = \frac{1}{2}$
$e^{\frac{1}{2}} = 1 + \frac{1}{2} + \frac{1}{(2^2)2!} + \frac{1}{(2^3)3!} + .. +  \frac{e ^{c}}{(2^{(n+1)})(n+1)!}$
For any n, the error term equals $E_{n} = \frac{e ^{c}}{(2^{(n+1)})(n+1)!}$ but $c$ is between $x$ and $x_{0}$, i.e., $0$ and $\frac{1}{2}$ which means $e^c \le e^{\frac{1}{2}} \le \sqrt{3}$ 
$E_{n} \le \frac{\sqrt{3}}{(2^{(n+1)})(n+1)!}$
For $n = 6$, $E_{6} \le \frac{\sqrt{3}}{46,080} = 3.76 \left( 10^{-5} \right)$. Then to have an error of magnitude at most $10^{-6}$, we should use $n = 7$?
 [/INST] $$E_7  \leq \frac{\sqrt{3}}{2^8 \cdot 8!} \approx 1.67803161 \times 10^{-7} < 10^{-6}$$
$n=7$ suffices.
Note that Showing $n =6$ is not sufficient doesn't imply that $n=7 $ is sufficient, we still have to verify that it works.
Code to check that $n=7$ is indeed the smallest number of terms required.
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