Science:Math Exam Resources/Courses/MATH101 A/April 2024/Question 18 (c)/Solution 1

By answering the question in the hint (or by differentiating the power series in (a) and evaluating at 0), we find

$$A_{n}={\frac {f^{(n)}(0)}{n!}}.$$

Rearranging, we find

$$f^{(n)}(0)=\left({\frac {(-1)^{n}}{2^{n}}}+2\right)(n!).$$

Let us now compute the limits (i) and (ii). For the first, we have

$$\lim _{n\to \infty }|f^{(n)}(0)-2(n!)|=\lim _{n\to \infty }\left|{\frac {(-1)^{n}\cdot (n!)}{2^{n}}}\right|=\lim _{n\to \infty }{\frac {n!}{2^{n}}}.$$

To evaluate this last limit, we need to remember $n!=n\cdot (n-1)\cdot (n-2)\cdots 3\cdot 2\cdot 1$, which is a product of $n$ integers. We see then that

$${\frac {n!}{2^{n}}}={\frac {n\cdot (n-1)\cdot (n-2)\dots 3\cdot 2\cdot 1}{2^{n}}}={\frac {n}{2}}\cdot {\frac {n-1}{2}}\cdots {\frac {2}{2}}\cdot {\frac {1}{2}}.$$

Except for the last ${\frac {1}{2}}$ factor, all of the numbers in the above product are at least 1, so we have

$${\frac {n}{2}}\cdot {\frac {n-1}{2}}\cdots {\frac {2}{2}}\cdot {\frac {1}{2}}\geq {\frac {n}{2}}\cdot 1\cdots 1\cdot {\frac {1}{2}}={\frac {n}{4}}.$$

Putting it all together,

$$|A_{n}-2(n!)|={\frac {n!}{2^{n}}}\geq {\frac {n}{4}}.$$

Since $\lim _{n\to \infty }{\frac {n}{4}}=+\infty$, it follows that $\lim _{n\to \infty }|A_{n}-2(n!)|=+\infty$ as well. Finally, the relative discrepancy (ii) is

$${\frac {|f^{(n)}(0)-2(n!)|}{2(n!)}}={\frac {\frac {n!}{2^{n}}}{2(n!)}}={\frac {1}{2\cdot 2^{n}}},$$

which converges to 0 as $n\to \infty$.