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 $$f^{(n)}(0)=(\frac{(-1{)}^n}{2^n}+2)(n!).$$ Let us now compute the limits (i) and (ii). For the first, we have $$\underset{n\to \mathrm{\infty }}{\lim }|f^{(n)}(0)-2(n!)|=\underset{n\to \mathrm{\infty }}{\lim }\left|\frac{(-1{)}^n\cdot (n!)}{2^n}\right|=\underset{n\to \mathrm{\infty }}{\lim }\frac{n!}{2^n}.$$ 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 $$\frac{n}{2}\cdot \frac{n-1}{2}\cdots \frac{2}{2}\cdot \frac{1}{2}\ge \frac{n}{2}\cdot 1\cdots 1\cdot \frac{1}{2}=\frac{n}{4}.$$ Putting it all together, $$|A_n-2(n!)|=\frac{n!}{2^n}\ge \frac{n}{4}.$$ Since $\underset{n\to \mathrm{\infty }}{\lim }\frac{n}{4}=+\mathrm{\infty }$, it follows that $\underset{n\to \mathrm{\infty }}{\lim }|A_n-2(n!)|=+\mathrm{\infty }$ 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 $n\to \mathrm{\infty }$.