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

MATH101 A April 2024

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[hide] Question 18 (c)
Let $f(x)={\frac {6}{2-x-x^{2}}}.$ (c) A colleague suggests approximating $f^{(n)}(0)\approx 2(n!)$ for large n. To assess this idea, find $f^{(n)}(0)$ exactly and then calculate the limit as $n\to \infty$ for both (i) the absolute difference, $\left|f^{(n)}(0)-2(n!)\right|$ and (ii) the relative discrepancy, ${\frac {\left|f^{(n)}(0)-2(n!)\right|}{2(n!)}}$.

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[show] Hint
What is the relationship between the power series representation from part (a) $f(x)=\sum _{n=0}^{\infty }A_{n}x^{n}$ and the Taylor series representation $f(x)=\sum _{n=0}^{\infty }{\frac {f^{(n)}(0)}{n!}}x^{n}$?

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[show] Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. 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 $n\to \infty$.