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

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MATH101 A April 2024

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Other MATH101 A Exams

• April 2024 •

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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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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Solution
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 $n\to \infty$ .