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

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MATH101 A April 2024

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Other MATH101 A Exams

• April 2024 •

Question 18 (a)
Let $f(x)={\frac {6}{2-x-x^{2}}}.$ (a) Find the coefficients $A_{0},\;A_{1},...$ so that, for some $R>0,$ $f(x)=A_{0}+A_{1}x+A_{2}x^{2}+...=\sum _{k=0}^{\infty }A_{k}x^{k},\quad -R<x<R.\quad (1)$ Hint: Start by expressing $f(x)$ as a sum of simpler functions.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
The denominator of $f(x)$ factors as $(2+x)(1-x)$ . Can you write $f(x)$ as a sum ${\frac {A}{2+x}}+{\frac {B}{1-x}}$ , for some choices of $A,B$ ? Do these summands look familiar?

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Solution
Following the hint, let's look for $A,B$ such that $f(x)={\frac {A}{2+x}}+{\frac {B}{1-x}}.$ Following steps similar to those of Question 15, we find $A=B=2$ . We must now notice that each summand on the right-hand side of $f(x)={\frac {2}{2+x}}+{\frac {2}{1-x}}$ is the closed form of a geometric series $\sum _{n=0}^{\infty }ax^{n}$ . Recall that this series converges to ${\frac {a}{1-x}}$ , as long as $x\in (-1,1)$ . Let's figure out how to express each summand as a series separately, since we are allowed to recombine them using Theorem 3.5.13 of [CLP]. For the first, we have ${\frac {2}{2+x}}={\frac {1}{1+x/2}}={\frac {1}{1-(-x/2)}}=\sum _{n=0}^{\infty }\left({\frac {-x}{2}}\right)^{n}.$ For the second, we have ${\frac {2}{1-x}}=\sum _{n=0}^{\infty }2x^{n},$ and we will analyse both radii of convergence in part (b) . It follows then that $f(x)=\sum _{n=0}^{\infty }\left({\frac {-x}{2}}\right)^{n}+\sum _{n=0}^{\infty }2x^{n}=\sum _{n=0}^{\infty }\left({\frac {(-1)^{n}}{2^{n}}}+2\right)x^{n}.$ Therefore the sequence of coefficients is $A_{n}=\left({\frac {(-1)^{n}}{2^{n}}}+2\right)$ .