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

MATH101 A April 2024

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[hide] 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.

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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!

[show] 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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[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. 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 $A_{n}=\left({\frac {(-1)^{n}}{2^{n}}}+2\right)$.