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

MATH101 A April 2024

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[hide] Question 18 (b)
Let $f(x)={\frac {6}{2-x-x^{2}}}.$ (b) Find the largest value of R for which the identity (1), introduced in part (a), holds.

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 hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

[show] Hint 1
In part (a) we found an expression of $f(x)$ as a sum of two geometric series. Can you find the radii of convergence of the two series? What do they say about the radius of convergence of the sum?

[show] Hint 2
Use the ratio test to determine which values $x$ make the series for $f(x)$ converge.

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[show] Solution 1

We saw in part (a) that $$f(x)={\frac {2}{2+x}}+{\frac {2}{1-x}}=\sum _{n=0}^{\infty }\left({\frac {-x}{2}}\right)^{n}+\sum _{n=0}^{\infty }2x^{n},$$ where the first series converges if and only if $-1<-x/2<1$, and the second series converges if and only if $-1<x<1$. Since have the following sequence of equivalences $$-1<-x/2<1\iff 1>x/2>-1\iff 2>x>-2,$$ we see that the first series has radius of convergence $2$. We see then that, if $-1<x<1$, then both series converge, and so their sum converges as well. This suggest that the radius of convergence is $R=1,$ but we still need to check that the radius of convergence is not larger. Indeed, the example $$\sum _{n=0}^{\infty }x^{n}+\sum _{n=0}^{\infty }-x^{n}=0$$ shows that it is possible a sum of series to converge with a larger radius than the radii of the summand series. To see that this does not happen for us, we must check how the series for $f(x)$ behaves at $x=\pm 1$. At $x=-1$, we find the partial sum: $$S_{N}=\sum _{n=0}^{N}\left(\left({\frac {-x}{2}}\right)^{n}+2x^{n}\right)=\sum _{n=0}^{N}{\frac {1}{2^{n}}}+\sum _{n=0}^{N}2(-1)^{n}={\frac {1-(1/2)^{N+1}}{1-(1/2)}}+2\cdot {\begin{cases}1&{\text{ if }}N{\text{ is even}}\\0&{\text{ if }}N{\text{ is odd }}\end{cases}},$$ which may be computed using the formula $\sum _{n=0}^{N}x^{n}={\frac {1-x^{N+1}}{1-x}}$. The point is that, because of the second term in $S_{N}$, the limit $\lim _{N\to \infty }S_{N}$ does not exist, so our series for $f(x)$ does not converge for $x=-1$. At $x=1$, the $n^{\text{th}}$ summand of the series for $f(x)$ is $\left({\frac {-x}{2}}\right)^{n}+2x^{n}=\left({\frac {-1}{2}}\right)^{n}+2,$ which is positive and does not converge to 0 as $n$ tends to infinity. Therefore the series does not converge at $x=1$ either. This completes the proof that the radius of convergence is indeed 1.

[show] Solution 2

By the Ratio Test, the interval of convergence is the interval on which $\lim _{n\to \infty }\left|{\frac {A_{n+1}}{A_{n}}}x\right|<1$. Calculating this limit, $${\begin{aligned}\lim _{n\to \infty }\left|{\frac {A_{n+1}}{A_{n}}}x\right|=\lim _{n\to \infty }\left|{\frac {2+(-1/2)^{n+1}}{2+(-1/2)^{n}}}x\right|=\left|{\frac {2}{2}}x\right|=|x|,\end{aligned}}$$ which is less than $1$ for $x$ in $(-1,1)$. Thus, the radius of convergence is $1$.