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I know that you can determine radius of convergence of a function $f(x)$ by factoring the function to look like $1/(1-x)$. But when I tried to do that here, it didn't work. This is my work:my work

But the book says that the interval of convergence is (-1,1). They used a completely different method so I can't verify mine, other than knowing that something must be wrong because I got a different answer, which seems to be a mix of the first interval and second interval of convergence.

book solution

Alex G
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2 Answers2

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What your work shows is that the denominator in the expression for $f(x)$ is $0$ when $x=-1$ or $x=2.$ The numerator is never $0,$ so $f(x)$ has a non-removable discontinuity at $x=-1.$ That is, if we assign any value to $f(-1)$ then $f(x)$ is still discontinuous at $x=-1.$ So the radius of convergence cannot be greater than $|-1|=1.$ Otherwise, if the radius is $1+r$ with $r>0,$ the power series for $f$ would converge to a continuous function on the closed disc centered at $0$ with radius $1+r/2,$ and with the sum $S(x)$ of the series equalling $f(x)$ for $|x|<1$. Then we could assign $f(-1)=S(-1).$ But that would remove the discontinuity at $x=-1.$

DanielWainfleet
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  • So, in doing these kinds of problems in the future, how would I detect this sort of thing? I take it this issue really only crops up when the function can be split into two or more sums. If I calculate the domain of x by looking at the denominator, can I just check my answer against that and adjust if my answer $I$ spans an interval that breaches a discontinuity? – Alex G May 10 '16 at 05:24
  • It's not that the function can be split into a sum but whether the closure of the domain includes a non-removable discontinuity. Example 1: $f(x)=1/(x+1).$ Example 2: $f(x)=(e^{x-1}-1)/(x-1)$ for $x\ne 0$ and if $ f(1 )$ is not defined, we can assign $f(1)=1$,The point $x=1$ is a removable discontinuity. The radius of convergence for $f$ is $\infty.$ Caution: There are examples where a point $x$ may be a removable discontinuity for $f$ but not for its derivative. Being equal to a power series is a much stronger condition than being continuous. – DanielWainfleet May 10 '16 at 05:39
  • Wait, so I don't understand. First when you say $x != 0$, that seems like a non-issue. The function is still defined, no? Second, when you say that because $f(1)$ is undefined, we can assign it. Why? How? $f(1)$ would be $½$ in ex1 and $undefined$ in ex2 (so I guess you're redefining for $x=1$?) Finally, the radius of convergence doesn't need to start from zero as its center, right? What's to stop the radius of convergence from being centered at ½, so that it can have a radius of $3/2$ and still not overlap with the discontinuous points $x=-1,x=2$? – Alex G May 10 '16 at 05:46
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    In Ex. 1. the point is x=-1, not x=+1. In Ex.2 we have f(x) converging to 1 as x tends to 1. I was talking in all cases about a circle centered at 0. Of course we may take a different center, which may result in a different radius of convergence. For f(x)=1/(x+1) the power series centered at 1000 has a radius of convergence of 999. – DanielWainfleet May 10 '16 at 13:04
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You simply worked out the power series of your function around the point $\;x=\frac12\;$ and you got (correctly, in my opinion. Observe your interval of convergence avoids, as it should, both discontinuity points of the function...) what you got.

What you show "they did" is the development of the function as power series about $\;x=0\;$ ...Perhaps this is what was required, I can't tell.

Observe the function is

$$\frac3{x^2-x-2}=\frac3{(x-2)(x+1)}\;$$

and since they apparently wanted a McClaurin series they didn't choose the point you did to develop around it the function.

By the way, what book is that?

DonAntonio
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  • The only directions were "Express the function as the sum of a power series by first using partial fractions. Find the interval of convergence." Granted, I didn't use partial fractions, but I was asking out of curiosity how my answer didn't match up. The part I don't get is, does their answer actually miss the mark somewhat? Because by using 0 as the centerpoint, they miss an entire unit's worth of valid values (from $x=1 to x=2$). As far as valid answers go, what is the significance of the centerpoint that one chooses? And it's James Stewart's Calculus 7th Ed. Great textbook. – Alex G May 10 '16 at 07:20
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    @AlexG I see. Well, the partial fractions thing was a huge clue about what to do, yet without being accurate about the center point of the series the exercise remains open. In my opinion what you did is better in the sense that if gives a valid power series for the function using the maximal interval possible for it: you cannot go over $;x=-1,,2;$, and they "missed" a lot by centering their power series around zero. Yet it all depends on the purpose of the development: around zero it's easier to grasp and substitute. The answers thus couldn't possibly match as the centers didn't. – DonAntonio May 10 '16 at 07:32
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    Thank you for the clear explanation. I've been returning to this problem over the course of the last few hours trying to see if there's a new insight that showed why mine might be wrong or something. Like you said the partial fractions for sure, but I figured that my way should be just as valid ignoring that part. Either way, this seems like a good problem to have, as I feel like I understand the whole convergence issue better -- especially because before this, I wasn't even thinking about where the "center" would be! Thanks again. – Alex G May 10 '16 at 07:36
  • @AlexG My pleasure. When we learn something, and this happens to me all the time in this site, that's the best we can expect for. – DonAntonio May 10 '16 at 16:30