Rewriting functions as Power Series (Power Series Representation)

Learn how to rewrite functions as power series, find intervals and radii of convergence using geometric series techniques.

Key Takeaways

Any function of the form a / (1 - r) can be rewritten as a geometric power series.
The radius of convergence is found by ensuring |r| < 1 and solving for x accordingly.
Factoring and rewriting denominators can help convert more complex functions into geometric series form.
The interval of convergence for geometric series excludes endpoints -1 and 1 unless otherwise tested.
Given initial terms, you can identify the common ratio r and first term a to express the series and find convergence.

What the video covers

Introduction to rewriting functions as power series using geometric series representation.
Example 1: Converting f(x) = 8 / (1 - x) into a power series and finding its interval and radius of convergence.
Explanation of geometric series sum formula and identifying a and r in the function.
Example 2: Handling a more complex function f(x) = -1 / (2 + x^2) by factoring and rewriting in geometric form.
Deriving the power series representation and solving inequalities to find interval and radius of convergence.
Example 3: Given the first few terms of a series, identifying a and r to write the power series and find convergence properties.
Step-by-step solving of absolute value inequalities to determine intervals of convergence.
Clarification on when endpoints are included or excluded in intervals of convergence for geometric series.
Summary of methods to quickly find radius of convergence based on interval endpoints and center.
Encouragement to practice and apply these techniques to similar problems.

Chapters

How do you convert a function like f(x) = 8 / (1 - x) into a power series?

You recognize it as a geometric series with a = 8 and r = x, then write it as the sum from n=1 to infinity of 8 * x^(n-1).

How do you find the interval of convergence for a geometric power series?

Set the absolute value of the ratio r less than 1, solve the inequality for x, and express the solution as an interval.

What do you do if the function denominator is not in the form 1 - r?

Factor the denominator to isolate a term in the form 1 - r, possibly moving constants to the numerator, to rewrite the function as a geometric series.

Full Transcript — Download SRT & Markdown

Speaker A

Hello everyone, how's it going? My name is Dan the Tutor. In today's video, we're going to be looking at more power series, specifically problems that look like this. So let's say they give you a function f(x) = 8 / (1 - x), and the question

Speaker A

goes something like, turn this function into a power series and write the interval and radius of convergence. And you look at this and you think to yourself, this isn't a power series, this is just a normal function. Well, the

Speaker A

thing is, this can be turned into a power series, specifically the geometric form, because let me remind you, the sum of a geometric series is a over 1 - r. So in other words, in this problem, a = 8 and r

Speaker A

would just be x. And so therefore, you can actually write it as a geometric series, which hopefully we remember how to write a geometric series. It's a * r^(n - 1), assuming the series starts at n = 1. So

Speaker A

then this can be written as the power series from n = 1 to infinity of 8 * x^(n - 1). And there, now it's a power series. And because it's a power series, I can find the radius and

Speaker A

interval of convergence very easily. All I have to do is say r is x, and the absolute value of r has to be less than one. So in this case, the absolute value of x must be less than one. And then

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whenever I want to get rid of this absolute value sign, I'm going to sandwich it in between -1 and positive 1 like this. And then there, there's my interval of convergence, although I'll write it in interval notation: -1 to positive 1.

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That's it for the interval. We don't need to check the bounds like we would for a ratio test power series because geometric never includes the endpoints negative 1 and positive 1. And then for the radius of convergence, this is really

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easy. It's just one. And that's because the radius of convergence is always the high number minus the low number over two. In other words, 1 - (-1) / 2, which gets you 1. Or you can get it faster just by

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saying this power series is centered at zero because it's smack dab in the middle of negative 1 and one, which means the radius will always be the number right here, in this case one. So that's it for the first one. Now let's do another.

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Let's say this time I have f(x) = -1 / (2 + x^2) this time. So this one's less friendly because if I want the form a over 1 - r, this is not the form I want. This two stops it from happening. So what

Speaker A

am I going to do? I'm going to factor out a 2 out of my denominator, so it will be 2 * the quantity 1 + x^2 / 2. That's what you have to do. And then even this two

Speaker A

can be moved up to the numerator as -1/2 over 1 + x^2 / 2. And the reason why I write it like this is because now I clearly see a is -1/2 and r is x^2 / 2. Why negative? Because it's supposed to be 1

Speaker A

minus r. If it's plus, that means r is going to end up being kind of like a double negative to make that positive. So here's my value for a and r. If I want to write it as a series then series from n

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= 1 to infinity of -1/2 * (x^2 / 2)^n power. And once I'm here I get to say r is x^2 over 2, and to solve this I have to say the absolute value of x^2 over 2 has

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to be less than 1. The good news is the negative sign goes away because of the absolute value part. And once I get here I need to again sandwich it in between -1 and positive 1. Now I have to solve for x.

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That means multiplying both sides by 2, so -2 < x^2 < 2. And then here's the interesting thing: I have to take the square root of both sides. Now you can't take the square root of a

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negative number, so that side gets eliminated. But because whenever you take a square root it's technically plus or minus, here's what ends up happening every time you are now sandwiched in between this guy less than negative square root of 2 less than x less

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than positive square root of 2. So let me erase this just a little bit so it's less cluttered. We went from here, I took the square root of everything, this side cancelled, the plus or minus, the square root meant

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the negative square root went over here and the positive square root went over here. And again, I'm sandwiched in between them. My interval of convergence will be negative square root of 2 to positive square root of 2 like this. And the radius, the shortcut would just tell

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us it's square root of 2 because we're centered at zero again. So there, that's interval and radius of convergence for this one as well. Now the only other problem I'm worried that you would is something like this where they give

Speaker A

you the first few terms of the series. Let's say it's 2 - x + x^2 / 2 - x cubed over 4 + dot dot dot, and the question again would be find the radius and the interval of convergence and write this

Speaker A

as a power series. So first I would say that this is in the geometric form, or I'm not sure if you're familiar with this or not, but the geometric form when you write it out as terms is a + a r

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+ a r^2 + a r^3 + dot dot dot, meaning the first term is a, a is just two. And if I want to find out what r is, I just have to take this term, negative x, and

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divide it by the first term 2. In other words, I'm saying r = -x / 2, and that's true. So now the series is equal to n = 1 to infinity of 2 * (-x / 2)^n power.

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And if I want to find the interval and the radius again, you should probably try it on your own at this point. I think you can do it, so pause the video, give it a try, and unpause it when you're ready

Speaker A

to see my solution. So here's what we do: it's all about r, which is x / 2. I'm saying the absolute value of r has to be less than one. So in other words, absolute value of x / 2 less than 1. Again, negative

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sign goes away because of the absolute value. And now I'm going to sandwich it between -1 and positive 1 to get rid of the absolute value. And finally, I'm going to multiply both sides by 2, -2 less than x

Speaker A

less than positive 2. So there, there's my interval of convergence, -2 to positive 2 with parentheses. And the radius of convergence will just be positive 2. And there we go. So that's all the problems I wanted to look at today. Thank you all

Speaker A

for watching. If you have any questions, please post them in the comments. Have a great rest of your day, and I'll see you in the next video. Take care and bye-bye. [Music]

Topics:power seriesgeometric seriesradius of convergenceinterval of convergenceseries representationfunction rewritingcalculusDan the Tutormath tutorialinfinite series

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