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# Find the Taylor series for $f(x)$ centered at the given value of $a.$ [Assume that $f$ has a power series expansion. Do not show that $R_n (x) \to 0.$] Also find the associated radius of convergence.$f(x) = \sin x,$ $a = \pi$

## The Taylor series for $f(x) = sin(x)$ centered at $a = \pi$ is,$$f(x) = -\frac{1}{1!}(x-\pi) + \frac{1}{3!}(x-\pi)^3 - \frac{1}{5!}(x-\pi)^5 + \frac{1}{7!}(x-\pi)^7 + \cdots,$$ or alternatively,$$f(x) = \sum_{n=1}^{\infty}\left(\frac{(-1)^n}{(2n-1)!}\cdot(x-\pi)^{2n-1}\right).$$The radius of convergence is $R=\infty.$

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##### Heather Z.

Oregon State University

##### Kristen K.

University of Michigan - Ann Arbor

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Okay. The first step for finding a taylor series is find some derivatives. So we'll start with sine of X. And then it's first derivative coastline necks. Second derivative minus cynics, their derivative. When this goes my next fourth derivative back to syntax again then it will repeat. Okay now you have to find those function values at a so we're gonna plug in pi pi is zero F. Prime of Pie is -1. Double Prime Occupies zero. Triple Prime of Pie Positive one. Yeah. Next 10. And then they start over get -101 dot dot dot. Okay so now I'm going to write the taylor series. I'm gonna call T F. X. It's the constant zero plus the derivative times X minus A. Which is by to the first power over one factorial plus zero times X minus pi square over two factorial plus one. I'm on the third derivative now X minus pi cubed over three factorial plus zero x minus pi to the fourth over four factorial. Um Let's do one more minus one plus minus one plus minus one. X minus pi to the fifth over five factorial. Yeah. Okay so notice what we got is -1 X minus pi. It's the 1/1 factorial plus one. X minus pi to the third over three factorial minus juan X minus pi to the fifth over five factorial. Okay if you can guess what the next one is going to be then you totally understand the series. So it should be add one x minus pi to the seventh over seven factorial. All right, so now we just have to write it in an enclosed form. So notice that it alternates. So we need a -1. And we're going to start this at I don't know what letter you use. I'm gonna start it at K equals zero. So I don't want this to be K. Because then the first term which is -1 here would be positive. So I need this to be K plus one. So when you plug in zero you get -1. When you plug in one you get positive one. When you plug in to get -1. Okay then all of them have an X -9 but not to the K. Power. It's to the odd powers 135 and seven. So to get odd numbers you can always use two K plus one or two K minus one or two K plus three. Or to k minus three. So you have to pick which one works the best. Okay, so when we plug in zero we want to get one. When we plug in one we want to get three. When we plug in to we want to get five. So two K plus one. And then since this and the exponent and the number on the bottom match up, we know that that's two K plus one factorial. Okay, so that's the taylor series. Um now let's find its interval or a radius of convergence. So remember you do that by taking the limit as in RK whatever K goes to infinity of a K plus one over a K And setting it less than one. No. Okay, so since this is an absolute value, we don't have to worry about the -1. So we're going to take the limit as K goes to infinity X minus pi to the two K plus one in parentheses. Plus one Over to K-plus one plus one factorial. And then we're going to divide it by the ak term which means we can just multiply by its reciprocal to k plus one factorial and then x minus pi to the two K plus one. All right, so this thing to K plus one plus one is two K plus 2 +12 K plus three. We have the limit as K goes to infinity X plus one. To the two K plus three Over to K-plus three factorial. UK Plus one Factorial. An x minus pi to the two K plus one. Whoops. Okay, so let's look at that. I don't know why I put x plus one because I was being messy up above there. I think x minus pi. It's what I mean here, X minus pi. So let's look at those. We have x minus pi to the two K plus three Over X -9 To the two K Plus one. So to simplify Since it's a fraction you subtract the exponents. So you get X -9 to the to power. Can we still have the limit here And it's by to the to oh and the absolute value is still here too. Yeah. Okay now let's look at the factorial is we have two K plus one factorial on the top and two K plus three factorial on the bottom. So let's just look at an example. So for example if Kay was three then the top is Seven factorial and the bottom is nine factorial. So what you're left with is two of them on the bottom the last two. So if K was a 100 Then it's 201 factorial on the top 203 factorial on the bottom. So every time you're just left with two on the bottom it's the last one and the one right before it. So the last one is two K plus three. And so the one right before it is too K plus two. Mhm. All right. Now take the limit as K goes to infinity. Oh if k goes to infinity the bottoms infinity and the top is a number a number divided by infinity zero, zero is less than one. So this uh um it's convergent everywhere. So it's radius of convergence is infinity because we got zero when we took the limit and zero is less than one. That means convergent everywhere. Okay hope you loved it. I've ever given myself a smiling face

Oklahoma State University

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##### Heather Z.

Oregon State University

##### Kristen K.

University of Michigan - Ann Arbor

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