How To Find Interval Of Convergence Of Taylor Series
Starting from 6:xvi How would I find the radius of convergence for this series? The radius of convergence is half of the interval of convergence. In the video, the interval is -five to v, which is an interval of ten, so the radius of convergence is 5. shouldn't we test the stop points? or exercise we ever not include them? Yes, the end points demand to be individually tested. Sometimes they're included, sometimes they're not, sometimes just one is and the other is not. bully question! Think of it this way (sorry for the weird formatting): When n=1, 5^n is just 5, correct? Recall this rule of exponents: (X)^a•(X)^b = (Ten)^(a+b) So, if you contrary that rule, information technology works this way: If yous had a larger number, say 5^(n+7), then it devolves into 5•five^(n+6), considering five^1 = 5. In the terminate, (five^n)/(5^[n+1]) = (v^n)/(5•five^n). Hope this helps! Best of luck! This might be a stupid question merely are in that location whatsoever function series whose convergence intervals are not contiguous? If no, why? When i do the ratio exam and stop up with an answer that doesnt include n does that mean that it diverges and has no interval of convergence? Or does that mean that it converges over any interval to that answer? For instance If the answer is smaller than i, the serial e'er converges; if the reply is greater than one, the serial ever diverges. It is inconclusive if the respond is equal to one. In other words, if it doesn't include due north you nonetheless use the Ratio Test as usual. The limit as n approaches infinity won't change anything. If my limit evaluates to 0 does that mean that it converges over any interval? My reply is going to exist yes, just I also desire to say "for some cases" because I haven't seen enough cases. Take for instance the infinite series of (x^m)/chiliad!. You lot'll discover that |x| times the limit as k approaches infinity is equal to 0, pregnant for whatever value of x you pick, you'll ever find that r=0 for all real numbers, and then the IOC would be over all reals. Monotonically decreasing means strictly decreasing; there's no point where the n+1th term is greater than the nth term. Does the interval of convergence ever have to be symmetrical about c? That is, do the endpoints of the interval of convergence always take to exist equidistant from c? No. I have had a trouble earlier where I was given the job to find the interval of convergence for a Taylor series centered at one, only the interval of convergence 0<ten<four. Many factors affect the interval of convergance, and information technology is rarely symmetrical Why does the alternating harmonic series converge? Shouldn't it arroyo zero at the aforementioned speed every bit the regular harmonic series - only from both sides of the ten axis? The absolute value of the terms does go to nada at the same rate as the harmonic series, but every other term is being subtracted, not added. The terms beingness subtracted are one/2, i/4, 1/six, 1/8, 1/10, and so on. That is, half of the harmonic series. You lot tin can recall of it as subtracting a divergent serial from a divergent series; the effect may or may not diverge.Desire to join the conversation?
(This is unaffected by whether the endpoints of the interval are included or not) 
What well-nigh 5^(n+ane)? If n=one, wouldn't that be five^2 = 25?
Therefore: (v)^n•(five)^ane = (v)^(north+ane).
while n=one, five^(n+one) = 5•v^(due north)
ii^n*x^2n or (x^n)/(two^n)
Video transcript
- [Instructor] So nosotros have an space series here, and the goal of this video is to try to figure out the interval of convergence for this series. And that's another way of saying for what ten values, what range of x values, is this serial going to converge? And similar always intermission this video and see if you can figure it out. When you expect at this series, information technology doesn't fit cleanly into something like a geometric serial or an alternating serial. When I come across something like this, I think about the ratio test, considering information technology tends to be pretty general. To apply the ratio examination, we want to think about the limit, the limit as n approaches infinity of the n plus oneth term, divided by the nth term and the absolute value of that. If this thing is less than i, so when this thing is less than one, then we are going to converge. And the x values that brand this matter greater than 1, nosotros are going to diverge. And the 10 values that make this equal to one, well then we're going to be inconclusive so nosotros're going to accept to use other techniques to recall almost whether we're going to converge or diverge. And so permit's just think about this, let's but evaluate this, and so permit's practice that. Limit as n approaches infinity of the accented value of a sub-north plus one well that's going to be 10 to the northward plus one. Let me color code this, but so we know what we're doing. Then this thing right over here is going to be ten to the n plus one over n plus one times five to the n plus one. Nosotros're going to dissever that by the nth term. We're going to separate that by the nth term. And that'due south but going to be x to the due north over n times five to the n. We're going to take the accented value of this whole matter. Now allow me simplify this. I'll simplify it down here. So this is the same thing as x to the n plus i over northward plus one times 5 to the n plus ane, times the reciprocal of this. And then it'southward going to be due north times 5 to the northward, over x to the n. And we could simplify this. This is going to exist equal to, let'due south see, you divide numerator and denominator past 10 to the due north, y'all're left with simply an ten. Then divide numerator and denominator by five to the n. That is going to be, that's a one, this is a one and then this is just going to be, v to the n plus i divided by v to the n, that'due south but going to be a 5. And so what do nosotros have? Nosotros have 10 times n. x times n, over, distribute the five, five northward, five n plus ane. Oh, permit me be careful there. Allow me distribute the five. V northward plus five, 5 times n, five times one. Five north plus i. Five due north plus five (chuckles). All right, so let me just rewrite that. This is going to be equal to the limit as n approaches infinity of the absolute value of this thing. And actually to help us with this limit, let me rewrite it a little scrap. Let me split the numerator and the denominator both past n. I'm not changing the value, I'thou doing the same thing to the numerator and the denominator. I'1000 dividing it past the same value. So if I separate the numerator and the denominator by n, this is going to be the aforementioned thing as x over five plus v to the due north. And then when y'all divided the numerator and the denominator by n, information technology becomes very articulate what happens as due north approaches infinity, Equally due north approaches infinity, 10 doesn't change, 5 doesn't change, just five over due north goes to zero. And so this limit is going to be equal to 10 over five. So that's a pretty neat, clean thing. Now we tin can use this to think about, and really let me write this. This is going to be the absolute value of x over five. Now we can think most under which weather condition is the absolute value of x over 5 going to be less than one and we're definitely gonna converge? Under what weather are we going to be greater than i and definitely diverge? And then under what atmospheric condition is it inconclusive? So let'due south just see when nosotros know we tin converge. So the accented value of ten over five is less than one. This is our convergence situation. Well, that's the same affair as maxim that ten negative one is less than x over five, which is less than one. And you multiply all the sides by v. This is the same thing as negative five is less than x, which is less than five. So if we know that this is true, this is definitely going to be part of the interval of convergence. Nosotros know that if x meets these constraints, and then our serial is going to converge. But we're non done yet. We take to think about the inconclusive example. So let'southward call back near the scenario where the absolute value of x over five, accented value of x over five is equal to one. Or another way of thinking about this, that means ten over 5 is equal to one or x over five is equal to negative i. And this means that x is equal to five or x is equal to negative five. And so these are the 2 inconclusive cases using the ratio test. So allow'due south exam them out individually by looking back at the series and just substituting x equals 5 or ten equals negative five. So in the get-go scenario, let me find a new color hither, let me employ cerise. So the first scenario of 10 equals five, let's get to our series. And so the serial is going to be the sum from north equals one to infinity of five to the n over n times five to the north. Well, this is only going to be equal to the sum n equals 1 to infinity of one over northward. This is a harmonic serial. This is the p-series where p is equal to one. And we know our p-series of p is equal to one. That's going to diverge. And we know the harmonic series nosotros've done in other videos, this definitely diverges. So this diverges. You could practise that by p-series convergence exam. If the p for a p-series is one, well you're gonna diverge. Then at present let'southward think about, then five is definitely not part of our interval of convergence. Now allow's remember virtually x equals negative five. When 10 equals negative v, let me get a another color going hither. When x is equal to negative five, and then this matter is going to exist equal to the sum from n equals one to infinity of negative five to the north. Actually let me only write that as, I'll write it out negative five to the n over n times v to the n. This is the same thing as the sum from northward equals one to infinity. We could write this as negative one to the n times 5 to the, times 5 to the n, over n times 5 to the n. And at present this matter, this is an alternating harmonic series. And and then you lot could actually employ the, yous might already know that that converges, or yous could use the alternating series test. The alternating serial test, it might be a little bit clearer if I write it like this. That this is an alternating serial. So in an alternating series test, if nosotros run across that this thing is monotonically decreasing and the limit as due north approaches infinity is zero, this matter converges. The alternate harmonic series really converges. So this converges. So given that this converges, you could view this as this boundary hither. We would include that in our interval of convergence. So x doesn't just take to be strictly greater than negative five, information technology could exist greater than or equal to negative five, but it has to be less than five. This is our true interval of convergence.
Source: https://www.khanacademy.org/math/ap-calculus-bc/bc-series-new/bc-10-13/v/radius-convergence-ratio-test
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